Elementary lemma on modules over PID

Question

On page 36 of these notes on Class Field Theory, Lemma 3.3, $M_n$ is the kernel of the map $ \pi^n: M \to M $, where $ M $ is a $ \mathcal{O}_K $-module ($ K $ is a local field), and $ \pi \in K $ is a uniformizer. The proof is given by induction, and after noting that $ M_n $ must have $ q^n $ elements, the argument is that $ M_n $ must be cyclic, otherwise $ M_1 $ is not cyclic. I'm not sure how he arrives at this conclusion. Why is it that $ M_n $ not cyclic forces $ M_1 $ to be that as well? For example, why can't $ M_2 $ be $ A/(\pi) \oplus A/(\pi)$. This is probably very basic but I can't figure it out. Thanks for your help.

Answer 1

Since $\pi$ is surjective, there is a sequence $(e_1,e_2,\ldots)$ where $e_1$ generates $M_1$ and $\pi e_n = e_{n-1}$ for $n>1$. Obviously $e_n \in M_n$, but I claim that $e_n$ generates $M_n$.

So if $n>1$, suppose $e_{n-1}$ generates $M_{n-1}$. Given $x \in M_n$, since $\pi x \in M_{n-1}$ we can write $\pi x = a e_{n-1}$ where $a \in A$, so $\pi(x - ae_n) = 0$. Now $e_1$ generates $\ker \pi$, so there is $b \in A$ such that $x - ae_n = be_1$. Then $$x = (a + b\pi^{n-1})e_n$$ so $e_n$ generates $M_n$.