Finite Sum of Indecomposable Modules .... Bland, Proposition 4.2.10 .... ....

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 4.2: Noetherian and Artinian Modules and need some help to fully understand the proof of part of Proposition 4.2.10 ... ...

Proposition 4.2.10 reads as follows:View attachment 8213My questions are as follows:Question 1

In the above proof by Bland we read the following:

" ... ... If \(\displaystyle M\) is indecomposable, then we are done ... "

Is Bland arguing that if \(\displaystyle M\) is indecomposable then we can regard \(\displaystyle M\) itself as a "finite sum" of indecomposable R-modules ... ... can someone please confirm that this is the case ...

Question 2

In the above proof by Bland we read the following:

" ... ... Since \(\displaystyle M\) is not indecomposable, we may write \(\displaystyle M = X \bigoplus Y\). At least one of \(\displaystyle X\) and \(\displaystyle Y\) cannot be a finite direct sum of its indecomposable submodules. ... ... "

Can someone please explain why at least one of \(\displaystyle X\) and \(\displaystyle Y\) cannot be a finite direct sum of its indecomposable submodules ... ... ?

... indeed ... Bland is arguing the \(\displaystyle M\) is not indecomposable ... so \(\displaystyle M\) is decomposable ... so \(\displaystyle M = X \bigoplus Y\) ... but how does \(\displaystyle M\) being decomposable stop \(\displaystyle X\) and \(\displaystyle Y\) both being decomposable ... ?--------------------------------------------------------------------------------------------------------------------------------------------

***EDIT***

Regarding Question 2 ... I think I should have read the proof more carefully ... and noted that Bland is assuming not only that M is not indecomposable ... but also that \(\displaystyle M\) fails to have a decomposition of the form ...

\(\displaystyle M = M_1 \bigoplus M_2 \bigoplus \ ... \ ... \ \bigoplus M_n \) ... ... ... ... ... (1)

... so if both of \(\displaystyle X\) and \(\displaystyle Y\) were finite direct sums of indecomposable submodules then \(\displaystyle M\) would have a decomposition of the form (1) ... which violates the assumption that \(\displaystyle M\) fails to have a decomposition of the form ...

Is that correct ...?

----------------------------------------------------------------------------------------------------------------------------------------------Help will be appreciated ...

Peter=========================================================================Definition 4.2.9 is relevant to the above post so I am providing the text of Definition 4.2.9 ... as follows ...

View attachment 8214Hope that helps ...

Peter

 

[steenis]

$M$ can be decomposable or indecomposable.

If $M$ is indecomposable then we are ready, because then $M$ is a finite sum of $1$ indecomposable submodule of $M$, namely $M$ itself. That answers question 1.

If $M$ is decomposable, then we have to prove that $M$ is a finite direct sum of indecomposable submodules of $M$.

For the proof we suppose that there is no finite direct sum of indecomposable submodules of $M$, i.e., $M$ cannot be written as a finite direct sum of indecomposable submodules of $M$. This will turn out to be a contradiction.
Your edited answer of question 2 is correct.

See my answer in the other post.