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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.

Iif

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)?

2 added 47 characters in body; edited body

Let p be a prime, and let R(p) be the residue field at p. If R -> R(p) is not a surjection, then then R(p) is an R module whose endomorphism ring is R(p), but such that the image of R is a proper submodule.

It feels like this is the main way things can go wrong, though, so I might conjecture that if

Iif the map R -> 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 is an integral domain, the map R/p -> R(p) should be a surjection only if R/p was already a field; that is, if p was maximal. Therefore, R has Krull dimension 0. Is this enough to imply that the ring was semisimple (aside from some finite-generation concerns)?

1

Let p be a prime, and let R(p) be the residue field at p. If R -> R(p) is not a surjection, then then R(p) is an R module whose endomorphism ring is R(p), but such that the image of R is a proper submodule.

It feels like this is the main way things can go wrong, though, so I might conjecture that if the map R -> 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 is an integral domain, the map R/p -> R(p) should be a surjection only if R/p was already a field; that is, if p was maximal.