Integral element

Let $A\to B$ be a homomorphism of commutative rings (often viewed as an inclusion $A\subseteq B$). An element $b\in B$ is integral over $A$ if there exists a monic polynomial

such that $b$ is a root, i.e.

A fundamental equivalent criterion is:

Finite-module criterion. An element $b\in B$ is integral over $A$ if and only if the $A$-subalgebra $A[b]\subseteq B$ is finitely generated as an $A$-module.

This notion is the element-wise building block of an integral extension. It also underlies the definitions of integral closure and integrally closed domains.

Examples

1. Quadratic integers. In $B=\mathbb{Z}[\sqrt2]$, the element $\sqrt2$ is integral over $A=\mathbb{Z}$ because it satisfies the monic polynomial $T^2-2$.

1. A non-example: a localization element. In $B=\mathbb{Z}[1/2]$, the element $1/2$ is not integral over $\mathbb{Z}$. (Intuitively, integrality would force $\mathbb{Z}[1/2]$ to be a finite $\mathbb{Z}$-module via the finite-module criterion, which it is not.)

1. Integral over a subring generated by squares. Let $A=k[x^2]\subseteq B=k[x]$ for a field $k$. Then $x\in B$ is integral over $A$, since it satisfies the monic equation $T^2-x^2=0$ with coefficient $x^2\in A$.