Classification of indecomposable modules over a given ring Let $K$ be a field, $x$ an indeterminate and $n$ a positive integer. 
How can one classify all the indecomposable modules up to isomorphism over the ring $R=K[x]/(x^n)$ ?
 A: As pointed out by user26857, we know by the structure theorem for finitely generated modules over a principal ideal domain that any finitely generated indecomposable module over $R$ is isomorphic to $K[x]/(x^i)$ for some $i\in \{1,\ldots, n\}$.
By using the axiom of choice, we can prove that any indecomposable $R$-module is finitely generated, thus proving that the above classify all indecomposable modules.
Let $M$ be an indecomposable $R$-module.  It is useful to note that all elements of $R$ have the form $\lambda x^i$, with $\lambda$ invertible.  The annihilator of $M$ in $R$ has the form $(x^m)$ for some $m\in\{1, \ldots, n\}$.  Let $z\in M$ be such that $x^{m-1}z \neq 0$.  Let us prove that $z$ generates $M$.
Let $X$ be the set of submodules of $M$ which have trivial intersection with $Rz$.  Then any chain in $X$ has an upper bound, namely the union of the submodules of that chain.  By Zorn's lemma, $X$ admits a maximal element $V$.
Let us now prove that $M=Rz\oplus V$.  Assume the contrary, and let $w\in M\setminus (Rz\oplus V)$.
If $Rw\cap(Rz\oplus V)$ is trivial, then $(Rw+V)\cap Rz$ is trivial as well, contradicting the maximality of $V$.
Thus $Rw\cap(Rz\oplus V)$ is non-trivial; let $p$ be the smallest integer such that there exists a non-zero element $x^pw = \lambda x^qz+v$, with $\lambda\in R$ invertible and $v\in V$.  Note that $p\leq q$, since $0=x^{m-p}x^pw=\lambda x^{m-p+q}z + x^{m-p}v = 0$ implies that $x^{m-p+q}z=0$, which in turn implies that $m-p+q \geq m$.
Consider the element $w-\lambda x^{q-p}z$.  The claim is that $R(w-\lambda x^{q-p}z)+V$ has trivial intersection with $Rz$.  Indeed, let $x^t(w-\lambda x^{q-p}z) + u = \mu x^s z$ be an element of this intersection, with $u\in V$ and $\mu\in R$ invertible.  Then $x^tw \in Rz\oplus V$, so $t\geq p$ by minimality of $p$.  Thus $x^t(w-\lambda x^{q-p}z) + u = \mu x^s z$ is both an element of $V$ and of $Rz$, and so it is zero.
Therefore $R(w-\lambda x^{q-p}z)+V$ has trivial intersection with $Rz$, and by maximality of $V$, we have that $w-\lambda x^{q-p}z \in V$.  Hence $w\in Rz\oplus V$, contradicting the assumption that $w\in M \setminus (Rz\oplus V)$.
Hence $M= Rz\oplus V$, and since $M$ is indecomposable, $V=0$.  We have proved that any indecomposable module over $R$ is generated by one element, and thus that the list of indecomposable modules at the top of this post is the complete list of indecomposable $R$-modules.
