Elementary proof of a theorem of Köthe

Question

Let $K$ be a field, $X$ an indeterminate, $n$ a positive integer and $A$ the ring $K[X]/(X^n)$.

A theorem of Köthe in

Köthe G., Verallgemeinerte Abelsche Gruppen mit hyperkomplexem Operatorenring. Math. Z. 39, 31–44 (1935)

implies that

(1) any $A$-module is a direct sum of cyclic modules.

In this answer, Pierre-Guy Plamondon gives an elementary proof of the following corollary to (1):

(2) any indecomposable $A$-module is cyclic.

This prompts the question:

Is there an elementary proof of (1)?

EDIT. I think one can write down an elementary proof along the following lines. (Warning: what is below is a sketch with many gaps. I believe the gaps may be easily filled, but I may be wrong. I hope somebody will post an answer which either fill the gaps or use a better idea.)

We prove that any $A$-module is a direct sum of cyclic submodules by induction on $n\ge1$. The case $n=1$, being clear, we assume that $n\ge2$ and that the statement holds for $n-1$.

Let $V$ be a $K$-vector space and $x$ an endomorphism such that $x^n=0$. We regard $V$ as an $A$-module in the obvious way. Let $U$ be a $K$-linear subspace of $V$ such that $V=U\oplus\operatorname{Ker}x^{n-1}$.

(a) Prove that the natural map $A\otimes_KU\to V$ is injective and denote its image by $AU$. (In particular $AU$ is a free $A$-module of rank $\dim x^{n-1}V$.)

(b) Mimicking Pierre-Guy's argument, show that there is a sub-$A$-module $W$ of $V$ such that $V=AU\oplus W$.

(c) Prove $x^{n-1}W=0$.

(d) Conclude by using the induction hypothesis.

Let me insist on the fact that the point is not to convince ourselves that the statement is true. We know it is true! The point is rather to find a proof which would be as elementary and as complete as possible.

Answer 1

HINT: The proof of the existence of Jordan form for nilpotent operators on finite dimensional vector space works also for ifinite dimensional spaces. Let $T$ a linear on the vector space $V$ such that $T^n=0$. Consider the flag of subspaces: $$ \operatorname{Ker T}\cap \operatorname{Im} T^{n-1}\subset \operatorname{Ker T}\cap \operatorname{Im} T^{n-2} \subset\ldots \operatorname{Ker T} \cap \operatorname{Im} T \subset \operatorname{Ker T}$$

Take a basis of $\operatorname{Ker T}$ adapted to this flag $$ B_{n-1} \cup B_{n-2}\cup \ldots \cup B_0$$

Lift each $B_{n-k}$ basis $n-k$ times under $T$ to get $B_{ij}$, $n-1\ge i\ge j \ge 0$, and $T(B_{i,j+1})= B_{ij}$, $B_{i0} = B_i$. Now show that the union of the $B_{ij}$ is a basis of $V$. ( at this point it's useful to look at a Young diagram with cells filled with the $B_{ij}$ -- the Young diagram is of type $(1,2,\ldots, n)$). Once we have done that we see that the space $V$ is a direct sum of indécomposable cyclic $T$ modules. I leave the details to be filled in, one should work first with say $n=2$, then $n=3$, and then it becomes transparent.