jump to navigation

Integrality, invariant theory for finite groups, and more tools for Noetherian testing August 11, 2009

Posted by Akhil Mathew in algebra, commutative algebra.
Tags: , , ,
trackback

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 {A := k[X_1, X_2, \dots]} be the polynomial ring in infinitely many variables over a field {k}. Then {A} is not Noetherian because of the ascending chain

\displaystyle (X_0) \subset (X_0, X_1) \subset (X_0, X_1, X_2) \subset \dots.

However, the quotient field of {A} 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 {A_0} be a Noetherian ring and {A} a finitely generated {A_0}-algebra. Let {B \subset A} be a subalgebra such that {A} is integral over {B}. Then {B} is finitely generated, hence Noetherian.

 We know that {A} is finitely generated, say {A=A_0[x_1, \dots, x_n]}. Then, as is well-known, to say that {A} is integral over {B} is to say that each {x_i} satisfies an integral equation {P_i(x_i)=0} with {P_i \in B[X]} monic. But there are finitely many {P_i} and each has finitely many coefficients, all in {B}. We can take the subring {C \subset B \subset A} generated by all these coefficients. Then {C} is finitely generated over {A_0}, hence Noetherian. Also {A} is integral over {C}.

The following is also well-known: 

Proposition 2 If {R \subset S} is an extension of rings with {S} integral over {R} and finitely generated as an {R}-algebra, then {S} is finitely generated as an {R}-module.  

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

Back to the theorem. As above, {A} is a f.g. {C}-module. But {C} is Noetherian, so the submodule {B \subset A} is a f.g. {C}-module too. Taking the generators of {C} and the {C} module-generators of {B}, we find {B} finitely generated.

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

As an example, if we fix a field {k}, we can make the symmetric group {S_n} act on {k[x_1, \dots, x_n]} by permuting variables; explicitly

\displaystyle \sigma f(x_1, \dots, x_n) := f( x_{\sigma^{-1} 1}, \dots, x_{\sigma^{-1}n}).

The {S_n} 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 {T} in

\displaystyle \prod_i (T - x_i) \in k[x_1, \dots, x_n, T].

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

In general:

Corollary 3 If {A_0} is a Noetherian ring and {A} a finitely generated {A_0}-algebra acted on by {G} (by {A_0}-algebra automorphisms), then {A^G} is a finitely generated {A_0}-algebra too.  

This follows because {A} is integral over {A^G}; indeed, {x \in A} satisfies the polynomial equation

\displaystyle \prod_{\sigma \in G} (T - \sigma x);

this polynomial is {G}-invariant, and consequently so are the coefficients of {T}. 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.

About these ads

Comments»

1. hilbertthm90 - August 11, 2009

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.

2. Akhil Mathew - August 12, 2009

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.

3. Home Page - October 11, 2013

Hi, the whole thing is going sound here and ofcourse every one
is sharing facts, that’s actually good, keep up writing.

4. sethsnap.com - April 9, 2014

Do you mind if I quote a couple of your articles as
long as I provide credit and sources back to your site?
My blog is in the very same area of interest as yours
and my visitors would certainly benefit from a lot of the information you provide here.
Please let me know if this alright with you. Many thanks!

5. Nichole - May 19, 2014

Hey there outstanding website! Does running a blog such as this require a lot of work?
I have absolutely no understanding of computer programming but I
had been hoping to start my own blog in the near future.
Anyway, if you have any ideas or tips for new blog owners please share.

I know this is off subject however I simply needed to ask.
Thanks a lot!

6. Dennis - June 15, 2014

For both comments immediately above, you might want to contact Akhil directly. He has a new blog that you can find by just googling his name and wordpress. I would like to resurrect this blog, but it’s been dead and I think the rest of us don’t learn as well by blogging as he does.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 44 other followers

%d bloggers like this: