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PROBLEM: Let $R$ be a ring with $1$ and $M$ be a unital $R$-module (i.e. $1x=x$). Let there for each submodule $M_1\neq M$ exist a submodule $M_2\neq M$, such that $M_1\cap M_2=\{0\}$. How can I prove, that $M$ is semisimple?

DEFINITIONS: A module $M$ is semisimple iff $\exists$ simple submodules $M_i\leq M$, such that $M=\bigoplus_{i\in I}M_i$. A module $M_i$ is simple iff it has no submodules (other than $\{0\}$ and $M$).

KNOWN FACTS: $M$ is semisimple $\Leftrightarrow$ $\exists$ simple submodules $M_i\leq M$, such that $M=\sum_{i\in I}M_i$ (the sum need not be direct) $\Leftrightarrow\forall\!M_1\!\leq\!M\;\exists M_2\!\leq\!M$ such that $M_1\oplus M_2=M$ (i.e. every submodule is a direct sumand).

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    @Leon, when typing actual $\TeX$ (as opposed to writing on websites like this) you can tweak the spacing used by the typesetting engine, so in that context you'd never add manual spacing like that. On web pages like this, I think it is not worth the effort (and it may even be possible to do the same kind of tweaking, really: it is a matter to reading MathJax's docs)2011-03-07

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Let $M$ be your module, and let $M_1$ be a submodule. Consider the set $\mathcal S$ of all submodules $N$ of $M$ such that $M_1\cap N=0$, and order $\mathcal S$ by inclusion. It is easy to see that $\mathcal S$ satisfies the hypothesis of Zorn's Lemma, so there exists an element $M_2\in\mathcal S$ which is maximal.

We have $M_1\cap M_2=0$ and we want to show that $M_1+M_2=M$. If that were not the case, your hypothesis would provide a submodule $P\subseteq M$ such that $(M_1+M_2)\cap P=0$. Can you see how to reach a contradiction now?

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    @Mielu aha, you're right, thanks for correcting me, seems I was too hasty.2011-03-07