algebraic integer

Colloquially, an **algebraic integer** is a solution to an equation

$x^n + a_1 x^{n-1} + \ldots + a_n = 0 \qquad (1)$

where each $a_i$ is an integer. More precisely, an element $x$ belonging to an algebraic extension of $\mathbb{Q}$ is an (algebraic) integer, or more briefly is *integral*, if it satisfies an equation of the form (1). Equivalently, if $k$ is an algebraic extension of $\mathbb{Q}$ (e.g., if $k$ is a number field), an element $\alpha \in k$ is integral if the subring $\mathbb{Z}[\alpha] \subseteq k$ is finitely generated as a $\mathbb{Z}$-module.

This notion may be relativized as follows: given an integral domain in its field of fractions $A \subseteq E$ and a finite field extension $E \subseteq F$, an element $\alpha \in F$ is **integral** over $A$ if $A[\alpha] \subseteq F$ is finitely generated as an $A$-module.

If $\alpha, \beta$ are integral over $\mathbb{Z}$ (say), then $\alpha + \beta$ and $\alpha \cdot \beta$ are integral over $\mathbb{Z}$. For, if $\beta$ is integral over $\mathbb{Z}$, it is *a fortiori* integral over $\mathbb{Z}[\alpha]$, hence

$(\mathbb{Z}[\alpha])[\beta] = \mathbb{Z}[\alpha, \beta]$

is finitely generated over $\mathbb{Z}[\alpha]$ and therefore, since $\alpha$ is integral, also finitely generated over $\mathbb{Z}$. It follows that the submodules $\mathbb{Z}[\alpha + \beta]$ and $\mathbb{Z}[\alpha \cdot \beta]$ are therefore also finitely generated over $\mathbb{Z}$ (since $\mathbb{Z}$ is a Noetherian ring). Thus the integral elements form a ring. In particular, the integral elements in a number field $k$ form a ring often denoted by $\mathcal{O}_k$, usually called the **ring of integers** in $k$. This ring is a Dedekind domain?.

Revised on April 25, 2010 23:33:50
by Toby Bartels
(98.19.56.65)