# More on Integrality

Lemma 1.Let be an integral extension. If is an ideal and , the resulting injection is an integral extension.

**Proof**

Any element of can be written as , . Then *x* satisfies a monic polynomial relation:

.

Taking the relation modulo gives a monic polynomial relation for with coefficients in . ♦

Lemma 2.Let be an integral extension and suppose C is the integral closure of A in B. If is a multiplicative subset, then is the integral closure of in .

**Proof**

Suppose where and . Since *x* is integral over *A* we have for some . Then in we have the equality

so is integral over .

Conversely, suppose is integral over where and . Then is also integral over so

for some

Multiplying by , there exist and such that

Multiplying by , we see that is integral over *A* so and . ♦

Corollary 1.

- If is an integral extension, so is .
- If is a normal domain, so is .

**Proof**

For the second statement note that . ♦

**Exercise A**

Prove that an integral domain *A* is normal if and only if is normal for each maximal ideal . Thus normality is a local property.

# Spectra of Integral Extensions

The inclusion map induces . It turns out geometrically, such a map is like a finite-to-one map.

For example, let and . The inclusion corresponds to projection of the curve onto the *X*-axis. The map is generically two-to-one, except at the points *X* = -1, 0, +1 on the *Y*-axis. This corresponds to the following morphism .

[ Image edited from GeoGebra plot. ]

We start with the following.

Lemma 3.If is an integral extension of domains, then A is a field if and only if B is a field.

**Proof**

(⇒) Let *A* be a field and . We can find such that . If we assume *n* is minimal, then since *B* is a domain which gives

so *b* is a unit in *B*.

(⇐) Let *B* be a field and . Then exists in *B*. This is integral over *A* so we have for some . Multiplying throughout by gives us

so *a* is a unit in *A*. ♦

**Exercise B**

Find a counter-example when *B* is not an integral domain.

Corollary 2.Let be an integral extension of rings, and . Then is maximal if and only if is maximal.

**Proof**

We get an inclusion of domains . By lemma 3, is a field if and only if is a field. ♦

Definition.Let be a homomorphism of any rings, which inducees

.

We say that

pulls backto .

# Consequences

Proposition 1.If is integral, then is surjective.

**Proof**

Let be prime and ; we get the following diagram

where the rows are injective. Since is non-trivial it has a maximal ideal , which pulls back to a prime ideal . By corollary 2, pulls back to a maximal ideal of which must be ; this pulls back to . Hence pulls back to . ♦

Proposition 2.If is integral, then is a closed map, i.e. it takes closed subsets to closed subsets.

**Proof**

Let be a closed subset, for an ideal . Let so we get an injection . By proposition 1, we get a surjective map

so maps surjectively onto the closed subset . ♦

**Note**

It follows that if is a ring homomorphism such that is integral, then is a closed map, since .

# Going Up Theorem

Proposition 3 (Going Up).Let be an integral extension. Suppose are prime ideals of A and is a prime ideal of B which pulls back to .

Then there exists a prime ideal of B containing which pulls back to .

Note that we must have .

**Proof**

Since we have an integral extension of domains . By surjectivity of , there is a prime ideal of which pulls back to , where is a prime ideal of *B* containing . Then pulls back to . ♦

Corollary 3.Suppose we have a chain of prime ideals of A, and a prime ideal of B which pulls back to .

Then there is a chain of prime ideals of B such that each lies over .

**Proof**

Repeatedly apply proposition 3 to extend the bottom chain. ♦

The final piece of the puzzle is the following.

Proposition 4.If are prime ideals of B and for , then .

**Proof**

Suppose ; let . Again we have

Since , give us prime ideals of . They both pull back to and hence to , the unique maximal ideal. By corollary 2, this means and are both maximal, hence equal. So . ♦

As a result we have:

Main Theorem.Let be an integral extension of rings. Then .

**Proof**

By Going Up and surjectivity of (proposition 1), any prime chain in *A* lifts to a prime chain in *B*. Conversely, any prime chain in *B* maps to a prime chain in *A* of the same length by proposition 4. ♦

**Example**

Since is a finite extension we have

,

since is a PID.

**Exercise C**

1. Take where and . Lift the following prime chains of *A* to prime chains of *B*

2. Prove that if is a finite extension with induced , for any , is a finite subset of Spec *B*. We say that has **finite fibres**.

In the last example at the end “ is a finite extension”. shouldn’t it be ?

Oops. Thanks! 😀

The last statement “We say that f^* has finite fibres” should be ” …\phi … $

Thanks! Corrected. 🙂