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Noetherian module implies finite direct sum of indecomposables?

Let R be a ring and let M be a Noetherian R-module.

If M is indecomposable we are done. Otherwise, M is a direct sum of two proper and non-trivial submodules.

If M were also Artinian, I could use induction on the (finite) length of M and prove the result in the title.

I don't know how to proceed in the general case. Thanks in advance for any ideas!

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    Write M (or even a submodule of M if you don't want to prove you get all of M) as an infinite direct sum (by induction on its summands not being indecomposable), and then take an increasing sequence of summands to get the contradiction.2012-04-11
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    Right. Could you please clarify on the induction part? Thanks!2012-04-11
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    Write M = M1 + ... + Mn + A where A is not indecomposable (n=0 corresponds to A=M not indecomposable), since A is not indecomposable write A=M(n+1) + A' and so M = M1 + .... + Mn + M(n+1) + A' and the induction proceeds to create an increasing chain of submodules M1 + ... + Mn.2012-04-11
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    Thanks Jack! When I reread what you wrote before I got it. There is no formal induction but an inductive process to create the strict ascending chain.2012-04-11
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    @JackSchmidt Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-10

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