Let $R$ be a simple ring having a minimal left ideal $L$. Then every simple $R$-module is isomorphic to $L$.

by Jak   Last Updated January 14, 2018 13:20 PM

Lemma : Let $R$ be a simple ring having aminimal left ideal $L$. Then every simple $R$-module is isomorphic to $L$.

Proof : Let $M$ be a simple R-module. By simplicity of $R$ we have $LR = R$. Hence $LM \supseteq L(RM) = (LR)M = RM \neq 0.$ Choose $m \in M$ such that $Lm \neq 0$. Since $Lm$ is a submodule of $M$ it follows that $Lm = M$. The map $\varphi : L \longrightarrow M$ given by $ϕ(l) = \varphi m$ is therefore a surjective $R$-module homomorphism. Its kernel is a left ideal of $R$ properly contained in $L$, so it can only be $0$. Therefore $\varphi$ is an isomorphism.

In the first line, why can`t we say $LR = 0$? surjective

We know $ M = I m $ so $ \varphi $ is , why $\varphi$ is bijective ($ker \varphi = 0 $) ?



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