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 {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.