Let $R=\mathbb{C}[X]$ and $M_k=\mathbb C[X]/\langle X^k\rangle$ $(k\ge 0)$. This is an $R$-module.
So far I have shown that there are for $0\le a\le b$ well-defined $R$-module homomorphisms $$u:M_a\rightarrow M_b\quad\text{and}\quad v:M_b\rightarrow M_{b-a}$$ with $u(X^n+\langle X^a\rangle)=X^{n+b-a}+\langle X^b\rangle$ and $v(X^n+\langle X^b\rangle)=X^n+\langle X^{b-a}\rangle$.
In another question it was proved, that the following short sequence is exact.
Now the exercise is:
Show that for $0< a< b$ $$0\rightarrow M_a\overset{u}{\rightarrow}M_b\overset{v}{\rightarrow}M_{b-a}\rightarrow 0$$ does not split.
I tried using the splitting lemma: A function $t:M_b\rightarrow M_a$ with $t\circ u=\text{id}_{M_a}$ should not exist.
Best Answer
A module $M$ is indecomposable when, if $A$ and $B$ are submodules and $M=A\oplus B$, then either $A=\{0\}$ or $B=\{0\}$.
For $k>0$, the module $\mathbb{C}[X]/\langle X^k\rangle$ is indecomposable, because its (simple) submodule $S_k=\langle X^{k-1}\rangle/\langle X^k\rangle$ is contained in every non trivial submodule.
A slightly different approach. Suppose the map $t$ exists and $a>0$. Since $u$ is injective, it maps the simple submodule $S_a=\langle X^{a-1}\rangle/\langle X^a\rangle$ of $M_a$ onto the unique simple submodule $S_b=\langle X^{b-1}\rangle/\langle X^b\rangle$ of $M_b$.
Since $t\circ u$ is the identity, we deduce that $S_b$ is not contained in the kernel of $t$. Thus $t$ is injective.