# Modules of finite support

I'm reading Dwyer and Fried's paper "Homology of free abelian covers, I". In it, they make the following claim, which I'm having trouble verifying.

Let $F$ be a field and $A = F[x_1^{\pm 1},\ldots,x_{\beta}^{\pm 1}]$ be a ring of Laurent polynomials over $F$. Let $M$ be a finitely-generated module over $A$. Recall that the support of $M$ is the set of all prime ideals $p$ of $A$ such that $M_p \neq 0$, or equivalently which satisfy $\text{ann}(M) \subset p$. The claim that I'm having trouble verifying is that $M$ is finite-dimensional over $F$ if and only if the support of $M$ consists of finitely many prime ideals.

Thanks for any help!

If $M$ is finite-dimensional, the ring $A/\mathrm{ann}(M)$ is as well, so it is Artinian, hence it has only finitely many prime ideals, and we have $\mathrm{supp}(M)=\mathrm{Spec}(A/\mathrm{ann}(M))$.

Conversely, suppose the support of $M$ is finite. For any $\mathfrak p\in\mathrm{supp}(M)$, the subset $\mathrm{Spec}(A/\mathfrak p)\subseteq\mathrm{supp}(M)$ is finite, but since $A/\mathfrak p$ is finitely generated over $F$, this implies that $\mathfrak p$ is maximal. $M$ has a finite filtration $M=M^0\supseteq M^1\supseteq\dots$ such that each $M^i/M^{i+1}$ is isomorphic to $A/\mathfrak p$ for some $\mathfrak p\in\mathrm{supp}(A)$, and each $A/\mathfrak p$ is finite-dimensional since $\mathfrak p$ is maximal, so we conclude that $M$ itself is finite-dimensional.

• Can you please give an argument/reference for the existence of that finite filtration. (In Matsumura, Commutative Algebra it's (7.E) Theorem 10, but there M is required to be finitely generated). – Ralph Oct 30 '11 at 9:29
• $M$ is assumed to be finitely generated in the question. – user2035 Oct 30 '11 at 9:55

EDIT I've just realized that I totally misread the question, so here is an answer to what you actually asked. :)

It's a rather standard fact that if $M$ is a finitely generated module over a noetherian ring, $A$ then $\mathrm{Supp}M=V(\mathrm{ann} M)$ is a closed subset of $\mathrm{Spec} A$. The reason for this is that both being in the support of $M$ or containing $\mathrm{ann} M$ is equivalent to containing the annihilator of all of the (finitely many) generators.

This proves that if $M$ is finitely generated then there are only finitely many minimal primes in its support. In other words, $$\mathrm{Supp} M = \bigcup_{i=1}^r V(P_i).$$ for an appropriate set of primes.

So, the support consisting of finitely many primes is equivalent to all minimal primes being maximal. Since $M$ is finitely generated and the residue fields at maximal ideals are finitely generated field extensions of $F$, this is equivalent to $M$ being finite dimensional over $F$.