In a unital $R$-module $M$, if $\forall M_1\!\lneq\!M\;\;\exists M_2\!\lneq\!M$, such that $M_1\!\cap\!M_2\!=\!\{0\}$, then $M$ is semisimple

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).

• In the first paragraph, you mean $M_1\cap M_2={0}$. – Jonathan Beardsley Mar 7 '11 at 20:49
• 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) – Leon Mar 7 '11 at 20:52
• @Leon, he is talking about the indices... (And essentially all humans write $0$ instead of $\{0\}$ :) ) – Mariano Suárez-Álvarez Mar 7 '11 at 20:52
• oh, sorry, haven't noticed it before. – Leon Mar 7 '11 at 20:54
• Yeah, sorry, just commenting on the indices. I meant $\{0\}$. – Jonathan Beardsley Mar 7 '11 at 20:54

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?
• 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?)? – Jonathan Beardsley Mar 7 '11 at 20:56
• @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... – Mariano Suárez-Álvarez Mar 7 '11 at 21:01
• Neither $M_1$ nor $M_2$ are going to be always simple. Just consider the case of vector spaces for examples... – Mariano Suárez-Álvarez Mar 7 '11 at 21:04
• 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. – Leon Mar 7 '11 at 21:20
• @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 $m_1=m_2=0$. – Theta33 Mar 7 '11 at 22:04