FIELDS AND GALOIS THEORY

J.S. MILNE

August 21, 1996; v2.01

Abstract. Thes are the notes for the second part of Math 594, University of Michigan, Winter 1994, 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 “Math594” as the subject.

Contents

1. Extensions of Fields 1 1.1. Definitions 1 1.2. The characteristic of a field 1 1.3. The polynomial ring F [X] 2 1.4. Factoring polynomials 2 1.5. Extension fields; degrees 4 1.6. Construction of some extensions 5 1.7. Generators of extension fields 6 1.8. Algebraic and transcendental elements 7 1.9. Transcendental numbers 8 1.10. Constructions with straight-edge and compass. 9

2. Splitting Fields; Algebraic Closures 12 2.1. Maps from simple extensions. 12 2.2. Splitting fields 13 2.3. Algebraic closures 14

3. The Fundamental Theorem of Galois Theory 18 3.1. Multiple roots 18 3.2. Groups of automorphisms of fields 20 3.3. Separable, normal, and Galois extensions 21 3.4. The fundamental theorem of Galois theory 23 3.5. Constructible numbers revisited 26 3.6. Galois group of a polynomial 26

Copyright 1996 J.S. Milne. You may make one copy of these notes for your own personal use.

3.7. Solvability of equations 27

4. Computing Galois Groups. 28 4.1. When is G_{f} ⊂ A_{n}? 28 4.2. When is G_{f} transitive? 29 4.3. Polynomials of degree ≤ 3 29 4.4. Quartic polynomials 29 4.5. Examples of polynomials with S_{p} as Galois group over ℚ 31 4.6. Finite fields 32 4.7. Computing Galois groups over ℚ 33

5. Applications of Galois Theory 36 5.1. Primitive element theorem. 36 5.2. Fundamental Theorem of Algebra 38 5.3. Cyclotomic extensions 39 5.4. Independence of characters 41 5.5. Hilbert’s Theorem 90. 42 5.6. Cyclic extensions. 44 5.7. Proof of Galois’s solvability theorem 45 5.8. The general polynomial of degree n 46

Symmetric polynomials The general polynomial A brief history

5.9. Norms and traces 49 5.10. Infinite Galois extensions (sketch) 53

6. Transcendental Extensions 55

1. Extensions of Fields

1.1. Definitions. A field is a set F with two composition laws + and · such that

(a) (F, +) is an abelian group;

(b) let F^{×} = F − {0}; then (F^{×}, ·) is an abelian group;

(c) (distributive law) for all a, b, c ∈ F, (a + b)c = ac + bc (hence also a(b + c) = ab + ac).

Equivalently, a field is a nonzero commutative ring (meaning with 1) such that every nonzero element has an inverse. A field contains at least two distinct elements, 0 and 1. The smallest, and one of the most important, fields is 𝔽_{2} = ℤ/2ℤ = {0, 1}.

Lemma 1.1. A commutative ring R is a field if and only if it has no ideals other than (0) and R.

Proof. Suppose R is a field, and let I be a nonzero ideal in R. If a is a nonzero element of 𝔞, then 1 = a^{−1}a ∈ I, and so I = R. Conversely, suppose R is a commutative ring with no nontrivial ideals; if a ≠ 0, then (a) = R, which means that there is a b in F such that ab = 1.

Example 1.2. The following are fields: ℚ, ℝ, ℂ, 𝔽_{p} = ℤ/pℤ.

A homomorphism of fields α : F → F′ is simply a homomorphism of rings, i.e., it is a map with the properties

α(a + b) = α(a) + α(b), α(ab) = α(a)α(b), α(1) = 1, all a, b ∈ F.

Such a homomorphism is always injective, because the kernel is a proper ideal (it doesn’t contain 1), which must therefore be zero.

1.2. The characteristic of a field. The map

ℤ → F, n ↦ 1_{F} + 1_{F} + ⋯ + 1_{F} (ntimes),

is a homomorphism of rings.

Case 1: Kernel = (0); then n · 1_{F} = 0 ⟹ n = 0 (in ℤ). The map ℤ → F extends to a homomorphism ℚ ↪ F, \frac{m}{n} ↦ (m · 1_{F})(n · 1_{F})^{−1}. Thus F contains a copy of ℚ. In this case, we say that F has characteristic zero.

Case 2: Kernel ≠ (0), i.e., n · 1_{F} = 0 some n ≠ 1. The smallest such n will be a prime p (else F will have nonzero zero-divisors), and p generates the kernel. In this case, {m · 1_{F} | m ∈ ℤ} ≈ 𝔽_{p}, and F contains a copy of 𝔽_{p}. We say that F has characteristic p.

The fields 𝔽_{p}, p prime, and ℚ are called the prime fields. Every field contains a copy of one of them.

Remark 1.3. The binomial theorem

(a + b)^{m} = a^{m} + \binom{m}{1}a^{m−1}b + ⋯ + \binom{m}{r}a^{m−r}b^{r} + ⋯ + b^{m}

holds in any ring. If p is prime, then p|\binom{p}{r} for all r, 1 ≤ r ≤ p − 1. Therefore, when F has characteristic p, (a + b)^{p} = a^{p} + b^{p}. Hence a ↦ a^{p} is a homomorphism F → F, called the Frobenius endomorphism of F. When F is finite, it is an isomorphism, called the Frobenius automorphism.

1.3. The polynomial ring F [X]. I shall assume everyone knows the following (see Jacobson Chapter II, or Math 593).

(a) Let I be a nonzero ideal in F [X]. If f(X) is a nonzero polynomial of least degree in I, then I = (f(X)). When we choose f to be monic, i.e., to have leading coefficient one, it is uniquely determined by I. There is a one-to-one correspondence between the nonzero ideals of F [X] and the monic polynomials in F [X]. The prime ideals correspond to the irreducible monic polynomials.

(b) Division algorithm: given f(X) and g(X) ∈ F [X] with g ≠ 0, we can find q(X) and r(X) ∈ F [X] with deg(r) < g such that f = gq + r; moreover, q(X) and r(X) are uniquely determined. Thus the ring F [X] is a Euclidean domain.

(c) Euclid’s algorithm: Let f and g ∈ F [X] have gcd d(X); the algorithm gives polynomials a(X) and b(X) such that

a(X) · f(X) + b(X) · g(X) = d(X), deg(a) ≤ deg(g), deg(b) ≤ deg(f).

Recall how it goes. Using the division algorithm, we construct a sequence of quotients and remainders:

f = q_{0}g + r_{0}
g = q_{1}r_{0} + r_{1}
r_{0} = q_{2}r_{1} + r_{2}
⋯
r_{n−2} = q_{n}r_{n−1} + r_{n}
r_{n−1} = q_{n+1}r_{n}.

Then r_{n} = gcd(f, g), and

r_{n} = r_{n−2} − q_{n}r_{n−1} = r_{n−2} − q_{n}(r_{n−3} − q_{n−1}r_{n−2}) = ⋯ = af + bg.

Maple knows Euclid’s algorithm—to learn its syntax, type “?gcdex;”.

(d) Since F [X] is an integral domain, we can form its field of fractions F (X). It consists of quotients f(X)/g(X), f and g polynomials, g ≠ 0.

1.4. Factoring polynomials. It will frequently be important for us to know whether a polynomial is irreducible and, if it isn’t, what its factors are. The following results help.

Proposition 1.4. Suppose r = \frac{c}{d}, c, d ∈ ℤ, gcd(c, d) = 1, is a root of a polynomial

a_{m}X^{m} + a_{m−1}X^{m−1} + ⋯ + a_{0}, a_{i} ∈ ℤ.

Then c|a_{0} and d|a_{m}.

Proof. It is clear from the equation

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

that d|a_{m}c^{m}, and therefore, d|a_{m}. The proof that c|a_{0} is similar.

Example 1.5. The polynomial X^{3} − 3X − 1 is irreducible in ℚ[X] because its only possible roots are ±1 (and they aren’t).

Proposition 1.6. Let f(X) ∈ ℤ[X] be such that its coefficients have greatest common divisor 1. If f(X) factors nontrivially in ℚ[X], then it factors nontrivially in ℤ[X]; moreover, if f(X) ∈ ℤ[X] is monic, then any monic factor of f(X) in ℚ[X] lies in ℤ[X].

Proof. Use Gauss’s lemma (see Jacobson, 2.16, or Math 593).

Proposition 1.7. (Eisenstein criterion) Let

f = a_{m}X^{m} + a_{m−1}X^{m−1} + ⋯ + a_{0}, a_{i} ∈ ℤ;

suppose that there is a prime p such that:

p does not divide a_{m}, p divides a_{m−1}, ..., a_{0}, p^{2} does not divide a_{0}.

Then f is irreducible in ℚ[X].

Proof. We may remove any common factor from the coefficients f, and hence assume that they have gcd = 1. Therefore, if f(X) factors in ℚ[X], it factors in ℤ[X]:

a_{m}X^{m} + a_{m−1}X^{m−1} + ⋯ + a_{0} = (b_{n}X^{n} + ⋯ + b_{0})(c_{r}X^{r} + ⋯ + c_{0}), b_{i}, c_{i} ∈ ℤ, n, r < m.

Since p, but not p^{2}, divides a_{0} = b_{0}c_{0}, p must divide exactly one of b_{0}, c_{0}, say p divides b_{0}. Now from the equation

a_{1} = b_{0}c_{1} + b_{1}c_{0},

we see that p|b_{1}. Now from the equation

a_{2} = b_{0}c_{2} + b_{1}c_{1} + b_{2}c_{0},

we see that p|b_{2}. By continuing in this way, we find that p divides b_{0}, b_{1}, …, b_{n}, which contradicts the fact that p does not divide a_{m}.

The above three propositions hold with ℤ replaced by any unique factorization domain.

Proposition 1.8. There is an algorithm for factoring a polynomial in ℚ[X].

Proof. Consider f(X) ∈ ℚ[X]. Multiply f(X) by an integer, so that it is monic, and then replace it by D^{deg(f)}f(\frac{X}{D}), D = a common denominator for the coefficients of f, to obtain a monic polynomial with integer coefficients. Thus we need consider only polynomials

f(X) = X^{m} + a_{1}X^{m−1} + ⋯ + a_{m}, a_{i} ∈ ℤ.

From the fundamental theorem of algebra (see later), we know that f splits completely in ℂ[X]:

f(X) = ∏_{i=1}^{m}(X − α_{i}), α_{i} ∈ ℂ.

From the equation f(α_{i}) = 0, it follows that |α_{i}| is less than some bound M depending on a_{1}, …, a_{m}. Now if g(X) is a monic factor of f(X), then its roots in ℂ are certain of the α_{i}, and its coefficients are symmetric polynomials in its roots. Therefore the absolute values of the coefficients of g(X) are bounded. Since they are also integers (by 1.6), we see that there are only finitely many possibilities for g(X). Thus, to find the factors of f(X) we (better Maple) only have to do a finite amount of checking.

One other observation is sometimes useful: Suppose that the leading coefficient of f(X) ∈ ℤ[X] is not divisible by the prime p; if f(X) is irreducible in 𝔽_{p}[X], then it is irreducible in ℤ[X]. Unfortunately, this test is not always effective: for example, X^{4} − 10X^{2} + 1 is reducible modulo every prime, but it is irreducible in ℚ[X].

Maple knows how to factor polynomials in ℚ[X] and in 𝔽_{p}[X]. For example >factor(6*X^2+18*X-24); will find the factors of 6X^{2} + 18X − 24, and >Factor(X^2+3*X+3) mod 7; will find the factors of X^{2} + 3X + 3 modulo 7, i.e., in 𝔽_{7}[X].

Thus, we need not concern ourselves with the problem of factorizing polynomials in ℚ[X] or 𝔽_{p}[X].

1.5. Extension fields; degrees. A field E containing a field F is called an extension (field) of F. Such an E can be regarded (in an obvious fashion) as an F -vector space. We write [E : F] for the dimension (possibly infinite) of E as an F -vector space, and call [E : F] the degree of E over F. We often say that E is finite over F when it has finite degree over F.

Example 1.9. (a) The field of complex numbers ℂ has degree 2 over ℝ (basis {1, i}).

(b) The field of real numbers ℝ has infinite degree over ℚ. (We know ℚ is countable, which implies that any finite-dimensional vector space over ℚ is countable; but ℝ is not countable. More explicitly, one can find real numbers α such that 1, α, α^{2}, … are linearly independent (see section 1.9 below)).

(c) The field of Gaussian numbers ℚ(i) =_{df} {a + bi ∈ ℂ | a, b ∈ ℚ} has degree 2 over ℚ (basis {1, i}).

(d) The field F (X) has infinite degree over F. (It contains the F -subspace F [X], which has the infinite basis {1, X, X^{2}, … }.)

Proposition 1.10. Let L ⊃ E ⊃ F (all fields). Then L/F is of finite degree ⟺ L/E and E/F are both of finite degree, in which case

[L : F] = [L : E][E : F].

Proof. Assume that L/E and E/F are of finite degree, and let {e_{i}} be a basis for E/F and {ℓ_{j}} a basis for L/E. I claim that {e_{i}ℓ_{j}} is a basis for L over F. I first show that it spans L. Let γ ∈ L. Then, because {ℓ_{j}} spans L as an E-vector space,

γ = ∑ α_{j}ℓ_{j}, some α_{j} ∈ E,

and because {e_{i}} spans E as an F -vector space, for each j,

α_{j} = ∑ a_{ij}e_{i}, some a_{ij} ∈ F .

On putting these together, we find that

γ = ∑ a_{ij}e_{i}ℓ_{j}.

Next I show that {e_{i}ℓ_{j}} is linearly independent. A linear relation ∑ a_{ij}e_{i}ℓ_{j} = 0 can be rewritten ∑_{j}(∑_{i} a_{ij}e_{i})ℓ_{j} = 0. The linear independence of the ℓ_{j}’s now shows that ∑_{i} a_{ij}e_{i} = 0 for each j, and the linear independence of the e_{i}’s now shows that each a_{ij} = 0.

Conversely, if L is of finite degree over F, then it is certainly of finite degree over E. Moreover, E, being a subspace of a finite dimensional F -space, is also finite dimensional.

1.6. Construction of some extensions. Let f(X) ∈ F be an monic polynomial of degree m, and let (f) be the ideal generated by f. Consider the quotient ring F [X]/(f(X)), and write x for the image of X in F [X]/(f(X)), i.e., x is the coset X + (f(X)). Then:

(a) The map

P (X) ↦ P (x) : F [X] → F [x]

is a surjective homomorphism; we have f(x) = 0.

(b) From the division algorithm, we know each element g of F [X]/(f) is represented by a unique polynomial r of degree < m. Hence each element of F [x] can be written uniquely as a sum

a_{0} + a_{1}x + ⋯ + a_{m−1}x^{m−1},  a_{i} ∈ F,  (*).

(c) The addition of two elements, written in the form (*), is obvious.

(d) To multiply two elements in the form (*), multiply in the usual way, and use the relation f(x) = 0 to express the monomials of degree ≥ m in x in terms of lower degree monomials.

(e) Now assume f(X) is irreducible. To find the inverse of an element α ∈ F [x], write α in the form (*), i.e., set α = g(x) where g(X) is a polynomial of degree ≤ m − 1. Then use Euclid’s algorithm in F [X] to obtain polynomials a(X) and b(X) such that

a(X)f(X) + b(X)g(X) = d(X)

with d(X) the gcd of f and g. In our case, d(X) is 1 because f(X) is irreducible and deg g(X) < deg f(X). On replacing X with x in the equation, we find b(x)g(x) = 1. Hence b(x) is the inverse of g(x).

Conclusion: For any monic irreducible polynomial f(X) ∈ F [X], F [x] = F [X]/(f(X)) is a field of degree m over F. Moreover, if we know how to compute in F, then we know how to compute in F [x].

Example 1.11. Let f(X) = X^{2} + 1 ∈ ℝ[X]. Then ℝ[x] has:

elements: a + bx, a, b ∈ ℝ;

addition: obvious;

multiplication: (a + bx)(a′ + b′x) = (aa′ − bb′) + (ab′ + a′b)x. We usually write i for x and ℂ for ℝ[x].

Example 1.12. Let f(X) = X^{3} − 3X − 1 ∈ ℚ[X]. This is irreducible over ℚ, and so ℚ[x] has basis {1, x, x^{2}} as a ℚ-vector space. Let

β = x^{4} + 2x^{3} + 3 ∈ ℚ[x].

Then using that x^{3} − 3x − 1 = 0, we find that β = 3x^{2} + 7x + 5. Because X^{3} − 3X − 1 is irreducible,

gcd(X^{3} − 3X − 1, 3X^{2} + 7X + 5) = 1.

In fact, Euclid’s algorithm (courtesy of Maple) gives

(X^{3} − 3X − 1)(\frac{−7}{37}X + \frac{29}{111}) + (3X^{2} + 7X + 5)(\frac{7}{111}X^{2} − \frac{26}{111}X + \frac{28}{111}) = 1.

Hence

(3x^{2} + 7x + 5)(\frac{7}{111}x^{2} − \frac{26}{111}x + \frac{28}{111}) = 1;

we have found the inverse of β.

1.7. Generators of extension fields. Let E be an extension field of F, and let S be a subset of E. The intersection of all the subrings of E containing F and S is again a subring of E (containing F and S). We call it the subring of E generated by F and S, and we write it F [S].

Lemma 1.13. The ring F [S] consists of all the elements of E that can be written as finite sums of the form

∑ a_{i_{1}···i_{n}}α_{1}^{i_{1}} ⋯ α_{n}^{i_{n}}, a_{i_{1}···i_{n}} ∈ F, α_{i} ∈ S. (*)

Proof. Let R be the set of all such elements; it is easy to check that R is a ring containing F and S, and that any ring containing F and S contains R; therefore R equals F [S].

Note that the expression of an element in the form (*) will not be unique in general. When S = {α_{1}, ..., α_{n}}, we write F [α_{1}, ..., α_{n}] for F [S].

Lemma 1.14. Let E ⊃ R ⊃ F with E and F fields and R a ring. If R is finite-dimensional when regarded as an F -vector space, then it is field.

Proof. Let α be a nonzero element of R—we have to show that α is invertible. The map x ↦ αx : R → R is an injective F -linear map, and is therefore surjective. In particular, there is an element β ∈ R such that αβ = 1.

Example 1.15. An element of ℚ[π], π = 3.14159..., can be written uniquely as a finite sum

a_{0} + a_{1}π + a_{2}π^{2} + ⋯, a_{i} ∈ ℚ.

An element of ℚ[i] can be written uniquely in the form a + bi, a, b ∈ ℚ. (Everything considered in ℂ.)

Let E again be an extension field of F and S a subset of E. The subfield F (S) of E generated by F and S is the intersection of all subfields of E containing F and S. It is equal to the field of fractions of F[S] (since this is a field containing F and S, and is the smallest such field). Lemma 1.14 shows that F [S] is sometimes already a field, in which case F (S) = F [S]. We write F (α_{1}, ..., α_{n}) for F (S) when S = {α_{1}, ..., α_{n}}.

Thus: F [α_{1}, …, α_{n}] consists of all elements of E that can be expressed as polynomials in the α_{i} with coefficients in F, and F (α_{1}, …, α_{n}) consists of all elements of E that can be expressed as quotients of two such polynomials.

Example 1.16. An element of ℚ(π) can be expressed as a quotient

g(π)/h(π), g(X), h(X) ∈ ℚ[X], h(π) ≠ 0.

The ring ℚ[i] is already a field.

An extension E of F is said to be simple if E = F (α) some α ∈ E. For example, ℚ(π) and ℚ[i] are simple extensions of ℚ.

When F and F′ are subfields of E, then we write F · F′ for F (F′)(= F′(F)), and we call it the composite of F and F′. It is the smallest subfield of E containing both F and F′.

1.8. Algebraic and transcendental elements. Let E be an extension field of F, and let α ∈ E. Then we have a homomorphism

f(X) ↦ f(α) : F [X] → E.

There are two possibilites.

Case 1: The kernel of the map is (0), i.e.,

f(α) = 0, f(X) ∈ F [X] ⟹ f(X) = 0.

In this case we say that α transcendental over F. The isomorphism F [X] → F [α] extends to an isomorphism F (X) → F (α).

Case 2: The kernel is ≠ (0), i.e., g(α) = 0 for some nonzero g(X) ∈ F [X]. We then say that α is algebraic over F. Let f(X) be the monic polynomial generating the kernel of the map. It is irreducible (if f = gh is a proper factorization, then g(α)h(α) = f(α) = 0, but g(α) ≠ 0 ≠ h(α)). We call f the minimum polynomial of α over F. It is characterized as an element of F [X] by each of the following sets of conditions:

f is monic; f(α) = 0; g(α) = 0 and g ∈ F [X] ⟹ f|g;

f is the monic polynomial of least degree such f(α) = 0;

f is monic, irreducible, and f(α) = 0. Note that g(X) ↦ g(α) induces an isomorphism F [X]/(f) → F [α]. Since the first is a field, so also is the second: F (α) = F [α]. Moreover, each element of F [α] has a unique expression

a_{0} + a_{1}α + a_{2}α^{2} + ⋯ + a_{m−1}α^{m−1}, a_{i} ∈ F,

where m = deg(f). In other words, 1, α, …, α^{m−1} is a basis for F [α] over F. Hence [F (α) : F] = m. Since F [x] ≈ F [α], arithmetic in F [α] can be performed using the same rules as in F [x].

