Integrality, invariant theory for finite groups, and more tools for Noetherian testing August 11, 2009Posted by Akhil Mathew in algebra, commutative algebra.
Tags: algebra, integrality, invariant theory, Noetherian rings
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.
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 1 Let 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 2 If 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.
Corollary 3 If 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.