ALGEBRAIC GEOMETRY

J.S. Milne

August 24, 1996; v2.01

Abstract. These are the notes for Math 631, University of Michigan, Fall 1993, exactly as they were handed out during the course except for some minor corrections.

Please send comments and corrections to me at jmilne@umich.edu using “Math631” as the subject.

Contents

0. Algorithms for Polynomials . . . . . . . . . . . . 3

1. Algebraic Sets . . . . . . . . . . . . . . . . . . 12

2. Affine Algebraic Varieties . . . . . . . . . . . . 32

3. Algebraic Varieties . . . . . . . . . . . . . . . 47

4. Local Study: Tangent Spaces; Tangent Cones; Singularites . . . . . . . . . . . . . . . . . . . . . . 63

5. Projective Varieties; Complete Varieties . . . . . 86

6. Finite Maps . . . . . . . . . . . . . . . . . . . 109

7. Dimension Theory . . . . . . . . . . . . . . . . 117

8. Regular Maps and their Fibres . . . . . . . . . 126

Some notations. Recall that a field ]k is said to be algebraically closed if every polynomial f(X) with coefficients in k factors completely in k. Examples: ℂ, or the set of all complex numbers algebraic over ℚ. Every field k is contained in an algebraically closed field. A field of characteristic zero contains a copy of ℚ; a field of characteristic p contains a copy of 𝔽_{p} =_{df} ℤ/pℤ. The symbol ℕ denotes the natural numbers, ℕ = {0, 1, 2, … }. “Ring” will mean “commutative ring with 1”, and a homomorphism of rings will always carry 1 to 1. For a ring A, A^{×} is the group of units in A:

A^{×} = {a ∈ A | ∃b ∈ A such that ab = 1}.

\begin{matrix}c \\ ◯ \end{matrix}1996 J.S. Milne. You may make one copy of these notes for your own personal use.

We use Gothic (fraktur) letters for ideals:

𝔞 𝔟 𝔠 𝔪 𝔫 𝔭 𝔮 𝔄 𝔅 ℭ 𝔐 𝔑 𝔓 𝔔.

A more complete course would cover also at least the following topics:

9. Chow coordinates, Hilbert polynomials, moduli varieties. 10. Divisors, rational maps, and projective embeddings. 11. Intersection theory. 12. Differentials. 13. Algebraic varieties over nonalgebraically closed fields. 14. Algebraic varieties over ℂ. 15. A brief history. 16. Algebraic groups. 17. Abelian varieties. 18. Curves (Riemann Roch theorem). 19. Jacobian varieties. 20. Birational classification; surfaces. 21. Lefschetz pencils. 22. Cohomology.

And then it would do everything again, in the language of schemes.

Books

The following books will be most useful for the course. Atiyah, M.F and MacDonald, I.G.: Introduction to Commutative Algebra, Addison-Wesley

1969. When I need a reference for something in commutative algebra, this is what I use.

Mumford, David: The Red Book of Varieties and Schemes, Springer Lecture Notes in

Math. 1358, 1988, QA564.M97. The first chapter will be particularly useful. Shafarevich, I.: Basic Algebraic Geometry, Springer, 1974, QA564.S533. I don’t like his

approach to the foundations (he defines “affine algebraic variety”, “projective algebraic variety”,..., but never “algebraic variety”), but the book contains a lot of beautiful material and is easy to read.

Cox, Little, O’Shea: Ideals, Varieties, and Algorithms, Springer, 1992. This gives an algorithmic approach to algebraic geometry, which makes everything very downto-earth and computational, but the cost is that the book doesn’t get very far in 500pp.

Other books: Dieudonné, J.: Cours de Géometrie Algébrique, 1974, QA564.D51. Volume 2 is an intro-

duction to algebraic geometry; Volume 1 is a history of algebraic geometry up to about 1974.

Dieudonné, J.: History of Algebraic Geometry, Wadsworth, 1985. A translation of Volume

1 of the above work up-dated to 1985. Eisenbud, David, and Harris, Joe: Schemes: the language of modern algebraic geometry,

Wadsworth, 1992. An elementary introduction to scheme theory. Griffiths, Phillip and Harris, Joseph: Principles of Algebraic Geometry, Wiley, 1978. A

comprehensive study of algebraic varieties over ℂ using analytic methods. Grothendieck, A., and Dieudonné, J: Eléments de Géométrie Algébrique. Publ. Math.

IHES 1960–1967. This was intended to cover everything in algebraic geometry in 13 massive books, i.e., it was supposed to do for algebraic geometry what Euclid’s “Elements” did for geometry. Unlike Euclid, they quit after 4 books.

Harris, Joe: Algebraic Geometry: A first course, Springer, 1992. It contains much beautiful

mathematics, in particular, many beautiful examples, but it is disorganized. Hartshorne: Algebraic Geometry, Springer 1977. Chapters II and III give an excellent

account of scheme theory and cohomology, so good in fact, that no one seems willing to write a competitor. The first chapter on varieties is very sketchy. Every student of algebraic geometry must master this book. The sequel to this course covers most of chapters II and III.

Kunz, Ernst: Introduction to Commutative Algebra and Algebraic Geometry, Birkhaüser, 1985. Spends a little too much space on the author’s own interests, but otherwise it is a well-written book, whose approach is not very different from that in this course.

Iitaka: Algebraic Geometry: an introduction to birational geometry of algebraic varieties, Springer, 1982. Schemes. Not as good as Hartshorne, but is more elementary, and

0 J.S. MILNE

covers some things that Hartshorne doesn’t. Mumford, David: Introduction to Algebraic Geometry, Harvard notes, 1966. QA564.M948.

The “red book” is a reprint of this. Mumford, David: Algebraic Geometry I: Complex Projective Varieties. Algebraic geome-

try over ℂ—some things become easier when one has the complex topology. Reid, Miles: Undergraduate Algebraic Geometry. A brief, elementary introduction. The

final section contains an interesting, but idiosyncratic, account of algebraic geometry in the twentieth century.

Finally, one shouldn’t forget some of the older books such as Hodge and Pedoe—they contain many interesting things that have not yet made their way into more modern accounts.

Introduction

Just as the starting point of linear algebra is the study of the solutions of systems of linear equations,

∑_{j=1}^{n} a_{ij}X_{j} = d_{i}, i = 1, …, m,  (*)

the starting point for algebraic geometry is the study of the solutions of systems of polynomial equations:

f_{i}(X_{1}, …, X_{n}) = 0, i = 1, …, m, f_{i} ∈ k[X_{1}, …, X_{n}].

Note immediately one difference between linear equations and polynomial equations: theorems for linear equations don’t depend on which field k you are working over,^{1} but those for polynomial equations depend on whether or not k is algebraically closed and (to a lesser extent) whether k has characteristic zero. Since I intend to emphasize the geometry in this course, we will work over algebraically closed fields until towards the end.

A better description of algebraic geometry is that it is the study of polynomial functions and the spaces on which they are defined (algebraic varieties), just as topology is the study of continuous functions and the spaces on which they are defined (topological spaces), differential geometry (=advanced calculus) the study of differentiable functions and the spaces on which they are defined (differential manifolds), and complex analysis the study of holomorphic functions and the spaces on which they are defined (Riemann surfaces and complex manifolds). The approach adopted in this course makes plain the similarities between these different fields. Of course, the polynomial functions form a much less rich class than the others, but by restricting our study to polynomials we are able to do calculus over any field: we can simply define

\frac{d}{dX} ∑ a_{i}X^{i} = ∑ ia_{i}X^{i−1}.

Moreover, calculations (on a computer) with polynomials are easier than with more general functions.

Consider a differentiable function f(x, y, z). In calculus, we learn that the equation

f(x, y, z) = C  (**)

defines a surface S in ℝ^{3}, and that the tangent space to S at a point P = (a, b, c) has equation^{2}

(\frac{∂f}{∂x} )_{P} (x − a) + (\frac{∂f}{∂y} )_{P} (y − b) + (\frac{∂f}{∂z} )_{P} (z − c) = 0.  (***)

1For example, suppose that the system (*) has coefficients a_{ij} ∈ k, and that K is a field containing k. Then (*) has a solution in k^{n} if and only if it has a solution in K^{n}, and the dimension of the spaces of solutions is the same for both fields. (Exercise!)

2Think of S as a level surface for the function f, and note that the equation is that of a plane through (a, b, c) perpendicular to the gradient vector (∇f)_{P} at P.)

The inverse function theorem says that a differentiable map α: S → S′ of surfaces is a local isomorphism at a point P ∈ S if it maps the tangent space at P isomorphically onto the tangent space at P′ = α(P).

Consider a polynomial f(x, y, z) with coefficients in a field k. In this course, we shall learn that the equation (**) defines a surface in k^{3}, and we shall use the equation (***) to define the tangent space at a point P on the surface. However, and this is one of the essential differences between algebraic geometry and the other fields, the inverse function theorem doesn’t hold in algebraic geometry. One other essential difference: 1/X is not the derivative of rational function of X, and in characteristic p ≠ 0, nor is X^{np−1}.

0. Algorithms for Polynomials

In this chapter, we first review some basic definitions from commutative algebra, and then we derive some algorithms for working in polynomial rings. The symbol k denotes a field (not necessarily algebraically closed).

Rings. Recall that all rings are required to be commutative and have an identity element 1, and that a homomorphism of rings is required to map 1 to 1. A subset R of a ring A is a subring if it is closed under addition, multiplication, the formation of negatives, and contains the identity element.^{3}

Ideals. Let A be a ring. Recall that an ideal 𝔞 in A is a subset such that (a) 𝔞 is a subgroup of A regarded as a group under adddition;

(b) a ∈ 𝔞, r ∈ A ⟹ ra ∈ A. The ideal generated by a subset S of A is the intersection of all ideals A containing 𝔞—it is easy to verify that this is in fact an ideal, and that it consists of all finite sums of the form ∑ r_{i}s_{i} with r_{i} ∈ A, s_{i} ∈ S. When S = {s_{1}, …, s_{m}}, we shall write (s_{1}, …, s_{m}) for the ideal it generates.

Let 𝔞 and 𝔟 be ideals in A. The set {a + b | a ∈ 𝔞, b ∈ 𝔟} is an ideal, denoted by 𝔞 + 𝔟. The ideal generated by {ab | a ∈ 𝔞, b ∈ 𝔟} is denoted by 𝔞𝔟. Note that 𝔞𝔟 ⊂ 𝔞 ∩ 𝔟. Clearly 𝔞𝔟 consists of all finite sums ∑ a_{i}b_{i} with a_{i} ∈ 𝔞 and b_{i} ∈ 𝔟, and if 𝔞 = (a_{1}, …, a_{m}) and 𝔟 = (b_{1}, …, b_{m}), then 𝔞𝔟 = (a_{1}b_{1}, …, a_{i}b_{j}, …, a_{m}b_{n}).

Let 𝔞 be an ideal of A. The set of cosets of 𝔞 in A forms a ring A/𝔞, and a ↦ a + 𝔞 is a homomorphism φ : A → A/𝔞. The map 𝔟 ↦ φ^{−1}(𝔟) is a one-to-one correspondence between the ideals of A/𝔞 and the ideals of A containing 𝔞.

An ideal 𝔭 if prime if 𝔭 ≠ A and ab ∈ 𝔭 ⟹ a ∈ 𝔭 or b ∈ 𝔭. Thus 𝔭 is prime if and only if A/𝔭 is nonzero and has the property that

ab = 0, b ≠ 0 ⟹ a = 0,

i.e., A/𝔭 is an integral domain.

An ideal 𝔪 is maximal if 𝔪 ≠ A and there does not exist an ideal 𝔫 contained strictly between 𝔪 and A. Thus 𝔪 is maximal if and only if A/𝔪 has no proper nonzero ideals, and so is a field. Note that

𝔪 maximal ⟹ 𝔪 prime.

The ideals of A × B are all of the form 𝔞 × 𝔟, with 𝔞 and 𝔟 ideals in A and B. (Proof: If 𝔠 is an ideal in A × B and (a, b) ∈ 𝔠, then (a, 0) = (a, b)(1, 0) ∈ 𝔠 and (0, b) = (a, b)(0, 1) ∈ 𝔠. This shows that 𝔠 = 𝔞 × 𝔟 with 𝔞 = {a | (a, b) ∈ 𝔠 some b} and 𝔟 = {b | (a, b) ∈ 𝔠 some a}.)

3The definition on page 2 of Atiyah-MacDonald is incorrect, since it omits the condition that x ∈ R ⟹ −x ∈ R—the subset ℕ of ℤ satisfies their conditions, but it is not a subring of ℤ.

Proposition 0.1. The following conditions on a ring A are equivalent: (a) every ideal in A is finitely generated;

(b) every ascending chain of ideals 𝔞_{1} ⊂ 𝔞_{2} ⊂ ⋯ becomes stationary, i.e., for some m, 𝔞_{m} = 𝔞_{m+1} = 𝔞_{m+1} = ⋯;

(c) every nonempty set of ideals in A has maximal element, i.e., an element not properly contained in any other ideal in the set.

Proof. (a) ⟹ (b): If 𝔞_{1} ⊂ 𝔞_{2} ⊂ ⋯ is an ascending chain, then ∪𝔞_{i} is again an ideal, and hence has a finite set {a_{1}, …, a_{n}} of generators. For some m, all the a_{i} belong 𝔞_{m}, and then

𝔞_{m} = 𝔞_{m+1} = ⋯ = 𝔞.

(b) ⟹ (c): If (c) is false, then there exists a nonempty set S of ideals with no maximal element. Let 𝔞_{1} ∈ S; because 𝔞_{1} is not maximal in S, there exists an ideal 𝔞_{2} in S that properly contains 𝔞_{1}. Similarly, there exists an ideal 𝔞_{3} in S properly containing 𝔞_{2}, etc.. In this way, we can construct an ascending chain of ideals 𝔞_{1} ⊂ 𝔞_{2} ⊂ 𝔞_{3} ⊂ ⋯ in S that never becoms stationary.

(c) ⟹ (a): Let 𝔞 be an ideal, and let S be the set of ideals 𝔟 ⊂ 𝔞 that are finitely generated. Let 𝔠 = (a_{1}, …, a_{r}) be a maximal element of S. If 𝔠 ≠ 𝔞, so that there exists an element a ∈ 𝔞, a ∉ 𝔠, then 𝔠′ = (a_{1}, …, a_{r}, a) ⊂ 𝔞 and properly contains 𝔠, which contradicts the definition of 𝔠. □

A ring A is Noetherian if it satisfies the conditions of the proposition. Note that, in a Noetherian ring, every ideal is contained in a maximal ideal (consider the set of all proper ideals of A containing the given ideal). In fact, this is true in any ring, but the proof for non-Noetherian rings requires the axiom of choice (Atiyah-MacDonald, p3).

Algebras. Let A be a ring. An A-algebra is a ring B together with a homomorphism i_{B} : A → B. A homomorphism of A-algebras B → C is a homomorphism of rings φ : B → C such that φ(i_{B}(a)) = i_{C}(a) for all a ∈ A.

An A-algebra B is said to be finitely generated (or of finite-type over A) if there exist elements x_{1}, …, x_{n} ∈ B such that every element of B can be expressed as a polynomial in the x_{i} with coefficients in i(A), i.e., such that the homomorphism A[X_{1}, …, X_{n}] → B sending X_{i} to x_{i} is surjective.

A ring homomorphism A → B is finite, and B is a finite A-algebra, if B is finitely generated as an A-module (the term “module-finite” is used in this context only by the English-insensitive).

Let k be a field, and let A be a k-algebra. If 1 ≠ 0 in A, then the map k → A is injective, and we can identify k with its image, i.e., we can regard k as a subring of A. If 1 = 0 in a ring R, the R is the zero ring, i.e., R = {0}.

Polynomial rings. Let k be a field. A monomial in X_{1}, …, X_{n} is an expression of the form

X_{1}^{a_{1}} ⋯ X_{n}^{a_{n}}, a_{j} ∈ ℕ.

The total degree of the monomial is ∑ a_{i}. We sometimes abbreviate it by X^{α}, α = (a_{1}, …, a_{n}) ∈ ℕ^{n}.

The elements of the polynomial ring k[X_{1}, …, X_{n}] are finite sums

∑ c_{a_{1}···a_{n}}X_{1}^{a_{1}} ⋯ X_{n}^{a_{n}}, c_{a_{1}···a_{n}} ∈ k, a_{j} ∈ ℕ,

with the obvious notions of equality, addition, and multiplication. Thus the monomials from a basis for k[X_{1}, …, X_{n}] as a k-vector space.

The ring k[X_{1}, …, X_{n}] is an integral domain, and the only units (i.e., elements with inverses in k[X_{1}, …, X_{n}]) in it are the nonzero constant polynomials. A nonconstant polynomial f(X_{1}, …, X_{n}) is irreducible if, whenever f = gh, one of g or h is a constant.

Theorem 0.2. The ring k[X_{1}, …, X_{n}] is a unique factorization domain, i.e., each nonzero nonconstant polynomial f can be written as a finite product of irreducible polynomials, and the factorization is unique up to order and constants.

Proof. This sometimes proved in Math 593. There is a detailed proof in Herstein, Topics in Algebra, 3.11. It proceeds by induction on the number of variables: if R is a unique factorization domain, then so also is R[X]. □

Corollary 0.3. A principal ideal (f) in k[X_{1}, …, X_{n}] is prime if and only if it is irreducible.

Proof. Assume (f) is a prime ideal. Then f can’t be a unit (otherwise (f) is the whole ring), and if f = gh then gh ∈ (f), which, because (f) is prime, implies that g or h is in (f), i.e., that one is divisible by f. Conversely, assume f is irreducible. If gh ∈ (f), then f|gh, which implies that f|g or f|h (here is where we use that k[X_{1}, …, X_{n}] is a unique factorization domain), i.e., that g or h ∈ (f). □

The two main results of this section will be:

(a) (Hilbert basis theorem) Every ideal in k[X_{1}, …, X_{n}] has a finite set of generators (in fact, of a special sort).

(b) There exists an algorithm for deciding whether a polynomial f belongs to an ideal (g_{1}, …, g_{r}).

Division in k[X]. The division algorithm allows us to divide a nonzero polynomial into another: let f and g be polynomials in k[X] with g ≠ 0; then there exist unique polynomials q, r ∈ k[X] such that f = qg + r with either r = 0 or deg r < deg g. Moreover, there is an algorithm for deciding whether f ∈ (g), namely, find r and check whether it is zero.

In Maple, quo(f,g,X) computes q; rem(f,g,X) computes r.

The Euclidean algorithm allows you to pass from a finite set of generators for an ideal in k[X] to a single generator.

Orderings on monomials. Before we can describe an algorithm for dividing in k[X_{1}, …, X_{n}], we shall need to choose a way of ordering monomials. Essentially this amounts to defining and ordering on ℕ^{n}. There are two main systems, the first of which is preferred by humans, and the second by machines.

(Pure) lexicographic ordering (lex). Here monomials are orderd by lexicographic (dictionary) order. More precisely, let α = (a_{1}, …, a_{n}) and β = (b_{1}, …, b_{n}) be two elements of ℕ^{n}; then

α > β, and X^{α} > X^{β} (lexicographic ordering)

if, in the vector difference α − β ∈ ℤ, the left-most nonzero entry is positive. For example,

XY^{2} > Y^{3}Z^{4}; X^{3}Y^{2}Z^{4} > X^{3}Y^{2}Z.

Note that this isn’t quite how the dictionary would order them: it would put XXXYYZZZZ after XXXYYZ.

Graded reverse lexicographic order (grevlex). Here monomials are ordered by total degree, with ties broken by reverse lexicographic ordering. In more detail, α > β if ∑ a_{i} > ∑ b_{i}, or ∑ a_{i} = ∑ b_{i} and in α − β the right-most nonzero entry is negative. For example:

X^{4}Y^{4}Z^{7} > X^{5}Y^{5}Z^{4} (total degree greater)
XY^{5}Z^{2} > X^{4}Y Z^{3}, X^{5}Y Z > X^{4}Y Z^{2}.

Orderings on k[X_{1}, …, X_{n}]. Fix an ordering on the monomials in k[X_{1}, …, X_{n}]. Then we can write an element f of k[X_{1}, …, X_{n}] in a canonical fashion, by reordering its elements in decreasing order. For example, we would write

f = 4XY^{2}Z + 4Z^{2} − 5X^{3} + 7X^{2}Z^{2}

as

f = −5X^{3} + 7X^{2}Z^{2} + 4XY^{2}Z + 4Z^{2}  ( lex)

or

f = 4XY^{2}Z + 7X^{2}Z^{2} − 5X^{3} + 4Z^{2}  ( grevlex)
Let f = ∑ a_{α}X^{α} ∈ k[X_{1}, …, X_{n}]. Write it in decreasing order:
f = a_{α_{0}}X^{α_{0}} + a_{α_{1}}X^{α_{1}} + ⋯, α_{0} > α_{1} > ⋯, a_{α_{0}} ≠ 0.

Then we define:

(a) the multidegree of f to be multdeg(f) = α_{0};

(b) the leading coefficient of f to be LT(f) = a_{α_{0}};

(c) the leading monomial of f to be LM(f) = X^{α_{0}};

(d) the leading term of f to be a_{α_{0}}X^{α_{0}}. For example, for the polynomial f = 4XY^{2}Z + ⋯, the multidegree is (1, 2, 1), the leading coefficient is 4, the leading monomial is XY^{2}Z, and the leading term is 4XY^{2}Z.

The division algorithm in k[X_{1}, …, X_{n}]. Fix a monomial ordering in ℕ^{n}. Suppose given a polynomial f and an ordered set (g_{1}, …, g_{s}) of polynomials; the division algorithm then constructs polynomials a_{1}, …, a_{s} and r such that

f = a_{1}g_{1} + ⋯ + a_{s}g_{s} + r

where either r = 0 or no monomial in r is divisible by any of LT(g_{1}), …, LT(g_{s}). Step 1: If LT(g_{1})|LT(f), divide g_{1} into f to get

f = a_{1}g_{1} + h, a_{1} = \frac{LT(f)}{LT(g_{1})} ∈ k[X_{1}, …, X_{n}].

If LT(g_{1})|LT(h), repeat the process until

f = a_{1}g_{1} + f_{1}

(different a_{1}) with LT(f_{1}) not divisible by LT(g_{1}). Now divide g_{2} into f_{1}, and so on, until

f = a_{1}g_{1} + ⋯ + a_{s}g_{s} + r_{1}

with LT(r_{1}) not divisible by any of LT(g_{1}), …, LT(g_{s}).

Step 2: Rewrite r_{1} = LT(r_{1}) + r_{2}, and repeat Step 1 with r_{2} for f. Now

f = a_{1}g_{1} + ⋯ + a_{s}g_{s} + LT(r_{1}) + r_{3}

(different a_{i}’s).

Step 3: Rewrite r_{3} = LT(r_{3}) + r_{4}, and repeat Step 1 with r_{4} for f. Now

f = a_{1}g_{1} + ⋯ + a_{s}g_{s} + LT(r_{1}) + LT(r_{3}) + r_{3}

(different a_{i}’s).

Continue until you achieve a remainder with the required property. In more detail,^{4} after dividing through once by g_{1}, …, g_{s}, you repeat the process until no leading term of one of the g_{i}’s divides the leading term of the remainder. Then you discard the leading term of the remainder, and repeat....

Example 0.4. (a) Consider

f = X^{2}Y + XY^{2} + Y^{2}, g_{1} = XY − 1, g_{2} = Y^{2} − 1.

First, on dividing g_{1} into f, we obtain

X^{2}Y + XY^{2} + Y^{2} = (X + Y)(XY − 1) + X + Y^{2} + Y.

4This differs from the algorithm in Cox et al., p63, which says to go back to g_{1} after every successful division.

This completes the first step, because the leading term of Y^{2} − 1 does not divide the leading term of the remainder X + Y^{2} + Y. We discard X, and write

Y^{2} + Y = 1 · (Y^{2} − 1) + Y + 1.

Altogether

X^{2}Y + XY^{2} + Y^{2} = (X + Y) · (XY − 1) + 1 · (Y^{2} − 1) + X + Y + 1.

(b) Consider the same polynomials, but with a different order for the divisors

f = X^{2}Y + XY^{2} + Y^{2}, g_{1} = Y^{2} − 1, g_{2} = XY − 1.

In the first step,

X^{2}Y + XY^{2} + Y^{2} = (X + 1) · (Y^{2} − 1) + X · (XY − 1) + 2X + 1.

Thus, in this case, the remainder is 2X + 1.

Remark 0.5. (a) If r = 0, then f ∈ (g_{1}, …, g_{s}).

(b) Unfortunately, the remainder one obtains depends on the ordering of the g_{i}’s. For example, (lex ordering)

XY^{2} − X = Y · (XY + 1) + 0 · (Y^{2} − 1) + −X − Y

but

XY^{2} − X = X · (Y^{2} − 1) + 0 · (XY − 1) + 0.

Thus, the division algorithm (as stated) will not provide a test for f lying in the ideal generated by g_{1}, …, g_{s}.

Monomial ideals. In general, and ideal 𝔞 will contain a polynomial without containing the individual terms of the polynomial; for example, the ideal 𝔞 = (Y^{2} − X^{3}) doesn’t contain Y^{2} or X^{3}.

Definition 0.6. An ideal 𝔞 is monomial if

∑ c_{α}X^{α} ∈ 𝔞 ⟹ X^{α} ∈ 𝔞 all α with c_{α} ≠ 0.

Proposition 0.7. Let 𝔞 be a monomial ideal, and let A = {α | X^{α} ∈ 𝔞}. Then A satisfies the condition

α ∈ A, β ∈ ℕ^{n} ⟹ α + β ∈ A.  (*)

and 𝔞 is the k-subspace of k[X_{1}, …, X_{n}] generated by the X^{α}, α ∈ A. Conversely, if A is a subset of ℕ^{n} satisfying (*), then the k-subspace 𝔞 of k[X_{1}, …, X_{n}] generated by {X^{α} | α ∈ A} is a monomial ideal.

Proof. It is clear from its definition that a monomial ideal 𝔞 is the k-subspace of k[X_{1}, …, X_{n}] generated by the set of monomials it contains. If X^{α} ∈ 𝔞 and X^{β} ∈ k[X_{1}, …, X_{n}], then X^{α}X^{β} = X^{α+β} ∈ 𝔞, and so A satisfies the condition (*). Conversely,

(∑_{α∈A} c_{α}X^{α}) ( ∑_{β∈ℕ^{n}} d_{β}X^{β}) = ∑_{α,β} c_{α}d_{β}X^{α+β} (finite sums),

and so if A satisfies (*), then the subspace generated by the monomials X^{α}, α ∈ A, is an ideal. □

The proposition gives a classification of the monomial ideals in k[X_{1}, …, X_{n}]: they are in one-to-one correspondence with the subsets A of ℕ^{n} satisfying (*). For example, the monomial ideals in k[X] are exactly the ideals (X^{n}), n ≥ 1, and the zero ideal (corresponding to the empty set A). We write

< X^{α} | α ∈ A >

for the ideal corresponding to A (subspace generated by the X^{α}, α ∈ A).

Lemma 0.8. Let S be a subset of ℕ^{n}. Then the ideal 𝔞 generated by {X^{α} | α ∈ S} is the monomial ideal corresponding to

A =_{df} {β ∈ ℕ^{n} | β − α ∈ ℕ^{n}, some α ∈ S}.

Thus, a monomial is in 𝔞 if and only if it is divisible by one of the X^{α}, α ∈ S.

Proof. Clearly A satisfies (*), and 𝔞 ⊂ < X^{β} | β ∈ A >. Conversely, if β ∈ A, then β − α ∈ ℕ^{n} for some α ∈ S, and X^{β} = X^{α}X^{β−α} ∈ 𝔞. The last statement follows from the fact that X^{α}|X^{β} ⟺ β − α ∈ ℕ^{n}. □

Let A ⊂ ℕ^{2} satisfy (*). By studying the geometry of A, it is clear that there is a finite set of elements S = {α_{1}, …, α_{s}} of A such that

A = {β ∈ ℕ^{2} | β − α ∈ ℕ^{2}, some α ∈ S}.

(The α_{i}’s are the “corners” of A.) The 𝔞 is the ideal generated by the monomials X^{α_{i}}. This suggests the following result.

Theorem 0.9 (Dickson’s Lemma). Let 𝔞 be the monomial ideal corresponding to the subset A ⊂ ℕ^{n}. Then 𝔞 is generated by a finite subset of {X^{α} | α ∈ A}.

Proof. A formal proof, by induction on the number of variables, can be found in the book by Cox et al. p70. □

Hilbert basis theorem.

Definition 0.10. For a nonzero ideal 𝔞 in k[X_{1}, …, X_{n}], we let (LT(𝔞)) be the ideal generated by

{LT(f) | f ∈ 𝔞}.

Lemma 0.11. Let 𝔞 be a nonzero ideal in k[X_{1}, …, X_{n}]; then (LT(𝔞)) is a monomial ideal, and it equals (LT(g_{1}), …,LT(g_{n})) for some g_{1}, …, g_{n} ∈ 𝔞.

Proof. Since (LT(𝔞)) can also be described as the ideal generated by the leading monomials (rather than the leading terms) of elements of 𝔞, it follows from Lemma 0.8 that it is a monomial. Now Dickson’s lemma shows that it equals (LT(g_{1}), …, LT(g_{s})) for some g_{i} ∈ 𝔞. □

Theorem 0.12 (Hilbert Basis Theorem). Every ideal 𝔞 in k[X_{1}, …, X_{n}] is finitely generated; more precisely, 𝔞 = (g_{1}, …, g_{s}) where g_{1}, …, g_{s} are any elements of 𝔞 whose leading terms generate LT(𝔞).

Proof. Let f ∈ 𝔞. On applying the division algorithm, we find

f = a_{1}g_{1} + ⋯ + a_{s}g_{s} + r, a_{i}, r ∈ k[X_{1}, …, X_{n}],

where either r = 0 or no monomial occurring in it is divisible by any LT(g_{i}). But r = f − ∑ a_{i}g_{i} ∈ 𝔞, and therefore LT(r) ∈ LT(𝔞) = (LT(g_{1}), …, LT(g_{s})), which, according to Lemma 0.8, implies that every monomial occurring in r is divisible by one in LT(g_{i}). Thus r = 0, and g ∈ (g_{1}, …, g_{s}). □

Standard (Gröbner) bases. Fix a monomial ordering of k[X_{1}, …, X_{n}].

Definition 0.13. A finite subset S = {g_{1}, …, g_{s}} of an ideal 𝔞 is standard (Grobner, Groebner, Gröbner) bases for^{5} 𝔞 if

(LT(g_{1}), …, LT(g_{s})) = LT(𝔞).

In other words, S is a standard basis if the leading term of every element of 𝔞 is divisible by at least one of the leading terms of the g_{i}.

Theorem 0.14. Every ideal has a standard basis, and it generates the ideal; if {g_{1}, …, g_{s}} is a standard basis for 𝔞, the f ∈ 𝔞 ⟺ the remainder on division by the g_{i} is 0.

Proof. Our proof of the Hilbert basis theorem shows that 𝔞 has a standard basis. Let f ∈ 𝔞. The argument in the same proof, that the remainder of f on division by g_{1}, …, g_{s} is zero, used only that {g_{1}, …, g_{s}} is a standard basis for 𝔞. Thus f ∈ (g_{1}, …, g_{s}), and we see that 𝔞 = (g_{1}, …, g_{s}). □

Remark 0.15. The proposition shows that, for f ∈ 𝔞, the remainder of f on division by {g_{1}, …, g_{s}} is independent of the order of the g_{i} (in fact, it’s always zero). This is not true if f ∉ 𝔞—see the example using Maple at the end of this section.

5Standard bases we first introduced (under that name) by Hironaka in the mid-1960s, and independently, but slightly later, by Buchberger in his Ph.D. thesis. Buchberger named them after his thesis adviser Gröbner.

Let 𝔞 = (f_{1}, …, f_{s}). Typically, {f_{1}, …, f_{s}} will fail to be a standard basis because in some expression

cX^{α}f_{i} − dX^{β}f_{j}, c, d ∈ k,  (**)

the leading terms will cancel, and we will get a new leading term not in the ideal generated by the leading terms of the f_{i}. For example,

X^{2} = X · (X^{2}Y + X − 2Y^{2}) − Y · (X^{3} − 2XY)

is in the ideal generated by X^{2}Y + X − 2Y^{2} and X^{3} − 2XY but it is not in the ideal generated by their leading terms.

There is an algorithm for transforming a set of generators for an ideal into a standard basis, which, roughly speaking, makes adroit use of equations of the form (**) to construct enough new elements to make a standard basis—see Cox et al. pp80–87.

We now have an algorithm for deciding whether f ∈ (f_{1}, …, f_{r}). First transform {f_{1}, …, f_{r}} into a standard basis {g_{1}, …, g_{s}}, and then divide f by g_{1}, …, g_{s} to see whether the remainder is 0 (and hence lies in the ideal) or nonzero (and it doesn’t). This algorithm is implemented in Maple—see below.

A standard basis {g_{1}, …, g_{s}} is minimal if each g_{i} has leading coefficient 1 and, for all i, the leading term of g_{i} does not belong to the ideal generated by the leading terms of the remaining g’s. A standard basis {g_{1}, …, g_{s}} is reduced if each g_{i} has leading coefficient 1 and if, for all i, no monomial of g_{i} lies in the ideal generated by the leading terms of the remaining g’s. One can prove (Cox et al. p91) that every nonzero ideal has a unique reduced standard basis.

Remark 0.16. Consider polynomials f, g_{1}, …, g_{s} ∈ k[X_{1}, …, X_{n}]. The algorithm that replaces g_{1}, …, g_{s} with a standard basis works entirely within k[X_{1}, …, X_{n}], i.e., it doesn’t require a field extension. Likewise, the division algorithm doesn’t require a field extension. Because these operations give well-defined answers, whether we carry them out in k[X_{1}, …, X_{n}] or in K[X_{1}, …, X_{n}], K ⊃ k, we get the same answer. Maple appears to work in the subfield of ℂ generated over ℚ by all the constants occurring in the polynomials.

The above is, in large part, a summary of Cox et al. pp1–111, to which the reader is referred for more details.

The original notes conclude with the annotated transcript of a session on Maple applying the above algorithm to show that

q = 3x^{3}yz^{2} − xz^{2} + y^{3} + yz

doesn’t lie in the ideal

(x^{2} − 2xz + 5, xy^{2} + yz^{3}, 3y^{2} − 8z^{3}).

1. Algebraic Sets

We now take k to be an algebraically closed field.

Definition of an algebraic set. An algebraic subset V (S) of k^{n} is the set of common zeros of some set S of polynomials in k[X_{1}, …, X_{n}] :

V (S) = {(a_{1}, …, a_{n}) ∈ k^{n} | f(a_{1}, …, a_{n}) = 0 all f(X_{1}, …, X_{n}) ∈ S}.

Note that S ⊂ S′ ⟹ V (S) ⊃ V (S′); the more equations we have, the fewer the solutions.

Recall that the ideal 𝔞 generated by a set S consists of all finite sums

∑ f_{i}g_{i}, f_{i} ∈ k[X_{1}, …, X_{n}], g_{i} ∈ S.

Such a sum ∑ f_{i}g_{i} is zero at any point at which the g_{i} are zero, and so V (S) ⊂ V (𝔞), but the reverse conclusion is also true because S ⊂ 𝔞. Thus V (S) = V (𝔞)—the zero set of S is the same as that of the ideal generated by S. Hence the algebraic sets can also be described as the sets of the form V (𝔞), 𝔞 an ideal in k[X_{1}, …, X_{n}].

Example 1.1. (a) If S is a system of homogeneous linear equations, then V (S) is a subspace of k^{n}. If S is a system of nonhomogeneous linear equations, V (S) is either empty or is the translate of a subspace of k^{n}.

(b) If S consists of the single equation

Y^{2} = X^{3} + aX + b, 4a^{3} + 27b^{2} ≠ 0,

then V (S) is an elliptic curve. (For more on elliptic curves, and their relation to Fermat’s last theorem, see the New York Times.)

[Sketch the curve; explain that we generally picture algebraic sets as if the field k were ℝ, even though it isn’t, ℝ not being algebraically closed.]

(c) If S is the empty set, then V (S) = k^{n}.

(d) The algebraic subsets of k are the finite subsets (including ∅) and k itself.

(e) Some generating sets for an ideal will be more useful than others for determining what the algebraic set is. For example, a Gröbner basis for the ideal

𝔞 = (X^{2} + Y^{2} + Z^{2} − 1, X^{2} + Y^{2} − Y, X − Z)

is (according to Maple)

X − Z, Y^{2} − 2Y + 1, Z^{2} − 1 + Y.

The middle polynomial has (double) root 1, and it follows that V (𝔞) consists of the single point (0,1,0).

The Hilbert basis theorem. In our definition, we allowed an algebraic set to be the zero set of an infinite set of polynomials, but the Hilbert basis theorem shows that it will then automatically be the zero set of a finite set of polynomials. More precisely, the theorem shows that every ideal in k[X_{1}, …, X_{n}] can be generated by a finite set of elements, and we have already observed that any set of generators of an ideal has the same zero set as the ideal.

We sketched a proof of the Hilbert basis theorem in the last section. Here we give the slick nonalgorithmic proof.

Theorem 1.2 (Hilbert Basis Theorem). The ring k[X_{1}, …, X_{n}] is Noetherian, i.e., every ideal is finitely generated.

Proof. For n = 1, this is proved in Math 512 and Math 593: k[X] is a principal ideal domain, which means that every ideal is generated by a single element. We shall prove the theorem by induction on n. Note that the obvious map

k[X_{1}, …, X_{n−1}][X_{n}] → k[X_{1}, …, X_{n}]

is an isomorphism—this simply says that every polynomial f in n variables X_{1}, …, X_{n} can be expressed uniquely as a polynomial in X_{n} with coefficients in k[X_{1}, …, X_{n−1}] :

f(X_{1}, …, X_{n}) = a_{0}(X_{1}, …, X_{n−1})X_{n}^{r} + ⋯ + a_{r}(X_{1}, …, X_{n−1}).

Thus the next lemma will complete the proof.

Lemma 1.3. If A is Noetherian, then so also is A[X].

Proof. For a polynomial

f(X) = a_{0}X^{r} + a_{1}X^{r−1} + ⋯ + a_{r}, a_{i} ∈ A, a_{0} ≠ 0,

r is called the degree of f, and a_{0} is its leading coefficient. We call 0 the leading coefficient of the polynomial 0.

Let 𝔞 be an ideal in A[X]. The leading coefficients of the polynomials in 𝔞 form an ideal 𝔞′ in A, and since A is Noetherian, 𝔞′ will be finitely generated. Let g_{1}, …, g_{m} be elements of 𝔞 whose leading coefficients generate 𝔞′, and let r be the maximum degree of the g_{i}.

Now let f ∈ 𝔞, and suppose f has degree s > r, say, f = aX^{s} + · · ·. Then a ∈ 𝔞′, and so we can write

a = ∑ b_{i}a_{i}, b_{i} ∈ A, a_{i} = leading coefficient of g_{i}.

Now

f − ∑ b_{i}g_{i}X^{s−r_{i}}, r_{i} = deg(g_{i}),

has degree < deg(f). By continuing in this way, we find that

f ≡ f_{t} mod (g_{1}, …, g_{m})

with f_{t} a polynomial of degree t < r.

For each d < r, let 𝔞_{d} be the subset of A consisting of 0 and the leading coefficients of all polynomials in 𝔞 of degree d; it is again an ideal in A. Let g_{d,1}, …, g_{d,m_{d}} be polynomials of degree d whose leading coefficients generate 𝔞_{d}. Then the same argument as above shows that any polynomial f_{d} in 𝔞 of degree d can be written

f_{d} ≡ f_{d−1} mod (g_{d,1}, …, g_{d,m_{d}})

with f_{d−1} of degree ≤ d − 1. On applying this remark repeatedly we find that

f_{t} ∈ (g_{r−1,1}, …, g_{r−1,m_{r−1}}, …, g_{0,1}, …, g_{0,m_{0}}).

Hence

f ∈ (g_{1}, …, g_{m}, g_{r−1,1}, …, g_{r−1,m_{r−1}}, …, g_{0,1}, …, g_{0,m_{0}}),

and so the polynomials g_{1}, …, g_{0,m_{0}} generate 𝔞.

Aside 1.4. One may ask how many elements are needed to generate an ideal 𝔞 in k[X_{1}, …, X_{n}], or, what is not quite the same thing, how many equations are needed to define an algebraic set V. When n = 1, we know that every ideal is generated by a single element. Also, if V is a linear subspace of k^{n}, then linear algebra shows that it is the zero set of n − dim(V) polynomials. All one can say in general, is that at least n − dim(V) polynomials are needed to define V (see §6), but often more are required. Determining exactly how many is an area of active research. Chapter V of the book by Kunz contains a good discussion of this problem.

The Zariski topology.

Proposition 1.5. There are the following relations:

(i) 𝔞 ⊂ 𝔟 ⟹ V (𝔞) ⊃ V (𝔟); (ii) V (0) = k^{n}; V (k[X_{1}, …, X_{n}]) = ∅; (iii) V (𝔞𝔟) = V (𝔞 ∩ 𝔟) = V (𝔞) ∪ V (𝔟); (iv) V (∑ 𝔞_{i}) = ∩V (𝔞_{i}).

Proof. The first two statements are obvious. For (iii), note that

𝔞𝔟 ⊂ 𝔞 ∩ 𝔟 ⊂ 𝔞, 𝔟 ⟹ V (𝔞𝔟) ⊃ V (𝔞 ∩ 𝔟) ⊃ V (𝔞) ∪ V (𝔟).

For the reverse inclusions, observe that if a ∉ V (𝔞) ∪ V (𝔟), then there exist f ∈ 𝔞, g ∈ 𝔟 such that f(a) ≠ 0, g(a) ≠ 0; but then (fg)(a) ≠ 0, and so a ∉ V (𝔞𝔟). For (iv) recall that, by definition, ∑ 𝔞_{i} consists of all finite sums of the form ∑ f_{i}, f_{i} ∈ 𝔞_{i}—it is the smallest ideal containing all the 𝔞_{i}’s. Thus (iv) is obvious. □

Statements (ii), (iii), and (iv) show that the algebraic subsets of k^{n} satisfy the axioms to be the closed subsets for a topology on k^{n}: both the whole space and the empty set are closed; a finite union of closed sets is closed; an arbitrary intersection of closed sets is closed. This topology is called the Zariski topology. It has many strange properties (for example, already on k one sees that it not Hausdorff), but it is nevertheless of great importance.

We know that the proper closed subsets of k are just the finite sets and k. Call a curve in k^{2} the set of solutions of an equation

f(X, Y) = 0,

with f(X, Y) a nonzero irreducible polynomial. Then we shall see in (1.24) below that, apart from k^{2} itself, the closed sets in k^{2} are finite unions of (isolated) points and curves. Note that the Zariski topology on ℂ^{n} is much coarser than the complex topology.

The Hilbert Nullstellensatz. We wish to examine the relation between the algebraic subsets of k^{n} and the ideals of k[X_{1}, …, X_{n}], but first we consider the question of when a set of polynomials has a common zero, i.e., when the equations

g(X_{1}, …, X_{n}) = 0, g ∈ 𝔞,

are “consistent”. Obviously, the equations

g_{i}(X_{1}, …, X_{n}) = 0, i = 1, …, m

are inconsistent if there exist f_{i} ∈ k[X_{1}, …, X_{n}] such that

∑ f_{i}g_{i} = 1,

i.e., if 1 ∈ (g_{1}, …, g_{m}) or, equivalently, (g_{1}, …, g_{m}) = k[X_{1}, …, X_{n}]. The next theorem provides a converse to this.

Theorem 1.6 (Hilbert Nullstellensatz). Every proper ideal 𝔞 in k[X_{1}, …, X_{n}] has a zero in k^{n}.

Proof. A point a ∈ k^{n} defines a homomorphism “evaluate at a”

k[X_{1}, …, X_{n}] → k, f(X_{1}, …, X_{n}) ↦ f(a_{1}, …, a_{n}),

and clearly a ∈ V (𝔞) ⟺ 𝔞 ⊂ kernel of this map.

Conversely, if φ: k[X_{1}, …, X_{n}] → k is a homomorphism of k-algebras such that Ker(φ) ⊃ 𝔞, then

(a_{1}, …, a_{n}) =_{df} (φ(X_{1}), …, φ(X_{n}))

lies in V (𝔞). Thus, to prove the theorem, we prove that there exists a k-algebra homomorphism k[X_{1}, …, X_{n}]/𝔞 → k.

Since every proper ideal 𝔞 is contained in a maximal ideal 𝔪, and V (𝔞) ⊃ V (𝔪), it suffices to prove the theorem for a maximal ideal 𝔪. Then K =_{df} k[X_{1}, …, X_{n}]/𝔪 is a field, and it is finitely generated as an algebra over k (by X_{1} + 𝔪, …, X_{n} + 𝔪). To complete the proof, we must show K = k. The next lemma accomplishes this.

Although we shall apply the lemma only in the case that k is algebraically closed, in order to make the induction in its proof work, we need to allow arbitrary k’s in the statement.

Lemma 1.7 (Zariski’s Lemma). Let k ⊂ K be fields (k not necessarily algebraically closed). If K is finitely generated^{6} as an algebra over k, then K is algebraic over k. (Hence K = k if k is algebraically closed.)

Proof. We shall prove this by induction on r, the minimum number of elements required to generate K as a k-algebra. Suppose first that r = 1, so that K = k[x] for some x ∈ K. Write k[X] for the polynomial ring over k in the single variable X, and consider the homomorphism of k-algebras k[X] → K, X ↦ x. If x is not algebraic over k, then this is an isomorphism k[X] → K, which contradicts the condition that K be a field. Therefore x is algebraic over k, and this implies that every element of K = k[x] is algebraic over k (see Math 594).

For the general case, we need to use results about integrality (see the Appendix to this Section). Consider an integral domain A with field of fractions K, and a field L containing K. An element of L is said to be integral over A if it satisfies an equation of the form

X^{n} + a_{1}X^{n−1} + ⋯ + a_{n} = 0, a_{i} ∈ A.

We shall need three facts:

(a) The set of elements of L integral over A is a subring of L.

(b) If β ∈ L is algebraic over K, then aβ is integral over A for some a ∈ A.

(c) If A is a unique factorization domain, then an element of K that is integral

over A must lie in A. Now suppose that K can be generated (as a k-algebra) by r elements, say, K = k[x_{1}, …, x_{r}]. If the conclusion of the lemma is false for K/k, then at least one x_{i}, say x_{1}, is not algebraic over k. Thus, as before, k[x_{1}] is a polynomial ring in one variable over k (≈ k[X]), and its field of fractions k(x_{1}) is a subfield of K. Clearly K is generated as a k(x_{1})-algebra by x_{2}, …, x_{r}, and so the induction hypothesis implies that x_{2}, …, x_{r} are algebraic over k(x_{1}). From (b) we deduce that there exists a d_{i} ∈ k[x_{1}] such that d_{i}x_{i} is integral over k[x_{1}]. Write d = ∏ d_{i}.

Let f ∈ K; by assumption, f is a polynomial in the x_{i} with coefficients in k. For a sufficiently large N, d^{N}f will be a polynomial in the d_{i}x_{i}. Then (a) implies that d^{N}f is integral over k[x_{1}]. When we apply this to an element f of k(x_{1}), (c) shows that d^{N}f ∈ k[x_{1}]; i.e., k(x_{1}) = ∪_{N}d^{−N}k[x_{1}]. But this is absurd, because k[x_{1}] (≈ k[X]) has infinitely many distinct irreducible polynomials^{7} that can occur as denominators of elements of k(x_{1}). □

6See p4 for a definition of this notion. Warning: internal references to page numbers will be slightly off.

The correspondence between algebraic sets and ideals. For a subset W of k^{n}, we write I(W) for the set of polynomials that are zero on W:

I(W) = {f ∈ k[X_{1}, …, X_{n}] | f(a) = 0 all a ∈ W }.

It is an ideal in k[X_{1}, …, X_{n}]. There are the following formulas:

(i) V ⊂ W ⟹ I(V) ⊃ I(W); (ii) I(∅) = k[X_{1}, …, X_{n}]; I(k^{n}) = 0; (iii) I(∪W_{i}) = ∩I(W_{i}).

Only the statement I(k^{n}) = 0, i.e., that every nonzero polynomial is nonzero at some point of k^{n}, is nonobvious. It is not difficult to prove this directly—in fact it’s true for any infinite field k—but it also follows easily from the Nullstellensatz (see (1.11a) below).

Example 1.8. Let P be the point (a_{1}, …, a_{n}). Clearly I(P) ⊃ (X_{1} − a_{1}, …, X_{n} − a_{n}), but (X_{1} − a_{1}, …, X_{n} − a_{n}) is a maximal ideal, because “evaluation at (a_{1}, …, a_{n})” defines an isomorphism

k[X_{1}, …, X_{n}]/(X_{1} − a_{1}, …, X_{n} − a_{n}) → k.

As I(P) ≠ k[X_{1}, …, X_{n}], we must have I(P) = (X_{1} − a_{1}, …, X_{n} − a_{n}). The radical rad(𝔞) of an ideal 𝔞 is defined to be

{f | f^{r} ∈ 𝔞, some r ∈ ℕ, r > 0}.

It is again an ideal, and rad(rad(𝔞)) = rad(𝔞).

An ideal is said to be radical if it equals its radical, i.e., f^{r} ∈ 𝔞 ⟹ f ∈ 𝔞. Equivalently, 𝔞 is radical if and only if A/𝔞 is a reduced ring, i.e., a ring without nonzero nilpotent elements (elements some power of which is zero). Since an integral domain is reduced, a prime ideal (a fortiori a maximal ideal) is radical.

