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This is related to another solved question submodule of indecomposable module

It has shown some indecomposable modules have decomposable modules. Now my question is,

Suppose M is an injective and indecomposable module, can it have an decomposable submodule?

Thanks in advance.

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For finite length modules over Artinian rings it can not have such a module, since an indecomposable injective module has simple socle. –  Julian Kuelshammer Dec 2 '12 at 21:12
    
@Julian, can you give some reference? –  ougao Dec 2 '12 at 21:15
    
Auslander, Reiten, Smalo: Representation theory of artin algebras states in Proposition 4.1 (d): Let $\Lambda$ be an artin algebra. Then a module of finite length is injective if and only if its socle is simple. –  Julian Kuelshammer Dec 2 '12 at 21:44
    
@YACP, sorry for the typoes, I have fixed it. It should be decomposable submodule. –  ougao Dec 16 '12 at 19:41
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1 Answer

up vote 3 down vote accepted
+50

This is impossible. I will prove that if an injective module $M$ contains a decomposable module, then $M$ is decomposable.

By hypothesis, $M$ has a submodule of the form $N_1 \oplus N_2$, where each $N_i$ is nonzero. Because $M$ is injective, it contains a submodule isomorphic to the injective hull $E(N_1 \oplus N_2) = E(N_1) \oplus E(N_2)$ (for this equality, see Lam's Lectures on Modules and Rings, equation (3.39)). Because this is an injective submodule of $M$, we have $M = E(N_1) \oplus E(N_2) \oplus P$ for some submodule $P$. Because both of the injective hulls $E(N_i)$ are nonzero, it follows that $M$ is decomposable.

The positive way to state this result is that an injective module is indecomposable if and only if it is uniform. You can find this statement in the same book by Lam, Theorem 3.52.

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