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!



Modules of finite support
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.


