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Let $R$ be a unital ring and $M$ a left $R$-module. Let $\Gamma$ be the set of all maximal left ideal of $R$.

I want to prove $J(M)=\cap_{I\in \Gamma}IM$.

If $R$ is commutative, then $J(M)=\cap_{I\in \Gamma}IM$.

proof: For every maximal submodule $N$ of $M$(otherwise $J(M)=M$), there exists a maximal left ideal $I$ such that $M/N\cong R/I$. Thus $IM\subset N$. An then, we have $J(M)\supset \cap_{I\in \Gamma}IM$. For every $I\in \Gamma$, $R/I$ is a field. Consequently, $M/IM$ is a left semisimple $R/I$-module. So, $J(_{R/I}M/IM)=J(_{R}M/IM)=\{0\}$. Thus $J(M)\subset IM$.

But I don't know whether this result is still true when $R$ is non-commutative.

Mod.esty
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  • So to be clear, an answer to your question would be proving or disproving the noncommutative version? I think so after reading your last line, but then again it may be not apparent to some that you're asking a question straightforwardly. – rschwieb Jun 05 '22 at 23:39
  • @rschwieb Yse. I don't know whether the non-commutative version is true or false... – Mod.esty Jun 05 '22 at 23:50

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