If 𝔞 and 𝔟 are radical, then 𝔞∩𝔟 is radical, but 𝔞+𝔟 need not be. (Consider, for example, 𝔞 = (X^{2} − Y) and 𝔟 = (X^{2} + Y); they are both prime ideals in k[X, Y], but X^{2} ∈ 𝔞 + 𝔟, X ∉ 𝔞 + 𝔟.)

As f^{r}(a) = f(a)^{r}, f^{r} is zero wherever f is zero, and so I(W) is radical. In particular, IV (𝔞) ⊃ rad(𝔞). The next theorem states that these two ideals are equal.

7If k is infinite, then consider the polynomials X − a, and if k is finite, consider the minimum polynomials of generators of the extension fields of k. Alternatively, and better, adapt Euclid’s proof that there are infinitely many prime numbers.

Theorem 1.9 (Strong Hilbert Nullstellensatz). (a) The ideal IV (𝔞) is the radical of 𝔞; in particular, IV (𝔞) = 𝔞 if 𝔞 is a radical ideal.

(b) The set V I(W) is the smallest algebraic subset of k^{n} containing W; in particular, V I(W) = W if W is an algebraic set.

Proof. (a) We have already noted that IV (𝔞) ⊃ rad(𝔞). For the reverse inclusion, consider h ∈ IV (𝔞); we have to show that some power of h belongs to 𝔞. We may assume h ≠ 0 as 0 ∈ 𝔞. We are given that h is identically zero on V (𝔞), and we have to show that h^{N} ∈ 𝔞 for some N > 0. Let g_{1}, …, g_{m} be a generating set for 𝔞, and consider the system of m + 1 equations in n + 1 variables, X_{1}, …, X_{n}, Y,

\{ \begin{matrix}g_{i}(X_{1}, …, X_{n}) & = 0, i = 1, …, m, \\ 1 − Y h(X_{1}, …, X_{n}) & = 0. \end{matrix}

If (a_{1}, …, a_{n}, b) satisfies the first m equations, then (a_{1}, …, a_{n}) ∈ V (𝔞); consequently, h(a_{1}, …, a_{n}) = 0, and (a_{1}, …, a_{n}, b) doesn’t satisfy the last equation. According to the original Nullstellensatz, there therefore exist f_{i} ∈ k[X_{1}, …, X_{n}, Y] such that

1 = ∑_{i=1}^{m} f_{i}g_{i} + f_{m+1} · (1 − Y h).

Regard this as an identity in the field k(X_{1}, …, X_{n}, Y) and substitute 1/h for Y. We obtain the identity

1 = ∑_{i=1}^{m} f_{i}(X_{1}, …, X_{n}, \frac{1}{h}) · g_{i}(X_{1}, …, X_{n})

in k(X_{1}, …, X_{n}). Clearly

f_{i}(X_{1}, …, X_{n}, \frac{1}{h}) = \frac{polynomial in X_{1}, …, X_{n}}{h^{N_{i}}}

for some N_{i}. Let N be the largest of the N_{i}. On multiplying the identity by h^{N} we obtain an equation

h^{N} = ∑ (polynomial in X_{1}, …, X_{n}) · g_{i}(X_{1}, …, X_{n}),

which shows that h^{N} ∈ 𝔞.

(b) Let V be an algebraic set containing W, and write V = V (𝔞). Then 𝔞 ⊂ I(W), and so V (𝔞) ⊃ V I(W). □

Corollary 1.10. The map 𝔞 ↦ V (𝔞) defines a one-to-one correspondence between the set of radical ideals in k[X_{1}, …, X_{n}] and the set of algebraic subsets of k^{n}; its inverse is I.

Proof. We know that IV (𝔞) = 𝔞 if 𝔞 is a radical ideal, and that V I(W) = W if W is an algebraic set. □

Remark 1.11. (a) Note that V (0) = k^{n}, and so

I(k^{n}) = IV (0) = rad(0) = 0,

as claimed above.

(b) The one-to-one correspondence in the corollary is order inverting. Therefore the maximal proper radical ideals correspond to the minimal nonempty algebraic sets. But the maximal proper radical ideals are simply the maximal ideals in k[X_{1}, …, X_{n}], and the minimal nonempty algebraic sets are the one-point sets. As I((a_{1}, …, a_{n})) = (X_{1} − a_{1}, …, X_{n} − a_{n}), this shows that the maximal ideals of k[X_{1}, …, X_{n}] are precisely the ideals of the form (X_{1} − a_{1}, …, X_{n} − a_{n}).

(c) The algebraic set V (𝔞) is empty if and only if 𝔞 = k[X_{1}, …, X_{n}] (because V (𝔞) empty ⟹ rad(𝔞) = k[X_{1}, …, X_{n}] ⟹ 1 ∈ rad(𝔞) ⟹ 1 ∈ 𝔞).

(d) Let W and W′ be algebraic sets. Then W ∩W′ is the largest algebraic subset contained in both W and W′, and so I(W ∩ W′) must be the smallest radical ideal containing both I(W) and I(W′). Hence I(W ∩ W′) = rad(I(W) + I(W′)).

For example, let W = V (X^{2} − Y), W′ = V (X^{2} + Y); then I(W ∩ W′) = rad(X^{2}, Y) = (X, Y) (assuming characteristic ≠ 2). Note that W ∩ W′ = {(0, 0)}, but when realized as the intersection of Y = X^{2} and Y = −X^{2}, it has “multiplicity 2”. Picture.

Finding the radical of an ideal. Typically, an algebraic set V will be defined by a finite set of polynomials {g_{1}, …, g_{s}}, and then we shall need to find I(V) = rad((g_{1}, …, g_{s})).

Proposition 1.12. The polynomial h ∈ rad(𝔞) if and only if 1 ∈ (𝔞, 1 − Y h) (the ideal in k[X_{1}, …, X_{n}, Y] generated by the elements of 𝔞 and 1 − Y h).

Proof. We saw that 1 ∈ (𝔞, 1 − Y h) implies h ∈ rad(𝔞) in the course of proving (1.9). Conversely, if h^{N} ∈ 𝔞, then

\begin{aligned}1 & = Y^{N}h^{N} + (1 − Y^{N}h^{N}) \\ & = Y^{N}h^{N} + (1 − Y h) · (1 + Y h + ⋯ + Y^{N−1}h^{N−1}) ∈ 𝔞 + (1 − Y h). □ \end{aligned}

Thus we have an algorithm for deciding whether h ∈ rad(𝔞), but not yet an algorithm for finding a set of generators for rad(𝔞). There do exist such algorithms (see Cox et al. p177 for references), and one has been implemented in Macaulay. To start Macaulay on the Suns, type: “Macaulay”; type “<radical” to find out the syntax for finding radicals. I hope to include a brief primer on using Macaulay in the notes sometime.

The Zariski topology on an algebraic set. We now examine the Zariski topology on k^{n} and an on an algebraic subset of k^{n}, more closely. The Zariski topology on ℂ^{n} is much coarser than the complex topology. Part (b) of (1.9) says that, for each subset W of k^{n}, V I(W) is the closure of W, and (1.10) says that there is a one-to-one correspondence between the closed subsets of k^{n} and the radical ideals of k[X_{1}, …, X_{n}].

Let V be an algebraic subset of k^{n}, and let I(V) = 𝔞. Then the algebraic subsets of V correspond to the radical ideals of k[X_{1}, …, X_{n}] containing 𝔞.

Proposition 1.13. Let V be an algebraic subset of k^{n}.

(a) The points of V are closed for the Zariski topology (thus V is a T_{1}-space).

(b) Every descending chain of closed subsets of V becomes constant, i.e., given

V_{1} ⊃ V_{2} ⊃ V_{3} ⊃ ⋯ (closed subsets of V),

eventually V_{N} = V_{N+1} = . . .. Alternatively, every ascending chain of open sets becomes constant.

(c) Every open covering of V has a finite subcovering.

Proof. (a) We have already observed that {(a_{1}, …, a_{n})} is the algebraic set defined by the ideal (X_{1} − a_{1}, …, X_{n} − a_{n}).

(b) A sequence V_{1} ⊃ V_{2} ⊃ ⋯ gives rise to a sequence of radical ideals I(V_{1}) ⊂ I(V_{2}) ⊂ …, which eventually becomes constant because k[X_{1}, …, X_{n}] is Noetherian.

(c) Let V = ⋃_{i∈I} U_{i} with each U_{i} open. Choose an i_{0} ∈ I; if U_{i_{0}} ≠ V, then there exists an i_{1} ∈ I such that U_{i_{0}} ⫋ U_{i_{0}} ∪ U_{i_{1}}. If U_{i_{0}} ∪ U_{i_{1}} ≠ V, then there exists an i_{2} ∈ I etc.. Because of (b), this process must eventually stop. □

A topological space having the property (b) is said to be Noetherian. The condition is equivalent to the following: every nonempty set of closed subsets of V has a minimal element. A space having property (c) is said to be quasi-compact (by Bourbaki at least; others call it compact).

The coordinate ring of an algebraic set. Let V be an algebraic subset of k^{n}, and let I(V) = 𝔞. An element f(X_{1}, …, X_{n}) of k[X_{1}, …, X_{n}] defines a mapping k^{n} → k, a ↦ f(a) whose restriction to V depends only on the coset f + 𝔞 of f in the quotient ring

k[V] = k[X_{1}, …, X_{n}]/𝔞 = k[x_{1}, …, x_{n}].

Moreover, two polynomials f_{1}(X_{1}, …, X_{n}) and f_{2}(X_{1}, …, X_{n}) restrict to the same function on V only if they define the same element of k[V]. Thus k[V] can be identified with a ring of functions V → k.

We call k[V] the ring of regular functions on V, or the coordinate ring of V. It is a finitely generated reduced k-algebra (because 𝔞 is radical), but need not be an integral domain.

For an ideal 𝔟 in k[V], we set

V (𝔟) = {a ∈ V | f(a) = 0, all f ∈ 𝔟}.

Let W = V (𝔟). The maps

k[X_{1}, …, X_{n}] → k[V] = \frac{k[X_{1}, …, X_{n}]}{𝔞} → k[W] = \frac{k[V]}{𝔟}

should be regarded as restricting a function from k^{n} to V, and then restricting that function to W.

Write π for the map k[X_{1}, …, X_{n}] → k[V]. Then 𝔟 ↦ π^{−1}(𝔟) is a bijection from the set of ideals of k[V] to the set of ideals of k[X_{1}, …, X_{n}] containing 𝔞 under which radical, prime, and maximal ideals correspond to radical, prime, and maximal ideals (each of these conditions can be checked on the quotient ring, and k[X_{1}, …, X_{n}]/π^{−1}(𝔟) ≈ k[V]/𝔟). Clearly

V (π^{−1}(𝔟)) = V (𝔟),

and so 𝔟 ↦ V (𝔟) gives a bijection between the set of radical ideals of k[V] and the set of algebraic sets contained in V.

For h ∈ k[V], we write

D(h) = {a ∈ V | h(a) ≠ 0}.

It is an open subset of V, because it is the complement of V ((h)).

Proposition 1.14. (a) The points of V are in one-to-one correspondence with the maximal ideals of k[V].

(b) The closed sets of V are in one-to-one correspondence with the radical ideals of k[V].

(c) The sets D(h), h ∈ k[V], form a basis for the topology of V, i.e., each D(h) is open, and each open set is a union (in fact, a finite union) of D(h)’s.

Proof. (a) and (b) are obvious from the above discussion. For (c), we have already observed that D(h) is open. Any other open set U ⊂ V is the complement of a set of the form V (𝔟), 𝔟 an ideal in k[V]. If f_{1}, …, f_{m} generate 𝔟, then U = ∪D(f_{i}). □

The D(h) are called the basic (or principal) open subsets of V. We sometimes write V_{h} for D(h). Note that D(h) ⊂ D(h′) ⟺ V (h) ⊃ V (h′) ⟺ rad((h)) ⊂ rad((h′)) ⟺ h^{r} ∈ (h′) some r ⟺ h^{r} = h′g, some g.

Note that some at least of this looks familiar. If V is a topological space, then the zero set of a family of functions f: V → ℝ is closed, and the set where such a function is nonzero is open.

Irreducible algebraic sets. A nonempty subset W of a topological space V is said to be irreducible if it satisfies any one of the following equivalent conditions:

(a) W is not the union of two proper closed subsets;

(b) any two nonempty open subsets of W have a nonempty intersection;

(c) any nonempty open subset of W is dense.

The equivalences (a) ⟺ (b) and (b) ⟺ (c) are obvious. Also, one sees that if W is irreducible, and W = W_{1} ∪ … ∪ W_{r} with each W_{i} closed, then W = W_{i} for some i.

This notion is not useful for Hausdorff topological spaces, because such a space is irreducible only if it consists of a single point—otherwise any two points have disjoint open neighbourhoods, and so (b) fails.

Proposition 1.15. An algebraic set W is irreducible and only if I(W) is prime.

Proof. ⟹ : Suppose fg ∈ I(W). At each point of W, either f or g is zero, and so W ⊂ V (f) ∪ V (g). Hence

W = (W ∩ V (f)) ∪ (W ∩ V (g)).

As W is irreducible, one of these sets, say W ∩ V (f), must equal W. But then f ∈ I(W). Thus I(W) is prime.

⟸ : Suppose W = V (𝔞) ∪ V (𝔟) with 𝔞 and 𝔟 radical ideals—we have to show that W = V (𝔞) or V (𝔟). Recall that V (𝔞) ∪ V (𝔟) = V (𝔞 ∩ 𝔟), and that 𝔞 ∩ 𝔟 is radical; hence I(W) = 𝔞 ∩ 𝔟. If W ≠ V (𝔞), then there is an f ∈ 𝔞, f ∉ I(W). But fg ∈ 𝔞 ∩ 𝔟 = I(W) for all g ∈ 𝔟, and, because f ∉ I(W) and I(W) is prime, this implies that 𝔟 ⊂ I(W); therefore W ⊂ V (𝔟). □

Thus, there are one-to-one correspondences

radical ideals ↔ algebraic subsets
prime ideals ↔ irreducible algebraic subsets
maximal ideals ↔ one-point sets.

These correspondences are valid whether we mean ideals in k[X_{1}, …, X_{n}] and algebraic subsets of k^{n}, or ideals in k[V] and algebraic subsets of V. Note that the last correspondence implies that the maximal ideals in k[V] are of the form (x_{1} − a_{1}, …, x_{n} − a_{n}), (a_{1}, …, a_{n}) ∈ V.

Example 1.16. Let f ∈ k[X_{1}, …, X_{n}]. As we noted in §0, k[X_{1}, …, X_{n}] is a unique factorization domain, and so (f) is a prime ideal ⟺ f is irreducible. Thus

V (f) is irreducible ⟺ f is irreducible.

On the other hand, suppose f factors, f = ∏ f_{i}^{m_{i}}, with the f_{i} distinct irreducible polynomials. Then (f) = ∩(f_{i}^{m_{i}}), rad((f)) = (∏ f_{i}) = ∩(f_{i}), and V (f) = ∪V (f_{i}) with V (f_{i}) irreducible.

Proposition 1.17. Let V be a Noetherian topological space. Then V is a finite union of irreducible closed subsets, V = V_{1}∪…∪V_{m}. Moreover, if the decomposition is irredundant in the sense that there are no inclusions among the V_{i}, then the V_{i} are uniquely determined up to order.

Proof. Suppose the first assertion is false. Then, because V is Noetherian, there will be a closed subset W of V that is minimal among those that cannot be written as a finite union of irreducible closed subsets. But such a W cannot itself be irreducible, and so W = W_{1} ∪ W_{2}, with each W_{i} a proper closed subset of W. From the minimality of W, it follows that each W_{i} is a finite union of irreducible closed subsets, and so therefore is W. We have arrived at a contradiction.

Suppose that V = V_{1} ∪ … ∪ V_{m} = W_{1} ∪ … ∪ W_{n} are two irredundant decompositions. Then V_{i} = ∪_{j}(V_{i} ∩ W_{j}), and so, because V_{i} is irreducible, V_{i} ⊂ V_{i} ∩ W_{j} for some j. Consequently, there is a function f: {1, …, m} → {1, …, n} such that V_{i} ⊂ W_{f(i)} for each i. Similarly, there is a function g: {1, …, n} → {1, …, m} such that W_{j} ⊂ V_{g(j)}. Since V_{i} ⊂ W_{f(i)} ⊂ V_{gf(i)}, we must have gf(i) = i and V_{i} = W_{f(i)}; similarly fg = id. Thus f and g are bijections, and the decompositions differ only in the numbering of the sets. □

The V_{i} are called the irreducible components of V. They are the maximal closed irreducible subsets of V. In the above example, the V (f_{i}) are the irreducible components of V (f). A Hausdorff space is Noetherian if and only if it is finite, in which case the irreducible components are the one-point sets.

Corollary 1.18. A radical ideal 𝔞 of k[X_{1}, …, X_{n}] is a finite intersection of prime ideals, 𝔞 = 𝔭_{1} ∩ … ∩ 𝔭_{n}; if there are no inclusions among the 𝔭_{i}, then they are uniquely determined up to order.

Proof. Write V (𝔞) = ∪V_{i}, and take 𝔭_{i} = I(V_{i}). □

Remark 1.19. (a) In a Noetherian ring, every ideal 𝔞 has a decomposition into primary ideals: 𝔞 = ∩𝔮_{i} (see Atiyah-MacDonald, IV, VII). For radical ideals, this becomes a much simpler decomposition into prime ideals, as in the corollary.

(b) In k[X], (f(X)) is radical if and only if f is square-free, in which case f is a product of distinct irreducible polynomials, f = p_{1} … p_{r}, and (f) = (p_{1}) ∩ … ∩ (p_{r}) (a polynomial is divisible by f if and only if it is divisible by each p_{i}).

Dimension. We briefly introduce the notion of the dimension of an algebraic variety. Later we shall discuss this in more detail.

Let V be an irreducible algebraic subset. Then I(V) is a prime ideal, and so k[V] is an integral domain. Let k(V) be its field of fractions—k(V) is called the field of rational functions on V. The dimension of V is defined to be the transcendence degree of k(V) over k.

[For those who know some commutative algebra, according to the last theorem in Atiyah-MacDonald, this is equal to the Krull dimension of k[V]; we shall prove this later.]

Example 1.20. (a) Let V = k^{n}; then k(V) = k(X_{1}, …, X_{n}), and so dim(V) = n. Later we shall see that the Noether normalization theorem implies that V has dimension n if and only if there is a surjective finite-to-one map V → k^{n}.

(b) If V is a linear subspace of k^{n} (or a translate of such a subspace), then it an easy exercise to show that the dimension of V in the above sense is the same as its dimension in the sense of linear algebra (in fact, k[V] is canonically isomorphic to k[X_{i_{1}}, …, X_{i_{d}}] where the X_{i_{j}} are the “free” variables).

In linear algebra, we justify saying V has dimension n by pointing out that its elements are parametrized by n-tuples; unfortunately, it is not true in general that the points of an algebraic set of dimension n are parametrized by n-tuples; the most one can say is that there is a finite-to-one map to k^{n}.

(c) An irreducible algebraic set has dimension 0 if and only if it consists of a single point. (Because k(V) has transcendence degree 0 over k if and only if it equals k, k being algebraically closed; thus k ⊂ k[V] ⊂ k(V) = k implies that k[V] is a field, and so 𝔭 is maximal. Now apply (1.11b))

The zero set of a single nonconstant nonzero polynomial f(X_{1}, …, X_{n}) is called a hypersurface in k^{n}.

Proposition 1.21. An irreducible hypersurface in k^{n} has dimension n − 1.

Proof. Let k[x_{1}, …, x_{n}] = k[X_{1}, …, X_{n}]/(f), and let k(x_{1}, …, x_{n}) be the field of fractions of k[x_{1}, …, x_{n}]. Since x_{1}, …, x_{n} generate k(x_{1}, …, x_{n}) and they are algebraically dependent, the transcendence degree must be < n (because 1.42 shows that {x_{1}, …, x_{n}} contains a transcendence basis). To see that it is not < n − 1, note that if X_{n} occurs in f, then it occurs in all nonzero multiples of f, and so no nonzero polynomial in X_{1}, …, X_{n−1} belongs to (f). This means that x_{1}, …, x_{n−1} are algebraically independent. □

For a reducible algebraic set V, we define the dimension of V to be the maximum of the dimensions of its irreducible components. When these all have the same dimension d, then we say that V has pure dimension d.

Proposition 1.22. If V is irreducible and Z is a proper closed subvariety of V, then dim(Z) < dim(V).

Proof. We may assume that Z is irreducible. Then Z corresponds to a nonzero prime ideal 𝔭 in k[V], and k[Z] = k[V]/𝔭.

Suppose V ⊂ k^{n}, so that k[V] = k[X_{1}, …, X_{n}]/I(V) = k[x_{1}, …, x_{n}]. If X_{i} is regarded as a function on k^{n}, then its image x_{i} in k[V] is the restriction of this function to V.

Let f ∈ k[V]. The image f̄ of f in k[V]/𝔭 = k[Z] can be regarded as the restriction of f to Z. With this notation, k[Z] = k[x̄_{1}, …, x̄_{n}]. Suppose that dim Z = d and that x̄_{1}, …, x̄_{d} are algebraically independent. I will show, for any nonzero f ∈ 𝔭, the d+1 elements x_{1}, …, x_{d}, f are algebraically independent, which implies that dim V ≥ d + 1.

Suppose otherwise. Then there is a nontrivial algebraic relation among the x_{i} and f, which we can write

a_{0}(x_{1}, …, x_{d})f^{m} + a_{1}(x_{1}, …, x_{d})f^{n−1} + ⋯ + a_{m}(x_{1}, …, x_{d}) = 0,

with a_{i}(x_{1}, …, x_{d}) ∈ k[x_{1}, …, x_{d}]. Because the relation is nontrivial, at least one of the a_{i} is nonzero (in the polynomial ring k[x_{1}, …, x_{d}]). After cancelling by a power of f if necessary, we can assume a_{m}(x_{1}, …, x_{d}) ≠ 0 (in this step, we use that k[V] is an integral domain). On restricting the functions in the above equality to Z, i.e., applying the homomorphism k[V] → k[Z], we find that

a_{m}(x̄_{1}, …, x̄_{d}) = 0,

which contradicts the algebraic independence of x̄_{1}, …, x̄_{d}. □

Example 1.23. Let F (X, Y) and G(X, Y) be nonconstant polynomials with no common factor. Then V (F (X, Y)) has dimension 1 by (1.21), and so V (F (X, Y)) ∩ V (G(X, Y)) must have dimension zero; it is therefore a finite set.

Remark 1.24. Later we shall show that if, in the situation of (1.21), Z is a maximal proper irreducible subset of V, then dim Z = dim V − 1. This implies that the dimension of an algebraic set V is the maximum length of a chain

V_{0} ⫌ V_{1} ⫌ ⋯ ⫌ V_{d}

with each V_{i} closed and irreducible and V_{0} an irreducible component of V. Note that this description of dimension is purely topological—it makes sense for any Noetherian topological space.

On translating the description in terms of ideals, we see immediately that the dimension of V is equal to the Krull dimension of k[V]—the maximal length of a chain of prime ideals,

𝔭_{d} ⫌ 𝔭_{d−1} ⫌ ⋯ ⫌ 𝔭_{0}.

Example 1.25. We classify the irreducible closed subsets V of k^{2}. If V has dimension 2, then (by 1.21) it can’t be a proper subset of k^{2}, so it is k^{2}. If V has dimension 1, then V ≠ k^{2}, and so I(V) contains a nonzero polynomial, and hence a nonzero irreducible polynomial f (being a prime ideal). Then V ⊃ V (f), and so equals V (f). Finally, if V has dimension zero, it is a point. Correspondingly, we can make a list of all the prime ideals in k[X, Y]: they have the form (0), (f) (with f irreducible), or (X − a, Y − b).

Appendix A: Integrality

Let A be an integral domain, and let L be a field containing A. An element α of L is said to be integral over A if it is a root of a monic polynomial with coefficients in A, i.e., if it satisfies an equation

α^{n} + a_{1}α^{n−1} + … + a_{n} = 0, a_{i} ∈ A.

Before proving that the elements of L integral over A form a ring, we need to review symmetric polynomials.

Symmetric polynomials. A polynomial P (X_{1}, ..., X_{r}) ∈ A[X_{1}, …, X_{r}] is said to be symmetric if it is unchanged when its variables are permuted, i.e., if

P (X_{σ(1)}, …, X_{σ(r)}) = P (X_{1}, …, X_{r}), all σ ∈ Sym_{r}.

For example

S_{1} = ∑ X_{i}, S_{2} = ∑_{i<j} X_{i}X_{j}, …, S_{r} = X_{1} ⋯ X_{r},

are all symmetric. These particular polynomials are called the elementary symmetric polynomials.

Theorem 1.26. (Symmetric function theorem) Let A be a ring. Every symmetric polynomial P (X_{1}, ..., X_{r}) in A[X_{1}, ..., X_{r}] is equal to a polynomial in the symmetric elementary polynomials with coefficients in A, i.e., P ∈ A[S_{1}, ..., S_{r}].

Proof. We define an ordering on the monomials in the X_{i} by requiring that

X_{1}^{i_{1}}X_{2}^{i_{2}} ⋯ X_{r}^{i_{r}} > X_{1}^{j_{1}}X_{2}^{j_{2}} ⋯ X_{r}^{j_{r}}

if either

i_{1} + i_{2} + ⋯ + i_{r} > j_{1} + j_{2} + ⋯ + j_{r}

or equality holds and, for some s,

i_{1} = j_{1}, …, i_{s} = j_{s}, but i_{s+1} > j_{s+1}.

Let X_{1}^{k_{1}} ⋯ X_{r}^{k_{r}} be the highest monomial occurring in P with a coefficient c ≠ 0. Because P is symmetric, it contains all monomials obtained from X_{1}^{k_{1}} ⋯ X_{r}^{k_{r}} by permuting the X’s. Hence k_{1} ≥ k_{2} ≥ ⋯ ≥ k_{r}.

Clearly, the highest monomial in S_{i} is X_{1} ⋯ X_{i}, and it follows that the highest monomial in S_{1}^{d_{1}} ⋯ S_{r}^{d_{r}} is

X_{1}^{d_{1}+d_{2}+···+d_{r}}X_{2}^{d_{2}+···+d_{r}} ⋯ X_{r}^{d_{r}}.

Therefore

P (X_{1}, …, X_{r}) − cS_{1}^{k_{1}−k_{2}}S_{2}^{k_{2}−k_{3}} ⋯ S_{r}^{k_{r}} < P (X_{1}, …, X_{r}).

We can repeat this argument with the polynomial on the left, and after a finite number of steps, we will arrive at a representation of P as a polynomial in S_{1}, …, S_{r}. (For more details, see Jacobson, Basic Algebra I, 2.20, p139.) □

Let f(X) = X^{n} + a_{1}X^{n−1} + ⋯ + a_{n} ∈ A[X], and let α_{1}, …, α_{n} be the roots of f(X) in some ring containing A, i.e., f(X) = ∏(X − α_{i}). Then

a_{1} = −S_{1}(α_{1}, …, α_{n}), a_{2} = S_{2}(α_{1}, …, α_{n}), …, a_{n} = ±S_{n}(α_{1}, …, α_{n}).

Thus the elementary symmetric polynomials in the roots of f(X) lie in A, and so the theorem implies that every symmetric polynomial in the roots of f(X) lie in A.

Integral elements.

Theorem 1.27. The set of elements of L integral over A forms a ring.

Proof. Let α and β be integral over A; we have to show that α ± β and αβ are integral over A. Let Ω be an algebraically closed field containing L.

We are given that α is a root of a polynomial f(X) = X^{m} +a_{1}X^{m−1} +⋯+a_{m}, a_{i} ∈ A. Write

f(X) = ∏(X − α_{i}), α_{i} ∈ Ω.

Similarly, β is a root of polynomial g(X) = X^{n} + b_{1}X^{n−1} + ... +b_{n}, b_{i} ∈ A, and we write

f(X) = ∏(X − β_{i}), β_{i} ∈ Ω.

Let γ_{1}, γ_{2}, ..., γ_{mn} be the family of numbers of the form α_{i} +β_{j} (or α_{i} −β_{j}, or α_{i}β_{j}). I claim that h(X) =_{df} ∏_{\begin{matrix}1≤i≤m \\ 1≤j≤n \end{matrix}} (X − γ_{ij}) has coefficients in A. This will prove that

α + β is integral over A because h(α + β) = 0.

The coefficients of h are symmetric in the α_{i} and β_{j}. Let P (α_{1}, ..., α_{m}, β_{1}, ..., β_{n}) be one of these coefficients, and regard it as a polynomial Q(β_{1}, ..., β_{n}) in the β’s with coefficients in A[α_{1}, ..., α_{m}]; then its coefficients are symmetric in the α_{i}, and so lie in A. Thus P (α_{1}, ..., α_{m}, β_{1}, ..., β_{n}) is a symmetric polynomial in the β’s with coefficients in A—it therefore lies in A, as claimed. □

Definition 1.28. The ring of elements of L integral over A is called the integral closure of A in L.

Proposition 1.29. Let K be the field of fractions of A, and let L be a field containing K. If α ∈ L is algebraic over K, then there exists a d ∈ A such that dα is integral over A.

Proof. By assumption, α satisfies an equation

α^{m} + a_{1}α^{m−1} + ⋯ + a_{m} = 0, a_{i} ∈ K.

Let d be a common denominator for the a_{i} (this means that da_{i} ∈ A, all i), and multiply through the equation by d^{m} :

d^{m}α^{m} + a_{1}d^{m}α^{m−1} + ⋯ + a_{m}d^{m} = 0.

We can rewrite this as

(dα)^{m} + a_{1}d(dα)^{m−1} + ⋯ + a_{m}d^{m} = 0.

As a_{1}d, …, a_{m}d^{m} ∈ A, this shows that dα is integral over A. □

Corollary 1.30. Let A be an integral domain with field of fractions K, and let L be an algebraic extension of K. If B is the integral closure of A in L, then L is the field of fractions of B.

Proof. The proposition shows that every α ∈ L can be written α = β/d with β ∈ B, d ∈ A. □

Definition 1.31. A ring A is integrally closed if it is its own integral closure in its field of fractions K, i.e., if

α ∈ K, α integral over A ⟹ α ∈ A.

Proposition 1.32. A unique factorization domain (e.g. a principal ideal domain) is integrally closed.

Proof. Suppose a/b, a, b ∈ A, is an element of the field of fractions of A that is integral over A. If b is a unit, then a/b ∈ A. Otherwise we may suppose that there is an irreducible element p of A dividing b but not a. As a/b is integral over A, it satisifies an equation

(a/b)^{n} + a_{1}(a/b)^{n−1} + ⋯ + a_{n} = 0, a_{i} ∈ A.

On multiplying through by b^{n}, we obtain the equation

a^{n} + a_{1}a^{n−1}b + ... + a_{n}b^{n} = 0.

The element p then divides every term on the left except a^{n}, and hence must divide a^{n}. Since it doesn’t divide a, this is a contradiction. □

Proposition 1.33. Let K be the field of fractions of A, and let L be an extension of K of finite degree. Assume A is integrally closed. An element α of L is integral over A if and only if its minimum polynomial over K has coefficients in A.

Proof. Assume α is integral over A, so that

α^{m} + a_{1}α^{m−1} + ... + a_{m} = 0, some a_{i} ∈ A.

Let α′ be a conjugate of α, i.e., a root of the minimum polynomial of α over K. Then there is an K-isomorphism

σ: K[α] ⟶ K[α′], σ(α) = α′;

see^{8} Math 594. On applying σ to the above equation we obtain the equation

α′^{m} + a_{1}α′^{m−1} + ... + a_{m} = 0,

which shows that α′ is integral over A. Hence all the conjugates of α are integral over A, and it follows from (1.25) that the coefficients of f(X) are integral over A. They lie in K, and A is integrally closed, and so they lie in A. This proves the “only if” part of the statement, and the “if” part is obvious. □

8If f(X) is the minimum polynomial of α, hence also of α′, over K, then the map h(X) ↦ h(α): K[X] → K[α] induces an isomorphism τ: K[X]/(f(X)) → K[α]. Similarly, h(X) ↦ h(α′): K[X] → K[α′] induces an isomorphism τ′: K[X]/(f(X)) → K[α′], and we set σ = τ′ ∘ τ^{−1}.

Appendix B: Transcendence degree

This topic is part of the syllabus for Math 594 (and the QR exams), but is often omitted.

We consider fields Ω ⊃ k, with Ω to be thought of as being much bigger than k; e.g., ℂ ⊃ ℚ.

Elements α_{1}, …, α_{n} of Ω are algebraically dependent over k if they satisfy an algebraic relation with coefficients in k, i.e., if there exists a nonzero polynomial f(X_{1}, …, X_{n}) with coefficients in k such that f(α_{1}, …, α_{n}) = 0. Otherwise, the elements are algebraically independent. Note the similarity with linear independence: if f is required to be homogeneous of degree 1, then the definition becomes that of linear independence.

Example 1.34. (a) A single element α is algebraically independent if and only if it is transcendental over F, i.e., it is not a root of a polynomial in one variable with coefficients in F.

(b) The complex numbers π and e are probably algebraically independent over ℚ, but this has not been proved (of course, it is known that each is transcendental).

An infinite set A is algebraically independent if every finite subset of A is algebraically independent.

Remark 1.35. If α_{1}, …, α_{n} are algebraically independent over k, then the map

f(X_{1}, …, X_{n}) ↦ f(α_{1}, …, α_{n}): k[X_{1}, …, X_{n}] → k[α_{1}, …, α_{n}],

is an injection, and hence is an isomorphism. This isomorphism extends to the fields of fractions,

X_{i} ↦ α_{i}: k(X_{1}, …, X_{n}) → k(α_{1}, …, α_{n}).

In this case, k(α_{1}, …, α_{n}) is called a pure transcendental extension of k.

If A is a subset of Ω, then we write k(A) for the field generated by A over k, i.e., it is the smallest subfield of Ω containing k and A. It consists of all quotients of polynomials in the elements of A with coefficients in k.

An element β is algebraically dependent on a set A (over k) if it is algebraic over k(A); i.e., if there exist α_{1}, α_{2}, ..., α_{n} ∈ k(A) such that β^{n}+α_{1}β^{n−1}+…+α_{n} = 0. A set B is algebraically dependent on A if each element of B is algebraically dependent on A.

Theorem 1.36 (Fundamental Result). Let A = {α_{1}, …, α_{m}} and B = {β_{1}, …, β_{n}} be two subsets of Ω. Assume:

(a) A is algebraically independent;

(b) A is algebraically dependent on B. Then m ≤ n.

Proof. We first prove a lemma.

Lemma 1.37 (The Exchange Property). Let {α_{1}, …, α_{n}} be a subset of Ω; if β is algebraically dependent on {α_{1}, …, α_{m}} but not on {α_{1}, …, α_{m−1}}, then α_{m} is algebraically dependent on {α_{1}, …, α_{m−1}, β}.

Proof. From the assumption, we know there is a polynomial f(X_{1}, …, X_{m}, Y) with coefficients in k such that

f(α_{1}, …, α_{m}, Y) ≠ 0, f(α_{1}, …, α_{m}, β) = 0.

Write

f(X_{1}, …, X_{m}, Y) = ∑ a_{i}(X_{1}, …, X_{m−1}, Y)X_{m}^{i}.

and observe that the polynomials a_{i}(X_{1}, …, X_{m−1}, Y) are not all zero (because f is not the zero polynomial). Since β is not algebraically dependent on {α_{1}, …, α_{m−1}}, it follows therefore that not all the elements a_{i}(α_{1}, …, α_{m−1}, β) are zero. Therefore, f(α_{1}, …, α_{m−1}, X_{m}, β) is not the zero polynomial. Since f(α_{1}, …, α_{m}, β) = 0, this shows that α_{m} is algebraically dependent on {α_{1}, …, α_{m−1}, β}. □

Lemma 1.38 (Transitivity of Algebraic Dependence). If γ is algebraically dependent on B, and B is algebraically dependent on A, then γ is algebraically dependent on A.

Proof. A standard result in field theory says that if γ is algebraic over K, and K is algebraic over k, then γ is algebraic over k. Apply this with k(A ∪ B) for K and k(A) for k. □

We now prove the theorem. Let r be the number of elements that A and B have in common. If r = m, then A ⊂ B, and our result is obvious. Suppose that r < m, and write B = {α_{1}, …, α_{r}, β_{r+1}, …, β_{n}}. Since α_{r+1} is algebraically dependent on {α_{1}, …, α_{r}, β_{r+1}, …, β_{n}} but not on {α_{1}, …, α_{r}}, there will be a β_{j}, r + 1 ≤ j ≤ n, such that α_{r+1} is algebraically dependent on {α_{1}, …, α_{r}, β_{r+1}, …, β_{j}} but not {α_{1}, …, α_{r}, β_{r+1}, …, β_{j−1}}. Then the exchange lemma then shows that β_{j} is algebraically dependent on B_{1} =_{df} B ∪ {α_{r+1}} − {β_{j}}. Therefore B is algebraically dependent on B_{1}, and so A is algebraically dependent on B_{1} (by the last lemma). If r + 1 < m, repeat the argument with A and B_{1}. Eventually we’ll achieve r = m, and m ≤ n. □

Definition 1.39. A transcendence basis for Ω over k is an algebraically independent set A such that E is algebraic over k(A).

Corollary 1.40. If Ω has a finite transcendence basis over k, then all transcendence bases are finite and have the same number of elements.

Proof. This is obvious from the theorem. □

It is possible to show also that any two (possibly infinite) transcendence bases for Ω over k have the same cardinality. The cardinality of a transcendence basis for Ω over k is called the transcendence degree, tr deg_{F} Ω of Ω over k.

Example 1.41. Let Ω =field of meromorphic functions on a compact complex manifold M.

(a) If M = Riemann sphere, then Ω is a pure transcendental extension of ℂ of transcendence degree 1; the only meromorphic functions are the rational functions.

(b) If M = a Riemann surface, then tr deg_{ℂ}Ω = 1 (Ω is pure transcendental ⟺ M ≈ the Riemann sphere).

(c) If has complex dimension n, then tr deg_{ℂ}Ω ≤ n; equality holds if M is embeddable in some projective space.

Proposition 1.42. Let K be a field containing k, and assume K is algebraic over k[β_{1}, …, β_{n}]. Then {β_{1}, …, β_{n}} contains a transcendence basis for K over k.

Proof. Let B be a maximal algebraically independent subset of {β_{1}, …, β_{n}}. I claim that B is a transcendence basis. If not, there will be an element {β_{1}, …, β_{n}} transcendental over k(B), and the next lemma shows that B ∪ {β} will be algebraically independent, contradicting the maximality of B. Hence B is a transcendence basis for Ω over k. □

Lemma 1.43. Suppose that A is algebraically independent, and that A ∪ {β} is algebraically dependent. Then β is algebraic over k(A).

Proof. We are given that there exists a nonzero polynomial f(X_{1}, …, X_{n}, X) ∈ k[X_{1}, …, X_{n}, X] such that f(α_{1}, …, α_{n}, β) = 0, some distinct α_{1}, …, α_{n} ∈ A. The symbol X must occur in f, for otherwise the equation would show that α_{1}, …, α_{n} are algebraically independent, contrary to assumption. We can write f as a polynomial in X:

g_{0}X^{n} + g_{1}X^{n−1} + ⋯ + g_{n}, g_{i} ∈ k[X_{1}, …, X_{n}], g_{0} ≠ 0.

As g_{0} ≠ 0 and the α_{i} are algebraically independent, g_{0}(α_{1}, …, α_{n}) ≠ 0. Since β is a root of

f = g_{0}(α_{1}, …, α_{n})X^{n} + g_{1}(α_{1}, …, α_{n})X^{n−1} + ... + g_{n}(α_{1}, …, α_{n}),

it is algebraic over k(α_{1}, …, α_{n}) ⊂ k(A). □

2. Affine Algebraic Varieties

In this section we define on an algebraic set the structure of a ringed space, and then we define the notion of affine algebraic variety—roughly speaking, this is an algebraic set with no preferred embedding into k^{n}. This is in preparation for §3, where we define an algebraic variety to be a ringed space that is a finite union of affine algebraic varieties satisfying a natural separation axiom (in the same way that a topological manifold is a union of open subsets of ℝ^{n} satisfying the Hausdorff axiom).

Ringed spaces. Let V be a topological space and k a field. Suppose that for every open subset U of V we have a set O_{V} (U) of functions U → k. Then O_{V} is called a sheaf of k-algebras if it satisfies the following conditions:

(2.1a) O_{V} (U) is an k-subalgebra of the algebra of all functions U → k, i.e., for each c ∈ k, the constant function c is in O_{V} (U), and if f, g ∈ O_{V} (U), then so also do f ± g, and fg.

(2.1b) If U′ is an open subset of U and f ∈ O_{V} (U), then f|U′ ∈ O_{V} (U′).

(2.1c) Let U = ∪U_{α} be an open covering of an open subset U of V; then a function f : U → k is in O_{V} (U) if f|U_{α} ∈ O_{V} (U_{α}) for all α (i.e., the condition for f to be in O_{V} (U) is local).

Example 2.2. (a) Let V be any topological space, and for each open subset U of V let O_{V} (U) be the set of all continuous real-valued functions on U. Then O_{V} is a sheaf of ℝ-algebras.

(b) Recall that a function f: U → ℝ, where U is an open subset of ℝ^{n}, is said to be C^{∞} (or infinitely differentiable) if its partial derivatives of all orders exist and are continous. Let V be an open subset of ℝ^{n}, and for each open subset U of V let O_{V} (U) be the set of all infinitely differentiable functions on U. Then O_{V} is a sheaf of ℝ-algebras.

(c) Recall that a function f: U → ℂ, where U is an open subset of ℂ^{n}, is said to be analytic if it has a power series expansion in a neighbourhood of each point of U. Let V be an open subset of ℂ^{n}, and for each open subset U of V let O_{V} (U) be the set of all analytic functions on U. Then O_{V} is a sheaf of ℂ-algebras.

(d) Nonexample: let V be a topological space, and for each open subset U of V let O_{V} (U) be the set of all real-valued constant functions on U; then O_{V} is not a sheaf, unless V is irreducible! If “constant” is replaced with “locally constant”, then O_{V} becomes a sheaf of ℝ-algebras (in fact, the smallest such sheaf).

A pair (V, O_{V}) consisting of a topological space V and a sheaf of k-algebras will be called a ringed space. For historical reasons, we often write Γ(U, O_{V}) for O_{V} (U), and call its elements sections of O_{V} over U.

Let (V, O_{V}) be a ringed space. For any open subset U of V, the restriction O_{V} |U of O_{V} to U, defined by

Γ(U′, O_{V} |U) = Γ(U′, O_{V}), all open U′ ⊂ U,

is a sheaf again.

Let (V, O_{V}) be ringed space, and let P ∈ V. Consider pairs (f, U) consisting of an open neighbourhood U of P and f ∈ O_{V} (U); we write (f, U) ∼ (f′, U′) if f|U = f′|U for some U′′ ⊂ U ∩ U′. An equivalence class of pairs is called a germ of a function at P. The set of equivalence classes of such pairs forms a k-algebra denoted O_{V,P} or O_{P}. In all the interesting cases, it is a local ring with maximal ideal the set of germs that are zero at P.

In a fancier terminology,

O_{P} = \underrightarrow{lim} O_{V} (U), (direct limit over open neighbourhoods U of P).

Example 2.3. Let V = ℂ, and let O be the sheaf of holomorphic functions on ℂ. Call a power series ∑_{n≥0} a_{n}(z − c)^{n}, a_{n} ∈ ℂ, convergent if it converges on some neighbourhood of c. The set of such power series is a ℂ-algebra, and I claim that it is canonically isomorphic to the ring of germs of functions O_{c}. From Math 596, we know that if f is a holomorphic function on a neighbourhood U of c, then f has a power series expansion f = ∑ a_{n}(z −c)^{n} in some (possibly smaller) neighbourhood. Moreover another pair (g, U′) will define the same power series if and only if g agrees with f on some neighbourhood of c contained in U ∩ U′. Thus we have injective map from the ring of germs of holomorphic functions at c to the ring of convergent power series, and it is obvious that it is an isomorphism.

Review of rings of fractions. Before defining the sheaf of regular functions on an algebraic set, we need to review some of the theory of rings of fractions. When the initial ring is an integral domain (the most important case), the theory is very easy because all the rings are subrings of the field of fractions.

A multiplicative subset of a ring A is a subset S with the property:

1 ∈ S, a, b ∈ S ⟹ ab ∈ S.

Define an equivalence relation on A × S by

(a, s) ∼ (b, t) ⟺ u(at − bs) = 0 for some u ∈ S.

Write \frac{a}{s} for the equivalence class containing (a, s), and define addition and multiplication in the obvious way:

\frac{a}{s} + \frac{b}{t} = \frac{at + bs}{st}, \frac{a}{s} \frac{b}{t} = \frac{ab}{st}.

We then obtain a ring S^{−1}A = \{\frac{a}{s} | a ∈ A, s ∈ S\}, and a canonical homomorphism a ↦ \frac{a}{1} : A → S^{−1}A, not necessarily injective. For example, if S contains 0, then S^{−1}A is the zero ring.

Write i for the homomorphism a ↦ \frac{a}{1} : A → S^{−1}A. Then (S^{−1}A, i) has the following universal property: every element s ∈ S maps to a unit in S^{−1}A, and any other homomorphism α: A → B with this property factors uniquely through i:

α = (A ⟶^{i} S^{−1}A ⟶^{unique} B).

The uniqueness is obvious—the map S^{−1}A → B must be \frac{a}{s} ↦ α(a) · α(s)^{−1} —and it is easy to check that this formula does define a homomorphism S^{−1}A → B. For example, to see that it is well-defined, note that

\frac{a}{c} = \frac{b}{d} ⟹ s(ad − bc) = 0 some s ∈ S ⟹ α(a)α(d) − α(b)α(c) = 0,

because α(s) is a unit in B, and so

α(a)α(c)^{−1} = α(b)α(d)^{−1}.

As usual, this universal property determines the pair (S^{−1}A, i) uniquely up to a unique isomorphism.

In the case that A is an integral domain we can form the field of fractions F = S^{−1}A, S = A − {0}, and then for any other multiplicative subset S of A not containing 0, S^{−1}A can be identified with \{\frac{a}{s} ∈ F |a ∈ A, s ∈ S\}.

We shall be especially interested in the following examples.

(i) Let h ∈ A. Then S_{h} =_{df} {1, h, h^{2}, … } is a multiplicative subset of A, and we write A_{h} = S_{h}^{−1}A. Thus every element of A_{h} can be written in the form a/h^{m}, a ∈ A, and

\frac{a}{h^{m}} = \frac{b}{h^{n}} ⟺ h^{N}(ah^{n} − bh^{m}) = 0, some N.

In the case that A is an integral domain, with field of fractions F, A_{h} is the subring of F of elements of the form a/h^{m}, a ∈ A, m ∈ ℕ.

(ii) Let 𝔭 be a prime ideal in A. Then S_{𝔭} =_{df} A − 𝔭 is a multiplicative subset of A, and we write A_{𝔭} = S_{𝔭}^{−1}A. Thus each element of A_{𝔭} can be written in the form \frac{a}{c}, c ∉ 𝔭, and

\frac{a}{c} = \frac{b}{d} ⟺ s(ad − bc) = 0, some s ∉ 𝔭.

The subset 𝔪 = \{\frac{a}{s}|a ∈ 𝔭, s ∉ S\} is a maximal ideal in A_{𝔭}, and it is the only maximal ideal^{9}. Therefore A_{𝔭} is a local ring.

Lemma 2.4. For any ring A, the map ∑ a_{i}X^{i} ↦ ∑ \frac{a_{i}}{h^{i}} defines an isomorphism

A[X]/(1 − hX) ⟶^{≈} A_{h}.

Proof. In the ring A[x] = A[X](1 − hX), 1 = hx, and so h is a unit. Consider a second homomorphism of rings α: A → B such that α(h) is a unit in B. Then α extends to a homomorphism ∑ a_{i}X^{i} ↦ ∑ α(a_{i})α(h)^{−i}: A[X] → B. Under this homomorphism 1−hX ↦ 1−α(h)α(h)^{−1} = 0, and so the map factors through A[x].

9First check 𝔪 is an ideal. Next, if 𝔪 = A_{𝔭}, then 1 ∈ 𝔪; but 1 = \frac{a}{s} , a ∈ 𝔭, s ∉ 𝔭 means u(s − a) = 0 some u ∉ 𝔭, and so a = us ∉ 𝔭. Finally, 𝔪 is maximal, because any element of A_{𝔭} not in 𝔪 is a unit.

The resulting homomorphism γ: A[x] → B has the property that its composite with A → A[x] is α, and (because hx = 1 in A[x]) it is the unique homomorphism with this property. Therefore A[x] has the same universal property as A_{h}, and so the two are (uniquely) isomorphic by an isomorphism that makes h^{−1} correspond to x.

For more details on rings of fractions, see Atiyah-MacDonald Chapt 3.

The ringed space structure on an algebraic set. Let V be an algebraic subset of k^{n}. An element h of k[V] defines a function V → k, and h^{−1} defines a function

D(h) → k, a ↦ 1/h(a).

Thus a pair of elements g, h ∈ k[V] with h ≠ 0 define a function

a ↦ \frac{g(a)}{h(a)} : D(h) → k.

We define a function f: U → k on an open subset U of V to be regular if it is of this form in a neighbourhood of each of its points, i.e., if for all a ∈ U, there exist g, h ∈ k[V] with h(a) ≠ 0 such that the functions f and \frac{g}{h} agree in a neighbourhood of a. Write O_{V} (U) for the set of regular functions on U.

For example, if V = k^{n}, then a function f: U → k is regular at a point a ∈ U if there are polynomials g(X_{1}, …, X_{n}) and h(X_{1}, …, X_{n}) with h(a) ≠ 0 and f(b) = \frac{g(b)}{h(b)} for all b such that the expression on the right is defined.

Propostion 2.5. The map U ↦ O_{V} (U) defines a sheaf of k-algebras on V.

