Rings of Algebraic Integers

Fix an algebraic closure $\overline{\mathbf{Q}}$ of $\mathbf{Q}$. For example, $\overline{\mathbf{Q}}$ could be the subfield of the complex numbers $\mathbf{C}$ generated by all roots in $\mathbf{C}$ of all polynomials with coefficients in $\mathbf{Q}$.

Much of this course is about algebraic integers.

Definition 5.1.1 (Algebraic Integer)   An element $\alpha\in\overline{\mathbf{Q}}$ is an if it is a root of some monic polynomial with coefficients in $\mathbf{Z}$.

Definition 5.1.2 (Minimal Polynomial)   The of $\alpha\in\mathbf{Q}$ is the monic polynomial $f\in\mathbf{Q}[x]$ of least positive degree such that $f(\alpha)=0$.

The minimal polynomial of $\alpha$ divides any polynomial $h$ such that $h(\alpha)=0$, for the following reason. 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 polynomial of least positive degree with root $\alpha$, so $r=0$.

Lemma 5.1.3   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 $\alpha$ and $g\in\mathbf{Z}[x]$ is a monic integral polynomial such that $g(\alpha)=0$. As mentioned after the definition of minimal polynomial, we have $g=fh$, for some $h\in\mathbf{Q}[x]$. 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 $g$. Then $p^{i+j}g = (p^if)(p^j g)$, 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 5.1.4   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 5.1.3). 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 degree of any $f_i$. Then there exist integers $a_i$ such that $\alpha^d = \sum a_i f_i(\alpha)$, hence $\alpha$ satisfies the monic polynomial $x^d - \sum a_i f_i(x) \in \mathbf{Z}[x]$, so $\alpha$ is integral. $\qedsymbol$

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 $G/2G=0$.

Proposition 5.1.5   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 $\alpha, \beta\in \mathbf{Z}$, 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. $\qedsymbol$

Recall that a is a subfield $K$ of $\overline{\mathbf{Q}}$ such that the degree $[K:\mathbf{Q}] := \dim_\mathbf{Q}(K)$ is finite.

Definition 5.1.6 (Ring of Integers)   The of a number field $K$ is the ring

$$\O _K = K \cap \overline{\mathbf{Z}}= \{x \in K :$$    $x$ is an algebraic integer$$\}.$$

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 $x^2-x-1$. According to , 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 5.1.7 (Order)   An 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 will frequently consider orders in practice 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 $\mathbf{Z}[\alpha]\neq \O _K$.

Lemma 5.1.8   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 lowest terms and $b>0$. The monic minimal polynomial of $\alpha$ is $bx-a\in\mathbf{Z}[x]$, so if $b\neq 1$ then Lemma 5.1.3 implies that $\alpha$ is not an algebraic integer, a contradiction.

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 $\mathbf{Q}\O _K = K$. $\qedsymbol$

In the next two sections we will develop some basic properties of norms and traces, and deduce further properties of rings of integers.