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.

  • $\begingroup$ Wow, talk about vagueness... there are at least five books with that title. And even if you specified the authors, not everyone will have the book available. So, if you really want an answer, write down the proof, or sketch of the proof. $\endgroup$
    – Ennar
    Dec 8, 2016 at 9:10
  • $\begingroup$ Sorry, I forgot to add the author. Now I will add the proof. $\endgroup$
    – iam_agf
    Dec 8, 2016 at 9:13
  • $\begingroup$ Corrected. Added the proof. $\endgroup$
    – iam_agf
    Dec 8, 2016 at 9:25
  • $\begingroup$ Thank you. This proof is standard and I'm not aware of shorter one. Note that you have to argue at some point that every nonzero submodule has irreducible submodule, I don't think that there is a way around this (how would you get decomposition otherwise?). $\endgroup$
    – Ennar
    Dec 8, 2016 at 9:30

1 Answer 1


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.


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