There are quite a few more tools to tell whether a ring is Noetherian. In this post, I’ll discuss another basic tool: integrality. I’ll discuss the application to invariant theory for finite groups.

**Subrings **

In general, it is **not** true that a subring of a Noetherian ring is Noetherian. For instance, let be the polynomial ring in infinitely many variables over a field . Then is not Noetherian because of the ascending chain

However, the quotient field of is Noetherian. This applies to any non-Noetherian integral domain.

There are special cases where we can conclude a subring of a Noetherian ring is Noetherian.

For instance, if integrality holds:

Theorem 1Let be a Noetherian ring and a finitely generated -algebra. Let be a subalgebra such that is integral over . Then is finitely generated, hence Noetherian.

We know that is finitely generated, say . Then, as is well-known, to say that is integral over is to say that each satisfies an integral equation with monic. But there are finitely many and each has finitely many coefficients, all in . We can take the subring generated by all these coefficients. Then is finitely generated over , hence Noetherian. Also is integral over .

The following is also well-known:

Proposition 2If is an extension of rings with integral over and finitely generated as an -algebra, then is finitely generated as an -module.

The proof is basically induction—if for integral over , then looking at polynomial equations one can see that is a f.g. module. Then one inducts on the number of generators.

Back to the theorem. As above, is a f.g. -module. But is Noetherian, so the submodule is a f.g. -module too. Taking the generators of and the module-generators of , we find finitely generated.

This result provides a quick application to “invariant theory.” The idea here is that if we have a group (assumed finite in this post) acting on an algebra by algebra automorphisms, then the subset of fixed points is actually an algebra, so it’s of interest to check whether, say, finitely generated implies finitely generated.

As an example, if we fix a field , we can make the symmetric group act on by permuting variables; explicitly

The invariants in this case are just **symmetric polynomials**. There is a theorem that these are generated in the **elementary symmetric functions** which are the coefficients of in

So in this case, the ring of invariants is finitely generated.

In general:

Corollary 3If is a Noetherian ring and a finitely generated -algebra acted on by (by -algebra automorphisms), then is a finitely generated -algebra too.

This follows because is **integral** over ; indeed, satisfies the polynomial equation

this polynomial is -invariant, and consequently so are the coefficients of . Now apply the previous theorem.

In general, I’m doing these commutative algebra posts from memory, but I’ll pause here to cite Bourbaki, whose elegant treatment of such matters still lingers in my mind.

So, next time: We’ll show that if each prime ideal of a ring is finitely generated., the ring is Noetherian; as David Eisenbud points out in his (excellent) book *Commutative Algebra: With a View Towards Algebraic Geometry*, ideals maximal to some property (e.g. not being finitely generated) tend to be prime. EGA 0 has some elaboration on these kinds of properties, which I would like to discuss.

August 11, 2009 at 11:42 pm

My class got a big kick out of these theorems. My professor wrote a handout that was a big “meta-theorem” about maximal with respect to some property being prime. It was nearly impossible to follow since there were * and ** and *** that you had to keep looking up what was each property. I particularly like the maximal with respect to not being principal is prime.

August 12, 2009 at 10:37 am

Hm, do you have a reference for this big “meta-theorem”? I learned about several special cases (like the one you mentioned about not being principal), but never a generalization of all this, which sounds interesting.