Rings of Algebraic Integers

In this section we will learn about rings of algebraic integers and discuss some of their properties. We will prove that the ring of integers $\O_K$ of a number field is noetherian.

Fix an algebraic closure $\overline{\mathbf{Q}}$ of $\mathbf {Q}$. Thus $\overline{\mathbf{Q}}$ is an infinite field extension of $\mathbf {Q}$ with the property that every polynomial $f\in\mathbf{Q}[x]$ splits as a product of linear factors in $\overline{\mathbf{Q}}[x]$. One choice of $\overline{\mathbf{Q}}$ is the subfield of the complex numbers $\mathbf{C}$ generated by all roots in $\mathbf{C}$ of all polynomials with coefficients in  $\mathbf {Q}$. Note that any two choices of  $\overline{\mathbf{Q}}$ are isomorphic, but there will be many isomorphisms between them.

An algebraic integer is an element of $\overline{\mathbf{Q}}$.

Definition 2.3.1 (Algebraic Integer)   An element $\alpha\in\overline{\mathbf{Q}}$ is an algebraic integer if it is a root of some monic polynomial with coefficients in  $\mathbf {Z}$.

For example, $\sqrt{2}$ is an algebraic integer, since it is a root of $x^2-2$, but one can prove $1/2$ is not an algebraic integer, since one can show that it is not the root of any monic polynomial over  $\mathbf {Z}$. Also $\pi$ and $e$ are not algebraic numbers (they are transcendental).

Example 2.3.2   We compute some minimal polynomials in SAGE. The minimal polynomial of $1/2$:

sage: (1/2).minpoly()
x - 1/2

We construct a root $a$ of $x^2-2$ and compute its minimal polynomial:

sage: k.<a> = NumberField(x^2 - 2)
sage: a^2 - 2
0
sage: a.charpoly()
x^2 - 2

Finally we compute the minimal polynomial of $\sqrt{2}/2 + 3$, which is not integral:

sage: (a/2 + 3).charpoly()
x^2 - 6*x + 17/2

The only elements of $\mathbf {Q}$ that are algebraic integers are the usual integers $\mathbf {Z}$. However, there are elements of $\overline{\mathbf{Q}}$ that have denominators when written down, but are still algebraic integers. For example,

$$\alpha = \frac{1+\sqrt{5}}{2}$$

is an algebraic integer, since it is a root of the monic polynomial $x^2 - x - 1$. We verify this using SAGE below, though of course this is easy to do by hand (you should try much more complicated examples in SAGE).

sage: k.<a> = QuadraticField(5)
sage: a^2
5
sage: alpha = (1 + a)/2
sage: alpha.charpoly()
x^2 - x - 1
sage: alpha.is_integral()
True

Definition 2.3.3 (Minimal Polynomial)   The minimal polynomial of $\alpha\in\overline{\mathbf{Q}}$ is the monic polynomial $f\in\mathbf{Q}[x]$ of least positive degree such that $f(\alpha)=0$.

It is a consequence of Lemma 2.3.5 that the minimal polynomial $\alpha$ is unique. The minimal polynomial of $1/2$ is $x-1/2$, and the minimal polynomial of $\sqrt[3]{2}$ is $x^3-2$.

Example 2.3.4   We compute the minimal polynomial of a number expressed in terms of $\sqrt[4]{2}$:

sage: k.<a> = NumberField(x^4 - 2)
sage: a^4
2
sage: (a^2 + 3).minpoly()
x^2 - 6*x + 7

Lemma 2.3.5   Suppose $\alpha\in\overline{\mathbf{Q}}$. Then the minimal polynomial of $\alpha$ divides any polynomial $h$ such that $h(\alpha)=0$.

Proof. Let $f$ be a minimal polynomial of $\alpha$. If $h(\alpha)=0$, use the division algorithm to write $h=qf + r$, where $0\leq \deg(r) < \deg(f)$. We have

$$r(\alpha) = h(\alpha) - q(\alpha) f(\alpha) = 0,$$

so $\alpha$ is a root of $r$. However, $f$ is the monic polynomial of least positive degree with root $\alpha$, so $r=0$. $\qedsymbol$

Lemma 2.3.6   If $\alpha$ is an algebraic integer, then the minimal polynomial of $\alpha$ has coefficients in  $\mathbf {Z}$.

