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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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    In the first paragraph, you mean $M_1\cap M_2={0}$.2011-03-07
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    umm, no, we suppose the submodules $M_1$ and $M_2$ have only the "vector" $0$ in common. (if by $0$ you mean $\{0\}$, then its a matter of notation)2011-03-07
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    @Leon, he is talking about the indices... (And essentially all humans write $0$ instead of $\{0\}$ :) )2011-03-07
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    oh, sorry, haven't noticed it before.2011-03-07
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    Yeah, sorry, just commenting on the indices. I meant $\{0\}$.2011-03-07
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    Slightly off topic... but why do you write all those `\!` in your latex, @Leon? Fighting $\TeX$ is the silliest way to use $\TeX$!2011-03-07
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    Hmm, well, it seems to me that $\TeX$ makes spaces in math mode too large. I thought it would be easier to read, but hey, if people prefer it that way, less work for me :).2011-03-07
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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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    Is there any way to do this without Zorn's Lemma? Also, does this show anything about the simplicity of $M_1$ and $M_2$ (or do we just iterate the process?)?2011-03-07
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    @JBeardz: notice that when $R$ is a field, then the hypothesis holds trivially. This means that the result implies that vector spaces are semisimple, so that they have bases (because simple modules in this case are one-dimensional). The statement "vector spaces have bases" is *equivalent* to the Axiom of Choice so, in particular, it is pretty pointless to try to prove Leon's statement without using some form of choice...2011-03-07
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    Neither $M_1$ nor $M_2$ are going to be always simple. Just consider the case of vector spaces for examples...2011-03-07
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    Aha, if $M_1\cap M_2=0$ and $(M_1+M_2)\cap P=0$, then $M_1\cap (M_2+P)$ $=(M_1\cap M_2)+(M_1\cap P)$ $=0$ so we have found a strictly larger submodule $M_2+P$ that intersects $M_1$ only in $0$. This means that $M_2$ is not maximal, contradiction.2011-03-07
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    @Leon Lampert I don't think $M_1\cap (M_2+P)\subset (M_1\cap M_2)+(M_1\cap P)$ is true. You can do it like that: take $m_1=m_2+p\in M_1\cap (M_2+P)$, then $m_1-m_2=p$. By the second condition, $m_1-m_2=p=0$, while from the first condition $m_1=m_2=0$.2011-03-07
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    @Mielu aha, you're right, thanks for correcting me, seems I was too hasty.2011-03-07