Proof. We have to check the conditions (2.1).

(a) Clearly, a constant function is regular. Suppose f and f′ are regular on U, and let a ∈ U. By assumption, there exist g, g′, h, h′ ∈ k[V], with h(a) ≠ 0 ≠ h′(a) such that f and f′ agree with \frac{g}{h} and \frac{g′}{h′} respectively near a. Then ff′ agrees with \frac{gh′+g′h}{hh′} near a, and so ff′ is regular on U. Similarly f ± f′ are regular on U. Thus O_{V} (U) is a k-algebra.

(b) It is clear from the definition that the restriction of a regular function to an open subset is again regular.

(c) The condition for f to be regular is obviously local. □

Lemma 2.6. The element g/h^{m} of k[V]_{h} defines the zero function on D(h) if and only if gh = 0 (in k[V]) (and hence g/h^{m} = 0 in k[V]_{h}).

Proof. If g/h^{m} is zero on D(h), then gh is zero on V because h is zero on the complement of D(h). Therefore gh is zero in k[V]. Conversely, if gh = 0, then g(a)h(a) = 0 for all a ∈ k^{n}, and so g(a) = 0 for all a ∈ D(h). □

Proposition 2.7. (a) The canonical map k[V]_{h} → O_{V} (D(h)) is an isomorphism.

(b) For any a ∈ V, there is a canonical isomorphism O_{a} → k[V]_{𝔪_{a}}, where 𝔪_{a} is the maximal ideal (x_{1} − a_{1}, …, x_{n} − a_{n}).

Proof. (a) The preceding lemma shows that k[V]_{h} → O_{V} (D(h)) is injective, and so it remains to show that every regular function f on V arises from an element of k[V]_{h}.

By definition, we know that there is an open covering V = ∪V_{i} and elements g_{i}, h_{i} ∈ k[V] with h_{i} nowhere zero on V_{i} such that f|V_{i} = \frac{g_{i}}{h_{i}}. Since the sets of the form D(a) form a basis for the topology on V, we can assume that V_{i} = D(a_{i}), some a_{i} ∈ k[V]. By assumption D(a_{i}) ⊂ D(h_{i}), and so a_{i}^{N} = h_{i}g′_{i} for some h′_{i} ∈ k[V] (see p22). On D(a_{i}), f = \frac{g_{i}}{h_{i}} = \frac{g_{i}g′_{i}}{h_{i}g′_{i}} = \frac{g_{i}g′_{i}}{a_{i}^{N}}. Note that D(a_{i}^{N}) = D(a_{i}). Therefore, after

replacing g_{i} with g_{i}g′_{i} and h_{i} with a_{i}^{N}, we can suppose that V_{i} = D(h_{i}).

We now have that D(h) = ∪D(h_{i}) and that f|D(h_{i}) = \frac{g_{i}}{h_{i}}. Because D(h) is quasicompact^{10}, we can assume that the covering is finite. As \frac{g_{i}}{h_{i}} = \frac{g_{j}}{h_{j}} on D(h_{i}) ∩ D(h_{j}) = D(h_{i}h_{j}), we have (by the Lemma) that

h_{i}h_{j}(g_{i}h_{j} − g_{j}h_{i}) = 0.  (2.8.1)

Because D(h) = ∪D(h_{i}) = ∪D(h_{i}^{2}), V ((h)) = V ((h_{1}^{2}, …, h_{m}^{2})), and so there exist a_{i} ∈ k[V] such that

h^{N} = ∑ a_{i}h_{i}^{2}.  (2.8.2)

I claim that f is the function on D(h) defined by \frac{∑ a_{i}g_{i}h_{i}}{h^{N}} .

Let a be a point of V. Then a will be in one of the D(h_{i}), say D(h_{j}). We have the following equalities in k[V]:

\begin{aligned}h_{j}^{2} ∑_{i=1}^{n} a_{i}g_{i}h_{i} & = ∑_{i=1}^{n} a_{i}g_{j}h_{i}^{2}h_{j}  by (2.8.1) \\ & = g_{j}h_{j}h^{N}  by (2.8.2). \end{aligned}

But f|D(h_{j}) = \frac{g_{j}}{h_{j}}, i.e., fh_{j} and g_{j} agree as functions on D(h_{j}). Therefore we have the following equality of functions on D(h_{j}):

h_{j}^{2} ∑ a_{i}g_{i}h_{i} = fh_{j}^{2}h^{N}.

Since h_{j}^{2} is never zero on D(h_{j}), we can cancel it, to find that, as claimed, the function fh^{N} on D(h_{j}) equals that defined by ∑ a_{i}g_{i}h_{i}.

(b) First a general observation: in the definition of the germs of a sheaf at a, it suffices to consider pairs (f, U) with U lying in a fixed basis for the neighbourhoods of a. Thus each element of O_{a} is represented by a pair (f, D(h)) where h(a) ≠ 0 and f ∈ k[V]_{h}, and two pairs (f_{1}, D(h_{1})) and (f_{2}, D(h_{2})) represent the same element of O_{a} if and only if f_{1} and f_{2} restrict to the same function on D(h) for some a ∈ D(h) ⊂ D(h_{1}h_{2}).

10Recall (1.13) that V is Noetherian, i.e., has the ascending chain condition on open subsets.

This implies that any open subset of V is also Noetherian, and hence is quasi-compact.

For each h ∉ 𝔭, there is a canonical homomorphism α_{h}: k[V]_{h} → k[V]_{𝔭}, and we map the element of O_{a} represented by (f, D(h)) to α_{h}(f). It is now an easy exercise to check that this map is well-defined, injective, and surjective. □

The proposition gives us an explicit description of the value of O_{V} on any basic open set and of the ring of germs at any point a of V. When V is irreducible, this becomes a little simpler because all the rings are subrings of k(V). We have:

Γ(D(h), O_{V}) = \{\frac{g}{h^{N}} ∈ k(V) | g ∈ k[V], N ∈ ℕ\};
O_{a} = \{\frac{g}{h} ∈ k(V)|h(a) ≠ 0\};
Γ(U, O_{V}) = ∩O_{a} (intersection over all a ∈ U)
= ∩Γ(D(h_{i}), O_{V}) if U = ∪D(h_{i}).

Note that every nonzero element of k(V) defines a function on some open subset of V —we call such an object a rational function on V. The last equality then says that the regular functions on U are the rational functions on V that are defined at each point of U.

Example 2.9. (a) Let V = k^{n}. Then the ring of regular functions on V, Γ(V, O_{V}), is k[X_{1}, …, X_{n}]. For any nonzero polynomial h(X_{1}, …, X_{n}), the ring of regular functions on D(h) is

\{\frac{g}{h^{N}} ∈ k(X_{1}, …, X_{n})|g, h ∈ k[X_{1}, …, X_{n}]\}.

For any point a = (a_{1}, …, a_{n}), the ring of germs of functions at a is

O_{a} = \{\frac{g}{h} ∈ k(X_{1}, …, X_{n})|h(a) ≠ 0\} = k[X_{1}, …, X_{n}]_{(X_{1}−a_{1},... ,X_{n}−a_{n})},

and its maximal ideal consists of those g/h with g(a) = 0.

(b) Let U = {(a, b) ∈ k^{2} | (a, b) ≠ (0, 0)}. It is an open subset of k^{2}, but it is not a basic open subset, because its complement {(0, 0)} has dimension 0, and therefore can’t be of the form V ((f)) (see 1.21). Since U = D(X) ∪ D(Y), the ring of regular functions on U is

Γ(D(X), O) ∩ Γ(D(Y), O) = k[X, Y]_{X} ∩ k[X, Y]_{Y} .

Thus (as an element of k(X, Y)), it can be written

f = \frac{g(X, Y)}{X^{N}} = \frac{h(X, Y)}{Y^{M}}.

Since k[X, Y] is a unique factorization domain, we can assume that the fractions are in their lowest terms. On multiplying through by X^{N}Y^{M}, we find that

g(X, Y)Y^{M} = h(X, Y)X^{N}.

Because X doesn’t divide the left hand side, it can’t divide the right either, and so N = 0. Hence f ∈ k[X, Y], and every regular function on U extends to k^{2}: Γ(U, O) = k[X, Y].

Morphisms of ringed spaces. A morphism of ringed spaces (V, O_{V}) → (W, O_{W}) is a continuous map φ: V → W such that

f ∈ O_{W} (U) ⟹ f ∘ φ ∈ O_{V} (φ^{−1}U)

for all open subsets U of W. Sometimes we write φ^{∗}(f) for f ∘ φ. If U is an open subset of V, then the inclusion (U, O_{V} |U) ↪ (V, O_{V}) is a morphism of ringed spaces. A morphism of ringed spaces is an isomorphism if it is bijective and its inverse is also a morphism of ringed spaces.

Example 2.10. (a) Let V and V′ be topological spaces endowed with their sheaves O_{V} and O_{V′} of continuous real valued functions. Then any continuous map φ: V → V′ is automatically a morphism of ringed structures (V, O_{V}) → (V′, O_{V′}).

(b) Let U and U′ be open subsets of ℝ^{n} and ℝ^{m} respectively. Recall from advanced calculus that a mapping

φ = (φ_{1}, …, φ_{m}): U ⟶ U′ ⊂ ℝ^{m}

is said to be infinitely differentiable (or C^{∞}) if each φ_{i} is infinitely differentiable, in which case f ∘ φ is infinitely differentiable for all infinitely differentiable functions f: U′ → ℝ. Note that φ_{i} = x_{i} ∘ φ, where x_{i} is the coordinate function (a_{1}, …, a_{n}) ↦ a_{i}.

Let V and V′ be open subsets of ℝ^{n} and ℝ^{m} respectively, endowed with their sheaves of infinitely differentiable functions O_{V} and O_{V′}. The above statement shows that a continuous mapping φ: V → V′ is infinitely differentiable if and only if it is a morphism of ringed spaces.

(c) Same as (b), but replace ℝ with ℂ and “infinitely differentiable” with “analytic”.

Remark 2.11. A morphism of ringed spaces maps germs of functions to germs of functions. More precisely, a morphism φ: (V, O_{V}) → (V′, O_{V′}) induces a map

O_{V,P} ⟵ O_{V′,φ(P)},

namely^{11}, [(f, U)] ↦ [(f ∘ φ, φ^{−1}(U))].

11I sometimes use [*] to denote an equivalence class containing *.

Affine algebraic varieties. We have just seen that every algebraic set gives rise to a ringed space (V, O_{V}). We define an affine algebraic variety over k to be a ringed space that is isomorphic to a ringed space of this form. A morphism of affine algebraic varieties is a morphism of ringed spaces; we often call it a regular map V → W or a morphism V → W, and we write Mor(V, W) for the set of such morphisms. With these definitions, the affine algebraic varieties become a category. Since we consider no nonalgebraic affine varieties, we shall often drop the “algebraic”.

In particular, every algebraic set has a natural structure of an affine variety. We usually write 𝔸^{n} for the k^{n} regarded as an affine algebraic variety. Note that the affine varieties we have constructed so far have all been embedded in 𝔸^{n}. We shall now see how to construct “unembedded” affine varieties.

A reduced finitely generated k-algebra is called and affine k-algebra. Thus for such an algebra A, there exist x_{i} ∈ A (not necessarily algebraically independent), such that A = k[x_{1}, …, x_{n}], and the kernel of the homomorphism

X_{i} ↦ x_{i}: k[X_{1}, …, X_{n}] → A

is a radical ideal. Zariski’s Lemma 1.7 implies that, for any maximal ideal 𝔪 ∈ A, the map k → A → A/𝔪 is an isomorphism. Thus we can identify A/𝔪 with k. For f ∈ A, we write f(𝔪) for the image of f in A/𝔪 = k, i.e., f(𝔪) = f (mod 𝔪).

We can associate with any affine k-algebra A a ringed space (V, O_{V}). First, V is the set of maximal ideals in A. For h ∈ A, h ≠ 0, let

D(h) = {𝔪 | h(𝔪) ≠ 0, i.e., h ∉ 𝔪},

and endow V with the topology for which the D(h) are a basis. A pair of elements g, h ∈ A, h ≠ 0, define a function

𝔪 ↦ \frac{g(𝔪)}{h(𝔪)} : D(h) → k,

and we define a function f: U → k on an open subset U of V to be regular if it is of this form on a neighbourhood of each point of U. Write O_{V} (U) for the set of regular functions on U.

Proposition 2.12. The pair (V, O_{V}) is an affine variety with Γ(V, O_{V}) = A.

Proof. Represent A as a quotient k[X_{1}, …, X_{n}]/𝔞 = k[x_{1}, …, x_{n}]. Then the map

(a_{1}, …, a_{n}) ↦ (x_{1} − a_{1}, …, x_{n} − a_{n}) (ideal in A)

is a bijection φ: V (𝔞) → V with inverse

𝔪 ↦ (x_{1}(𝔪), …, x_{n}(𝔪)): V → V (𝔞) ⊂ k^{n}.

It is easy to check that this is a homeomorphism, and that a function f on an open subset of V is regular (according to the above definition) if and only if f ∘ φ is regular. □

We write specm(A) for the topological space V, and Specm(A) for the ringed space (V, O_{V}). If we start with an affine variety V and let A = Γ(V, O_{V}), then the Specm(A) ≈ (V, O_{V}) (canonically). We again write k[V] for Γ(V, O_{V}), the ring of functions regular on the whole of V.

Thus, for each affine k-algebra A, we have an affine variety Specm(A), and conversely, for each affine variety (V, O_{V}), we have an affine k-algebra Γ(V, O_{V}). We now make this correspondence into an equivalence of categories.

Remark 2.13. I claim that a radical ideal 𝔞 in k[X_{1}, …, X_{n}] is equal to the intersection of the maximal ideals containing it. Indeed, the maximal ideals in k[X_{1}, …, X_{n}] are all of the form 𝔪_{a} = (X_{1} − a_{1}, …, X_{n} − a_{n}), and f ∈ 𝔪_{a} ⟺ f(a) = 0. Thus 𝔪_{a} ⊃ 𝔞 ⟺ a ∈ V (𝔞), and if f ∈ 𝔪_{a} for all a ∈ V (𝔞), then f is zero on V (𝔞), i.e., f ∈ IV (𝔞) = 𝔞.

This remark implies that, for any affine k-algebra A, the intersection of the maximal ideals of A is zero, because A is isomorphic to a k-algebra k[X_{1}, …, X_{n}]/𝔞 and we can apply the remark to 𝔞. Hence the map that associates with f ∈ A the map specm A → k, 𝔪 ↦ f(𝔪), is injective: A can be identified with a ring of functions on specm A.

The category of affine algebraic varieties. Let α: A → B be a homomorphism of affine k-algebras. For any h ∈ A, α(h) is invertible in B_{α(h)}, and so the homomorphism A → B → B_{α(h)} extends to a homomorphism

\frac{g}{h^{m}} ↦ \frac{α(g)}{α(h)^{m}} : A_{h} → B_{α(h)}.

For any maximal ideal 𝔫 of B, 𝔪 =_{df} α^{−1}(𝔫) is maximal in A, because A/𝔪 → B/𝔫 = k is an injective map of k-algebras and this implies A/𝔪 = k. Thus α defines a map

φ: specm B ⟶ specm A, φ(𝔫) = α^{−1}(𝔫) = 𝔪.

For 𝔪 = α^{−1}(𝔫) = φ(𝔫), we have a commutative diagram:

A ⟶^{α} B
↓  ↓
A/𝔪 ⟶^{=} A/𝔫.

Recall that the image of an element f of A in A/𝔪 = k is denoted f(𝔪). Therefore, the commutativity of the diagram means that, for f ∈ A,

f(φ(𝔫)) = α(f)(𝔫), i.e., f ∘ φ = α.  (*)

Since φ^{−1}D(f) = D(f ∘ φ) (obviously), it follows from (*) that

φ^{−1}(D(f)) = D(α(f)),

and so φ is continuous.

Let f be a regular function on D(h), and write f = g/h^{m}, g ∈ A. Then, from (*) we see that f ∘ φ is the function on D(α(h)) defined by α(g)/α(h)^{m}. In particular, it is regular, and so f ↦ f ∘φ maps regular functions on D(h) to regular functions on D(α(h)). It follows that f ↦ f ∘φ sends regular functions on any open subset of specm(A) to regular functions on the inverse image of the open subset. Thus α defines a morphism Specm(B) → Specm(A).

Conversely, by definition, a morphism of φ: (V, O_{V}) → (W, O_{W}) of affine algebraic varieties defines a homomorphism of the associated affine k-algebras k[W] → k[V]. Since these maps are inverse, we have shown:

Proposition 2.14. For any affine algebras A and B,

Hom_{k-alg}(A, B) ⟶^{≈} Mor(Specm(B), Specm(A));

for any affine varieties V and W,

Mor(V, W) ⟶^{≈} Hom_{k-alg}(k[W], k[V]).

A covariant functor F: A → B of categories is defined to be an equivalence of categories if

(a) it is fully faithful, i.e., for all objects A, A′ of A,

Hom(A, A′) → Hom(F (A), F (A′))

is a bijection;

(b) it is essentially surjective, i.e., every object of B is isomorphic to an object of the form F (A), A ∈ Ob(A).

One can show that such a functor F has a quasi-inverse, i.e., there is a functor G: B → A, which is also an equivalence, and is such that G(F (A)) ≈ A, F (G(B)) ≈ B for all objects A of A and B of B. Hence the relation of equivalence is an equivalence relation. (In fact one can do better—see Bucur and Deleanu (1968), I.6, or any other book on category theory.)

Similarly one defines the notion of a contravariant functor being an equivalence of categories. Proposition 2.14 can now be restated in stronger form as:

Proposition 2.15. The functor A ↦ Specm A is a (contravariant) equivalence from the category of affine k-algebras to that of affine varieties with quasi-inverse (V, O_{V}) ↦ Γ(V, O_{V}).

Explicit description of morphisms of affine varieties.

Proposition 2.16. Let V = V (𝔞) ⊂ k^{m}, W = V (𝔟) ⊂ k^{n}. The following conditions on a continuous map φ: V → W are equivalent:

(i) φ is regular; (ii) the components φ_{1}, …, φ_{m} of φ are all regular; (iii) f ∈ k[W] ⟹ f ∘ φ ∈ k[V].

Proof. (i) ⟹ (ii). By definition φ_{i} = y_{i} ∘ φ where y_{i} is the coordinate function (b_{1}, …, b_{n}) ↦ b_{i}: W → k. Hence this implication follows directly from the definition of a regular map.

(ii) ⟹ (iii). The map f ↦ f ∘ φ is a k-algebra homomorphism from the ring of all functions W → k to the ring of all functions V → k, and (ii) says that the map sends the coordinate functions y_{i} on W into k[V]. Since the y_{i}’s generate k[W] as a k-algebra, this implies that this map sends k[W] into k[V].

(iii) ⟹ (i). The map f ↦ f ∘ φ is a homomorphism α: k[W] → k[V]. It therefore defines a map specm k[V] → specm k[W], and it remains to show that this coincides with φ when we identify specm k[V] with V and specm k[W] with W. Let a ∈ V, let b = φ(a), and let 𝔪_{a} and 𝔪_{b} be the ideals of elements of k[V] and k[W] that are zero at a and b respectively. Then, for f ∈ k[W],

α(f) ∈ 𝔪_{a} ⟺ f(φ(a)) = 0 ⟺ f(b) = 0 ⟺ f ∈ 𝔪_{b}.

Therefore α^{−1}(𝔪_{a}) = 𝔪_{b}, which is what we needed to show. □

Remark 2.17. For all a ∈ V, f ↦ f ∘ φ maps germs of regular functions at φ(a) to germs of regular functions at a; in fact, it induces a local homomorphism O_{V ,φ(a)} → O_{V,a}.

Now consider equations

Y_{1} = P_{1}(X_{1}, …, X_{m})
…
Y_{n} = P_{n}(X_{1}, …, X_{m}).

On the one hand, they define a mapping φ: k^{m} → k^{n}, namely,

(a_{1}, …, a_{m}) ↦ (P_{1}(a_{1}, …, a_{m}), …, P_{n}(a_{1}, …, a_{m})).

On the other, they define a homomorphism of k-algebras α: k[Y_{1}, …, Y_{n}] → k[X_{1}, …, X_{n}], namely, that sending

Y_{i} ↦ P_{i}(X_{1}, …, X_{n}).

This map coincides with f ↦ f ∘ φ, because

α(f)(a) = f(…, P_{i}(a), …) = f(φ(a)).

Now consider closed subsets V (𝔞) ⊂ k^{m} and V (𝔟) ⊂ k^{n} with 𝔞 and 𝔟 radical ideals. I claim that φ maps V (𝔞) into V (𝔟) if and only if α(𝔟) ⊂ 𝔞. Indeed, suppose φ(V (𝔞)) ⊂ V (𝔟), and let f ∈ 𝔟; for b ∈ V (𝔟),

α(f)(b) = f(φ(b)) = 0,

and so α(f) ∈ IV (𝔟) = 𝔟. Conversely, suppose α(𝔟) ⊂ 𝔞, and let a ∈ V (𝔞); for f ∈ 𝔞,

f(φ(a)) = α(f)(a) = 0,

and so φ(a) ∈ V (𝔞). When these conditions hold, φ is the morphism of affine varieties V (𝔞) → V (𝔟) corresponding to the homomorphism k[Y_{1}, …, Y_{m}]/𝔟 → k[X_{1}, …, X_{n}]/𝔞 defined by α.

Thus, we see that the morphisms

V (𝔞) ⟶ V (𝔟)

are all of the form

a ↦ (P_{1}(a), …, P_{m}(a)), P_{i} ∈ k[X_{1}, …, X_{n}].

Example 2.18. (a) Consider a k-algebra R. From a k-algebra homomorphism α: k[X] → R, we obtain an element α(X) ∈ R, and α(X) determines α completely. Moreover, α(X) can be any element of R. Thus

α ↦ α(X): Hom_{k−alg}(k[X], R) ⟶^{≈} R.

According to (2.14)

Mor(V, 𝔸^{1}) = Hom_{k-alg}(k[X], k[V]).

Thus the regular maps V → 𝔸^{1} are simply the regular functions on V (as we would hope).

(b) Define 𝔸^{0} to be the ringed space (V_{0}, O_{V_{0}}) with V_{0} consisting of a single point, and Γ(V_{0}, O_{V_{0}}) = k. Equivalently, 𝔸^{0} = Specm k. Then, for any affine variety V,

Mor(𝔸^{0}, V) ≈ Hom_{k-alg}(k[V], k) ≈ V

where the last map sends α to the point corresponding to the maximal ideal Ker(α).

(c) Consider t ↦ (t^{2}, t^{3}): 𝔸^{1} → 𝔸^{2}. This is bijective onto its image, the variety V : Y^{2} = X^{3}, but it is not an isomorphism onto its image, i.e., the inverse map is not a morphism. Because of (2.14), it suffices to show that it doesn’t induce an isomorphism on the rings of regular functions. We have k[𝔸^{1}] = k[T] and k[V] = k[X, Y]/(Y^{2} − X^{3}) = k[x, y]. The map on rings is

x ↦ T^{2}, y ↦ T^{3}, k[x, y] → k[T],

which is injective, but the image is k[T^{2}, T^{3}] ≠ k[T]. In fact, k[x,y] is not integrally closed: (y/x)^{2} − x = 0, and so (y/x) is integral over k[x, y], but y/x ∉ k[x, y] (it maps to T under the inclusion k(x, y) ↪ k(T)).

(d) Consider x ↦ x^{p} : 𝔸^{n} → 𝔸^{n}, p = characteristic of k. This is a bijection, but it is not an isomorphism: the map on rings is k[X_{1}, …, X_{n}] → k[X_{1}, …, X_{n}], X_{i} ↦ X_{i}^{p}.

This map is the famous Frobenius map. Take k to be the algebraic closure of 𝔽_{p}, the field with p elements, and write F^{m} for the map. Then the fixed points of F^{m} are precisely the points of 𝔸^{n} with coordinates in 𝔽_{p^{m}}, the field with p^{m}- elements (recall from Math 594 that 𝔽_{p^{m}} is precisely the subfield of k consisting of those elements satisfying the equation X^{p^{m}} = X). Let P (X_{1}, …, X_{n}) be a polynomial with coefficients in 𝔽_{p^{m}}, P = ∑ c_{α}X^{α}, c_{α} ∈ 𝔽_{p^{m}}. If P (a) = 0, a ∈ k^{n}, i.e., ∑ c_{α}a_{1}^{i_{1}} ⋯ a_{n}^{i_{n}} = 0, then

0 = (∑ c_{α}a_{1}^{i_{1}} ⋯ a_{n}^{i_{n}})^{p^{m}} = ∑ c_{α}a_{1}^{p^{m}i_{1}} ⋯ a_{n}^{p^{m}i_{n}},

and so P (F^{m}a) = 0. Thus F^{m} maps V (P) into V (P), and its fixed points are the solutions of

P (X_{1}, …, X_{n}) = 0

in 𝔽_{p^{m}}.

In one of the most beautiful pieces of mathematics of the last fifty years, Grothendieck defined a cohomology theory (the étale cohomology) that allowed him to obtain an expression for the number of solutions in terms of a Lefschetz fixed point formula, and Deligne used the theory to obtain very precise estimates for the number of solutions.

Subvarieties. For any ideal 𝔞 in A, we define

V (𝔞) = {P ∈ specm A | f(P) = 0 all f ∈ 𝔞} = {𝔪 maximal ideal in A | 𝔞 ⊂ 𝔪}.

This is a closed subset of specm A, and every closed subset is of this form.

Now assume 𝔞 is radical, so that A/𝔞 is again reduced. Corresponding to the homomorphism A → A/𝔞, we get a regular map

Specm A/𝔞 → Specm A

The image is V (𝔞), and specm A/𝔞 → V (𝔞) is a homeomorphism. Thus every closed subset of specm A has a natural ringed structure making it into an affine algebraic variety. We call V (𝔞) with this structure a closed subvariety of V.

Aside 2.19. If (V, O_{V}) is a ringed space, and Z is a closed subset of V, we can define a ringed space structure on Z as follows: let U be an open subset of Z, and let f be a function U → k; then f ∈ Γ(U, O_{Z}) if for each P ∈ U there is a germ (U′, f′) of a function at P (regarded as a point of V) such that f′|Z ∩ U′ = f. One can check that when this construction is applied to Z = V (𝔞), the ringed space structure is that given in the above example.

Proposition 2.20. Let (V, O_{V}) be an affine variety and let h ∈ k[V], h ≠ 0. Then (D(h), O_{V} |D(h)) is an affine variety; in fact if V = specm(A), then D(h) = specm(A_{h}). More explicitly, if V = V (𝔞) ⊂ k^{n}, then

(a_{1}, …, a_{n}) ↦ (a_{1}, …, a_{n}, h(a_{1}, …, a_{n})^{−1}): D(h) → k^{n+1},

defines an isomorphism of D(h) onto V (𝔞, 1 − hX_{n+1}).

Proof. The map A → A_{h} defines a morphism specm A_{h} → specm A. The image is D(h), and it is routine (using (2.4)) to verify the rest of the statement. □

For example, there is an isomorphism of affine varieties

x ↦ (x, 1/x) : 𝔸^{1} − {0} → V ⊂ 𝔸^{2},

where V is the subvariety XY = 1 of 𝔸^{2}. [Picture]

Remark 2.21. We have seen that all closed subsets of an affine variety V, and all basic open subsets, are again affine varieties, but it need not be true that (U, O_{V} |U) is an affine variety, U open in V. Note that if a ringed space (U, O_{U}) is an affine variety, then we must have (U, O_{U}) ≈ Specm(A), A = Γ(U, O_{U}). In particular, the map

P ↦ 𝔪_{P} = {f ∈ A|f(P) = 0}

will be a bijection from U onto specm(A).

Consider U ⊂ 𝔸^{2} − (0, 0) = D(X) ∪ D(Y). We saw in (2.9b) that Γ(U, O_{𝔸^{2}}) = k[X, Y]. Now U → specm k[X, Y] is not a bijection, because the ideal (X, Y) is not in the image.

However, U is clearly a union of affine algebraic varieties — it is an algebraic variety — see the next section.

Properties of specm(α).

Proposition 2.22. Let α : A → B be a homomorphism of affine k-algebras, and let φ: Specm(B) → Specm(A) be the corresponding morphism of affine varieties (so that α(f) = φ ∘ f).

(a) The image of φ is dense if and only if α it is injective.

(b) φ defines an isomorphism of Specm(B) onto a closed subvariety of Specm(A) if and only if α is surjective.

Proof. (a) Let f ∈ A. If the image of φ is dense, then

f ∘ φ = 0 ⟹ f = 0.

Conversely, if the image of φ is not dense, there will be a nonzero function f ∈ A that is zero on its image, i.e., such that f ∘ φ = 0.

(b) If α is surjective, then it defines an isomorphism A/𝔞 → B where 𝔞 is the kernel of α. This induces an isomorphism of Specm(B) with its image in Specm(A). □

A regular map φ : V → W of affine algebraic varieties is said to be a dominating if the image is dense in W. The proposition then says that:

φ is dominating ⟺ f ↦ f ∘ φ: Γ(W, O_{W}) → Γ(V, O_{V}) is injective.

A little history. We have associated with any affine k-algebra A an affine variety whose underlying topological space is the set of maximal ideals in A. It may seem strange to be describing a topological space in terms of maximal ideals in a ring, but the analysts have been doing this for 50 years. Gel’fand and Kolmogorov in 1939 proved that if S and T are completely regular topological spaces, and the rings of real-valued continuous functions on S and T are isomorphic (just as rings), then S and T are homeomorphic. The first step in the proof showed that, for such a space S, the map

P ↦ 𝔪_{P} = {f: S → ℝ | f(P) = 0}

defines a one-to-one correspondence between the points in the space and maximal ideals in the ring. (See Shields’s article in Math. Intelligencer, Summer 1989, 15-17.) (A space S is completely regular if it is T_{1} and for every closed subset C and point P ∉ C, there is a real-valued continuous function f on the space such that f(P) = 0 and f is identically 1 on C.)

3. Algebraic Varieties

An algebraic variety is a ringed space that is locally isomorphic to an affine algebraic variety, just as a topological manifold is a ringed space that is locally isomorphic to an open subset of ℝ^{n}; both are required to satisfy a separation axiom.

Algebraic prevarieties. As motivation, recall the following definitions.

Definition 3.1. (a) A topological manifold is a ringed space (V, O_{V}) such that V is Hausdorff and every point of V has an open neighbourhood U for which (U, O_{V} |U) is isomorphic to the ringed space of continuous functions on an open subset of ℝ^{n} (cf. (2.2a)).

(b) A differentiable manifold is a ringed space such that V is Hausdorff and every point of V has an open neighbourhood U for which (U, O_{V} |U) is isomorphic to a ringed space as in (2.2b).

(c) A complex manifold is a ringed space such that V is Hausdorff and every point of V has an open neighbourhood U for which (U, O_{V} |U) is isomorphic to a ringed space as in (2.2c).

Often one imposes additional conditions on V, for example that it is second countable or connected. The above definitions are easily seen to be equivalent to the more classical definitions in terms of charts and atlases.

Definition 3.2. An algebraic prevariety is a ringed space (V, O_{V}) such that V is quasi-compact and every point of V has an open neighbourhood U such that (V, O_{V} |U) is an affine algebraic variety. Equivalently, a ringed space (V, O_{V}) is an algebraic prevariety if there is a finite open covering V = ∪V_{i} such that (V_{i}, O_{V} |V_{i}) is an affine algebraic variety for all i.

An algebraic variety will be defined to be an algebraic prevariety satisfying a certain separation condition.

An open subset U of an algebraic prevariety V such that (U, O_{V} |U) is an affine algebraic variety is called an open affine (subvariety) in V.

Let (V, O_{V}) be an algebraic variety, and let U be an open subset of V. The functions f: U → k lying in Γ(U, O_{V}) are called regular. Note that if (U_{i}) is an open covering of V by affine varieties, then f: U → k is regular if and only if f|U_{i} ∩ U is regular for all i (this is just a special case of condition (c) to be a sheaf, p12). Thus understanding the regular functions on open subsets of V amounts to understanding the regular functions on the open affine subvarieties and how these subvarieties fit together to form V.

Example 3.3. Any open subset of an affine variety together with its induced ringed structure is an algebraic prevariety (in fact variety). For example, 𝔸^{2} − {(0, 0)} is an algebraic variety.

Example 3.4. (Projective space). Let

ℙ^{n} = k^{n+1} − {(0, …, 0)}/∼

where (a_{0}, …, a_{n}) ∼ (b_{0}, …, b_{n}) if there is a c ∈ k^{×}, such that (a_{0}, …, a_{n}) = (cb_{0}, …, cb_{n}). Thus the equivalence classes are the lines through the origin in k^{n+1}. We write (a_{0} : …: a_{n}) for the equivalence class containing (a_{0}, …, a_{n}). For each i, let

U_{i} = {(a_{0} : …: a_{i} : …: a_{n}) ∈ ℙ^{n} | a_{i} ≠ 0}.

Then ℙ^{n} = ∪U_{i}. The map u_{i}

(a_{1}, …, a_{n}) ↦ (a_{0} : …: a_{i−1} : 1 : a_{i+1}, …: a_{n}): k^{n} → U_{i}

is a bijection. We use this map to endow U_{i} with the Zariski topology, and we endow ℙ^{n} with the topology such that U ⊂ ℙ^{n} is open if and only if U ∩ U_{i} is open in U_{i} for all i. Define a function f: U → k on an open subset U to be f regular if f ∘ u_{i} is a regular function on k^{n} for all i. These definitions endow ℙ^{n} with the structure of a ringed space, and each map u_{i} is an isomorphism of ringed spaces (𝔸^{n}, O_{𝔸^{n}}) → (U_{i}, O_{V} |U_{i}). Thus ℙ^{n} is an algebraic prevariety. Later, in §5, we shall study ℙ^{n} in detail; in particular, we shall show that ℙ^{n} is separated.

Algebraic varieties. In the study of topological manifolds, the Hausdorff condition eliminates such bizarre possibilities as the line with the origin doubled, where a sequence tending to the origin has two different limits.

It is not immediately obvious how to impose a separation axiom on our algebraic varieties, because even affine algebraic varieties are not Hausdorff. The key is to restate the Hausdorff condition. Intuitively, the significance of this condition is that it implies that a sequence in the space can have at most one limit. Thus a map into the space should be determined by its values on a dense subset, i.e., if φ and ψ are maps Z → U that agree on a dense subset of Z then they should agree on the whole of Z. Equivalently, the set where φ and ψ agree should be closed. Surprisingly, affine varieties have this property, provided φ and ψ are required to be regular maps.

Lemma 3.5. Let φ and ψ be regular maps of affine algebraic varieties Z → V. The subset of Z on which φ and ψ agree is closed.

Proof. There are regular functions x_{i} on V such that P ↦ (x_{1}(P), …, x_{n}(P)) identifies V with a closed subset of 𝔸^{n} (take the x_{i} to be any set of generators for k[V] as a k-algebra). Now x_{i} ∘ φ and x_{i} ∘ φ are regular functions on Z, and the set where φ and ψ agree is ⋂_{i=1}^{n} V (x_{i} ∘ φ − x_{i} ∘ ψ), which is closed. □

Definition 3.6. An algebraic prevariety is said to be separated, or to be an algebraic variety, if it satisfies the following additional condition:

separation axiom: for every pair of morphisms of ringed spaces φ, ψ: Z ⇉ V with Z an algebraic prevariety, the set {z ∈ Z | φ(z) = ψ(z)} is closed in Z.

The is terminology not completely standardized: often one requires a variety to be irreducible, and sometimes one calls a prevariety a variety.

Regular maps. In each of the examples (3.1a,b,c), a morphism manifolds (continuous map, differentiable map, analytic map respectively) is just a morphism of ringed spaces. This motivates the following definition.

Let (V, O_{V}) and (W, O_{W}) be algebraic (pre)varieties. A map φ: V → W is said to be regular if it is a morphism of ringed spaces. A composite of regular maps is again regular (this is a general fact about morphisms of ringed spaces).

Note that we have four categories:

(Affine varieties) ⊂ (Alg. varieties) ⊂ (Alg. prevarieties) ⊂ (ringed spaces)

Each subcategory is full (i.e., the morphisms Mor(V, W) are the same in the four categories).

Proposition 3.7. Let (V, O_{V}) and (W, O_{W}) be prevarieties, and let φ: V → W be a continuous map (of topological spaces). Let W = ∪W_{i} be a covering of W by open affines, and let φ^{−1}(W_{j}) = ∪V_{ji} be a covering of φ^{−1}(W_{j}) by open affines. Then φ is regular if and only if its restrictions

φ|V_{ji}: V_{ji} ⟶ W_{j}

are regular for all i, j.

Proof. We assume that φ satisfies this condition, and prove that it is regular. Let f be a regular function on an open subset U of W. Then f|U ∩ W_{j} is regular for each W_{j} (because the regular functions form a sheaf), and so f ∘ φ|φ^{−1}(U) ∩ V_{ji} is regular for each j, i (this is our assumption). It follows that f ∘ φ is regular on φ^{−1}(U) (sheaf condition). Thus φ is regular. The converse is equally easy. □

Remark 3.8. In order to check that a prevariety V is separated, it suffices to show that for every pair of regular maps φ, ψ: Z → V with Z an affine algebraic variety {z ∈ Z | φ(z) = ψ(z)} is closed in Z. To prove this remark, cover Z with open affines. Thus (3.5) shows that affine varieties are separated.

Example 3.9. (The affine line with the origin doubled.) Let V_{1} and V_{2} be copies of 𝔸^{1}. Let V^{∗} = V_{1} ⨿ V_{2} (disjoint union), and give it the obvious topology. Define an equivalence relation on V^{∗} by

x (in V_{1}) ∼ y (in V_{2}) ⟺ x = y and x ≠ 0.

Let V be the quotient space V = V^{∗}/∼ with the quotient topology (a set is open if and only if its inverse image in V^{∗} is open). Then V_{1} and V_{2} are open subspaces of V, V = V_{1} ∪ V_{2}, and V_{1} ∩ V_{2} = 𝔸^{1} − {0}. Define a function on an open subset to be regular if its restriction to each V_{i} is regular. This makes V into a prevariety, but not a variety: it fails the separation axiom because the two maps 𝔸^{1} ⟶^{=} V_{i} ⟶ V, i = 1, 2, agree exactly on 𝔸^{1} − {0}, which is not closed in 𝔸^{1}.

Aside 3.10. A differentiable manifold of dimension d is locally isomorphic to an open subset of ℝ^{d}. In particular, all manifolds of the same dimension are locally isomorphic. This is not true for algebraic varieties, for two reasons:

(a) We are not assuming our varieties are nonsingular (see the next section).

(b) The inverse function theorem fails in our context. If P is a nonsingular point on variety of dimension d, we shall see (in the next section) that there is a neighbourhood U of P and a regular map φ: U → 𝔸^{d} such that map (dφ)_{P} : T_{P} → T_{φ(P)} on the tangent spaces is an isomorphism. If the inverse function theorem were true in our context, it would tell us that an open neighbourhood of P is isomorphic to an open neighbourhood of φ(P).

Subvarieties. Let (V, O_{V}) be a prevariety. Then V is a finite union of open affines, and in each open affine the open affines (in fact the basic open subsets) form a basis for the topology. From this it follows the open affines form a basis for the topology on V, i.e., every open subset U of V is a union of open affines (of V). It follows that, for any open subset U of V, (U, O_{V} |U) is a prevariety. Obviously the inclusion U ↪ V is regular. A regular map φ: W → V is an open immersion if φ(W) is open in V and φ defines an isomorphism W → φ(W) (of prevarieties).

Any closed subset Z in V has a canonical structure of an algebraic prevariety: endow it with the induced topology, and say that a function f on an open subset of Z is regular if each point P in the open subset has an open neighbourhood U in V such that f extends to a regular function on U. To show that Z, with this ringed space structure is a prevariety, check that for every open affine U ⊂ V, the ringed space (U ∩ Z, O_{Z}|U ∩ Z) is isomorphic to U ∩ Z with its ringed space structure acquired as a closed subset of U (see p45). A regular map φ: W → V is a closed immersion if φ(W) is closed in V and φ defines an isomorphism W → φ(W) (of prevarieties).

A subset W of a topological space V is said to be locally closed if every point P in W has an open neighbourhood U in V such that W ∩U is closed in U; equivalently, W is the intersection of an open and a closed subset of V. A locally closed subset W of a prevariety V acquires a natural structure as a prevariety: write it as the intersection W = U ∩ Z of an open and a closed subset; Z is a prevariety, and W (being open in Z) therefore acquires the structure of a prevariety. This structure on W has the following characterization: the inclusion map W ↪ V is regular, and a map φ: V′ → W with V′ a prevariety is regular if and only if it is regular when regarded as a map into V. With this structure, W is called a sub(pre)variety of V. A morphism φ : V′ → V is called an immersion if it induces an isomorphism of V′ onto a subvariety of V. Every immersion is the composite of an open immersion with a closed immersion (in both orders).

A subprevariety of a variety is automatically separated.

Proposition 3.11. A prevariety V is separated if and only if it has the following property: if a two regular maps φ, ψ: Z ⇉ V agree on a dense subset of Z, then they agree on the whole of Z.

Proof. If V is separated, then the set where φ and ψ agree is closed, and so must be the whole of Z.

Conversely, consider a pair of maps φ, ψ: Z ⇉ V, and let S be the subset of Z on which they agree. We assume V has the property in the statement of the lemma, and show that S is closed. Let S̄ be the closure of S in Z. According to the above discussion, S̄ has the structure of a closed prevariety of Z, and the maps φ|S̄ and ψ|S̄ are regular. Because they agree on a dense subset of S̄ they agree on the whole of S̄, and so S = S̄ is closed. □

Prevarieties obtained by patching. Let V = ∪V_{i} (finite union), and suppose that each V_{i} has the structure of an algebraic prevariety satisfying the following condition: for all i, j, V_{i} ∩ V_{j} is open in both V_{i} and V_{j} and the structures of an algebraic prevariety induced on it by V_{i} and V_{j} are equal. Then we can define the structure of a ringed space on V as follows: U ⊂ V is open if and only if U ∩ V_{i} is open for all i, and f: U → k is regular if and only if f|U ∩ V_{i} is regular for all i. It is straightforward to check that this does make V into a ringed space (V, O_{V}).

Proposition 3.12. The ringed space (V, O_{V}) is a prevariety, and the inclusions V_{i} ↪ V are regular maps.

Proof. One only has to check that the ringed space structure on each V_{i} induced by that of V is the original one. □

Products of varieties. Let V and W be objects in a category C. A triple

(V × W, p: V × W → V, q: V × W → W)

is said to be the product of V and W if, for all objects Z in C, the map φ ↦ (p ∘ φ, q ∘ φ) is a bijection

Hom(Z, V × W) → Hom(Z, V) × Hom(Z, W),

i.e., if every pair of morphisms Z → V, Z → W factors uniquely through V × W :

Z
⋮
↙  ↓  ↘
V ← V × W → W

Clearly, the product, if it exists, is uniquely determined up to a unique isomorphism.^{12}

12If (P, p′: P → V, q′: P → W) also has this property, then there exists a unique morphism γ: P → V × W such that p ∘ γ = p′ and q ∘ γ = q′ (universal property of V × W), and there exists a unique morphism γ′: V × W → P such that p′ ∘ γ′ = p and q′ ∘ γ′ = q (universal property of P). The composite γ ∘ γ′ is the unique morphism V × W → V × W such that p ∘ γ ∘ γ′ = p and q ∘ γ ∘ γ′ = q. But we already know one such morphism, namely, the identity morphism, and so γ ∘ γ′ = id. Similarly γ′ ∘ γ = id, and so γ and γ′ are inverse isomorphisms.

For example, the product of two sets (in the category of sets) is the usual cartesion product of the sets, and the product of two topological spaces (in the category of topological spaces) is the cartesian product of the spaces (as sets) with the usual product topology.

We shall show that products exist in the category of algebraic varieties. Suppose, for the moment, that V × W exists. It follows from (2.18b) that for any prevariety Z, Mor(𝔸^{0}, Z) is the underlying set of Z, i.e., for any z ∈ Z, the map 𝔸^{0} → Z with image z is regular, and these are all the regular maps. Thus, from the definition of products we have

(underlying set of V × W) = Mor(𝔸^{0}, V × W)

= Mor(𝔸^{0}, V) × Mor(𝔸^{0}, W)

= (underlying set of V) × (underlying set of W).

Thus our problem can be restated as follows: given two prevarieties V and W, define on the set V × W the structure of a prevariety such that the projection maps p, q: V × W ⇉ V, W are regular, and such that a map φ: T → V × W of sets (with T an algebraic prevariety) is regular if and only if its components p ∘ φ, q ∘ φ are regular. Clearly, there can be at most one such structure on the set V ×W (because the identity map will identify any two structures having these properties).

Before we can define products of algebraic varieties, we need to review tensor products.

Review of tensor products. Let A and B be k-algebras. A k-algebra C together with homomorphisms i: A → C and j: B → C is called the tensor product of A and B if it has the following universal mapping property: for every pair of homomorphisms (of k-algebras) α: A → R and β: B → R, there is a unique homomorphism γ: C → R such that γ ∘ i = α and γ ∘ j = β:

i  j
A → C ← B
α ↘ ⋮ ↙ β
↓
R

Clearly, if the tensor product exists, it is uniquely determined up to a unique isomorphism (same argument as in the above footnote). We write it A ⊗_{k} B.

Construction. Let C^{∗} be the k-vector space with basis A×B. Thus the elements of C^{∗} are finite sums ∑ c_{i}(a_{i}, b_{i}) with c_{i} ∈ k, a_{i} ∈ A, b_{i} ∈ B. Let D be the subspace of C^{∗} generated by the following elements,

\begin{matrix}(a + a′, b) − (a, b) − (a′, b), & a, a′ ∈ A, b ∈ B, \\ (a, b + b′) − (a, b) − (a, b′), & a ∈ A, b, b′ ∈ B, \\ (ca, b) − c(a, b), & a ∈ A, b ∈ B, c ∈ k, \\ (a, cb) − c(a, b), & a ∈ A, b ∈ B, c ∈ k, \end{matrix}

and define C = C^{∗}/D. Write a ⊗ b for the class of (a, b) in C—we have imposed the fewest conditions forcing (a, b) ↦ a ⊗ b to be k-bilinear. Every element of C can be written as a finite sum, ∑ a_{i} ⊗ b_{i}, a_{i} ∈ A, b_{i} ∈ B, and the map

A × B → C, (a, b) ↦ a ⊗ b

is k-bilinear. By definition, C is a k-vector space, and there is a product structure on C such that (a ⊗ b)(a′ ⊗ b′) = aa′ ⊗ bb′—for this one has to check that the map

C^{∗} × C^{∗} → C, ((a, b), (a′, b′)) ↦ aa′ ⊗ bb′

factors through C × C. It becomes a k-algebra by means of the homomorphism c ↦ c(1 ⊗ 1) = c ⊗ 1 = 1 ⊗ c. The maps

a ↦ a ⊗ 1: A → C and b ↦ 1 ⊗ b: B → C

are homomorphisms, and it is routine to check that they make C into the tensor product of A and B in the above sense.

Example 3.13. The algebra B, together with the given map k → B and the identity map B → B, satisfies the condition to be k ⊗_{k} B.

Example 3.14. (a) The ring k[X_{1}, …, X_{m}, X_{m+1}, …, X_{m+n}], together with the maps

k[X_{1}, …, X_{m}] ⟶_{inclusion}^{obvious} k[X_{1}, …, X_{m+n}] ⟵_{inclusion}^{obvious} k[X_{m+1}, …, X_{m+n}]

is the tensor product of k[X_{1}, …, X_{m}] and k[X_{m+1}, …, X_{m+n}]. To verify this we only have to check that, for every k-algebra R, the map

Hom_{k-alg}(k[X_{1}, …, X_{m+n}], R) → Hom_{k-alg}(k[X_{1}, …], R)×Hom_{k-alg}(k[X_{m+1}, …], R)

induced by the inclusions is a bijection. But this map can be identified with the bijection

R^{m+n} → R^{m} × R^{n}.

In terms of the constructive definition of tensor products, we have shown that there is an isomorphism

f ⊗ g ↦ fg: k[X_{1}, …, X_{m}] ⊗_{k} k[X_{m+1}, …, X_{m+n}] → k[X_{1}, …, X_{m+n}].

(b) Let 𝔞 and 𝔟 be ideals in k[X_{1}, …, X_{m}] and k[X_{m+1}, …, X_{m+n}] respectively, and let (𝔞, 𝔟) be the ideal in k[X_{1}, …, X_{m+n}] generated by the elements of 𝔞 and 𝔟. Then there is an isomorphism

f ⊗ g ↦ fg: \frac{k[X_{1}, …, X_{m}]}{𝔞} ⊗_{k} \frac{k[X_{m+1}, …, X_{m+n}]}{𝔟} ⟶ \frac{k[X_{1}, …, X_{m+n}]}{(𝔞, 𝔟)}.

Again this comes down to checking that the map Hom_{k-alg}(k[X_{1}, …, X_{m+n}]/(𝔞, 𝔟), R) →

Hom_{k-alg}(k[X_{1}, …, X_{m}]/𝔞, R) × Hom_{k-alg}(k[X_{m+1}, …, X_{m+n}]/𝔟, R)

is a bijection. But the three sets are respectively V (𝔞, 𝔟) = zero-set of (𝔞, 𝔟) in R^{m+n}, V (𝔞) = zero-set of 𝔞 in R^{m}, V (𝔟) = zero-set of 𝔟 in R^{n}, and so this is obvious.

Remark 3.15. (a) If (b_{α}) is a family of generators (resp. basis) for B as a k-vector space, then (1⊗b_{α}) is a family of generators (resp. basis) for A⊗_{k}B as an A-module.

