Classification of $k[x]/(x^2)$-modules

Question

Let $k$ be a field. How to prove that every finitely generated $k[x]/(x^2)$-module is a direct sum of copies of $k[x]/(x^2)$ and $k[x]/(x)$?

This follows from a more general result by Nakayama on modules over uniserial Artinian rings, but I wonder if there is a direct proof which does not use much theory. Actually the theorem applies to all modules, but I am only interested in finitely generated modules here. The tag "representation theory" comes from the interpretation $k[x]/(x^2) \cong k[\Sigma_2]$ when $\mathrm{char}(k)=2$. Thus, the goal is to classify all representations of $\Sigma_2$ in characteristic $2$.

Edit: Sorry, this was a silly question. Since $k[x]$ is a PID, finitely generated $k[x]$-modules have a well-known classification. One just has to look at those killed by $x^2$, and we are done.

Answer 1

I am not sure what you are allowed to use. This is a simple consequence of the fact that $A=k[x]/(x^2)$ is an injective $A$-module. If $M$ is any finitely generated $A$-module and if it has an element $m$ with $\mathrm{Ann}\, m=0$, then we get an inclusion $A\to M$, which splits. Thus we are reduced to the case when all elements in $M$ have non-zero annihilator. But, then $xM=0$ and the rest is clear.

Answer 2

Here's an elementary linear algebra proof.

Let $M$ be a $k[x]/(x^2)$-module, and let $T:M\to M$ be multiplication by $x$.

Let $U=T(M)$. Since $x^2=0$, $T(M)\leq\ker(T)$, and so we can choose a subspace $V$ of $M$ such that $\ker(T)=U\oplus V$.

Then we can choose $W$ so that $M=U\oplus V\oplus W$.

Then $T$ induces an isomorphism from $W$ to $U$. Let $\{w_i\}$ be a basis of $W$, so $\{w_ix\}$ is a basis of $U$. Let $\{v_j\}$ be a basis of $V$.

Then $M$ is the direct sum of submodules $V_j=\langle v_j\rangle$, which are isomorphic to $k[x]/(x)$, and $L_i=\langle w_i,w_ix\rangle$, which are isomorphic to $k[x]/(x^2)$.