Let $R$ be a ring, and let $M$ be a simple $R$-Module, meaning that it only has the trivial submodules {0} and $M$. Show that there's a maximal ideal $I \subset R$ so that $M \cong R/I$.
Thanks in advance! I thought that if $I$ is a maximal ideal, then $R/I$ must be a field and therefore only has the trivial ideals, which makes it a simple module. But I'm stuck at that point.
Your statement about $I$ maximal is true, and so we see that $R/I$ is simple as an $R$-module.
Also, note that if $J$ is any non-maximal ideal, say $J\subset N\subset R$. Then $N/J$ is a submodule of $R/J$, so $R/J$ is not simple.
Let $M$ be a simple $R$-module. Let $x\in M$ such that there exists $r\in R$ such that $rx \neq 0$. This exists since $M$ is not trivial as an $R$-module.
Since $0\neq Rx\subset M$ is a submodule, then simplicity forces $M = Rx$. Let $I =\{ r\in R : rx = 0\}$ be the annihilator of $x$. Then $Rx \cong R/I$ as $R$-modules.
Thus, $M\cong R/I$, and since $M$ is simple, so is $R/I$, hence $I$ is maximal.