(b) Let k ↪ Ω be fields. Then

Ω ⊗_{k} k[X_{1}, …, X_{n}] = Ω[1 ⊗ X_{1}, …, 1 ⊗ X_{n}] = Ω[X_{1}, …, X_{n}]

(here, as elsewhere in these notes, “=” means “canonically isomorphic”). If A = k[X_{1}, …, X_{n}]/(g_{1}, …, g_{m}), then

Ω ⊗_{k} A = Ω[X_{1}, …, X_{n}]/(g_{1}, …, g_{m}).

For more details on tensor products, see Atiyah-MacDonald, Chapter 2 (but note that the definition there (p31) of the homomorphism A → D making the tensor product into an A-algebra is incorrect).

Products of affine varieties. The tensor product of two k-algebras A and B has the universal property to be a product, but with the arrows reversed. Because of the category anti-equivalence (2.14), this will show that Specm A ⊗_{k} B is the product of Specm A and Specm B in the category of affine algebraic varieties once we have shown that A ⊗_{k} B is an affine k-algebra.

Proposition 3.16. Let A and B be finitely generated k-algebras; if A and B are reduced, then so also is A ⊗_{k} B; if A and B are integral domains, then so also is A ⊗_{k} B.

Proof. Assume A and B to be reduced, and let α ∈ A⊗_{k} B. Then α = ∑_{i=1}^{n} a_{i} ⊗b_{i}, some a_{i} ∈ A, b_{i} ∈ B. If one of the b_{i}’s is a linear combination of the remaining b’s, say, b_{n} = ∑_{i=1}^{n−1} c_{i}b_{i}, c_{i} ∈ k, then, using the bilinearity of ⊗, we find that

α = ∑_{i=1}^{n−1} a_{i} ⊗ b_{i} + ∑_{i=1}^{n−1} c_{i}a_{n} ⊗ b_{i} = ∑_{i=1}^{n−1}(a_{i} + c_{i}a_{n}) ⊗ b_{i}.

Thus we can suppose that in the original expression of α, the b_{i}’s are linearly independent over k.

Now suppose that α is nilpotent, and let 𝔪 be a maximal ideal in A. The homomorphism a ↦ ā: A → A/𝔪 = k induces a homomorphism

a ⊗ b ↦ ā ⊗ b ↦ āb: A ⊗_{k} B ⟶ k ⊗_{k} B ⟶^{≈} B.

The image ∑ ā_{i}b_{i} of α under this homomorphism is a nilpotent element of B, and hence is zero (because B is reduced). As the b_{i}’s are linearly independent over k, this means that the ā_{i} are all zero. Thus, for all i, a_{i} lies in every maximal ideal 𝔪 of A, and so is zero (by 2.13). Hence α = 0. This shows that A ⊗_{k} B is reduced.

Assume A and B to be integral domains, and let α, α′ ∈ A ⊗ B be such that αα′ = 0. As before, we can write α = ∑ a_{i} ⊗ b_{i} and α′ = ∑ a′_{i} ⊗ b′_{i} with the sets {b_{1}, b_{2}, … } and {b′_{1}, b′_{2}, … } linearly independent over k. For each maximal ideal 𝔪 of A, we know (∑ ā_{i}b_{i})(∑ ā′_{i}b′_{i}) = 0 in B, and so either (∑ ā_{i}b_{i}) = 0 or (∑ ā′_{i}b′_{i}) = 0. Thus either all the a_{i} ∈ 𝔪 or all the a′_{i} ∈ 𝔪. This shows that

specm(A) = V (a_{1}, …, a_{m}) ∪ V (a′_{1}, …, a′_{n}).

Since specm(A) is irreducible (see 1.15), we must have specm(A) = V (a_{1}, …, a_{m}) or V (a′_{1}, …, a′_{n}). In the first case α = 0, and in the second α′ = 0. □

Example 3.17. The statement of (3.16) requires that k be algebraically closed.

(a) Suppose k is nonperfect of characteristic p. To say that k is not perfect means that there is an element α in an algebraic closure of k such that α ∉ k but α^{p} ∈ k. Let k′ = k[α], α^{p} = a ∈ k, α ∉ k. Then (α ⊗ 1 − 1 ⊗ α) ≠ 0 in k′ ⊗_{k} k′ (in fact, the elements α^{i} ⊗ α^{j}, 0 ≤ i, j ≤ p − 1, form a basis for k′ ⊗_{k} k′ as a k-vector space), but

(α ⊗ 1 − 1 ⊗ α)^{p} = (a ⊗ 1 − 1 ⊗ a) = (1 ⊗ a − 1 ⊗ a) = 0.

Thus k′ ⊗_{k} k′ is not reduced, even though k′ is a field.

(b) Let K be a finite separable extension of k and let Ω be a “big” field containing k (for example an algebraic closure of k). Write K = k[α] = k[X]/(f(X)), and assume f(X) splits in Ω[X], say, f(X) = ∏ X − α_{i}. Then

K ⊗_{k} Ω ≈ Ω[X]/(f(X)) ≈ ∏ Ω[X]/(X − α_{i}),

and so it is not an integral domain. (The second isomorphism follows from the Chinese remainder theorem.)

Having (3.16), we can make the following definition: let V and W be affine varieties, and let Γ(V, O_{V}) = A and Γ(W, O_{W}) = B; then V × W = Specm(A ⊗_{k} B) with the projection maps p: V × W → V and q: V × W → W defined by the maps A → A ⊗_{k} B and B → A ⊗_{k} B.

Proposition 3.18. Let V and W be affine varieties; the projection maps p: V × W → V, q: V × W → W are regular, and a map φ: U → V × W is regular if and only if p ∘ φ and q ∘ φ are regular. Therefore (V × W, p, q) is the product of V and W in the category of algebraic prevarieties. If V and W are irreducible, then so also is V × W.

Proof. : The projection maps are regular because they correspond to the k-algebra homomorphisms k[V] → k[V]⊗_{k}k[W] and k[W] → k[V]⊗_{k}k[W]. Let φ: U → V ×W be a map (of sets) such that p ∘ φ and q ∘ φ are regular. If U is affine, then φ corresponds to the map k[V] ⊗ k[W] → k[U] induced by

f ↦ f ∘ (p ∘ φ): k[V] → k[U] and f ↦ f ∘ (q ∘ φ): k[W] → k[U],

and so is regular. This shows that, for a general U, the restriction of φ to every open affine of U is regular, and this implies that φ is regular (see 3.7).

The final statement follows from the second statement in 3.16. □

Example 3.19. (a) It follows from (3.14a) that

𝔸^{m} ⟵^{p} 𝔸^{m+n} ⟶^{q} 𝔸^{n},

where p(a_{1}, …, a_{m+n}) = (a_{1}, …, a_{m}), q(a_{1}, …, a_{m+n}) = (a_{m+1}, …, a_{m+n}), is the product of 𝔸^{m} and 𝔸^{n}.

(b) It follows from (3.14b) that

V (𝔞) ⟵^{p} V (𝔞, 𝔟) ⟶^{q} V (𝔟)

is the product of V (𝔞) and V (𝔟).

Warning! The topology on V × W is not the product topology; for example, the topology on 𝔸^{2} = 𝔸^{1} × 𝔸^{1} is not the product topology.

Products in general. Now let V and W be two algebraic prevarieties V and W. We define their product as follows: As a set, we take V × W. Now write V and W as unions of open affines, V = ∪V_{i}, W = ∪W_{j}. Then V × W = ∪V_{i} × W_{j}, and we give V × W the topology for which U ⊂ V × W is open if and only if U ∩ (V_{i} × W_{j}) is open for all i and j. Finally, we define a ringed space structure by saying that a function f : U → k on an open subset U is regular if its restriction to U ∩ (U_{i} × V_{j}) is regular for all i and j.

Proposition 3.20. With the above structure, V ×W is a prevariety, the projection maps

p: V × W → V, q: V × W → W

are regular, and a map φ: U → V × W is regular if and only if p ∘ φ and q ∘ φ are regular. It is therefore the product of V and W in the category of prevarieties.

Proof. Straightforward. □

Proposition 3.21. If V and W are separated, then so also is V × W.

Proof. Straightforward.

Example 3.22. An algebraic group is a variety G together with regular maps

mult: G × G → G, inverse: G → G,

and an element e ∈ G that make G into a group in the usual sense. For example, SL_{n} and GL_{n} are algebraic groups, and any finite group can be regarded as an algebraic group. Connected affine algebraic groups are called linear algebraic groups because they can all be realized as closed subgroups of GL_{n} for some n, and connected algebraic groups that can be realized as closed algebraic subvarieties of a projective space are called abelian because they are related to the integrals studied by Abel.

Coarse Classification: every algebraic group contains a sequence of normal subgroups G ⊃ G^{0} ⊃ G_{1} ⊃ {e} with G/G^{0} a finite group, G^{0}/G_{1} an abelian variety, and G_{1} a linear algebraic group.

The separation axiom. Now that we have the notion of the product of varieties, we can restate the separation axiom in terms of the diagonal.

By way of motivation, consider a topological space V and the diagonal Δ ⊂ V × V, Δ = {(x, x) | x ∈ V }. If Δ is closed (for the product topology), then every pair of points (x, y) ∉ Δ has a neighbourhood U × U′ such that U × U′ ∩ Δ = ∅, i.e., if x ≠ y, then there are neighbourhoods U and U′ of x and y respectively such that U ∩ U′ = ∅. Thus V is Hausdorff. Conversely, if V is Hausdorff, the reverse argument shows that Δ is closed.

For a variety V, we let Δ = Δ_{V} (the diagonal) be the subset {(v, v) | v ∈ V } of V × V.

Proposition 3.23. An algebraic prevariety V is separated if and only if Δ_{V} is closed.

Proof. Assume Δ to be closed, and let φ and ψ be regular maps Z → V. The map

(φ, ψ): Z → V × V, z ↦ (φ(z), ψ(z))

is regular, because its composites with the projections to V are φ and ψ. In particular, it is continuous, and so (φ, ψ)^{−1}(Δ) is closed. But this is precisely the subset on which φ and ψ agree.

Conversely, suppose V is separated. By definition, this means that for any prevariety Z and regular maps φ, ψ: Z → V, the set on which φ and ψ agree is closed in Z. Apply this with φ and ψ the two projection maps V × V → V, and note that the set on which they agree is Δ. □

Corollary 3.24. For any prevariety V, the diagonal is a locally closed subset of V × V.

Proof. Let P ∈ V, and let U be an open affine neighbourhood of P. Then U × U is a neighbourhood of (P, P) in V × V, and Δ_{V} ∩ (U × U) = Δ_{U}, which is closed in U × U because U is separated. □

Thus Δ_{V} can be regarded as a subvariety of V × V. The graph Γ_{φ} of a regular map φ : V → W is defined to be {(v, φ(v)) | v ∈ V }. [Usual picture.]

Corollary 3.25. For any morphism φ: V → W of prevarieties, the graph Γ_{φ} of φ is locally closed in V ×W, and it is closed if W is separated. The map v ↦ (v, φ(v)) is an isomorphism of V onto Γ_{φ}.

Proof. The first statement follows from the preceding corollary because the graph is the inverse image of the diagonal of W × W under the regular map

(v, w) ↦ (φ(v), w): V × W ⟶ W × W.

The second follows from the fact that the regular map Γ_{φ} ↪ V × W ⟶^{p} V is an inverse to v ↦ (v, φ(v)): V → Γ_{φ}. □

Theorem 3.26. The following three conditions on a prevariety are equivalent: (a) V is separated;

(b) for every pair of open affines U and U′ in V, U ∩ U′ is an open affine, and Γ(U ∩ U′, O_{V}) is generated by the functions P ↦ f(P)g(P), f ∈ Γ(U, O_{V}), g ∈ Γ(U′, O_{V}), i.e., the map k[U] ⊗_{k} k[U′] → k[U ∩ U′] is surjective;

(c) the condition in (b) holds for the sets in some open affine covering of V.

Proof. Let U_{i} and U_{j} be open affines in V. We shall prove:

(i) Δ closed ⟹ U_{i} ∩ U_{j} affine. (ii) If U_{i} ∩ U_{j} is affine, then

(U_{i} × U_{j}) ∩ Δ is closed ⟺ the map k[U_{i}] ⊗_{k} k[U_{j}] → k[U_{i} ∩ U_{j}] is surjective.

Since {U_{i} × U_{j}} is an open covering of V × V, Δ is closed in V × V ⟺ Δ ∩ (U_{i} × U_{j}) is closed in U_{i} × U_{j} for each pair (i, j). Thus these statements show that (a) ⟹ (b) and (c) ⟹ (a). Since the implication (b) ⟹ (c) is trivial, it remains to prove (i) and (ii).

Proof of (i): The graph of the inclusion ι: U_{i} ∩ U_{j} ↪ V is Γ_{ι} = (U_{i} × U_{j}) ∩ Δ ⊂ (U_{i} ∩U_{j})×V. If Δ is closed, (U_{i} ×U_{j})∩Δ is a closed subvariety of an affine variety, and hence is affine (see p45). Since U_{i} ∩ U_{j} ≈ Γ_{ι}, it also is affine.

Proof of (ii): Now assume that U_{i} ∩U_{j} is affine. Then (U_{i} ×U_{j})∩Δ_{V} is closed in U_{i}×U_{j} ⟺ v ↦ (v, v): U_{i}∩U_{j} → U_{i}×U_{j} is a closed immersion ⟺ the morphism k[U_{i} × U_{j}] → k[U_{i} ∩ U_{j}] is surjective (see 2.22b). Since k[U_{i} × U_{j}] = k[U_{i}] ⊗_{k} k[U_{j}], this completes the proof of (ii). □

Example 3.27. (a) Let V = ℙ^{1}, and let U_{0} and U_{1} be the standard open subsets (U_{i} = 𝔸^{1}). Then U_{0} ∩ U_{1} = 𝔸^{1} − {0}, and the maps on rings corresponding to the inclusions U_{i} ↪ U_{0} ∩ U_{1} are k[X] → k[X, X^{−1}], X ↦ X, and k[X] → k[X, X^{−1}], X ↦ X^{−1}. Thus the sets U_{0} and U_{1} satisfy the condition in (b).

(b) Let V be 𝔸^{1} with the origin doubled (see 3.9), and let U and U′ be the upper and lower copies of 𝔸^{1} in V. Then U ∩ U′ is affine, but k[U] ⊗ k[U′] → k[U ∩ U′] is not surjective. In fact the map is

k[X] ⊗ k[Y] = k[X, Y] ⟶ k[X, X^{−1}], X ↦ X, Y ↦ X.

(c) Let V be 𝔸^{2} with the origin doubled, and let U and U′ be the upper and lower copies of 𝔸^{2} in V. Then U ∩ U′ is not affine (see 2.21).

Dimension. Let V be an irreducible algebraic variety. Then every open subset of V is dense, and is irreducible. If U ⊃ U′ are open affines in V, then we have

k[U] ⊂ k[U′] ⊂ k(U).

Therefore k(U) is also the field of fractions of k[U′]. This remark shows that we can attach to V a field k(V), called the field of rational functions on V, such that for every open affine U in V, k(V) is the field of fractions of k[U]. The dimension of V is defined to be the transcendence degree of k(V) over k. Note the dim(V) = dim(U) for any open subset U of V. In particular, dim(V) = dim(U) for U an open affine in V. From this it follows that some of the results in §1 carry over—for example, if Z is a proper closed subvariety of V, then dim(Z) < dim(V).

Proposition 3.28. Let V and W be irreducible varieties. Then

dim(V × W) = dim(V) + dim(W).

Proof. We can assume V and W to be affine, and write k[V] = k[x_{1}, …, x_{m}] and k[W] = k[y_{1}, …, y_{n}] where {x_{1}, …, x_{d}} and {y_{1}, …, y_{e}} are maximal algebraically independent sets of elements of k[V] and k[W]. Thus d = dim(V) and e = dim(W). Then^{13}

k[V ×W] = k[V]⊗_{k}k[W] ⊃ k[x_{1}, …, x_{d}]⊗_{k}k[y_{1}, …, y_{e}] ≈ k[x_{1}, …, x_{d}, y_{1}, …, y_{e}].

Therefore {x_{1} ⊗ 1, …, x_{d} ⊗ 1, 1 ⊗ y_{1}, …, 1 ⊗ y_{e}} will be algebraically independent in k[V] ⊗_{k} k[W]. Obviously k[V × W] is generated as a k-algebra by the elements x_{i} ⊗ 1, 1 ⊗ y_{j}, 1 ≤ i ≤ m, 1 ≤ j ≤ n, and all of them are algebraic over

k[x_{1}, …, x_{d}] ⊗_{k} k[y_{1}, …, y_{e}].

Thus the transcendence degree of k(V × W) is d + e. □

Let V be an algebraic variety. Then V is a finite union of Noetherian topological spaces, and so is Noetherian. Consequently (see 1.17), V is a finite union V = ∪V_{i} of its irreducible components, and we define dim(V) = max dim(V_{i}).

13In general, it is not true that if M′ and N′ are R-submodules of M and N, then M′ ⊗_{R} N′ is an R-submodule of M ⊗_{R} N. However, this is true if R is a field, because then M′ and N′ will be direct summands of M and N, and tensor products preserve direct summands.

An algebraic variety as a functor of affine k-algebras. Let A be an affine k-algebra, and let V be an algebraic variety. We define a point of V with coordinates in A to be a regular map Specm(A) → V. For example, if V = V (𝔞) ⊂ k^{n}, then

V (A) = {(a_{1}, …, a_{n}) ∈ A^{n} | f(a_{1}, …, a_{n}) = 0 all f ∈ 𝔞},

which is what you expect. In particular V (k) = V (as a set), i.e., V (as a set) can be identified with the set of points of V with coordinates in k. Note that (V × W)(A) = V (A) × W (A).

Theorem 3.29. A regular map φ: V → W of algebraic varieties defines a family of maps of sets, φ(A): V (A) → W (A), one for each affine k-algebra A, such that for every homomorphism α: A → B of k-algebras,

\begin{matrix}A \\ ↓ & α \\ B \end{matrix} \begin{matrix}V (A) & ⟶^{φ(A)} & W (A) \\ ↓^{V (α)} & ↓^{W (α)} \\ V (B) & ⟶^{φ(B)} & W (B) \end{matrix}  (*)

commutes. Every family of maps with this property arises from a unique morphism of algebraic varieties.

The proof is trivial, once one has made the correct definitions, which we do in the next subsection.

Categories and functors. A category C consists of (a) a class of objects ob(C);

(b) for each pair (A, B) of objects, a set Mor(A, B), whose elements are called morphisms from A to B, and are written α: A → B;

(c) for each triple of objects (A, B, C) a map (called composition)

(α, β) ↦ β ∘ α: Mor(A, B) × Mor(B, C) ⟶ Mor(A, C).

Composition is required to be associative, i.e., (γ ∘ β) ∘ α = γ ∘ (β ∘ α), and for each object A there is required to be an element id_{A} ∈ Mor(A, A) such that id_{A} ∘α = α, β ∘ id_{A} = β, for all (appropriate) α and β.

Example 3.30. (a) There is a category of sets, Sets, whose objects are the sets and whose morphisms are the usual maps of sets.

(b) There is a category Aff_{k} of affine k-algebras, whose objects are the affine k-algebras and whose morphisms are the homomorphisms of k-algebras.

(c) There is a category Var_{k} of algebraic varieties over k, whose objects are the algbraic varieties over k and whose morphisms are the regular maps.

(d) The objects in a category need not be sets with structure, and the morphisms need not be maps.

Exercise 3.31. List twenty more examples of categories.

Let C and D be categories. A covariant functor F from C to D consists of (a) a map A ↦ F (A), sending each object of C to an object of D, and,

(b) for each pair of objects A, B of C, a map

α ↦ F (α): Mor(A, B) ⟶ Mor(F (A), F (B))

such that F (id_{A}) = id_{F (A)} and F (β ∘ α) = F (β) ∘ F (α). A contravariant functor is defined similarly, except that the map on morphisms

is

α ↦ F (α): Mor(A, B) ⟶ Mor(F (B), F (A))

A functor F: C → D is fully faithful if, for all objects A and B of C, the map

Mor(A, B) ⟶ Mor(F (A), F (B))

is a bijection. Then F defines an equivalence of C with the full subcategory of D whose objects are isomorphic to F (A) for some object A of C (see p42). For example, the functor A ↦ Specm A is fully faithful contravariant functor Aff_{k} → Var_{k}, and defines an equivalence of the first category with the subcategory of the second whose objects are the affine algebraic varieties.

Example 3.32. (a) For any object V of a category C, we have a contravariant functor

h_{V} : C ⟶ Sets,

which sends an object A to the set Mor(A, V) and sends a morphism α: A → B to

φ ↦ φ ∘ α: h_{V} (B) → h_{V} (A),

i.e., h_{V} (∗) = Mor(∗, V) and h_{V} (α) = ∗ ∘ α. [Check the conditions!] (b) We have a contravariant functor

V ↦ Γ(V, O_{V}): Var_{k} ⟶ Aff_{k}.

Let F and G be two functors C → D. A morphism α: F → G is a collection of morphisms α(A): F (A) → G(A), one for each object A of C, such that, for every morphism u: A → B in C, the following diagram commutes:

\begin{matrix}A \\ ↓ & u \\ B \end{matrix} \begin{matrix}F (A) & ⟶^{α(A)} & G(A) \\ ↓^{F (u)} & ↓^{G(u)} \\ F (B) & ⟶^{α(B)} & G(B). \end{matrix}  (**)

Example 3.33. Let α: V → W be a morphism in C. The collection of maps

h_{α}(A): h_{V} (A) ⟶ h_{W} (A), φ ↦ α ∘ φ

is a morphism of functors. [Check!]

With this notion of morphism, the functors C ⟶ D form a category Fun(C, D) (except possibly for some foundational set-theoretic difficulties).

Proposition 3.34 (Yoneda Lemma). The functor

V ↦ h_{V} : C ⟶ Fun(C, Sets)

is fully faithful.

Proof. Let A, B be objects of C. We construct an inverse to

α ↦ h_{α}: Mor(A, B) ⟶ Mor(h_{A}, h_{B}).

For a morphism of functors γ: h_{A} → h_{B}, define β(γ) = γ(id_{A})—it is morphism A → B. Then

β(h_{α}) =_{def} h_{α}(id_{A}) =_{def} α ∘ id_{A} = α,

and

h_{β(γ)}(α) =_{def} β(γ) ∘ α =_{def} γ(id_{A}) ∘ α = γ(α)

because of the commutativity of (**):

\begin{matrix}A \\ ↓ & α \\ B \end{matrix} \begin{matrix}h_{A}(A) & ⟶^{γ} & h_{B}(A) \\ ↓^{∗∘α} & & ↓^{∗∘α} \\ h_{A}(B) & ⟶^{γ} & h_{B}(B). \end{matrix}

Thus α → h_{α} and γ ↦ β(γ) are inverse maps. □

Algebraic varieties as functors (continued). The Yoneda lemma shows that the functor V ↦ h_{V} embeds the category of affine algebraic varieties as a full subcategory of the category of covariant functors Aff_{k} → Sets, and it is not difficult to deduce that it embeds the category of all algebraic varieties in to the category of such functors (use 3.12 for example). This proves (3.29).

It is not unusual for a variety to be most naturally defined in terms of its points functor. For example, for any affine k-algebra, let SL_{n}(A) be the group of n × n matrices with coefficients in A having determinant 1. A homomorphism A → B induces a homomorphism SL_{n}(A) → SL_{n}(B), and so SL_{n}(A) is a functor. In fact, it is the points functor of the affine variety:

Specm k[X_{11}, …, X_{nn}]/(det(X_{ij}) − 1).

Matrix multiplication defines a morphism of functors

SL_{n} × SL_{n} ⟶ SL_{n}

which, because of (3.29), arises from a morphism of algebraic varieties. In fact, SL_{n} is an algebraic group.

Instead of defining varieties to be ringed spaces, it is possible to define them to be functors Aff_{k} → Sets satisfying certain conditions.

Dominating maps. A regular map α: V → W is said to be dominating if the image of α is dense in W. Suppose V and W are irreducible. If V′ and W′ are open affine subsets of V and W such that φ(V′) ⊂ W′, then (2.22) implies that the map f ↦ f ∘ φ: k[W′] → k[V′] is injective. Therefore it extends to a map on the fields of fractions, k(W) → k(V), and this map is independent of the choice of V′ and W′.

4. Local Study: Tangent Planes, Tangent Cones, Singularities

In this section, we examine the structure of a variety near a point. I begin with the case of a curve, since the ideas in the general case are the same, but the formulas are more complicated.

Tangent spaces to plane curves. We first consider the case of a curve

V : F (X, Y) = 0

in the plane 𝔸^{2} defined by a nonconstant polynomial F (X, Y). We assume that F (X, Y) has no multiple factors, so that (F (X, Y)) is a radical ideal and I(V) = (F (X, Y)). We can factor F into a product of irreducible polynomials, F (X, Y) = ∏ F_{i}(X, Y), and then V = ∪V (F_{i}) expresses V as a union of its irreducible components. Each component V (F_{i}) has dimension 1 (see 1.21) and so V has pure dimension 1. More explicitly, suppose for simplicity that F (X, Y) itself is irreducible, so that k[V] = k[X, Y]/(F (X, Y)) = k[x, y] is an integral domain. If F ≠ X − c, then x is transcendental over k and y is algebraic over k(x), and so x is a transcendence basis for k(V) over k. Similarly, if F ≠ Y − c, then y is a transcendence basis for k(V) over k.

Let (a, b) be a point on V. In Math 215, the equation of the tangent at P = (a, b) is defined to be

\frac{∂F}{∂X} (a, b)(X − a) + \frac{∂F}{∂Y} (a, b)(Y − b) = 0.  (4.0.0)

This is the equation of a line unless both \frac{∂F}{∂X} (a, b) and \frac{∂F}{∂Y} (a, b) are zero, in which case it is the equation of a plane.

Definition 4.1. The tangent space T_{P} V to V at P = (a, b) is the space defined by equation (4.0.0).

When \frac{∂F}{∂X} (a, b) and \frac{∂F}{∂Y} (a, b) are not both zero, T_{P} (V) is a line, and we say that P is a nonsingular or smooth point of V. Otherwise, T_{P} (V) has dimension 2, and we say that P is singular or multiple. The curve V is said to be nonsingular or smooth when all its points are nonsingular.

We regard T_{P} (V) as a subspace of the two-dimensional vector space T_{P} (𝔸^{2}), which is the two-dimensional space of vectors with origin P.

Example 4.2. (a) X^{m} +Y^{m} = 1. All points are nonsingular unless the characteristic divides m (in which case X^{m} + Y^{m} − 1 has multiple factors).

(b) Y^{2} = X^{3}. Here only (0,0) is singular. [Picture]

(c) Y^{2} = X^{2}(X + 1). Here again only (0,0) is singular. [Picture.]

(d) Y^{2} = X^{3} + aX + b. In this case, V is singular ⟺ Y^{2} − X^{3} − aX − b, 2Y, and 3X^{2} + a have a common zero ⟺ X^{3} + aX + b and 3X^{2} + a have a common zero. Since the latter is the derivative of the former, we see that V is singular if and only if X^{3} + aX + b has a multiple root.

(e) (X^{2} + Y^{2})^{2} + 3X^{2}Y − Y^{3} = 0. The origin is (very) singular. [Picture.]

(f) (X^{2} + Y^{2})^{3} − 4X^{2}Y^{2} = 0. The origin is (even more) singular. [Picture.]

(g) V = V (F G) where F G has no multiple factors and F and G are relatively prime. Then V = V (F) ∪ V (G), and a point (a, b) is singular if and only if it is a singular point of V (F), a singular point of V (G), or a point of V (F) ∩ V (G). This follows immediately from the equations given by the product rule:

\frac{∂(F G)}{∂X} = F · \frac{∂G}{∂X} + \frac{∂F}{∂X} · G, \frac{∂(F G)}{∂Y} = F · \frac{∂G}{∂Y} + \frac{∂F}{∂Y} · G.

Proposition 4.3. Let V be the curve defined by a nonconstant polynomial F without multiple factors. The set of nonsingular points is an open dense subset V.^{14}

Proof. We can assume that F is irreducible. We have to show that the set of singular points (the singular locus) is a proper closed subset. Since it is defined by the equations

F = 0, \frac{∂F}{∂X} = 0, \frac{∂F}{∂Y} = 0,

it is obviously closed. It will be proper unless ∂F/∂X and ∂F/∂Y are identically zero on V, and are therefore both multiples of F, but, since they have lower degree, this is impossible unless they are both zero. Clearly ∂F/∂X = 0 if and only if F is a polynomial in Y (k of characteristic zero) or is a polynomial in X^{p} and Y (k of characteristic p). A similar remark applies to ∂F/∂Y. Thus if ∂F/∂X and ∂F/∂Y are both zero, then F is constant (characteristic zero) or a polynomial in X^{p}, Y^{p}, and hence a p^{th} power (characteristic p). These are contrary to our assumptions. □

Tangent cones to plane curves. Note that if P = (0, 0), then the equation defining the tangent space is the linear term of F : since (0,0) is on V,

F = aX + bY + terms of higher degree,

and the equation of the tangent space is F_{ℓ}(X, Y) =_{df} aX + bY = 0.

In general a polynomial F (X, Y) can be written (uniquely) as a finite sum

F = F_{0} + F_{1} + F_{2} + … + F_{m} + …

where F_{m} is a homogeneous polynomial of degree m (i.e., a sum of terms aX^{i}Y^{j} with i + j = m). The first nonzero term on the right (the homogeneous summand of F of least degree) will be written F_{∗} and called the leading form of F.

14The word “singular” means uncommon or extraordinary as in, for example, he spoke with singular shrewdness. Thus the proposition says that singular points (mathematical sense) are singular (usual sense).

Definition 4.4. Suppose (0, 0) ∈ V (F), F a polynomial without square factors. The geometric tangent cone at (0,0) is the zero set of F_{∗}. The tangent cone is the pair (V (F_{∗}), F_{∗}). To obtain the tangent cone at any other point, translate to the origin, and then translate back.

Example 4.5. (a) Y^{2} = X^{3} : the geometric tangent cone at (0,0) is given by Y^{2} = 0.

(b) Y^{2} = X^{2}(X + 1) : the geometric tangent cone at (0,0) is given by Y^{2} = X^{2}, i.e., Y = ±X.

(c) (X^{2} + Y^{2})^{2} + 3X^{2}Y − Y^{3} = 0 : the geometric tangent cone at (0,0) is given by 3X^{2}Y − Y^{3} = 0, i.e, it is the union of the lines Y = 0, Y = ±\sqrt{3X}.

(d) (X^{2} + Y^{2})^{3} − 4X^{2}Y^{2} = 0 : the geometric tangent cone at (0,0) is given by 4X^{2}Y^{2} = 0, i.e., it is the union of the x and y axes (each doubled).

In general we can factor F_{∗} as

F_{∗}(X, Y) = ∏ X^{r_{0}}(Y − a_{i}X)^{r_{i}}.

Then deg F_{∗} = ∑ r_{i} is called the multiplicity of the singularity, mult_{P} (V). A multiple point is ordinary if its tangents are nonmultiple, i.e., r_{i} = 1 all i. An ordinary double point is called a node, and a nonordinary double point is called a cusp. (There are many names for special types of singularities—see any book, especially an old book, on curves.)

The local ring at a point on a curve.

Proposition 4.6. Let P be a point on a curve V, and let 𝔪 be the corresponding maximal ideal in k[V]. If P is nonsingular, then dim_{k} 𝔪/𝔪^{2} = 1, and otherwise dim_{k} 𝔪/𝔪^{2} = 2.

Proof. Assume first that P = (0, 0). Then 𝔪 = (x, y) in k[V] = k[X, Y]/(F (X, Y)) = k[x, y]. Note that 𝔪^{2} = (x^{2}, xy, y^{2}), and

𝔪/𝔪^{2} = (X, Y)/𝔪^{2} + (F (X, Y)) = (X, Y)/(X^{2}, XY, Y^{2}, F (X, Y)).

In this quotient, every element is represented by a linear polynomial cx+dy, and the only relation is F_{ℓ}(x, y) = 0. Clearly dim 𝔪/𝔪^{2} = 1 if F_{ℓ} ≠ 0, and dim 𝔪/𝔪^{2} = 2 otherwise.

The same argument works in the general case, except that you have work with the variables X′ = X − a and Y′ = Y − b, i.e., translate the point to the origin. □

We explain what the condition dim_{k}(𝔪/𝔪^{2}) = 1 means for the local ring O_{P} = k[V]_{𝔪}—see later for the more details. Let 𝔫 be the maximal ideal 𝔪k[V]_{𝔪} of this local ring. Then the map 𝔪 ↦ 𝔫 induces an isomorphism 𝔪/𝔪^{2} → 𝔫/𝔫^{2}, and so we have

P nonsingular ⟺ dim_{k} 𝔪/𝔪^{2} = 1 ⟺ dim_{k} 𝔫/𝔫^{2} = 1.

Nakayama’s lemma shows that the last condition is equivalent to 𝔫 being a principal ideal. Since O_{P} is of dimension 1, 𝔫 being principal means O_{P} is a regular local ring of dimension 1, and hence a discrete valuation ring, i.e., a principal ideal domain with exactly one prime element (up to associates). Thus

P nonsingular ⟺ O_{P} regular ⟺ O_{P} is a discrete valuation ring.

Tangent spaces of subvarieties of 𝔸^{m}. Before defining tangent spaces at points of closed subvarietes of 𝔸^{m} we review some terminology from linear algebra.

Linear algebra. For a vector space k^{m}, let X_{i} be the i^{th} coordinate function a ↦ a_{i}. Thus X_{1}, …, X_{m} is the dual basis to the standard basis for k^{m}. A linear form ∑ a_{i}X_{i} can be regarded as an element of the dual vector space (k^{m})^{∨} = Hom(k^{m}, k).

Let A = (a_{ij}) be an n × m matrix. It defines a linear map α: k^{m} → k^{n}, by

( a_{1} ; ⋮ ; a_{m} ) ↦ A ( a_{1} ; ⋮ ; a_{m} ) .

Thus, if α(a) = b, then

b_{i} = ∑_{j=1}^{m} a_{ij}a_{j}.

Write X_{1}, …, X_{m} for the coordinate functions on k^{m} and Y_{1}, …, Y_{n} for the coordinate functions on k^{n}. Then the last equation can be rewritten as:

Y_{i} ∘ α = ∑_{j=1}^{m} a_{ij}X_{j}.

This says that, when we apply α to a, then the i^{th} coordinate of the result is ∑_{j=1}^{m} a_{ij}(X_{j}a) = ∑_{j=1}^{m} a_{ij}a_{j}.

Tangent spaces. Consider an affine variety V ⊂ k^{m}, and let 𝔞 = I(V). The tangent space T_{a}(V) to V at a = (a_{1}, …, a_{m}) is the subspace of the vector space with origin a cut out by the linear equations

∑_{i=1}^{m} \frac{∂F}{∂X_{i}} |_{a} (X_{i} − a_{i}) = 0,  F ∈ 𝔞.  (*)

Thus T_{a}(𝔸^{m}) is the vector space of dimension m with origin a, and T_{a}(V) is the subspace of T_{a}(𝔸^{m}) defined by the equations (*).

Write (dX_{i})_{a} for (X_{i} − a_{i}); then the (dX_{i})_{a} form a basis for the dual vector space T_{a}(𝔸^{m})^{∨} to T_{a}(𝔸^{m})—in fact, they are the coordinate functions on T_{a}(𝔸^{m}). As in advanced calculus, for a function F ∈ k[X_{1}, …, X_{m}], we define the differential of F at a by the equation:

(dF)_{a} = ∑ \frac{∂F}{∂X_{i}} |_{a} (dX_{i})_{a}.

It is again a linear form on T_{a}(𝔸^{m}). In terms of differentials, T_{a}(V) is the subspace of T_{a}(𝔸^{m}) defined by the equations:

(dF)_{a} = 0, F ∈ 𝔞.  (**)

In (*) and (**) it suffices to take the F in a generating subset for 𝔞. The product rule for differentiation shows that if G = ∑_{j} H_{j}F_{j}, then

(dG)_{a} = ∑_{j} H_{j}(a) · (dF_{j})_{a} + F_{j}(a) · (dG_{j})_{a}.

If F_{1}, …, F_{r} generate 𝔞 and a ∈ V (𝔞), so that F_{j}(a) = 0, this equation becomes

(dG)_{a} = ∑_{j} H_{j}(a) · (dF_{j})_{a}.

Thus (dG)_{a}(t) = 0 if (dF_{j})_{a}(t) = 0 for all j.

When V is irreducible, a point a on V is said to be nonsingular ( or smooth) if the dimension of the tangent space at a is equal to the dimension of V; otherwise it is singular ( or multiple). When V is reducible, we say a is nonsingular if dim T_{a}(V) is equal to the maximum dimension of an irreducible component of V passing through a. It turns out then that a is singular precisely when it lies on more than one irreducible component, or when it lies on only one but is a singular point of that component.

Let 𝔞 = (F_{1}, …, F_{r}), and let

J = Jac(F_{1}, …, F_{r}) = (\frac{∂F_{i}}{∂X_{j}} ) = ( \begin{matrix}\frac{∂F_{1}}{∂X_{1}} , & …, & \frac{∂F_{1}}{∂X_{m}} \\ ⋮ & & ⋮ \\ \frac{∂F_{r}}{∂X_{1}} , & …, & \frac{∂F_{r}}{∂X_{m}} \end{matrix} ) .

Then the equations defining T_{a}(V) as a subspace of T_{a}(𝔸^{m}) have matrix J(a). Therefore (see Math 513),

dim_{k} T_{a}(V) = m − rankJ(a),

and so a is nonsingular if and only if the rank of Jac(F_{1}, …, F_{r})(a) is equal to m − dim(V). For example, if V is a hypersurface, say I(V) = (F (X_{1}, …, X_{m})), then

Jac(F)(a) = (\frac{∂F}{∂X_{1}} (a), …, \frac{∂F}{∂X_{m}} (a)) ,

and a is nonsingular if and only if not all of the partial derivatives \frac{∂F}{∂X_{i}} vanish at a. We can regard J as a matrix of regular functions on V. For each r,

{a ∈ B | rankJ(a) ≤ r}

is closed in V, because it the set where certain determinants vanish (cf. the solution to Homework §2, #1). Therefore, there is an open subset U of V on which rankJ(a) attains its maximum value, and the rank jumps on closed subsets. Later we shall show that the maximum value of rankJ(a) is m − dim V, and so the nonsingular points of V form a nonempty open subset of V.

The differential of a map. Consider a regular map

α: 𝔸^{m} → 𝔸^{n}, a ↦ (P_{1}(a_{1}, …, a_{m}), …, P_{n}(a_{1}, …, a_{m})).

We think of α as being given by the equations

Y_{i} = P_{i}(X_{1}, …, X_{m}), i = 1, … n.

It corresponds to the map of rings α^{∗}: k[Y_{1}, …, Y_{n}] → k[X_{1}, …, X_{m}] sending Y_{i} to P_{i}(X_{1}, …, X_{m}), i = 1, … n.

Define (dα)_{a} : T_{a}(𝔸^{m}) → T_{b}(𝔸^{n}) to be the map such that

(dY_{i})_{b} ∘ (dα)_{a} = ∑ \frac{∂P_{i}}{∂X_{j}} |_{a} (dX_{j})_{a},

i.e., relative to the standard bases, (dα)_{a} is the map with matrix

Jac(P_{1}, …, P_{n})(a) = ( \begin{matrix}\frac{∂P_{1}}{∂X_{1}} (a), & …, & \frac{∂P_{1}}{∂X_{m}} (a) \\ ⋮ & & ⋮ \\ \frac{∂P_{n}}{∂X_{1}} (a), & …, & \frac{∂P_{n}}{∂X_{m}} (a) \end{matrix} ) .

For example, suppose a = (0, …, 0) and b = (0, …, 0), so that T_{a}(𝔸^{m}) = k^{m} and T_{b}(𝔸^{n}) = k^{n}, and

P_{i} = ∑_{j=1}^{m} c_{ij}X_{j} + (higher terms), i = 1, …, n.

Then Y_{i} ∘ (dα)_{a} = ∑_{j} c_{ij}X_{j}, and the map on tangent spaces is given by the matrix (c_{ij}), i.e., it is simply t ↦ (c_{ij})t.

Let F ∈ k[X_{1}, …, X_{m}]. We can regard F as a regular map 𝔸^{m} → 𝔸^{1}, whose differential will be a linear map

(dF)_{a} : T_{a}(𝔸^{m}) → T_{b}(𝔸^{1}), b = F (a).

When we identify T_{b}(𝔸^{1}) with k, we obtain an identification of the differential of F (F regarded as a regular map) with the differential of F (F regarded as a regular function).

Lemma 4.7. Let α: 𝔸^{m} → 𝔸^{n} be as at the start of this subsection. If α maps V = V (𝔞) ⊂ k^{m} into W = V (𝔟) ⊂ k^{n}, then (dα)_{a} maps T_{a}(V) into T_{b}(W), b = α(a).

Proof. We are given that

f ∈ 𝔟 ⟹ f ∘ α ∈ 𝔞,

and have to prove that

f ∈ 𝔟 ⟹ (df)_{b} ∘ (dα)_{a} is zero on T_{a}(V).

The chain rule holds in our situation:

\frac{∂f}{∂X_{i}} = ∑_{i=1}^{n} \frac{∂f}{∂Y_{j}} \frac{∂Y_{j}}{∂X_{i}} , Y_{j} = P_{j}(X_{1}, …, X_{m}), f = f(Y_{1}, …, Y_{n}).

If α is the map given by the equations

Y_{j} = P_{j}(X_{1}, …, X_{m}), j = 1, …, m,

then the chain rule implies

d(f ∘ α)_{a} = (df)_{b} ∘ (dα)_{a}, b = α(a).

Let t ∈ T_{a}(V); then

(df)_{b} ∘ (dα)_{a}(t) = d(f ∘ α)_{a}(t),

which is zero if f ∈ 𝔟 because then f ∘ α ∈ 𝔞. Thus (dα)_{a}(t) ∈ T_{b}(W). □

We therefore get a map (dα)_{a} : T_{a}(V) → T_{b}(W). The usual rules from advanced calculus (and differential geometry) hold. For example,

(dβ)_{b} ∘ (dα)_{a} = d(β ∘ α)_{a}, b = α(a).

Example 4.8. Let V be the union of the coordinate axes in 𝔸^{3}, and let W be V (XY (X −Y)) ⊂ 𝔸^{2}. Then V is not isomorphic to W because T_{o}(V) has dimension 3, but T_{o}(W) has dimension 2. (Note that V = V (XY, Y Z, XZ), from which it is clear that the origin o is the only singular point on V, and that the tangent space there has dimension 3. An isomorphism V → W would have to send the singular point to the singular point, i.e., o ↦ o, and map T_{o}(V) isomorphically onto T_{0}(W).)

Etale maps. Let V and W be smooth varieties. A regular map α: V → W is étale at a if (dα)_{a} : T_{a}(V) → T_{b}(W) is an isomorphism; α is étale if it is étale at all points of V.

Example 4.9. (a) A regular map α = (P_{1}, …, P_{n}): 𝔸^{n} → 𝔸^{n} is étale at a if and only if rank Jac(P_{1}, …, P_{n})(a) = n, i.e, det (\frac{∂P_{i}}{∂X_{j}} (a)) ≠ 0 (because the map on the tangent spaces has matrix Jac(P_{1}, …, P_{n})(a)).

(b) Consider a map,

W = Specm(A[X]/(f(X)) → Specm(A) = V.

Then P (X) = ∑ c_{i}X^{i} where each c_{i} is a function on V, and the points of W lying over a point a ∈ V correspond to the roots of ∑ c_{i}(a)X^{i}. The map is étale at the point (a, b) if and only if b is a simple root of ∑ c_{i}(a)X^{i}.

Proof: Write A = Specm k[X_{1}, …, X_{n}]/𝔞, 𝔞 = (f_{1}, …, f_{r}), so that A[X]/(f(X)) = k[X_{1}, …, X_{n}]/(f_{1}, …, f_{r}, f). The tangent spaces to W and V at (a, b) and a respectively are the null spaces of the matrices

( \begin{matrix}\frac{∂f_{1}}{∂X_{1}} (a) & … & \frac{∂f_{1}}{∂X_{m}} (a) & 0 \\ ⋮ & & ⋮ \\ \frac{∂f_{n}}{∂X_{1}} (a) & … & \frac{∂f_{n}}{∂X_{m}} (a) & 0 \\ \frac{∂f}{∂X_{1}} (a) & … & \frac{∂f}{∂X_{m}} (a) & \frac{∂f}{∂X} (a, b) \end{matrix} )  ( \begin{matrix}\frac{∂f_{1}}{∂X_{1}} (a) & … & \frac{∂f_{1}}{∂X_{m}} (a) \\ ⋮ & & ⋮ \\ \frac{∂f_{n}}{∂X_{1}} (a) & … & \frac{∂f_{n}}{∂X_{m}} (a) \end{matrix} )

and the map T_{(a,b)}(W) → T_{a}(V) is induced by the projection map k^{n+1} → k^{n} omitting the last coordinate. This map is an isomorphism if and only if \frac{∂f}{∂X} (a, b)≠ 0, because then any solution to the smaller set of equations extends uniquely to a solution of the larger set. But \frac{∂f}{∂X} (a, b) = ^{\frac{d(∑_{i} c_{i}(a)X^{i})}{dX}}(b), which is zero if and only if b is a multiple root of ∑_{i} c_{i}(a)X^{i}.

(c) Consider a dominating map α: W → V of smooth affine varieties, corresponding to a map A → B of rings. Suppose B can be written B = A[Y_{1}, …, Y_{n}]/(P_{1}, …, P_{n}). A similar argument to the above shows that α is étale if and only if det (\frac{∂P_{i}}{∂X_{j}} (a)) ≠ 0.

(d) The example in (b) is typical; in fact every étale map is locally of this form, provided V is normal (in the sense defined below). More precisely, let α: W → V be étale at P ∈ V, and assume V to normal; then there exist a map α′: W′ → V′ with k[W′] = k[V′][X]/(f(X)), and a commutative diagram

W ⊃ U_{1} ⟶^{≈} U′_{1} ⊂ W′
↓ ↓ ↓ ↓
V ⊃ U_{2} ⟶^{≈} U′_{2} ⊂ V′

with the U’s all open subvarieties and P ∈ U_{1}.

Warning! In advanced calculus (or differential geometry, or the theory of complex manifolds), the inverse function theorem says that a map α that is étale at a point a is a local isomorphism there, i.e., there exists open neighbourhoods U and U′ of a and α(a) such that α induces an isomorphism U → U′. This is not true in algebraic geometry, at least not for the Zariski topology: a map can be étale at a point without being a local isomorphism. Consider for example the map

α: 𝔸^{1} − {0} → 𝔸^{1} − {0}, a ↦ a^{2}.

This is étale if the characteristic is ≠ 2, because the Jacobian matrix is (2X), which has rank one for all X ≠ 0 (alternatively, it is of the form (4.9b) with f(X) = X^{2} − T, where T is the coordinate function on 𝔸^{1}, and X^{2} − c has distinct roots for c ≠ 0). Nevertheless, I claim that there do not exist nonempty open subsets U and U′ of 𝔸^{1} − {0} such that α defines an isomorphism U → U′. If there did, then α would define an isomorphism k[U′] → k[U] and hence an isomorphism on the fields of fractions k(𝔸^{1}) → k(𝔸^{1}). But on the fields of fractions, α defines the map k(X) → k(X), X ↦ X^{2}, which is not an isomorphism.

Aside 4.10. There is a conjecture that any étale map α: 𝔸^{n} → 𝔸^{n} is an isomorphism. If we write α = (P_{1}, …, P_{n}), then this becomes the statement

det (\frac{∂P_{i}}{∂X_{j}} (a)) ≠ 0 all a ⟹ α has a inverse.

The condition, det (\frac{∂P_{i}}{∂X_{j}} (a)) ≠ 0 all a, implies that det (\frac{∂P_{i}}{∂X_{j}} ) is a nonzero constant. This conjecture, which is known as the Jacobian problem, has not been solved in general as far as I know. It has caused many mathematicians a good deal of grief. It is probably harder than it is interesting. See Bass et al, Bull. AMS 7 (1982), 287-330.

Intrinsic definition of the tangent space. The definition we have given of the tangent space at a point requires the variety to be embedded in affine space. In this subsection, we give a more intrinsic definition.

By a linear form in X_{1}, …, X_{n} we mean an expression ∑ c_{i}X_{i}, c_{i} ∈ k. The linear forms form a vector space of dimension n which is naturally dual to k^{n}.

Lemma 4.11. Let 𝔠 be an ideal in k[X_{1}, …, X_{n}] generated by linear forms, ℓ_{1}, …, ℓ_{r}, which we may assume to be linearly independent. Let X_{i_{1}}, …, X_{i_{n−r}} be such that {ℓ_{1}, …, ℓ_{r}, X_{i_{1}}, …, X_{i_{n−r}}} is a basis for the linear forms in X_{1}, …, X_{n}. Then k[X_{1}, …, X_{n}]/𝔠 = k[X_{i_{1}}, …, X_{i_{n−r}}].

Proof. This is obvious if the linear forms ℓ_{1}, …, ℓ_{r} are X_{1}, …, X_{r}. Because {X_{1}, …, X_{n}} and {ℓ_{1}, …, ℓ_{r}, X_{i_{1}}, …, X_{i_{n−r}}} are both bases for the linear forms, each element of one set can be expressed as a linear combination of the elements of the second set. Therefore

k[X_{1}, …, X_{n}] = k[ℓ_{1}, …, ℓ_{r}, X_{i_{1}}, …, X_{i_{n−r}}]

and so

k[X_{1}, …, X_{n}]/𝔠 = k[ℓ_{1}, …, ℓ_{r}, X_{i_{1}}, …, X_{i_{n−r}}]/(ℓ_{1}, …, ℓ_{r}) = k[X_{i_{1}}, …, X_{i_{n−r}}].

