Integral extension

An extension $R\subseteq A$ is integral when every element of $A$ is integral over $R$.

If $A\subseteq B$ is integral and $\mathfrak{n}$ is a maximal ideal of $B$, then $\mathfrak{n}\cap A$ is maximal in $A$. Indeed, $B/\mathfrak{n}$ is an integral domain integral over $A/(\mathfrak{n}\cap A)$, and a subring over which a field is integral is itself a field.

For an integral extension $A\subseteq B$ and $\mathfrak{p}\in \mathrm{Spec}A$, localization gives a bijectionIt combines the prime ideal correspondence for localization with the contraction of a maximal ideal under an integral extension.

If $a_1,\dots ,a_n$ are integral over $R$, then $R[a_1,\dots ,a_n]$ is a finitely generated $R$-module. In particular, every element of this subalgebra is integral over $R$.

An integral domain is integrally closed when every element of its fraction field that is integral over the domain already belongs to the domain.

The integral closure of $R$ in an $R$-algebra $A$ is the subring of elements of $A$ integral over $R$.

If a graded ring $A$ is a graded extension of a graded subring $R$, then the integral closure of $R$ in $A$ is graded: every homogeneous component of an integral element is integral. This follows by separating the extremal degrees in a monic integral equation and inducting on the number of nonzero homogeneous components.

Let $R\subseteq A$ be an integral extension of domains with $R$ integrally closed. Given primes ${\mathfrak{p}}_0\subseteq {\mathfrak{p}}_1$ of $R$ and a prime ${\mathfrak{q}}_1$ of $A$ over ${\mathfrak{p}}_1$, there is a prime ${\mathfrak{q}}_0\subseteq {\mathfrak{q}}_1$ over ${\mathfrak{p}}_0$.