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.

$m, xm, x^2m,...$ 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 $p_m(x)m=0$