Example 1.17. Let α ∈ ℂ be such that α^{3} − 3α − 1 = 0. The minimum polynomial of α over ℚ is X^{3} − 3X − 1 (because this polynomial is monic, irreducible, and has α as a root). The set {1, α, α^{2}} is a basis for ℚ[α] over ℚ. The calculations in an example above show that if β is the element α^{4} + 2α^{3} + 3 of ℚ[α], then β = 3α^{2} + 7α + 5, and

β^{−1} = \frac{7}{111}α^{2} − \frac{26}{111}α + \frac{28}{111}.

Remark 1.18. Maple knows how to compute in ℚ[α]. For example, factor(X^4+4); returns the factorization

(X^{2} − 2X + 2)(X^{2} + 2X + 2).

Now type: alias(c=RootOf(X^2+2*X+2);. Then

factor(X^4+4,c); returns the factorization

(X + c)(X − 2 − c)(X + 2 + c)(X − c),

i.e., Maple has factored X^{4} + 4 in ℚ[c] where c has minimum polynomial X^{2} + 2X + 2.

An extension E/F is algebraic if all elements of E are algebraic over F; otherwise it is transcendental over F.

Proposition 1.19. (a) If [E : F] is finite, then E is algebraic over F.

(b) If E is algebraic over F and finitely generated (as a field), then [E : F] is finite.

Proof. (a) If α were transcendental over F, then 1, α, α^{2}, … would be linearly independent over F.

(b) Let E = F [α_{1}, ..., α_{n}]; then F [α_{1}] is finite over F (because α_{1} is algebraic over F); F [α_{1}, α_{2}] is finite over F [α_{1}] (because α_{2} is algebraic over F, and hence F [α_{1}]). Hence F [α_{1}, α_{2}] is finite over F. This argument can be continued.

Corollary 1.20. If E is algebraic over F then any subring R of E containing F is a field.

Proof. Let α ∈ R; then F [α] is a field and F [α] ⊂ R. Therefore α has an inverse in R.

A field F is said to be algebraically closed if E algebraic over F implies E = F. Equivalent condition: the only irreducible polynomials in F [X] are of degree one; every nonconstant polynomial in F [X] has a root in F.

Example 1.21. The field of complex numbers ℂ is algebraically closed. The set of all complex numbers algebraic over ℚ is an algebraically closed field. Every field F has an algebraically closed algebraic extension field (which is unique up to a nonunique isomorphism). All these statements will be proved later.

1.9. Transcendental numbers. A complex number is said to be algebraic or transcendental according as it is algebraic or transcendental over ℚ. First some history:

1844: Liouville showed that certain numbers (now called Liouville numbers) are transcendental.

1873: Hermite showed that e is transcendental.

1873: Cantor showed that the set of algebraic numbers is countable, but that ℝ is not countable. [Thus almost all numbers are transcendental, but it is usually very difficult to prove that a particular number is transcendental.]

1882: Lindemann showed that π is transcendental.

1934: Gelfond-Schneider showed that if α and β are algebraic, α ≠ 0, 1, and β ∉ ℚ, then α^{β} is transcendental. (This was one of Hilbert’s famous problems)

1994: Euler’s constant

γ = lim_{n→∞}(∑_{k=1}^{n} 1/k − log n)

has not yet been proven to be transcendental.

1994: The numbers e + π and e − π are surely transcendental, but they have not even been proved to be irrational!

Proposition 1.22. The set of algebraic numbers is countable.

Proof. Define the height h(r) of a rational number to be max(|m|, |n|), where r = m/n is the expression of r in its lowest terms. There are only finitely many rational numbers with height less than a fixed number N. Let A(N) be the set of algebraic numbers whose minimum equation over ℚ is of degree ≤ N and has coefficients of height < N. Then A(N) is finite for each N. Count the elements of A(10); then count the elements of A(100); then count the elements of A(1000), and so on.

A typical Liouville number is ∑_{n=0}^{∞} \frac{1}{10^{n!}} —in its decimal expansion there are increasingly long strings of zeros. We prove that the analogue of this number in base 2 is transcendental.

Theorem 1.23. The number α = ∑ \frac{1}{2^{n!}} is transcendental.

Proof. Suppose not, and let

f(X) = X^{d} + a_{1}X^{d−1} + ⋯ + a_{d}, a_{i} ∈ ℚ,

be the minimum polynomial of α over ℚ. Thus [ℚ[α] : ℚ] = d. Let

f(X) = ∏_{i=1}^{d}(X − α_{i}), α_{i} ∈ ℂ, α_{1} = α,

and choose a nonzero integer D such that Df(X) ∈ ℤ[X]. Let Σ_{N} = ∑_{n=0}^{N} \frac{1}{2^{n!}} , so that Σ_{N} → α as N → ∞, and let x_{N} = f(Σ_{N}).

Because f(X) is irreducible in ℚ[X], it has no rational root, except possibly α; but Σ_{N} ≠ α, and so x_{N} ≠ 0. (In fact α is obviously nonrational because its expansion to base 2 is not periodic.)

Clearly x_{N} ∈ ℚ; in fact (2^{N!})^{d}Dx_{N} ∈ ℤ, and so

|(2^{N!})^{d}Dx_{N}| ≥ 1.

On the other hand,

|x_{N}| = ∏ |Σ_{N} − α_{i}| ≤ |α_{1} − Σ_{N}|(M + Σ_{N})^{d−1}, where M = max_{i≠1} |α_{i}|,

and

|α_{1} − Σ_{N}| = ∑_{n=N+1}^{∞} \frac{1}{2^{n!}} ≤ \frac{2}{2^{(N+1)!}}

Hence

|(2^{N!})^{d}Dx_{N}| ≤ 2 · \frac{2^{d·N!}}{2^{(N+1)!}} · (M + Σ_{N})^{d−1} → 0 as N → ∞

because \frac{2^{d·N!}}{2^{(N+1)!}} = (\frac{2^{d}}{2^{N+1}} )^{N!} → 0. We have a contradiction.

1.10. Constructions with straight-edge and compass. The Greeks understood that integers and the rational numbers. They were surprised to find that the length of the diagonal of a square of side 1, namely \sqrt{2}, is not rational. They thus realized that they needed to extend their number system. They then hoped that the “constructible” numbers would suffice. Suppose we are given a length, which we call 1, a straight-edge, and a compass (device for drawing circles). A number (better a length) is constructible if it can be constructed by forming successive intersections of

• lines drawn through two points already constructed, and

• circles with centre a point already constructed and radius a constructed length.

This led them to three famous problems that they were unable to solve: is it possible to duplicate the cube, trisect an angle, or square the circle by straight-edge and compass constructions? We’ll see that the answer to all three is negative.

Let F be a subfield of ℝ. The F -plane is F × F ⊂ ℝ × ℝ. We make the following definitions:

A line in the F -plane is a line through two points in the F -plane. Such a line is given by an equation:

ax + by + c = 0, a, b, c ∈ F.

A circle in the F -plane is a circle with centre an F -point and radius an element of F. Such a circle is given by an equation:

(x − a)^{2} + (y − b)^{2} = c^{2}, a, b, c ∈ F.

Lemma 1.24. Let L ≠ L′ be F -lines, and let C ≠ C′ be F -circles.

(a) L ∩ L′ = ∅ or consists of a single F -point.

(b) L ∩ C = ∅ or consists of one or two points in the F [\sqrt{e}]-plane, some e ∈ F.

(c) C ∩ C′ = ∅ or consists of one or two points in the F [\sqrt{e}]-plane, some e ∈ F.

Proof. The points in the intersection are found by solving the simultaneous equations, and hence by solving (at worst) a quadratic equation with coefficients in F.

Lemma 1.25. (a) If c and d are constructible, then so also are c ± d, cd, and \frac{c}{d} (d ≠ 0). (b) If c > 0 is constructible, then so also is \sqrt{c}.

Proof. First show that it is possible to construct a line perpendicular to a given line through a given point, and then a line parallel to a given line through a given point. Hence it is possible to construct a triangle similar to a given one on a side with given length. By an astute choice of the triangles, one constructs cd and c^{−1}. For (b), draw a circle of radius \frac{c+1}{2} about (\frac{c+1}{2}, 0), and draw a vertical line through the point A = (1, 0) to meet the circle at P. The length AP is \sqrt{c}. (For more details, see for example, Rotman, Galois Theory, Appendix 3.)

Theorem 1.26. (a) The set of constructible numbers is field.

(b) A number α is constructible if and only if it is contained in field of the form

ℚ[\sqrt{a_{1}}, …, \sqrt{a_{r}}], a_{i} ∈ ℚ[\sqrt{a_{1}}, …, \sqrt{a_{i−1}}].

Proof. (a) Immediate from (a) of Lemma 1.25.

(b) From (a) we know that the set of constructible numbers is a field containing ℚ, and it follows from (a) and Lemma 1.25 that every number in ℚ[\sqrt{a_{1}}, …, \sqrt{a_{r}}] is constructible. Conversely, it follows from Lemma 1.24 that every constructible number is in a field of the form ℚ[\sqrt{a_{1}}, …, \sqrt{a_{r}}].

Now we can apply the (not quite elementary) result Proposition 1.10 to obtain:

Corollary 1.27. If α is constructible, then α is algebraic over ℚ, and [ℚ[α] : ℚ] is a power of 2.

Proof. We know that [ℚ[α] : ℚ] divides [ℚ[\sqrt{a_{1}}, …, \sqrt{a_{r}}] : ℚ] = 2^{r}.

Corollary 1.28. It is impossible to duplicate the cube by straight-edge and compass constructions.

Proof. The problem is to construct a cube with volume 2. This requires constructing a root of the polynomial X^{3} − 2 = 0. But this polynomial is irreducible (by Eisenstein’s criterion for example), and so [ℚ[\sqrt[3]{α}] : ℚ] = 3.

Corollary 1.29. In general, it is impossible to trisect an angle by straight-edge and compass constructions.

Proof. Knowing an angle is equivalent to knowing the cosine of the angle. Therefore, to trisect 3α, we have to construct a solution to

cos 3α = 4 cos^{3} α − 3 cos α.

For example, take 3α = 60; to construct α, we have to solve 8x^{3} − 6x − 1 = 0, which is irreducible.

Corollary 1.30. It is impossible to square the circle by straight-edge and compass constructions.

Proof. A square with the same area as a circle of radius r has side \sqrt{π}r. Since π is transcendental, so also is \sqrt{π}.

We now consider another famous old problem, that of constructing a regular polygon. Note that X^{m} − 1 is not irreducible; in fact

X^{m} − 1 = (X − 1)(X^{m−1} + X^{m−2} + ⋯ + 1).

Lemma 1.31. If p is prime then X^{p−1} + ⋯ + 1 is irreducible; hence ℚ[e^{2πi/p}] has degree p − 1 over ℚ.

Proof. Consider

f(X + 1) = \frac{(X + 1)^{p} − 1}{X} = X^{p−1} + ⋯ + a_{2}X^{2} + a_{1}X + p,

with a_{i} = \binom{p}{i+1}. Since p|a_{i}, i = 1, ..., p−2, f(X+1) is irreducible by Eisenstein’s criterion.

In order to construct a regular p-gon, p an odd prime, we need to construct cos \frac{2π}{p}. But

ℚ[e^{\frac{2πi}{p}}] ⊃ ℚ[cos \frac{2π}{p}] ⊃ ℚ. The degree of ℚ[e^{\frac{2πi}{p}}] over ℚ[cos \frac{2π}{p}] is 2—the equation

α^{2} − 2 cos \frac{2π}{p} + 1 = 0, α = e^{\frac{2πi}{p}} ,

shows that it is ≤ 2, and it is not 1 because ℚ[e^{\frac{2πi}{p}}] is not contained in ℝ. Hence [ℚ[cos \frac{2π}{p}] : ℚ] = \frac{p−1}{2}.

Thus if the regular p-gon is constructible, then (p − 1)/2 = 2^{k} some k (later, we shall see a converse), which imples p = 2^{k+1} + 1. But 2^{r} + 1 can only be prime if r is a power of 2, because otherwise r has an odd factor t, and for t odd,

Y^{t} + 1 = (Y + 1)(Y^{t−1} − Y^{t−2} + ⋯ + 1).

Thus if the regular p-gon is constructible, then p = 2^{2^{k}} + 1 for some k. Fermat conjectured that all numbers of the form 2^{2^{k}} + 1 are prime, and claimed to show that this is true for k ≤ 5—for this reason primes of this form are called Fermat primes. For 0 ≤ k ≤ 4, the numbers p = 3, 5, 17, 257, 65537, are prime but Euler showed that 2^{32} + 1 = 641 · 6700417, and we don’t know of any more Fermat primes.

Gauss showed that

cos \frac{2π}{17} = −\frac{1}{16} + \frac{1}{16}\sqrt{17} + \frac{1}{16}\sqrt{34 − 2\sqrt{17}} + \frac{1}{8}\sqrt{17 + 3\sqrt{17} − \sqrt{34 − 2\sqrt{17}} − 2\sqrt{34 + 2\sqrt{17}}}

when he was 18 years old. This success encouraged him to become a mathematician.

2. Splitting Fields; Algebraic Closures

2.1. Maps from simple extensions. Let E and E′ be fields containing F. An F -homomorphism is a homomorphism φ : E → E′ such that φ(a) = a for all a ∈ F. Thus an F -homorphism maps a polynomial ∑ a_{i_{1}···i_{m}}α_{1}^{i_{1}} ⋯ α_{m}^{i_{m}}, a_{i_{1}···i_{m}} ∈ F, to

∑ a_{i_{1}···i_{m}}φ(α_{1})^{i_{1}} ⋯ φ(α_{m})^{i_{m}}.

An F -isomorphism is a bijective F -homomorphism. Note that if E and E′ have the same finite degree over F, then an F -homomorphism is automatically an F -isomorphism.

Proposition 2.1. Let F (α) be a simple field extension of a field F, and let Ω be a second field containing F.

(a) Assume α is transcendental over F; then for any F -homomorphism φ : F (α) → Ω, φ(α) is transcendental over F, and the map φ ↦ φ(α) defines a one-to-one correspondence

{F -homomorphisms φ : F (α) → Ω} ↔ { elements of Ω transcendental over F }.

(b) Assume α is algebraic over F, with minimum polynomial f(X); then for any F - homomorphism φ : F [α] → Ω, φ(α) is a root of f(X) in Ω, and the map φ ↦ φ(α) defines a one-to-one correspondence

{F -homomorphisms φ : F [α] → Ω} ↔ { distinct roots of f(X) in Ω}.

In particular, the number of such maps is the number of distinct roots of f in Ω.

Proof. (a) Let γ ∈ Ω. To say that α is transcendental over F means that F [α] is the ring of polynomials in α (as variable). By the universal property of polynomial rings, there is a unique F -homomorphism φ : F [α] → Ω sending α to γ. This extends to F (α) if and only if all nonzero elements of F [α] are sent to invertible (i.e., nonzero) elements of Ω, which is so if and only if γ is transcendental.

(b) Let f(X) = ∑ a_{i}X^{i}, and consider an F -homomorphism φ : F [α] → Ω. On applying φ to the equation ∑ a_{i}α^{i} = 0, we obtain the equation ∑ a_{i}φ(α)^{i} = 0, which shows that γ =_{df} φ(α) is a root of f(X) in Ω. Conversely, let γ ∈ Ω be a root of f(X). The map F [X] → Ω, g(X) ↦ g(γ), factors through F [X]/(f(X)). When composed with the inverse of the isomorphism F [X]/(f(X)) → F [α], it becomes a homomorphism F [α] → Ω sending α to γ.

We shall need a slight generalization of this result.

Proposition 2.2. Let F (α) be a simple field extension of a field F, and let φ_{0} : F → Ω be a homomorphism of F into a second field Ω.

(a) Assume α is transcendental over F; then the map φ ↦ φ(α) defines a one-to-one correspondence

{extensions φ : F (α) → Ω of φ_{0}} ↔ {elements of Ω transcendental over φ_{0}(F)}.

(b) Assume α is algebraic over F, with minimum polynomial f(X); then the map φ ↦ φ(α) defines a one-to-one correspondence

{extensions φ : F [α] → Ω of φ_{0}} ↔ { distinct roots of (φ_{0}f)(X)in Ω}.

In particular, the number of such maps is the number of distinct roots of φ_{0}f in Ω.

Proof. The proof is essentially the same as that of the preceding proposition.

By φ_{0}f we mean the polynomial obtained by applying φ_{0} to the coefficients of f, i.e., f = ∑ a_{i}X^{i} ⟹ φ_{0}f = ∑ φ(a_{i})X^{i}.

2.2. Splitting fields. Let f be a polynomial with coefficients in F. A field E containing F is said to split f if f splits in E[X], i.e., if f(X) = ∏(X − α_{i}) with α_{i} ∈ E. If E is also generated by the α_{i}, then it is called a splitting field for f.

Note that if f(X) = ∏ f_{i}(X)^{m_{i}}, then a splitting field for ∏ f_{i}(X) is also a splitting field for f (and conversely).

Example 2.3. (a) Let f(X) = aX^{2} + bX + c ∈ ℚ[X] be irreducible, and let α = \sqrt{b^{2} − 4ac}; then the subfield ℚ[α] of ℂ generated by α is a splitting field for f.

(b) Let f(X) = X^{3} + aX^{2} + bX + c ∈ ℚ[X] be irreducible, and let α_{1}, α_{2}, α_{3} be its roots in ℂ. Then ℚ[α_{1}, α_{2}, α_{3}] = ℚ[α_{1}, α_{2}] is a splitting field for f(X). Note that [ℚ[α_{1}] : ℚ] = 3 and that [ℚ[α_{1}, α_{2}] : ℚ[α_{1}]] = 1 or 2, and so [ℚ[α_{1}, α_{2}] : ℚ] = 3 or 6. We’ll see later that the degree is 3 if and only if the discriminant of f(X) is a square in F. For example, the discriminant of X^{3} + bX + c is −4b^{3} − 27c^{2}, and so the splitting field of X^{3} + 10X + 1 has degree 6 over ℚ.

Proposition 2.4. Every polynomial has a splitting field.

Proof. Let f ∈ F [X]. Let g_{1} be an irreducible factor of f(X), and let F_{1} = F [X]/(g_{1}(X)) = F [α_{1}], α_{1} = X + (g_{1}). Then α_{1} is a root of f(X) in F_{1}, and we define f_{1}(X) to be the quotient f(X)/(X − α_{1}) (in F_{1}[X]). Then f_{1} ∈ F_{1}[X], and the same construction gives us a field F_{2} = F_{1}[α_{2}] with α_{2} a root of f_{1}. By continuing in this fashion, we obtain a splitting field.

Remark 2.5. Let n = deg f. In the proof, [F_{1} : F] ≤ n, [F_{2} : F_{1}] ≤ n − 1, ..., and so the degree of the splitting field over F is ≤ n!. Whether or not there exist polynomials of degree n in F [X] whose splitting field has degree n! depends on F. For example, there don’t for n > 1 if F = ℂ or 𝔽_{p}, nor for n > 2 if F = ℝ. However, later we shall see how to write down large numbers (in fact infinitely many) polynomials of degree n in ℚ[X] whose splitting fields have degree n!.

Example 2.6. (a) Let f = X^{p} − 1/X − 1; any field generated by a root of f is a splitting field (if ζ is one root, the remainder are ζ^{2}, ζ^{3}, …, ζ^{p−1}).

(b) Suppose F is of characteristic p, and let f = X^{p} − X − a; any field generated by a root of f is a splitting field (if α is one root, the remainder are α + 1, ..., α + p − 1).

(c) If α is one root of X^{n}−a, then the remaining roots are all of the form ζα, where ζ^{n} = 1. Therefore, if F contains all the nth roots of 1, i.e., if X^{n} − 1 splits in F [X], then F [α] is a splitting field for X^{n} − a. Note that if p is the characteristic of F, then X^{p} − 1 = (X − 1)^{p}, and so F automatically contains all the pth roots of 1.

Proposition 2.7. Let f ∈ F [X], and let E be a splitting field for f, and let Ω ⊃ F be a second field splitting f.

(a) There exists at least one F -homomorphism φ : E → Ω.

(b) The number of F -homomorphisms E → Ω is ≤ [E : F], and = [E : F] if f has deg(f) distinct roots in Ω.

(c) If Ω is also a splitting field for f, then each F -homomorphism E → Ω is an isomorphism. In particular, any two splitting fields for f are F -isomorphic.

Proof. Write E = F [α_{1}, ..., α_{m}], m ≤ deg(f), with the α_{i} the distinct roots of f(X). The minimum polynomial of α_{1} is an irreducible polynomial f_{1} dividing f. As f (hence f_{1}) splits in Ω, Proposition 2.1 shows that there exists an F -homomorphism φ_{1} : F [α_{1}] → Ω, and the number of φ_{1}’s is ≤ deg(f_{1}) = [F [α_{1}] : F], with equality holding when f has distinct roots in Ω.

Next, the minimum polynomial of α_{2} over F [α_{1}] is an irreducible factor f_{2} of f(X) in F [α_{1}][X]. According to Proposition 2.2, each φ_{1} extends to a homomorphism φ_{2} : F [α_{1}, α_{2}] → Ω, and the number of extensions is ≤ deg(f_{2}) = [F [α_{1}, α_{2}] : F [α_{1}]], with equality holding when f has distinct roots in Ω.

On combining these statements we conclude that there exists an F -homomorphism φ : F [α_{1}, α_{2}] → Ω, and the number of such homomorphisms is ≤ [F [α_{1}, α_{2}] : F], with equality holding when f has deg(f) distinct roots in Ω.

After repeating the argument m times, we obtain (a) and (b). For (c), note that, because an F -homomorphism E → Ω is injective, we must have [E : F] ≤ [Ω : F]. If Ω is also a splitting field, then we obtain the reverse inequality also. We therefore have equality, and so any F -homomorphism E → Ω is an isomorphism.

