Noetherian module implies finite direct sum of indecomposables? 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!
 A: 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.