□

Let V = V (𝔞) ⊂ k^{n}, and assume the origin P ∈ V. Let 𝔞_{ℓ} be the ideal generated by the linear terms f_{ℓ} of the f ∈ 𝔞. By definition, T_{P} (V) = V (𝔞_{ℓ}). Let A_{ℓ} = k[X_{1}, …, X_{n}]/𝔞_{ℓ}, and let 𝔪 be the maximal ideal in k[V] corresponding to the origin; thus 𝔪 = (x_{1}, …, x_{n}).

Proposition 4.12. There are canonical isomorphisms

Hom_{k-linear}(𝔪/𝔪^{2}, k) ⟶^{≈} Hom_{k-alg}(A_{ℓ}, k) ⟶^{≈} T_{P} (V).

Proof. First isomorphism. Let 𝔫 = (X_{1}, …, X_{n}) be the maximal ideal at the origin in k[X_{1}, …, X_{n}]. We have 𝔪/𝔪^{2} = 𝔫/(𝔫^{2} + 𝔞), and as f − f_{ℓ} ∈ 𝔫^{2} for any f ∈ 𝔞, we have 𝔪/𝔪^{2} = 𝔫/(𝔫^{2} + 𝔞_{ℓ}). Let f_{1,ℓ}, …, f_{r,ℓ} be a basis for the vector space 𝔞_{ℓ}; there are n − r indeterminates X_{i_{1}} …, X_{i_{n−r}} forming with the f_{i,ℓ} a basis for the linear forms on k^{n}. Then X_{i_{1}}, …, X_{i_{n}} form a basis for 𝔪/𝔪^{2} as a k-vector space, and the lemma shows that A_{ℓ} = k[X_{i_{1}} …, X_{i_{n−r}}]. Any homomorphism α: A_{ℓ} → k of k-algebras is determined by its values α(X_{i_{1}}), …, α(X_{i_{n−r}}), and they can be arbitrarily given. Since the k-linear maps 𝔪/𝔪^{2} → k have a similar description, the first isomorphism is now obvious.

Second isomorphism. To give a k-algebra homomorphism A_{ℓ} → k is the same as to give an element (a_{1}, …, a_{n}) ∈ k^{n} such that f(a_{1}, …, a_{n}) = 0 for all f ∈ A_{ℓ}, which is the same as to give an element of T_{P} (V). □

Lemma 4.13. Let 𝔪 be a maximal ideal of a ring A, and let 𝔫 = 𝔪A_{𝔪}. For all n, the map

a + 𝔪^{n}: A/𝔪^{n} → A_{𝔪}/𝔫^{n}

is an isomorphism. Moreover, it induces isomorphisms

𝔪^{r}/𝔪^{n} ⟶ 𝔫^{r}/𝔫^{n}

for all r < n.

Proof. The second statement follows from the first, because of the exact commutative diagram:

0 ⟶ 𝔪^{r}/𝔪^{n} ⟶ A/𝔪^{n} ⟶ A/𝔪^{r} ⟶ 0
↓  ↓^{≈}  ↓^{≈}
0 ⟶ 𝔫^{r}/𝔫^{n} ⟶ A_{𝔪}/𝔫^{n} ⟶ A_{𝔪}/𝔫^{r} ⟶ 0.

To simplify the exposition, in proving that the first map is an isomorphism, I’ll assume A ⊂ S^{−1}A. In order to show that the map A/𝔪^{n} → A_{𝔫}/𝔫^{n} is injective, we have to show that 𝔫^{m} ∩ A = 𝔪^{m}. But 𝔫^{m} = S^{−1}𝔪^{m}, S = A − 𝔪, and so we have to show that 𝔪^{m} = (S^{−1}𝔪^{m}) ∩ A. An element of (S^{−1}𝔪^{m}) ∩ A can be written a = b/s with b ∈ 𝔪^{m}, s ∈ S, and a ∈ A. Then sa ∈ 𝔪^{m}, and so sa = 0 in A/𝔪^{m}. The only maximal ideal containing 𝔪^{m} is 𝔪 (because 𝔪′ ⊃ 𝔪^{m} ⟹ 𝔪′ ⊃ 𝔪), and so the only maximal ideal in A/𝔪^{m} is 𝔪/𝔪^{m}; in particular, A/𝔪^{m} is a local ring. As s is not in 𝔪/𝔪^{m}, it is a unit in A/𝔪^{m}, and so sa = 0 in A/𝔪^{m} implies a = 0 in A/𝔪^{m}, i.e., a ∈ 𝔪^{m}.

We now prove that the map is surjective. Let \frac{a}{s} ∈ A_{𝔪}. Because s ∉ 𝔪 and 𝔪 is maximal, we have that (s) + 𝔪 = A, i.e., (s) and 𝔪 are relatively prime. Therefore (s) and 𝔪^{m} are relatively prime (no maximal ideal contains both of them), and so there exist b ∈ A and q ∈ 𝔪^{m} such that bs + q = 1. Then b maps to s^{−1} in A_{𝔪}/𝔫^{m} and so ba maps to \frac{a}{s}. More precisely: because s is invertible in A_{𝔪}/𝔫^{m}, \frac{a}{s} is the unique element of this ring such that s\frac{a}{s} = a; since s(ba) = a(1 − q), the image of ba in A_{𝔪} also has this property and therefore equals \frac{a}{s}. □

Therefore, we also have a canonical isomorphism

T_{P} (V) ⟶^{≈} Hom_{k-lin}(𝔫_{P} /𝔫_{P}^{2} , k),

where 𝔫_{P} is now the maximal ideal in O_{P} (= A_{𝔪}).

Definition 4.14. The tangent space T_{P} (V) at a point P of a variety V is Hom_{k-lin}(𝔫_{P} /𝔫_{P}^{2} , k), where 𝔫_{P} the maximal ideal in O_{P}.

When V is embedded in affine space, the above remarks show that this definition agrees with the more explicit definition on p68. The advantage of this definition is that it depends only on a (small) neighourhood of P; in particular, it doesn’t depend on an affine embedding of V.

A regular map α: V → W sending P to Q defines a local homomorphism O_{Q} → O_{P}, which induces maps 𝔪_{Q} → 𝔪_{P}, 𝔪_{Q}/𝔪_{Q}^{2} → 𝔪_{P} /𝔪_{P}^{2} , and T_{P} (V) → T_{Q}(W). The last map is written (dα)_{P}. When some open neighbourhoods of P and Q are realized as closed subvarieties of affine space, then (dα)_{P} becomes identified with the map defined earlier.

In particular, if f ∈ 𝔪_{P}, then f is represented by a regular map U → 𝔸^{1}, P ↦ 0, and hence defines a linear map (df)_{P} : T_{P} (V) → k. This is just the map sending a tangent vector (element of Hom_{k-lin}(𝔪_{P} /𝔪_{P}^{2} , k)) to its value at f mod 𝔪_{P}^{2} . Again, in the concrete situation V ⊂ 𝔸^{m} this agrees with the previous definition. In general, for f ∈ O_{P}, i.e., for f a germ of a function at P, we define

(df)_{P} = f − f(P) mod 𝔪^{2}.

The tangent space at P and the space of differentials at P are dual vector spaces—in contrast to the situation in advanced calculus, for us it is easier to define first the space of differentials, and then define the tangent space to be its dual.

Consider for example, a ∈ V (𝔞) ⊂ 𝔸^{n}, with 𝔞 a radical ideal. For f ∈ k[𝔸^{n}] = k[X_{1}, …, X_{n}], we have (trivial Taylor expansion)

f = f(P) + ∑ c_{i}(X_{i} − a_{i}) + terms of degree ≥ 2 in the X_{i} − a_{i},

that is,

f − f(P) ≡ ∑ c_{i}(X_{i} − a_{i}) mod 𝔪_{P}^{2} .

Therefore (df)_{P} can be identified with

∑ c_{i}(X_{i} − a_{i}) = ∑ \frac{∂f}{∂X_{i}} |_{a} (X_{i} − a_{i}),

which is how we originally defined the differential.^{15} The tangent space T_{a}(V (𝔞)) is the zero set of the equations

(df)_{P} = 0, f ∈ 𝔞,

and set {(df)_{P} |_{T_{a}(V)} | f ∈ k[X_{1}, …, X_{n}]} is the dual space to T_{a}(V).

The dimension of the tangent space. In this section we show that the dimension of the tangent space is at least that of the variety. First we review some commutative algebra.

Some commutative algebra. Let S be a multiplicative subset of a ring A, and let S^{−1}A be the corresponding ring of fractions. Any ideal 𝔞 in A, generates an ideal S^{−1}𝔞 in S^{−1}A. If 𝔞 contains an element of S, then S^{−1}𝔞 contains a unit, and so is the whole ring. Thus some of the ideal structure of A is lost in the passage to S^{−1}A, but, as the next lemma shows, some is retained.

Proposition 4.15. Let S be a multiplicative subset of the ring A. The map 𝔭 ↦ S^{−1}𝔭 = 𝔭(S^{−1}A) is a bijection from the set of prime ideals of A that are disjoint from S to the set of prime ideals of S^{−1}A.

Proof. It is straightforward to verify that

𝔮 ↦ (inverse image of 𝔮 in A)

provides an inverse to 𝔭 ↦ S^{−1}𝔭. (See Atiyah and MacDonald, p41–42.) □

For example, let V be an affine variety and P a point on V. The proposition shows that there is a one-to-one correspondence between the prime ideals of k[V] contained in 𝔪_{P} and the prime ideals of O_{P}. In geometric terms, this says that there is a one-to-one correspondence between the prime ideals in O_{P} and the irreducible closed subvarieties of V passing through P.

Now let A be a local Noetherian ring with maximal ideal 𝔪. Then 𝔪 is an A-module, and the action of A on 𝔪/𝔪^{2} factors through k =_{df} A/𝔪.

Proposition 4.16. The elements a_{1}, …, a_{n} of 𝔪 generate 𝔪 as an ideal if and only if their residues modulo 𝔪^{2} generate 𝔪/𝔪^{2} as a vector space over k. In particular, the minimum number of generators for the maximal ideal is equal to the dimension of the vector space 𝔪/𝔪^{2}.

Proof. If a_{1}, …, a_{n} generate 𝔪, it is obvious that their residues generate 𝔪/𝔪^{2}. Conversely, suppose that their residues generate 𝔪/𝔪^{2}, so that 𝔪 = (a_{1}, …, a_{n}) + 𝔪^{2}. Since A is Noetherian and (hence) 𝔪 is finitely generated, Nakayama’s lemma, applied with M = 𝔪 and N = (a_{1}, …, a_{n}), shows that 𝔪 = (a_{1}, …, a_{n}). □

15The same discussion applies to any f ∈ O_{P}. Such an f is of the form \frac{g}{h} with h(a) ≠ 0, and has a (not quite so trivial) Taylor expansion of the same form, but with an infinite number of terms, i.e., it lies in the power series ring k[[X_{1} − a_{1}, …, X_{n} − a_{n}]].

Lemma 4.17 (Nakayama’s Lemma). Let A be a local Noetherian ring, and let M be a finitely generated A-module. If N is a submodule of M such that M = N +𝔪M, then M = N.

Proof. After replacing M with the quotient module M/N, we can assume that N = 0. Thus we have to show that if M = 𝔪M, then M = 0. Let x_{1}, …, x_{n} generate M, and write

x_{i} = ∑_{j} a_{ij}x_{j}

for some a_{ij} ∈ 𝔪. We see that x_{1}, …, x_{n} can be considered to be solutions to the system of n equations in n variables

∑_{j}(δ_{ij} − a_{ij})x_{j} = 0, δ_{ij} = Kronecker delta,

and so Cramer’s rule tells us that det(δ_{ij} −a_{ij})·x_{i} = 0 for all i. But on expanding it out, we find that det(δ_{ij} −a_{ij}) = 1+m with m ∈ 𝔪. In particular, det(δ_{ij} −a_{ij}) ∉ 𝔪, and so it is a unit. We deduce that all the x_{i} are zero, and that M = 0. □

A Noetherian local ring A of Krull dimension d is said to be regular if its maximal ideal can be generated by d elements. Thus A is regular if and only if its Krull dimension is equal to the dimension of 𝔪/𝔪^{2}.

Two results from §7. We shall need to use two results that won’t be proved until §7.

(4.18). For any irreducible variety V and regular functions f_{1}, …, f_{r} on V, the irreducible components of V (f_{1}, …, f_{r}) have codimension ≤ r.

Note that for polynomials of degree 1 on k^{n}, this is familiar from linear algebra: A system of r linear equations in n variables either has no solutions (the equations are inconsistent) or has a family of solutions of at least dimension n − r.

The Krull dimension of a Noetherian local ring A is the maximum length of a chain of prime ideals:

𝔪 = 𝔭_{0} ⫌ 𝔭_{1} ⫌ ⋯ ⫌ 𝔭_{d}.

In §7, we shall prove:

(4.19). If V is an irreducible variety of dimension d, then the local ring at each point P of V has dimension d.

The height of a prime ideal 𝔭 in a Noetherian ring A, is the maximum length of a chain of prime ideals:

𝔭 = 𝔭_{0} ⫌ 𝔭_{1} ⫌ ⋯ ⫌ 𝔭_{d}.

Because of (4.15), the height of 𝔭 is the Krull dimension of A_{𝔭}. Thus the above result can be restated as: If V is an irreducible affine variety of dimension d, then every maximal ideal in k[V] has height d.

Sketch of proof of (4.19): If V = 𝔸^{d}, then A = k[X_{1}, …, X_{d}], and all maximal ideals in this ring have height d, for example,

(X_{1} − a_{1}, …, X_{d} − a_{d}) ⊃ (X_{1} − a_{1}, …, X_{d−1} − a_{d−1}) ⊃ … ⊃ (X_{1} − a_{1}) ⊃ 0

is a chain of prime ideals of length d that can’t be refined. In the general case, the Noether normalization theorem says that k[V] is integral over a polynomial ring k[x_{1}, …, x_{d}], x_{i} ∈ k[V]; then clearly x_{1}, …, x_{d} is a transcendence basis for k(V), and the going up and down theorems (see Atiyah and MacDonald, Chapt 5) show that the local rings of k[V] and k[x_{1}, …, x_{d}] have the same dimension.

The dimension of the tangent space. Note that (4.16) implies that the dimension of T_{P} (V) is the minimum number of elements needed to generate 𝔫_{P} ⊂ O_{P}.

Theorem 4.20. Let V be irreducible; then dim T_{P} (V) ≥ dim(V), and equality holds if and only if O_{P} is regular.

Proof. Suppose f_{1}, …, f_{r} generate the maximal ideal 𝔫_{P} in O_{P}. Then f_{1}, …, f_{r} are all defined on some open affine neighbourhood U of P, and I claim that P is an irreducible component of V (f_{1}, …, f_{r}) ⊂ U. If not, there will be some irreducible component Z ≠ P of V (f_{1}, …, f_{r}) passing through P. Write Z = V (𝔭) with 𝔭 a prime ideal in k[U]. Because V (𝔭) ⊂ V (f_{1}, …, f_{r}) and because Z contains P and is not equal to it, we have

(f_{1}, …, f_{r}) ⊂ 𝔭 ⫋ 𝔪_{P} (ideals in k[U]).

On passing to the local ring O_{P} = k[U]_{𝔪_{P}}, we find (using 4.15) that

(f_{1}, …, f_{r}) ⊂ 𝔭O_{P} ⫋ 𝔫_{P} (ideals in O_{P}).

This contradicts the assumption that the f_{i} generate 𝔪_{P}. Hence P is an irreducible component of V (f_{1}, …, f_{r}), and (4.18) implies that

r ≥ codimP = dim V.

Since the dimension of T_{P} (V) is the minimum value of r, this implies that dim T_{P} (V) ≥ dim V. If equality holds, then 𝔪_{P} can be generated by dim V elements, which (because of 4.19) implies that O_{P} is regular. Conversely, if O_{P} is regular, then the minimum value of r is dim V, and so equality holds. □

As in the affine case, we define a point P to be nonsingular if dim T_{P} (V) = dim V. Thus a point P is nonsingular if and only if O_{P} is a regular local ring. In more geometric terms, we can say that a point P on a variety V of dimension d is nonsingular if and only if it can be defined by d equations in some neighbourhood. More precisely, P is nonsingular if there exists an open neighbourhood U of P and d regular functions f_{1}, …, f_{d} on U that generate the ideal 𝔪_{P}.

According to (Atiyah and MacDonald 11.23), a regular local ring is an integral domain. This provides another explanation of why a point on the intersection of two irreducible components of a variety can’t be nonsingular: the local ring at such a point in not an integral domain. (Suppose P ∈ Z_{1} ∩ Z_{2}, with Z_{1} ∩ Z_{2} ≠ Z_{1}, Z_{2}. Since Z_{1} ∩ Z_{2} ≠ Z_{1}, there is a nonzero regular function f_{1} defined on an open neighbourhood U of P in Z_{1} that is zero on U ∩ Z_{1} ∩ Z_{2}. Extend f_{1} to a neighbourhood of P in Z_{1} ∪Z_{2} by setting f_{1}(Q) = 0 for all Q ∈ Z_{2}. Then f_{1} defines a germ of regular function at P. Similarly construct function f_{2} that is zero on Z_{1}. Then f_{1} and f_{2} define nonzero germs of functions at P, but their product is zero.)

An integral domain that is integrally closed in its field of fractions is also called a normal ring.

An algebraic variety is normal if O_{P} is normal for all P ∈ V; equivalently (see [A-M, 5.13]) if for all open affines U ⊂ V, k[U] is a finite product of normal rings. Since, as we just noted, the local ring at a point lying on two irreducible components can’t be an integral domain, a normal variety is a disjoint union of irreducible varieties.

A regular local Noetherian ring is always normal (cf. A-M p123); conversely, a normal local integral domain of dimension one is regular (ibid.). Thus nonsingular varieties are normal, and normal curves are nonsingular. However, a normal surface need not be nonsingular: the cone

X^{2} + Y^{2} − Z^{2} = 0

is normal, but is singular at the origin. However, it is true that the singular locus of a normal variety must have codimension ≥ 2. For example, a normal surface can only have isolated singularities, i.e., the singular locus can’t contain a curve.

Singular points are singular. The set of singular points on a variety is called the singular locus of the variety.

Theorem 4.21. The nonsingular points of a variety V form a dense open subset.

Proof. We have to show that the singular points form a proper closed subset of every irreducible component of V.

Closed: We can assume that V is affine, say V = V (𝔞) ⊂ 𝔸^{n}. Let P_{1}, …, P_{r} generate 𝔞. Then the set of singular points is the zero set of the ideal generated by the (n − d) × (n − d) minors of the matrix

Jac(P_{1}, …, P_{r})(a) = ( \begin{matrix}\frac{∂P_{1}}{∂X_{1}} (a) & … & \frac{∂P_{1}}{∂X_{m}} (a) \\ ⋮ & & ⋮ \\ \frac{∂P_{r}}{∂X_{1}} (a) & … & \frac{∂P_{r}}{∂X_{m}} (a) \end{matrix} )

Proper: Suppose first that V is an irreducible hypersurface in 𝔸^{d+1}, i.e., that it is the zero set of a single nonconstant irreducible polynomial F (X_{1}, …, X_{d+1}). By (1.21), dim V = d. In this case, the proof is the same as that of (4.3): if \frac{∂F}{∂X_{1}} is identically zero on V (F), then \frac{∂F}{∂X_{1}} must be divisible by F, and hence zero. Thus F must be a polynomial in X_{2}, … (characteristic zero) or in X_{1}^{p}, X_{2}, … (characteristic p). Therefore, if all the points of V are singular, then F is constant (characteristic 0) or a p^{th} power (characteristic p) which contradict the hypothesis.

We shall complete the proof by showing that there is a nonempty open subset of V that is isomorphic to a nonempty open subset of an irreducible hypersurface in 𝔸^{d+1}.

Two irreducible varieties V and W are said to be birationally equivalent if k(V) ≈ k(W).

Lemma 4.22. Two irreducible varieties V and W are birationally equivalent if and only if there are open subsets U and U′ of V and W respectively such that U ≈ U′.

Proof. Assume that V and W are birationally equivalent. We may suppose that V and W are affine, corresponding to the rings A and B say, and that A and B have a common field of fractions K. Write B = k[x_{1}, …, x_{n}]. Then x_{i} = a_{i}/b_{i}, a_{i}, b_{i} ∈ A, and B ⊂ A_{b_{1}...b_{r}}. Since Specm(A_{b_{1}...b_{r}}) is a basic open subvariety of V, we may replace A with A_{b_{1}...b_{r}}, and suppose that B ⊂ A. The same argument shows that there exists a d ∈ B ⊂ A such A ⊂ B_{d}. Now

B ⊂ A ⊂ B_{d} ⟹ B_{d} ⊂ A_{d} ⊂ (B_{d})_{d} = B_{d},

and so A_{d} = B_{d}. This shows that the open subvarieties D(b) ⊂ V and D(b) ⊂ W are isomorphic. This proves the “only if” part, and the “if” part is obvious. □

Lemma 4.23. Let V be an irreducible algebraic variety of dimension d; then there is a hypersurface H in 𝔸^{d+1} birationally equivalent to V.

Proof. Let K = k(x_{1}, …, x_{n}), and assume n > d + 1. After renumbering, we may suppose that x_{1}, …, x_{d} are algebraically independent. Then f(x_{1}, …, x_{d+1}) = 0 for some nonzero irreducible polynomial f(X_{1}, …, X_{d+1}) with coefficients in k. Not all ∂f/∂X_{i} are zero, for otherwise k will have characteristic p ≠ 0 and f will be a p^{th} power. After renumbering, we may suppose that ∂f/∂X_{d+1} ≠ 0. Then k(x_{1}, …, x_{d+1}) is separable and algebraic over k(x_{1}, …, x_{d}), and x_{d+2} is algebraic over k(x_{1}, …, x_{d}). Therefore, by Abel’s theorem on the primitive element (my Math 594 notes, p36, or van der Waerden , §40), there is an element y such that k(x_{1}, …, x_{d+2}) = k(x_{1}, …, x_{d}, y). Thus K is generated by n − 1 elements (as a field containing k). After repeating the process, possibly several times, we will have K = k(z_{1}, …, z_{d+1}) with z_{d+1} separable over k(z_{1}, …, z_{d}). Now take f to be an irreducible polynomial satisfied by z_{1}, …, z_{d+1} and define H to be the hypersurface f = 0.

Corollary 4.24. Any algebraic group G is nonsingular.

Proof. From the theorem we know that there is an open dense subset U of G of nonsingular points. For any g ∈ G, a ↦ ga is an isomorphism G → G, and so gU consists of nonsingular points. Clearly G = ∪gU. □

In fact, any variety on which a group acts transitively by regular maps will be nonsingular.

Aside 4.25. If V has codimension 1 in 𝔸^{d+1}, then I(V) = (f) for some polynomial f.

Proof. We know I(V) = ∩I(V_{i}) where the V_{i} are the irreducible components of V, and so if we can prove I(V_{i}) = (f_{i}) then I(V) = (f_{1} ⋯ f_{r}). Thus we may suppose that V is irreducible. Let 𝔭 = I(V); it is a prime ideal, and it is nonzero because otherwise dim(V) = d+1. Therefore it contains an irreducible polynomial f . From (0.3) we know (f) is prime. If (f) ≠ 𝔭 , then we have

V = V (𝔭) ⫋ V ((f)) ⫋ 𝔸^{d+1},

and dim(V) < dim(V (f)) < d + 1 (see 1.22), which contradicts the fact that V has dimension d. □

Aside 4.26. Lemma 4.18 can be improved as follows: if V and W are irreducible varieties, then every inclusion k(W) ⊂ k(V) is defined by a regular surjective map α: U → U′, with U and U′ open subsets of V and W.

Aside 4.27. An irreducible variety V of dimension d is said to rational if it is birationally equivalent to 𝔸^{d}. It is said to be unirational if k(V) can be embedded in k(𝔸^{d}); geometrically, this means that there are open subsets U′ of 𝔸^{d} and U of V and a regular surjective map U′ → U. Lüroth’s theorem (which sometimes used to be taught in Math 594) says that a unirational curve is rational, that is, a subfield of k(X) not equal to k is a pure transcendental extension of k. It was proved by Castelnuovo that when k has characteristic zero every unirational surface is rational. Only in the seventies was it shown that this is not true for three dimensional varieties (Artin, Mumford, Clemens, Griffiths, Manin,...). When k has characteristic p ≠ 0, Zariski showed that there exist nonrational unirational surfaces, and P. Blass (UM thesis 1977) showed that there exist infinitely many surfaces V, no two birationally equivalent, such that k(X^{p}, Y^{p}) ⊂ k(V) ⊂ k(X, Y).

Aside 4.28. Note that , if V is irreducible, then

dim V = min_{P} dim T_{P} (V)

This formula can be useful in computing the dimension of a variety.

Etale neighbourhoods. Recall that a regular map α: W → V is said to be étale at a nonsingular point P of W if the map (dα)_{P} : T_{P} (W) → T_{α(P)}(V) is an isomorphism.

Let P be a nonsingular point on a variety V of dimension d. A local system of parameters at P is a family {f_{1}, …, f_{d}} of germs of regular functions at P generating 𝔫_{P} ⊂ O_{P}. Equivalent conditions: the images of f_{1}, …, f_{d} in 𝔫_{P} /𝔫_{P}^{2} generate it as a k-vector space (see 4.16); or (df_{1})_{P} , …, (df_{d})_{P} is a basis for dual space to T_{P} (V).

Proposition 4.29. Let {f_{1}, …, f_{d}} be a local system of parameters at a nonsingular point P of V. Then there is a nonsingular open neighbourhood U of P such that f_{1}, f_{2}, …, f_{d} are represented by pairs (f̃_{1}, U), …, (f̃_{d}, U) and the map (f̃_{1}, …, f̃_{d}): U → 𝔸^{d} is étale.

Proof. Obviously, the f_{i} are represented by regular functions f̃_{i} defined on a single open neighbourhood U′ of P, which, because of (4.21), we can choose to be nonsingular. The map α = (f̃_{1}, …, f̃_{d}): U′ → 𝔸^{d} is étale at P, because the dual map to (dα)_{a} is (dX_{i})_{0} ↦ (df̃_{i})_{a}. The next lemma then shows that α is étale on an open neighbourhood U of P. □

Lemma 4.30. Let W and V be nonsingular varieties. If α: W → V is étale at P, then it is étale at all points in an open neighbourhood of P.

Proof. The hypotheses imply that W and V have the same dimension d, and that their tangent spaces all have dimension d. We may assume W and V to be affine, say W ⊂ 𝔸^{m} and V ⊂ 𝔸^{n}, and that α is given by polynomials P_{1}(X_{1}, …, X_{m}), …, P_{n}(X_{1}, …, X_{m}). Then (dα)_{a} : T_{a}(𝔸^{m}) → T_{α(a)}(𝔸^{n}) is a lin-

ear map with matrix (\frac{∂P_{i}}{∂X_{j}} (a)), and α is not étale at a if and only if the kernel of this map contains a nonzero vector in the subspace T_{a}(V) of T_{a}(𝔸^{n}). Let f_{1}, …, f_{r} generate I(W). Then α is not étale at a if and only if the matrix

( \begin{matrix}\frac{∂f_{i}}{∂X_{j}} (a) \\ \frac{∂P_{i}}{∂X_{j}} (a) \end{matrix} )

has rank less than m. This is a polynomial condition on a, and so it fails on a closed subset of W, which doesn’t contain P. □

Let V be a nonsingular variety, and let P ∈ V. An étale neighbourhood of a point P of V is an étale map π: U → V from a nonsingular variety U to V whose image contains P.

Corollary 4.31. Let V be a nonsingular variety of dimension d, and let P ∈ V. There is an open Zariski neighbourhood U of P and an étale neighbourhood U′ of (0, …, 0) ∈ 𝔸^{d} such that U ≈ U′.

Proof. This is a restatement of the Proposition. □

Aside 4.32. Note the analogy with the definition of a differentiable manifold. [Discuss the étale “topology”. Note that, although two nonsingular algebraic varieties of the same dimension need not be locally isomorphic for the Zariski topology, they are locally isomorphic for the étale topology.]

Dual numbers and derivations. In general, if A is a k-algebra and M is an A-module, then a k-derivation is a map D: A → M such that (4.33.1) D(c) = 0 for all c ∈ k; (4.33.2) D(a + b) = D(a) + D(b); (4.33.3) D(a · b) = a · Db + b · Da (Leibniz rule). Note that the conditions imply that D is k-linear (but not A-linear). We write Der_{k}(A, M) for the space of all k-derivations A → M.

For example, the map f ↦ (df)_{P} =_{df} f − f(P) mod 𝔫_{P}^{2} is a k-derivation O_{P} → 𝔫_{P} /𝔫_{P}^{2} .

Proposition 4.34. There are canonical isomorphisms

Der_{k}(O_{P} , k) ⟶^{≈} Hom_{k-lin}(𝔫_{P} /𝔫_{P}^{2} , k) ⟶^{≈} T_{P} (V).

Proof. Note that

O_{P} = k ⊕ 𝔫_{P} , f ↔ (f(P), f − f(P)).

A derivation D: O_{P} → k is zero on k and on 𝔫_{P}^{2} (Leibniz’s rule). It therefore defines a linear map 𝔫_{P} /𝔫_{P}^{2} → k, and all such linear maps arise in this way, by composition

O_{P} ⟶^{f↦(df)_{P}} 𝔫_{P} /𝔫_{P}^{2} ⟶ k.

□

The ring of dual numbers is k[ε] = k[X]/(X^{2}). As a k-vector space it has a basis {1, ε}.

Proposition 4.35. The tangent space

T_{P} (V) = Hom(O_{P} , k[ε]) (local homomorphisms of local k-algebras).

Proof. Let α: O_{P} → k[ε] be a local homomorphism of k-algebras, and write α(a) = a_{0} + D_{α}(a)ε. Because α is a homomorphism of k-algebras, a ↦ a_{0} is the quotient map O_{P} → O_{P} /𝔪 = k. We have

\begin{aligned}α(ab) & = (ab)_{0} + D_{α}(ab)ε, and \\ α(a)α(b) & = (a_{0} + D_{α}(a)ε)(b_{0} + D_{α}(b)ε) = a_{0}b_{0} + (a_{0}D_{α}(b) + b_{0}D_{α}(a))ε. \end{aligned}

On comparing these expressions, we see that D_{α} satisfies Leibniz’s rule, and therefore is a k-derivation O_{P} → k. All such derivations arise in this way. □

For an affine variety V and a k-algebra A (not necessarily an affine k-algebra), we define V (A), the set of points of V with coordinates in A, to be Hom_{k-alg}(k[V], A). For example, if V = V (𝔞) ⊂ 𝔸^{n}, then

V (A) = {(a_{1}, …, a_{n}) ∈ A^{n} | f(a_{1}, …, a_{n}) = 0 all f ∈ 𝔞}.

Consider an α ∈ V (k[ε]), i.e., a k-algebra homomorphism α: k[V] → k[ε]. The composite k[V] → k[ε] → k is a point P of V, and

𝔪_{P} = Ker(k[V] → k[ε] → k) = α^{−1}((ε)).

Therefore elements of k[V] not in 𝔪_{P} map to units in k[ε], and so α extends to a homomorphism α′: O_{P} → k[ε]. By construction, this is a local homomorphism of local k-algebras, and every such homomorphism arises in this way. In this way we get a one-to-one correspondence between the local homomorphisms of k-algebras O_{P} → k[ε] and the set

{P′ ∈ V (k[ε]) | P′ ↦ P under the map V (k[ε]) → V (k)}.

This gives us a new interpretation of the tangent space at P.

Consider, for example, V = V (𝔞) ⊂ 𝔸^{n}, 𝔞 a radical ideal in k[X_{1}, …, X_{n}], and let a ∈ V. In this case, it is possible to show directly that

T_{a}(V) = {a′ ∈ V (k[ε]) | a′ maps to a under V (k[ε]) → V (k)}

Note that when we write a polynomial F (X_{1}, …, X_{n}) in terms of the variables X_{i} − a_{i}, we obtain a formula (trivial Taylor formula)

F (X_{1}, …, X_{n}) = F (a_{1}, …, a_{n}) + ∑ \frac{∂F}{∂X_{i}} |_{a} (X_{i} − a_{i}) + R

with R a finite sum of products of at least two terms (X_{i} − a_{i}). Now let a ∈ k^{n} be a point on V, and consider the condition for a + εb ∈ k[ε]^{n} to be a point on V. When we substitute a_{i} + εb_{i} for X_{i} in the above formula and take F ∈ 𝔞, we obtain:

F (a_{1} + εb_{1}, …, a_{n} + εb_{n}) = ε(∑ \frac{∂F}{∂X_{i}} |_{a} b_{i}).

Consequently, (a_{1} + εb_{1}, …, a_{n} + εb_{n}) lies on V if and only if (b_{1}, …, b_{n}) ∈ T_{a}(V) (original definition p68).

[Discuss the geometric significance of all this.]

Remark 4.36. The description of the tangent space in terms of dual numbers is particularly convenient when our variety is given to us in terms of its points functor. For example, let M_{n} be the set of n × n matrices, and let I be the identity matrix. Write e for I when it is to be regarded as the identity element of GL_{n}. Then we have

T_{e}(GL_{n}) = {I + εA | A ∈ M_{n}} ≈ M_{n};
T_{e}(SL_{n}) = {I + εA | det(I + εA) = I} = {I + εA | trace(A) = 0}.

Assume the characteristic ≠ 2, and let O_{n} be orthogonal group:

O_{n} = {A ∈ GL_{n} | AA^{tr} = I}.

(tr=transpose). This is the group of matrices preserving the quadratic form X_{1}^{2} + ⋯ + X_{n}^{2}. Then det : O_{n} → {±1} is a homomorphism, and the special orthogonal group SO_{n} is defined to be the kernel of this map. We have

T_{e}(O_{n}) = T_{e}(SO_{n})
= {I + εA ∈ M_{n} | (I + εA)(I + εA)^{tr} = I}
= {I + εA ∈ M_{n} | A is skew-symmetric}.

Note that, because an algebraic group is nonsingular, dim T_{e}(G) = dim G—this is a very convenient way to compute the dimension of an algebraic group.

On the tangent space T_{e}(GL_{n}) = M_{n} of GL_{n}, there is a bracket operation

[M, N] = MN − NM

which makes T_{e}(GL_{n}) into a Lie algebra. For any closed algbraic subgroup G of GL_{n}, T_{e}(G) is stable under the bracket operation on T_{e}(GL_{n}) and is a sub-Lie-algebra of M_{n}, which we denote Lie(G). The Lie algebra structure on Lie(G) is independent of the embedding of G into GL_{n} (in fact, it has an intrinsic definition), and G ↦ Lie(G) is functor from the category of linear algebraic groups to that of Lie algebras.

This functor is not fully faithful, for example, any étale homomorphism G → G′ will define an isomorphism Lie(G) → Lie(G′), but is nevertheless very useful.

Assume k has characteristic zero. A connected algebraic group G is said to be semisimple if it has no closed connected commutative normal subgroup (except {e}). Such a group G may have a finite nontrivial centre Z(G), and we call two semisimple groups G and G′ locally isomorphic if G/Z(G) ≈ G′/Z(G′). For example, SL_{n} is semisimple, with centre μ_{n}, the set of diagonal matrices diag(ζ, …, ζ), ζ^{n} = 1, and SL_{n} /μ_{n} = PSL_{n}. A Lie algebra is semisimple if it has no commutative ideal (except {0}). One can prove that

G is semisimple ⟺ Lie(G) is semisimple,

and the map G ↦ Lie(G) defines a one-to-one correspondence between the set of local isomorphism classes of semisimple algebraic groups and the set of isomorphism classes of Lie algebras. The classification of semisimple algebraic groups can be deduced from that of semisimple Lie algebras and a study of the finite coverings of semisimple algebraic groups. (Interestingly, the relation between Lie groups and Lie algebras is quite similar.)

Tangent cones. In this section, I assume familiarity with parts of Atiyah-MacDonald, Chapters 11, 12.

Let V = V (𝔞) ⊂ k^{m}, 𝔞 = rad(𝔞), and let P = (0, …, 0) ∈ V. Define 𝔞_{∗} to be the ideal generated by the polynomials F_{∗} for F ∈ 𝔞, where F_{∗} is the leading form of F (see p66). The geometric tangent cone at P, C_{P} (V) is V (𝔞_{∗}), and the tangent cone is the pair (V (𝔞_{∗}), k[X_{1}, …, X_{n}]/𝔞_{∗}). Obviously, C_{P} (V) ⊂ T_{P} (V).

Computing the tangent cone. If 𝔞 is principal, 𝔞 = (F), then 𝔞_{∗} = (F_{∗}), but if 𝔞 = (F_{1}, …, F_{r}), then it need not be true that 𝔞_{∗} = (F_{1∗}, …, F_{r∗}). Consider for example 𝔞 = (XY, XZ + Z(Y^{2} − Z^{2})). One can show that this is a radical ideal either by asking Macaulay (assuming you believe Macaulay), or by following the method suggested in Cox et al, p474, prob 3 to show that it is an intersection of prime ideals. Since

Y Z(Y^{2} − Z^{2}) = Y · (XZ + Z(Y^{2} − Z^{2})) − Z · (XY) ∈ 𝔞

and is homogeneous, it is in 𝔞_{∗}, but it is not in the ideal generated by XY, XZ. In fact, 𝔞_{∗} is the ideal generated by

XY, XZ, Y Z(Y^{2} − Z^{2}).

This raises the following question: given a set of generators for an ideal 𝔞, how do you find a set of generators for 𝔞_{∗}? There is an algorithm for this in Cox et al. p467. Let 𝔞 be an ideal (not necessarily radical) such that V = V (𝔞), and assume the origin is in V. Introduce an extra variable T such that T “>” the remaining variables. Make each generator of 𝔞 homogeneous by multiplying its monomials by appropriate (small) powers of T, and find a Gröbner basis for the ideal generated by these homogeneous polynomials. Remove T from the elements of the basis, and then the polynomials you get generate 𝔞_{∗}.

Intrinsic definition of the tangent cone. Let A be a local ring with maximal ideal 𝔫. The associated graded ring is

gr(A) = ⊕𝔫^{i}/𝔫^{i+1}.

Note that if A = B_{𝔪}, 𝔫 = 𝔪A, then gr(A) = ⊕𝔪^{i}/𝔪^{i+1} (because of (4.13)).

Proposition 4.37. The map k[X_{1}, …, X_{m}]/𝔞_{∗} → gr(O_{P}) sending the class of X_{i} in k[X_{1}, …, X_{m}]/𝔞_{∗} to the class of X_{i} in gr(O_{P}) is an isomorphism.

Proof. Let 𝔪 be the maximal ideal in k[X_{1}, …, X_{m}]/𝔞 corresponding to P. Then

gr(O_{P}) = ∑ 𝔪^{i}/𝔪^{i+1}
= ∑(X_{1}, …, X_{m})^{i}/(X_{1}, …, X_{m})^{i+1} + 𝔞 ∩ (X_{1}, …, X_{m})^{i}
= ∑(X_{1}, …, X_{m})^{i}/(X_{1}, …, X_{m})^{i+1} + 𝔞_{i}

where 𝔞_{i} is the homogeneous piece of 𝔞_{∗} of degree i (that is, the subspace of 𝔞_{∗} consisting of homogeneous polynomials of degree i). But

(X_{1}, …, X_{m})^{i}/(X_{1}, …, X_{m})^{i+1}+𝔞_{i} = i^{th} homogeneous piece of k[X_{1}, …, X_{m}]/𝔞_{∗}.

□

For a general variety V and P ∈ V, we define the geometric tangent cone C_{P} (V) of V at P to be Specm(gr(O_{P})_{red}), where gr(O_{P})_{red} is the quotient of gr(O_{P}) by its nilradical.

Recall (A-M, 11.21) that dim(A) = dim(gr(A)). Therefore the dimension of the geometric tangent cone at P is the same as the dimension of V (in contrast to the dimension of the tangent space).

Recall (A-M, 11.22) that gr(O_{P}) is a polynomial ring in d variables (d = dim V) if and only if O_{P} is regular. Therefore, P is nonsingular if and only if gr(O_{P}) is a polynomial ring in d variables, in which case C_{P} (V) = T_{P} (V).

Using tangent cones, we can extend the notion of an étale morphism to singular varieties. Obviously, a regular map α: V → W induces a homomorphism gr(O_{α(P)}) → gr(O_{P}). We say that α is étale at P if this is an isomorphism. Note that then there is an isomorphism of the geometric tangent cones C_{P} (V) → C_{α(P)}(W), but this map may be an isomorphism without α being étale at P. Roughly speaking, to be étale at P, we need C_{P} (V) → C_{α(P)}(W) to be an isomorphism and to preserve the “multiplicities” of the components.

It is a fairly elementary result that a local homomorphism of local rings α: A → B induces an isomorphism on the graded rings if and only if it induces an isomorphism on the completions. Thus α: V → W is étale at P if and only if the map is Ô_{α(P)} → Ô_{P} an isomorphism. Thus (4.29) shows that the choice of a local system of parameters f_{1}, …, f_{d} at P determines an isomorphism Ô_{P} → k[[X_{1}, …, X_{d}]].

We can rewrite this as follows: let t_{1}, …, t_{d} be a local system of parameters at P; then there is a canonical isomorphism Ô_{P} → k[[t_{1}, …, t_{d}]]. For f ∈ Ô_{P}, the image of f ∈ k[[t_{1}, …, t_{d}]] can be regarded as the Taylor series of f.

For example, let V = 𝔸^{1}, and let P be the point a. Then t = X − a is a local parameter at a, O_{P} consists of quotients f(X) = g(X)/h(X) with h(a) ≠ 0, and the coefficients of the Taylor expansion ∑_{n≥0} a_{n}(X −a)^{n} of f(X) can be computed as in Math 115, that is, a_{n} = f^{(n)}(a)/n!.

5. Projective Varieties; Complete Varieties

Recall that we defined

ℙ^{n} = k^{n+1} − {origin}/∼,

where (a_{0}, …, a_{n}) ∼ (b_{0}, …, b_{n}) if and only if there exists a c ≠ 0 in k such that (a_{0}, …, a_{n}) = c(b_{0}, …, b_{n}). Write (a_{0} : …: a_{n}) for the equivalence class of (a_{0}, …, a_{n}), and π for the map k^{n+1} − {origin}/∼ → ℙ^{n}. Let U_{i} be the set of (a_{0} : …: a_{n}) ∈ ℙ^{n} such that a_{i} ≠ 0. Then (a_{0}, …, a_{n}) ↦ (a_{0} : …: 1 : ⋯: a_{n}) is a one-to-one correspondence k^{n} ↔ U_{i}, and we used these correspondences to define the structure of a ringed space on ℙ^{n}; specifically, we said that U ⊂ ℙ^{n} is open if and only if U ∩ U_{i} is open for each i, and f: U → k is regular if and only if f|U ∩ U_{i} is regular for all i.

In this chapter, we shall first derive another description of the topology on ℙ^{n}, and then we shall show that the ringed space structure makes ℙ^{n} into a separated algebraic variety. A closed subvariety of ℙ^{n} (or any variety isomorphic to such a variety) is called a projective variety, and a locally closed subvariety of ℙ^{n} (or any variety isomorphic to such a variety) is called a quasi-projective variety. Note that every affine variety is quasi-projective, but there are many varieties that are not quasi-projective. We study morphisms between (quasi-) projective varieties. Finally, we show that a projective variety is “complete”, that is, it has the analogue of a property that distinguishes compact topological spaces among locally compact spaces.

Projective varieties are important for the same reason compact manifolds are important: results are often simpler when stated for projective varieties, and the “part at infinity” often plays a role, even when we would like to ignore it. For example, a famous theorem of Bezout says that a curve of degree m in the projective plane^{16} intersects a curve of degree n in exactly mn points (counting multiplicities). For affine curves, one has only an inequality.

Algebraic subsets of ℙ^{n}. A polynomial F (X_{0}, …, X_{n}) is said to be homogeneous of degree d if it is a sum of terms a_{i_{0},... ,i_{n}}X_{0}^{i_{0}}···X_{n}^{i_{n}} with i_{0}+⋯+i_{n} = d; equivalently,

F (tX_{0}, …, tX_{n}) = t^{d}F (X_{0}, …, X_{n})

for all t ∈ k. Write k[X_{0}, …, X_{n}]_{d} for the subspace of k[X_{0}, …, X_{n}] of polynomials of degree d. Then

k[X_{0}, …, X_{n}] = ⨁_{d≥0} k[X_{0}, …, X_{n}]_{d};

that is, each polynomial F can be written uniquely as a sum F = ∑ F_{d} with F_{d} of degree d.

Let P = (a_{0} : …: a_{n}) ∈ ℙ^{n}. Then P can also be written (ca_{0} : …: ca_{n}) for any c ∈ k^{×}, and so we can’t speak of the value of a polynomial F (X_{0}, …, X_{n}) at P. However, if F is homogeneous, then F (ca_{0}, …, ca_{n}) = c^{d}F (a_{0}, …, a_{n}), and so it does make sense to say that F is zero or not zero at P. We define a projective algebraic set to be the set of common zeros in ℙ^{n} of a collection of homogeneous polynomials.

16This means that it is defined by a homogeneous polynomial F (X, Y, Z) of degree m.

Example 5.1. Consider the projective algebraic subset E of ℙ^{2} defined by the homogeneous equation

Y^{2}Z = X^{3} + aXZ^{2} + bZ^{3}  (*)

where X^{3} + aX + b is assumed not to have multiple roots. It consists of the points (x : y : 1) on the affine curve E_{aff}

Y^{2} = X^{3} + aX + b,

together with the point “at infinity” (0:1:0).

Poincaré is usually credited (incorrectly!) that E is an algebraic group, with the group law such that P + Q + R = 0 if and only if P, Q, and R lie on a straight line. The zero for the group is the point at infinity.

Curves defined by an equation (*) are called elliptic curves. They can also be described as the curves of genus one, or as the abelian varieties of dimension one.

In the case k = ℂ, there is a lattice L ⊂ ℂ and a function ℘ (the Weierstrass ℘-function) that is analytic on ℂ − L and doubly periodic for L such that

℘^{′2} = ℘^{3} + a℘ + b.

The map z ↦ (℘(z) : ℘′(z) : 1): ℂ/L − {0} → ℙ^{2} is a bijection from ℂ/L − {0} onto E_{aff}. This map extends to an isomorphism ℂ/L ⟶^{≈} E by sending 0 to (0 : 1 : 0). (See Math 604, or any good course on complex analysis.)

In the case that a, b ∈ ℚ, we can speak of the zeros of (*) with coordinates in ℚ. They also form a group E(ℚ), which Mordell showed to be finitely generated. It is easy to compute the torsion subgroup of E(ℚ), but there is at present no known algorithm for computing the rank of E(ℚ). (More precisely, there is an “algorithm” which always works, but which has not been proved to terminate after a finite amount of time, at least not in general.) There is a very beautiful theory surrounding elliptic curves over ℚ (and other number fields), which in some sense goes back 1,800 years to Diophantus. (See Math 679 for all of this.)

An ideal 𝔞 ⊂ k[X_{0}, …, X_{n}] is said to be homogeneous if it contains with any polynomial F all the homogeneous components of F, i.e., if F ∈ 𝔞 ⟹ F_{d} ∈ 𝔞, all d. Such an ideal is generated by homogeneous polynomials (obviously), and conversely, an the ideal generated by a set of homogeneous polynomials is homogeneous. The radical of a homogeneous ideal is homogeneous, the intersection of two homogeneous ideals is homogeneous, and a sum of homogeneous ideals is homogeneous.

For a homogeneous ideal 𝔞, we write V (𝔞) for the set of common zeros of the (homogeneous) polynomials in 𝔞. If F_{1}, …, F_{r} are homogeneous generators for 𝔞, then clearly V (𝔞) is the set of common zeros of the F_{i}. The sets V (𝔞) have similar properties to their namesakes in 𝔸^{n} :

𝔞 ⊂ 𝔟 ⟹ V (𝔞) ⊃ V (𝔟);

V (0) = ℙ^{n}; V (𝔞) = ∅ ⟺ rad(𝔞) ⊃ (X_{0}, …, X_{n});

V (𝔞𝔟) = V (𝔞 ∩ 𝔟) = V (𝔞) ∪ V (𝔟);

V (∑ 𝔞_{i}) = ∩V (𝔞_{i}).

The first statement is obvious. For the second, let V^{aff}(𝔞) be the zero set of 𝔞 in k^{n+1}. It is a cone (that is, it contains together with any point P the line through P and the origin), and

V (𝔞) = (V^{aff}(𝔞) − (0, …, 0))/∼ .

We have V (𝔞) = ∅ ⟺ V^{aff}(𝔞) ⊂ {(0, …, 0)} ⟺ rad(𝔞) ⊃ (X_{0}, …, X_{n}), by the Hilbert Nullstellensatz. The remaining statements can be proved directly, or by using the relation between V (𝔞) and V^{aff}(𝔞).

Let C be a cone in k^{n+1}; then I(C) is a homogeneous ideal in k[X_{0}, …, X_{n}], because

F (ca_{0}, …, ca_{n}) = ∑ c^{d}F_{d}(a_{0}, …, a_{n}),

and so, if F (ca_{0}, …, ca_{n}) = 0 for all c ∈ k^{×}, we must also have F_{d}(a_{0}, …, a_{n}) = 0. For any S ⊂ ℙ^{n}, C = π^{−1}(S) ∪ {origin} is a cone in k^{n+1}, and we define I(S) = I^{aff}(C).

Proposition 5.2. The maps V and I define a bijection between the set of algebraic subsets of ℙ^{n} and the set of homogeneous radical ideals of k[X_{0}, …, X_{n}], except that V maps both the ideals (X_{0}, …, X_{n}) and k[X_{0}, …, X_{n}] to the empty set. An algebraic set V in ℙ^{n} is irreducible if and only if I(V) is prime; in particular, ℙ^{n} is irreducible.