Corollary 2.8. Let E and L be extension fields of F, with E finite over F; then there exists an extension field Ω of L and an F -homomorphism E → Ω.

Proof. Write E = F [α_{1}, …, α_{m}], and let f_{i} be the minimum polynomial of α_{i} over F. Let E′ be a splitting field of f =_{df} ∏ f_{i} regarded as an element of E[X], and replace E with the subfield of E′ generated by F and all the roots of f(X). Thus E is now the splitting field of f(X) ∈ F [X]. Let Ω be a splitting field for f regarded as an element of L[X]. The proposition shows that there is an F -homomorphism E → Ω.

Remark 2.9. After replacing E by its (isomorphic) image in Ω, we will have that E and L are subfields of Ω. This will allow us to assume that E and L are subfields of a common field.

Warning! If E and E′ are splitting fields of f(X) ∈ F [X], then we know there is an F -isomorphism E → E′, but there will in general be no preferred such isomorphism. Error and confusion can result if you simply identify the fields.

2.3. Algebraic closures. Recall that Ω is said to be algebraically closed if every nonconstant polynomial f(X) ∈ Ω[X] has a root in Ω (and hence splits in Ω[X]); equivalently, if the only irreducible polynomials in Ω[X] are those of degree 1. Recall also that a field Ω containing F is said to be an algebraic closure of F if it is algebraic over F and it is algebraically closed. We want to show that (assuming the axiom of choice) every field has an algebraic closure. The following criterion suggests how this might be done.

Lemma 2.10. Suppose that Ω is algebraic over F and every polynomial f ∈ F [X] splits in Ω[X]; then Ω is the algebraic closure of F.

Proof. Let f ∈ Ω[X]. We know (see §1.6) how to construct a finite extension E of Ω containing a root α of f. We want to show that α in fact lies in Ω. Write f = a_{n}X^{n}+⋯+a_{0}, a_{i} ∈ Ω, and consider the sequence of fields F ⊂ F [a_{1}, …, a_{n}] ⊂ F [a_{1}, …, a_{n}, α]. Because each a_{i} is algebraic over F, F [a_{1}, …, a_{n}] is a finite field extension of F, and because f ∈ F [a_{1}, …, a_{n}][X], α is algebraic over F [a_{1}, …, a_{n}]. Therefore α lies in a finite extension of F, and is therefore algebraic over F, i.e., it is the root of a polynomial with coefficients in F. But, by assumption, this polynomial splits in Ω[X], and so all its roots lie in Ω. In particular, α ∈ Ω.

Lemma 2.11. Let Ω ⊃ F; then

E = {α ∈ Ω | α algebraic over F }

is a field.

Proof. If α and β are algebraic over F, then F [α, β] is of finite degree over F, and so is a field (see 1.14). Every element of F [α, β] is algebraic over F, including α ± β, α/β, αβ, . . ..

The field E constructed in the lemma is called the algebraic closure of F in E. The preceding lemma shows that if every polynomial in F [X] splits in Ω[X], then E is an algebraic closure of F. Thus to construct an algebraic closure of F, it suffices to construct an extension in which every polynomial in F [X] splits. We know how to do this for a single polynomial, but passing from there to all polynomials causes set-theoretic problems.

Theorem 2.12 (*). ^{1}Every field has an algebraic closure.

Once we prove the fundamental theorem of algebra, we will know that the algebraic closure in ℂ of any subfield F of ℂ is an algebraic of F. We sketch three proofs of the general result. The first doesn’t assume the axiom of choice, but does assume that F is countable.

Proof. (First proof of 2.12) Because F is countable, it follows that F [X] is countable, i.e., we can list its elements f_{1}(X), f_{2}(X), . . .. Define the fields E_{i} inductively as follows: E_{0} = F; E_{i} is the splitting field of f_{i} over E_{i−1}. Note that E_{0} ⊂ E_{1} ⊂ E_{2} ⊂ · · ·. Define Ω = ∪E_{i}; it is obviously an algebraic closure of F.

Remark 2.13. Since the E_{i} are not subsets of a fixed set, forming the union requires explanation: define Ω^{∗} to be the disjoint union of the E_{i}; let a, b ∈ Ω^{∗}, say a ∈ E_{i} and b ∈ E_{j}; write a ∼ b if a = b when regarded as elements of the larger of E_{i} or E_{j}; verify that ∼ is an equivalence relation, and let Ω = Ω^{∗}/ ∼.

1Results marked with an asterisk require the axiom of choice for their proof.

Proof. (Second proof of 2.12) If A and B are rings containing a field F, then A ⊗_{F} B is a ring containing F, and there are F -homomorphisms A, B → A ⊗_{F} B. More generally, if (A_{i})_{i∈I} is some family of rings each of which contains F, then ⊗_{F} A_{i} is a ring containing F, and there are F -homomorphisms A_{j} → ⊗_{F} A_{i} for each j ∈ I. It is defined to be the quotient of the F -vector space with basis ΠA_{i} by the subspace generated by elements of the form:

• (x_{i}) + (y_{i}) − (z_{i}) with x_{j} + y_{j} = z_{j} for one j ∈ I and x_{i} = y_{i} = z_{i} for all i ≠ j.
• (x_{i}) − a(y_{i}) with x_{j} = ay_{j} for one j ∈ I and x_{i} = y_{i} for all i ≠ j.

It can be made into a ring in an obvious fashion (see Bourbaki, Algèbre, Chapt 3, Appendix).

For each polynomial f ∈ F [X], choose a splitting field E_{f}, and let Ω = (⊗_{f}E_{f})/M where M is a maximal ideal in ⊗_{f}E_{f}—Zorn’s lemma implies that M exists (see below). Then Ω is a field (see 1.1), and there are F -homomorphisms E_{f} → Ω (which must be injective) for each f ∈ F [X]. Since f splits in E_{f}, it must also split in the larger field Ω. The algebraic closure of F in Ω is therefore an algebraic closure of F. (Actually, Ω itself is an algebraic closure of F.)

Lemma 2.14 (Zorn’s). Let (S, ≤) be a nonempty partially ordered set (reflexive, transitive, anti-symmetric, i.e., a ≤ b and b ≤ a ⟹ a = b). Suppose that every totally ordered subset T of S (i.e., for all s, t ∈ T, either s ≤ t or t ≤ s) has an upper bound in S (i.e., there exists an s ∈ S such that t ≤ s for all t ∈ T). Then S has a maximal element (i.e., an element s such that s ≤ s′ ⟹ s = s′).

Zorn’s lemma is equivalent to the Axiom of Choice.

Lemma 2.15 (*). Every nonzero commutative ring A has a maximal ideal.

Proof. Let S be the set of all proper ideals in A, partially ordered by inclusion. If T is a totally ordered set of ideals, then J = ⋃_{I∈T} I is again an ideal, and it is proper because if 1 ∈ J then 1 ∈ I for some I in T. Thus J is an upper bound for T. Now Zorn’s lemma implies that S has a maximal element, which is a maximal ideal in A.

Proof. (Third proof of 2.12) First show that the cardinality of any field algebraic over F is the same as that of F. Next choose an uncountable set Ξ of cardinality greater than that of F, and identify F with a subset of Ξ. Let S be the set triples (E, +, ·) with E ⊂ S and (+, ·) a field structure on E such that (E, +, ·) contains F as a subfield and is algebraic over it. Write (E, +, ·) ≤ (E′, +′, ·′) if the first is a subfield of the second. Apply Zorn’s lemma to show that S has maximal elements, and then show that a maximal element is algebraically closed. (See Jacobson, Lectures in Algebra, III, p144 for the details.)

There do exist naturally occurring fields, not contained in ℂ, that are uncountable. For example, for any field F there is a ring F [[T]] of formal power series ∑_{i≥0} a_{i}T^{i}, a_{i} ∈ F, and its field of fractions is uncountable even if F is finite.

Theorem 2.16 (*). Let Ω be an algebraic closure of F, and let E be an algebraic extension of F; then there is an F -homomorphism E → Ω. If E is also an algebraic closure of F, then any such map is an isomorphism.

Proof. Suppose first that E is countably generated over F, i.e., E = F [α_{1}, ..., α_{n}, …]. Then we can extend the inclusion map F → Ω to F [α_{1}] (map α_{1} to any root of its minimal polynomial in Ω), then to F [α_{1}, α_{2}], and so on.

The uncountable case is a straightforward application of Zorn’s lemma.

Let S be the set of pairs (M, φ_{M}) with M a field F ⊂ M ⊂ E and φ_{M} an F -homomorphim M → Ω. Write (M, φ_{M}) ≤ (N, φ_{N}) if M ⊂ N and φ_{N}|M = φ_{M}. This makes S into a partially ordered subset. Let T be a totally ordered subset of S. Then M′ = ∪_{M∈T} M is a subfield of E, and we can define a homomorphism φ′ : M′ → Ω by requiring that φ′(x) = φ_{M}(x) if x ∈ M. The pair (M′, φ′) is an upper bound for T in S. Hence Zorn’s lemma gives us a maximal element (M, φ) in S. Suppose that M ≠ E. Then there exists an element α ∈ E, α ∉ M. Since α is algebraic over M, we can apply (2.2) to extend φ to M[α], contradicting the maximality of M. Hence M = E, and the proof of the first statement is complete.

If E is algebraically closed, then every polynomial f ∈ F [X] splits in E and hence in φ(E), i.e., f(X) = ∏(X − α_{i}), α_{i} ∈ φ(E). Let α ∈ Ω, and let f(X) be the minimum polynomial of α. Then X −α is a factor of f(X) in Ω[X], but, as we just observed, f(X) splits in φ(E)[X]. Because of unique factorization, this implies that α ∈ φ(E).

The above proof is a typical application of Zorn’s lemma: once we know how to do something in a finite (or countable) situation, Zorn’s lemma allows us to do it in general.

Remark 2.17. Even for a finite field F, there will exist uncountably many isomorphisms from one algebraic closure to a second, none of which is to be preferred over any other. Thus it is (uncountably) sloppy to say that the algebraic closure of F is unique. All one can say is that, given two algebraic closures Ω, Ω′ of F, then, thanks to the axiom of choice, there exists an F -isomorphism Ω → Ω′.

3. The Fundamental Theorem of Galois Theory

In this section, we prove the fundamental theorem of Galois theory, which gives a one-to-one correspondence between the subfields of the splitting field of a separable polynomial and the subgroups of the Galois group of f.

3.1. Multiple roots. Let f, g ∈ F [X]. Even when f and g have no common factor in F [X], you might expect that they could acquire a common factor in Ω[X] for some Ω ⊃ F. In fact, this doesn’t happen—gcd’s don’t change when the field is extended.

Proposition 3.1. Let f and g be polynomials in F [X], and let Ω ⊃ F. If r(X) is the gcd of f and g computed in F [X], then it is also the gcd of f and g in Ω[X]. In particular, if f and g are monic and irreducible and f ≠ g, then they do not have a common root in any extension field of F.

Proof. Let r_{F} (X) and r_{Ω}(X) be the greatest common divisors of f and g in F [X] and Ω[X] respectively. Certainly r_{F} (X)|r_{Ω}(X) in Ω[X]. The Euclidean algorithm shows that there are polynomials a and b in F [X] such that

a(X)f(X) + b(X)g(X) = r_{F} (X).

Since r_{Ω}(X) divides f and g in Ω[X], it divides the left-hand side of the equation, and therefore also the right. Hence r_{Ω} = r_{F}.

For the second statement, note that the hypotheses imply that gcd(f, g) = 1 (in F [X]). Hence they can’t have a common factor X − α in any extension field.

The proposition allows us to write gcd(f, g), without reference to a field.

Let f ∈ F [X], and let f(X) = ∏(X − α_{i})^{m_{i}}, α_{i} distinct, be a splitting of f over some large field Ω ⊃ F. We then say that α_{i} is a root of multiplicity m_{i}. A root of multiplicity one is said to be simple.

We say that f has multiple roots if it has roots of multiplicity > 1 in some big field Ω. It then has multiple roots in the subfield of Ω generated by its roots, and because any two splitting fields are F -isomorphic, this shows that f will have roots of multiplicity > 1 in every field containing F in which it splits.

If f has multiple factors in F [X], say f = ∏ f_{i}(X)^{m_{i}} with some m_{i} > 1, then obviously it will have multiple roots. If f = ∏ f_{i} with the f_{i} distinct monic irreducible polynomials, then the proposition shows that f can only have multiple roots if one of the f_{i} has multiple roots. Thus it remains to examine irreducible polynomials for multiple roots.

Example 3.2. Let F be of characteristic p, and assume that F has an element a that is not a pth-power (e.g., F = 𝔽_{p}(T); a = T). Then X^{p} − a is irreducible in F [X], but X^{p} − a = (X − α)^{p} in its splitting field. Thus an irreducible polynomial can have multiple roots.

We define the derivative f′(X) of a polynomial f(X) = ∑ a_{i}X^{i} to be ∑ ia_{i}X^{i−1}. When F = ℝ, this agrees with the usual definition. The usual rules for differentiating sums and products still hold, but note that the derivative of X^{p} is zero in characteristic p.

Proposition 3.3. Let f be a (monic) irreducible polynomial in F [X]. The following statements are equivalent:

(a) f has at least one multiple root (in a splitting field);

(b) gcd(f, f′) ≠ 1;

(c) F has characteristic p ≠ 0 and f(X) = g(X^{p}), some g ∈ F [X];

(d) all the roots of f are multiple.

Proof. (a) ⟹ (b). Let α be a multiple root of f, and write f = (X − α)^{m}g(X), m > 1, in some splitting field. Then

f′(X) = m(X − α)^{m−1}g(X) + (X − α)^{m}g′(X).

Hence f′(α) = 0, and so gcd(f, f′) ≠ 1.

(b) ⟹ (c). Since f is irreducible and deg(f′) < deg(f),

gcd(f, f′) ≠ 1 ⟹ f′ = 0 ⟹ f = g(X^{p}).

(c) ⟹ (d). Suppose f(X) = g(X^{p}), and let g(X) = ∏(X − a_{i})^{m_{i}} in some splitting field. Then

f(X) = g(X^{p}) = ∏(X^{p} − a_{i})^{m_{i}} = ∏(X − α_{i})^{pm_{i}}

where α_{i}^{p} = a_{i} (in some big field). Hence every root of f(X) has multiplicity at least p. (d) ⟹ (a). Every root multiple ⟹ at least one root multiple (I hope).

Definition 3.4. A polynomial f ∈ F [X] is said to be separable if all its irreducible factors have simple roots.

Note that the preceding discussion shows that f is not separable if and only if

(a) the characteristic of F is p ≠ 0, and

(b) at least one of the irreducible factors of f is a polynomial in X^{p}.

A field F is said to be perfect if all polynomials in F [X] are separable.

Proposition 3.5. A field F is perfect if and only if it either

• has characteristic 0, or

• it has characteristic p and F = F^{p} (i.e., every element of F is a pth power).

Proof. ⟹ : If char F = p and it contains an element a that is not a pth power, then F [X] contains a nonseparable polynomial, namely, X^{p} − a.

⟸ : If char F = p and F = F^{p}, then every polyonomial in X^{p} is a pth power— ∑ a_{i}X^{p} = (∑ b_{i}X)^{p} if a_{i} = b_{i}^{p}—and so can’t be irreducible.

Example 3.6. (a) All finite fields are perfect (because a ↦ a^{p} is an injective homomorphism F → F, which must be surjective if F is finite). In fact, any field algebraic over 𝔽_{p} is perfect.

(b) If F_{0} has characteristic p, then F = F_{0}(X) is not perfect (because X is not a pth power).

3.2. Groups of automorphisms of fields. Consider fields E ⊃ F. We write Aut(E/F) for the group of F -automorphisms of E, i.e., automorphisms σ : E → E such that σ(a) = a for all a ∈ F.

Example 3.7. (a) There are two obvious automorphisms of ℂ, namely, the identity map and complex conjugation. We’ll see later (last section) that by using the Axiom of Choice, one can construct uncountably many more. They are all noncontinuous and (I’ve been told) nonmeasurable—hence they require the Axiom of Choice for their construction.

(b) Let E = ℂ(X). Then Aut(E/ℂ) consists of the maps X ↦ \frac{aX+b}{cX+d}, ad − bc ≠ 0 (Jacobson, Lectures III, p158), and so Aut(E/ℂ) = PGL_{2}(ℂ). Analysts will note that this is the same as the automorphism group of the Riemann sphere. This is not a coincidence: the field of meromorphic functions on the Riemann sphere ℙ_{ℂ}^{1} is ℂ(z) ≈ ℂ(X), and so there is a map Aut(ℙ_{ℂ}^{1}) → Aut(ℂ(z)/ℂ), which one can show is an isomorphism.

(c) The group Aut(ℂ(X_{1}, X_{2})/ℂ) is quite complicated—there is a map

PGL_{3}(ℂ) = Aut(ℙ_{ℂ}^{2}) ↪ Aut(ℂ(X_{1}, X_{2})/ℂ),

but this is very far from being surjective. When there are more X’s, the group is unknown. (The group Aut(ℂ(X_{1}, …, X_{n})/ℂ) is the group of birational automorphisms of ℙ_{ℂ}^{n}. It is called the Cremona group. Its study is part of algebraic geometry.)

In this section, we shall be concerned with the groups Aut(E/F) when E is a finite extension of F.

Proposition 3.8. If E is a splitting field of a monic separable polynomial f ∈ F [X], then Aut(E/F) has order [E : F].

Proof. Let f = ∏ f_{i}^{m_{i}}, with the f_{i} monic irreducible and distinct. The splitting field of f is the same as the splitting field of ∏ f_{i}. Hence we may assume f is a product of distinct monic separable irreducible polynomials, and hence has deg f distinct roots in E. Now Proposition 2.7b shows that there are [E : F] distinct F -homomorphisms E → E; they are automatically isomorphisms.

Example 3.9. (a) Let E = F [α], f(α) = 0; if f has no other root in E than α, then Aut(E/F) = 1. For example, if \sqrt[3]{2} denotes the real cube root of 2, then Aut(ℚ[\sqrt[3]{2}]/ℚ) = 1. Thus, in the proposition, it is essential that E be a splitting field.

(b) Let F be a field of characteristic p ≠ 0, and let a be an element of F that is not a pth power. The splitting field of f = X^{p} − a is F [α] where α is the unique root of f. Then Aut(E/F) = 1. Thus, in the proposition, it is essential that E be the splitting field of a separable polynomial.

When G is a group of automorphisms of a field E, we write

E^{G} = Inv(G) = {α ∈ E | σα = α, all σ ∈ G}.

It is a subfield of E, called the subfield of G-invariants of E or the subfield of E fixed by G.

We have maps

G ↦ Inv(G) F ↦ Aut(E/F).

Goal: Show that when E is the splitting field of a separable polynomial in F [X] and G = Aut(E/F), then

H ↦ Inv(H), M ↦ Aut(E/M)

give a one-to-one correspondence between the set of intermediate fields M, F ⊂ M ⊂ E, and the set of subgroups H of G.

Lemma 3.10 (E. Artin). Let G be a finite group of automorphisms of a field E, and let F = E^{G}; then [E : F] ≤ (G : 1).

Proof. Let G = {σ_{1} = 1, …, σ_{m}}, and let α_{1}, …, α_{n} be n > m elements of E. We shall show that the α_{i} are linearly dependent over F. In the system

σ_{1}(α_{1})x_{1} + ⋯ + σ_{1}(α_{n})x_{n} = 0
⋯  ⋯
σ_{m}(α_{1})x_{1} + ⋯ + σ_{m}(α_{n})x_{n} = 0

there are m equations and n > m unknowns, and hence there are nontrivial solutions (in E). Choose a nontrivial solution (c_{1}, …, c_{n}) with the fewest nonzero elements. After renumbering the α_{i}’s, we may suppose that c_{1} ≠ 0, and then (after multiplying by a scalar) that c_{1} = 1. With these normalizations, we’ll see that all c_{i} ∈ F. Hence the first equation (recall σ_{1} = 1)

α_{1}c_{1} + ⋯ + α_{n}c_{n} = 0

shows that the α_{i} are linearly dependent over F.

If not all c_{i} are in F, then σ_{k}(c_{i}) ≠ c_{i} for some i, k. On apply σ_{k} to the equations

σ_{1}(α_{1})c_{1} + ⋯ + σ_{1}(α_{n})c_{n} = 0
⋯  ⋯
σ_{m}(α_{1})c_{1} + ⋯ + σ_{m}(α_{n})c_{n} = 0

and using that {σ_{k}σ_{1}, …, σ_{k}σ_{m}} is a permutation of {σ_{1}, …, σ_{m}}, we find that (1, …, σ_{k}(c_{i}), …) is also a solution to the system of equations. On subtracting it from the first, we obtain a solution (0, …, c_{i} − σ_{k}(c_{i}), …), which is nonzero (look at the ith coordinate), but has more zeros than the first solution (look at the first coordinate)— contradiction.

3.3. Separable, normal, and Galois extensions. An algebraic extension E/F is said to be separable if the minimum polynomial of every element of E is separable, i.e., doesn’t have multiple roots (in a splitting field); equivalently, if every irreducible polynomial in F [X] having a root in E is separable. Thus E/F is inseparable if and only if

(a) F is nonperfect, and in particular has characteristic p ≠ 0, and

(b) there is an element α of E whose minimal polynomial is of the form g(X^{p}), g ∈ F [X].

For example, E = 𝔽_{p}(T) is an inseparable extension of 𝔽_{p}(T^{p}).

An algebraic extension E/F is normal if the minimum polynomial of every element of E splits in E; equivalently, if every irreducible polynomial f ∈ F [X] having a root in E splits in E.

