Assume $M$ is both noetherian and artinian and fix $S_0\subseteq M$ a simple submodule. How to prove that $S_0$ is contained in some indecomposable direct summand of $M$?
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$\begingroup$ "noetherian + artinian = finite length" is a good start. $\endgroup$– Dag Oskar MadsenCommented Mar 15, 2017 at 18:49
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$\begingroup$ Guess you have some ground ring (associative unital, not necessarily commutative). The generic tag for this is ra. If you mean commutative this should be said explicitly (and then the generic tag would be ac) $\endgroup$– YCorCommented Mar 17, 2017 at 19:35
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2 Answers
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It's not true.
Consider representations of the quiver $$\bullet\stackrel{\alpha}{\rightarrow}\bullet\stackrel{\beta}{\leftarrow}\bullet.$$
The representation $k \to k^2 \leftarrow k$, where the arrows map onto distinct one-dimensional subspaces of $k^2$, has a unique decomposition into indecomposable summands $k\to k\leftarrow0$ and $0\to k\leftarrow k$, neither of which contains the simple subrepresentation generated by an element of $k^2$ that is neither in the image of $\alpha$ nor the image of $\beta$.
answered Mar 16, 2017 at 13:20
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$\begingroup$ I was actually reading the proof of Gabriel's Theorem in Schiffler's book where I found the paragraph $\endgroup$ Commented Mar 28, 2017 at 16:44
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$\begingroup$ I was actually reading the proof of Gabriel's Theorem in Schiffler's book where I found a sentece that I don't understand (see in bold at the botton), and I thought it was equivalent to may question $\endgroup$ Commented Mar 28, 2017 at 16:51
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$\begingroup$ ".. Suppose that $End( M )\notcong k$. Since M is indecomposable every endomorphism of M is of the form $\lambda id+g$ for some $\lambda\in k$ and some nilpotent endomorphism g. Then, since $End M \notcong k$, there exists a nonzero nilpotent endomorphism $g\in End M$ . Thus $g ^m=0$, for some $m\geq 2$. We may suppose without loss of generality that $m = 2$; otherwise, it suffices to replace g by the endomorphism $g^{m-1}$ . $\endgroup$ Commented Mar 28, 2017 at 16:52
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$\begingroup$ Moreover, among all nonzero endomorphisms whose square is zero, we choose g such that the image of g is of minimal dimension. Since $g^2=0$, we have $im g \subset ker g$, and hence there exists an indecomposable summand L of ker g such that $im g \cap L$ is nonzero. $\endgroup$ Commented Mar 28, 2017 at 16:52
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$\begingroup$ @MarcoFarinati Unless there are extra conditions on the algebra or module, my example can be adjusted to give a counterexample to that. However, you can then change $g$ to another endomorphism with the same kernel, along the lines of tj_'s answer, so that it does work. $\endgroup$ Commented Mar 30, 2017 at 9:10
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As shown by Jeremy Rickard's answer, $S := S_0$ is usually not contained in an indecomposable direct summand. The purpose of this answer is to show the weaker statement
$S$ can be embedded into an indecomposable direct summand of $M$.
Proof: WLOG assume $S \neq 0$. Since $M$ is artinian, it is a direct sum of indecomposable submodules $M_1,..., M_m$. Let $n \le m$ be minimal such that there is an embedding (i.e. an injective hom. of modules) $S \hookrightarrow \oplus_{i=1}^n M_i$. If $n=1$ we are done. If $n > 1$ consider the composition $$S \hookrightarrow \bigoplus_{i=1}^n M_i \twoheadrightarrow \bigoplus_{i=1}^{n-1}M_i$$ If it's kernel is zero, $S$ embedds into $\oplus_{i=1}^{n-1}M_i$, in contradiction to the minimality of $n$. Hence, the kernel is non-zero and by simplicity of $S$, it's $S$, i.e. the composition is the zero map. Hence $$\text{im}(S \hookrightarrow \bigoplus_{i=1}^n M_i) \subseteq \ker(\bigoplus_{i=1}^n M_i \twoheadrightarrow \bigoplus_{i=1}^{n-1}M_i) = M_n$$ Thus the composition $S \hookrightarrow \bigoplus_{i=1}^n M_i \twoheadrightarrow M_n$ is injective. QED.
Edit: Simpler proof: From $M=\oplus_{i=1}^m M_i$ we have $Hom(S,M)\cong \oplus_i Hom(S,M_i)$ and since the LHS is non-zero (it has the inclusion map), there is $i$ such that $Hom(S,M_i) \neq 0$. Let $0 \neq f: S \to M_i$ be a hom. By simplicity of $S$ we conclude $\ker f = 0$, i.e. $f$ is an embedding.
