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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.2012-10-04

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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.2012-10-04
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    @DavidWard The connection is not very strong. Do you have a more specific question that this is just a step toward?2012-10-04
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    @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.2012-10-04
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    @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 $A$.2012-10-04
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    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.2012-10-04
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    @DavidWard: Take $G$ to be simple of order 7920, $k$ of characteristic 11. Then there are modules $M$ simple of dimension 1, $N$ simple of dimension 10, and $A$ indecomposable of dimension 20 with socle simple of dimension 9 such that there is a surjection from $A$ to $M \oplus N$. Basically, I just looked for a group with cyclic Sylow subgroup whose Brauer tree was not a star.2012-10-05
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    Thanks for all the responses. There must be some more properties than I can exploit in my setting that I have not yet realised.2012-10-05
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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.