Integral extension - Knowlpedia

Let $A\to B$ be a homomorphism of commutative rings. The map (or extension) is integral if every $b\in B$ is integral over $A$ : for each $b$ , some monic polynomial in $A[T]$ has $b$ as a root.

Properties

If $B$ is finitely generated as an $A$ -module, then $A\to B$ is integral. Localizing an integral extension at a multiplicative set preserves integrality (compare localization). Integral extensions also satisfy lying over and going up.

Examples

Adjoining a root of a monic polynomial. For any commutative ring $A$ and monic $f(T)\in A[T]$ , the quotient $B := A[T]/(f)$ is integral over $A$ : the class of $T$ in $B$ satisfies $f(T)=0$ .

Classical quadratic extensions. The inclusion $\mathbb{Z}\subseteq \mathbb{Z}[i]$ is integral because $\mathbb{Z}[i]$ is generated as a $\mathbb Z$ -module by $1$ and $i$ .

A cusp subring inside a polynomial ring. Let $k$ be a field and $A=k[x^2,x^3]\subseteq B=k[x]$ . Then $B=A[x]$ is integral over $A$ because $x$ satisfies $T^2-x^2=0$ , whose constant coefficient $x^2$ belongs to $A$ .

Non-example: polynomial extensions are not integral. If $A\ne 0$ , the inclusion $A\subseteq A[x]$ is not integral: the indeterminate $x$ satisfies no monic polynomial with coefficients in $A$ .