Why a particular module over a DVR is cyclic if it's quotient and kernel are cyclic?

Question

I'm looking at this lemma below. Here $A$ is a DVR with uniformizer $\pi$.

I'm not sure how to justify the underlined step. Why does $M_n$ have to be cyclic if $M_1$ is cyclic?

Answer 1

By the classification of finitely generated $A$-modules, $M_n$ is isomorphic to $\bigoplus_{i=1}^m A/(\pi^{k_i})$ for some $m$ and some $k_1,\dots,k_m\geq 1$. Now notice that the kernel of $\pi$ on $\bigoplus_{i=1}^m A/(\pi^{k_i})$ is just $\bigoplus_{i=1}^m (\pi^{k_i-1})/(\pi^{k_i})\cong \bigoplus_{i=1}^m A/(\pi)$, and in particular it has $q^m$ elements. But the kernel of $\pi$ on $M_n$ is just $M_1$ which by assumption has $q$ elements. Thus we must have $m=1$, so that $M_n\cong A/(\pi^{k_1})$ is cyclic.

(As a side note, the use of finite cardinalities here, which is only valid if you know that $A/(\pi)$ is finite, can be avoided by talking about the dimension of $A/(\pi)$-vector spaces, or more generally the length of $A$-modules. So, this result is actually valid over any DVR.)