Length of a module over a PID

Question

In my lecture notes there is an exercise that I wanted to solve but wasn't able to so far. It goes as follows:

Let $A$ be a PID and $p \in A$ be irreducible (hence prime). Show that $A/(p^n)$ has finite length as $A$-module for all $n \in \mathbb{N}$ and $l(A/(p^n)) = n$.

So I was able to show that $A/(p^n)$ is Noetherian and Artinian, thus has finite length and (by Jordan-Hölder) it suffices to find any composition series of length $n$. However, I'm stuck here. I tried induction over $n$ (since it is trivial that $l(A/(p))=1$ but didn't get far and I think it might be easier without an induction argument. But how can I find a composition series? How do submodules and their quotients in that ring even look like?

Answer 1

Hint:

The chain of submodules of $A/(p^n)$: $$\{0\}\subset (p^{n-1})/(p^n)\subset\dots\subset (p^{n-k+1})/(p^{n})\subset (p^{n-k })/(p^{n})\subset\dots\subset (p)/(p^n)\subset A/(p^n)$$ is a composition series, by the third isomorphism theorem.

Answer 2

Maybe I made a mistake but I was able to prove this using this idea: Consider

$$0 \rightarrow R/pR \rightarrow R/p^nR \rightarrow R/p^{n-1}R \rightarrow 0$$

Find suitable morphisms such that this sequence becomes exact, and then one can apply the length formula and get $l(R/p^nR) = l(R/pR) + l(R/p^{n-1}R)$, and if $l(R/pR) = 1$ was shown the claim follows by induction.