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 $v(X^n+\langle X^b\rangle)=X^n+\langle X^{b-a}\rangle$.
Now the exercise is:
Show that for $0\le a\le b$ $$0\rightarrow M_a\overset{u}{\rightarrow}M_b\overset{v}{\rightarrow}M_{b-a}\rightarrow 0$$ is a short exact sequence.
I need help with all of the three things which I have to show, ie.
- $u$ is injective
- $v$ is surjective
- $\text{im}(u)=\text{ker}(v)$
For example:
$$X^{n+b-a}\in\langle X^b\rangle\iff n-a\ge0\iff n\ge a\iff \;X^n\in\langle X^a\rangle$$
and you already have injectivity of $\;u\;$ , and as surjectivity of $\;v\;$ is almost completely trivial, you're left only with Im$\,u=\ker v\;$, but also $$\left(X^{n+b-a}+\langle X^b\rangle\right):= X^{n+b-a}+\langle X^{b-a}\rangle=0 \;\text{ (in the quotient}\;\;M_{b-a})$$
and you have already Im $\,u\subset\ker v\;$ . Now you try what is left.