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Let $M\neq \{0\}$ be a semi-simple left $R$ module .Prove that it contains a simple sub-module.

An $R-$ module $M$ is said to be semi-simple if every submodule of $M$ is a direct summand of M My solution

Since $M\neq \{0\}$; $\exists m\in M$ such that $m\neq 0$.Then I can consider the left $R-$ module $Rm$ ;By hypothesis it is a direct summand of $M$.Thus $M=N+Rm$ where $N$ is a sub module of $M$

How to proceed next?

Learnmore
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    Can you clarify that the definition of semisimple you're using is that every submodule is a direct summand? There are several equivalent definitions, one of which makes this a tautology (being a direct sum of simple modules). – Qiaochu Yuan Mar 21 '15 at 07:45
  • @QiaochuYuan I have edited it – Learnmore Mar 21 '15 at 08:06

3 Answers3

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Your definition is equivalent to "An $R$-module $M$ is semisimple if it can be written as a direct sum of family of simple modules" for example see here page 2 for the proof, in which your method is subsidised by the proof of the theorem.

Also what you want is a direct consequence of the above definition.

Math137
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First show that every submodule of a semisimple module is also semisimple.

Let $M$ be semisimple, $N$ a submodule of $M$ and $L$ a submodule of $N$. Since $M$ is semisimple, we have $N\oplus N'=M$ and also $L\oplus L'=M$. Consider then $L''=L'\cap N$. Then $L\cap N=L\cap L'\cap N=\{0\}$. If $x\in N$, we have $x=y+z$, with $y\in L$ and $z\in L'$. But $z=x-y$, so $z\in L'\cap N=L''$. Therefore $L\oplus L''=N$.

So, let $M\ne0$ be semisimple and take $x\ne0$, $x\in M$. Then $Rx$ is semisimple and it's sufficient to show that $Rx$ has a simple submodule.

Consider $I=\operatorname{Ann}_R(x)=\{r\in R:rx=0\}$. Then $Rx\cong R/I$ and $I\ne R$. Let $M$ be a maximal left ideal containing $I$; then $L=Mx$ is a direct summand of $Rx$ and $L/Mx\cong (R/I)/(M/I)\cong R/M$ is simple, so a complement of $L$ is a simple submodule of $Rx$.

egreg
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If $M\not=0$ is a semisimple $R-$module then there exists a simple $R-$module $N\leq M$

Let $x\in M-\{0\}$ and $\mathcal{X}=\{M':\ M'\leq M \text{ submodule with } x\not\in M'\}$. It is $\mathcal{X}\not=\emptyset$ since $0\in\mathcal{X}$. If $\mathcal{Y}\subseteq\mathcal{X}$ is a chain then the $R-$submodule $M_0=\cup\mathcal{Y}$ does not contain $x$ (o $M_0\in\mathcal{X}$) and it is "bigger" than every $M'\in\mathcal{Y}$. Hence (Zorn's Lemma) there is a maximal $M'\in\mathcal{X}$. Since $M$ is semisimple $\exists\ N\leq M$ st $M=M'\oplus N$.

$N$ is a simple $R-$module. It is $x\notin M'$ so $M'\subsetneq M\Rightarrow N\not=0$. Suppose that there is a submodule $0\subsetneq L\subsetneq N$. Since $N$ is semisimple we can find $L'\subseteq N$ st $N=L\oplus L'$. Then $0\subsetneq L'\subsetneq N$. So $M'\subsetneq M'\oplus L\Rightarrow$ $\fbox{$x\in M'\oplus L$}$. It is also, $M'\subsetneq M'\oplus L'\Rightarrow \fbox{$x\in M'\oplus L'$}$ and so $$x\in (M'\oplus L)\cup (M'\oplus L')\subseteq M=M'\oplus N=M'\oplus L\oplus L'$$ But it is $(M'\oplus L)\cap(M'\oplus L')=M'\Rightarrow x\in M'$ contradiction since $M'\in\mathcal{X}$

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