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This question is motivated by attempting to find the indecomposable modules over $R=\mathbb{K}[X]/(X^n)$. This can be done using the structure theorem assuming we know that the indecomposable modules are all finitely generated. This in fact is the case (see Classification of indecomposable modules over a given ring). The proof in this post uses the fact that given some indecomposable $R$-module $M$, there exists $m \in M$ such that $X^{i-1}m \neq 0$ where $Ann_R(M)=(X^i)$. It is then proven that $M \cong Rm$.

In looking for an alternative proof, I managed to prove that $R$ is quasi-Frobenius. This follows from the fact that the ideals of $R$ are of the form $([X^i])$ where $0 \leq i \leq n$. Now, given that $R$ is quasi-Frobenius, we know that every projective $R$-module is injective. In particular, $Rm$ is injective (where we have defined $m$ above). So we have reduced our original problem to constructing a short exact sequence of the following form. $$0 \to Rm \to M \xrightarrow{f} f(M) \to 0 $$

Does such a short exact sequence exist?

cofnmarol
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It's not necessarily the case that $Rm$ is injective (or projective) as an $R$-module. For example, take $M=R/(X)$.

However, you can make your proof work by considering $M$ as a module for $S=\mathbb{K}[X]/(X^i)$, where $\text{Ann}_R(M)=(X^i)$. If $m\in M$ is an element with $X^{i-1}m\neq0$, then there is a short exact sequence of $S$-modules $$0\to S\to M\stackrel{f}{\to}f(M)\to0,$$ where the map $S\to M$ is $s\mapsto sm$, and $S$ is quasi-Frobenius and so this sequence splits.

Jeremy Rickard
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