Thus if f ∈ F [X] is irreducible of degree m and has a root in E, then

\begin{matrix}E/F separable & ⟹ & roots of f distinct \\ E/F normal & ⟹ & f splits in E \end{matrix} \} ⟹ f has m distinct roots in E.

Therefore, E/F is normal and separable if and only if, for each α ∈ E, the minimum polynomial of α has [F [α] : F] distinct roots in E.

Example 3.11. (a) The field ℚ[\sqrt[3]{2}], where \sqrt[3]{2} is the real cube root of 2, is separable but not normal over ℚ (X^{3} − 2 doesn’t split in ℚ[α]).

(b) The field 𝔽_{p}(T) is normal but not separable over 𝔽_{p}(T^{p})—it is the splitting field of the inseparable polynomial X^{p} − T^{p}.

Theorem 3.12. Let E be an extension field of F. The following statements are equivalent:

(a) E is the splitting field of a separable polynomial f ∈ F [X];

(b) F = E^{G} for some finite group of automorphisms of E;

(c) E is normal and separable, and of finite degree, over F.

Moreover, if E is as in (a), then F = E^{Aut(E/F)}; if G and F are as in (b) then G = Aut(E/F).

Proof. (a) ⟹ (b). Let G = Aut(E/F), and let F′ = E^{G} ⊃ F. Then E is also the splitting field of f ∈ F′[X], and f is still separable when regarded as a polynomial over F′. Hence Proposition 3.8 shows that

[E : F′] = # Aut(E/F′)
[E : F] = # Aut(E/F).

Since Aut(E/F′) = Aut(E/F) = G, we conclude that F = F′, and so F = E^{Aut(E/F)}.

(b) ⟹ (c). By Artin’s lemma, we know that [E : F] ≤ (G : 1); in particular, it is finite. Let α ∈ E and let f be the minimum polynomial of α; we have to prove that f splits into distinct factors in E. Let {α_{1} = α, ..., α_{m}} be the orbit of α under G, and let

g(X) = ∏(X − α_{i}) = X^{m} + a_{1}X^{m−1} + ⋯ + a_{m}.

Any σ ∈ G merely permutes the α_{i}. Since the a_{i} are symmetric polynomials in the α_{i}, we find that σa_{i} = a_{i} for all i, and so g(X) ∈ F [X]. It is monic, and g(α) = 0, and so f(X)|g(X) (see p7). But also g(X)|f(X), because each α_{i} is a root of f(X) (if α_{i} = σα, then applying σ to the equation f(α) = 0 gives f(α_{i}) = 0). We conclude that f(X) = g(X), and so f(X) splits into distinct factors in E.

(c) ⟹ (a). Because E has finite degree over F, it is generated over F by a finite number of elements, say, E = F [α_{1}, ..., α_{m}], α_{i} ∈ E, α_{i} algebraic over F. Let f_{i} be the minimum polynomial of α_{i} over F. Because E is normal over F, each f_{i} splits in E, and so E is the splitting field of f = ∏ f_{i}. Because E is separable over F, f is separable.

Finally, we have to show that if G is a finite group acting on a field E, then G = Aut(E/E^{G}). We know that:

• [E : E^{G}] ≤ (G : 1) (Artin),

• G ⊂ Aut(E/E^{G}), and,

• E is the splitting field of a separable polynomial in E^{G}[X] (because b ⟹ a), and so (by 3.8) the order of Aut(E/E^{G}) is [E : E^{G}].

Now the inequalities

[E : E^{G}] ≤ (G : 1) ≤ (Aut(E/E^{G}) : 1) = [E : E^{G}]

must be equalities, and so G = Aut(E/E^{G}).

An extension of fields E ⊃ F satisfying the equivalent conditions of the proposition is called a Galois extension, and Aut(E/F) is called the Galois group Gal(E/F) of E over F. Note that we have shown that F = E^{Gal(E/F)}.

Remark 3.13. Let E be Galois over F with Galois group G, and let α ∈ E. The elements α_{1} = α, α_{2}, ..., α_{m} of the orbit of α are called the conjugates of α. In the course of the proof of the the above theorem we showed that the minimum polynomial of α is ∏(X − α_{i}).

Corollary 3.14. Every finite separable extension E of F is contained in a finite Galois extension.

Proof. Let E = F [α_{1}, ..., α_{m}]. Let f_{i} = minimum polynomial of α_{i} over F, and take E′ to be the splitting field of ∏ f_{i} over F.

Corollary 3.15. Let E ⊃ M ⊃ F; if E is Galois over F, then it is Galois over M.

Proof. We know E is the splitting field of some f ∈ F [X]; it is also the splitting field of f regarded as an element of M[X].

Remark 3.16. When we drop the assumption that E is separable over F, we can still say something. Let E be a finite extension of F. An element α ∈ E is said to be separable over F if its minimum polynomial over F is separable. The elements of E separable over F form a subfield E′ of E that is separable over F; write [E : F]_{sep} = [E′ : F] (separable degree of E over F). If Ω is an algebraically closed field containing F, then there are exactly [E : F]_{sep} F -homomorphisms E → Ω. When E ⊃ M ⊃ F (finite extensions),

[E : F]_{sep} = [E : M]_{sep}[M : F]_{sep}.

In particular,

E is separable over F ⟺ E is separable over M and M is separable over F.

3.4. The fundamental theorem of Galois theory.

Theorem 3.17 (Fundamental theorem of Galois theory). Let E be a Galois extension of F, and let G = Gal(E/F). The maps H ↦ E^{H} and M ↦ Gal(E/M) are inverse bijections between the set of subgroups of G and the set of intermediate fields between E and F:

{subgroups of G} ↔ {intermediate fields F ⊂ M ⊂ E}.

Moreover:

(a) The correspondence is inclusion-reversing, i.e., H_{1} ⊃ H_{2} ⟺ E^{H_{1}} ⊂ E^{H_{2}}.

(b) Indexes equal degrees, i.e., (H_{1} : H_{2}) = [E^{H_{2}} : E^{H_{1}}].

(c) The group σHσ^{−1} ↔ σE, i.e., E^{σHσ^{−1}} = σ(E^{H}); Gal(E/σM) = σ Gal(E/M)σ^{−1}.

(d) The group H is normal in G ⟺ E^{H} is normal (hence Galois) over F, in which

case

Gal(E^{H}/F) = G/H.

Proof. Let H be a subgroup of G. We first have to show that Gal(E/E^{H}) = H. But we have already observed that E is Galois over E^{H}, and Theorem 3.12 shows that Gal(E/E^{H}) = H.

Next let M be an intermediate field, and let H = Gal(E/M). We have to show that E^{H} = M, but this is again proved in Theorem 3.12.

Thus we have proved that Inv(·) and Gal(E/·) are inverse bijections.

(a) We have the obvious implications:

H_{1} ⊃ H_{2} ⟹ E^{H_{1}} ⊂ E^{H_{2}} ⟹ Gal(E/E^{H_{1}}) ⊃ Gal(E/E^{H_{2}}).

But Gal(E/E^{H_{i}}) = H_{i}.

(b) In the case H_{2} = 1, the first equality follows from (3.8) and (3.12). The general case follows, using that

(H_{1} : 1) = (H_{1} : H_{2})(H_{2} : 1) and [E : E^{H_{1}}] = [E : E^{H_{2}}][E^{H_{2}} : E^{H_{1}}].

(c) If H = {τ ∈ G | τα = α, all α ∈ M}, i.e., H = Gal(E/M), then σHσ^{−1} = (τ ∈ G | τσα = σα, all α ∈ M}, i.e., σHσ^{−1} = Gal(E/σM).

(d) Assume H to be normal in G, and let M = E^{G}. Because σHσ^{−1} = H for all σ ∈ G, we must have σM = M for all σ ∈ G, i.e., the action of G on E stabilizes M. We therefore have a homomorphism

σ ↦ σ|M : G → Aut(M/F)

with kernel H. Let G′ be the image. Then F = M^{G′}, and so M is Galois over F with Galois group G′ (by Theorem 3.12).

Conversely, assume that M is normal over F, and write E = F [α_{1}, ..., α_{m}]. For σ ∈ G, σα_{i} is a root of the minimum polynomial of α_{i} over F, and so lies in M. Hence σM = M, and this implies that σHσ^{−1} = H (by (c)).

Remark 3.18. The theorem shows that there is an order reversing bijection between the intermediate fields of E/F and the subgroups of G. Using this we can read off more results. For example let M_{1}, M_{2}, …, M_{r} be intermediate fields, and let H_{i} be the subgroup corresponding to M_{i} (i.e., H_{i} = Gal(E/M_{i})). Then (by definition) M_{1}M_{2} ⋯ M_{r} is the smallest field containing all M_{i}; hence it must correspond to the largest subgroup contained in all H_{i}, which is ⋂ H_{i}. Therefore

Gal(E/M_{1} ⋯ M_{r}) = H_{1} ∩ ... ∩ H_{r}.

Example 3.19. We analyse the extension ℚ[ζ]/ℚ, where ζ is the primitive 7th root of 1, say ζ = e^{2πi/7}. Then ℚ[ζ] is the splitting field of the irreducible polynomial

X^{6} + X^{5} + X^{4} + X^{3} + X^{2} + X + 1

(see 1.31), and so is Galois of degree 6 over ℚ. For any σ ∈ G, σζ = ζ^{i}, some i, 1 ≤ i ≤ 6, and the map σ ↦ i defines an isomorphism Gal(ℚ[ζ]/ℚ) → (ℤ/7ℤ)^{×}. Let σ be the element of Gal(ℚ[ζ]/ℚ) such that σζ = ζ^{3}. Then σ generates Gal(ℚ[ζ]/ℚ) because the class of 3 in (ℤ/7ℤ)^{×} generates it (the powers of 3 mod 7 are 3, 2, 6, 4, 5, 1). We investigate the subfields of ℚ[ζ] corresponding to the subgroups < σ^{3} > and < σ^{2} >.

Note that σ^{3}ζ = ζ^{6} = ζ̄ (complex conjugate of ζ). The subfield of ℚ[ζ] corresponding to < σ^{3} > is ℚ[ζ + ζ̄], and ζ + ζ̄ = 2 cos \frac{2π}{7}. Since < σ^{3} > is a normal subgroup of < σ >, ℚ[ζ + ζ̄] is Galois over ℚ, with Galois group < σ > / < σ^{3} > . The conjugates of α_{1} = ζ + ζ̄ are α_{3} = ζ^{3} + ζ^{−3}, α_{2} = ζ^{2} + ζ^{−2}. Direct calculation shows that

∑ α_{i} = ∑_{i=1}^{6} ζ^{i} = −1,

α_{1}α_{2}α_{3} = (ζ+ζ^{6})(ζ^{2}+ζ^{5})(ζ^{3}+ζ^{4}) = (ζ+ζ^{3}+ζ^{4}+ζ^{6})(ζ^{3}+ζ^{4}) = (ζ^{4}+ζ^{6}+1+ζ^{2}+ζ^{5}+1+ζ+ζ^{3}) = 1. α_{1}α_{2} + α_{1}α_{3} + α_{2}α_{3} = −2.

Hence the minimum polynomial of ζ + ζ̄ is

g(X) = X^{3} + X^{2} − 2X − 1.

The minimum polynomial of cos \frac{2π}{7} = \frac{α_{1}}{2} is therefore

\frac{g(2X)}{8} = X^{3} + X^{2}/2 − X/2 − 1/8.

The subfield of ℚ[ζ] corresponding to < σ^{2} > is generated by β = ζ +ζ^{2} +ζ^{4}. Let β′ = σβ. Then (β − β′)^{2} = −7. Hence the field fixed by < σ^{2} > is ℚ[\sqrt{−7}].

Example 3.20. We compute the Galois group of the splitting field E of X^{5} − 2 ∈ ℚ[X]. Recall (from the Homework) that E = ℚ[ζ, α] where ζ is a primitive 5th root of 1, and α is a root of X^{5} − 2. For example, we could take E to be the splitting field of X^{5} − 2 in ℂ, with ζ = e^{2πi/5} and α equal to the real 5th root of 2. We have the picture:

ℚ[ζ, α]
N ⧸ ⧹ H
ℚ[ζ]  ℚ[α]
G/N ⧹ ⧸
ℚ

The degrees

[ℚ[ζ] : ℚ] = 4, [ℚ[α] : ℚ] = 5, [ℚ[ζ, α] : ℚ] = 20.

Hence G = Gal(ℚ[ζ, α]/ℚ) has order 20, and the subgroups N and H corresponding to ℚ[ζ] and ℚ[α] have orders 5 and 4 respectively (because N = Gal(ℚ[ζ, α]/ℚ[ζ] …). Because ℚ[ζ] is normal over ℚ (it is the splitting field of X^{5} − 1), N is normal in G. Because ℚ[ζ] · ℚ[α] = ℚ[ζ, α], we have H ∩ N = 1 (see 3.18), and so G = N ⋊_{θ} H. We have H ≈ G/N ≈ (ℤ/5ℤ)^{×}, which is cyclic, being generated by the class of 2. Let τ be the generator of H corresponding to 2 under this isomorphism, and let σ be a generator of N. Thus σ(α) is another root of X^{5} − 2, which we can take to be ζα (after possibly replacing σ by a power). Hence:

\{ \begin{matrix}τζ & = & ζ^{2} \\ τα & = & α \end{matrix} \{ \begin{matrix}σζ & = & ζ \\ σα & = & ζα. \end{matrix}

Note that τστ^{−1}(α) = τσα = τ(ζα) = ζ^{2}α and it fixes ζ; therefore τστ^{−1} = σ^{2}. Thus G has generators σ and τ and defining relations

σ^{5} = 1, τ^{4} = 1, τστ^{−1} = σ^{2}.

The subgroup H has five conjugates, which correspond to the five fields ℚ[ζ^{i}α],

σ^{i}Hσ^{−i} ↔ σ^{i}ℚ[α] = ℚ[ζ^{i}α], 1 ≤ i ≤ 5.

Definition 3.21. An extension E ⊃ F is called a cyclic, abelian, ..., solvable extension if it is Galois with cyclic, abelian, ..., solvable Galois group.

3.5. Constructible numbers revisited. Earlier, we showed that a number α is constructible if and only if it is contained in a field ℚ[\sqrt{a_{1}}] ⋯ [\sqrt{a_{r}}]. In particular

α constructible ⟹ [ℚ[α] : ℚ] = 2^{s} some s.

Now we can prove a partial converse to this last statement.

Theorem 3.22. If α is contained in a Galois extension of ℚ of degre 2^{r} then it is constructible.

Proof. Suppose α ∈ E where E is Galois over ℚ of degree 2^{r}, and let G = Gal(E/ℚ). From a theorem on the structure of p-groups, we know there will be a sequence of groups

{1} ⊂ G_{1} ⊂ G_{2} ⊂ ⋯ ⊂ G_{r} = G

with G_{i}/G_{i−1} of order 2. Correspondingly, there will be a sequence of fields,

ℚ ⊂ E_{1} ⊂ E_{2} ⊂ ⋯ ⊂ E_{r} = E

with E_{i} of degree 2 over E_{i−1}.

But (see below), every quadratic extension is obtained by extracting a square root, and we know that square roots can be constructed using only a ruler and compass. This proves the theorem.

Lemma 3.23. Let E/F be a quadratic extension of fields of characteristic ≠ 2. Then E = F [\sqrt{d}] for some d ∈ F.

Proof. Let α ∈ E, α ∉ F, and let X^{2} + bX + c be the minimum polynomial of α. The α = \frac{−b±\sqrt{b^{2}−4c}}{2}, and so E = F [\sqrt{b^{2} − 4c}].

Corollary 3.24. If p is a prime of the form 2^{k} + 1, then cos \frac{2π}{p} is constructible.

Proof. The field ℚ[e^{2πi/p}] is Galois over ℚ with Galois group G ≈ (ℤ/pℤ)^{×}, which has order p − 1 = 2^{k}.

Thus a regular p-gon, p prime, is constructible if and only if p is a Fermat prime, i.e., of the form 2^{2^{r}} + 1. For example, we have proved that the regular 65537-polygon is constructible, without (happily) having to exhibit an explicit formula for cos \frac{2π}{65537}.

3.6. Galois group of a polynomial. If the polynomial f ∈ F [X] is separable, then its splitting field E is Galois over F, and we call Gal(E/F) the Galois group G_{f} of f.

Let f = ∏_{i=1}^{n}(X − α_{i}) in the splitting field E. We know elements of Gal(E/F) map roots of f to roots of f, i.e., they map the set {α_{1}, α_{2}, …, α_{n}} into itself. Since they are automorphisms, they define permutations of {α_{1}, α_{2}, …, α_{n}}. As E = F [α_{1}, ..., α_{n}], an element of Gal(E/F) is uniquely determined by its action on {α_{1}, α_{2}, …, α_{n}}. Thus G_{f} can be identified with a subset of Sym({α_{1}, α_{2}, …, α_{n}}) ≈ S_{n}. From the definitions, one sees that G_{f} consists of the permutations σ of {α_{1}, α_{2}, …, α_{n}} with the property

P ∈ F [X_{1}, …, X_{m}], P (α_{1}, …, α_{m}) = 0 ⟹ P (σα_{1}, …, σα_{m}) = 0.

This gives a description of G_{f} without mentioning fields or abstract groups (neither of which were available to Galois).

Note that (G_{f} : 1) ≤ deg(f)!.

3.7. Solvability of equations.

Let f be a polynomial. We say the equation f(X) = 0 is solvable (by extracting radicals) if there is a tower

F = F_{0} ⊂ F_{1} ⊂ F_{2} ⊂ ⋯ ⊂ F_{m}

such that

(a) F_{i} = F_{i−1}[α_{i}], α_{i}^{m_{i}} ∈ F_{i−1};

(b) F_{m} contains a splitting field for f.

Theorem 3.25. (Galois, 1832) Let F be a field of characteristic zero. The equation f = 0 is solvable if and only if the Galois group of f is solvable.

We shall prove this later. Also we shall exhibit polynomials f(X) ∈ ℚ[X] with Galois group S_{n}, which therefore are not solvable when n ≥ 5.

Remark 3.26. If F has characteristic p, then the theorem fails for two reasons:

(i) f may not be separable, and so not have a Galois group; (ii) X^{p} − X − a is not solvable by radicals.

If the definition of solvable is changed to allow extensions of the type in (ii) in the chain, and f is required to be separable then the theorem becomes true in characteristic p.

4. Computing Galois Groups.

In this section, we investigate general methods for computing Galois groups.

4.1. When is G_{f} ⊂ A_{n}? Consider a polynomial

f(X) = X^{n} + a_{1}X^{n−1} + ⋯ + a_{n}

and let f(X) = ∏_{i=1}^{n}(X − α_{i}) in some splitting field. Set

Δ(f) = ∏_{1≤i<j≤n}(α_{i} − α_{j}), D(f) = Δ(f)^{2} = ∏_{1≤i<j≤n}(α_{i} − α_{j})^{2}.

Note that D(f) ≠ 0 if f has a only simple roots, i.e., if f is separable with no multiple factors. Identify G_{f} with a subgroup of Sym({α_{1}, …, α_{n}}) (as in §3.6).

Proposition 4.1. Assume f is separable, and let σ ∈ G_{f}.

(a) σΔ(f) = sign(σ)Δ(f), where sign(σ) is the signature of σ.

(b) σD(f) = D(f).

Proof. The first equation follows immediately from the definition of the signature of σ (see Groups, p31), and the second equation is obtained by squaring the first.

Corollary 4.2. Let f(X) ∈ F [X] be of degree n and have only simple roots. Let F_{f} be a splitting field for f, so that G_{f} = Gal(F_{f}/F).

(a) The discriminant D(f) ∈ F.

(b) The subfield of F_{f} corresponding to A_{n} ∩ G_{f} is F [Δ(f)]. Hence

G_{f} ⊂ A_{n} ⟺ Δ(f) ∈ F ⟺ D(f) is a square in F.

Proof. (a) We know that D(f) is an element of F_{f} fixed by G_{f} =_{df} Gal(F_{f}/F). Therefore it lies in F (by the Fundamental Theorem of Galois Theory).

(b) Because f has simple roots, Δ(f) ≠ 0, and so the formula σΔ(f) = sign(σ)Δ(f) shows that σ fixes Δ(f) ⟺ σ ∈ A_{n}. Therefore G_{f} ∩ A_{n} is the subgroup of G_{f} corresponding to F [Δ(f)], and so G_{f} ∩ A_{n} = G_{f} ⟺ F [Δ(f)] = F.

The discriminant of f can be expressed as a universal polynomial in the coefficients of f—we shall prove this later. For example:

D(aX^{2} + bX + c) = b^{2} − 4ac
D(X^{3} + bX + c) = −4b^{3} − 27c^{2}.

By completing the cube, one can put any cubic polynomial in this form.

The formulas for the discriminant rapidly become very complicated, for example, that for X^{5} + aX^{4} + bX^{3} + cX^{2} + dX + e has about 60 terms. Fortunately, Maple knows them: the syntax is “discrim(f,X);” where f is a polynomial in the variable X.

Remark 4.3. Suppose F ⊂ ℝ. Then D(f) will not be a square if it is negative. It is known that the sign of D(f) is (−1)^{s} where 2s is the number of nonreal roots of f in ℂ. Thus if s is odd, then G_{f} is not contained in A_{n}. This can be proved more directly by noting that complex conjugation will act on the roots as the product of s transpositions (cf. the proof of Proposition 4.13). Of course the converse is not true: when s is even, G_{f} is not necessarily contained in A_{n}.

4.2. When is G_{f} transitive?

Proposition 4.4. Let f(X) ∈ F [X] have only simple roots. Then f(X) is irreducible if and only if G_{f} permutes the roots of f transitively.

