Integral extension of a ring

An extension $B$ of a commutative ring $A$ with unit element such that every element $x \in B$ is integral over $A$, that is, satisfies an equation of the form

$$x ^ {n} + a _ {n - 1 } x ^ {n - 1 } + \dots + a _ {0} = 0,$$

where $a _ {i} \in A$, a so-called equation of integral dependence.

An element $x$ is integral over $A$ if and only if one of the following two equivalent conditions is satisfied: 1) $A [ x]$ is an $A$- module of finite type; or 2) there exists a faithful $A [ x]$- module that is an $A$- module of finite type. An integral element is algebraic over $A$. If $A$ is a field, the converse assertion holds. Elements of the field $\mathbf C$ of complex numbers that are integral over $\mathbf Z$ are called algebraic integers. If a ring $B$ is a module of finite type over $A$, then every element $x \in B$ is integral over $A$( the converse need not be true).

Suppose that $R \supset A$ is a commutative ring, and let $x$ and $y$ be elements of $R$ that are integral over $A$. Then $x + y$ and $xy$ are also integral over $A$, and the set of all elements of $R$ that are integral over $A$ forms a subring, called the integral closure of $A$ in $R$. All rings considered below are assumed to be commutative.

If $B$ is integral over $A$ and $A ^ \prime$ is some $A$- algebra, then $B \otimes A ^ \prime$ is integral over $A ^ \prime$. If $B$ is an integral extension of $A$ and $S$ is some multiplicative subset of $A$, then the ring $S ^ {-} 1 B$ is integral over $S ^ {-} 1 A$. An integral domain $A$ is said to be integrally closed if the integral closure of $A$ in its field of fractions is $A$. A factorial ring is integrally closed. A ring $A$ is integrally closed if and only if for every maximal ideal $\mathfrak p \subset A$ the local ring $A _ {\mathfrak p }$ is integrally closed.

Let $B$ be an integral extension of $A$ and let $\mathfrak p$ be a prime ideal of $A$. Then $\mathfrak p B \neq B$ and there exists a prime ideal $\mathfrak P$ of $B$ that lies above $\mathfrak p$( that is, $\mathfrak P$ is such that $\mathfrak p = \mathfrak P \cap A$). The ideal $\mathfrak P$ is maximal if and only if $\mathfrak p$ is maximal. If $L$ is a finite extension of the field of fractions of a ring $A$ and $B$ is the integral closure of $A$ in $L$, then there are only finitely-many prime ideals of $B$ that lie above a given prime ideal of $A$.

Suppose that $C \supset B \supset A$; then $C \supset A$ is an integral extension if and only if both $C \supset B$ and $B \supset A$ are integral extensions.

References

[1]	S. Lang, "Algebra" , Addison-Wesley (1974)
[2]	N. Bourbaki, "Elements of mathematics. Commutative algebra" , Addison-Wesley (1972) (Translated from French)