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

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

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.

William Stein 2004-05-06