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algebraic integer

Colloquially, an algebraic integer is a solution to an equation

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

$$(\mathbb{Z}[\alpha ])[\beta ]=\mathbb{Z}[\alpha ,\beta ]$$(\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 ${𝒪}_k$, usually called the ring of integers in $k$. This ring is a Dedekind domain?.