Let p $p$ be a prime, and let R(p) $R(p)$ be the residue field at p. $p$. If $R -> R(p) \to R(p)$ is not a surjection, then then R(p) $R(p)$ is an R $R$ module whose endomorphism ring is R(p), $R(p)$, but such that the image of R $R$ is a proper submodule.
If the map $R -> R(p) \to R(p)$ is a surjection for all primes, then having a field as an endomorphism ring should imply that a module is simple. Because R/p $R/p$ is an integral domain, the map $R/p -> R(p) \to R(p)$ should be a surjection only if R/p $R/p$ was already a field; that is, if p $p$ was maximal. Therefore, R $R$ has Krull dimension 0. Is this enough to imply that the ring was semisimple (aside from some finite-generation concerns)?