Proof. Note that we have bijections

{algebraic subsets of ℙ^{n}, ≠ ∅} ⟶

{closed cones in k^{n+1}, ≠ {(0, …, 0), ∅} ⟶^{I^{aff}}

{homogeneous radical ideals in k[X_{0}, …, X_{n}], ≠ (X_{0}, …, X_{n}), k[X_{0}, …, X_{n}]} ⟶^{V}

{algebraic subsets of ℙ^{n}, ≠ ∅}.

Here the first map sends V to π^{−1}(V)∪{origin}, which is also the closure of π^{−1}(V), and the third map is V in the sense of projective geometry. The composite of any three of these maps is the identity map. Obviously, V is irreducible if and only if the closure of π^{−1}(V) is irreducible, which is true if and only if I(V) is a prime ideal. □

The Zariski topology on ℙ^{n}. The statements above show that projective algebraic sets are the closed sets for a topology on ℙ^{n}. In this section, we verify that it agrees with that defined in the first paragraph of this section. For a homogeneous polynomial F, let

D(F) = {P ∈ ℙ^{n} | F (P) ≠ 0}.

Then, just as in the affine case, D(F) is open, and the sets of this type form a basis for the topology of ℙ^{n}.

With each polynomial f(X_{1}, …, X_{n}), we associate the homogeneous polynomial of the same degree

f^{∗}(X_{0}, …, X_{n}) = X_{0}^{deg(f)}f (\frac{X_{1}}{X_{0}} , …, \frac{X_{n}}{X_{0}} ) ,

and with each homogeneous polynomial F (X_{0}, …, X_{n}) we associate the polynomial

F_{∗}(X_{1}, …, X_{n}) = F (1, X_{1}, …, X_{n}).

Proposition 5.3. For the topology on ℙ^{n} just defined, each U_{i} is open, and when we endow it with the induced topology, the bijection

U_{i} ↔ 𝔸^{n}, (a_{0} : …: 1 : …: a_{n}) ↔ (a_{0}, …, a_{i−1}, a_{i+1}, …, a_{n})

becomes a homeomorphism.

Proof. It suffices to prove this with i = 0. The set U_{0} = D(X_{0}), and so it is a basic open subset in ℙ^{n}. Clearly,

D(F (X_{0}, …, X_{n})) ∩ U_{0} = D(F (1, X_{1}, …, X_{n})) = D(F_{∗})

and, for f ∈ k[X_{1}, …, X_{n}],

D(f) = D(f^{∗}) ∩ U_{0}.

Thus, under U_{0} ↔ 𝔸^{n}, the basic open subsets of 𝔸^{n} correspond to the intersections with U_{i} of the basic open subsets of ℙ^{n}, which proves that the bijection is a homeomorphism. □

Remark 5.4. It is possible to use this to give a different proof that ℙ^{n} is irreducible. We apply the criterion that a space is irreducible if and only if every nonempty open subset is dense (see p22). Note that each U_{i} is irreducible, and that U_{i} ∩ U_{j} is open and dense in each of U_{i} and U_{j} (as a subset of U_{i}, it is the set of points (a_{0} : …: 1 : …: a_{j} : …: a_{n}) with a_{j} ≠ 0). Let U be a nonempty open subset of ℙ^{n}; then U ∩ U_{i} is open in U_{i}. For some i, U ∩ U_{i} is nonempty, and so must meet U_{i} ∩ U_{j}. Therefore U meets every U_{j}, and so is dense in every U_{j}. It follows that its closure is all of ℙ^{n}.

We identify 𝔸^{n} with U_{0}, and examine the closures in ℙ^{n} of closed subsets of 𝔸^{n}.

With each ideal 𝔞 in k[X_{1}, …, X_{n}], we associate the homogeneous ideal 𝔞^{∗} in k[X_{0}, …, X_{n}] generated by {f^{∗} | f ∈ 𝔞}. For a closed subset V of 𝔸^{n}, set V^{∗} = V (𝔞^{∗}) with 𝔞 = I(V).

With each homogeneous ideal 𝔞 in k[X_{0}, X_{1}, …, X_{n}], we associate the ideal 𝔞_{∗} in k[X_{1}, …, X_{n}] generated by {F_{∗} | F ∈ 𝔞}. When V is a closed subset of ℙ^{n}, we set V_{∗} = V (𝔞_{∗}) with 𝔞 = I(V).

Proposition 5.5. (a) For V a closed algebraic subset of 𝔸^{n}, V^{∗} is the closure of V in ℙ^{n}, and (V^{∗})_{∗} = V. If V = ∪V_{i} is the decomposition of V into its irreducible components, then V^{∗} = ∪V_{i}^{∗} is the decomposition of V^{∗} into its irreducible components.

(b) For V a closed algebraic subset of ℙ^{n}, V_{∗} = V ∩ 𝔸^{n}. If no irreducible component of V lies in H_{∞} or contains H_{∞}, then V_{∗} is a proper subset of 𝔸^{n}, and (V_{∗})^{∗} = V.

Proof. Straightforward. □

The hyperplane at infinity. It is often convenient to think of ℙ^{n} as being 𝔸^{n} = U_{0} with a hyperplane added “at infinity”. More precisely, identify the U_{0} with 𝔸^{n}. The complement of U_{0} in ℙ^{n} is H_{∞} = {(0 : a_{1} : …: a_{n}) ⊂ ℙ^{n}}, which can be identified with ℙ^{n−1}.

For example, ℙ^{1} = 𝔸^{1} ∪ H_{∞} (disjoint union), with H_{∞} consisting of a single point, and ℙ^{2} = 𝔸^{2} ∪ H_{∞} with H_{∞} a projective line. Consider the line

aX + bY + 1 = 0

in 𝔸^{2}. Its closure in ℙ^{2} is the line

aX + bY + Z = 0.

It intersects the hyperplane H_{∞} = V (Z) at the point (−b : a : 0), which equals (1 : −a/b : 0) when b ≠ 0. Note that −a/b is the slope of the line aX + bY + 1 = 0, and so the point at which a line intersects H_{∞} depends only on the slope of the line: parallel lines meet in exactly one point at infinity. We can think of the projective plane ℙ^{2} as being the affine plane 𝔸^{2} with one point added at infinity for each direction in 𝔸^{2}.

A similar discussion applies to lines in ℙ^{n}.

Note that the point at infinity on the elliptic curve Y^{2} = X^{3} + aX + b is the intersection of the closure of any vertical line with H_{∞}.

ℙ^{n} is an algebraic variety. For each i, write O_{i} for the sheaf on U_{i} defined by the bijection 𝔸^{n} ↔ U_{i} ⊂ ℙ^{n}.

Lemma 5.7. Write U_{ij} = U_{i} ∩ U_{j}; then O_{i}|U_{ij} = O_{j}|U_{ij}. When endowed with this sheaf U_{ij} is an affine variety; moreover, Γ(U_{ij}, O_{i}) is generated as a k-algebra by the functions (f|U_{ij})(g|U_{ij}) with f ∈ Γ(U_{i}, O_{i}), g ∈ Γ(U_{j}, O_{j}).

Proof. It suffices to prove this for (i, j) = (0, 1). All rings occurring in the proof will be identified with subrings of the field k(X_{0}, X_{1}, …, X_{n}).

Recall that

U_{0} = \{(a_{0} : a_{1} : …: a_{n}) | a_{0} ≠ 0\}; (a_{0} : a_{1} : …: a_{n}) ↔ (\frac{a_{1}}{a_{0}} , \frac{a_{2}}{a_{0}} , …, \frac{a_{n}}{a_{0}} ) ∈ 𝔸^{n}.

Let k[\frac{X_{1}}{X_{0}} , \frac{X_{2}}{X_{0}} , …, \frac{X_{n}}{X_{0}} ] be the subring of k(X_{0}, X_{1}, …, X_{n}) generated by the quotients \frac{X_{i}}{X_{0}} —it is the polynomial ring in the n variables \frac{X_{1}}{X_{0}} , …, \frac{X_{n}}{X_{0}} . An element f(\frac{X_{1}}{X_{0}} , …, \frac{X_{n}}{X_{0}} ) ∈ k[\frac{X_{1}}{X_{0}} , …, \frac{X_{n}}{X_{0}} ] defines the map

(a_{0} : a_{1} : …: a_{n}) ↦ f(\frac{a_{1}}{a_{0}} , …, \frac{a_{n}}{a_{0}} ): U_{0} → k,

and in this way k[\frac{X_{1}}{X_{0}} , \frac{X_{2}}{X_{0}} , …, \frac{X_{n}}{X_{0}} ] becomes identified with the ring of regular functions on U_{0}, and U_{0} with Specm k[\frac{X_{1}}{X_{0}} , …, \frac{X_{n}}{X_{0}} ].

Next consider the open subset of U_{0},

U_{01} = {(a_{0} : …: a_{n}) | a_{0} ≠ 0, a_{1} ≠ 0}.

It is D(\frac{X_{1}}{X_{0}} ), and is therefore an affine subvariety of (U_{0}, O_{0}). The inclusion U_{01} ↪ U_{0} corresponds to the inclusion of rings k[\frac{X_{1}}{X_{0}} , …, \frac{X_{n}}{X_{0}} ] ↪ k[\frac{X_{1}}{X_{0}} , …, \frac{X_{n}}{X_{0}} , \frac{X_{0}}{X_{1}} ]. An element f(\frac{X_{1}}{X_{0}} , …, \frac{X_{n}}{X_{0}} , \frac{X_{0}}{X_{1}} ) of k[\frac{X_{1}}{X_{0}} , …, \frac{X_{n}}{X_{0}} , \frac{X_{0}}{X_{1}} ] defines the function (a_{0} : …: a_{n}) ↦ f(\frac{a_{1}}{a_{0}} , …, \frac{a_{n}}{a_{0}} , \frac{a_{0}}{a_{1}} ) on U_{01}.

Similarly,

U_{1} = \{(a_{0} : a_{1} : …: a_{n}) | a_{1} ≠ 0\}; (a_{0} : a_{1} : …: a_{n}) ↔ (\frac{a_{0}}{a_{1}} , …, \frac{a_{n}}{a_{1}} ) ∈ 𝔸^{n},

and we identify U_{1} with Specm k[\frac{X_{0}}{X_{1}} , \frac{X_{2}}{X_{0}} , …, \frac{X_{n}}{X_{1}} ]. An element f(\frac{X_{0}}{X_{1}} , …, \frac{X_{n}}{X_{1}} ) ∈ k[\frac{X_{0}}{X_{1}} , …, \frac{X_{n}}{X_{1}} ] defines the map (a_{0} : …: a_{n}) ↦ f(\frac{a_{0}}{a_{1}} , …, \frac{a_{n}}{a_{1}} ): U_{1} → k.

When regarded as an open subset of U_{1},

U_{01} = {(a_{0} : …: a_{n}) | a_{0} ≠ 0, a_{1} ≠ 0},

is D(\frac{X_{0}}{X_{1}} ), and is therefore an affine subvariety of (U_{1}, O_{1}), and the inclusion U_{01} ↪ U_{1} corresponds to the inclusion of rings k[\frac{X_{0}}{X_{1}} , …, \frac{X_{n}}{X_{1}} ] ↪ k[\frac{X_{0}}{X_{1}} , …, \frac{X_{n}}{X_{1}} , \frac{X_{1}}{X_{0}} ]. An element f(\frac{X_{0}}{X_{1}} , …, \frac{X_{n}}{X_{1}} ) of k[\frac{X_{0}}{X_{1}} , …, \frac{X_{n}}{X_{1}} , \frac{X_{1}}{X_{0}} ] defines the function (a_{0} : …: a_{n}) ↦ f(\frac{a_{0}}{a_{1}} , …, \frac{a_{n}}{a_{1}} , \frac{a_{1}}{a_{0}} ) on U_{01}.

The two rings k[\frac{X_{1}}{X_{0}} , …, \frac{X_{n}}{X_{0}} , \frac{X_{0}}{X_{1}} ], k[\frac{X_{0}}{X_{1}} , …, \frac{X_{n}}{X_{1}} , \frac{X_{1}}{X_{0}} ] are equal as subrings of k(X_{0}, X_{1}, …, X_{n}), and an element of this ring defines the same function on U_{01} regardless of which of the two rings it is considered an element. Therefore, whether we regard U_{01} as a subvariety of U_{0} or of U_{1} it inherits the same structure as an affine algebraic variety. This proves the first two assertions, and the third is obvious: k[\frac{X_{1}}{X_{0}} , …, \frac{X_{n}}{X_{0}} , \frac{X_{0}}{X_{1}} ] is generated by its subrings k[\frac{X_{1}}{X_{0}} , …, \frac{X_{n}}{X_{0}} ] and k[\frac{X_{0}}{X_{1}} , \frac{X_{2}}{X_{1}} , …, \frac{X_{n}}{X_{1}} ]. □

Write u_{i} for the map 𝔸^{n} → U_{i} ⊂ ℙ^{n}. For any open subset U of ℙ^{n}, we define f: U → k to be regular if and only if f ∘ u_{i} is a regular function on u_{i}^{−1}(U) for all i. This obviously defines a sheaf O of k-algebras on ℙ^{n}.

Proposition 5.8. For each i, the bijection 𝔸^{n} → U_{i} is an isomorphism of ringed spaces, 𝔸^{n} → (U_{i}, O|U_{i}); therefore (ℙ^{n}, O) is a prevariety. It is in fact a variety.

Proof. Let U be an open subset of U_{i}. Then f: U → k is regular if and only if (a) it is regular on U ∩ U_{i}, and

(b) it is regular on U ∩ U_{j} for all j ≠ i. But the last lemma shows that (a) implies (b) because U ∩ U_{j} ⊂ U_{ij}. To prove that ℙ^{n} is separated, apply the criterion (3.26c) to the covering {U_{i}} of ℙ^{n}. □

Example 5.9. Assume k does not have characteristic 2, and let C be the plane projective curve: Y^{2}Z = X^{3}. For each a ∈ k^{×}, there is an automorphism

φ_{a}: C → C, (x : y : z) ↦ (ax : y : a^{3}z).

Patch two copies of C × 𝔸^{1} together along C × (𝔸^{1} − {0}) by identifying (P, u) with (φ_{u}(P), u^{−1}), P ∈ C, u ∈ 𝔸^{1} − {0}. One obtains in this way a singular 2dimensional variety that is not quasi-projective (see Hartshorne, p171). (It is even complete (see below), and so if it were quasi-projective, it would be projective. It is known that every irreducible separated curve is quasi-projective, and every nonsingular complete surface is projective, and so this is an example of minimum dimension. In Shafarevich p287-292 there is an example of a nonsingular complete variety of dimension 3 that is not projective.)

The field of rational functions of a projective variety. Recall (p60) that we attached to each irreducible variety V a field k(V) with the property that k(V) is the field of fractions of k[U] for any open affine U ⊂ V. We now describe this field in the case that V = ℙ^{n}. Recall that k[U_{0}] = k[\frac{X_{1}}{X_{0}} , …, \frac{X_{n}}{X_{0}} ]. We regard this as a subring of k(X_{0}, …, X_{n}), and wish to identify the field of fractions of k[U_{0}] as a subfield of k(X_{0}, …, X_{n}). Any nonzero F ∈ k[U_{0}] can be written

F (\frac{X_{1}}{X_{0}} , …, \frac{X_{n}}{X_{0}} ) = \frac{F^{∗}(X_{0},... ,X_{n})}{X_{0}^{deg(F)}},

and it follows that the field of fractions of k[U_{0}] is

k(U_{0}) = \{\frac{G(X_{0}, …, X_{n})}{H(X_{0}, …, X_{n})} | G, H homogeneous of the same degree\} ∪ \{0\}.

Write k(X_{0}, …, X_{n})_{0} for this field (the subscript 0 is short for “subfield of elements of degree 0”), so that k(ℙ^{n}) = k(X_{0}, …, X_{n})_{0}. Note that an element F = \frac{G}{H} in k(X_{0}, …, X_{n})_{0} defines a well-defined function

D(H) → k, (a_{0} : …: a_{n}) ↦ \frac{G(a_{0}, …, a_{n})}{H(a_{0}, …, a_{n})},

which is obviously regular (look at its restriction to U_{i}).

We now extend this discussion to any irreducible projective variety V. Such a V can be written V = V (𝔭), where 𝔭 is a homogeneous ideal in k[X_{0}, …, X_{n}]. Let k_{h}[V] = k[X_{0}, …, X_{n}]/𝔭—it is called the homogeneous coordinate ring of V. (Note that k_{h}[V] is the ring of regular functions on the affine cone over V; therefore its dimension is dim(V) + 1. It depends, not only on V, but on the embedding of V into ℙ^{n}—it is not intrinsic to V (see 5.19 below).) We say that a nonzero f ∈ k_{h}[V] is homogeneous of degree d if it can be represented by a homogeneous polynomial F of degree d in k[X_{0}, …, X_{n}]. We give 0 degree 0.

Lemma 5.10. Each element of k_{h}[V] can be written uniquely in the form

f = f_{0} + ⋯ + f_{d}

with f_{i} homogeneous of degree i.

Proof. Let F represent f; then F can be written F = F_{0} + ⋯ + F_{d} with F_{i} homogeneous of degree i, and when reduced modulo 𝔭, this gives a decomposition of f of the required type. Suppose f also has a decomposition f = ∑ g_{i}, with g_{i} represented by the homogeneous polynomial G_{i} of degree i. Then F − G ∈ 𝔭, and the homogeneity of 𝔭 implies that F_{i} − G_{i} = (F − G)_{i} ∈ 𝔭. Therefore f_{i} = g_{i}. □

It therefore makes sense to speak of homogeneous elements of k[V]. For such an element h, we define D(h) = {P ∈ V | h(P) ≠ 0}.

Since k_{h}[V] is an integral domain, we can form its field of fractions k_{h}(V). Define

k_{h}(V)_{0} = \{\frac{g}{h} ∈ k_{h}(V) | g and h homogeneous of the same degree\} ∪ \{0\}.

Proposition 5.11. The field of rational functions on V is k_{h}(V)_{0}.

Proof. Consider V_{0} =_{df} U_{0} ∩ V. As in the case of ℙ^{n}, we can identify k[V_{0}] with a subring of k_{h}[V], and then the field of fractions of k[V_{0}] becomes identified with k_{h}(V)_{0}.

Regular functions on a projective variety. Again, let V be an irreducible projective variety. Let f ∈ k(V)_{0}, and let P ∈ V. If we can write f = \frac{g}{h} with g and h homogeneous of the same degree and h(P) ≠ 0, then we define f(P) = \frac{g(P)}{h(P)}. By g(P) we mean the following: let P = (a_{0} : …: a_{n}); represent g by a homogeneous G ∈ k[X_{0}, …, X_{n}], and write g(P) = G(a_{0}, …, a_{n}); this is independent of the choice of G, and if (a_{0}, …, a_{n}) is replaced by (ca_{0}, …, ca_{n}), then g(P) is multiplied by c^{deg(g)} = c^{deg(h)}. Thus the quotient \frac{g(P)}{h(P)} is well-defined.

Note that we may be able to write f as \frac{g}{h} with g and h homogeneous polynomials of the same degree in many essentially different ways (because k_{h}[V] need not be a unique factorization domain), and we define the value of f at P if there is one such representation with h(P) ≠ 0. The value f(P) is independent of the representation f = \frac{g}{h} (write P = (a_{0} : …: a_{n}) = a; if \frac{g}{h} = \frac{g′}{h′} in k_{h}(V)_{0}, then gh′ = g′h in k_{h}[V], which is the ring of regular functions on the affine cone over V; hence g(a)h′(a) = g′(a)h(a), which proves the claim).

Proposition 5.12. For each f ∈ k(V) =_{df} k_{h}(V)_{0}, there is an open subset U of V where f(P) is defined, and P ↦ f(P) is a regular function on U. Every regular function φ on an open subset of V is defined by some f ∈ k(V).

Proof. Straightforward from the above discussion. Note that if the functions defined by f_{1} and f_{2} agree on an open subset of V, then f_{1} = f_{2} in k(V).

Remark 5.13. (a) The elements of k(V) = k_{h}(V)_{0} should be thought of as the analogues of meromorphic functions on a complex manifold; the regular functions on an open subset U of V are the “meromorphic functions without poles” on U. [In fact, when k = ℂ, this is more than an analogy: a nonsingular algebraic variety over ℂ defines a complex manifold, and the meromorphic functions on the manifold are precisely the rational functions on the variety. For example, the meromorphic functions on the Riemann sphere are the rational functions in z.]

(b) We shall see presently (5.20) that, for any nonzero homogeneous h ∈ k_{h}[V], D(h) is an open affine subset of V. The ring of regular functions on it is

k[D(h)] = {g/h^{m} | g homogeneous of degree m deg(h)} ∪ {0}.

We shall also see that the ring of regular functions on V itself is just k, i.e., any regular function on an irreducible (connected will do) projective variety is constant. However, if U is an open nonaffine subset of V, then the ring Γ(U, O_{V}) of regular functions can be almost anything—it needn’t even be a finitely generated k-algebra!

Morphisms from projective varieties. We describe the morphisms from a projective variety to another variety.

Proposition 5.14. The map

π: 𝔸^{n+1} − {origin} → ℙ^{n}, (a_{0}, …, a_{n}) ↦ (a_{0} : …: a_{n})

is a morphism of algebraic varieties; it is open, that is, U is open in 𝔸^{n+1} − {0} implies π(U) open in ℙ^{n}; a map α: ℙ^{n} → V with V a prevariety is regular if and only if α ∘ π is regular.

Proof. The restriction of π to D(X_{i}) is the projection

(a_{0}, …, a_{n}) ↦ (\frac{a_{0}}{a_{i}} : …: \frac{a_{n}}{a_{i}}): k^{n+1} − V (X_{i}) → U_{i},

which is the regular map of affine varieties corresponding to the map of k-algebras

k [\frac{X_{0}}{X_{i}} , …, \frac{X_{n}}{X_{i}} ] → k[X_{0}, …, X_{n}][X_{i}^{−1}].

It now follows from (3.7) that π is regular.

Let U be an open subset of k^{n+1} − {origin}, and let U′ be the union of all the lines through the origin that meet U, that is, U′ = π^{−1}π(U). Then U′ is again open in k^{n+1} − {origin}, because U′ = ∪cU, c ∈ k^{×}, and x ↦ cx is an automorphism of k^{n+1} − {origin}. The complement Z of U′ in k^{n+1} − {origin} is a closed cone, and the proof of (5.2) shows that its image is closed in ℙ^{n}; but π(U) is the complement of π(Z).

The rest of the proof is straightforward. □

Thus, the regular maps ℙ^{n} → V are just the regular maps 𝔸^{n+1} −{origin} → V factoring through ℙ^{n} (as maps of sets).

Remark 5.15. Consider polynomials F_{0}(X_{0}, …, X_{m}), …, F_{n}(X_{0}, …, X_{m}) of the same degree. The map

(a_{0} : …: a_{m}) ↦ (F_{0}(a_{0}, …, a_{m}) : …: F_{n}(a_{0}, …, a_{m}))

obviously defines a regular map to ℙ^{n} on the open subset of ℙ^{m} where not all F_{i} vanish, that is, on the set ∪D(F_{i}) = ℙ^{n} \ V (F_{1}, …, F_{n}). Its restriction to any subvariety V of ℙ^{m} will also be regular. It may be possible to extend the map to a larger set by representing it by different polynomials. Conversely, every such map arises in this way, at least locally. More precisely, there is the following result.

Proposition 5.16. Let V = V (𝔞) ⊂ ℙ^{m}, W = V (𝔟) ⊂ ℙ^{n}. A map φ: V → W is regular if and only if, for every P ∈ V, there exist polynomials F_{0}(X_{0}, …, X_{m}), …, F_{n}(X_{0}, …, X_{m}), homogeneous of the same degree, such that

Q = (b_{0} : …: b_{n}) ↦ (F_{0}(b_{0}, …, b_{m}) : …: F_{n}(b_{0}, …, b_{m}))

for all points Q = (b_{0} : …: b_{m}) in some neighbourhood of P in V (𝔞).

Proof. Straightforward. □

Example 5.17. We prove that the circle X^{2} + Y^{2} = Z^{2} is isomorphic to ℙ^{1}. After an obvious change of variables, the equation of the circle becomes C : XZ = Y^{2}. Define

φ: ℙ^{1} → C, (a : b) ↦ (a^{2} : ab : b^{2}).

For the inverse, define

ψ : C → ℙ^{1} by \{ \begin{matrix}(a : b : c) ↦ (a : b) & if a ≠ 0 \\ (a : b : c) ↦ (b : c) & if b ≠ 0 \end{matrix}.

Note that,

a ≠ 0 ≠ b, ac = b^{2} ⟹ \frac{c}{b} = \frac{b}{a}

and so the two maps agree on the set where they are both defined. Clearly, both φ and ψ are regular, and one checks directly that they are inverse.

Examples of regular maps of projective varieties. We list some of the classic maps.

Example 5.18. Let L = ∑ c_{i}X_{i} be a nonzero linear form in n + 1 variables. Then the map

(a_{0} : …: a_{n}) ↦ (\frac{a_{0}}{L(a)}, …, \frac{a_{n}}{L(a)})

is a bijection of D(L) ⊂ ℙ^{n} onto the hyperplane L(X_{1}, …, X_{n}) = 1 of 𝔸^{n+1}, with inverse

(a_{0}, …, a_{n}) ↦ (a_{0} : …: a_{n}).

Both maps are regular—for example, the components of the first map are the regular functions \frac{X_{j}}{∑ c_{i}X_{i}}. As V (L − 1) is affine, so also is D(L), and its ring of regular functions is k[\frac{X_{0}}{∑ c_{i}X_{i}} , …, \frac{X_{n}}{∑ c_{i}X_{i}}]. (This is really a polynomial ring in n variables— any (one) variable X_{j}/ ∑ c_{i}X_{i} for which c_{j} ≠ 0 can be omitted—see Lemma 4.11.)

Example 5.19. (The Veronese mapping.) Let

I = {(i_{0}, …, i_{n}) ∈ ℕ^{n+1} | ∑ i_{j} = m}.

Note that I indexes the monomials of degree m in n + 1 variables. It has \binom{m+n}{m} elements^{17}. Write ν_{n,m} = \binom{m+n}{m}−1, and consider the projective space ℙ^{ν_{n,m}} whose coordinates are indexed by I; thus a point of ℙ^{ν_{n,m}} can be written (…: b_{i_{0}...i_{n}} : …). Write ν for the map

ℙ^{n} → ℙ^{ν_{n,m}}, (a_{0} : …: a_{n}) ↦ (…: b_{i_{0}...i_{n}} : …), b_{i_{0}...i_{n}} = a_{0}^{i_{0}} … a_{n}^{i_{n}}.

For example, when n = 1 and m = 2, the Veronese map is

ℙ^{1} → ℙ^{2}, (a_{0} : a_{1}) ↦ (a_{0}^{2} : a_{0}a_{1} : a_{1}^{2}).

Its image is the curve ν(ℙ^{1}) : X_{0}X_{2} = X_{1}^{2}, and the map

(b_{2,0} : b_{1,1} : b_{0,2}) ↦ (_{(b_{1,1}:b_{0,2}) if b_{0,2}≠0}^{(b_{2,0}:b_{1,1}) if b_{2,0}≠0}

\binom{m + n}{n} = \binom{m + n − 1}{m} + \binom{m + n − 1}{m − 1}

(X + 1)^{m+n} = (X + 1)(X + 1)^{m+n−1},

17This can be proved by induction on m + n. If m = 0 = n, then \binom{0}{0} = 1, which is correct.

A general homogeneous polynomial of degree m can be written uniquely as

F (X_{0}, X_{1}, …, X_{n}) = F_{1}(X_{1}, …, X_{n}) + X_{0}F_{2}(X_{0}, X_{1}, …, X_{n})

with F_{1} homogeneous of degree m and F_{2} homogeneous of degree m − 1. But because they are the coefficients of X^{m} in and this proves what we want.

is an inverse ν(ℙ^{1}) → ℙ^{1}. (Cf. Example 5.17.)^{18} When n = 1 and m is general, the Veronese map is

ℙ^{1} → ℙ^{m}, (a_{0} : a_{1}) ↦ (a_{0}^{m} : a_{0}^{m−1}a_{1} : …: a_{1}^{m}).

I claim that, in the general case, the image of ν is a closed subset of ℙ^{ν_{n,m}} and that ν defines an isomorphism of projective varieties ν: ℙ^{n} → ν(ℙ^{n}).

First note that the map has the following interpretation: if we regard the coordinates a_{i} of a point P of ℙ^{n} as being the coefficients of a linear form L = ∑ a_{i}X_{i} (well-defined up to multiplication by nonzero scalar), then the coordinates of ν(P) are the coefficients of the homogeneous polynomial L^{m} with the binomial coefficients omitted.

As L ≠ 0 ⟹ L^{m} ≠ 0, the map ν is defined on the whole of ℙ^{n}, that is,

(a_{0}, …, a_{n}) ≠ (0, …, 0) ⟹ (…, b_{i_{0}...i_{n}}, …) ≠ (0, …, 0).

Moreover, L_{1} ≠ cL_{2} ⟹ L_{1}^{m} ≠ cL_{2}^{m}, because k[X_{0}, …, X_{n}] is a unique factorization domain, and so ν is injective. It is clear from its definition that ν is regular.

We shall see later in this chapter that the image of any projective variety under a regular map is closed, but in this case one can prove directly that ν(ℙ^{n}) is defined by the system of equations:

b_{i_{0}...i_{n}}b_{j_{0}...j_{n}} = b_{k_{0}...k_{n}}b_{ℓ_{0}...ℓ_{n}},  i_{h} + j_{h} = k_{h} + ℓ_{h}.  (*)

Obviously ℙ^{n} maps into the algebraic set defined by these equations. Conversely, let

V_{i} = {(…. : b_{i_{0}...i_{n}} : …) | b_{0...0m0...0} ≠ 0}.

Then ν(U_{i}) ⊂ V_{i} and ν^{−1}(V_{i}) = U_{i}. It is possible to write down a regular map V_{i} → U_{i} inverse to ν|U_{i} : for example, define V_{0} → ℙ^{n} to be

(…: b_{i_{0}...i_{n}} : …) ↦ (b_{m,0,... ,0} : b_{m−1,1,0,... ,0} : b_{m−1,0,1,0,... ,0} : …: b_{m−1,0,... ,0,1}).

Finally, one checks that ν(ℙ^{n}) ⊂ ∪V_{i}.

For any closed variety W ⊂ ℙ^{n}, ν|W is an isomorphism of W onto a closed subvariety ν(W) of ν(ℙ^{n}) ⊂ ℙ^{ν_{n,m}}.

Remark 5.20. The Veronese mapping has a very important property. If F is a nonzero homogeneous form of degree m ≥ 1, then V (F) ⊂ ℙ^{n} is called a hypersurface of degree m and V (F) ∩ W is called a hypersurface section of the projective variety W. When m = 1, “surface” is replaced by “plane”.

18Note that, although ℙ^{1} and ν(ℙ^{1}) are isomorphic, their homogeneous coordinate rings are not. In fact k_{h}[ℙ^{1}] = k[X_{0}, X_{1}], which is the affine coordinate ring of the smooth variety 𝔸^{2}, whereas k_{h}[ν(ℙ^{1})] = k[X_{0}, X_{1}, X_{2}]/(X_{0}X_{2} −X_{1}^{2}) which is the affine coordinate ring of the singular variety X_{0}X_{2} − X_{1}^{2}.

Now let H be the hypersurface in ℙ^{n} of degree m

∑ a_{i_{0}...i_{n}}X_{0}^{i_{0}} ⋯ X_{n}^{i_{n}} = 0,

and let L be the hyperplane in ℙ^{ν_{n,m}} defined by

∑ a_{i_{0}...i_{n}}X_{i_{0}...i_{n}}.

Then ν(H) = ν(ℙ^{n}) ∩ L, i.e.,

H(a) = 0 ⟺ L(ν(a)) = 0.

Thus for any closed subvariety W of ℙ^{n}, ν defines an isomorphism of the hypersurface section W ∩ H of V onto the hyperplane section ν(W) ∩ L of ν(W). This observation often allows one to reduce questions about hypersurface sections to questions about hyperplane sections.

As one example of this, note that ν maps the complement of a hypersurface section of W isomorphically onto the complement of a hyperplane section of ν(W), which we know to be affine. Thus the complement of any hypersurface section of a projective variety is an affine variety—we have proved the statement in (5.13b).

Example 5.21. An element A = (a_{ij}) of GL_{n+1} defines an automorphism of ℙ^{n} :

(x_{0} : …: x_{n}) ↦ (…: ∑ a_{ij}x_{j} : …);

clearly it is a regular map, and the inverse matrix gives the inverse map. Scalar matrices act as the identity map.

Let PGL_{n+1} = GL_{n+1} /k^{×}I, where I is the identity matrix, that is, PGL_{n+1} is the quotient of GL_{n+1} by its centre. Then PGL_{n+1} is the complement in ℙ^{(n+1)^{2}−1} of the hypersurface det(X_{ij}) = 0, and so it is an affine variety with ring of regular functions

k[PGL_{n+1}] = {F (…, X_{ij}, …)/ det(X_{ij})^{m} | deg(F) = m · (n + 1)} ∪ {0}.

It is an affine algebraic group.

The homomorphism PGL_{n+1} → Aut(ℙ^{n}) is obviously injective. It is also surjective. (The only reference I know for this is Mumford, Geometric Invariant Theory, p20, but there must be other more elementary references.)

Example 5.22. (The Segre mapping). This is the mapping

((a_{0} : …: a_{m}), (b_{0} : …: b_{n})) ↦ ((…: a_{i}b_{j} : …)): ℙ^{m} × ℙ^{n} → ℙ^{mn+m+n}.

The index set for ℙ^{mn+m+n} is {(i, j) | 0 ≤ i ≤ m, 0 ≤ j ≤ n}. Note that if we interprete the tuples on the left as being the coefficients of two linear forms L_{1} = ∑ a_{i}X_{i} and L_{2} = ∑ b_{j}Y_{j}, then the image of the pair is the set of coefficients of the homogeneous form of degree 2, L_{1}L_{2}. From this observation, it is obvious that the map is defined on the whole of ℙ^{m} × ℙ^{n} (L_{1} ≠ 0 ≠ L_{2} ⟹ L_{1}L_{2} ≠ 0) and is injective. On any subset of the form U_{i} × U_{j} it is defined by polynomials, and so it is regular. Again one can show that it is an isomorphism onto its image, which is the closed subset of ℙ^{mn+m+n} defined by the equations

w_{ij}w_{kl} − w_{il}w_{kj} = 0.

(See Shafarevich, I, §5, 1.) For example, the map

((a_{0} : a_{1}), (b_{0} : b_{1})) ↦ (a_{0}b_{0} : a_{0}b_{1} : a_{1}b_{0} : a_{1}b_{1}): ℙ^{1} × ℙ^{1} → ℙ^{3}

has image the hypersurface

H : W Z = XY.

The map

(w : x : y : z) ↦ ((w : y), (w : x))

is an inverse on the set where it is defined. [Incidentally, ℙ^{1} × ℙ^{1} is not isomorphic to ℙ^{2}, because in the first variety there are closed curves, e.g., two vertical lines, that don’t intersect.]

If V and W are closed subvarieties of ℙ^{m} and ℙ^{n}, then the map sends V × W isomorphically onto a closed subvariety of ℙ^{mn+m+n}. Thus products of projective varieties are projective.

There is an explicit description of the topology on ℙ^{m} × ℙ^{n} : the closed sets are the sets of common solutions of families of equations

F (X_{0}, …, X_{m}; Y_{0}, …, Y_{n}) = 0

with F separately homogeneous in the X’s and in the Y ’s.

Example 5.23. Let L_{1}, …, L_{n−d} be linearly independent linear forms in n + 1 variables; their zero set E in k^{n+1} has dimension d + 1, and so their zero set in ℙ^{n} is a d-dimensional linear space. Define π: ℙ^{n} − E → ℙ^{n−d−1} by π(a) = (L_{1}(a) : …: L_{n−d}(a)); such a map is called a projection with centre E. If V is a closed subvariety disjoint from E, then π defines a regular map V → ℙ^{n−d−1}. More generally, if F_{1}, …, F_{r} are homogeneous forms of the same degree, and Z = V (F_{1}, …, F_{r}), then a ↦ (F_{1}(a) : …: F_{r}(a)) is a morphism ℙ^{n} − Z → ℙ^{r−1}.

By carefully choosing the centre E, it is possible to project any smooth curve in ℙ^{n} isomorphically onto a curve in ℙ^{3}, and nonisomorphically (but bijectively on an open subset) onto a curve in ℙ^{2} with only nodes as singularities.^{19} For example, suppose we have a nonsingular curve C in ℙ^{3}. To project to ℙ^{2} we need three linear forms L_{0}, L_{1}, L_{2} and the centre of the projection is the point where all forms are zero. We can think of the map as projecting from the centre P_{0} onto some (projective) plane by sending the point P to the point where P_{0}P intersects the plane. To project C to a curve with only ordinary nodes as singularities, one needs to choose P_{0} so that it doesn’t lie on any tangent to C, any trisecant (line crossing the curve in 3 points), or any chord at whose extremities the tangents are coplanar. See for example Samuel, P., Lectures on Old and New Results on Algebraic Curves, Tata Notes, 1966.

19A nonsingular curve of degree d in ℙ^{2} has genus \frac{(d−1)(d−2)}{2} . Thus, if g is not of this form, a curve of genus g can’t be realized as a nonsingular plane curve.

Proposition 5.24. Let V be a projective variety, and let S be a finite set of points of V. Then S is contained in an open affine subset of V.

Proof. Find a hyperplane passing through at least one point of V but missing the elements of S, and apply 5.18. (See the Exercises.)

Remark 5.25. There is a converse: let V be a nonsingular complete (see below) irreducible variety; if every finite set of points in V is contained in an open affine subset of V then V is projective. (Conjecture of Chevalley; proved by Kleiman about 1966.)

Complete varieties. Complete varieties are the analogues in the category of varieties of compact topological spaces in the category of Hausdorff topological spaces. Recall that the image of a compact space under a continuous map is compact, and hence is closed if the image space is Hausdorff. Moreover, a Hausdorff space V is compact if and only if, for all topological spaces W, the projection q: V × W → W is closed, i.e., maps closed sets to closed sets (see Bourbaki, Topologie Générale, I, §10).

Definition 5.26. An algebraic variety V is said to be complete if for all algebraic varieties W, the projection q: V × W → W is closed.

Note that a complete variety is required to be separated, i.e., we really mean it to be a variety and not a prevariety.

Example 5.27. Consider the projection

(x, y) ↦ y: 𝔸^{1} × 𝔸^{1} → 𝔸^{1}

This is not closed; for example, the variety V : XY = 1 is closed in 𝔸^{2} but its image in 𝔸^{1} omits the origin. However, if we replace V with its closure in ℙ^{1} × 𝔸^{1}, then its projection is the whole of 𝔸^{1}.

Proposition 5.28. Let V be complete.

(a) A closed subvariety of V is complete.

(b) If V′ is complete, so also is V × V′.

(c) For any morphism α : V → W, α(V) is closed and complete; in particular, if V is a subvariety of W, then it is closed in W.

(d) If V is connected, then any regular map α: V → ℙ^{1} is either constant or onto.

(e) If V is connected, then any regular function on V is constant.

Proof. (a) Let Z be a closed subvariety of a complete variety V. Then for any variety W, Z × W is closed in V × W, and so the restriction of the closed map q: V × W → W to Z × W is also closed.

(b) The projection V × V′ × W → W is the composite of the projections

V × V′ × W → V′ × W → W,

both of which are closed.

(c) Let Γ_{α} = {(v, α(v))} ⊂ V × W be the graph of α. It is a closed subset of V × W (because W is a variety, see 3.25), and α(V) is the projection of Γ_{α} onto W. Since V is complete, the projection is closed, and so α(V) is closed, and hence is a subvariety of W. Consider

Γ_{α} × W ⟶ α(V) × W ⟶ W.

We have that Γ_{α} is complete (because it is isomorphic to V, see 3.25), and so the mapping Γ_{α} × W → W is closed. As Γ_{α} → α(V) is surjective, it follows that α(V) × W → W is also closed.

(d) Recall that the only proper closed subsets of ℙ^{1} are the finite sets, and such a set is connected if and only if it consists of a single point. Because α(V) is connected and closed, it must either be a single point (and α is constant) or ℙ^{1} (and α is onto).

(e) A regular function on V is a regular map f: V → 𝔸^{1} ⊂ ℙ^{1}. Regard it as a map into ℙ^{1}. If it isn’t constant, it must be onto, which contradicts the fact that it maps into 𝔸^{1}. □

Corollary 5.29. Consider a regular map α : V → W; if V is connected complete and W is affine, then the image of α is a point.

Proof. Embed W as a closed subvariety of 𝔸^{n}, and write α = (α_{1}, …, α_{n}). Then each α_{i} is a regular function on V, and hence is constant. □

Remark 5.30. The statement that a complete variety V is closed in any larger variety W perhaps explains the name: if V is complete, W is irreducible, and dim V = dim W, then V = W. (Contrast 𝔸^{n} ⊂ ℙ^{n}.)

Theorem 5.31. A projective variety is complete.

Lemma 5.32. A variety V is complete if and only if q: V × W → W is a closed mapping for all irreducible affine varieties W.

Proof. Straightforward. □

After (5.28a), it suffices to prove the Theorem for projective space ℙ^{n} itself; thus we have to prove that the projection W × ℙ^{n} → W is a closed mapping in the case that W is an affine variety. Note that W × ℙ^{n} is covered by the open affines W × U_{i}, 0 ≤ i ≤ n, and that a subset U of W × ℙ^{n} is closed if and only if its intersection with each W × U_{i} is closed. We shall need another more explicit description of the topology on W × ℙ^{n}.

Let A = k[W], and let B = A[X_{0}, …, X_{n}]. Note that B = A⊗_{k} k[X_{0}, …, X_{n}], and so we can view it as the ring of regular functions on W × 𝔸^{n+1}: f ⊗ g takes the value f(w) · g(a) at the point (w, a) ∈ W × 𝔸^{n+1}. The ring B has an obvious grading—a monomial aX_{0}^{i_{0}} … X_{n}^{i_{n}}, a ∈ A, has degree ∑ i_{j}—and so we have the notion of a homogeneous ideal 𝔟 ⊂ B. It makes sense to speak of the zero set V (𝔟) ⊂ W × ℙ^{n} of such an ideal. For example, for any ideal 𝔞 ⊂ A, 𝔞B is homogeneous, and V (𝔞B) = V (𝔞) × ℙ^{n}.

Lemma 5.33. (i) For each homogeneous ideal 𝔟 ⊂ B, the set V (𝔟) is closed, and every closed subset of W × ℙ^{n} is of this form.

(ii) The set V (𝔟) is empty if and only if operatornamerad(𝔟) ⊃ (X_{0}, …, X_{n}). (iii) If W is irreducible, then W = V (𝔟) for some homogeneous prime ideal 𝔟.

Proof. In the case that A = k, we proved all this on pp 90–92, and the same arguments apply in the present more general situation. For example, to see that V (𝔟) is closed, apply the criterion stated above.

The set V (𝔟) is empty if and only if the cone V^{aff}(𝔟) ⊂ W × 𝔸^{n+1} defined by 𝔟 is contained in W × {origin}. But ∑ a_{i_{0}...i_{n}}X_{0}^{i_{0}} … X_{n}^{i_{n}}, a_{i_{0}...i_{n}} ∈ k[W], is zero on W × {origin} if an only if its constant term is zero, and so

I^{aff}(W × {origin}) = (X_{0}, X_{1}, …, X_{n}).

Thus, the Nullstellensatz shows that V (𝔟) = ∅ ⟹ operatornamerad(𝔟) = (X_{0}, …, X_{n}). Conversely, if X_{i}^{N} ∈ 𝔟 for all i, then obviously V (𝔟) is empty.

For the final statement, note that if V (𝔟) is irreducible, then the closure of its inverse image in W × 𝔸^{n+1} is also irreducible, and so the ideal of functions zero on it prime. □

Proof (of 5.31). Write p for the projection W × ℙ^{n} → W. We have to show that Z closed in W ×ℙ^{n} implies p(Z) closed in W. If Z is empty, this is true, and so we can assume it to be nonempty. Then Z is a finite union of irreducible closed subsets Z_{i} of W × ℙ^{n}, and it suffices to show that each p(Z_{i}) is closed. Thus we may assume that Z is irreducible, and hence that Z = V (𝔟) with 𝔟 a prime homogeneous ideal in B = A[X_{0}, …, X_{n}].

Note that if p(Z) ⊂ W′, W′ a closed subvariety of W, then Z ⊂ W′ × ℙ^{n}—we can then replace W with W′. This allows us to assume that p(Z) is dense in W, and we now have to show that p(Z) = W.

Because p(Z) is dense in W, the image of the cone V^{aff}(𝔟) under the projection W × 𝔸^{n+1} → W is also dense in W, and so (see 2.22a) the map A → B/𝔟 is injective.

Let w ∈ W: we shall show that if w ∉ p(Z), i.e., if there does not exist a P ∈ ℙ^{n} such that (w, P) ∈ Z, then p(Z) is empty, which is a contradiction.

Let 𝔪 ⊂ A be the maximal ideal corresponding to w. Then 𝔪B + 𝔟 is a homogeneous ideal, and V (𝔪B + 𝔟) = V (𝔪B) ∩ V (𝔟) = (w × ℙ^{n}) ∩ V (𝔟), and so w will be in the image of Z unless V (𝔪B + 𝔟) ≠ ∅. But if V (𝔪B + 𝔟) = ∅, then 𝔪B + 𝔟 ⊃ (X_{0}, …, X_{n})^{N} for some N (by 5.33ii), and so 𝔪B + 𝔟 contains the set B_{N} of homogeneous polynomials of degree N. Because 𝔪B and 𝔟 are homogeneous ideals,

B_{N} ⊂ 𝔪B + 𝔟 ⟹ B_{N} = 𝔪B_{N} + B_{N} ∩ 𝔟.

In detail: the first inclusion says that an f ∈ B_{N} can be written f = g + h with g ∈ 𝔪B and h ∈ 𝔟. On equating homogeneous components, we find that f_{N} = g_{N} + h_{N}. Moreover: f_{N} = f; if g = ∑ m_{i}b_{i}, m_{i} ∈ 𝔪, b_{i} ∈ B, then g_{N} = ∑ m_{i}b_{iN}; and h_{N} ∈ 𝔟 because 𝔟 is homogeneous. Together these show f ∈ 𝔪B_{N} + B_{N} ∩ 𝔟.

Let M = B_{N}/B_{N} ∩ 𝔟, regarded as an A-module. The displayed equation says that M = 𝔪M. The argument in the proof of Nakayama’s lemma (4.17) shows that (1 + m)M = 0 for some m ∈ 𝔪. Because A → B/𝔟 is injective, the image of 1 + m in B/𝔟 is nonzero. But M = B_{N}/B_{N} ∩ 𝔟 ⊂ B/𝔟, which is an integral domain, and so the equation (1 + m)M = 0 implies that M = 0. Hence B_{N} ⊂ 𝔟, and so X_{i}^{N} ∈ 𝔟 for all i, which contradicts the assumption that Z = V (𝔟) is nonempty. □

[[See my Math 679 notes for Resultants.]]

Elimination theory. We have shown that, for any closed subset Z of ℙ^{m}×W, q(Z) is closed in W. Elimination theory^{20} is concerned with providing an algorithm for passing from the equations defining Z to the equations defining q(Z). We illustrate this in one case.

Let P = s_{0}X^{m} + s_{1}X^{m−1} + ⋯ + s_{m} and Q = t_{0}X^{n} + t_{1}X^{n−1} + ⋯ + t_{n} be polynomials. The resultant of P and Q is defined to be the determinant

| \begin{matrix}s_{0} & s_{1} & … & s_{m} \\ & s_{0} & … & & s_{m} \\ & & … & & & … \\ t_{0} & t_{1} & … & t_{n} \\ & t_{0} & … & & t_{n} \\ & & … & & & … \end{matrix} |  n-rows ; m-rows

There are n rows of s’s and m rows of t’s, so that the matrix is (m + n) × (m + n); all blank spaces are to be filled with zeros. The resultant is a polynomial in the coefficients of P and Q.

20Elimination theory became unfashionable several decades ago—one prominent algebraic geometer went so far as to announce that Theorem 5.31 eliminated elimination theory from mathematics, provoking Abhyankar, who prefers equations to abstractions, to start the chant “eliminate the eliminators of elimination theory”. With the rise of computers, it has become fashionable again.

Proposition 5.34. The resultant Res(P, Q) = 0 if and only if (i) both s_{0} and t_{0} are zero; or (ii) the two polyomials have a common root.

Proof. If (i) holds, then certainly Res(P, Q) = 0. Suppose that α is a common root of P and Q, so that there exist polynomials P_{1} and Q_{1} of degrees m − 1 and n − 1 respectively such that

P (X) = (X − α)P_{1}(X), Q(X) = (X − α)Q_{1}(X).

From these equations we find that

P (X)Q_{1}(X) − Q(X)P_{1}(X) = 0.  (*)

On equating the coefficients of X^{m+n−1}, …, X, 1 in (*) to zero, we find that the coefficients of P_{1} and Q_{1} are the solutions of a system of m + n linear equations in m + n unknowns. The matrix of coefficients of the system is the transpose of the matrix

( \begin{matrix}s_{0} & s_{1} & … & s_{m} \\ & s_{0} & … & & s_{m} \\ & & … & & & … \\ t_{0} & t_{1} & … & t_{n} \\ & t_{0} & … & & t_{n} \\ & & … & & & … \end{matrix} )

The existence of the solution shows that this matrix has determinant zero, which implies that Res(P, Q) = 0.

