Let $R$ be a ring with $1$. It's well known that if $M$ an indecomposable injective right $R$-module, then $M\cong E(R/\mathfrak{p})$ for some prime ideal $p\subset R$ where $E(R/\mathfrak{p})$ is the injective envelope of $R/\mathfrak{p}$. I know that if $M$ is an indecomposable injective module, then $M=E(A)$ for every nonzero submodule $A\subseteq M$. In particular, $M=E(xR)$ for every $0\neq x\in M$. But $xR\cong R/I$, as right $R$-modules, for some right ideal $I\subset R$. In fact, $I$ is the right annihilator of $x$. I wonder if we can take $x$ so that $I$ is a prime ideal.
Thanks in advance.
Note: I'm not assuming that the ring is commutative. However, is the statement true for commutative rings?!.