Proof. Suppose $f\in\mathbf{Q}[x]$ is the minimal polynomial of $\alpha$. Since $\alpha$ is an algebraic integer, there is a polynomial $g\in\mathbf{Z}[x]$ that is monic such that $g(\alpha)=0$. By Lemma 2.3.5, we have $g=fh$, for some monic $h\in\mathbf{Q}[x]$. If $f\not\in\mathbf{Z}[x]$, then some prime $p$ divides the denominator of some coefficient of $f$. Let $p^i$ be the largest power of $p$ that divides some denominator of some coefficient $f$, and likewise let $p^j$ be the largest power of $p$ that divides some denominator of a coefficient of $h$. Then $p^{i+j}g = (p^if)(p^j h)$, and if we reduce both sides modulo $p$, then the left hand side is 0 but the right hand side is a product of two nonzero polynomials in $\mathbf{F}_p[x]$, hence nonzero, a contradiction. $\qedsymbol$

Proposition 2.3.7   An element $\alpha\in\overline{\mathbf{Q}}$ is integral if and only if $\mathbf{Z}[\alpha]$ is finitely generated as a $\mathbf {Z}$-module.

Proof. Suppose $\alpha$ is integral and let $f\in\mathbf{Z}[x]$ be the monic minimal polynomial of $\alpha$ (that $f\in\mathbf{Z}[x]$ is Lemma 2.3.6). Then $\mathbf{Z}[\alpha]$ is generated by $1,\alpha,\alpha^2,\ldots,\alpha^{d-1}$, where $d$ is the degree of $f$. Conversely, suppose $\alpha\in\overline{\mathbf{Q}}$ is such that $\mathbf{Z}[\alpha]$ is finitely generated, say by elements $f_1(\alpha), \ldots, f_n(\alpha)$. Let $d$ be any integer bigger than the degrees of all $f_i$. Then there exist integers $a_i$ such that $\alpha^d = \sum_{i=1}^n a_i f_i(\alpha)$, hence $\alpha$ satisfies the monic polynomial $x^d - \sum_{i=1}^n a_i f_i(x) \in \mathbf{Z}[x]$, so $\alpha$ is integral. $\qedsymbol$

Example 2.3.8   The rational number $\alpha=1/2$ is not integral. Note that $G=\mathbf{Z}[1/2]$ is not a finitely generated $\mathbf {Z}$-module, since $G$ is infinite and $G/2G=0$. (You can see that $G/2G=0$ implies that $G$ is not finitely generated, by assuming that $G$ is finitely generated, using the structure theorem to write $G$ as a product of cyclic groups, and noting that $G$ has nontrivial $2$-torsion.)

Proposition 2.3.9   The set $\overline{\mathbf{Z}}$ of all algebraic integers is a ring, i.e., the sum and product of two algebraic integers is again an algebraic integer.

Proof. Suppose $\alpha, \beta\in \overline{\mathbf{Z}}$, and let $m, n$ be the degrees of the minimal polynomials of $\alpha, \beta$, respectively. Then $1,\alpha,\ldots,\alpha^{m-1}$ span $\mathbf{Z}[\alpha]$ and $1,\beta,\ldots,\beta^{n-1}$ span $\mathbf{Z}[\beta]$ as $\mathbf {Z}$-module. Thus the elements $\alpha^i\beta^j$ for $i \leq m, j\leq n$ span $\mathbf{Z}[\alpha, \beta]$. Since $\mathbf{Z}[\alpha + \beta]$ is a submodule of the finitely-generated module $\mathbf{Z}[\alpha, \beta]$, it is finitely generated, so $\alpha+\beta$ is integral. Likewise, $\mathbf{Z}[\alpha\beta]$ is a submodule of $\mathbf{Z}[\alpha, \beta]$, so it is also finitely generated and $\alpha\beta$ is integral. $\qedsymbol$

Example 2.3.10   We illustrate an example of a sum and product of two algebraic integers being an algebraic integer. We first make the relative number field obtained by adjoining a root of $x^3 - 5$ to the field $\mathbf{Q}(\sqrt{2})$:

sage: k.<a, b> = NumberField([x^2 - 2, x^3 - 5])
sage: k
Number Field in a with defining polynomial x^2 + -2 over its base field