Conversely, suppose that Res(P, Q) = 0 but neither s_{0} nor t_{0} is zero. Because the above matrix has determinant zero, we can solve the linear equations to find polynomials P_{1} and Q_{1} satisfying (*). If α is a root of P, then it must also be a root of P_{1} or Q. If the former, cancel X − α from the left hand side of (*) and continue. As deg P_{1} < deg P, we eventually find a root of P that is not a root of P_{1}, and so must be a root of Q. □

The proposition can be restated in projective terms. We define the resultant of two homogeneous polynomials

P (X, Y) = s_{0}X^{m} + s_{1}X^{m−1}Y + ⋯ + s_{m}Y^{m}, Q(X, Y) = t_{0}X^{n} + ⋯ + t_{n}Y^{n},

exactly as in the nonhomogeneous case.

Proposition 5.35. The resultant Res(P, Q) = 0 if and only if P and Q have a common zero in ℙ^{1}.

Proof. The zeros of P (X, Y) in ℙ^{1} are of the form:

(i) (a : 1) with a a root of P (X, 1), or (ii) (1 : 0) in the case that s_{0} = 0.

Thus (5.35) is a restatement of (5.34). □

Now regard the coefficients of P and Q as indeterminants. Thus the pairs of polynomials (P, Q) are parametrized by the space 𝔸^{m+1} × 𝔸^{n+1} = 𝔸^{m+n+2}. Consider the closed subset V (P, Q) in 𝔸^{m+n+2} × ℙ^{1}. The proposition shows that its projection on 𝔸^{m+n+2} is the set defined by Res(P, Q) = 0. Thus in this case we have not only given a new proof of the theorem, but we have given an algorithm for passing from the polynomials defining the closed set to those defining its projection.

Elimination theory does this in general. Given P_{i}(T_{1}, …, T_{m}; X_{0}, …, X_{n}), homogeneous in the X_{i}, elimination theory gives an algorithm for finding polynomials R_{j}(T_{1}, …, T_{n}) such that the P_{i}(a_{1}, …, a_{m}; X_{0}, …, X_{n}) have a common zero if and only if R_{j}(a_{1}, …, a_{n}) = 0 for all j. (Our theorem only shows that the R_{j} exist.) See Cox et al, p388.

Maple can find the resultant of two polynomials in one variable: for example, entering “resultant((x + a)^{5}, (x + b)^{5}, x)” gives the answer (−a + b)^{25}. Explanation: the polynomials have a common root if and only if a = b, and this can happen in 25 ways. Macaulay doesn’t seem to know how to do more.

The rigidity theorem. The paucity of maps between projective varieties has some interesting consequences. First an observation: for any point w ∈ W, the projection map V × W → V defines an isomorphism V × {w} → V with inverse v ↦ (v, w): V → V × W (this map is regular because its components are).

Theorem 5.36 (Rigidity Theorem). Let α: V × W → U be a regular map, and assume that V is complete, that V and W are irreducible, and that U is separated. If there are points u_{0} ∈ U, v_{0} ∈ V, and w_{0} ∈ W such that

α(V × {w_{0}}) = {u_{0}} = α({v_{0}} × W)

then α(V × W) = {u_{0}}.

Proof. Let U_{0} be an open affine neighbourhood of u_{0}. Because the projection map q: V × W → W is closed, Z =_{df} q(α^{−1}(U − U_{0})) is closed in W. Note that a point w of W lies outside Z if and only α(V × {w}) ⊂ U_{0}. In particular w_{0} ∈ W − Z, and so W − Z is dense in W. As V × {w} is complete and U_{0} is affine, α(V × {w}) must be a point whenever w ∈ W − Z : in fact, α(V × {w}) = α(v_{0}, w) = {u_{0}}. Thus α is constant on the dense subset V × (W − Z) of V × W, and so is constant. □

An abelian variety is defined to be a complete connected group variety. (It is then automatically smooth, projective, and irreducible.)

Corollary 5.37. Every regular map α : A → B of abelian varieties is the composite of a homomorphism with a translation; in particular, a regular map α: A → B such that α(0) = 0 is a homomorphism.

Proof. After composing α with a translation, we may assume that α(0) = 0. Consider the map

φ: A × A → B, φ(a, a′) = α(a + a′) − α(a) − α(a′).

Then φ(A × 0) = 0 = φ(0 × A) and so φ = 0. This means that α is a homomorphism. □

Corollary 5.38. The group law on an abelian variety is commutative.

Proof. Commutative groups are distinguished by the fact that the map taking an element to its inverse is a homomorphism: if (gh)^{−1} = g^{−1}h^{−1}, then, on taking inverses, we find that gh = hg. Since the negative map, a ↦ −a: A → A, takes the identity element to itself, the preceding corollary shows that it is a homomorphism. □

Projective space without coordinates. Let E be a vector space over k of dimension n+1. Then the set ℙ(E) of lines through zero in E has a natural structure of an algebraic variety: the choice of a basis for E defines an bijection ℙ(E) → ℙ^{n}, and the inherited structure of an algebraic variety on ℙ(E) is independent of the choice of the basis. Note that in contrast to ℙ^{n}, which has n + 1 distinguished hyperplanes, namely, X_{0} = 0, …, X_{n} = 0, all hyperplanes in ℙ(E) are equal.

One can also define the structure of an algebraic variety on the set G_{n+1,r}(E) of r-dimensional subspaces in E. The resulting varieties are called Grassmanians. They are projective.

Bezout’s theorem. Let V be a hypersurface in ℙ^{n} (that is, a closed subvariety of codimension 1). For such a variety, I(V) = (F (X_{0}, …, X_{n})) with F a homogenous polynomial without repeated factors. We define the degree of V to be the degree of F.

The next theorem is one of the oldest, and most famous, in algebraic geometry.

Theorem 5.39 (Bezout). Let C and D be curves in ℙ^{2} of degrees m and n respectively. If C and D have no irreducible component in common, then they intersect in exactly mn points, counted with appropriate multiplicities.

Proof. Decompose C and D into their irreducible components. Clearly it suffices to prove the theorem for each irreducible component of C and each irreducible component of D. We can therefore assume that C and D are themselves irreducible.

We know from (1.22) that C ∩ D is of dimension zero, and so is finite. After a change of variables, we can all assume that a ≠ 0 for all points (a : b : c) ∈ C ∩ D.

Let F (X, Y, Z) and G(X, Y, Z) be the polynomials defining C and D, and write

F = s_{0}Z^{m} + s_{1}Z^{m−1} + ⋯ + s_{m}, G = t_{0}Z^{n} + t_{1}Z^{n−1} + ... + t_{n}

with s_{i} and t_{j} polynomials in X and Y of degrees i and j respectively. Clearly s_{m} ≠ 0 ≠ t_{n}, for otherwise F and G would have Z as a common factor. Let R be the resultant of F and G, regarded as polynomials in Z. It is a homogeneous polynomial of degree mn in X and Y, or else it is identically zero. If the latter occurs, then for every (a, b) ∈ k^{2}, F (a, b, Z) and G(a, b, Z) have a common zero, which contradicts the finiteness of C ∩D. Thus R is a nonzero polynomial of degree mn. Write R(X, Y) = X^{mn}R_{∗}(\frac{Y}{X}) where R_{∗}(T) is a polynomial of degree ≤ mn in T = \frac{Y}{X}.

Suppose first that deg R_{∗} = mn, and let α_{1}, …, α_{mn} be the roots of R_{∗} (some of them may be multiple). Each such root can be written α_{i} = \frac{b_{i}}{a_{i}}, and R(a_{i}, b_{i}) = 0. According to (5.35) this means that the polynomials F (a_{i}, b_{i}, Z) and G(a_{i}, b_{i}, Z) have a common root c_{i}. Thus (a_{i} : b_{i} : c_{i}) is a point on C ∩ D, and conversely, if (a : b : c) is a point on C ∩ D (so a ≠ 0), then \frac{b}{a} is a root of R_{∗}(T). Thus we see in this case, that C ∩ D has precisely mn points, provided we take the multiplicity of (a : b : c) to be the multiplicity of \frac{b}{a} as a root of R_{∗}.

Now suppose that R_{∗} has degree r < mn. Then R(X, Y) = X^{mn−r}P (X, Y) where P (X, Y) is a homogeneous polynomial of degree r not divisible by X. Obviously R(0, 1) = 0, and so there is a point (0 : 1 : c) in C ∩ D, in contradiction with our assumption. □

Remark 5.39. The above theorem has the defect that the notion of multiplicity has been too obviously chosen to make the theorem come out right. It is possible to show that the theorem holds with the following more natural definition of multiplicity. Let P be an isolated point of C ∩ D. There will be an affine neighbourhood U of P and regular functions f and g on U such that C ∩ U = V (f) and D ∩ U = V (g). We can regard f and g as elements of the local ring O_{P}, and clearly operatornamerad(f, g) = 𝔪, the maximal ideal in O_{P}. It follows that O_{P} /(f, g) is finite-dimensional over k, and we define the multiplicity of P in C ∩ D to be dim_{k}(O_{P} /(f, g)). For example, if C and D cross transversely at P, then f and g will form a system of local parameters at P, that is, (f, g) = 𝔪, and so the multiplicity is one.

The attempt to find good notions of multiplicities in very general situations has motivated much of the most interesting work in commutative algebra over the last 20 years.

Hilbert polynomials. Recall that for a projective variety V ⊂ ℙ^{n},

k_{h}[V] = k[X_{0}, …, X_{n}]/𝔟 = k[x_{0}, …, x_{n}],

where 𝔟 = I(V). We observed that 𝔟 is homogeneous, and therefore k_{h}[V] is a graded ring:

k_{h}[V] = ⊕_{n≥0}k_{h}[V]_{m},

where k_{h}[V]_{m} is the subspace generated by the monomials in the x_{i} of degree m. Clearly k_{h}[V]_{m} is a finite-dimensional k-vector space.

Theorem 5.40. There is a unique polynomial P (V, T) such that P (V, m) = dim_{k} k[V]_{m} for all m sufficiently large.

Proof. Omitted. □

Example 5.41. For V = ℙ^{n}, k_{h}[V] = k[X_{0}, …, X_{n}], and (see the footnote on p98), dim k_{h}[V]_{m} = \binom{m+n}{m} = \binom{m+n}{n}, and so

P (ℙ^{n}, T) = \binom{T + n}{n} = \frac{(T + n) ⋯ (T + 1)}{n!}.

The polynomial P (V, T) is called the Hilbert polynomial of V. Despite the notation, it depends not just on V but also on its embedding in projective space.

Theorem 5.41. Let V be a projective variety of dimension d and degree δ; then

P (V, T) = \frac{δ}{d!}T^{d} + terms of lower degree.

Proof. Omitted. □

The degree of a projective variety is the number of points in the the intersection of the variety and of a linear variety of complementary dimension (see later).

Example 5.42. Let V be the image of the Veronese map

(a_{0} : a_{1}) ↦ (a_{0}^{d} : a_{0}^{d−1}a_{1} : …: a_{1}^{d}): ℙ^{1} ⟶ ℙ^{d}.

Then k_{h}[V]_{m} can be identified with the set of homogeneous polynomials of degree m · d in two variables (look at the map 𝔸^{2} → 𝔸^{d+1} given by the same equations), which is a space of dimension dm + 1, and so

P (V, T) = dT + 1.

Thus V has dimension 1 (which we certainly knew) and degree d.

As explained in the notes, Macaulay knows how to compute Hilbert polynomials.

References: Hartshorne, I, §7; Atiyah-Macdonald, Chapter 11; Harris, Lecture

6. Finite Maps

Recall that an A-algebra B is said to be finite if it is finitely generated as an Amodule; this is equivalent to B being finitely generated as an A-algebra and integral over A.

Definition 6.1. A regular map φ: W → V is said to be finite if for all open affine subsets U of V, φ^{−1}(U) is an affine variety, and k[φ^{−1}(U)] is a finite k[U]-algebra.

Proposition 6.2. It suffices to check the condition in the definition for all subsets in an open affine covering (U_{i}) of V.

Proof. Omitted. (See Mumford, IAG, III.1, Pptn 5). □

Hence a map φ: Specm(B) → Specm(A) of affine varieties is finite if and only if B is a finite A-algebra.

Proposition 6.3. (a) For any closed subvariety Z of V, the inclusion Z ↪ V is finite.

(b) The composite of two finite morphisms is finite.

(c) The product of two finite morphisms is finite.

Proof. (a) Let U be an open affine subvariety of V. Then Z∩U is a closed subvariety of U. It is therefore affine, and the map Z ∩U → U corresponds to a map A → A/𝔞 of rings, which is obviously finite.

(b) If B is a finite A-algebra and C is a finite B-algebra, then C is a finite A-algebra: indeed, if {b_{i}} is a set of generators for B as an A-module, and {c_{j}} is a set of generators for C as a B-module, then {b_{i}c_{j}} is a set of generators for C as an A-module.

(c) If B and B′ are respectively finite A and A′-algebras, then B ⊗_{k} B′ is a finite A ⊗_{k} A′-algebra: indeed, if {b_{i}} is a set of generators for B as an A-module, and {b′_{j}} is a set of generators for B′ as an A-module, the {b_{i} ⊗ b′_{j}} is a set of generators for B ⊗_{A} B′ as an A-module. □

By way of contrast, an open immersion is rarely finite. For example, the inclusion 𝔸^{1} − {0} ↪ 𝔸^{1} is not finite because the ring k[T, T^{−1}] is not finitely generated as a k[T]-module. (Any finite set of elements has a power of T as a common denominator.)

A regular map φ: W → V is quasi-finite if it has finite fibres, that is, if φ^{−1}(P) is finite for all P ∈ V. (In general, the subvarieties φ^{−1}(P), P ∈ V, are called the fibres of the map φ.)

Proposition 6.4. A finite map φ: W → V is quasi-finite.

Proof. Let P ∈ V; we wish to show φ^{−1}(P) is finite. After replacing V with an affine neighbourhood of P, we can suppose that it is affine, and then W will be affine also. The map φ then corresponds to a map α: A → B of affine k-algebras, and a point Q of W maps to P if and only α^{−1}(𝔪_{Q}) = 𝔪_{P}. But this holds if and only if^{21} 𝔪_{Q} ⊃ α(𝔪_{P}), and so the points of W mapping to P are in one-to-one correspondence with the maximal ideals of B/α(𝔪)B. Clearly B/α(𝔪)B is generated as a k-vector space by the image of any generating set for B as an A-module, and the next lemma shows that it has only finitely many maximal ideals. □

Lemma 6.5. A finite k-algebra A has only finitely many maximal ideals.

Proof. Let 𝔪_{1}, …, 𝔪_{n} be maximal ideals in A. They are obviously coprime in pairs, and so the Chinese Remainder Theorem (see below) shows that the map

A → A/𝔪_{1} × ⋯ × A/𝔪_{n}, a ↦ (…, a mod 𝔞_{i}, …),

is surjective. It follows that dim_{k} A ≥ ∑ dim_{k}(A/𝔪_{i}) ≥ n (dimensions as k-vector spaces). □

Lemma 6.6. (Chinese Remainder Theorem). Let 𝔞_{1}, …, 𝔞_{n} be ideals in a ring A. If 𝔞_{i} is coprime to 𝔞_{j} (i.e., 𝔞_{i} + 𝔞_{j} = A) whenever i ≠ j, then the map

A → A/𝔞_{1} × ⋯ × A/𝔞_{m}

is surjective, with kernel ∏ 𝔞_{i} = ∩𝔞_{i}.

Proof. The proof is elementary (see Atiyah-MacDonald, 1.10). □

Theorem 6.7. A finite map φ: W → V is closed.

Proof. Again we can assume V and W to be affine. Let Z be a closed subset of W. The restriction of φ to Z is finite (by 6.3a and (b)), and so we can replace W with Z; we then we have to show that Im(φ) is closed. The map corresponds to a finite map of rings A → B. This will factor, A → A/𝔞 ↪ B, from which we obtain maps

Specm(B) → Specm(A/𝔞) ↪ Specm(A).

The second map identifies Specm(A/𝔞) with the closed subvariety V (𝔞) of Specm(A), and so it remains to show that the first map is surjective. This is a consequence of the next lemma. □

Lemma 6.8 (Going-Up Theorem). Let A ⊂ B be rings with B integral over A. (a) For every prime ideal 𝔭 of A, there is a prime ideal 𝔮 of B such that 𝔮∩A = 𝔭.

(b) Let 𝔭 = 𝔮 ∩ A; then 𝔭 is maximal if and only if 𝔮 is maximal.

Proof. (a) If S is a multiplicative subset of a ring A, then the prime ideals of S^{−1}A are in one-to-one correspondence with the prime ideals of A not meeting S (see 4.15). It therefore suffices to prove (a) after A and B have been replaced by S^{−1}A and S^{−1}B, where S = A − 𝔭. Thus we may assume that A is local, and that 𝔭 is its unique maximal ideal. In this case, for all proper ideals 𝔟 of B, 𝔟∩A ⊂ 𝔭 (otherwise 𝔟 ⊃ A ∋ 1). To complete the proof of (a), I shall show that for all maximal ideals 𝔫 of B, 𝔫 ∩ A = 𝔭.

21Clearly then α^{−1}(𝔪_{Q}) ⊃ 𝔪_{P}, and we know it is a maximal ideal.

Consider B/𝔫 ⊃ A/(𝔫∩A). Here B/𝔫 is a field, which is integral over its subring A/(𝔫 ∩ A), and 𝔫 ∩ A will be equal to 𝔭 if and only if A/(𝔫 ∩ A) is a field. Thus the claim follows from the next lemma.

Lemma 6.9. Let A be a subring of a field K. If K is integral over A, then A also is a field.

Proof. Let a ∈ A, a ≠ 0. Then a^{−1} ∈ K, and it is integral over A:

(a^{−1})^{n} + a_{1}(a^{−1})^{n−1} + ⋯ + a_{n} = 0, a_{i} ∈ A.

On multiplying through by a^{n−1}, we find that

a^{−1} + a_{1} + ⋯ + a_{n}a^{n−1} = 0,

from which it follows that a^{−1} ∈ A. □

Proof (of 6.8b). The ring B/𝔮 contains A/𝔭, and it is integral over A/𝔭. If 𝔮 is maximal, then (6.9) shows that 𝔭 is also. For the converse, note that any integral domain algebraic over a field is a field (it is a union of integral domains finite over k; see p5 of my Math 594 notes). □

Corollary 6.10. Let φ: W → V be finite; if V is complete, then so also is W.

Proof. Consider

W × T → V × T → T, (w, t) ↦ (φ(w), t) ↦ t.

Because W × T → V × T is finite (see 6.3c), it is closed, and because V is complete, V × T → T is closed. A composite of closed maps is closed, and therefore the projection W × T → T is closed. □

Example 6.11. (a) Project XY = 1 onto the X axis. This map is quasi-finite but not finite, because k[X, X^{−1}] is not finite over k[X].

(b) The map 𝔸^{2} − {origin} ↪ 𝔸^{2} is quasi-finite but not finite, because the inverse image of 𝔸^{2} is not affine (2.21).

(c) Let V = V (X^{n} + T_{1}X^{n−1} + ⋯ + T_{n}) ⊂ 𝔸^{n+1}, and consider the projection map

(a_{1}, …, a_{n}, x) ↦ (a_{1}, …, a_{n}): V → 𝔸^{n}.

The fibre over any point (a_{1}, …, a_{n}) ∈ 𝔸^{n} is the set of solutions of

X^{n} + a_{1}X^{n−1} + ⋯ + a_{n} = 0,

and so it has exactly n points, counted with multiplicities. The map is therefore quasi-finite; it is also finite because it corresponds to the finite map of k-algebras,

k[T_{1}, …, T_{n}] → k[T_{1}, …, T_{n}, X]/(X^{n} + T_{1}X^{n−1} + ⋯ + T_{n}).

(d) Let V = V (T_{0}X^{n} + T_{1}X^{n−1} + ⋯ + T_{n}) ⊂ 𝔸^{n+2}. The projection φ: V → 𝔸^{n+1} has finite fibres except for the fibre above (0, …, 0), which is 𝔸^{1}. The restriction φ | V − φ^{−1}(origin) is quasi-finite, but not finite. Above points of the form (0, …, 0, ∗, …, ∗) some of the roots “vanish off to ∞”. (Example (a) is a special case of this.)

(e) Let P (X, Y) = T_{0}X^{n} + T_{1}X^{n−1}Y + ... +T_{n}Y^{n}, and let V be its zero set in ℙ^{1} × (𝔸^{n+1} − {origin}). In this case, the projection map V → 𝔸^{n+1} − {origin} is finite. (Prove this directly, or apply 6.24 below.)

(f) The morphism 𝔸^{1} → 𝔸^{2}, t ↦ (t^{2}, t^{3}) is finite because the image of k[X, Y] in k[T] is k[T^{2}, T^{3}], and {1, T } is a set of generators for k[T] over this subring.

(g) The morphism 𝔸^{1} → 𝔸^{1}, a ↦ a^{m} is finite (special case of (c)).

(h) The obvious map

(𝔸^{1} with the origin doubled) → 𝔸^{1}

is quasi-finite but not finite (the inverse image of 𝔸^{1} is not affine).

Noether Normalization Theorem. This theorem sometimes allows us to reduce the proofs of statements about affine varieties to the case of 𝔸^{n}.

Theorem 6.12. For any irreducible affine algebraic variety V of a variety of dimension d, there is a finite surjective map φ: V → 𝔸^{d}.

Proof. This is a geometric re-statement of the original theorem. □

Theorem 6.13 (Noether Normalization Theorem). Let A be a finitely generated k-algebra, and assume that A is an integral domain. Then there exist elements y_{1}, …, y_{d} ∈ A that are algebraically independent over k and such that A is integral over k[y_{1}, …, y_{d}].

Proof. Let x_{1}, …, x_{n} generate A as a k-algebra. We can renumber the x_{i} so that x_{1}, …, x_{d} are algebraically independent and x_{d+1}, …, x_{n} are algebraically dependent on x_{1}, …, x_{d} (see 1.42).

Because x_{n} is algebraically dependent on x_{1}, …, x_{d}, there exists a nonzero polynomial f(X_{1}, …, X_{d}, T) such that f(x_{1}, …, x_{d}, x_{n}) = 0. Write

f(X_{1}, …, X_{d}, T) = a_{0}T^{m} + a_{1}T^{m−1} + ⋯ + a_{m}

with a_{i} ∈ k[X_{1}, …, X_{d}] (≈ k[x_{1}, …, x_{d}]). If a_{0} is a nonzero constant, we can divide through by it, and then x_{n} will satisfy a monic polynomial with coefficients in k[x_{1}, …, x_{d}], that is, x_{n} will be integral (not merely algebraic) over k[x_{1}, …, x_{d}]. The next lemma suggest how we might achieve this happy state by making a linear change of variables. □

Lemma 6.14. If F (X_{1}, …, X_{d}, T) is a homogeneous polynomial of degree r, then

F (X_{1} + λ_{1}T, …, X_{d} + λ_{d}T, T) = F (λ_{1}, …, λ_{d}, 1)T^{r} + terms of degree < r in T.

Proof. The polynomial F (X_{1}+λ_{1}T, …, X_{d}+λ_{d}T, T) is still homogeneous of degree r (in X_{1}, …, X_{d}, T), and the coefficient of the monomial T^{r} in it can be obtained by substituting 0 for each X_{i} and 1 for T. □

Proof (continued). Note that unless F (X_{1}, …, X_{d}, T) is the zero polynomial, it will always be possible to choose (λ_{1}, …, λ_{d}) so that F (λ_{1}, …, λ_{d}, 1) ≠ 0 —substituting T = 1 merely dehomogenizes the polynomial (no cancellation of terms occurs), and a nonzero polynomial can’t be zero on all of k^{n} (this can be proved by induction on the number of variables; it uses only that k is infinite).

Let F be the homogeneous part of highest degree of f, and choose (λ_{1}, …, λ_{d}) so that F (λ_{1}, …, λ_{d}, 1) ≠ 0. The lemma then shows that

f(X_{1} + λ_{1}T, …, X_{d} + λ_{d}T, T) = cT^{r} + b_{1}T^{r−1} + ⋯ + b_{0},

with c = F (λ_{1}, …, λ_{d}, 1) ∈ k^{×}, b_{i} ∈ k[X_{1}, …, X_{d}], deg b_{i} < r. On substituting x_{n} for T and x_{i} − λ_{i}x_{n} for X_{i} we obtain an equation demonstrating that x_{n} is integral over k[x_{1} − λ_{1}x_{n}, …, x_{d} − λ_{d}x_{n}]. Put x′_{i} = x_{i} − λ_{i}x_{n}, 1 ≤ i ≤ d. Then x_{n} is integral over the ring k[x′_{1}, …, x′_{d}], and it follows that A is integral over A′ = k[x′_{1}, …, x′_{d}, x_{d+1}, …, x_{n−1}]. Repeat the process for A′, and continue until the theorem is proved. □

Remark 6.15. The above proof uses only that k is infinite, not that it is algebraically closed (that’s all one needs for a nonzero polynomial not to be zero on all of k^{n}). There are other proofs that work also for finite fields (see Mumford, IAG, p4-6), but the above proof gives us the additional information that the y_{i}’s can be chosen to be linear combinations of the x_{i}. This has the following geometric interpretation: let V be a closed subvariety of 𝔸^{n} of dimension d; then there exists a linear map 𝔸^{n} → 𝔸^{d} whose restriction to V is a finite map V ↠ 𝔸^{d}.

Zariski’s main theorem. An obvious way to construct a nonfinite quasi-finite map W → V is to take a finite map W′ → V and remove a closed subset of W′. Zariski’s Main Theorem show that, when W and V are separated, every quasi-finite map arises in this way.

Theorem 6.16. (Zariski’s Main Theorem). Any quasi-finite map of varieties

φ: W → V factors into W ↪^{ι} W′ ⟶^{φ′} V with φ′ finite and ι an open immersion.

Proof. Omitted. □

Remark 6.17. Assume (for simplicity) that V and W are irreducible. The proof of the theorem provides the following description for W′: let U be an open affine subset of V; then, because φ′ is finite, φ′^{−1}(U) will be an open affine subset of W′, and, in fact, the ring k[φ′^{−1}(U)] is the integral closure of k[U] in k(W).

A regular map φ: W → V of irreducible varieties is said to be birational if it induces an isomorphism k(V) → k(W) on the fields of rational functions (that is, if it demonstrates that W and V are birationally equivalent—see p80).

Remark 6.18. One may ask how a birational regular map φ: W → V can fail to be an isomorphism. Here are three examples.

(a) The inclusion of an open subset into a variety is birational.

(b) The map 𝔸^{1} → C, t ↦ (t^{2}, t^{3}), is birational. Here C is the cubic Y^{2} = X^{3}, and the map k[C] → k[𝔸^{1}] = k[T] identifies k[C] with the subring k[T^{2}, T^{3}] of k[T]. Both rings have k(T) as their fields of fractions.

(c) For any smooth variety V and point P ∈ V, there is a regular birational map φ: V′ → V such that the restriction of φ to V′ − φ^{−1}(P) is an isomorphism onto V − P, but φ^{−1}(P) is the projective space attached to the vector space T_{P} (V). See the section on “Blowing up points” in §10.

The next result says that, if we require the target variety to be normal (thereby excluding example (b)), and we require the map to be quasi-finite (thereby excluding example (c)), then we are left with (a).

Corollary 6.19. Let φ: W → V be a birational regular map of irreducible varieties. Assume

(a) V is normal, and

(b) φ is quasi-finite. Then φ is an isomorphism of W onto an open subset of V.

Proof. Factor φ as in the theorem. For each open affine subset U of V, k[φ′^{−1}(U)] is the integral closure of k[U] in k(W). But k(W) = k(V) (because φ is birational), and k[U] is integrally closed in k(V) (because V is normal), and so U = φ′^{−1}(U) (as varieties). It follows that W′ = V. □

Remark 6.20. Let W and V be irreducible varieties, and let φ: W → V be a dominating map. It induces a map k(V) ↪ k(W), and if dim W = dim V, then k(W) is a finite extension of k(V). We shall see later that, if n is the separable degree of k(V) over k(W), then there is an open subset U of W such that φ is n : 1 on U, i.e., for P ∈ φ(U), φ^{−1}(P) has exactly n points.

Now suppose that φ is a bijective regular map W → V. We shall see later that this implies that W and V have the same dimension. Assume:

(i) k(W) is separable over k(V); (ii) V is normal.

From (i) and the preceding remark, we find that φ is birational, and from (ii) and the corollary, we find that φ is an isomorphism of W onto an open subset of V; as it is surjective, it must be an isomorphism of W onto V. We conclude: a bijective regular map φ: W → V satisfying the conditions (i) and (ii) is an isomorphism.

Remark 6.21. The full name of the theorem is “the main theorem of Zariski’s paper Trans. AMS, 53 (1943), 490-532”. Zariski’s original statement is that in (6.19). Grothendieck proved it in the stronger form (6.16) for all schemes. There is a good discussion of the theorem in Mumford, IAG, III.9.

Fibre products. Consider a variety S and two regular maps φ: V → S and ψ: W → S. Then the set

V ×_{S} W =_{df} {(v, w) ∈ V × W | φ(v) = ψ(w)}

is a closed subvariety of V × W, called the fibred product of V and W over S. Note that if S consists of a single point, then V ×_{S} W = V × W.

Write φ′ for the map (v, w) ↦ w: V ×_{S} W → W and ψ′ for the map (v, w) ↦ v: V ×_{S} W → V. We then have a commutative diagram:

V ×_{S} W ⟶^{ψ′} W
↓^{φ′}  ↓^{φ}
W ⟶^{ψ} S.

The fibred product has the following universal property: consider a pair of regular maps α: T → V, β: T → W; then

(α, β) = t ↦ (α(t), β(t)): T → V × W

factors through V ×_{S} W (as a map of sets) if and only if φα = ψβ, in which case (α, β) is regular (because it is regular as a map into V × W).

Suppose V, W, and S are affine, and let A, B, and R be their rings of regular functions. Then A ⊗_{R} B has the same universal property as V ×_{S} W, except with the directions of the arrows reversed. Since both objects are uniquely determined by their universal properties, this shows that k[V ×_{S} W] = A ⊗_{R} B/𝔑, where 𝔑 is the nilradical of A ⊗_{R} B (that is, the set of nilpotent elements of A ⊗_{R} B).

The map φ′ in the above diagram is called the base change of φ with respect to ψ. For any point P ∈ S, the base change of φ: V → S with respect to P ↪ S is the map φ^{−1}(P) → P induced by φ.

Proposition 6.22. The base change of a finite map is finite.

Proof. We may assume that all the varieties concerned are affine. Then the statement becomes: if A is a finite R-algebra, then A ⊗_{R} B/𝔑 is a finite B-algebra, which is obvious. □

Proper maps. A regular map φ: V → S of varieties is said to be proper if it is “universally closed”, that is, if for all maps T → S, the base change φ′: V ×_{S} T → T of φ is closed. Note that a variety V is complete if and only if the map V → {point} is proper. From its very definition, it is clear that the base change of a proper map is proper. In particular, if φ: V → S is proper, then φ^{−1}(P) is a complete variety for all P ∈ S.

Proposition 6.23. A finite map of varieties is proper.

Proof. The base change of a finite map is finite, and hence closed. □

The next result (whose proof requires Zariski’s Main Theorem) gives a purely geometric criterion for a regular map to be finite.

Proposition 6.24. A proper quasi-finite map φ: W → V of varieties is finite.

Proof. Factor φ into W ↪^{ι} W′ ⟶^{α} W with α finite and ι an open immersion. Factor ι into

W ⟶^{w↦(w,ιw)} W ×_{V} W′ ⟶^{(w,w′)↦w′} W′.

The image of the first map is Γ_{ι}, which is closed because W′ is a variety (see 3.25; W′ is separated because it is finite over a variety — see the Homework). Because φ is proper, the second map is closed. Hence ι is an open immersion with closed image. It follows that its image is a connected component of W′, and that W is isomorphic to that connected component. □

If W and V are curves, then any surjective map W → V is closed. Thus it is easy to give examples of closed surjective quasi-finite nonfinite maps. For example, the map

a ↦ a^{n}: 𝔸^{1} − {1} → 𝔸^{1},

which corresponds to the map on rings

k[T] → k[T, T^{−1}], T ↦ T^{n},

is such a map.

7. Dimension Theory

Recall that to an irreducible variety V, we attach a field k(V) that is the field of fractions of k[U] for any open affine subvariety U of V; it is also the field of fractions of O_{P} for any point P in V. We defined the dimension of V to be the transcendence degree of k(V) over k. Note that, directly from this definition, dim V = dim U for any open subvariety U of V. Also, if W → V is a finite surjective map, then dim W = dim V (because k(W) is a finite field extension of k(V)).

When V is not irreducible, we defined the dimension of V to be the maximum dimension of an irreducible component of V, and we said that V is pure of dimension d if the dimensions of the irreducible components are all equal to d.

In §1 and §3 we proved the following results: (7.1a) The dimension of a linear subvariety of 𝔸^{n} (that is, a subvariety defined by

linear equations) has the value predicted by linear algebra (see 1.20d, 4.11). In particular, dim 𝔸^{n} = n. As a consequence, dim ℙ^{n} = n.

(7.1b) Let Z be a proper closed subset of 𝔸^{n}; then Z has pure codimension one in 𝔸^{n} if and only if I(Z) is generated by a single nonconstant polynomial. Such a variety is called an affine hypersurface (see 1.21 and 4.25)^{22}.

(7.1c) If V is irreducible and Z is a proper closed subset of V, then dim Z < dim V (see 1.22).

Affine varieties. The fundamental additional result that we need is that, when we impose additional polynomial conditions on an algebraic set, the dimension doesn’t go down by more than linear algebra would suggest.

Theorem 7.2. Let V be an irreducible affine variety, and let f ∈ k[V]. If f is not zero or a unit in k[V], then V (f) is pure of dimension dim(V) − 1.

Alternatively we can state this as follows: let V be a closed subvariety of 𝔸^{n} and let F ∈ k[X_{1}, …, X_{n}]; then

V ∩ V (F) = V (case that F is identically zero on V); or V ∩ V (F) = ∅ (case that F has no zeros on V); or V ∩ V (F) has pure codimension 1 in V.

We can also state it in terms of the algebras: let A be an affine k-algebra; let f ∈ A be neither zero nor a unit, and let 𝔭 be a prime ideal that is minimal among those containing (f); then

tr deg_{k}A/𝔭 = tr deg_{k}A − 1.

Proof (of 7.2). We begin the proof of the Theorem. Note that we know it already in the case that V = 𝔸^{n} (see 7.1b).

We first show that it suffices to prove the theorem in the case that V (f) is irreducible.

22The cautious will check that we didn’t use 4.18 or 4.19 in the proof of 4.25

Suppose Z_{0}, …, Z_{n} are the irreducible components of V (f). We can choose a point P ∈ Z_{0} that does not lie on any other Z_{i} (otherwise the decomposition V (f) = ∪Z_{i} would be redundant). As Z_{1}, …, Z_{n} are closed, there is an open neighbourhood U of P, which we can take to be affine, that does not meet any Z_{i} except Z_{0}. Now V (f|U) = Z_{0} ∩ U, which is irreducible.

As V (f) is irreducible, operatornamerad(f) is a prime ideal 𝔭 ⊂ k[V]. According to the Noether normalization theorem (6.12), there is a finite surjective map π: V → 𝔸^{d}, which realizes k(V) is a finite extension of the field k(𝔸^{d}). The idea of the proof is to show that π(V) is the zero set of a single element f_{0} ∈ k[𝔸^{d}], and to use that we already know the theorem for 𝔸^{d}.

Lemma 7.3. Let A be an integral domain, and let L be a finite extension of the field of fractions K of A. If α ∈ L is integral over A, then so also is Nm_{L/K}α. Hence, if A is integrally closed (e.g., if A is a unique factorization domain), then Nm_{L/K}α ∈ A. In this last case, α divides Nm_{L/K} α in the ring A[α].

Proof. Let g(X) be the minimum polynomial of α over K,

g(X) = X^{r} + a_{r−1}X^{r−1} + ⋯ + a_{0}.

Then Nmα = a0^{\frac{n}{r}}, where n = [L: K]. In some extension field E of L, g(X) will split

g(X) = ∏(X − α_{i}), α_{1} = α, ∏ α_{i} = ±a_{0}.

Because α is integral over A, g(X) has coefficients in A (see 1.33), and so each α_{i} is integral over A. Since the set of elements of E integral over A is a subring of E, it follows that Nm α is integral over A.

Now suppose A is integrally closed, so that a = Nm α ∈ A. From the equation

0 = α(α^{r−1} + a_{r−1}α^{r−2} + ⋯ + a_{1}) + a_{0}

we see that α divides a_{0} in A[α], and therefore it also divides a = a0^{\frac{n}{r}}. □

Proof (of 7.2 continued). Let f_{0} = Nm_{k(V)/k(𝔸^{d})} f. According to the lemma, f_{0} lies in k[𝔸^{d}], and I claim that 𝔭 ∩ k[𝔸^{d}] = operatornamerad(f_{0}). The lemma shows that f divides f_{0} in k[V], and so f_{0} ∈ (f) ⊂ 𝔭. Hence operatornamerad(f_{0}) ⊂ 𝔭 ∩ k[𝔸^{d}]. For the reverse inclusion, suppose that g ∈ 𝔭 ∩ k[𝔸^{d}]. Then g ∈ operatornamerad(f), and so g^{m} = fh for some h ∈ k[V], m ∈ ℕ. Taking norms, we find that g^{me} = Nm(fh) = f_{0} · Nm(h) ∈ (f_{0}), where e = [k(V): k(𝔸^{n})], which proves the claim.

The inclusion k[V] ↪ k[𝔸^{d}] therefore induces an inclusion

k[𝔸^{d}]/operatornamerad(f_{0}) = k[𝔸^{d}]/𝔭 ∩ k[𝔸^{d}] ↪ k[V]/𝔭,

which makes k[V]/𝔭 into a finite algebra over k[𝔸^{d}]/operatornamerad(f_{0}). Hence

dim V (𝔭) = dim V (f_{0}).

Clearly f ≠ 0 ⟹ f_{0} ≠ 0, and f_{0} ∈ 𝔭 ⟹ f_{0} is not a nonzero constant. Therefore dim V (f_{0}) = d − 1 by (7.1b). □

Corollary 7.4. Let V be an irreducible variety, and let Z be a maximal proper closed irreducible subset of V. Then dim(Z) = dim(V) − 1.

Proof. For any open affine subset U of V meeting Z, dim U = dim V and dim U ∩ Z = dim Z. We may therefore assume that V itself is affine. Let f be a nonzero regular function on V vanishing on Z, and let V (f) be the set of zeros of f (in V). Then Z ⊂ V (f) ⊂ V, and Z must be an irreducible component of V (f) for otherwise it wouldn’t be maximal in V. Thus we can apply the theorem to obtain that dim Z = dim V − 1. □

Corollary 7.5 (Topological Characterization of Dimension). Suppose V is irreducible and that

V ⫌ V_{1} ⫌ ⋯ ⫌ V_{d} ≠ ∅

is a maximal chain of closed irreducible subsets of V. Then dim(V) = d. (Maximal means that the chain can’t be refined.)

Proof. From (7.4) we know that

dim V = dim V_{1} + 1 = dim V_{2} + 2 = ⋯ = dim V_{d} + d = d. □

Remark 7.6. (a) Recall that the Krull dimension of a ring A is the sup of the lengths of chains of prime ideals in A. It may be infinite, even when A is Noetherian (for an example of this, see Nagata, Local Rings, Appendix A.1). However a local Noetherian ring has finite Krull dimension, and so

Krull dim A = sup Krull dim A_{𝔪} (sup over maximal ideals of A). In Nagata’s nasty example, there is a sequence of maximal ideals 𝔪_{1}, 𝔪_{2}, 𝔪_{3}, ... in A such that Krull dim A_{𝔪_{i}} tends to infinity.

The corollary shows that, when V is affine, dim V = Krull dim k[V], but it shows much more. Note that each V_{i} in a maximal chain (as above) has dimension d − i, and that any closed irreducible subset of V of dimension d − i occurs as a V_{i} in a maximal chain. These facts translate into statements about ideals in affine k-algebras that do not hold for all Noetherian rings. For example, we have that Krull dim A_{𝔪} is the same for all maximal ideals of A, that is, all maximal ideals in A have the same height (we have proved 4.19). Moreover, if 𝔭 is an ideal in k[V] with height i, then there is a maximal (i.e., nonrefinable) chain of prime ideals

(0) ⫋ 𝔭_{1} ⫋ 𝔭_{2} ⫋ ⋯ ⫋ 𝔭_{d} ⫋ k[V]

with 𝔭_{i} = 𝔭.

(b) Now that we know that the two notions of dimension coincide, we can restate (7.2) as follows: let A be an affine k-algebra; let f ∈ A be neither zero nor a unit, and let 𝔭 be a prime ideal that is minimal among those containing (f); then

Krull dim A/𝔭 =Krull dim A − 1.

This statement does hold for all Noetherian local rings (see A-M 11.18), and is called Krull’s principal ideal theorem.

Corollary 7.7. Let V be an irreducible variety, and let Z be an irreducible component of V (f_{1}, … f_{r}), where the f_{i} are regular functions on V. Then codim(Z) ≤ r.

Proof. As in the proof of (7.4), we can assume V to be affine. We use induction on r. Because Z is a closed irreducible subset of V (f_{1}, … f_{r−1}), it is contained in some irreducible component Z′ of V (f_{1}, … f_{r−1}). By induction, codim(Z′) ≤ r − 1. Also Z is an irreducible component of Z′ ∩ V (f_{r}) because

Z ⊂ Z′ ∩ V (f_{r}) ⊂ V (f_{1}, …, f_{r})

and Z is a maximal closed irreducible subset of V (f_{1}, …, f_{r}). If f_{r} vanishes identically on Z′, then Z = Z′ and codim(Z) = codim(Z′) ≤ r−1; otherwise, the theorem shows that Z has codimension 1 in Z′, and codim(Z) = codim(Z′) + 1 ≤ r. □

Proposition 7.8. Let V and W be closed subvarieties of 𝔸^{n}; for any (nonempty) irreducible component Z of V ∩ W,

dim(Z) ≥ dim(V) + dim(W) − n;

that is, codim(Z) ≤ codim(V) + codim(W).

Proof. In the course of the proof of (3.26), we showed that V ∩ W is isomorphic to Δ ∩ (V × W), and this is defined by the n equations X_{i} = Y_{i} in V × W. Thus the statement follows from (7.7). □

Remark 7.9. (a) The example

\{ \begin{matrix}X^{2} + Y^{2} & = Z^{2} \\ Z & = 0 \end{matrix}

shows that Proposition 7.8 becomes false if one only looks at real points. Also, that the pictures we draw can mislead.

(b) The statement of (7.8) is false if 𝔸^{n} is replaced by an arbitrary affine variety. Consider for example the affine cone V

X_{1}X_{4} − X_{2}X_{3} = 0.

It contains the planes,

Z : X_{2} = 0 = X_{4}; Z = {(∗, 0, ∗, 0)}
Z′ : X_{1} = 0 = X_{3}; Z′ = {(0, ∗, 0, ∗)}

and Z ∩ Z′ = {(0, 0, 0, 0)}. Because V is a hypersurface in 𝔸^{4}, it has dimension 3, and each of Z and Z′ has dimension 2. Thus

codim Z ∩ Z′ = 3 ≦̸ 1 + 1 = codim Z + codim Z′.

The proof of (7.14) fails because the diagonal in V × V cannot be defined by 3 equations (it takes the same 4 that define the diagonal in 𝔸^{4})—thus the diagonal is not a set-theoretic complete intersection.

Remark 7.10. In (7.7), the components of V (f_{1}, …, f_{r}) need not all have the same dimension, and it is possible for all of them to have codimension < r without any of the f_{i} being redundant.

For example, let V be the same affine cone as in the above remark. Note that V (X_{1}) ∩ V is a union of the planes:

V (X_{1}) ∩ V = {(0, 0, ∗, ∗)} ∪ {(0, ∗, 0, ∗)}.

Both of these have codimension 1 in V (as required by (7.2)). Similarly, V (X_{2}) ∩ V is the union of two planes,

V (X_{2}) ∩ V = {(0, 0, ∗, ∗)} ∪ {(∗, 0, ∗, 0)},

but V (X_{1}, X_{2}) ∩ V consists of a single plane {(0, 0, ∗, ∗)}: it is still of codimension 1 in V, but if we drop one of two equations from its defining set, we get a larger set.

Proposition 7.11. Let Z be a closed irreducible subvariety of codimension r in an affine variety V. Then there exist regular functions f_{1}, …, f_{r} on V such that Z is an irreducible component of V (f_{1}, …, f_{r}) and all irreducible components of V (f_{1}, …, f_{r}) have codimension r.

Proof. We know that there exists a chain of closed irreducible subsets

V ⊃ Z_{1} ⊃ ⋯ ⊃ Z_{r} = Z

with codim Z_{i} = i. We shall show that there exist f_{1}, …, f_{r} ∈ k[V] such that, for all s ≤ r, Z_{s} is an irreducible component of V (f_{1}, …, f_{s}) and all irreducible components of V (f_{1}, …, f_{s}) have codimension s.

We prove this by induction on s. For s = 1, take any f_{1} ∈ I(Z_{1}), f_{1} ≠ 0, and apply Theorem 7.2. Suppose f_{1}, …, f_{s−1} have been chosen, and let Y_{1} = Z_{s−1}, …, Y_{m}, be the irreducible components of V (f_{1}, …, f_{s−1}). We seek an element f_{s} that is identically zero on Z_{s} but is not identically zero on any Y_{i}—for such an f_{s}, all irreducible components of Y_{i} ∩ V (f_{s}) will have codimension s, and Z_{s} will be an irreducible component of Y_{1} ∩ V (f_{s}). But Y_{i} ⊈ Z_{s} for any i (Z_{s} has smaller dimension than Y_{i}), and so I(Z_{s}) ⊈ I(Y_{i}). Now the prime avoidance lemma (see below) tells us that there is an element f_{s} ∈ I(Z_{s}) such that f_{s} ∉ I(Y_{i}) for any i, and this is the function we want. □

Lemma 7.12 (Prime Avoidance Lemma). If an ideal 𝔞 of a ring A is not contained in any of the prime ideals 𝔭_{1}, …, 𝔭_{r}, then it is not contained in their union.

Proof. We may assume that none of the prime ideals is contained in a second, because then we could omit it. Fix an i_{0} and choose elements

f_{i} ∈ 𝔭_{i}, f_{i} ∉ 𝔭_{i_{0}}, i ≠ i_{0}, g ∈ 𝔞, g ∉ 𝔭_{i_{0}}.

Then

h_{i} =_{df} f_{1}f_{2} … f_{i−1}f_{i+1} … f_{r}g

lies in 𝔭_{i} for i ≠ i_{0} and in 𝔞, but it is not in 𝔭_{i_{0}} (here we use that 𝔭_{i_{0}}is prime). The element ∑ h_{i} is therefore in 𝔞 but not in any 𝔭_{i}. □

Remark 7.13. The proposition shows that for a prime ideal 𝔭 in an affine k-algebra, if 𝔭 has height r, then there exist elements f_{1}, …, f_{r} ∈ A such that 𝔭 is minimal among the prime ideals containing (f_{1}, …, f_{r}). This statement is true for all Noetherian local rings.

Remark 7.14. The last proposition shows that a curve C in 𝔸^{3} is an irreducible component of V (f_{1}, f_{2}) for some f_{1}, f_{2} ∈ k[X, Y, Z]. In fact C = V (f_{1}, f_{2}, f_{3}) for suitable polynomials f_{1}, f_{2}, and f_{3}—this is an exercise in Shafarevich (I, §7, Ex. 8), where it is stated that it is not known whether two polynomials always suffice (see also Hartshorne, I, exercise 2.17). Apparently this is still not known in general, although progress has been made (see Kunz, Introduction to Commutative Algebra and Algebraic Geometry, p136). The union of two skew lines in ℙ^{3} can’t be defined by two polynomials (ibid. p140), but it is unknown whether all connected curves in ℙ^{3} can be defined by two polynomials. Macaulay (the man, not the program) showed that for every r ≥ 1, there is a curve C in 𝔸^{3} such that I(C) requires at least r generators (see the same exercise in Hartshorne for a curve whose ideal can’t be generated by 2 elements).

In general, a closed variety V of codimension r in 𝔸^{n} (resp. ℙ^{n}) is said to be a set-theoretic complete intersection if there exist r polynomials f_{i} ∈ k[X_{1}, …, X_{n}] (resp. homogeneous polynomials f_{i} ∈ k[X_{0}, …, X_{n}]) such that

V = V (f_{1}, …, f_{r}).

Such a variety is said to be an ideal-theoretic complete intersection if the f_{i} can be chosen so that I(V) = (f_{1}, …, f_{r}). Chapter V of Kunz’s book is concerned with the question of when a a variety is a complete intersection. Obviously there are many ideal-theoretic complete intersections, but most of the varieties one happens to be interested in turn out not to be. For example, no abelian variety of dimension > 1 is an ideal-theoretic complete intersection (being an ideal-theoretic complete intersection imposes constraints on the cohomology of the variety, which are not fulfilled in the case of abelian varieties).

Let P be a point on an irreducible variety V ⊂ 𝔸^{n}. Then (7.11) shows that there is a neighbourhood U of P in 𝔸^{n} and functions f_{1}, …, f_{r} on U such that U ∩ V = V (f_{1}, …, f_{r}) (zero set in U). Thus U ∩ V is a set-theoretic complete intersection in U. One says that V is a local complete intersection at P ∈ V if there is an open affine neighbourhood U of P in 𝔸^{n} such that I(V ∩ U) can be generated by r regular functions on U. Note that

ideal-theoretic complete intersection ⟹ local complete intersection at all P. It is not difficult to show that a variety is a local complete intersection at every nonsingular point.