Proof. ⟹ : If α and β are two roots of f(X) in a splitting field F_{f} for f, then they both have f(X) as their minimum polynomial, and so there is a natural F -isomorphism F [α] → F [β], namely,

F [α] ≈ F [X]/(f(X)) ≈ F [β], α ↔ X ↔ β.

Write F_{f} = F [α_{1}, α_{2}, ...] with α_{1} = α and α_{2}, α_{3}, … the other roots of f(X). Then the F -isomorphism F [α] → F [β] extends (step by step) to a homomorphism F [α_{1}, α_{2}, ...] → F_{f} (see 2.7), which must be an isomorphism.

⟸ : Let g(X) ∈ F [X] be an irreducible factor of f, and let α be one of its roots. If β is a second root of f, then (by assumption) β = σα for some σ ∈ G_{f}. Now the equation

0 = σg(α) =^{g(X)∈F [X]} g(σα)

shows that β is also a root of g, and we see that we must have f(X) = g(X).

Note that when f(X) is irreducible of degree n, then n|(G_{f} : 1) because [F [α] : F] = n and [F [α] : F]|[F_{f} : F] = (G_{f} : 1). Thus G_{f} is a transitive subgroup of S_{n} whose order is divisible by n.

4.3. Polynomials of degree ≤ 3.

Example 4.5. Let f(X) ∈ F [X] be a polynomial of degree 2. Then f is inseparable ⟺ F has characteristic 2 and f(X) = X^{2} − a for some a ∈ F \ F^{2}. If f is separable, then G_{f} = 1(= A_{2}) or S_{2} according as D(f) is a square in F or not.

Example 4.6. Let f(X) ∈ F [X] be a polynomial of degree 3. We can assume f to be irreducible, for otherwise we are essentially back in the previous case. Then f is inseparable ⟺ F has characteristic 3 and f(X) = X^{3} − a some a ∈ F \ F^{3}. If f is separable, then G_{f} is a transitive subgroup of S_{3} whose order is divisible by 3. There are only two possibilities: G_{f} = A_{3}(=< (123) >) or S_{3} according as D(f) is a square in F or not.

For example, X^{3} − 3X + 1 ∈ ℚ[X] is irreducible (apply 1.4), its discriminant is −4(−3)^{3} − 27 = 81 = 9^{2}, and so its Galois group is A_{3}.

On the other hand, X^{3}+3X +1 ∈ ℚ[X] is also irreducible (apply 1.4), but its discriminant is −135 which is not a square in ℚ, and so its Galois group is S_{3}.

4.4. Quartic polynomials. Let f(X) be a quartic polynomial, and assume that the roots of f are simple. In order to determine G_{f} we shall exploit the fact that S_{4} has

V = {1, (12)(34), (13)(24), (14)(23)}

as a normal subgroup—it is normal because it contains all elements of type 2+2—see Groups p34. Let E be the splitting field of f, and let f(X) = ∏(X − α_{i}) in E. We identify the Galois group G_{f} of f with a subgroup of the symmetric group S_{4} = Sym({α_{1}, α_{2}, α_{3}, α_{4}}). Consider the partially symmetric elements

α = α_{1}α_{2} + α_{3}α_{4}
β = α_{1}α_{3} + α_{2}α_{4}
γ = α_{1}α_{4} + α_{2}α_{3}.

They are distinct elements of E because the α_{i} are distinct, e.g.,

α − β = α_{1}(α_{2} − α_{3}) + α_{4}(α_{3} − α_{2}) = (α_{1} − α_{4})(α_{2} − α_{3}).

The group Sym({α_{1}, α_{2}, α_{3}, α_{4}}) permutes {α, β, γ} transitively. The stabilizer of each of α, β, γ must therefore be a subgroup of index 3 in S_{4}, and hence has order 8. For example, the stabilizer of β is < (1234), (13) >. Groups of order 8 in S_{4} are Sylow 2-subgroups. There are three of them, all isomorphic to D_{4}. By the Sylow theorems, V is contained in a Sylow 2subgroup, and, because they are conjugate and it is normal, it must be contained in all three. It follows that V is the intersection of the three Sylow 2-subgroups. Each Sylow 2-subgroup stabilizes exactly one of α, β, or γ, and therefore their intersection V is the subgroup of S_{4} fixing α, β, and γ.

Lemma 4.7. The field M = F [α, β, γ] corresponds to G_{f} ∩ V. Hence M is Galois over F, with Galois group G/G ∩ V.

Proof. The first statement follows from the above discussion, and the second follows from the Fundamental Theorem of Galois Theory.

Picture:

E  1
G ∩ V | |
M G ∩ V
G/G ∩ V |  |
F  G

Let g(X) = (X − α)(X − β)(X − γ) ∈ M[X]—it is called the resolvant cubic of f. Any permutation of the α_{i} (a fortiori, any element of G_{f}) merely permutes α, β, γ, and so fixes g(X). Therefore (by the Fundamental Theorem) g(X) has coefficients in F. More explicitly, we have:

Lemma 4.8. If f = X^{4}+bX^{3}+cX^{2}+dX +e, then g = X^{3}−cX^{2}+(bd−4e)X −b^{2}e+4ce−d^{2}. The discriminants of f and g are equal.

Proof. Compute everything in terms of the α_{i}’s. (Cf. Hungerford, V.4.10.)

Now let f be an irreducible separable quartic. Then G = G_{f} is a transitive subgroup of S_{4} whose order is divisible by 4. There are the following possibilities:

\begin{matrix}G & (G ∩ V : 1) & (G : V ∩ G) \\ S_{4} & 4 & 6 \\ A_{4} & 4 & 3 \\ V & 4 & 1 \\ D_{4} & 4 & 2 \\ C_{4} & 2 & 2 \end{matrix}

(G ∩ V : 1) = [E : M], (G : V ∩ G) = [M : F].

Note that G can’t, for example, be the group generated by (12) and (34) because this is not transitive. The groups of type D_{4} are the Sylow 2-subgroups discussed above, and the groups of type C_{4} are those generated by cycles of length 4.

We can compute (G : V ∩ G) from the resolvant cubic g, because G/V ∩ G = Gal(M/F), and M is the splitting field of g. Once we know (G : V ∩ G), we can deduce G except in the case that it is 2. If [M : F] = 2, then G∩V = V or C_{2}. Only the first group acts transitively on the roots of f, and so (from 4.4) we see that (in this case) G = D_{4} or C_{4} according as f is irreducible or not in M[X].

Example 4.9. Consider f(X) = X^{4} + 4X^{2} + 2 ∈ ℚ[X]. It is irreducible by Eisenstein’s criterion, and its resolvant cubic is (X − 4)(X^{2} − 8); thus M = ℚ[\sqrt{2}]. Note that f, when regarded as a polynomial in X^{2}, factors over M; hence G_{f} = C_{4}.

Example 4.10. Consider f(X) = X^{4} − 10X^{2} + 4 ∈ ℚ[X]. One can check directly (using 1.6) that it is irreducible, and its resolvant cubic is (X + 10)(X + 4)(X − 4). Hence G_{f} = V.

Example 4.11. Consider f(X) = X^{4} − 2 ∈ ℚ[X]. It is irreducible by Eisenstein’s criterion, and its resolvant cubic is g(X) = X^{3} + 8X. Hence M = ℚ[i\sqrt{2}]. One can check that f is irreducible over M, and so its Galois group is D_{4}.

Alternatively, analyze the equation as in (3.20).

Maple knows how to factor polynomials over ℚ and over ℚ[α] where α is a root of an irreducible polynomial. To learn the syntax, type: ?Factor.

4.5. Examples of polynomials with S_{p} as Galois group over ℚ. The next lemma gives a criterion for a subgroup of S_{p} to be the whole of S_{p}.

Lemma 4.12. Let p be a prime number. Then S_{p} is generated by any transposition and any p-cycle.

Proof. After renumbering, we may assume that the transposition is τ = (12). Let the p-cycle be σ = (i_{1} ⋯ i_{p}); we may choose to write σ so that 1 occurs in the first position, σ = (1 i_{2} ⋯ i_{p}). Now some power of σ will map 1 to 2 and will still be a p-cycle (here is where we use that p is prime). After replacing σ with the power, we may suppose σ = (1 2 j_{3} … j_{p}), and after renumbering again, we may suppose σ = (1 2 3 … p). Then we’ll have (2 3), (3 4), (4 5), … in the group generated by σ and τ, and these elements generated S_{p}.

Proposition 4.13. Let f be an irreducible polynomial of prime degree p in ℚ[X]. If f splits in ℂ and has exactly two nonreal roots, then G_{f} = S_{p}.

Proof. Let E ⊂ ℂ be the splitting field of f, and let α ∈ E be a root of f. Because f is irreducible, [ℚ[α] : ℚ] = deg f = p, and so p|[E : ℚ] = (G_{f} : 1). Therefore G_{f} contains an element of order p (Cauchy’s theorem), but the only elements of order p in S_{p} are p-cycles (here we use that p is prime again).

Let σ be complex conjugation on ℂ. Then σ transposes the two nonreal roots of f(X) and fixes the rest. Therefore G_{f} ⊂ S_{p} contains a transposition and a p-cycle, and so is the whole of S_{p}.

It remains to construct polynomials satisfying the conditions of the Proposition.

Example 4.14. Let p≥ 5 be a prime number. Choose a positive even integer m and even integers

n_{1} < n_{2} < ⋯ < n_{p−2}.

Let f(X) = g(X) − 2, where

g(X) = (X^{2} + m)(X − n_{1})...(X − n_{p−2}).

When we write f(X) = X^{p}+a_{1}X^{p−1}+⋯+a_{p}, then all a_{i} are even, and a_{p} = −(m ∏ n_{i})−2 is not divisible by 4. Hence Eisenstein’s criterion implies that f(X) is irreducible.

The polynomial g(X) certainly has exactly two nonreal roots. Its graph crosses the x-axis exactly p − 2 times, and its maxima and minima all have absolute value > 2 (because its values at odd integers have absolute value > 2). Hence the graph of f(X) = g(X) − 2 also crosses the x-axis exactly p − 2 times.

4.6. Finite fields. Let 𝔽_{p} = ℤ/pℤ, the field of p elements. As we noted in §1.2, any other field E of characteristic p contains a copy of 𝔽_{p}, namely, {m1_{E} | m ∈ ℤ}. No harm results if we identify 𝔽_{p} with this subfield of E.

Let E be a field of degree n over 𝔽_{p}. Then E has q = p^{n} elements, and so E^{×} is a group of order q − 1. Hence the nonzero elements of E are roots X^{q−1} − 1, and all elements of E (including 0) are roots of X^{q} − X. Hence E is a splitting field for X^{q} − X, and so any two fields with q elements are isomorphic.

Now let E be the splitting field of f(X) = X^{q} − X, q = p^{n}. The derivative f′(X) = −1, which is relatively prime to f(X) (in fact, to every polynomial), and so f(X) has q distinct roots in E. Let S be the set of its roots. Then S is obviously closed under multiplication and the formation of inverses, but it is also closed under subtraction: if a^{q} − a = 0 and b^{q} − b = 0, then

(a − b)^{q} = a^{q} − b^{q} = a − b.

Hence S is a field, and so S = E. In particular, E has p^{n} elements.

Proposition 4.15. For each power q = p^{n} there is a field 𝔽_{q} with q elements. It is the splitting field of X^{q} − X, and hence any two such fields are isomorphic. Moreover, 𝔽_{q} is Galois over 𝔽_{p} with cyclic Galois group generated by the Frobenius automorphism σ(a) = a^{p}.

Proof. Only the final statement remains to be proved. The field 𝔽_{q} is Galois over 𝔽_{p} because it is the splitting field of a separable polynomial. We noted in (1.3) that σ = (x ↦ x^{p}) is an automorphism of 𝔽_{q}. It has order n, and a ∈ 𝔽_{q} is fixed by σ if and only if a^{p} = a. But 𝔽_{p} consists exactly of such elements, and so the fixed field of < σ > is 𝔽_{p}. This proves that < σ >= Gal(𝔽_{q}/𝔽_{p}).

Corollary 4.16. Let E be a field with p^{n} elements. Then E contains exactly one field with p^{m} elements for each m|n, m ≥ 0, and E is Galois over that field.

Proof. We know that E is Galois over 𝔽_{p} and that Gal(E/𝔽_{p}) is the cyclic group of order n generated by σ. The subgroups of < σ > are the groups < σ^{m} > with m|n. The fixed field of < σ^{m} > is 𝔽_{p^{m}}.

Corollary 4.17. Every extension of finite fields is simple.

Proof. Consider E ⊃ F. Then E^{×} is a finite subgroup of the multiplicative group of a field, and hence is cyclic (see Exercise 3). If ζ generates E^{×} as a multiplicative group, then clearly E = 𝔽_{p}[ζ].

Corollary 4.18. Each monic irreducible polynomial of degree d|n in 𝔽_{p}[X] occurs exactly once as a factor of X^{p^{n}} − X.

Proof. First, the factors of X^{p^{n}} − X are distinct because it has no common factor with its derivative. If f(X) is irreducible of degree d, then f(X) has a root in a field of degree d over 𝔽_{p}. But the splitting field of X^{p^{n}} − X contains a copy of every field of degree d over 𝔽_{p} with d|n. Hence some root of X^{p^{n}} − X is also a root of f(X), and therefore f(X)|X^{p^{n}} − X.

Maple factors polynomials modulo p very quickly. The syntax is “Factor(f(X)) mod p;”. Thus, for example, to obtain a list of all monic polynomials of degree 1, 2, or 4 over 𝔽_{5}, ask Maple to factor X^{625} − X.

Let 𝔽 be an algebraic closure of 𝔽_{p}. Then 𝔽 contains one field 𝔽_{p^{n}} for each integer n ≥ 1— it consists of all roots of X^{p^{n}} − X—and 𝔽_{p^{m}} ⊂ 𝔽_{p^{n}} ⟺ m|n. The partially ordered set of finite subfields of 𝔽 is isomorphic to the set of integers n ≥ 1 partially ordered by divisibility.

Finite fields were sometimes called Galois fields, and 𝔽_{q} used to be denoted GF (q) (it still is in Maple). Maple contains a “Galois field package” to do computations in finite fields. For example, it can find a primitive element for 𝔽_{q} (i.e., a generator for 𝔽_{q}^{×}). To start it, type:

readlib(GF);.

4.7. Computing Galois groups over ℚ. We sketch a practical method for computing Galois groups over ℚ and similar fields. Our first result generalizes Proposition 4.4.

Proposition 4.19. Let f(X) be a monic separable polynomial in F [X] of degree m with distinct roots, and suppose that G_{f} ⊂ S_{m} has r orbits with m_{1}, …, m_{r} elements respectively (so that m = m_{1} + ⋯ + m_{r}); then f factors as f = f_{1} ⋯ f_{r} with f_{i} irreducible of degree m_{i}.

Proof. Let α_{1}, …, α_{m} be the distinct roots of f(X). For S ⊂ {1, 2, …, m}, consider f_{S} = ∏_{i∈S}(X − α_{i}). This polynomial divides f(X) in F_{f}[X], and it is fixed under the action of G_{f} (and hence has coefficients in F) if and only if S is stable under G_{f}. Therefore the irreducible factors are the polynomials f^{S} corresponding to minimal subsets S of {1, …, m} stable under G, but such sets S are precisely the orbits of G in {1, …, m}.

Now suppose F is finite, with p^{n} elements say, and let E be the splitting field of f. The Galois group of E over F is generated by the Frobenius automorphism σ : x ↦ x^{p^{n}}. When we regard σ as a permutation of the roots of f, then its factors in the cycle decomposition of σ correspond to the distinct orbits of σ. Hence, if the degrees of the distinct irreducible factors of f are m_{1}, m_{2}, …, m_{r}, then σ has a cycle decompostion of type

m_{1} + ⋯ + m_{r} = m.

Lemma 4.20. Let R be a unique factorization domain with field of fractions F, and let f be a monic polynomial in R[X]. Let P be a prime ideal in R, and let f̄ be the image of f in (R/P)[X]. Assume neither f nor f̄ has a multiple root. Then the roots α_{1}, …, α_{m} of f lie in R, and their reductions ᾱ_{i} modulo P are the roots of f̄. Moreover Gf̄ ⊂ G_{f} when both are identified with subgroups of Sym{α_{1}, …, α_{m}} = Sym{ᾱ_{1}, …, ᾱ_{m}}.

Proof. Omitted—see van der Waerden, Modern Algebra, I, §61 (second edition) or Math 676 (Algebraic Number Theory).

On combining these results, we obtain the following theorem.

Theorem 4.21 (Dedekind). Let f(X) ∈ ℤ[X] be a monic polynomial of degree m, and let p be a prime such that f mod p has simple roots (equivalently, D(f) is not divisible by p). Suppose that f̄ = ∏ f_{i} with f_{i} irreducible of degree m_{i} in 𝔽_{p}[X]. Then G_{f} contains an element whose cycle decomposition corresponds to the partition:

m = m_{1} + ⋯ + m_{r}.

Example 4.22. Consider X^{5} −X −1. Modulo 2, this factors as (X^{2} +X +1)(X^{3} +X^{2} +1), and modulo 3 it is irreducible. Hence G_{f} contains (12345) and (ik)(lmn), and hence also ((ik)(lmn))^{3} = (ik). Therefore G_{f} = S_{5}.

Lemma 4.23. A transitive subgroup of H ⊂ S_{n} containing a transposition and an (n − 1)- cycle is equal to S_{n}.

Proof. Let (123 … n−1) be the (n−1)-cycle. By virtue of the transitivity, the transposition can be transformed into (in), some 1 ≤ i ≤ n − 1. Now the (n − 1)-cycle and its powers will transform this into (1n), (2n), …, (n − 1 n), and these elements obviously generate S_{n}.

Example 4.24. Select monic polynomials of degree n, f_{1}, f_{2}, f_{3} with coefficients in ℤ such that:

(a) f_{1} is irreducible modulo 2;

(b) f_{2} = (degree 1)(irreducible of degree n − 1) mod 3;

(c) f_{3} = (irreducible of degree 2)(product of 1 or 2 irreducible polys of odd degree) mod 5.

We choose them to have distinct roots. Take

f = −15f_{1} + 10f_{2} + 6f_{3}.

Then

(i) G_{f} is transitive (it contains an n-cycle because f ≡ f_{1} mod 2); (ii) G_{f} contains a cycle of length n − 1 (because f ≡ f_{2} mod 3); (iii) G_{f} contains a transposition (because f ≡ f_{3} mod 5, and so it contains the product

of a transposition with a commuting element of odd order; on raising this to an appropriate odd power, we are left with the transposition). Hence G_{f} is S_{n}.

This gives the following strategy for computing Galois groups over ℚ. Factor f modulo a sequence of primes p not dividing D(f) to determine the cycle types of the elements in G_{f}—a difficult theorem in number theory, the effective Chebotarev density theorem, says that if a cycle type occurs in G_{f}, then this will be seen by looking modulo a set of prime numbers of positive density, and will occur for a prime less than some bound. Now look up a table of transitive subgroups of S_{n} with order divisible by n and their cycle type. If this doesn’t suffice to determine the group, then look at its action on the set of subsets of r roots for some r.

See, Butler and McKay, The transitive groups of degree up to eleven, Comm. Algebra 11 (1983), 863–911. This lists all transitive subgroups of S_{n}, n ≤ 11, and gives the cycle types of their elements and the orbit lengths of the subgroup acting on the r-sets of roots; with few exceptions, these invariants are sufficient to determine the subgroup up to isomorphism.

Maple can compute Galois groups for polynomials of degree ≤ 7 over ℚ. To learn the syntax, type ?galois;. Magma (the replacement for Cayley) probably knows much more, but my efforts to obtain a manual for it have been unsuccessful.

See also, Soicher and McKay, Computing Galois groups over the rationals, J. Number Theory, 20 (1985) 273–281.

5. Applications of Galois Theory

In this section, we apply the Fundamental Theorem of Galois Theory to obtain other results about polynomials and extensions of fields.

5.1. Primitive element theorem. Recall that a finite extension of fields E/F is simple if E = F [α] for some element α of E. Such an α is called a primitive element of E. We shall show that (at least) all separable extensions have primitive elements.

Consider for example ℚ[\sqrt{2}, \sqrt{3}]/ℚ. We know (see Exercise 13) that its Galois group over ℚ is a 4-group < σ, τ >, where

\{ \begin{matrix}σ\sqrt{2} & = & −\sqrt{2} \\ σ\sqrt{3} & = & \sqrt{3} \end{matrix} , \{ \begin{matrix}τ\sqrt{2} & = & \sqrt{2} \\ τ\sqrt{3} & = & −\sqrt{3}. \end{matrix}

Note that

\begin{aligned}σ(\sqrt{2} + \sqrt{3}) & = & −\sqrt{2} + \sqrt{3}, \\ τ(\sqrt{2} + \sqrt{3}) & = & \sqrt{2} − \sqrt{3}, \\ (στ)(\sqrt{2} + \sqrt{3}) & = & −\sqrt{2} − \sqrt{3}. \end{aligned}

These all differ from \sqrt{2} + \sqrt{3}, and so only the identity element of Gal(ℚ[\sqrt{2}, \sqrt{3}]/ℚ) fixes the elements of ℚ[\sqrt{2} + \sqrt{3}]. According to the Fundamental Theorem, this implies that \sqrt{2} + \sqrt{3} is a primitive element:

ℚ[\sqrt{2}, \sqrt{3}] = ℚ[\sqrt{2} + \sqrt{3}].

It is clear that this argument should work much more generally.

We say that an element α algebraic over a field F is separable over F if its minimum polynomial over F has no multiple roots. Thus a finite extension E of F is separable if and only if all its elements are separable over F.

