I had a post a few days back on why simple representations of algebras over a field {k} which are finitely generated over their centers are always finite-dimensional, where I covered some of the basic ideas, without actually finishing the proof; that is the purpose of this post.

So, let’s review the notation: {k} is our ground field, which we no longer assume algebraically closed (thanks to a comment in the previous post), {A} is a {k}-algebra, {Z} its center. We assume {Z} is a finitely generated ring over {k}, so in particular Noetherian: each ideal of {Z} is finitely generated.

Theorem 1 (Dixmier, Quillen) If {A} is a finite {Z}-module, then any simple {A}-module is a finite-dimensional {k}-vector space.


We know from the previous post that a simple representation of {A} is just {A/M} for {M \subset A} a maximal left ideal; then {M \cap Z \subset Z} is a left {Z}-ideal, hence a two-sided {Z}-ideal. Although {M} was maximal, we don’t necessarily have {M \cap Z} maximal in {Z}. So choose {I \subset Z} to be maximal and containing {M \cap Z}.

Claim 1 {I A \subset M}.

First of all, {IA} is a left ideal in {A}, because {I} consists of central elements. Now, if {IA \not\subset M}, we would have by maximality

\displaystyle  M + IA = A

or

\displaystyle   A/M \otimes_{Z} Z/I = A/(M+IA) = 0. \ \ \ \ \ (1)

In other words, we are viewing {A/M} as a {Z}-module.

Now we use:

Lemma 2 (Nakayama) Let {R} be a commutative ring, {J \subset R} be an ideal, and {N} a finitely generated {R}-module such that

\displaystyle  N / JN = 0;

then there exists {x \in 1 + N} such that {xN=0}.

The proof uses the Cayley-Hamilton theorem, and is given here.  It’s essentially equivalent to other versions of Nakayama’s lemma.

So now, back to the claim. In (1), by Nakayama, we would get an element {x \in 1 + I \subset Z} such that {x(A/M) = 0}, i.e. {xA \in M}, or {x \in M}. But then {x \in M \cap Z = I}, contradiction. We’ve thus proved our claim.

Now, onto the theorem itself: since {IA \subset M}, we can consider

\displaystyle  A/M

as both a {Z}-module and a {Z/I}-module. As a {Z}-module it is finitely generated by assumption, so as a {Z/I}-module it is finitely generated.

But we can now invoke the following:

Theorem 3 (Generalized Nullstellensatz) Let {k} be a field, {R} a finitely generated commutative ring over {k}, and {J \subset R} a maximal ideal. Then {R/J} is a finite extension of {k}.

When {k} is algebraically closed, any finite extension of {k} is just {k} itself, so this becomes a result from our prevoius post.

So, we see that {A/M} is a finite-dimensional {L := Z/I}-vector space for {L} a finite extension of {k}. Thus {A/M} is a finite-dimensional {k}-vector space.

Advertisements