Proposition 7.15. Let Z be a closed subvariety of codimension r in variety V, and let P be a point of Z that is nonsingular when regarded both as a point on Z and as a point on V. Then there is an open affine neighbourhood U of P and regular functions f_{1}, …, f_{r} on U such that Z ∩ U = V (f_{1}, …, f_{r}).

Proof. By assumption

dim_{k} T_{P} (Z) = dim Z = dim V − r = dim_{k} T_{P} (V) − r.

There exist functions f_{1}, …, f_{r} contained in the ideal of O_{P} corresponding to Z such that T_{P} (Z) is the subspace of T_{P} (V) defined by the equations

(df_{1})_{P} = 0, …, (df_{r})_{P} = 0.

All the f_{i} will be defined on some open affine neighbourhood U of P (in V), and clearly Z is the only component of Z′ =_{df} V (f_{1}, …, f_{r}) (zero set in U) passing through P. After replacing U by a smaller neighbourhood, we can assume that Z′ is irreducible. As f_{1}, …, f_{r} ∈ I(Z′), we must have T_{P} (Z′) ⊂ T_{P} (Z), and therefore dim Z′ ≤ dim Z. But I(Z′) ⊂ I(Z ∩ U), and so Z′ ⊃ Z ∩ U. These two facts imply that Z′ = Z ∩ U. □

Proposition 7.16. Let V be an affine variety such that k[V] is a unique factorization domain. Then every pure closed subvariety Z of V of codimension one is principal, i.e., I(Z) = (f) for some f ∈ k[V].

Proof. In (4.25) we proved this in the case that V = 𝔸^{n}, but the argument only used that k[𝔸^{n}] is a unique factorization domain. □

Example 7.17. The condition that k[V] is a unique factorization domain is definitely needed. Again let V be the cone

X_{1}X_{4} − X_{2}X_{3} = 0

in 𝔸^{4} and let Z and Z′ be the planes

Z = {(∗, 0, ∗, 0)} Z′ = {(0, ∗, 0, ∗)}.

Then Z ∩ Z′ = {(0, 0, 0, 0)}, which has codimension 2 in Z′. If Z = V (f) for some regular function f on V, then V (f|Z′) = {(0, …, 0)}, which is impossible (because it has codimension 2, which violates 7.2). Thus Z is not principal, and so

k[X_{1}, X_{2}, X_{3}, X_{4}]/(X_{1}X_{4} − X_{2}X_{3})

is not a unique factorization domain.

Projective varieties. The results for affine varieties extend to projective varieties with one important simplification: if V and W are projective varieties of dimensions r and s in ℙ^{n} and r + s ≥ n, then V ∩ W ≠ ∅.

Theorem 7.16. Let V = V (𝔞) ⊂ ℙ^{n} be a projective variety of dimension ≥ 1, and let f ∈ k[X_{0}, …, X_{n}] be homogeneous, nonconstant, and ∉ 𝔞; then V ∩ V (f) is nonempty and of pure codimension 1.

Proof. Since the dimension of a variety is equal to the dimension of any dense open affine subset, the only part that doesn’t follow immediately from (7.1) is the fact that V ∩ V (f) is nonempty. Let V^{aff}(𝔞) be the zero set of 𝔞 in 𝔸^{n+1} (that is, the affine cone over V). Then V^{aff}(𝔞) ∩ V^{aff}(f) is nonempty (it contains (0, …, 0)), and so it has codimension 1 in V^{aff}(𝔞). Clearly V^{aff}(𝔞) has dimension ≥ 2, and so V^{aff}(𝔞) ∩ V^{aff}(f) has dimension ≥ 1. This implies that the polynomials in 𝔞 have a zero in common with f other than the origin, and so V (𝔞) ∩ V (f) ≠ ∅. □

Corollary 7.17. Let f_{1}, ⋯, f_{r} be homogeneous nonconstant elements of k[X_{0}, …, X_{n}]; and let Z be an irreducible component of V ∩ V (f_{1}, … f_{r}). Then codim(Z) ≤ r, and if dim(V) ≥ r, then V ∩ V (f_{1}, … f_{r}) is nonempty.

Proof. Induction on r, as before. □

Corollary 7.18. Let α: ℙ^{n} → ℙ^{m} be regular; if m < n, then α is constant.

Proof. Let π: 𝔸^{n+1} − {origin} → ℙ^{n} be the map (a_{0}, …, a_{n}) ↦ (a_{0} : …: a_{n}). Then απ is regular, and one sees easily that there exist polynomials F_{0}, …, F_{m} ∈ k[X_{0}, …, X_{n}] such that απ is the map

(a_{0}, …, a_{n}) ↦ (F_{0}(a) : …: F_{m}(a)).

As απ factors through ℙ^{n}, the F_{i} must be homogeneous of the same degree. Note that

α(a_{0} : …: a_{n}) = (F_{0}(a) : …: F_{m}(a)).

If m < n and the F_{i} are nonconstant, then (7.17) shows they have a common zero and so α is not defined on all of ℙ^{n}. Hence the F′_{i}s must be constant. □

Proposition 7.19. Let Z be a closed irreducible subvariety of V; if codim(Z) = r, then there exist homogeneous polynomials f_{1}, …, f_{r} in k[X_{0}, …, X_{n}] such that Z is an irreducible component of V ∩ V (f_{1}, …, f_{r}).

Proof. Use the same argument as in the proof (7.9). □

Proposition 7.20. Every pure closed subvariety Z of ℙ^{n} of codimension one is principal, i.e., I(Z) = (f) for some f homogeneous element of k[X_{0}, …, X_{n}].

Proof. Follows from the affine case. □

Corollary 7.21. Let V and W be closed subvarieties of ℙ^{n}; if dim(V)+dim(W) ≥ n, then V ∩ W ≠ ∅, and every irreducible component of it has codim(Z) ≤ codim(V) + codim(W).

Proof. Write V = V (𝔞) and W = V (𝔟), and consider the affine cones V′ = V (𝔞) and W′ = W (𝔟) over them. Then

dim(V′) + dim(W′) = dim(V) + 1 + dim(W) + 1 ≥ n + 2.

As V′ ∩ W′ ≠ ∅, V′ ∩ W′ has dimension ≥ 1, and so it contains a point other than the origin. Therefore V ∩ W ≠ ∅. The rest of the statement follows from the affine case. □

Proposition 7.22. Let V be a closed subvariety of ℙ^{n} of dimension r < n; then there is a linear projective variety E of dimension n − r − 1 (that is, E is defined by r + 1 independent linear forms) such that E ∩ V = ∅.

Proof. Induction on r. If r = 0, then V is a finite set, and the next lemma shows that there is a hyperplane in k^{n+1} not meeting V. □

Lemma 7.23. Let W be a vector space of dimension d over an infinite field k, and let E_{1}, …, E_{r} be a finite set of nonzero subspaces of W. Then there is a hyperplane H in W containing none of the E_{i}.

Proof. Pass to the dual space V of W. The problem becomes that of showing V is not a finite union of proper subspaces E_{i}^{∨}. Replace each E_{i}^{∨} by a hyperplane H_{i} containing it. Then H_{i} is defined by a nonzero linear form L_{i}. We have to show that ∏ L_{j} is not identically zero on V. But this follows from the statement that a polynomial in n variables, with coefficients not all zero, can not be identically zero on k^{n}. (See the first homework exercise.)

Suppose r > 0, and let V_{1}, …, V_{s} be the irreducible components of V. By assumption, they all have dimension ≤ r. The intersection E_{i} of all the linear projective varieties containing V_{i} is the smallest such variety. The lemma shows that there is a hyperplane H containing none of the nonzero E_{i}; consequently, H contains none of the irreducible components V_{i} of V, and so each V_{i} ∩ H is a pure variety of dimension ≤ r−1 (or is empty). By induction, there is an linear subvariety E′ not meeting V ∩ H. Take E = E′ ∩ H. □

Let V and E be as in the theorem. If E is defined by the linear forms L_{0}, …, L_{r} then the projection a ↦ (L_{0}(a): ⋯: L_{r}(a)) defines a map V → ℙ^{r}. We shall see later that this map is finite, and so it can be regarded as a projective version of the Noether normalization theorem.

8. Regular Maps and Their Fibres.

Consider again the regular map φ: 𝔸^{2} → 𝔸^{2}, (x, y) ↦ (x, xy). We have seen that its image

C = (𝔸^{2} − {y-axis}) ∪ {(0, 0)}

is neither open nor closed, and, in fact, is not even locally closed. The fibre

φ^{−1}(x, y) = \{ \begin{matrix}single point & if x ≠ 0 \\ empty set & if x = 0, y ≠ 0 \\ \{0\} × 𝔸^{1} & if x = 0, y = 0. \end{matrix}

From this unpromising example, it would appear that it is not possible to say anything about the image of a regular map, nor about the dimension or number of elements in its fibres. However, it turns out that (almost) everything that can go wrong already goes wrong for this map. We shall show:

(a) the image of a regular map is a finite union of locally closed sets;

(b) the dimensions of the fibres can jump only on closed subsets;

(c) the number of elements (if finite) in the fibres can drop only on closed subsets, provided the map is finite, the target variety is normal, and k has characteristic zero.

Constructible sets. Let W be a topological space. A subset C of W is said to constructible if it is a finite union of sets of the form U ∩ Z with U open and Z closed. Obviously, if C is constructible and V ⊂ W, then C ∩ V is constructible. A constructible set in 𝔸^{n} is definable by a finite number of polynomials; more precisely, it is defined by a finite number of statements of the form

f(X_{1}, ⋯, X_{n}) = 0, g(X_{1}, ⋯, X_{n}) ≠ 0

combined using only “and” and “or” (or, better, statements of the form f = 0 combined using “and”, “or”, and “not”). The next proposition shows that a constructible set C that is dense in an irreducible variety V must contain a nonempty open subset of V. Contrast ℚ, which is dense in ℝ (real topology), but does not contain an open subset of ℝ, or any infinite subset of 𝔸^{1} that omits an infinite set.

Proposition 8.1. Let C be a constructible set whose closure C̄ is irreducible. Then C contains a nonempty open subset of C̄.

Proof. We are given that C = ∪(U_{i} ∩ Z_{i}) with each U_{i} open and each Z_{i} closed. We may assume that each set U_{i} ∩ Z_{i} in this decomposition is nonempty. Clearly C̄ ⊂ ∪Z_{i}, and as C̄ is irreducible, it must be contained in one of the Z_{i}. For this i

C ⊃ U_{i} ∩ Z_{i} ⊃ U_{i} ∩ C̄ ⊃ U_{i} ∩ C ⊃ U_{i} ∩ (U_{i} ∩ Z_{i}) = U_{i} ∩ Z_{i}.

Thus U_{i} ∩ Z_{i} = U_{i} ∩ C̄ is a nonempty open subset of C̄ contained in C. □

Theorem 8.2. A regular map φ: W → V maps constructible sets to constructible sets. In particular, if U is a nonempty open subset of W, then φ(U) contains a nonempty open subset of its closure in V.

The key result we shall need from commutative algebra is the following. (In the next two results, A and B are arbitrary commutative rings—they need not be k-algebras.)

Proposition 8.3. Let A ⊂ B be integral domains with B finitely generated as an algebra over A, and let b be a nonzero element of B. Then there exists an element a ≠ 0 in A with the following property: every homomorphism α: A → Ω from A into an algebraically closed field Ω such that α(a) ≠ 0 can be extended to a homomorphism β: B → Ω such that β(b) ≠ 0.

Consider, for example, the rings k[X] ⊂ k[X, X^{−1}]. A homomorphism α: k[X] → k extends to a homomorphism k[X, X^{−1}] → k if and only if α(X) ≠ 0. Therefore, for b = 1, we can take a = X. In the application we make of Proposition 8.3, we only really need the case b = 1, but the more general statement is needed so that we can prove it by induction.

Lemma 8.4. Let B ⊃ A be integral domains, and assume B = A[t] ≈ A[T]/𝔞. Let 𝔠 ⊂ A be the set of leading coefficients of the polynomials in 𝔞. Then every homomorphism α: A → Ω from A into an algebraically closed field Ω such that α(𝔠) ≠ 0 can be extended to a homomorphism of B into Ω.

Proof. Note that 𝔠 is an ideal in A. If 𝔞 = 0, then 𝔠 = 0, and there is nothing to prove (in fact, every α extends). Thus we may assume 𝔞 ≠ 0. Let f = a_{m}T^{m} + ⋯ + a_{0} be a nonzero polynomial of minimum degree in 𝔞 such that α(a_{m}) ≠ 0. Because B ≠ 0, we have that m ≥ 1.

Extend α to a homomorphism α̃: A[T] → Ω[T] by sending T to T. The Ω-submodule of Ω[T] generated by α̃(𝔞) is an ideal (because T · ∑ c_{i}α̃(g_{i}) = ∑ c_{i}α̃(g_{i}T)). Therefore, unless α̃(𝔞) contains a nonzero constant, it generates a proper ideal in Ω[T], which will have a zero c in Ω. The homomorphism

A[T] \underrightarrow{α̃} Ω[T] ⟶ Ω, T ↦ T ↦ c

then factors through A[T]/𝔞 = B and extends α.

In the contrary case, 𝔞 contains a polynomial

g(T) = b_{n}T^{n} + ⋯ + b_{0}, α(b_{i}) = 0 (i > 0), α(b_{0}) ≠ 0.

On dividing f(T) into g(T) we find that

a_{m}^{d}g(T) = q(T)f(T) + r(T), d ∈ ℕ, q, r ∈ A[T], deg r < m.

On applying α̃ to this equation, we obtain

α(a_{m})^{d}α(b_{0}) = α̃(q)α̃(f) + α̃(r).

Because α̃(f) has degree m > 0, we must have α̃(q) = 0, and so α̃(r) is a nonzero constant. After replacing g(T) with r(T), we may assume n < m. If m = 1, such a g(T) can’t exist, and so we may suppose m > 1 and (by induction) that the Lemma holds for smaller values of m.

For h(T) = c_{r}T^{r} + c_{r−1}T^{r−1} + ⋯ + c_{0}, let h′(T) = c_{r} + ⋯ + c_{0}T^{r}. Then the A-module generated by the polynomials T^{s}h′(T), s ≥ 0, h ∈ 𝔞, is an ideal 𝔞′ in A[T]. Moreover, 𝔞′ contains a nonzero constant if and only if 𝔞 contains a nonzero polynomial cT^{r}, which implies t = 0 and A = B (since B is an integral domain).

If 𝔞′ does not contain nonzero constants, then set B′ = A[T]/𝔞′ = A[t′]. Then 𝔞′ contains the polynomial g′ = b_{n} + ⋯ + b_{0}T^{n}, and α(b_{0})≠ 0. Because deg g′ < m, the induction hypothesis implies that α extends to a homomorphism B′ → Ω. Therefore, there is a c ∈ Ω such that, for all h(T) = c_{r}T^{r} + c_{r−1}T^{r−1} + ⋯ + c_{0} ∈ 𝔞,

h′(c) = α(c_{r}) + α(c_{r−1})c + ⋯ + c_{0}c^{r} = 0.

On taking h = g, we see that c = 0, and on taking h = f, we obtain the contradiction α(a_{m}) = 0. □

Proof (of 8.2). Suppose that we know the proposition in the case that B is generated by a single element, and write B = A[x_{1}, …, x_{n}]. Then there exists an element b_{n−1} such that any homomorphism α: A[x_{1}, …, x_{n−1}] → Ω such that α(b_{n−1}) ≠ 0 extends to a homomorphism β: B → Ω such that β(b) ≠ 0. Continuing in this fashion, we obtain an element a ∈ A with the required property.

Thus we may assume B = A[x]. Let 𝔞 be the kernel of the homomorphism X ↦ x, A[X] → A[x].

Case (i). The ideal 𝔞 = (0). Write

b = f(x) = a_{0}x^{n} + a_{1}x^{n−1} + ⋯ + a_{n}, a_{i} ∈ A,

and take a = a_{0}. If α: A → Ω is such that α(a_{0}) ≠ 0, then there exists a c ∈ Ω such that f(c) ≠ 0, and we can take β to be the homomorphism ∑ d_{i}x^{i} ↦ ∑ α(d_{i})c^{i}.

Case (ii). The ideal 𝔞 ≠ (0). Let f(T) = a_{m}T^{m} + ⋯, a_{m} ≠ 0, be an element of 𝔞 of minimum degree. Let h(T) ∈ A[T] represent b. Since b ≠ 0, h ∉ 𝔞. Because f is irreducible over the field of fractions of A, it and h are coprime over that field. Hence there exist u, v ∈ A[T] and c ∈ A − {0} such that

uh + vf = c.

It follows now that ca_{m} satisfies our requirements, for if α(ca_{m}) ≠ 0, then α can be extended to β: B → Ω by the previous lemma, and β(u(x) · b) = β(c) ≠ 0, and so β(b) ≠ 0. □

Aside 8.5. In case (ii) of the above proof, both b and b^{−1} are algebraic over A, and so there exist equations

a_{0}b^{m} + ⋯ + a_{m} = 0, a_{i} ∈ A, a_{0} ≠ 0;

a′_{0}b^{−n} + ⋯ + a′_{n} = 0, a′_{i} ∈ A, a′_{0} ≠ 0.

One can show that a = a_{0}a′_{0} has the property required by the Proposition—see Atiyah and MacDonald, 5.23.

Proof (of 8.2). We first prove the “in particular” statement of the Theorem. By considering suitable open affine coverings of W and V, one sees that it suffices to prove this in the case that both W and V are affine. If W_{1}, …, W_{r} are the irreducible components of W, then the closure of φ(W) in V, φ(W)^{−} = φ(W_{1})^{−}∪…∪φ(W_{r})^{−}, and so it suffices to prove the statement in the case that W is irreducible. We may also replace V with φ(W)^{−}, and so assume that both W and V are irreducible. Then φ corresponds to an injective homomorphism A → B of affine k-algebras. For some b ≠ 0, D(b) ⊂ U. Choose a as in the lemma. Then for any point P ∈ D(a), the homomorphism f ↦ f(P): A → k extends to a homomorphism β: B → k such that β(b) ≠ 0. The kernel of β is a maximal ideal corresponding to a point Q ∈ D(b) lying over P.

We now prove the theorem. Let W_{i} be the irreducible components of W. Then C ∩W_{i} is constructible in W_{i}, and φ(W) is the union of the φ(C ∩W_{i}); it is therefore constructible if the φ(C ∩ W_{i}) are. Hence we may assume that W is irreducible. Moreover, C is a finite union of its irreducible components, and these are closed in C; they are therefore constructible. We may therefore assume that C also is irreducible; C̄ is then an irreducible closed subvariety of W.

We shall prove the theorem by induction on the dimension of W. If dim(W) = 0, then the statement is obvious because W is a point. If C̄ ≠ W, then dim(C̄) < dim(W), and because C is constructible in C̄, we see that φ(C) is constructible (by induction). We may therefore assume that C̄ = W. But then C̄ contains a nonempty open subset of W, and so the case just proved shows that φ(C) contains an nonempty open subset U of its closure. Replace V be the closure of φ(C), and write

φ(C) = U ∪ φ(C ∩ φ^{−1}(V − U)).

Then φ^{−1}(V −U) is a proper closed subset of W (the complement of V −U is dense in V and φ is dominating). As C ∩ φ^{−1}(V − U) is constructible in φ^{−1}(V − U), the set φ(C ∩ φ^{−1}(V − U)) is constructible in V by induction, which completes the proof. □

The fibres of morphisms. We wish to examine the fibres of a regular map φ: W → V. Clearly, we can replace V by the closure of φ(W) in V and so assume φ to be dominating.

Theorem 8.6. Let φ: W → V be a dominating regular map of irreducible varieties. Then

(a) dim(W) ≥ dim(V);

(b) if P ∈ φ(W), then

dim(φ^{−1}(P)) ≥ dim(W) − dim(V)

for every P ∈ V, with equality holding exactly on a nonempty open subset U of V.

(c) The sets

V_{i} = {P ∈ V | dim(φ^{−1}(P)) ≥ i}

are closed φ(W).

Example 8.7. Consider the subvariety W ⊂ V × 𝔸^{m} defined by r linear equations

∑_{j=1}^{m} a_{ij}X_{j} = 0, a_{ij} ∈ k[V], i = 1, …, r,

and let φ be the projection W → V. For P ∈ V, φ^{−1}(P) is the set of solutions of

∑_{j=1}^{m} a_{ij}(P)X_{j} = 0, a_{ij}(P) ∈ k, i = 1, …, r,

and so its dimension is m − rank(a_{ij}(P)). Since the rank of the matrix (a_{ij}(P)) drops on closed subsets, the dimension of the fibre jumps on closed subsets.

Proof. (a) Because the map is dominating, there is a homomorphism k(V) ↪ k(W), and obviously tr deg_{k}k(V) ≤ tr deg_{k}k(W) (an algebraically independent subset of k(V) remains algebraically independent in k(W)).

(b) In proving the first part of (b), we may replace V by any open neighbourhood of P. In particular, we can assume V to be affine. Let m be the dimension of V. From (7.11) we know that there exist regular functions f_{1}, …, f_{m} such that P is an irreducible component of V (f_{1}, …, f_{m}). After replacing V by a smaller neighbourhood of P, we can suppose that P = V (f_{1}, …, f_{m}). Then φ^{−1}(P) is the zero set of the regular functions f_{1} ∘ φ, …, f_{m} ∘ φ, and so (if nonempty) has codimension ≤ m in W (see 7.7). Hence

dim φ^{−1}(P) ≥ dim W − m = dim(W) − dim(V).

In proving the second part of (b), we can replace both W and V with open affine subsets. Since φ is dominating, k[V] → k[W] is injective, and we may regard it as an inclusion (we identify a function x on V with x ∘ φ on W). Then k(V) ⊂ k(W). Write k[V] = k[x_{1}, …, x_{M}] and k[W] = k[y_{1}, …, y_{N}], and suppose V and W have dimensions m and n respectively. Then k(W) has transcendence degree n − m over k(V), and we may suppose that y_{1}, …, y_{n−m} are algebraically independent over k[x_{1}, …, x_{m}], and that the remaining y_{i} are algebraic over k[x_{1}, …, x_{m}, y_{1}, …, y_{n−m}]. There are therefore relations

F_{i}(x_{1}, …, x_{m}, y_{1}, …, y_{n−m}, y_{i}) = 0, i = n − m + 1, …, N.  (*)

with F_{i}(X_{1}, …, X_{m}, Y_{1}, …, Y_{n−m}, Y_{i}) a nonzero polynomial. We write ȳ_{i} for the restriction of y_{i} to φ^{−1}(P). Then

k[φ^{−1}(P)] = k[ȳ_{1}, …, ȳ_{N}].

The equations (*) give an algebraic relation among the functions x_{1}, …, y_{i} on W. When we restrict them to φ^{−1}(P), they become equations:

F_{i}(x_{1}(P), …, x_{m}(P), ȳ_{1}, …, ȳ_{n−m}, ȳ_{i}) = 0, i = n − m + 1, …, N.  (**)

If these are nontrivial algebraic relations, i.e., if none of the polynomials

F_{i}(x_{1}(P), …, x_{m}(P), Y_{1}, …, Y_{n−m}, Y_{i})

is identically zero, then the transcendence degree of k(ȳ_{1}, …, ȳ_{N}) over k will be ≤ n − m.

Thus, regard F_{i}(x_{1}, …, x_{m}, Y_{1}, …, Y_{n−m}, Y_{i}) as a polynomial in the Y ’s with coefficients polynomials in the x’s. Let V_{i} be the closed subvariety of V defined by the simultaneous vanishing of the coefficients of this polynomial—it is a proper closed subset of V. Let U = V −∪V_{i}—it is a nonempty open subset of V. If P ∈ U, then none of the polynomials F_{i}(x_{1}(P), …, x_{m}(P), Y_{1}, …, Y_{n−m}, Y_{i}) is identically zero, and so for P ∈ U, the dimension of φ^{−1}(P) is ≤ n − m, and hence = n − m by (a).

Finally, if for a particular point P, dim φ^{−1}(P) = n − m, then one can modify the above argument to show that the same is true for all points in an open neighbourhood of P.

(c) We prove this by induction on the dimension of V —it is obviously true if dim V = 0. We know from (b) that there is an open subset U of V such that

dim φ^{−1}(P) = n − m ⟺ P ∈ U.

Let Z be the complement of U in V; thus Z = V_{n−m+1}. Let Z_{1}, …, Z_{r} be the irreducible components of Z. On applying the induction to the restriction of φ to the map φ^{−1}(Z_{j}) → Z_{j} for each j, we obtain the result. □

Proposition 8.8. Let φ: W → V be a regular surjective closed mapping of varieties (e.g., W complete or φ finite). If V is irreducible and all the fibres φ^{−1}(P) are irreducible of dimension n, then W is irreducible of dimension dim(V) + n.

Proof. Let Z be a closed irreducible subset of W, and consider the map φ|Z: Z → V; it has fibres (φ|Z)^{−1}(P) = φ^{−1}(P) ∩ Z. There are three possibilities.

(i) φ(Z) ≠ V. Then φ(Z) is a proper closed subset of V.

(ii) φ(Z) = V, dim(Z) < n + dim(V). Then (b) of (8.6) shows that there is a nonempty open subset U of V such that for P ∈ U,

dim(φ^{−1}(P) ∩ Z) = dim(Z) − dim(V) < n;

thus for P ∈ U, φ^{−1}(P) ⊈ Z. (iii) φ(Z) = V, dim(Z) ≥ n + dim(V). Then (b) of (8.6) shows that

dim(φ^{−1}(P) ∩ Z) ≥ dim(Z) − dim(V) ≥ n

for all P; thus φ^{−1}(P) ⊂ Z for all P ∈ V, and so Z = W; moreover dim Z = n.

Now let Z_{1}, …, Z_{r} be the irreducible components of W. I claim that (iii) holds for at least one of the Z_{i}. Otherwise, there will be an open subset U of V such that for P in U, φ^{−1}(P) ⊈ Z_{i} for any i, but φ^{−1}(P) is irreducible and φ^{−1}(P) = ∪(φ^{−1}(P)∪Z_{i}), and so this is impossible. □

The fibres of finite maps. Let φ: W → V be a finite dominating morphism of irreducible varieties. Then dim(W) = dim(V), and so k(W) is a finite field extension of k(V). Its degree is called the degree of the map φ.

Lemma 8.9. An integral domain A is integrally closed if and only if A_{𝔪} is integrally closed for all maximal ideals 𝔪 of A.

Proof. ⟹ : If A is integrally closed, then so is S^{−1}A for any multiplicative subset S (not containing 0), because if

b^{n} + c_{1}b^{n−1} + ⋯ + c_{n} = 0, c_{i} ∈ S^{−1}A,

then there is an s ∈ S such that sc_{i} ∈ A for all i, and then

(sb)^{n} + (sc_{1})(sb)^{n−1} + ⋯ + s^{n}c_{n} = 0,

demonstrates that sb ∈ A, whence b ∈ S^{−1}A.

⇐: If c is integral over A, it is integral over each A_{𝔪}, hence in each A_{𝔪}, and A = ∩A_{𝔪} (if c ∈ ∩A_{𝔪}, then the set of a ∈ A such that ac ∈ A is an ideal in A, not contained in any maximal ideal, and therefore equal to A itself). □

Thus the following conditions on an irreducible variety V are equivalent:

(a) for all P ∈ V, O_{P} is integrally closed;

(b) for all irreducible open affines U of V, k[U] is integrally closed;

(c) there is a covering V = ∪V_{i} of V by open affines such that k[V_{i}] is integrally

closed for all i. An irreducible variety V satisfying these conditions is said to be normal. We also call a disjoint union of such varieties normal. Thus a variety V is normal if and only if O_{P} is an integrally closed integral domain for all P ∈ V.

Theorem 8.10. Let φ: W → V be a finite surjective regular map of irreducible varieties, and assume that V is normal.

(a) For all P ∈ V, #φ^{−1}(P) ≤ deg(φ).

(b) The set of points P of V such that #φ^{−1}(P) = deg(φ) is an open subset of V, and it is nonempty if k(W) is separable over k(V).

(c) If k has characteristic zero, then the sets V_{i} = {P ∈ V | #φ^{−1}(P) ≤ i} are closed in V.

The next examples show that we need W to be separated in (a), that we need k(W) to be separable over k(V) in (b), and that we need V to be normal in (a).

Example 8.11. (a) Consider the map

{𝔸^{1} with origin doubled } → 𝔸^{1}.

The degree is one and that map is one-to-one except at the origin where it is twoto-one.

(b) Consider the Frobenius map φ: 𝔸^{n} → 𝔸^{n}, (a_{1}, …, a_{n}) ↦ (a_{1}^{p}, …, a_{n}^{p}), where p = chark. This map has degree p^{n} but it is one-to-one. The field extension corresponding to the map is

k(X_{1}, …, X_{n}) ⊃ k(X_{1}^{p}, …, X_{n}^{p})

which is purely inseparable.

(c) Let C be the curve Y^{2} = X^{3} + X^{2}, and let φ: 𝔸^{1} → C be the map t ↦ (t^{2} − 1, t(t^{2} − 1)). The map corresponds to the inclusion k[x, y] ↪ k[T] and is of degree one. The map is one-to-one except that the points t = ±1 both map to 0. The ring k[x, y] is not integrally closed; in fact k[T] is its integral closure in its field of fractions.

Lemma 8.12. Let Q_{1}, …, Q_{r} be distinct points on an affine variety V. Then there is a regular function f on V taking distinct values at the Q_{i}.

Proof. We can embed V into 𝔸^{n}, and then it suffices to prove the statement with V = 𝔸^{n}. This we leave as an exercise to the reader. □

Proof of 8.10. In proving (a) of the theorem, we may assume that V and W are affine, and so the map corresponds to a finite map of k-algebras, k[V] → k[W]. Let φ^{−1}(P) = {Q_{1}, …, Q_{r}}. According to the lemma, there exists an f ∈ k[W] taking distinct values at the Q_{i}. Let

F (T) = T^{m} + a_{1}T^{m−1} + ⋯ + a_{m}

be the minimum polynomial of f over k(V). It has degree m ≤ [k(W): k(V)] = deg φ, and it has coefficients in k[V] because V is normal (see 1.33). Now F (f) = 0 implies F (f(Q_{i})) = 0, i.e.,

f(Q_{i})^{m} + a_{1}(P) · f(Q_{i})^{m−1} + ⋯ + a_{m}(P) = 0.

Therefore the f(Q_{i}) are all roots of a single polynomial of degree m, and so r ≤ m ≤ deg(φ).

In order to prove the first part of (b), we show that, if there is a point P ∈ V such that φ^{−1}(P) has deg(φ) elements, then the same is true for all points in an open neighbourhood of P. Choose f etc. as in the last paragraph. Then the polynomial

T^{m} + a_{1}(P) · T^{m−1} + ⋯ + a_{m}(P) = 0  (*)

has r = deg φ distinct roots, and so r = m = deg φ. Consider the discriminant disc F of F. Because (*) has distinct roots, disc(F)(P) ≠ 0, and so disc(F) is nonzero on an open neighbourhood U of P. The factorization

k[V] → k[V][T]/(F) ⟶^{T ↦f} k[W]

gives a factorization

W → Specm(k[V][T]/(F)) → V.

Each point P′ ∈ U has exactly m inverse images under the second map, and the first map is finite and dominating, and therefore surjective (recall that a finite map is closed). This proves that φ^{−1}(P′) has at least deg(φ) points for P′ ∈ U, and part (a) of the theorem then implies that it has exactly deg(φ) points.

We now show that if the field extension is separable, then there exists a point such that #φ^{−1}(P) has deg φ elements. Because k(W) is separable over k(V), there exists a f ∈ k[W] such that k(V)[f] = k(W). Its minimum polynomial F has degree deg(φ) and its discriminant is a nonzero element of k[V]. The diagram

W → Specm(A[T]/(F)) → V

shows that #φ^{−1}(P) ≥ deg(φ) for P a point such that disc(f)(P) ≠ 0. □

When k(W) is separable over k(V), then φ is said to be separable.

Correction: We have proved, under the hypotheses of the theorem, that V_{d−1}, d = deg φ, is a closed subset of V, and is a proper closed subset when φ is separable, but it doesn’t seem to be true that V_{i} is always closed, i < d − 1. The obvious induction argument fails because V_{d−1} need not be normal, and Example 8.11c shows that this condition is essential. It may be true if φ is assumed to be flat.

Lines on surfaces. As an application of some of the above results, we consider the problem of describing the set of lines on a surface of degree m in ℙ^{3}. To avoid possible problems, we assume for the rest of this chapter that k has characteristic zero.

We first need a way of describing lines in ℙ^{3}. Recall that we can associate with each projective variety V ⊂ ℙ^{n} an affine cone over Ṽ in k^{n+1}. This allows us to think of points in ℙ^{3} as being one-dimensional subspaces in k^{4}, and lines in ℙ^{3} as being two-dimensional subspaces in k^{4}. To such a subspace W ⊂ k^{4}, we can attach a one-dimensional subspace Λ^{2}W in Λ^{2}k^{4} ≈ k^{6}, that is, to each line L in ℙ^{3}, we can attach point p(L) in ℙ^{5}. Not every point in ℙ^{5} should be of the form p(L)— heuristically, the lines in ℙ^{3} should form a four-dimensional set. (Fix two planes in ℙ^{3}; giving a line in ℙ^{3} corresponds to choosing a point on each of the planes.) We shall show that there is natural one-to-one correspondence between the set of lines in ℙ^{3} and the set of points on a certain hyperspace Π ⊂ ℙ^{5}. Rather than using exterior algebras, I shall usually give the old-fashioned proofs.

Let L be a line in ℙ^{3} and let x = (x_{0} : x_{1} : x_{2} : x_{3}) and y = (y_{0} : y_{1} : y_{2} : y_{3}) be distinct points on L. Then

p(L) = (p_{01} : p_{02} : p_{03} : p_{12} : p_{13} : p_{23}) ∈ ℙ^{5}, p_{ij} = | \begin{matrix}x_{i} & x_{j} \\ y_{i} & y_{j} \end{matrix} | ,

depends only on L. The p_{ij} are called the Plücker coordinates of L, after Plücker (1801-1868).

In terms of exterior algebras, write e_{0}, e_{1}, e_{2}, e_{3} for the canonical basis for k^{4}, so that x, regarded as a point of k^{4} is ∑ x_{i}e_{i}, and y = ∑ y_{i}e_{i}; then ∧^{2}k^{4} is a 6-dimensional vector space with basis e_{i}^{∧}e_{j}, 0 ≤ i < j ≤ 3, and x^{∧}y = ∑ p_{ij}e_{i}^{∧}e_{j} with p_{ij} given by the above formula.

We define p_{ij} for all i, j, 0 ≤ i, j ≤ 3. Thus p_{ij} = −p_{ji}.

Lemma 8.9. The line L can be recovered from p(L) as follows:

L = {(∑_{j} a_{j}p_{0j} : ∑_{j} a_{j}p_{1j} : ∑_{j} a_{j}p_{2j} : ∑_{j} a_{j}p_{3j}) | (a_{0} : a_{1} : a_{2} : a_{3}) ∈ ℙ^{3}}.

Proof. Let L̃ be the cone over L in k^{4}—it is a two-dimensional subspace of k^{4}—and let x = (x_{0}, x_{1}, x_{2}, x_{3}) and y = (y_{0}, y_{1}, y_{2}, y_{3}) be two linearly independent vectors in L̃. Then

L̃ = {f(y)x − f(x)y | f: k^{4} → k linear}.

Write f = ∑ a_{j}X_{j}; then

f(y)x − f(x)y = (∑ a_{j}p_{0j}, ∑ a_{j}p_{1j}, ∑ a_{j}p_{2j}, ∑ a_{j}p_{3j}).

□

Lemma 8.10. The point p(L) lies on the quadric Π ⊂ ℙ^{5} defined by the equation

X_{01}X_{23} − X_{02}X_{13} + X_{03}X_{12} = 0.

Proof. Direct calculation, or (when char ≠ 2) note that

0 = | \begin{matrix}x_{0} & x_{1} & x_{2} & x_{3} \\ y_{0} & y_{1} & y_{2} & y_{3} \\ x_{0} & x_{1} & x_{2} & x_{3} \\ y_{0} & y_{1} & y_{2} & y_{3} \end{matrix} | = 2(p_{01}p_{23} − p_{02}p_{13} + p_{03}p_{12})

(expansion in terms of 2 × 2 minors). □

Lemma 8.11. Every point of Π is of the form p(L) for a unique line L.

Proof. Assume p_{03} ≠ 0; then the line through the points (0 : p_{01} : p_{02} : p_{03}) and (p_{03} : p_{13} : p_{23} : 0) has Plücker coordinates

\begin{matrix}(−p_{01}p_{03} : −p_{02}p_{03} : −p_{03}^{2} : \underbrace{p_{01}p_{23} − p_{02}p_{13}}_{−p_{03}p_{12}} : −p_{03}p_{13} : −p_{03}p_{23}) \\ = (p_{01} : p_{02} : p_{03} : p_{12} : p_{13} : p_{23}). \end{matrix}

A similar construction works when one of the other coordinates is nonzero, and this way we get inverse maps. □

Thus we have a canonical one-to-one correspondence

{lines in ℙ^{3}} ↔ {points on Π};

that is, we have identified the set of lines in ℙ^{3} with the points of an algebraic variety. We may now use the methods of algebraic geometry to study the set. [This is a special case of the Grassmanians mentioned on p108.]

We next consider the set of homogeneous polynomials of degree m in 4 variables,

F (X_{0}, X_{1}, X_{2}, X_{3}) = ∑_{i_{0}+i_{1}+i_{2}+i_{3}=m} a_{i_{0}i_{1}i_{2}i_{3}}X_{0}^{i_{0}} … X_{3}^{i_{3}}.

We don’t distinguish two polynomials if one is a nonzero multiple of the other.

Lemma 8.12. The set of homogeneous polynomials of degree m in 4 variables is a vector space of dimension \binom{3+m}{m}.

Proof. See a previous footnote. □

Let ν = \binom{3+m}{m} − 1 = \frac{(m+1)(m+2)(m+3)}{6} − 1; then we have a surjective map
ℙ^{ν} → {surfaces of degree m in ℙ^{3}},
(…: a_{i_{0}i_{1}i_{2}i_{3}} : …) ↦ V (F), F = ∑ a_{i_{0}i_{1}i_{2}i_{3}}X_{0}^{i_{0}} ⋯ X_{3}^{i_{3}}.

The map is not quite injective—for example, X^{2}Y and XY^{2} define the same surface—but nevertheless, we can (somewhat loosely) think of the points of ℙ^{ν} as being (possible degenerate) surfaces of degree m in ℙ^{3}.

Let Γ_{m} ⊂ Π × ℙ^{ν} ⊂ ℙ^{5} × ℙ^{ν} be the set of pairs (L, F) consisting of a line L in ℙ^{3} lying on the surface F (X_{0}, X_{1}, X_{2}, X_{3}) = 0.

Theorem 8.13. The set Γ_{m} is a closed irreducible subset of Π × ℙ^{ν}; it is therefore a projective variety. The dimension of Γ_{m} is \frac{m(m+1)(m+5)}{6} + 3.

Example 8.14. Take m = 1. Then the dimension of Γ_{1} is 5. Since there are ∞^{3} planes in ℙ^{3}, and each has ∞^{2} lines on it, this seems to be correct.

Proof. We first show that Γ_{m} is closed. Let

p(L) = (p_{01} : p_{02} : …) F = ∑ a_{i_{0}i_{1}i_{2}i_{3}}X_{0}^{i_{0}} ⋯ X_{3}^{i_{3}}.

From (8.13) we see that L lies on the surface F (X_{0}, X_{1}, X_{2}, X_{3}) = 0 if and only if

F (∑ b_{j}p_{0j} : ∑ b_{j}p_{1j} : ∑ b_{j}p_{2j} : ∑ b_{j}p_{3j}) = 0, all (b_{0}, …, b_{3}) ∈ k^{4}.

Expand this out as a polynomial in the b_{j}’s with coefficients polynomials in the a_{i_{0}i_{1}i_{2}i_{3}} and p_{ij}’s. Then the condition is satisfied by all b ∈ k^{4} if and only if the coefficients of the polynomial are all zero. But each coefficient is of the form

P (…, a_{i_{0}i_{1}i_{2}i_{3}}, …; p_{01}, p_{02} : …)

with P homogeneous separately in the a’s and p’s, and so the set is closed in Π×ℙ^{ν} (cf. the discussion in 5.22).

To complete the proof, we shall apply Proposition 8.8 to the projection map

(L, F)  Γ_{m} ⊂ Π × ℙ^{ν}
↓  ↓ φ
L  Π

For L ∈ Π, φ^{−1}(L) consists of the homomogeneous polynomials of degree m such that L ⊂ V (F) (taken up to nonzero scalars). After a change of coordinates, we can assume that L is the line

\{ \begin{matrix}X_{0} & = 0 \\ X_{1} & = 0, \end{matrix}

i.e., L = {(0, 0, ∗, ∗)}. Then L lies on F (X_{0}, X_{1}, X_{2}, X_{3}) = 0 if and only if X_{0} or X_{1} occurs in each nonzero monomial term in F, i.e.,

F ∈ φ^{−1}(L) ⟺ a_{i_{0}i_{1}i_{2}i_{3}} = 0 whenever i_{0} = 0 = i_{1}.

Thus φ^{−1}(L) is a linear subspace of ℙ^{ν}; in particular, it is irreducible. We now compute its dimension. Recall that F has ν + 1 coefficients altogether; the number with i_{0} = 0 = i_{1} is m + 1, and so φ^{−1}(L) has dimension

\frac{(m + 1)(m + 2)(m + 3)}{6} − 1 − (m + 1) = \frac{m(m + 1)(m + 5)}{6} − 1.

We can now deduce from (8.10) that Γ_{m} is irreducible and that

dim(Γ_{m}) = dim(Π) + dim(φ^{−1}(L)) = \frac{m(m + 1)(m + 5)}{6} + 3,

as claimed. □

Now consider the other projection

(L, F)  Γ_{m} ⊂ Π × ℙ^{ν}
↓  ↓ ψ
F  ℙ^{ν}

By definition

ψ^{−1}(F) = {L | L lies on V (F)}.

Example 8.18. Let m = 1. Then ν = 3 and dim Γ_{1} = 5. The projection ψ: Γ_{1} → ℙ^{3} is surjective (every plane contains at least one line), and (8.6) tells us that dim ψ^{−1}(F) ≥ 2. In fact of course, the lines on any plane form a 2-dimensional family, and so ψ^{−1}(F) = 2 for all F.

Theorem 8.19. When m > 3, the surfaces of degree m containing no line correspond to an open subset of ℙ^{ν}.

Proof. We have

dim Γ_{m}−dim ℙ^{ν} = \frac{m(m + 1)(m + 5)}{6} +3− \frac{(m + 1)(m + 2)(m + 3)}{6} +1 = 4−(m+1).

Therefore, if m > 3, then dim Γ_{m} < dim ℙ^{ν}, and so ψ(Γ_{m}) is a proper closed subvariety of ℙ^{ν}. This proves the claim. □

We now look at the case m = 2. Here dim Γ_{m} = 10, and ν = 9, which suggests that ψ should be surjective and that its fibres should all have dimension ≥ 1. We shall see that this is correct.

A quadric is said to be nondegenerate if it is defined by an irreducible polynomial of degree 2. After a change of variables, any nondegenerate quadric will be defined by an equation

XW = Y Z.

This is just the image of the Segre mapping (see 5.22)

(a_{0} : a_{1}), (b_{0} : b_{1}) ↦ (a_{0}b_{0} : a_{0}b_{1} : a_{1}b_{0} : a_{1}b_{1}): ℙ^{1} × ℙ^{1} → ℙ^{3}.

There are two obvious families of lines on ℙ^{1} × ℙ^{1}, namely, the horizontal family and the vertical family; each is parametrized by ℙ^{1}, and so is called a pencil of lines. They map to two families of lines on the quadric:

\{ \begin{matrix}t_{0}X & = t_{1}Z \\ t_{0}Y & = t_{1}W \end{matrix} and \{ \begin{matrix}t_{0}X & = t_{1}Y \\ t_{0}Z & = t_{1}W. \end{matrix}

Since a degenerate quadric is a surface or a union of two surfaces, we see that every quadric surface contains a line, that is, that ψ: Γ_{2} → ℙ^{9} is surjective. Thus (8.4) tells us that all the fibres have dimension ≥ 1, and the set where the dimension is > 1 is a proper closed subset. In fact the dimension of the fibre is > 1 exactly on the set of reducible F ’s, which we know to be closed (see the solution to Homework 9, Problem 1).

It follows from the above discussion that if F is nondegenerate, then ψ^{−1}(F) is isomorphic to the disjoint union of two lines, ψ^{−1}(F) ≈ ℙ^{1} ∪ ℙ^{1}. Classically, one defines a regulus to be a nondegenerate quadric surface together with a choice of a pencil of lines. One can show that the set of reguli is, in a natural way, an algebraic variety R, and that, over the set of nondegenerate quadrics, ψ factors into the composite of two regular maps:

Γ_{2} − ψ^{−1}(S) = pairs, (F, L) with L on F;
↓
R = set of reguli;
↓
ℙ^{9} − S  = set of nondegenerate quadrics.

The fibres of the top map are connected, and of dimension 1 (they are all isomorphic to ℙ^{1}), and second map is finite and two-to-one. Factorizations of this type occur quite generally (see the Stein factorization theorem (8.23) below).

We now look at the case m = 3. Here dim Γ_{3} = 19; ν = 19 : we have a map

ψ: Γ_{3}^{dim 19} ⟶ ℙ^{19}.

Theorem 8.20. The set of cubic surfaces containing exactly 27 lines corresponds to an open subset of ℙ^{19}; the remaining surfaces either contain an infinite number of lines or a nonzero finite number ≤ 27.

Example 8.21. (a) Consider the Fermat surface

X_{0}^{3} + X_{1}^{3} + X_{2}^{3} + X_{3}^{3} = 0.

Let ζ be a primitive cube root of one. There are the following lines on the surface:

\{ \begin{matrix}X_{0} + ζ^{i}X_{1} & = 0 \\ X_{2} + ζ^{j}X_{3} & = 0 \end{matrix}  \{ \begin{matrix}X_{0} + ζ^{i}X_{2} & = 0 \\ X_{1} + ζ^{j}X_{3} & = 0 \end{matrix}  \{ \begin{matrix}X_{0} + ζ^{i}X_{3} & = 0 \\ X_{1} + ζ^{j}X_{2} & = 0 \end{matrix}

There are three sets, each with nine lines, for a total of 27 lines.

(b) Consider the surface

X_{1}X_{2}X_{3} = X_{0}^{3}.

In this case, there are exactly three lines. To see this, look first in the affine space where X_{0} ≠ 0—here we can take the equation to be X_{1}X_{2}X_{3} = 1. A line in 𝔸^{3} can be written in parametric form X_{i} = a_{i}t + b_{i}, but a direct inspection shows that no such line lies on the surface. Now look where X_{0} = 0, that is, in the plane at infinity. The intersection of the surface with this plane is given by X_{1}X_{2}X_{3} = 0 (homogeneous coordinates), which is the union of three lines, namely,

X_{1} = 0; X_{2} = 0; X_{3} = 0.

Therefore, the surface contains exactly three lines.

(c) Consider the surface

X_{1}^{3} + X_{2}^{3} = 0.

Here there is a pencil of lines:

\{ \begin{matrix}t_{0}X_{1} & = t_{1}X_{0} \\ t_{0}X_{2} & = −t_{1}X_{0}. \end{matrix}

(In the affine space where X_{0} ≠ 0, the equation is X^{3} + Y^{3} = 0, which contains the line X = t, Y = −t, all t.)

We now discuss the proof of (8.20). If ψ: Γ_{3} → ℙ^{19} were not surjective, then ψ(Γ_{3}) would be a proper closed subvariety of ℙ^{19}, and the nonempty fibres would all have dimension ≥ 1 (by 8.6), which contradicts two of the above examples. Therefore the map is surjective^{23}, and there is an open subset U of ℙ^{19} where the fibres have dimension 0; outside U, the fibres have dimension > 0.

Given that every cubic surface has at least one line, it is not hard to show that there is an open subset U′ where the cubics have exactly 27 lines (see Reid, 1988, pp106–110); in fact, U′ can be taken to be the set of nonsingular cubics. According to (6.24), the restriction of ψ to ψ^{−1}(U) is finite, and so we can apply (8.10) to see that all cubics in U − U′ have fewer than 27 lines.

Remark 8.22. The twenty-seven lines on a cubic surface were discovered in 1849 by Salmon and Cayley, and have been much studied—see A. Henderson, The Twenty-Seven Lines Upon the Cubic Surface, Cambridge University Press, 1911. For example, it is known that the group of permutations of the set of 27 lines preserving intersections (that is, such that L ∩ L′ ≠ ∅ ⟺ σ(L) ∩ σ(L′) ≠ ∅) is isomorphic to the Weyl group of the root system of a simple Lie algebra of type E_{6}, and hence has 25920 elements.

It is known that there is a set of 6 skew lines on a nonsingular cubic surface V. Let L and L′ be two skew lines. Then “in general” a line joining a point on L to a point on L′ will meet the surface in exactly one further point. In this way one obtains an invertible regular map from an open subset of ℙ^{1} × ℙ^{1} to an open subset of V, and hence V is biregularly equivalent to ℙ^{2}.

Stein factorization. The following important theorem shows that the fibres of a proper map are disconnected only because the fibres of finite maps are disconnected.

Theorem 8.23. Let φ: W → V be a proper morphism of varieties. It is possible to factor φ into W ⟶^{φ_{1}} W′ with φ_{1} proper with connected fibres and φ_{2} finite.

Proof. This is usually proved at the same time as Zariski’s main theorem (if W and V are irreducible, and V is affine, then W′ is the affine variety with k[W′] the integral closure of k[V] in k(W)). □

23According to Miles Reid, 1988, p126, every adult algebraic geometer knows this proof that every cubic contains a line.