I have this problem
M is semisimple if and only if every submodule is a direct summand.
I found its proof in the book An introduction to homological algebra of Rotman, but I consider it large. My question is: does anyone knows a shorter proof?
The proof shown in this book is:
If $M$ is semisimple, then $M =\bigoplus_{j\in J} S_j$ , where every $S_j$ is simple. Given a subset $I \subseteq J$ , define $S_I =\bigoplus_{j\in I} S_j$. If $N$ is a submodule of $M$, we see, using Zorn’s Lemma, that there exists a subset $I$ of $J$ maximal with $S_I \cap N = \{0\}$. We claim that $M = N \oplus S_I$ , which will follow if we prove that
$S_j \subseteq N + S_I$ for all $j \in J$. This inclusion holds, obviously, if $j \in I$ . If $j\notin I$ , then the maximality of $I$ gives $(S_j + S_I )\cap N \neq \{0\}$. Thus, $s_j +s_I = n \neq 0$ for
some $s_j \in S_j$ , $s_I \in S_I$ , and $n \in N$ , so that $s_j = n − s_I \in (N + S_I ) \cap S_j$. Now $s_j = 0$, lest $s_I \in S_I \cap N = \{0\}$. Since $S_j$ is simple, we have $(N+S_I)\cap S_j=S_j$ ; that is, $S_j \subseteq N + S_I$.Suppose, conversely, that every submodule of $M$ is a direct summand. We
begin by showing that each nonzero submodule $N$ contains a simple submodule. Let $x \in N$ be nonzero; by Zorn’s Lemma, there is a submodule $Z \subseteq N$ maximal with $x \notin Z$ . Now $Z$ is a direct summand of $M$, by hypothesis, and so $Z$ is a direct summand of $N$ , by Corollary 2.24; say, $N = Z \oplus Y$ . We claim that $Y$ is simple. If $Y$ is a proper nonzero submodule of $Y$ , then $Y = Y \oplus Y$ and $N = Z \oplus Y = Z \oplus Y \oplus Y$ . Either $Z \oplus Y$ or $Z \oplus Y$ does not contain $x$ [lest $x\in (Z \oplus Y ) \cap (Z \oplus Y ) = Z$ ], contradicting the maximality
of $Z$ . Next, we show that $M$ is semisimple. By Zorn’s Lemma, there is a
family $(S_k )_{k\in K}$ of simple submodules of M maximal with the property that they generate their direct sum $D = \bigoplus_{k\in K} S_k$ . By hypothesis, $M = D \oplus E$ for some submodule $E$. If $E = \{0\}$, we are done. Otherwise, $E = S \oplus E$ for some simple submodule $S$, by the first part of our argument. But now the
family $\{S\} \cup (S_k )_{k\in K}$ violates the maximality of $(S_k )_{k\in K}$ , a contradiction.
Best Answer
Assume first that A is semisimple, and that B be a submodule of A. By Zorn's Lemma there is a submodule C of A maximal with respect to the property that $B\cap C =0$. If $B \oplus C < A$, there is a simple submodule $ S\leq A$ such that $S\nleq B\oplus C$. Also $S\cap (B\oplus C)=0$, so {$B,C,S$} is independent. But $B\cap(C\oplus S)=0$, contradicting the maximality of $C$. So $B \oplus C=A$. Hope it helps.