jump to navigation

Why simple modules are often finite-dimensional, I July 19, 2009

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

Today I want to talk (partially) about a general fact, that first came up as a side remark in the context of my project, and which Dustin Clausen, David Speyer, and I worked out a few days ago.  It was a useful bit of algebra for me to think about.

Theorem 1 Let {A} be an associative algebra with identity over an algebraically closed field {k}; suppose the center {Z \subset A} is a finitely generated ring over {k}, and {A} is a finitely generated {Z}-module. Then: all simple {A}-modules are finite-dimensional {k}-vector spaces.

We’ll get to this after discussing a few other facts about rings, interesting in their own right.

Generalities

Recall an object {E} of an abelian category is simple if any subobject is either zero or isomorphic to {E}. In the category of (left) {R}-modules for a ring {R}, this means the only proper submodule is zero.

We prove a general fact:

Theorem 2 Let {R} be a ring (always with identity). Then any (nonzero) simple {R}-module is isomorphic to {R/M} for {M \subset R} a maximal left ideal.

Proof: A submodule of {R/M} would be of the form {L/M} for {M \subset L} and {L} a left ideal, by the isomorphism theorems. But {M} is maximal. So we get one direction.

For the other, a simple {R}-module {S} is generated by one element {v \in S}; just pick any nonzero {v}, and note that {0 \neq Rv \subset S}. So {S \simeq R/I} for some left ideal {I}, the kernel of the surjection {R \rightarrow S} given by {x \rightarrow xv}. If {I} isn’t maximal, it’s contained in a maximal left ideal {M}, and we have {0 \neq M/I \neq R/I}, so {R/I} isn’t simple. \Box

The Nullstellensatz

We want to consider the case {A=Z} of the initial fact. So we have a finitely generated commutative ring {A}, and we will to show that its simple modules are one-dimensional. In other words, for any maximal ideal {M}, we have {A/M \simeq k}.

Theorem 3 Hypotheses as above, we have {A/M \simeq k} for any maximal ideal {M}. In detail, if {A} is a finitely generated commutative ring over the algebraically closed field {k}, then {A/M \simeq k} for any maximal ideal {M}.

I will discuss a more concrete setting that may clarify it:

Since {A} is commutative and finitely generated, we can write {A = k[x_1, \dots, x_n]/I} for some maximal ideal {I}; let {f: k[x_1, \dots, x_n] \rightarrow A} be the reduction map, and let {N = f^{-1}(M)}; then we have, by the isomorphism theorems

\displaystyle   A/M \simeq k[x_1, \dots, x_n]/(N+I);\ \ \ \ \ (1)

this is a field, so {N+I} is actually maximal. So we only need to consider the right hand side, i.e. find the maximal ideals in {k[x_1, \dots, x_n]}. The following two results will tells us what they are:

Theorem 4 (Hilbert’s Basis Theorem) The ring {k[x_1, \dots, x_n]} is Noetherian, i.e. each ideal {J \subset x_1, \dots, x_n} is finitely generated: {J} can be written as {J = (P_1, \dots, P_r)} for some polynomials {P_1, \dots, P_r}.

This might actually be a useful topic for a future post, but for now, I’ll simply quote it without proof.

Theorem 5 (The Weak Nullstellensatz) Let {f_1, \dots, f_k} be polynomials in {n} variables {x_1, \dots, x_n}, over the fixed algebraically closed field {k}. Suppose {f_1, \dots, f_k} have no common zero. Then the ideal {(f_1, \dots, f_k)=(1)}, i.e. there are polynomials {g_1, \dots, g_k} such that

\displaystyle  \sum g_i f_i = 1.

So let’s see how Theorem 3 follows from these two results. Indeed, I claim that a maximal ideal {J} of {k[x_1, \dots, x_n]} is of the form {(x_1-a_1, \dots, x_n-a_n)}; then in (1) we will see {A/M \simeq k}. First of all, we have {J = (f_1, \dots, f_k)} for some polynomials {f_1, \dots, f_k} by the basis theorem. Next {f_1, \dots, f_k} must have a common zero, otherwise {J=(1)}. Let the common zero be {a_1, \dots, a_n}. Then each {f_j \subset (x_1-a_1, \dots, x_n-a_n)}. This proves the claim.

Actually proving the Nullstellensatz could make another interesting algebraic post. For now, I’ll recommend David Speyer’s interpretation of it, and Terence Tao’s elementary proof.

In the next post, I’ll apply what we’ve discussed so far to prove our aims.

About these ads

Comments»

1. Steven Sam - July 22, 2009

Assuming that what I’ve thought of is a correct proof for the rest of Theorem 1, I don’t think algebraic closure is necessary: in Theorem 3, we just need that A/M is a finite extension of k, which is certainly true in general.

2. Akhil Mathew - July 22, 2009

Thanks for the comment! Yes, this should be true, using the extended Nullstellensatz (or some general algebraic fact); I believe someone else indicated it to me, in fact, so I should probably restate Theorem 1 in that form when I finish this loose end (ideally soon).

3. Why simple modules are often finite-dimensional II « Delta Epsilons - July 22, 2009

[...] vector spaces, Nakayama's lemma, representation theory, simple modules trackback I had a post a few days back on why simple representations of algebras over a field which are finitely generated [...]


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 43 other followers

%d bloggers like this: