More on Integrality
Let be an integral extension. If is an ideal and , the resulting injection is an integral extension.
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 . ♦
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 .
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
Multiplying by , there exist and such that
Multiplying by , we see that is integral over A so and . ♦
- If is an integral extension, so is .
- If is a normal domain, so is .
For the second statement note that . ♦
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.
If is an integral extension of domains, then A is a field if and only if B is a field.
(⇒) 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. ♦
Find a counter-example when B is not an integral domain.
Let be an integral extension of rings, and . Then is maximal if and only if is maximal.
We get an inclusion of domains . By lemma 3, is a field if and only if is a field. ♦
Let be a homomorphism of any rings, which inducees
We say that pulls back to .
If is integral, then is surjective.
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 . ♦
If is integral, then is a closed map, i.e. it takes closed subsets to closed subsets.
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 . ♦
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 .
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 . ♦
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 .
Repeatedly apply proposition 3 to extend the bottom chain. ♦
The final piece of the puzzle is the following.
If are prime ideals of B and for , then .
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:
Let be an integral extension of rings. Then .
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. ♦
Since is a finite extension we have
since is a PID.
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.