# Quotients of Indecomposable Modules

This is probably a really basic problem, but I am looking at the homomorphic image of an indecomposable module. Are there any standard results which help to determine when such an image will itself be indecomposable?

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I wonder which modules are quotients of indecomposables. Free modules of rank at least 2 are not. I wonder if there is a ring in which every "projective-free" (no projective summand) module is a quotient of an indecomposable module. Certainly for any $n$ there is a ring in which the direct sum of $n$ copies of a simple module is a quotient of an indecomposable. –  Jack Schmidt Oct 4 '12 at 17:54

If $M$ is any module and $\phi(M)$ is a homomorphic image, then the isomorphism theorems say that direct-indecomposability of $\phi(M)$ amounts to the nonexistence of submodules $N$, $N'$ of $M$ such that $N\cap N'=\ker(\phi)$ and $N+N'=M$.

You could exploit this by checking to see if $\ker(\phi)$ is meet-irreducible. (If it is, $0$ is meet irreducible in the image, precluding nontrivial decompositions.)

So for example, a module whose submodules form a chain always has indecomposable images.

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Thanks. I got as far as this when I was trying to come up with criterion, but couldn't get any further. –  David Ward Oct 4 '12 at 17:36
@DavidWard The connection is not very strong. Do you have a more specific question that this is just a step toward? –  rschwieb Oct 4 '12 at 17:39
@rschweib Yes, this is trying to prove a result concerning presheaves of abelian groups defined on simplicial complexes. Basically for the part I am looking at, I want to show that if $\varphi:A\rightarrow M\oplus N$ is a homomorphism of $kG$-modules ($k$ a finite field, $G$ a finite group), where $A$ is indecomposable, then the image of $A$ must lie in either $M$ or`$N$ or be equal to zero. –  David Ward Oct 4 '12 at 18:19
@DavidWard, that's false if all of $M$, $N$ and $A$ are isomorphic, for example —take $G$ the trivial group to get a concrete example— and will be false even if only both $M$ and $N$ have submodules isomorphic to $A$. –  Mariano Suárez-Alvarez Oct 4 '12 at 18:35
I believe one can find a group $G$ and field $k$ such that $A$ has simple socle, $M\not\cong N$ are simple, and $A$ mod its socle is $M\oplus N$. I have a 9-dimensional algebra with a 3-dimensional $A$ with these properties. –  Jack Schmidt Oct 4 '12 at 20:38

A quotient of an indecomposable module can be decomposable. As an example take as algebra $K[X,Y]$ and as module $M=K^3$ with $X$ acting as the elementary matrix $E_{1,2}$ and $Y$ as the elementary matrix $E_{1,3}$. Then $M$ is indecomposable but has an irreducible submodule $M_1$, spanned by the first basis vector of $K^3$, and $M/M_1$ is decomposable as you can easily check.

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Thanks for that. –  David Ward Oct 5 '12 at 8:20