Finitely generated module over $\mathbb{Q}$. Consider the following problem.
Problem: Suppose $M$ is a module over $\mathbb{Q}[x]$ such that $M$ is finitely generated over $\mathbb{Q}$. Prove that there is a non-zero polynomial $p(x) \in \mathbb{Q}[x]$ and a non-zero $m \in M$ such that $p(x)*m = 0$.
My attempt: Since $M$ is finitely generated over a field, $M \simeq \mathbb{Q}^n$ as a $\mathbb{Q}$-module. If $M$ is torsion-free over $\mathbb{Q}[x]$ too, then $M \simeq \mathbb{Q}[x]^m$ as $\mathbb{Q}[x]$-modules. Now restricting the action to $\mathbb{Q}$ again, we get $\mathbb{Q}^n \simeq \mathbb{Q}[x]^m$ as $\mathbb{Q}$-modules. At this point, I feel this is absurd, but can't quite explain why.
My questions: Is the above reasoning correct? How do I finish the argument? If not, what would be the right way to prove this?
Thanks.
 A: $m, xm, x^2m,...$ cannot all be linearly independent over $\mathbb{Q}$.  hence for every $m\in M$ there is a $p_m(x)\in\mathbb{Q}[x]$ with $p_m(x)m=0$
A: A variation on your attempt and other comments: the map $\mathbb Q[x]$ to $\mathbb Q$-linear endomorphisms of $M$ must have a non-trivial kernel, since $M$ is finite-dimensional as $\mathbb Q$-module and $\mathbb Q[x]$ is not. The kernel is thus a non-zero ideal in the PID $\mathbb Q[x]$, generated by the minimal polynomial of $x$ acting on $M$.
(Although it is interesting to understand Cayley-Hamilton and the characteristic polynomial, I think it is also useful to understand the minimal poly.)
Edit: In response to the comment, the map I'm thinking of just sends $x$ to the map $m\to x\cdot m$ for $m\in M$. 
A: You can build upon what you already wrote to finish:
Although it's not stated explicitly, you must be requiring that $M \neq 0,$
otherwise it contains no non-zero elements $m$ at all.
But if $M \neq 0$ and $M  = \mathbb Q[x]^m$, then $m \geq 1$,
and so $M$ contains a coyp of $\mathbb Q[x]$.  Now thinking, as you suggest, over $\mathbb Q$, you have an inclusion of $\mathbb Q$-vector spaces $\mathbb Q[x] \subset \mathbb Q^n$.  Is this possible?
[This is a minor variation of the reasoning suggested by Jack Schmidt in his comments.]

Incidentally, it's not only true that $M$ is not torsion-free; in fact if $M$ is finite-dimensional over $\mathbb Q$ it will be torsion over $\mathbb Q[x]$.  The same argument as above proves this (if $M$ is not torsion, it contains a copy
of $\mathbb Q[x]$), as do the arguments in the other answers.  
