Suppose $R$ is a PID, $p ∈ R$ is a prime element, and $s ≥ 1$. Prove the $R$-module $R/p^s$ has no proper submodule which is a summand.

Question

Suppose $R$ is a PID, $p ∈ R$ is a prime element, and $s ≥ 1$. Prove the $R$-module $R/p^s$ has no proper submodule which is a summand.

I would like to know how to prove this but I really do not know where to start. It makes intuitive sense to me because of the Theorem:

"Suppose $R$ is a PID and $d$ is a non-zero non-unit element of $R$. Assume $d = p_1 ^{s_1} p_2^{s_2}· · · p^{s_t}_t$ is the prime factorization of $d$. Then the natural map $R/d \rightarrow R/p^{s_1}_ 1 ⊕ · · · ⊕ R/p^{s_t}_ t$ is an isomorphism of $R$-modules"

However, I am not sure theorem would help in the proof.

Thanks.

Answer 1

There is a unique maximal submodule $M$ of $R/p^s$. Any two submodules of $R/p^s$ are contained therein and cannot add to $R/p^s$.