Here $a$ and $b$ are roots of $x^2-2$ and $x^3 - 5$, respectively.

sage: a^2
2
sage: b^3
5

We compute the minimal polynomial of the sum and product of $\sqrt[3]{5}$ and $\sqrt{2}$. The command absolute_minpoly gives the minimal polynomial of the element over the rational numbers.

sage: (a+b).absolute_minpoly()
x^6 - 6*x^4 - 10*x^3 + 12*x^2 - 60*x + 17
sage: (a*b).absolute_minpoly()
x^6 - 200

Of course the minimal polynomial of the product is $\sqrt[3]{5} \sqrt{2}$ is trivial to compute by hand. The minimal polynomial of $\alpha = \sqrt[3]{5} + \sqrt{2}$ could be computed by hand by computing the determinant of the matrix given by left multiplication of $\alpha$ on this basis:

$$1,\sqrt{2}, \sqrt[3]{5}, \sqrt[3]{5}\sqrt{2}, \sqrt[3]{5}^2, \sqrt[3]{5}^2\sqrt{2}.$$

The following is an alternative more symbolic way to compute the minimal polynomials above, though it is not provably correct. We compute $\alpha$ to 100 bits precision (via the n command), then use the LLL algorithm (via the algdep command) to heuristically find a linear relation between the first $6$ powers of $\alpha$.

sage: a = 5^(1/3); b = sqrt(2)
sage: c = a+b; c
5^(1/3) + sqrt(2)
sage: (a+b).n(100).algdep(6)
x^6 - 6*x^4 - 10*x^3 + 12*x^2 - 60*x + 17
sage: (a*b).n(100).algdep(6)
x^6 - 200

Definition 2.3.11 (Number field)   A number field is a field $K$ that contains the rational numbers $\mathbf {Q}$ such that the degree $[K:\mathbf{Q}] = \dim_\mathbf{Q}(K)$ is finite.

If $K$ is a number field, then by the primitive element theorem there is an $\alpha \in K$ so that $K = \mathbf{Q}(\alpha)$. Let $f(x) \in \mathbf{Q}[x]$ be the minimal polynomial of $\alpha$. For any fixed choice of $\overline{\mathbf{Q}}$, there is some $\alpha'\in\overline{\mathbf{Q}}$ such that $f(\alpha')=0$. The map $K\to \overline{\mathbf{Q}}$ that sends $\alpha$ to $\alpha'$ defines an embedding $K\hookrightarrow \overline{\mathbf{Q}}$. Thus any number field can be embedded (in $[K:\mathbf{Q}]$ possible ways) in any fixed choice $\overline{\mathbf{Q}}$ of an algebraic closure of $\mathbf {Q}$.

Definition 2.3.12 (Ring of Integers)   The ring of integers of a number field $K$ is the ring

$$\O_K = \{x \in K :$$    $x$ satisfies a monic polynomial with integer coefficients $$\}.$$

Note that $\O_K$ is a ring, because if we fix an embedding of $K$ into $\overline{\mathbf{Q}}$, then

$$\O_K = K \cap \overline{\mathbf{Z}}.$$

The field $\mathbf {Q}$ of rational numbers is a number field of degree $1$, and the ring of integers of $\mathbf {Q}$ is $\mathbf {Z}$. The field $K=\mathbf{Q}(i)$ of Gaussian integers has degree $2$ and $\O_K = \mathbf{Z}[i]$. The field $K=\mathbf{Q}(\sqrt{5})$ has ring of integers $\O_K = \mathbf{Z}[(1+\sqrt{5})/2]$. Note that the Golden ratio $(1+\sqrt{5})/2$ satisfies $x^2 - x - 1$. The ring of integers of $K=\mathbf{Q}(\sqrt[3]{9})$ is $\mathbf{Z}[\sqrt[3]{3}]$, where $\sqrt[3]{3}=\frac{1}{3}(\sqrt[3]{9})^2$.

Definition 2.3.13 (Order)   An order in $\O_K$ is any subring $R$ of $\O_K$ such that the quotient $\O_K/R$ of abelian groups is finite. (Note that $R$ must contain $1$ because it is a ring, and for us every ring has a $1$.)

