Let, $p \in R$(which is a Principal Ideal Domain), p is prime, $s > 0$. Show that $R \setminus p^{s}$ has no proper submodule which is a summand.
I have been told that there is a unique maximal submodule $M$ of $R\setminus p^{s}$. Any two submodules of $R\setminus p^{s}$ are contained therein and cannot add to $R\setminus p^{s}$. But how do I prove this claim? Any hints/suggestions?
Since $R$ is a principal ideal domain, the nonzero prime ideals are maximal, so the prime ideal $P=(p)$ is a maximal ideal.
In any ring $R$ with maximal ideal $P$, $R/P^n$ has a unique maximal ideal $P/P^n$. This is easy to see: if $M/P^n$ is any other maximal ideal, by primeness of $M$ you necessarily have $P\subseteq M$, but by maximality of $P$ you would then also have $P=M$.
Local rings only have trivial summands. Just think: if it had $A\oplus B=R$ where both $A,B$ were nonzero submodules, then they would both have to be contained in a maximal ideal, but there is only one, call it $M$. Then $A+B=R\subset M$, a contradiction.