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Can anyone give me a module $M$ over a ring $R$, such that $M$ is indecomposable, but $M$ has a submodule $N$ such that $N$ is decomposable?

In general, what further assumptions can we put on the ring $R$ or $M$ to guarantee that $M$ has no such decomposable submodule $N$?

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2 Answers 2

up vote 1 down vote accepted

If you take $R=\begin{pmatrix}k & 0 & 0\\k&k&0\\k&0&k\end{pmatrix}$, where $k$ is a field and the multiplication is given by matrix multiplication, then the direct summand of $R$ given by $\begin{pmatrix}k&0&0\\k&0&0\\k&0&0\end{pmatrix}$ is indecomposable, but has the submodule $\begin{pmatrix}0&0&0\\k&0&0\\k&0&0\end{pmatrix}$ decomposes into two one-dimensional modules.

I did not try for a rigorous proof, but an (at least sufficient) assumption I know in the context of finite dimensional $k$-algebras is that of left (or right, depending if you work with left or right modules) uniseriality as an assumption on $R$, that is the set of submodules of any indecomposable module is totally ordered by inclusion.

For an assumption on $M$ (again working with finite dimensional $k$-algebra) it is necessary and sufficient for $M$ to have simple socle, i.e. there is only one simple submodule of $M$.

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thanks, I also tried to seek an example over matrix ring, but have not figured it out. –  ougao Nov 27 '12 at 14:26

With regard to your first question, this is a problem in Rings and Categories of Modules by Anderson & Fuller. Check the preceding link for the question statement as well as a hint: namely, try a factor module of $_{R}R$ where $R = \mathbb{Q}[X, Y]$.

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@thanks, we can take $M=\frac{\mathbb{Q}[x,y]}{<x^2y>}$, which is indecomposable $R-$module, and it has a decomposable submodule $N=\frac{x^2\mathbb{Q}[x,y]+xy\mathbb{Q}[x,y]}{<x^2y>}$=$\frac{x^2\mathbb{Q}[x,y‌​]}{<x^2y>}\oplus\frac{xy\mathbb{Q}[x,y]}{<x^2y>}$. –  ougao Nov 28 '12 at 1:03

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