# Integrality

Throughout this article, *A* is a subring of *B*; we will also call *B* a **ring extension** of *A*.

Definition.An element is said to be

integralover A if we can find (where ) such thatin B.

For example, is integral over since . One intuitively guesses that is not integral over , but this is quite hard to show at this point.

Main Proposition.The following are equivalent for .

- is integral over .
- , the A-subalgebra of B generated by b, is finite over A.
- is contained in a ring C, where and C is finite over A.

**Proof**

(1) ⇒ (2). is the set of all polynomials in *b* with coefficients in *A*. Since *b* is integral over *A*, for some *n* > 0 we can express as an *A*-linear combination of . By induction, this holds for all . Hence is generated by as an *A*-module.

(2) ⇒ (3). Just take .

(3) ⇒ (1). Let generate *C* as an *A*-module; we can write each as an *A*-linear combination of . In matrix form,

where . Write *M* for the RHS matrix; denoting for the identity matrix, we get where is the column vector . Note that this relation holds in the ring *C*.

- Now for every square matrix
*M*with entries in any ring*C*, we can define its adjugate matrix*N*, which is a matrix of the same size with entries in*C*such that . All entries of*N*can be obtained by polynomials in the entries of*M*with integer coefficients so this holds in any ring.

Applying this to our matrix , we obtain . In particular . Expanding this determinant gives us a monic polynomial relation in *b* with coefficients in *A*. ♦

**Easy Exercise**

Prove that if are ring extensions such that *B* is finite over *A*, *C* is finite over *B*, then *C* is finite over *A*.

# Consequences

The above gives us:

Proposition 1.If are integral over A, so are .

**Proof**

Consider the extensions

Since is integral over *A*, it is also integral over , so both extensions are finite by the main proposition. Thus by the above exercise is also a finite extension of *A*.

Since , by the main proposition they are integral over *A*. ♦

**Example**

We see that and are integral over since each summand is clearly integral over .

Definition.By proposition 1, the set of integral over A forms a ring extension of A contained in B. We call this the

integral closureof A in B and denote it by .We say B is

integralover A if .We say A is

integrally closedin B if .

Next, we see that integrality is transitive.

Proposition 2.Let be ring extensions, where is integral.

- If is integral over B, then it is integral over A.
Hence if C is integral over B and B is integral over A, then C is integral over A.

**Proof**

Since *c* is integral over *B* there are such that

By the main proposition, we have a sequence of ring extensions, each finite over the previous.

so is finite over *A*; again by the main proposition, *c* is integral over *A*. ♦

Recall that for a field extension , a finite extension is algebraic but the converse may not be true; heuristically, this is because we can attach infinitely many algebraic elements to *k*. This inspires the following.

Proposition 3.If B is an A-algebra of finite type, and B is integral over A, then B is finite over A.

**Proof**

Write . In the extensions , each ring is finite over the previous. Hence *B* is finite over *A*. ♦

# Normal Domains

In this section, for an integral domain *A*, denotes its field of fractions.

Definition.We say A is a

normal domainif it is integrally closed in K.The

normalizationof A is the integral closure of A in K.

**Exercise A**

Prove that the normalization of *A* is a normal domain.

**Note**

In some books, *A* is called an *integrally closed domain*, but we consider it potentially confusing here. On the other hand, the term *normal* is used excessively in mathematics. However, in the context of commutative algebra this should not cause any confusion.

**Example**

Let where . Then . Now *A* is not a normal domain because is integral over *A* but does not lie in *A*.

One reason for defining normal domains lies in the following.

Proposition 4.Let be integral domains, where is normal. For an integral , consider it as an element of and take its minimal polynomial

.

Then .

**Proof**

Find a monic polynomial such that . Note that divides in . Now pick a field extension in which factors as a product of linear factors

.

Since each is integral over *A*. Thus the coefficients of are integral over *A*, being elementary symmetric polynomials in the . Hence the coefficients of lie in . ♦

However, to use the above result, we need a criterion for normal domains.

Proposition 5.A UFD is a normal domain.

**Proof**

Let *A* be a UFD and suppose is integral over *A *where *a* and *b* have no common prime factor. Pick such that

Then . Since *a* and *b* have no common prime factor, *b* is a unit so . ♦

**Examples**

1. Let , which has minimal polynomial over . Since is a UFD, it is a normal domain. Thus *b* is not integral over which solves our problem at the beginning of the article.

2. Let and . The minimal polynomial of over is , which does not lie in . On the other hand is integral over *A* because . This happens because *A* is not a normal domain.

**Exercise B**

1. Decide if each of the following is a normal domain. Find its normalization.

[ Hint: conjecture and prove a result about , where and *A* is a UFD in which 2 is invertible. ]

2. Consider where and . Is this an integral extension?

3. Let . Prove that there is a such that .

Please, can you give hints to (a) and (b) of Exercise B?

In part (1) I obtained that all except is normal. The latter one is not normal because is integral over . Is that right?

In part (2), B is not integral over A because Y is not integral over A: its minimal polynomial must be the integral equation of dependence. Is that right?

You got it right for (1), but I think you meant is integral over .

For (2), almost. But remember to use that you need the base ring to be a normal domain (see proposition 4). The way to overcome this is to extend and so that becomes a normal domain.