# 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) |

**How to Cite This Entry:**

Integral closure.

*Encyclopedia of Mathematics.*URL: http://encyclopediaofmath.org/index.php?title=Integral_closure&oldid=35202