Let $R$ be an integral domain which is not a field . Let $I$ be a non-zero ideal of $R$ such that for every proper ideal $J$ of $R$ containing $I$ , $J/I$ is a free $R$-submodule of $R/I$ . Then is $R/I$ a simple $R$-module ?
2026-03-29 08:18:21.1774772301
On modules of the form $R/I$ with every proper submodule free
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For any $i\in I$ and $x\in J/I$, $ix=0$. So if $J/I$ is a free $R$-module, we must have either $I=0$ or $J/I=0$. Since we are assuming $I\neq 0$, this means any proper ideal $J$ containing $I$ is equal to $I$. That is, $I$ is a maximal ideal so $R/I$ is a simple module.