Theorem 5.1. Let E = F [α_{1}, ..., α_{r}] be a finite extension of F, and assume that α_{2}, ..., α_{r} are separable over F (but not necessarily α_{1}). Then there is an element γ ∈ E such that E = F [γ].

Proof. For finite fields, we proved this in (4.16). Hence we may assume F to be infinite. It suffices to prove the statement for r = 2. Thus let E = F [α, β] with β separable over F [α]. Let f and g be the minimum polynomials of α and β over F. Let α_{1} = α, …, α_{s} be the roots of f in some field containing E, and let β_{1} = β, β_{2}, …, β_{t} be the roots of g. For j ≠ 1, β_{j} ≠ β_{1}, and so the the equation

α_{i} + Xβ_{j} = α_{1} + Xβ_{1}, j ≠ 1,

has exactly one solution, namely, X = \frac{α_{i}−α_{1}}{β_{1}−β_{j}}. If we choose a c different from any of these solutions (using that F is infinite), then

α_{i} + cβ_{j} ≠ α + cβ unless i = 1 = j.

I claim that γ = α + cβ generates E over F.

The polynomials g(X) and f(γ − cX) have coefficients in F [γ][X], and have β as a root:

g(β) = 0, f(γ − cβ) = f(α) = 0.

In fact, β is their only common root, because the roots of g are β_{1}, ..., β_{t}, and we chose c so that γ − cβ_{j} ≠ α_{i} unless i = 1 = j. Therefore gcd(g(X), f(γ − cX)) computed in some field splitting fg is X − β, but we have seen (Proposition 3.1) that the gcd of two polynomials has coefficients in the same field as the coefficients of the polynomials. Hence β ∈ F [γ], and then α = γ − cβ also lies in F [γ].

Remark 5.2. Assume F to be infinite. The proof shows that γ can be chosen to be of the form

γ = α_{1} + c_{2}α_{2} + ⋯ + c_{r}α_{r}, c_{i} ∈ F.

In fact, all but a finite number of elements of this form will serve. If E = F [α_{1}, …, α_{r}] is Galois over F, then an element of this form will be a primitive element provided it is moved by every element of Gal(E/F) except 1. These remarks make it very easy to write down primitive elements.

Our hypotheses are minimal: if two of the α’s are not separable, then the extension need not be simple. Before proving this, we need another result.

Proposition 5.3. Let E = F [γ] be a simple algebraic extension of F. Then there are only finitely many intermediate fields M,

F ⊂ M ⊂ E.

Proof. Let M be such a field, and let g(X) be the minimum polynomial of γ over M. Let M′ be the subfield of E generated over F by the coefficients of g(X). Clearly M′ ⊂ M, but (equally clearly) g(X) is the minimum polynomial of γ over M′. Hence

[E : M′] = deg g = [E : M],

and so M = M′: M is generated by the coefficients of g(X).

Let f(X) be the minimum polynomial of γ over F. Then g(X) divides f(X) in M[X], and hence also in E[X]. Therefore, there are only finitely many possible g’s, and consequently only finitely many possible M’s.

Remark 5.4. (a) Note that the proposition in fact gives a description of all the intermediate fields: each is generated over F by the coefficients of a factor g(X) of f(X) in E[X]. The coefficients of such a g(X) are partially symmetric polynomials in the roots of f(X) (i.e., fixed by some, but not necessarily all, of the permutations of the roots).

(b) The proposition has a converse: if E is a finite extension of F and there are only finitely many intermediate fields M, F ⊂ M ⊂ E, then E is a simple extension of F (see Dummit, p508). This gives another proof of the theorem when E is separable over F, because Galois theory shows that there are only finitely many intermediate fields in this case (embed E in a Galois extension of F).

(c) The simplest nonsimple extension is k(X, Y) ⊃ k(X^{p}, Y^{p}) = F, where k is an algebraically closed field of characteristic p. For any c ∈ k, we have

k(X, Y) = F [X, Y] ⊃ F [X + cY] ⊃ F

with the degree of each extension equal to p. If F [X + cY] = F [X + c′Y], c ≠ c′, then F [X + cY] would contain both X and Y, which is impossible because [k(X, Y) : F] = p^{2}. Hence there are infinitely many distinct intermediate fields.^{2}

5.2. Fundamental Theorem of Algebra.

We finally prove the misnamed^{3} fundamental theorem of algebra.

Theorem 5.5. The field ℂ of complex numbers is algebraically closed.

Proof. Define ℂ to be the splitting field of X^{2} + 1 ∈ ℝ[X], and let i be a root of X^{2} + 1 in ℂ; thus ℂ = ℝ[i]. We have to show (see 2.10) that every f(X) ∈ ℝ[X] has a root in ℂ.

The two facts we need to assume about ℝ are:

• Positive real numbers have square roots.

• Every polynomial of odd degree with real coefficients has a real root.

Both are immediate consequences of the Intermediate Value Theorem, which says that a continuous function on a closed interval takes every value between its maximum and minimum values (inclusive). (Intuitively, this says that, unlike the rationals, the real line has no “holes”.)

We first show that every element of ℂ has a square root. Write α = a + bi, with a, b ∈ ℝ, and choose c, d to be real numbers such that

c^{2} = \frac{(a + \sqrt{a^{2} + b^{2}})}{2}, d^{2} = \frac{(−a + \sqrt{a^{2} + b^{2}})}{2}.

Then c^{2} − d^{2} = a and (2cd)^{2} = b^{2}. If we choose the signs of c and d so that cd has the same sign as b, then (c + di)^{2} = α.

Let f(X) ∈ ℝ[X], and let E be a splitting field for f(X)(X^{2} + 1)—we have to show that E = ℂ. Since ℝ has characteristic zero, the polynomial is separable, and so E is Galois over ℝ. Let G be its Galois group, and let H be a Sylow 2-subgroup of G.

Let M = E^{H}. Then M is of odd degree over ℝ, and M = ℝ[α] some α (Theorem 5.1). The minimum polynomial of α over ℝ has odd degree, and so has a root in ℝ. It therefore has degree 1, and so M = ℝ and G = H.

We now have that Gal(E/ℂ) is a 2-group. If it is ≠ 1, then it has a subgroup N of index 2. The field E^{N} has degree 2 over ℂ, and can therefore be obtained by extracting the square root of an element of ℂ (see 3.23), but we have seen that all such elements already lie in ℂ. Hence E^{N} = ℂ, which is a contradiction. Thus E = ℂ.

Corollary 5.6. (a) The field ℂ is the algebraic closure of ℝ.

(b) The set of all algebraic numbers is an algebraic closure of ℚ.

2Zariski showed that there is even an intermediate field M that is not isomorphic to F (X, Y), and Piotr Blass showed in his UM thesis, 1977, using the methods of algebraic geometry, that there is an infinite sequence of intermediate fields, no two of which are isomorphic.

3Because it is not strictly a theorem in algebra: it is a statement about ℝ whose construction is part of analysis. In fact, I prefer the proof based on Liouville’s theorem in complex analysis to the more algebraic proof given in the text: if f(z) is a polynomial without a root in ℂ, then f(z)^{−1} will be bounded and holomorphic on the whole complex plane, and hence (by Liouville) constant. The Fundamental Theorem was quite a difficult theorem to prove. Gauss gave a proof in his doctoral dissertation in 1798 in which he used some geometric arguments which he didn’t justify. He gave the first rigorous proof in 1816. The elegant argument given here is a simplification by Emil Artin of earlier proofs.

Proof. Part (a) is obvious from the definition of “algebraic closure”, and (b) follows from the discussion on p15.

5.3. Cyclotomic extensions. A primitive n^{th} root of 1 in F is an element of order n in F^{×}. Such an element can exist only if F has characteristic 0 or characteristic p not dividing n.

Proposition 5.7. Let F be a field of characteristic 0 or characteristic p not dividing n. Let E be the splitting field of X^{n} − 1.

(a) There exists a primitive n^{th} root of 1 in E.

(b) If ζ is a primitive n^{th} root of 1 in E, then E = F [ζ].

(c) The field E is Galois over F, and the map

Gal(E/F) → (ℤ/nℤ)^{×}

sending σ to [i] if σζ = ζ^{i} is injective.

Proof. (a) The roots of X^{n} − 1 are distinct, because its derivative nX^{n−1} has only zero as a root (we use here the condition on the characteristic), and so E contains n distinct n^{th} roots of 1. The n^{th} roots of one form a finite subgroup of E^{×}, and so (see Exercise 3) they form a cyclic group. Any generator will have order n, and hence will be a primitive n^{th} root of 1.

(b) The roots of X^{n} − 1 are the powers of ζ, and F [ζ] contains them all.

(c) If ζ is one primitive n^{th} root of 1, then the remaining primitive n^{th} roots of 1 are the elements ζ^{i} with i relatively prime to n. Since σζ is again a primitive n^{th} root of 1 for any automorphism σ of E, it equals ζ^{i} for some i relatively prime to n, and the map σ ↦ i mod n is injective because ζ generates E over F. It obviously is a homomorphism (and is independent of the choice of ζ).

The map σ ↦ i : Gal(F [ζ]/F) → (ℤ/nℤ)^{×} need not be surjective. For example, if F = ℂ, then its image is {1}, and if F = ℝ, it is {±1} (n ≠ 2)—because F [ζ] = ℂ, Gal(ℂ/ℝ) is generated by complex conjugation ι, and ιζ = ζ̄ = ζ^{n−1}. On the other hand, when n = p is prime, we saw in (1.31) that [ℚ[ζ] : ℚ] = p − 1, and so the map is surjective. We shall prove that the map is surjective for all n when F = ℚ.

The polynomial X^{n} −1 has some obvious factors in ℚ[X], namely, the polynomials X^{d} −1 for any d|n. The quotient of X^{n} − 1 by all these factors for d < n is called the n^{th} cyclotomic polynomial Φ_{n}. Thus

Φ_{n} = ∏(X − ζ) (product over the primitive n^{th} roots of 1).

It has degree φ(n), the order of (ℤ/nℤ)^{×}. Since every n^{th} root of 1 is a primitive n^{th} root of 1 for exactly one d dividing n, we see that

X^{n} − 1 = ∏_{d|n} Φ_{d}(X).

For example, Φ_{1}(X) = X − 1, Φ_{2}(X) = X + 1, Φ_{3}(X) = X^{2} + X + 1, and

Φ_{6}(X) = \frac{X^{6} − 1}{(X − 1)(X + 1)(X^{2} + X + 1)} = X^{2} − X + 1.

This gives an easy inductive method of computing the cyclotomic polynomials. Alternatively ask Maple by typing: with(numtheory); cyclotomic(n,X);. Because X^{n} − 1 has coefficients in ℤ and is monic, any monic factor of it has coefficients in ℤ (see (1.6)). In particular, the cyclotomic polynomials lie in ℤ[X].

Lemma 5.8. Let F be a field of characteristic 0 or p not dividing n, and let ζ be a primitive n^{th} root of 1 in some extension field. The following are equivalent:

(a) the n^{th} cyclotomic polynomial Φ_{n} is irreducible;

(b) the degree [F [ζ] : F] = φ(n);

(c) the homomorphism

Gal(F [ζ]/F) → (ℤ/nℤ)^{×}

is an isomorphism.

Proof. Because ζ is a root of Φ_{n}, the minimum polynomial of ζ divides Φ_{n}. It is equal to it if and only if [F [ζ] : F] = φ(n), which is true if and only if the injection Gal(F [ζ]/F) ↪ (ℤ/nℤ)^{×} is onto.

Theorem 5.9. The n^{th} cyclotomic polynomial Φ_{n} is irreducible in ℚ[X].

Proof. Let f(X) be a monic irreducible factor of Φ_{n} in ℚ[X]. Its roots will be primitive n^{th} roots of 1, and we have to show they include all primitive n^{th} roots of 1. For this it suffices to show that

ζ a root of f(X) ⟹ ζ^{i} a root of f(X) for all i such that gcd(i, n) = 1.

Such an i is a product of primes not dividing n, and so it suffices to show that

ζ a root of f(X) ⟹ ζ^{p} a root of f(X) for all primes p ∤ n.

Write

Φ_{n}(X) = f(X)g(X).

Again (1.6) implies that f(X) and g(X) lie in ℤ[X]. Suppose ζ is a root of f, but that for some prime p not dividing n, ζ^{p} is not a root of f. Then ζ^{p} is a root g(X), which implies that ζ is a root of g(X^{p}). Since f(X) and g(X^{p}) have a common root, their greatest common divisor (in ℚ[X]) is ≠ 1 (see 3.1). Write h(X) ↦ h̄(X) for the map ℤ[X] ↦ 𝔽_{p}[X], and note that

gcd(f(X), g(X^{p})) ≠ 1 ⟹ gcd(f̄(X), ḡ(X^{p})) ≠ 1.

But ḡ(X^{p}) = ḡ(X)^{p} (use the mod p binomial theorem and that a^{p} = a for all a ∈ 𝔽_{p}), and so gcd(f̄(X), ḡ(X)^{p}) ≠ 1, which implies that f̄(X) and ḡ(X) have a common factor. Hence X^{n} − 1 (regarded as an element of 𝔽_{p}[X]) has multiple roots, but we saw in the proof of 5.7 that it doesn’t. Contradiction.

Remark 5.10. This proof is very old—in essence it goes back to Dedekind in 1857—but its general scheme has recently become very popular: take a statement in characteristic zero, reduce modulo p (where the statement may no longer be true), and exploit the existence of the Frobenius automorphism a ↦ a^{p} to obtain a proof of the original statement. For example, commutative algebraists use this method to prove results about commutative rings, and there are theorems about complex manifolds^{4} that have only been proved by reducing things to characteristic p.

There are some beautiful and mysterious relations between what happens in characteristic 0 and in characteristic p. For example, let f(X_{1}, ..., X_{n}) ∈ ℤ[X_{1}, ..., X_{n}]. We can (i) look at the solutions of f = 0 in ℂ, and so get a topological space; (ii) reduce mod p, and look at the solutions of f̄ = 0 in 𝔽_{p^{n}}.

4This is from my old notes—I no longer remember what I was thinking of.

The Weil conjectures (Weil 1949; proved by Grothendieck and Deligne 1973) assert that the Betti numbers of the space in (i) control the cardinalities of the sets in (ii).

Theorem 5.11. The regular n-gon is constructible if and only if n = 2^{k}p_{1} ⋯ p_{s} where the p_{i} are distinct Fermat primes.

Proof. The regular n-gon is constructible if and only if cos \frac{2π}{n} (or ζ = e^{2πi/n}) is constructible. We know that ℚ[ζ] is Galois over ℚ, and so (according to 1.27 and 3.22) ζ is constructible if and only if [ℚ[ζ] : ℚ] is a power of 2. But (see Groups 3.10)

φ(n) = ∏_{p|n}(p − 1)p^{n−1},

and this is a power of 2 if and only if n has the required form.

Remark 5.12. The final section of Gauss’s, Disquisitiones Arithmeticae (1801) is titled “Equations defining sections of a Circle”. In it Gauss proves that the n^{th} roots of 1 form a cyclic group, that X^{n} − 1 is solvable (this was before the theory of abelian groups had been developed, and before Galois), and that the regular n-gon is constructible when n is as in the Theorem. He also claimed to have proved the converse statement^{5}. This leads some people to credit him with the above proof of the irreducibility of Φ_{n}, but in the absence of further evidence, I’m sticking with Dedekind.

5.4. Independence of characters.

Theorem 5.13 (Dedekind’s theorem on the independence of characters). Let F be a field, and let G be a group (monoid will do). Then any finite set {χ_{1}, …, χ_{m}} of homomorphisms G → F^{×} is linearly independent over F, i.e.,

∑ a_{i}χ_{i} = 0 (as a function G → E) ⟹ a_{1} = 0, …, a_{m} = 0.

Proof. Induction on m. If m = 1, it’s obvious. Assume it for m − 1. We suppose

a_{1}χ_{1}(x) + a_{2}χ_{2}(x) + ⋯ + a_{m}χ_{m}(x) = 0 for all x ∈ G,

and show that this implies the a_{i} to be zero. Since χ_{1} ≠ χ_{2}, χ_{1}(g) ≠ χ_{2}(g) for some g ∈ G. On replacing x with gx in the equation, we obtain the equation

a_{1}χ_{1}(g)χ_{1}(x) + a_{2}χ_{1}(g)χ_{2}(x) + ⋯ + a_{m}χ_{1}(g)χ_{m}(x) = 0, all x ∈ G.

On multiplying the first equation by χ_{1}(g) and subtracting it from the second, we obtain the equation

a′_{2}χ_{2} + ⋯ + a′_{m}χ_{m} = 0, a′_{i} = a_{i}(χ_{i}(g) − χ_{1}(g)).

The induction hypothesis now shows that a′_{i} = 0 for all i ≥ 2. Since χ_{2}(g) − χ_{1}(g) ≠ 0, we must have a_{2} = 0, and the induction hypothesis shows that all the remaining a_{i}’s are also zero.

5“Whenever n − 1 involves prime factors other than 2, we are always led to equations of higher degree....WE CAN SHOW WITH ALL RIGOR THAT THESE HIGHER-DEGREE EQUATIONS CANNOT BE AVOIDED IN ANY WAY NOR CAN THEY BE REDUCED TO LOWER-DEGREE EQUATIONS. The limits of the present work exclude this demonstration here, but we issue this warning lest anyone attempt to achieve geometric constructions for sections other than the ones suggested by our theory...and so spend his time uselessly.”

Corollary 5.14. Let F_{1} and F_{2} be fields, and let σ_{1}, ..., σ_{m} be distinct homomorphisms F_{1} → F_{2}. Then σ_{1}, ..., σ_{m} are linearly independent over F_{2}.

Proof. Apply the theorem to χ_{i} = σ_{i}|F_{1}^{×}.

5.5. Hilbert’s Theorem 90. Let G be a finite group. A G-module is an abelian group M together with an action of G, i.e., a map G × M → M such that

(a) σ(m + m′) = σm + σm′ for all σ ∈ G, m, m′ ∈ M;

(b) (στ)(m) = σ(τm) for all σ, τ ∈ G, m ∈ M;

(c) 1m = m for all m ∈ M.

Thus, to give an action of G on M is the same as to give a homomorphism G → Aut(M) (automorphisms of M as an abelian group).

Example 5.15. Let E be a Galois extension of F, with Galois group G; then (E, +) and E^{×} are G-modules.

Let M be a G-module. A crossed homomorphism is a map f : G → M such that

f(στ) = f(σ) + σf(τ).

Note that the condition implies that f(1) = f(1 · 1) = f(1) + f(1), and so f(1) = 0.

Example 5.16. (a) Consider a crossed homomorphism f : G → M, and let σ ∈ G. Then

f(σ^{2}) = f(σ) + σf(σ),
f(σ^{3}) = f(σ · σ^{2}) = f(σ) + σf(σ) + σ^{2}f(σ)

and so on, until

f(σ^{n}) = f(σ) + σf(σ) + ⋯ + σ^{n−1}f(σ).

Thus, if G is a cyclic group of order n generated by σ, then a crossed homomorphism f : G → M is determined by f(σ) = x, and x satisfies the equation

x + σx + ⋯ + σ^{n−1}x = 0, (∗)

Conversely, if x ∈ M satisfies (*), then the formulas f(σ^{i}) = x + σx + ⋯ + σ^{i−1}x define a crossed homomorphism f : G → M. In this case we have a one-to-one correspondence

{crossed homs f : G → M} ↔^{f↦f(σ)} {x ∈ M satisfying (∗)}.

(b) For any x ∈ M, we obtain a crossed homomorphism by putting

f(σ) = σx − x, all σ ∈ G.

Such a crossed homomorphism is called a principal crossed homomorphism.

(c) If G acts trivially on M, i.e., σm = m for all σ ∈ G and m ∈ M, then a crossed homomorphism is simply a homomorphism, and there are no nontrivial principal crossed homomorphisms.

The sum of two crossed homomorphisms is again a crossed homomorphism, and the sum of two principal crossed homomorphisms is again principal. Thus we can define

H^{1}(G, M) = \frac{\{crossed homomorphisms\}}{\{principal crossed homomorphisms\}}.

The cohomology groups H^{n}(G, M) have been defined for all n ∈ ℕ, but since this was not done until the twentieth century, it will not be discussed in this course.

Example 5.17. Let π : X̃ → X be the universal covering space of a topological space X, and let Γ be the group of covering transformations. Under some fairly general hypotheses, a Γ-module M will define a sheaf M on X, and H^{1}(X, M) ≈ H^{1}(Γ, M). For example, when M = ℤ with the trivial action of Γ, this becomes the isomorphism H^{1}(X, ℤ) ≈ H^{1}(Γ, ℤ) = Hom(Γ, ℤ).

Theorem 5.18. Let E be a Galois extension of F with group G; then H^{1}(G, E^{×}) = 0, i.e., every crossed homomorphism G → E^{×} is principal.

Proof. Let f be a crossed homomorphism G → E^{×}. In multiplicative notation, this means,

f(στ) = f(σ) · σ(f(τ)), σ, τ ∈ G,

and we have to find a γ ∈ E^{×} such that f(σ) = σγ/γ for all σ ∈ G. Because the f(τ) are nonzero, Dedekind’s theorem implies that

∑ f(τ)τ : E → E

is not the zero map, i.e., there exists an α ∈ E such that

β = ∑_{τ∈G} f(τ)τα ≠ 0.

But then, for σ ∈ G,

σβ = ∑_{τ∈G} σ(f(τ)) · στ(α) = ∑_{τ∈G} f(σ)^{−1} f(στ) · στ(α) = f(σ)^{−1} ∑_{τ∈G} f(στ)στ(α) = f(σ)^{−1}β,

which shows that f(σ) = \frac{β}{σ(β)} and so we can take β = γ^{−1}.

