2024-08-16

Lemma (Artin-Tate)

Let be a Noetherian ring. Let be an -algebra homomorphism where is a finitely-generated -algebra and is a finitely-generated -module. Then is a finitely-generated -algebra.

Proof

Let be a set of -algebra generators for , and be a set of -module generators for . Therefore each

Let be the finitely-generated -subalgebra of generated by , so it is Noetherian by Proposition (Noetherian things) 3..

Claim: is a finitely-generated -module.
Proof. Why? If , then for some . By repeatedly applying the identities above, we see that . Hence, we are done, because and is finitely-generated as a -module. is Noetherian, so is finitely-generated as a -module. So since is a finitely-generated -algebra, is a finitely-generated -algebra. #TODO

Example

Let be a group acting on . Denote the -invariant subring by . Then for , we have (c.f. Galois theory)

Therefore every element of satisfies a monic polynomial with coefficients in . So is a finitely generated -module. By Lemma (Artin-Tate), is a finitely-generated -algebra.

Theorem (Zariski’s Lemma)

Let be a field. Let be a finitely-generated -algebra. If is a field then is a finite extension of .

Proof (countable case)

If is not finite and is countable, then we have a tower

where are algebraically independent, and is finite (due to field theory, c.f. transcendence bases). By Lemma (Artin-Tate), is a finitely-generated -algebra. So certainly is a finitely-generated -algebra. We now just need to show that if is any field, then is not a finitely-generated -algebra.

If there is an where this is false, then there exist rational functions

such that every is a polynomial in . This means that every with a non-trivial denominator has a denominator which is divisible by at least one of the . Take

which does not have that property. This is a contradiction.

Remark

  • Partial fractions things can’t be too big.
  • The argument above things must be of a certain size.

Corollary

Let be a field and let be a finitely-generated -algebra. If is an ideal, then

Remark

This tells us that varieties () are not missing a lot from the full prime ideal spectrum.

Proof

There exists and a surjection . So we can replace by by the Lattice Isomorphism Theorem. Also, because , we can assume is prime, i.e., it suffices to prove that

That is, if , then there exists maximal and .

Let . This is still a finitely-generated -algebra. Let be a maximal ideal of . Since is a unit in , . Let be the contraction of .

Claim: is maximal.
Proof. . is a field and is finitely-generated over . By Theorem (Zariski’s Lemma), is a finite extension. So is a domain that is finite-dimensional over . We now claim that if is a domain over and then is a field. This is enough because it implies that is a field, i.e., is maximal. □

Now and .

Example

Let . Then the conclusion of the previous corollary fails. This is because is a local ring: is the unique maximal ideal but .

Corollary (Strong Nullstellensatz)

Let be an algebraically closed field. Let be an ideal. If for all , then for some .

Recall that .

Proof

The above Corollary implies that

Recall that all the maximal ideals for some by Theorem (Weak Nullstellensatz).

Category theory

A category is a collection of objects together with arrows/maps/morphisms such that

  1. If and are objects, there is a set of morphisms .
  2. If , , are objects, there is an associative composition law between morphisms
  3. .

Examples

  1. .
  2. , finite sets.
  3. , the simplex category: non-empty finite sets with a total order and weakly order preserving functions, e.g., .
  4. Groups, abelian groups, modules over a ring , algebras over .
  5. .
  6. .
  7. Let be a set. We construct a category whose objects are elements of and if and otherwise.