As noted above, $\mathbf{Z}[i]$ is the ring of integers of $\mathbf{Q}(i)$. For every nonzero integer $n$, the subring $\mathbf{Z}+ni\mathbf{Z}$ of $\mathbf{Z}[i]$ is an order. The subring $\mathbf {Z}$ of $\mathbf{Z}[i]$ is not an order, because $\mathbf {Z}$ does not have finite index in $\mathbf{Z}[i]$. Also the subgroup $2\mathbf{Z}+ i\mathbf{Z}$ of $\mathbf{Z}[i]$ is not an order because it is not a ring.

We define the number field $\mathbf{Q}(i)$ and compute its ring of integers, which has discriminant $-4$.

sage: K.<i> = NumberField(x^2 + 1)
sage: OK = K.ring_of_integers(); OK
Order with module basis 1, i in Number Field in i with 
defining polynomial x^2 + 1
sage: OK.discriminant()
-4

Next we compute the order $\mathbf{Z}+ 3i \mathbf{Z}$.

sage: O3 = K.order(3*i); O3
Order with module basis 1, 3*i in Number Field in i with 
defining polynomial x^2 + 1
sage: O3.gens()
[1, 3*i]

Notice that the distriminant is $-36 = -4 \cdot 3^2$:

sage: O3.discriminant()
-36

We test whether certain elements are in the order.

sage: 5 + 9*i in O3
True
sage: 1 + 2*i in O3
False

We will frequently consider orders because they are often much easier to write down explicitly than $\O_K$. For example, if $K = \mathbf{Q}(\alpha)$ and $\alpha$ is an algebraic integer, then $\mathbf{Z}[\alpha]$ is an order in $\O_K$, but frequently $\mathbf{Z}[\alpha]\neq \O_K$.

Example 2.3.14   In this example $[\O_K : \mathbf{Z}[a]] = 2197$. First we define the number field $K=\mathbf{Q}(a)$ where $a$ is a root of $x^3 - 15 x^2 - 94 x - 3674$, then we compute the order $\mathbf{Z}[a]$ generated by $a$.

sage: K.<a> = NumberField(x^3 - 15*x^2 - 94*x - 3674)
sage: Oa = K.order(a); Oa
Order with module basis 1, a, a^2 in Number Field in a with defining 
polynomial x^3 - 15*x^2 - 94*x - 3674

Next we compute the maximal order $\O_K$ of $K$ with a basis, and compute that the index of $\mathbf{Z}[a]$ in $\O_K$ is $2197=13^3$.

sage: OK = K.maximal_order()
sage: OK.basis()
[25/169*a^2 + 10/169*a + 1/169, 5/13*a^2 + 1/13*a, a^2]
sage: Oa.index_in(OK)
2197

Lemma 2.3.15   Let $\O_K$ be the ring of integers of a number field. Then $\O_K\cap \mathbf{Q}=\mathbf{Z}$ and $\mathbf{Q}\O_K = K$.

Proof. Suppose $\alpha\in \O_K\cap\mathbf{Q}$ with $\alpha=a/b \in \mathbf{Q}$ in lowest terms and $b>0$. Since $\alpha$ is integral, $\mathbf{Z}[a/b]$ is finitely generated as a module, so $b=1$ (see Example 2.3.8).

To prove that $\mathbf{Q}\O_K = K$, suppose $\alpha \in K$, and let $f(x) \in \mathbf{Q}[x]$ be the minimal monic polynomial of $\alpha$. For any positive integer $d$, the minimal monic polynomial of $d\alpha$ is $d^{\deg(f)}f(x/d)$, i.e., the polynomial obtained from $f(x)$ by multiplying the coefficient of $x^{\deg(f)}$ by $1$, multiplying the coefficient of $x^{\deg(f)-1}$ by $d$, multiplying the coefficient of $x^{\deg(f)-2}$ by $d^2$, etc. If $d$ is the least common multiple of the denominators of the coefficients of $f$, then the minimal monic polynomial of $d\alpha$ has integer coefficients, so $d\alpha$ is integral and $d\alpha\in \O_K$. This proves that $\mathbf{Q}\O_K = K$. $\qedsymbol$