Let E be a Galois extension of F with Galois group G. We define the norm of an element α ∈ E to be

Nm α = ∏_{σ∈G} σα.

Then, for τ ∈ G,

τ(Nm α) = ∏_{σ∈G} τσα = Nm α,

and so Nm α ∈ F. The map α ↦ Nm α : E^{×} → F^{×} is a homomorphism. For example, the norm map ℂ^{×} → ℝ^{×} is α ↦ |α|^{2} and the norm map ℚ[\sqrt{d}]^{×} → ℚ^{×} is a + b\sqrt{d} ↦ a^{2} + db^{2}.

We are interested in determining the kernel of this homomorphism. Clearly if α is of the form \frac{β}{τβ}, then Nm(α) = 1. Our next result show that, for cyclic extensions, all elements with norm 1 are of this form.

Corollary 5.19 (Hilbert’s theorem 90). ^{6}Let E be a finite cyclic extension of F with Galois group < σ >; if Nm_{E/F} α = 1, then α = β/σβ for some β ∈ E.

Proof. Let m = [E : F]. The condition on α is that α · σα ⋯ σ^{m−1}α = 1, and so (see 5.16a) there is a crossed homomorphism f :<σ>→ E^{×} with f(σ) = α. The theorem now shows that f is principal, which means that there is a β with f(σ) = β/σβ.

5.6. Cyclic extensions. We are now able to classify the cyclic extensions of degree n of a field F in the case that F contains n n^{th} roots of 1.

Theorem 5.20. Let F be a field containing a primitive n^{th} root of 1.

(a) The Galois group of X^{n} − a is cyclic of order dividing n.

(b) Conversely, if E is cyclic of degree n over F, then there is an element β ∈ E such that E = F [β] and b =_{df} β^{n} ∈ F; hence E is the splitting field of X^{n} − b.

Proof. (a) If α is one root of X^{n} − a, then the other roots are the elements of the form ζα with ζ an n^{th} root of 1. Hence the splitting field of X^{n} − a is ℚ[α]. The map σ ↦ \frac{σα}{α} is an injective homomorphism of Gal(ℚ[α]/ℚ) into the cyclic group <ζ> .

(b) Let ζ be a primitive n^{th} root of 1 in F, and let σ generate Gal(E/F). Then Nm ζ = ζ^{n} = 1, and so, according to Hilbert’s Theorem 90, there is an element β ∈ E such that σβ = ζβ. Then σ^{i}β = ζ^{i}β, and so only the identity element of Gal(E/F [β]) fixes β—we conclude by the Fundamental Theorem of Galois Theory that E = F [β]. On the other hand σβ^{n} = ζ^{n}β^{n} = β^{n}, and so β^{n} ∈ F.

Remark 5.21. (a) Under the hypothesis of the theorem X^{n} −a is irreducible, and its Galois group is of order n, if

(i) a is not a p^{th} power for any p dividing n; (ii) if 4|n then a ∉ −4k^{4}.

See Lang, Algebra, VIII, §9, Theorem 16.

(b) If F has characteristic p (hence has no p^{th} roots of 1 other than 1), then X^{p} − X − a is irreducible in F [X] unless a = b^{p} − b for some b ∈ F, and when it is irreducible, its Galois group is cyclic of order p (generated by α ↦ α + 1 where α is a root). Moreover, every extension of F which is cyclic of degree p is the splitting field of such a polynomial.

Remark 5.22 (Kummer theory). Above we gave a description of all Galois extensions of F with Galois group cyclic of order n in the case that F contains a primitive n^{th} root of 1. Under the same assumption on F, it is possible to give a description of all the Galois extensions of F with abelian Galois group of exponent n, i.e., a quotient of (ℤ/nℤ)^{r} for some r.

Let E be such an extension of F, and let

S(E) = {a ∈ F^{×} | a becomes an n^{th} power in E};

6The theorem is Satz 90 in Hilbert’s book, Theorie der Algebraische Zahlkörper, 1897, which laid the foundations for modern algebraic number theory. Many point to it as a book that made a fundamental contribution to mathematical progress, but Emil Artin has been quoted as saying that it set number theory back thirty years—it wasn’t sufficiently abstract for his taste.

Then S(E) is a subgroup of F^{×} containing F^{×n}, and the map E ↦ S(E) defines a oneto-one correspondence between abelian extensions of E of exponent n and groups S(E), F^{×} ⊃ S(E) ⊃ F^{×n}, such that (S(E) : F^{×n}) < ∞. The field E is recovered from S(E) as the splitting field of ∏(X^{n} − a) (product over a set of representatives for S(E)/F^{×}^{n}). Moreover, there is a perfect pairing

(a, σ) ↦ \frac{σa}{a} : \frac{S(E)}{F^{×n}} × Gal(E/F) → μ_{n} (group of n^{th} roots of 1).

In particular, [E : F] = (S(E) : F^{×n}). (Cf. Exercise 5 for the case n = 2.)

5.7. Proof of Galois’s solvability theorem. Recall that a polynomial f(X) ∈ F [X] is said to be solvable if there is a tower of fields

F = F_{0} ⊂ F_{1} ⊂ F_{2} ⊂ ⋯ ⊂ F_{m}

such that

(a) F_{i} = F_{i−1}[α_{i}], where α_{i}^{m_{i}} ∈ F_{i−1} for some m_{i};

(b) F_{m} splits f(X).

Theorem 5.23. Let F be a field of characteristic 0. A polynomial f ∈ F [X] is solvable if and only if its Galois group G_{f} is solvable.

Before proving the sufficiency, we need a lemma.

Lemma 5.24. Let f ∈ F [X] be separable, and let F′ be an extension field of F. Then the Galois group of f as an element of F′[X] is a subgroup of that of f as an element of F [X].

Proof. Let E′ be a splitting field for f over F′, and let α_{1}, …, α_{m} be the roots of f(X) in E′. Then E = F [α_{1}, ..., α_{m}] is a splitting field of f over F. Any element of Gal(E′/F′) permutes the α_{i} and so maps E into itself. The map σ ↦ σ|E is an injection Gal(E′/F′) → Gal(E/F).

Proof. (G_{f} solvable ⟹ f solvable). Let f ∈ F [X] have solvable Galois group. Let F′ = F [ζ] where ζ is a primitive n^{th} root of 1 for some large n—for example, n = (deg f)! will do. The lemma shows that the Galois group G of f as an element of F′[X] is a subgroup of G_{f}, and hence is solvable. This means that there is a sequence of subgroups

G = G_{m} ⊃ G_{m−1} ⊃ ⋯ ⊃ G_{1} ⊃ G_{0} = {1}

such that each G_{i} is normal in G_{i+1} and G_{i+1}/G_{i} is cyclic (even of prime order, but we don’t need this). Let E be a splitting field of f(X) over F′, and let F_{i} = E^{G_{i}}. We have a sequence of fields

F ⊂ F [ζ] = F′ ⊂ F_{1} ⊂ F_{2} ⊂ ⋯ ⊂ F_{m} = F

with F_{i} Galois over F_{i−1} with cyclic Galois group. According to (5.20b), F_{i} = F_{i−1}[α_{i}] with α_{i}^{[F_{i}:F_{i−1}]} ∈ F_{i−1}. This shows that f is solvable.

Before proving the necessity, we need to make some observations. Let Ω be a Galois extension of F, and let E be an extension of F contained in Ω. The Galois closure Ẽ of E in Ω is the smallest subfield of Ω containing E that is Galois over F. Let G = Gal(Ω/F) and H = Gal(Ω/E). Then Ẽ will be the subfield of Ω corresponding to the largest normal subgroup of G containing H (Galois correspondence 3.17), but this is ⋂_{σ∈G} σHσ^{−1} (see Groups 4.10), and σHσ^{−1} corresponds to σE. Hence (see 3.18) Ẽ is the composite of the fields σE, σ ∈ G. In particular, we see that if E = F [α_{1}, …, α_{m}], then Ẽ is generated over F by the elements σα_{i}, σ ∈ G.

Proof. (f solvable ⟹ G_{f} solvable). It suffices to show that G_{f} is a quotient of a solvable group. Hence it suffices to find a Galois extension Ẽ of F with Gal(Ẽ/F) solvable and such that f(X) splits in Ẽ[X].

We are given that f splits in an extension F_{m} of F with the following property: F_{m} = F [α_{1}, …, α_{m}] and, for all i, there exists an m_{i} such that α_{i}^{m_{i}} ∈ F [α_{1}, …, α_{i−1}]. By (5.1) we know F_{m} = F [γ] for some γ. Let g(X) be the minimum polynomial of γ over F, and let Ω be a splitting field of g(X)(X^{n} − 1) for some suitably large n. We can identify F_{m} with a subfield of Ω. Let G = {σ_{1} = 1, σ_{2}, … } be the Galois group of Ω/F and let ζ be a primitive n^{th} root of 1 in Ω. Choose Ẽ to be the Galois closure of F_{m}[ζ] in Ω. According to the above remarks, Ẽ is generated over F by the elements

ζ, α_{1}, α_{2}, …, α_{m}, σ_{2}α_{1}, …, σ_{2}α_{m}, σ_{3}α_{1}, ….

When we adjoin these elements one by one, we get a sequence of fields

F ⊂ F [ζ] ⊂ F [ζ, α_{1}] ⊂ ⋯ ⊂ F′ ⊂ F′′ ⊂ ⋯ ⊂ Ẽ

such that each field F′′ is obtained from its predecessor F′ by adjoining an r^{th} root of an element of F′. According to (5.20a) and (5.7), each of these extensions is Galois with cyclic Galois group, and so G has a normal series with cyclic quotients. It is therefore solvable.

5.8. The general polynomial of degree n. When we say that the roots of

aX^{2} + bX + c

are

\frac{−b ± \sqrt{b^{2} − 4ac}}{2}

we are thinking of a, b, c as variables: for any particular values of a, b, c, the formula gives the roots of the particular equation. We shall prove in this section that there is no similar formula for the roots of the “general polynomial” of degree ≥ 5.

We define the general polynomial of degree n to be

f(X) = X^{n} − t_{1}X^{n−1} + ⋯ + (−1)^{n}t_{n} ∈ F [t_{1}, ..., t_{n}][X]

where the t_{i} are variables. We shall show that, when we regard f as a polynomial in X with coefficients in the field F (t_{1}, …, t_{n}), its Galois group is S_{n}. Then Theorem 5.23 proves the above remark (at least on characteristic zero).

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

P (X_{σ(1)}, …, X_{σ(n)}) = P (X_{1}, …, X_{n}), all σ ∈ S_{n}.

For example

p_{1} = ∑_{i} X_{i}  = X_{1} + X_{2} + ⋯ + X_{n},
p_{2} = ∑_{i<j} X_{i}X_{j}  = X_{1}X_{2} + X_{1}X_{3} + ⋯ + X_{1}X_{n} + X_{2}X_{3} + ⋯ + X_{n−1}X_{n},
p_{3} = ∑_{i<j<k} X_{i}X_{j}X_{k}, = X_{1}X_{2}X_{3} + ⋯
⋯
p_{r} = ∑_{i_{1}<···<i_{r}} X_{i_{1}}...X_{i_{r}}
⋯
p_{n} = X_{1}X_{2} ⋯ X_{n}

are all symmetric, because p_{r} is the sum of all monomials of degree r made up out of distinct X_{i}’s. These particular polynomials are called the elementary symmetric polynomials.

Theorem 5.25 (Symmetric polynomials theorem). Every  symmetric  polyno-

mial P (X_{1}, ..., X_{n}) in R[X_{1}, ..., X_{n}] is equal to a polynomial in the elementary symmetric polynomials with coefficients in R, i.e., P ∈ R[p_{1}, ..., p_{n}].

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

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

if either

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

or equality holds and, for some s,

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

For example,

X_{1}X_{2}^{3}X_{3} > X_{1}X_{2}^{2}X_{3} > X_{1}X_{2}X_{3}^{2}.

Let X_{1}^{k_{1}} ⋯ X_{n}^{k_{n}} 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_{n}^{k_{n}} by permuting the X’s. Hence k_{1} ≥ k_{2} ≥ ⋯ ≥ k_{n}.

The highest monomial in p_{i} is X_{1} ⋯ X_{i}, and it follows that the highest monomial in p_{1}^{d_{1}} ⋯ p_{n}^{d_{n}} is

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

Therefore

P (X_{1}, …, X_{n}) − cp_{1}^{k_{1}−k_{2}}p_{2}^{k_{2}−k_{3}} ⋯ p_{n}^{k_{n}} < P (X_{1}, …, X_{n}).

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 p_{1}, …, p_{n}.

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

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

Thus the elementary symmetric polynomials in the roots of f(X) lie in R, and so the theorem implies that every symmetric polynomial in the roots of f(X) lies in R. For example, the discriminant

D(f) = ∏_{i<j}(α_{i} − α_{j})^{2}

of f lies in R.

The general polynomial.

Theorem 5.26 (Symmetric functions theorem). When S_{n} acts on E = F (X_{1}, ..., X_{n}) by permuting the X_{i}’s, the field of invariants is F (p_{1}, ..., p_{n}).

Proof. Suppose f = \frac{g}{h}, g, h ∈ F [X_{1}, …, X_{n}], is symmetric, i.e., fixed by all σ ∈ S_{n}. Then H = ∏_{σ∈S_{n}} σh is symmetric, and so therefore is Hf. Both Hf and H are polynomials, and therefore lie in F [p_{1}, …, p_{n}]. Hence their quotient f = \frac{Hf}{H} lies in F (p_{1}, …, p_{n}).

Corollary 5.27. The field F (X_{1}, ..., X_{n}) is Galois over F (p_{1}, ..., p_{n}) with Galois group S_{n} (acting by permuting the X_{i}).

Proof. We have shown that F (p_{1}, …, p_{n}) = F (X_{1}, …, X_{n})^{S_{n}}, and so this follows from (3.12).

Theorem 5.28. The Galois group of the general polynomial of degree n is S_{n}.

Proof. Let f(X) be the general polynomial of degree n,

f(X) = X^{n} − t_{1}X^{n−1} + ⋯ + (−1)^{n}t_{n} ∈ F [t_{1}, ..., t_{n}][X].

Consider the homomorphism

F [t_{1}, …, t_{n}] → F [p_{1}, …, p_{n}], t_{i} ↦ p_{i}.

We shall prove shortly that this is an isomorphism, and therefore induces an isomorphism on the fields of fractions

F (t_{1}, …, t_{n}) → F (p_{1}, …, p_{n}), t_{i} ↦ p_{i}.

Under this isomorphism, f(X) corresponds to

g(X) = X^{n} − p_{1}X^{n−1} + ⋯ + (−1)^{n}p_{n}.

But g(X) = ∏(X − X_{i}) in F (X_{1}, …, X_{n})[X], and so F (X_{1}, …, X_{n}) is the splitting field of g(X) ∈ F (p_{1}, …, p_{n})[X]. Therefore the last corollary shows that the Galois group of g is S_{n}, which must also be the Galois group of f.

It remains to show that the homomorphism t_{i} ↦ p_{i} is an isomorphism. Let E ⊃ F (t_{1}, …, t_{n}) be a splitting field of f, and let α_{1}, ..., α_{n} be the roots of f in E. Consider the diagram

E ⊃ F [α_{1}, …, α_{n}] ⟵^{α_{i}←X_{i}} F [X_{1}, …, X_{n}]
∪  ∪
F [t_{1}, …, t_{n}]  ⟶^{t_{i}↦p_{i}} F [p_{1}, …, p_{n}].

The top and bottom maps are well-defined because F [X_{1}, ..., X_{n}] and F [t_{1}, ..., t_{n}] are polynomial rings. The diagram commutes because t_{i} = p_{i}(α_{1}, ..., α_{n}). Hence the lower horizontal map is injective, and, since it is obviously surjective, it is an isomorphism.

Remark 5.29. In the final section of this course, we’ll discuss algebraic independence. Then it will be obvious that the map t_{i} ↦ p_{i} : F [t_{1}, …, t_{n}] → F [p_{1}, …, p_{n}] is an isomorphism, which simplifies the proof.

Remark 5.30. Since S_{n} occurs as a Galois group over ℚ, and every finite group occurs as a subgroup of some S_{n}, it follows that every finite group occurs as a Galois group over some finite extension of ℚ, but does every finite Galois group occur as a Galois group over ℚ itself?

The Hilbert-Noether program for proving this was the following.

Hilbert proved that if G occurs as the Galois group of an extension E ⊃ ℚ(t_{1}, ..., t_{n}) (the t_{i} are variables), then it occurs infinitely often as a Galois group over ℚ. For the proof, realize E as the splitting field of a polynomial f(X) ∈ k[t_{1}, …, t_{n}][X] and prove that for infinitely many values of the t_{i}, the polynomial you obtain in ℚ[X] has Galois group G. (This is quite a difficult theorem—see Serre, Lectures on the Mordell-Weil Theorem, Chapter 9.)

Noether conjectured the following: Let G ⊂ S_{n} act on F (X_{1}, ..., X_{n}) by permuting the X_{i}; then F (X_{1}, …, X_{n})^{G} ≈ F (t_{1}, ..., t_{n}) (for variables t_{i}).

Unfortunately, Swan proved in 1969 that the conjecture is false for C_{47}. Hence this approach can not lead to a proof that all finite groups occur as Galois groups over ℚ, but it doesn’t exclude other approaches. [For more information on the problem, see Serre, ibid., Chapter 10, and Serre, Topics in Galois Theory, 1992.]

Remark 5.31. Take F = ℂ, and consider the subset of ℂ^{n+1} defined by the equation

X^{n} − T_{1}X^{n−1} + ⋯ + (−1)^{n}T_{n} = 0.

It is a beautiful complex manifold S of dimension n. Consider the projection

π : S → ℂ^{n}, (x, t_{1}, …, t_{n}) ↦ (t_{1}, …, t_{n}).

Its fibre over a point (a_{1}, …, a_{n}) is the set of roots of the polynomial

X^{n} − a_{1}X^{n−1} + ⋯ + (−1)^{n}a_{n}.

The discriminant of X^{n} − T_{1}X^{n−1} + ⋯ + (−1)^{n}T_{n}, regarded as a polynomial in X, is a polynomial D(f) ∈ ℂ[T_{1}, …, T_{n}]. Let Δ be the zero set of D(f) in ℂ^{n}. Then over each point of ℂ^{n} \ Δ, there are exactly n points of S, and S \ π^{−1}(Δ) is a covering space over ℂ^{n} \ Δ with group of covering transformations S_{n}.

A brief history. As far back as 1500 BC, the Babylonians (at least) knew a general formula for the roots of a quadratic polynomial. Cardan (about 1515 AD) found a general formula for the roots of a cubic polynomial. Ferrari (about 1545 AD) found a general formula for the roots of quartic polynomial (he introduced the resolvant cubic, and used Cardan’s result). Over the next 275 years there were many fruitless attempts to obtain similar formulas for higher degree polynomials, until, in about 1820, Ruffini and Abel proved that there are none.

5.9. Norms and traces. The trace of a square matrix is the sum of its diagonal elements, Tr(a_{ij}) = ∑_{i} a_{ii}. Since Tr UAU^{−1} = Tr(A), we can define the trace of an endomorphism α of a finite-dimensional vector space V to be the trace of the matrix of α with respect to any basis of V.

Similarly, we can define the determinant and characteristic polynomial of α to be the determinant and characteristic polynomial of the matrix of α with respect to any basis of V.

In a little more detail, a direct computation shows that Tr(AB) = Tr(BA), which shows that Tr(UAU^{−1}) = Tr(A) and hence Tr(α) is well-defined. The characteristic polynomial of α can be defined to be

c_{α}(X) = X^{n} + c_{1}X^{n−1} + ⋯ + c_{n}, c_{i} = (−1)^{i} Tr(α|Λ^{i}V), n = dim V;

in particular, c_{1} = − Tr(A) and c_{n} = (−1)^{n} det A. If A is the matrix of α with respect to some basis for V, then c_{α}(X) = det(XI_{n} − A).

For α and β endomorphisms of a finite-dimensional F -vector space V, we have

Tr(α) ∈ F; Tr(α + β) = Tr(α) + Tr(β);
det(α) ∈ F;  det(αβ) = det(α) det(β).

Now let E be a finite extension of F of degree n, and regard E as an F -vector space. Then α ∈ E defines an F -linear map α_{L} : E → E, x ↦ αx.

Define:

Tr_{E/F} (α) = Tr(α_{L}); Tr is a homomorphism (E, +) → (F, +);
Nm_{E/F} (α) = det(α_{L}); Nm is homomorphism (E^{×}, ×) → (F^{×}, ×);
c_{α}(X) = c_{α_{L}}(X).

Note that α ↦ α_{L} is an injective ring F -homomorphism from E into the ring of endomorphisms of E as a vector space over F, and so the minimum polynomial of α (in the sense of Section 1.8) is the same as the minimum polynomial of α_{L} (in the sense of linear algebra).

Example 5.32. (a) Consider the field extension ℂ ⊃ ℝ; the matrix of α_{L}, α = a + bi,

relative to the basis 1, i is ( \begin{matrix}a & −b \\ b & a \end{matrix} ), and so

Tr_{ℂ/ℝ}(α) = 2ℜ(α), Nm_{ℂ/ℝ}(α) = |α|^{2}.

(b) For α ∈ F, Tr(α) = rα, Nm(α) = α^{r}, r = [E : F].

(c) Let E = ℚ[α, i] be the splitting field of X^{8} − 2. What are the norm and the trace of α? The definition requires us to compute a 16 × 16 matrix. We shall see a quicker way of computing them presently.

