Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 3 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer
Thanks for that. – David Ward Oct 5 '12 at 8:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.