I just started studying modules and came across this problem in the notes: “Let R be a nonzero ring with unity and M a nontrivial initial left R module. Show M is a simple left R-module iff there exists a maximal left ideal $I_M$ of R such that M and $_RR/I_M$ are isomorphic as left R-modules.”
I believe I have the forward direction shown. Where I am struggling is the reverse direction. My attempt so far: Suppose such an $I_M$ exists. Let N be a nonzero sub module of M. By the second Isomorphism Theorem, ($N+I_M)$/$I_M$ is a nonzero sub module of $_RR$/$I_M$. Since $_RR$/$I_M$ is simple, we must have $_RR$/$I_M$= ($N+I_M)$/$I_M$. Thus $N+I_M=_RR$ and since $I_M$ is maximal, either $N=M$ or $N=${0}, thus M is simple.
Does this work? I’m particularly concerned about the part using the 2nd Iso. Theorem, as I’m not confident I used it properly. Any help or advice would be appreciated!
I think you are trying to prove that if $I$ is a maximal left ideal, then $R/I$ has no nontrivial left submodules.
Really the relevant proposition for you is the correspondence theorem , which says that the submodules of $R/I$ correspond to the submodules of $R$ containing $I$.
"$N+I_M$" doesn't even make sense because $I_M$ and $N$ are subsets of different sets. What you can do, though, is to say "let $f:R/I\to M$ be the isomorphism. Then $f^{-1}(N)$ is a nonzero submodule of $R/I$. As such, it has the form $I'/I$ where $I'$ is a left ideal of $R$ containing $I$. In fact since it's nonzero, it properly contains $I$. But then... " can you take it from here?