Artinian iff finite dimensional Let $F$ be a field and $M$ a finitely generated $F[x]$-module. Show that $M$ is Artinian iff the dimension of $M$ over $F$ is finite. 
Thoughts so far: For the backward direction, I assume for the contrapositive that $M$ is not Artinian, giving me a strictly decreasing chain of modules. Then I can form a infinite set of linearly independent elements in $M$ over $F$ by picking an element from each module in the chain that doesn't belong to the proceeding module in the chain. Does this reasoning make sense for this direction of double implication? 
For the other direction, I'm not sure how to begin. Any thoughts would be much appreciated. 
 A: Hint: Every submodule of M is a vector subspace, so a chain of submodules is in particular a chain of subspaces. This observation leads to a direct proof that a finite dimensional module is artinian.
Let us prove the converse. 


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*First, if u is an element of M, then there exists a polynomial in the ring such that pu=0, for otherwise $\{x^iF[x]u:i\geq0\}$ is an infinite descending chain of submodules of M all contained in F[x]u.

*From this, it follows that the module M is a torsion module. Since it is finitely generated, we can choose a finite sequence m_1, ..., m_k of elements in  which generates it. There are short exact sequences $$0\to (m_1,\dots,m_i)/(m_1,...,m_{i-1})\to M/(m_1,...,m_{i-1})\to M/(m_1,\dots,m_i)\to0$$ of artinian finitely generated torsion modules. An easy induction using these shows that to show that M is finite dimensional it is enough to show that all the modules $(m_1,\dots,m_i)/(m_1,...,m_{i-1})$ are finite dimensional, and for this more generally we can show that cyclic torsion modules are finite dimensional.

*This is clear, since all such modules are of the form $F[x]/(p)$ for a nonzero polynomial $p$.
A: One direction follows from the fact that a finite dimensional vector space over $F$ is artinian as $F$-module (chains of subspaces are finite).
Thus, an $F[x]$-module that's finite dimensional as $F$-vector space is artinian. This is true for modules $M$ over any $F$-algebra, because submodules are in particular subspaces.
The converse uses a particular property of $F[x]$ (it's not generally true that an artinian module over an $F$-algebra is necessarily finite dimensional).
Let $M$ be a finitely generated module over $F[x]$ and suppose $M$ is artinian. It's not restrictive to assume that $M$ is singly generated (why?), so let $mF[x]=M$. Then $F[x]/\!\operatorname{Ann}(m)\cong M$ is artinian.
In particular, $\operatorname{Ann}(m)$ is the ideal generated by nonzero polynomial (why?). Thus $F[x]/\!\operatorname{Ann}(m)$ is finite dimensional as vector space over $F$.
Notes


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*$\operatorname{Ann}(m)=\{f\in F[x]:mf=0\}$ is the annihilator of $m$.

*The structure theorem is not needed.
