I need to show that if $S$ is a multiplicatively closed subset of a ring $A$, $M$ is a finitely generated $A$-module, and $S^{-1}M = 0$, then there exists a single element $s$ in $S$ so that $sM = 0$.
This is what I have done so far:
If $S^{-1}M = 0$, then every element $m/t$ of $S^{-1}M$ is equivalent to $0/1$, so for each $m$ in $M$ and $t$ in $S$, there exists some $u$ in $S$ so that $u(mt + 0*1) = u(mt) = um(t) = 0$. Since $S$ is multiplicatively closed, if for some $u$ and $t$ in $S$, $ut = 0$, then $0$ is in $S$, and clearly $0M = 0$. So suppose that this is not the case.
I'm having trouble right here, since it seems like I just have a u for each combination of m and t, rather than a single element s that annihilates all of M. I feel like I should be using the fact that $M$ is finitely generated, somehow applying this fact to a set of generators, but I'm not quite sure how.
You actually didn't use that $M$ is finitely generated.
If $x_1,\dots,x_n$ is a system of generators of $M$, then $x_i/1=0/1$ in $S^{-1}M$, so there exists $s_i\in S$ such that $s_ix_i=0$. Set $s=s_1\cdots s_n$ and note that $sx_i=0$ for all $i=1,\dots,n$. Then $sM=0$ since every element of $M$ is a linear combination of the generators.