Any artinian module is a finite direct sum of indecomposable modules

54 Views Asked by At

Let $M_R$ be an artinian module. I proved that there exist $n\in \mathbb{N}$ and nonzero uniform submodules $U_1,\ldots,U_n$ of $M$ such that $U_1+\ldots+U_n$ is a direct sum and $\bigoplus_{i=1}^n U_i$ is an essential submodule of $M$. In fact, the direct sum $\bigoplus_{i=1}^n U_i$ is the maximal internal direct sum of nonzero uniform submodules of $M$.

How can we use this fact to prove the result stated in the title?. Or there is another way to prove it?.

Thanks in advance.

1

There are 1 best solutions below

0
On

An answer turns up as a byproduct in this solution written for the Noetherian version of this question. But we can probably argue more directly given the Artinian condition too.

Suppose $M$ is not a finite direct sum of indecomposable modules. Then $M$ itself is not indecomposable, yielding a proper decomposition $A\oplus B$. Clearly it can't be the case that both $A$ and $B$ decompose into finitely many indecomposable submodules, because taking them all together would make a finite decomposition for $M$. Let $C$ be one that doesn't decompose that way.

Notice what we have observed is that a module that doesn't decompose into finitely many indecomposables always has a proper submodule with the same property. Using this observation, we inductively define a properly descending chain $C_1\supset C_2\supset C_3\supset \cdots$, which obviously can't happen in an Artinian module. The module would have to be non-Artinian. This proves the contrapositive of the original statement to be proven.