# 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.

-
Consider the action of $x$ on a set of generators of $M$ over $\mathbb{Q}$ and write them as linear combinations of the generators. Form the matrix equation, then the characteristic polynomial will be the what you want. – Ehsan M. Kermani Jul 30 '12 at 20:47
In your attempt: before the second sentence, “Since $M$ is also finitely generated as a $\mathbb{Q}[x]$-module,” For the finish, calculate the vector space dimension of the two sides of your $\simeq$ and conclude there are very few solutions (just one by my count). – Jack Schmidt Jul 30 '12 at 20:55
Is there anything special about $\Bbb{Q}$ in this argument? It seems that using $\Bbb{R}$ would work as well.... – user38268 Jul 31 '12 at 0:31
@JackSchmidt Dear Jack, thanks for pointing that out. Could you elaborate on your suggestion? I can see that $m = n$, but how do I show then that this is impossible? Can't, for instance, $\mathbb{Q}$ act in a weird way on $\mathbb{Q}[x]$ such that they are isomorphic vector spaces? – student Jul 31 '12 at 17:50
@student: The dimension of the right hand side may be confusing: $$\dim\left({\mathbb{Q}[x]}^m\right) = \begin{cases} 0 & \text{ if } m = 0 \\ \infty & \text{ if } m > 0 \end{cases}$$ – Jack Schmidt Jul 31 '12 at 17:53

$m, xm, x^2m,...$ 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 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 $m\in M$.

-
Dear Paul, could you explain a little bit what is the map you're talking about? I can't see any natural map from $\mathbb{Q}[x]$ to $\mathbb{Q}$-linear endomorphisms of $M$, unless I'm given some fixed $T : M \rightarrow M$, and then interpret each $p(x) \in \mathbb{Q}[x]$ as $p(T)$ formally. – student Jul 31 '12 at 18:27

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.

-
Dear Matt, thanks for elaborating this point. I just have one question, if you allow me, which is the same objection to Jack's comments: in this reasoning, it seems we're assuming that $\mathbb{Q}$ "acts" on elements of $\mathbb{Q}[x]$ in the natural way; is this necessarily the case? Couldn't $\mathbb{Q}$ act on elements of $\mathbb{Q}[x]$ in some different way, thus enabling the dimension of $\mathbb{Q}[x]$ to be finite? – student Jul 31 '12 at 18:32
@student: Dear student, The embeding $\mathbb Q[x] \subset M$ is an inclusion of $\mathbb Q[x]$-modules, so you are giving your self that $\mathbb Q[x]$ acts on itself in the natural way (by multiplication of polynomials). In particular the $\mathbb Q$-action is given as the natural one. Regards, – Matt E Jul 31 '12 at 19:05