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?
Quotients of Indecomposable Modules
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0I 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
2 Answers
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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0Thanks 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
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.