Integrality, invariant theory for finite groups, and more tools for Noetherian testing August 11, 2009
Posted by Akhil Mathew in algebra, commutative algebra.
Tags: algebra, integrality, invariant theory, Noetherian rings
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]}](http://s0.wp.com/latex.php?latex=%7BA+%3A%3D+k%5BX_1%2C+X_2%2C+%5Cdots%5D%7D&bg=ffffff&fg=000000&s=0 "{A := k[X_1, X_2, \dots]}")
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 ![{A=A_0[x_1, \dots, x_n]}](http://s0.wp.com/latex.php?latex=%7BA%3DA_0%5Bx_1%2C+%5Cdots%2C+x_n%5D%7D&bg=ffffff&fg=000000&s=0 "{A=A_0[x_1, \dots, x_n]}")
. Then, as is well-known, to say that is integral over is to say that each 
satisfies an integral equation 
with ![{P_i \in B[X]}](http://s0.wp.com/latex.php?latex=%7BP_i+%5Cin+B%5BX%5D%7D&bg=ffffff&fg=000000&s=0 "{P_i \in B[X]}")
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 ![{S = R[a]}](http://s0.wp.com/latex.php?latex=%7BS+%3D+R%5Ba%5D%7D&bg=ffffff&fg=000000&s=0 "{S = R[a]}")
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 ![{k[x_1, \dots, x_n]}](http://s0.wp.com/latex.php?latex=%7Bk%5Bx_1%2C+%5Cdots%2C+x_n%5D%7D&bg=ffffff&fg=000000&s=0 "{k[x_1, \dots, x_n]}")
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
![\displaystyle \prod_i (T - x_i) \in k[x_1, \dots, x_n, T].](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cprod_i+%28T+-+x_i%29+%5Cin+k%5Bx_1%2C+%5Cdots%2C+x_n%2C+T%5D.&bg=ffffff&fg=000000&s=0 "\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 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.
Like this:
Like Loading...
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.
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.