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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. –  Jack Schmidt Apr 11 '12 at 18:14
    
Right. Could you please clarify on the induction part? Thanks! –  Rod Apr 11 '12 at 18:19
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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. –  Jack Schmidt Apr 11 '12 at 19:13
    
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. –  Rod Apr 11 '12 at 19:20
    
@JackSchmidt Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Jun 10 '13 at 19:03
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1 Answer

Suppose that $M$ cannot be written as a finite direct sum of indecomposable modules. We will show that $M$ is not noetherian by exhibiting an infinite ascending chain of submodules.

Since $M$ is not indecomposable (lest $M=M$ be the direct sum decomposition into indecomposables), we can write $M=M_1 \oplus A$ where $A$ is also not a finite direct sum of indecomposables. (If both $M_1$ and $A$ are finite direct sums of indecomposables, then evidently so is $M$, it is their direct sum, so one of them, WLOG $A$, is not a finite direct sum of indecomposables.

Suppose we have $M=M_1 \oplus \ldots M_n \oplus A$ where $A$ is not a finite direct sum of indecomposables. Then $A=M_{n+1} \oplus A'$ as above with $A'$ not a finite direct sum of indecomposables. Hence $M=M_1 \oplus \ldots \oplus M_n \oplus M_{n+1} \oplus A'$ and the induction proceeds to give us an ascending chain $$0 < M_1 < M_1 \oplus M_2 < \ldots < M_1 \oplus \ldots \oplus M_n < \ldots$$ of submodules. Hence $M$ is not noetherian.

In other words, any such module contains an infinite direct sum of submodules (its uniform dimension is infinite, also know as its Goldie dimension). Such a module cannot be artinian as you can keep removing a summand, but such a module cannot be noetherian as you can keep adding a summand. Modules of finite uniform dimension have many good properties described in section 6 of Lam's Lectures on Modules and Rings.

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