I'm really struggling with this excercise although it does not seem that hard. We are given a PID $R$ and a prime element $p \in R$. $M$ is an $R$-module and a short exact sequence is given by $0 \rightarrow R/(p) \rightarrow M \rightarrow R/(p) \rightarrow 0$. We need to show that either $M \cong R/(p) \oplus R/(p)$ or $M \cong R/(p^2)$.
Now I thought of using the fact that $M$ can be written as a direct sum of the form $M \cong R^{\rho} \oplus R/(p_1^{a_1})\oplus \dots \oplus R/(p_k^{a_k})$ where $p_i$ are prime and $a_i$ are positive integers. Also $\rho$ is unique. Now I want to prove that the only $\rho$ that could let $M$ fit in such an exact sequence is zero and next that only $p$ can be such prime element.
The second part "only $p$ itself can come in" seems the most doable but I cannot see how to start...
The given information is not sufficient to conclude. One needs to know the maps in the short exact sequence.
Case 1 corresponds to $$ 0 \to R/(p) \oplus 0 \to R/(p) \oplus R/(p) \to 0 \oplus R/p \to 0,$$ and Case 2 corresonds to $$ 0 \to (p)/(p^2) \to R/(p^2) \to R/(p) \to 0 $$ with the isomorphism $(p)/(p^2) \cong R/(p)$.