Proposition 5.33. Consider a finite field extension E/F, and let f(X) be the minimum polynomial of α ∈ E (in the sense of Section 1.8). Then

c_{α}(X) = f(X)^{[E:F [α]]}.

Proof. Suppose first that E = F [α]. In this case, we have to show that c_{α}(X) = f(X). But f(X)|c_{α}(X) because c_{α}(α_{L}) = 0 (Cayley-Hamilton theorem), and the injectivity of E → End_{F-linear}(E) then implies that c_{α}(α) = 0. Since the polynomials are monic of the same degree, they must be equal.

For the general case, write V for E regarded as an F -vector space. The endomorphism α_{L} of V defines an action of F [X] on V (see Math 593), and this action factors through F [X]/(f(X)) = F [α]. Because F [α] is a field, V is a free F [α]-module, and in fact, V ≈ F [α]^{m} with m = [E : F [α]] (count dimensions over F). Hence the characteristic polynomial of α acting on V is the m^{th} power of its characteristic polynomial acting on F [α], which, according to case already proved, is f(X).

Alternatively, we can be more explicit. Let β_{1}, ..., β_{n} be a basis for F [α] over F, and let γ_{1}, ..., γ_{m} be a basis for E over F. As we saw in the proof of (1.10), {β_{i}γ_{k}} is a basis for E over F. Write αβ_{i} = ∑ a_{ji}β_{j}; then A = (a_{ij}) has characteristic polynomial f(X) according to the first case proved. Note that αβ_{i}γ_{k} = ∑ a_{ji}β_{j}γ_{k}. Therefore the matrix of α_{L} in End(E) breaks up into n × n blocks with A’s down the diagonal and zero matrices elsewhere. Therefore its characteristic polynomial is f(X)^{m}.

Corollary 5.34. Suppose that the roots of the minimum polynomial of α are α_{1}, …, α_{n} (in some splitting field containing E), and that [E : F [α]] = m. Then

Tr(α) = m ∑_{i=1}^{n} α_{i}, Nm_{E/F} α = (∏_{i=1}^{n} α_{i})^{m} .

Proof. Write the minimum polynomial of α as

f(X) = X^{n} + a_{1}X^{n−1} + ⋯ + a_{n} = ∏(X − α_{i}).

Then

c_{α}(X) = (f(X))^{m} = X^{mn} + ma_{1}X^{mn−1} + ⋯ + a_{m}^{n},

and so

Tr_{E/F} (α) = −ma_{1} = m ∑ α_{i},

and

Nm_{E/F} (α) = (−1)^{mn}a_{n}^{m} = (∏ α_{i})^{m}.

Example 5.35. (a) Consider the extension ℂ ⊃ ℝ. If α ∈ ℂ \ ℝ, then

c_{α}(X) = f(X) = X^{2} − 2ℜ(α)X + |α|^{2}.

If α ∈ ℝ, then c_{α}(X) = (X − a)^{2}.

(b) Let E = ℚ[α, i] be the splitting field of X^{8} − 2 (see Exercise 16). The minimum polynomial of α = \sqrt[8]{2} is X^{8} − 2, and so

Tr_{ℚ[α]/ℚ} α = 0;  Tr_{E/ℚ} α = 0.
Nm_{ℚ[α]/ℚ} α = −2; Nm_{E/ℚ} α = 4.

Remark 5.36. Assume E is separable over F, and let Ω be an algebraic closure of F; let σ_{1}, ..., σ_{r} be the distinct embeddings of E into Ω. Then

Tr_{E/F} α = ∑ σ_{i}α
Nm_{E/F} α = ∏ σ_{i}α.

When E = F [α], this follows from the observation (cf. 2.1b) that the σ_{i}α are the roots of the minimum polynomial f(X) of α over F. In the general case, σ_{1}α, ..., σ_{r}α are still roots of f(X) in Ω, but now each root of f(X) occurs [E : F [α]] times (cf. the proof of 2.7).

For example, if E is Galois over F with Galois group G, then

Tr_{E/F} α = ∑_{σ∈G} σα
Nm_{E/F} α = ∏_{σ∈G} σα.

Proposition 5.37. For finite extensions E ⊃ M ⊃ F, we have

Tr_{E/M} ∘ Tr_{M/F} = Tr_{E/F} ,
Nm_{E/M} ∘ Nm_{M/F} = Nm_{E/F} .

Proof. If E is separable over F, then this can be proved fairly easily using the descriptions in the above remark. We omit the proof in the general case.

Proposition 5.38. Let f(X) ∈ F [X] factor as f(X) = ∏_{i=1}^{m}(X −α_{i}) in some splitting field, and let α = α_{1}. Then, with f′ = \frac{df}{dX}, we have

disc f(X) = (−1)^{m(m−1)/2} Nm_{F [α]/F} f′(α).

Proof. Compute that

disc f(X) =^{df} ∏_{i<j}(α_{i} − α_{j})^{2}
= (−1)^{m(m−1)/2} · ∏_{i}(∏_{j≠i}(α_{i} − α_{j}))
= (−1)^{m(m−1)/2} · ∏ f′(α_{j})
= (−1)^{m(m−1)/2} Nm_{F [α]/F} (f′(α)).

Example 5.39. We compute the discriminant of

f(X) = X^{n} + aX + b, a, b ∈ F,

assumed to be irreducible and separable. Let α be a root of f(X), and let γ = f′(α) = nα^{n−1} + a. We compute its norm. On multiplying the equation

α^{n} + aα + b = 0

by nα^{−1} and rearranging, we obtain the equation

nα^{n−1} = −na − nbα^{−1}.

Hence

γ = nα^{n−1} + a = −(n − 1)a − nbα^{−1}.

Solving for α gives

α = \frac{−nb}{γ + (n − 1)a},

from which it is clear that F [α] = F [γ], and so the minimum polynomial of γ over F has degree n also. If we write

f(\frac{−nb}{X + (n − 1)a}) = \frac{P (X)}{Q(X)},

then P (γ) = f(α) = 0. Since

P (X) = (X + (n − 1)a)^{n} − na(X + (n − 1)a)^{n−1} + (−1)^{n}n^{n}b^{n−1}

is monic of degree n, it must be the minimum polynomial of γ. Therefore Nm γ is (−1)^{n} times the constant term of this polynomial, and so we find that

Nm γ = n^{n}b^{n−1} + (−1)^{n−1}(n − 1)^{n−1}a^{n}.

Finally we obtain the formula,

disc(X^{n} + aX + b) = (−1)^{n(n−1)/2}(n^{n}b^{n−1} + (−1)^{n−1}(n − 1)^{n−1}a^{n}),

which is something Maple doesn’t know (because it doesn’t understand symbols as exponents). For example,

disc(X^{5} + aX + b) = 5^{5}b^{4} + 4^{4}a^{5}.

5.10. Infinite Galois extensions (sketch). Recall that we defined a finite extension Ω of F to be Galois over F if it is normal and separable, i.e., if every irreducible polynomial f ∈ F [X] having a root in Ω has deg f distinct roots in Ω. Similarly, we define an algebraic extension Ω of F to be Galois over F if it is normal and separable. Equivalently, a field Ω ⊃ F is Galois over F if it is a union of subfields E finite and Galois over F.

Let Gal(Ω/F) = Aut(Ω/F), and consider the map

σ ↦ (σ|E) : Gal(Ω/F) → ∏ Gal(E/F)

(product over the finite Galois extensions E of F contained in Ω). This map is injective, because Ω is a union of finite Galois extensions. We give each finite group Gal(E/F) the discrete topology and ∏ Gal(E/F) the product topology, and we give Gal(Ω/F) the subspace topology. Thus the subgroups Gal(Ω/E), [E : F] < ∞, form a fundamental system of neighbourhoods of 1 in Gal(Ω/F).

By the Tychonoff theorem, ∏ Gal(E/F) is compact, and it is easy to see that the image of Gal(Ω/F) is closed—hence it is compact and Hausdorff.

Theorem 5.40. Let Ω be Galois over F with Galois group G. The maps

H ↦ Ω^{H}, M ↦ Gal(Ω/M)

define a one-to-one correspondence between the closed subgroups of G and the intermediate fields M. A field M is of finite degree over F if and only if Gal(Ω/M) is open in Gal(Ω/F).

Proof. Omit—it is not difficult given the finite case. See for example, E. Artin, Algebraic Numbers and Algebraic Functions, p103.

Remark 5.41. The remaining assertions in the Fundamental Theorem of Galois Theory carry over to the infinite case provided that one requires the subgroups to be closed.

Example 5.42. Let Ω be an algebraic closure of a finite field 𝔽_{p}. Then G = Gal(Ω/𝔽_{p}) contains a canonical Frobenius element, σ = (a ↦ a^{p}), and it is generated by it as a topological group, i.e., G is the closure of <σ>. Endow ℤ with the topology for which the groups nℤ, n ≥ 1, form a fundamental system of neighbourhoods of 0. Thus two integers are close if their difference is divisible by a large integer.

As for any topological group, we can complete ℤ for this topology. A Cauchy seqence in ℤ is a sequence (a_{i})_{i≥1}, a_{i} ∈ ℤ, satisfying the following condition: for all n ≥ 1, there exists an N such that a_{i} ≡ a_{j} mod n for i, j > N. Call a Cauchy sequence in ℤ trivial if a_{i} → 0 as i → ∞, i.e., if for all n ≥ 1, there exists an N such that a_{i} ≡ 0 mod n. The Cauchy sequences form a commutative group, and the trivial Cauchy sequences form a subgroup. We can define ℤ̂ to be the quotient of the first group by the second. It has a ring structure, and the map sending m ∈ ℤ to the constant sequence m, m, m, … identifies ℤ with a subgroup of ℤ̂.

Let α ∈ ℤ̂ be represented by the Cauchy sequence (a_{i}). The restriction of σ to 𝔽_{p^{n}} has order n. Therefore (σ|𝔽_{p^{n}})^{a_{i}} is independent of i provided it is sufficiently large, and we can define σ^{α} ∈ Gal(𝔽/𝔽_{p}) to be such that, for each n, σ^{α}|𝔽_{p^{n}} = (σ|𝔽_{p^{n}})^{a_{i}} for all i sufficiently large (depending on n). The map α ↦ σ^{α} : ℤ̂ → Gal(𝔽/𝔽_{p}) is an isomorphism.

The group ℤ̂ is uncountable. To most analysts, it is a little weird—its connected components are one-point sets. To number theorists it will seem quite natural— the Chinese remainder theorem implies that it is isomorphic to ∏_{p prime} ℤ_{p} where ℤ_{p} is the ring of p-adic integers.

Example 5.43. Let Ω be the algebraic closure of ℚ in ℂ; then Gal(Ω/ℚ) is one of the most basic, and intractible, objects in mathematics. Note that, as far as we know, it could have every finite group as a quotient, and it certainly has S_{n} as a quotient group for every n (and every sporadic simple group, and every...). We do however understand Gal(F^{ab}/F) when F ⊂ ℂ is a finite extension of ℚ and F^{ab} is the union of all finite abelian extensions of F contained in ℂ. For example, Gal(ℚ^{ab}/ℚ) ≈ ℤ̂^{×}. (This is abelian class field theory—see Math 776.)

6. Transcendental Extensions

In this section we consider fields Ω ⊃ F with Ω much bigger than F. For example, we could have ℂ ⊃ ℚ.

Elements α_{1}, ..., α_{n} of Ω are said to be algebraically dependent over F if there is a nonzero polynomial f(X_{1}, ..., X_{n}) ∈ F [X_{1}, ..., X_{n}] such that f(α_{1}, ..., α_{n}) = 0. Otherwise, the elements are said to be algebraically independent over F. Thus they are algebraically independent if

a_{i_{1},...,i_{n}} ∈ F, ∑ a_{i_{1},...,i_{n}}α_{1}^{i_{1}}...α_{n}^{i_{n}} = 0 ⟹ a_{i_{1},...,i_{n}} = 0 all i_{1}, ..., i_{n}.

Note the similarity with linear independence. In fact, if f is required to be homogeneous of degree 1, then the definition becomes that of linear independence. The theory in this section is logically very similar to a part of linear algebra. It is useful to keep the following correspondences in mind:

Linear algebra  Transcendence
linearly independent  algebraically independent
A ⊂ span(B)  A algebraically dependent on B
basis  transcendence basis
dimension  transcendence degree

Example 6.1. (a) A single element α is algebraically independent over F if and only if it is transcendental over F.

(b) The complex numbers π and e are almost certainly algebraically independent over ℚ, but this has not been proved.

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

Remark 6.2. To say that α_{1}, ..., α_{n} are algebraically independent over F, is the same as to say that the map

f(X_{1}, ..., X_{n}) ↦ f(α_{1}, ..., α_{n}) : F [X_{1}, ..., X_{n}] → F [α_{1}, ..., α_{n}]

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

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

In this case, F (α_{1}, ..., α_{n}) is called a pure transcendental extension of F. Then (see 5.28) the polynomial

f(X) = X^{n} − α_{1}X^{n−1} + … (−1)^{n}α_{n}

has Galois group S_{n} over F (α_{1}, ..., α_{n}).

Let β ∈ Ω and let A ⊂ Ω. The following conditions are equivalent:

(a) β is algebraic over F (A);

(b) there exist α_{1}, …, α_{n} ∈ F (A) such that β^{n} + α_{1}β^{n−1} + ⋯ + α_{n} = 0;

(c) there exist α_{0}, …, α_{n} ∈ F [A] such that α_{0}β^{n} + ⋯ + α_{n} = 0;

(d) there exists a F (X_{1}, …, X_{m}, Y) ∈ F [X_{1} …, X_{m}, Y] and a_{1}, …, a_{m} ∈ F such that F (a_{1}, …, a_{m}, Y) ≠ 0 but F (a_{1}, …, a_{m}, β) = 0.

When these conditions hold, we say that β is algebraically dependent on A (over F). A set B is algebraically dependent on A if each element of B is algebraically dependent on A.

Theorem 6.3 (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 (over F);

(b) A is algebraically dependent on B (over F).

Then m ≤ n.

Proof. We first prove a lemma.

Lemma 6.4 (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. Because β is algebraically dependent on {α_{1}, …, α_{m}}, there exists a polynomial f(X_{1}, ..., X_{m}, Y) with coefficients in F such that

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

Write

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

and observe that, because f(α_{1}, …, α_{m}, Y) ≠ 0, at least one of the polynomials a_{i}(α_{1}, ..., α_{m−1}, Y), say a_{i_{0}}, is not the zero polynomial. Because β is not algebraically dependent on {α_{1}, ..., α_{m−1}}, a_{i_{0}}(α_{1}, ..., α_{m−1}, β) ≠ 0. 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 6.5 (Transitivity of algebraic dependence). If C is algebraically dependent on B, and B is algebraically dependent on A, then C is algebraically dependent on A.

Proof. The argument in the proof (2.10) shows that if γ is algebraic over a field E which is algebraic over a field F, then γ is algebraic over F (if a_{1}, …, a_{n} are the coefficients of the minimum polynomial of γ over E, then the field F [a_{1}, …, a_{n}, γ] has finite degree over F). Apply this with F (A ∪ B) for E and F (A) for F.

Proof. (of the theorem). We now prove the theorem. Let k be the number of elements that A and B have in common. If k = m, then A ⊂ B, and certainly m ≤ n. Suppose that k < m, and write B = {α_{1}, ..., α_{k}, β_{k+1}, ..., β_{n}}. Since α_{k+1} is algebraically dependent on {α_{1}, ..., α_{k}, β_{k+1}, ..., β_{n}} but not on {α_{1}, ..., α_{k}}, there will be a β_{j}, k + 1 ≤ j ≤ n, such that α_{k+1} is algebraically dependent on {α_{1}, ..., α_{k}, β_{k+1}, ..., β_{j}} but not {α_{1}, ..., α_{k}, β_{k+1}, ..., β_{j−1}}. The exchange lemma then shows that β_{j} is algebraically dependent on

B_{1} =_{df} B ∪ {α_{k+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 k + 1 < m, repeat the argument with A and B_{1}. Eventually we’ll achieve k = m, and m ≤ n.

Definition 6.6. A transcendence basis for Ω over F is an algebraically independent set A such that E is algebraic over F (A).

Lemma 6.7. If Ω is algebraic over F (A), and A is minimal among subsets of Ω with this property, then it is a transcendence basis for Ω over F.

Proof. If α_{1}, …, α_{m} ∈ A are not algebraically independent, then one is algebraically dependent on the remainder, and it follows from (6.5) that Ω will still be algebraic over F (A) after it has been dropped from A.

Theorem 6.8. If there is a finite subset A ⊂ Ω such that Ω is algebraic over F (A), then Ω has a finite transcendence basis over F. Moreover, every transcendence basis is finite, and they all have the same number of elements.

Proof. In fact, any minimal subset of A such that Ω is algebraic over F (A) will be a transcendence basis. The second statement follows from Theorem 6.3.

The cardinality of a transcendence basis for Ω over F is called the transcendence degree of Ω over F. For example, the pure transcendental extension F (X_{1}, …, X_{n}) has transcendence degree n over F.

Example 6.9. Let p_{1}, …, p_{n} be the elementary symmetric polynomials in X_{1}, …, X_{n}. The field F (X_{1}, …, X_{n}) is algebraic over F (p_{1}, …, p_{n}), and so {p_{1}, p_{2}, …, p_{n}} contains a transcendence basis for F (X_{1}, …, X_{n}). Because F (X_{1}, …, X_{n}) has transcendence degree n, the p_{i}’s must themselves be a transcendence basis.

Example 6.10. Let Ω be the field of meromorphic functions on a compact complex manifold M.

(a) The only meromorphic functions on the Riemann sphere are the rational functions in z. Hence, in this case, Ω is a pure transcendental extension of ℂ of transcendence degree 1.

(b) If M is a Riemann surface, then the transcendence degree of Ω over ℂ is 1, and Ω is a pure transcendental extension of ℂ ⟺ M is isomorphic to the Riemann sphere

(c) If M has complex dimension n, then the transcendence degree is ≤ n, with equality holding if M is embeddable in some projective space.

Lemma 6.11. Suppose that A is algebraically independent, but that A ∪ {β} is algebraically dependent. Then β is algebraic over F (A).

Proof. The hypothesis is that there exists a nonzero polynomial f(X_{1}, ..., X_{n}, Y) ∈ F [X_{1}, ..., X_{n}, Y] such that f(a_{1}, ..., a_{n}, β) = 0, some distinct a_{1}, ..., a_{n} ∈ A. Because A is algebraically independent, Y does occur in f. Therefore

f = g_{0}Y^{m} + g_{1}Y^{m−1} + ⋯ + g_{m}, g_{i} ∈ F [X_{1}, ..., X_{n}], g_{0} ≠ 0, m ≥ 1.

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

f = g_{0}(a_{1}, ..., a_{n})X^{m} + g_{1}(a_{1}, ..., a_{n})X^{m−1} + ⋯ + g_{m}(a_{1}, ..., a_{n}),

it is algebraic over F (a_{1}, ..., a_{n}) ⊂ F (A).

Proposition 6.12. Every maximal algebraically independent subset of Ω is a transcendence basis for Ω over F.

Proof. We have to prove that Ω is algebraic over F (A) if A is maximal among algebraically independent subsets. But the maximality implies that, for every β ∈ Ω, A ∪ {β} is algebraically dependent, and so the lemma shows that β is algebraic over F (A).

Theorem 6.13 (*). Every field Ω containing F has a transcendence basis over F.

Proof. Let S be the set of algebraically independent subsets of Ω. We can partially order it by inclusion. Let T be a totally ordered subset, and let B = ∪{A | A ∈ T }. I claim that B ∈ S, i.e., that B is algebraically independent. If not, there exists a finite subset B′ of B that is not algebraically independent. But such a subset will be contained in one of the sets in T, which is a contradiction. Now we can apply Zorn’s lemma to obtain a maximal algebraically independent subset A.

It is possible to show that any two (possibly infinite) transcendence bases for Ω over F have the same cardinality.

Proposition 6.14. Any two algebraically closed fields with the same transcendence degree over F are F -isomorphic.

Proof. Choose transcendence bases A and A′ for the two fields, and choose a bijection φ : A → A′. Then φ extends uniquely to an F -isomorphism φ : F [A] → F [A′], and hence to an isomorphism of the fields of fractions F (A) → F (A′). Use this isomorphism to identify F (A) with F (A′). Then the two fields in question are algebraic closures of the same field, and hence are isomorphic (Theorem 2.16).

Remark 6.15. Any two algebraically closed fields with the same uncountable cardinality and the same characteristic are isomorphic. The idea of the proof is as follows. Let F and F′ be the prime subfields of Ω and Ω′; we can identify F with F′. Then show that when Ω is uncountable, the cardinality of Ω is the same as the cardinality of a transcendence basis over F. Finally, apply the proposition.

Remark 6.16. What are the automorphisms of ℂ? If we assume the axiom of choice, then it is easy to construct many: choose any transcendence basis A for ℂ over ℚ, and choose any permutation φ of A; then α defines an isomorphism ℚ(A) → ℚ(A) that can be extended to an automorphism of ℂ. On the other hand, without the axiom of choice, there are probably only two, the identity map and complex conjugation. (I have been told that any other is nonmeasurable, and it is known that the axiom of choice is required to construct nonmeasurable functions.)

Theorem 6.17 (Lüroth’s theorem). Any subfield E of F (X) containing F but not equal to F is a pure transcendental extension of F.

Proof. See, Jacobson, Lectures in Abstract Algebra III, p157.

Remark 6.18. This fails when there is more than one variable—see the footnote on p38 and Noether’s conjecture 5.30. The best that is true is that if [F (X, Y) : E] < ∞ and F is algebraically closed of characteristic zero, then E is a pure transcendental extension of F (Theorem of Zariski, 1958).