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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|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|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====
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  S. Lang,   "Algebra" , Addison-Wesley  (1974)</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top">  N. Bourbaki,   "Elements of mathematics. Commutative algebra" , Addison-Wesley  (1972)  (Translated from French)</TD></TR></table>