February 5th, 2018, 04:20 PM  #1 
Senior Member Joined: Jan 2015 From: usa Posts: 104 Thanks: 1  Family of modules
Please help me to prove the following result: Let $R$ be a ring with $1$ and $\mathcal{F}$ a family of simple left $R$ modules. Let $M=\oplus_{S\in \mathcal{F}} S$ and suppose that $T$ is a simple submodule of $M$. Show that $T\cong S$ for some $S\in \mathcal{F}$. Thanks 
February 6th, 2018, 01:47 AM  #2 
Senior Member Joined: Oct 2009 Posts: 733 Thanks: 246 
Maybe do an easy case of $M=S_1\oplus S_2$ first? Take $T$ a submodule of $M$, you can distinguish some cases: 1) $T$ contains only elements from $S_1$. 2) $T$ contains only elements from $S_2$. 3) $T$ contains nonzero elements from both $S_1$ and $S_2$. Now continue. 
February 6th, 2018, 02:20 AM  #3 
Senior Member Joined: Jan 2015 From: usa Posts: 104 Thanks: 1 
Micrm@ss, In this case we need to show that $T\cong S_1$ or $T\cong S_2$ but i don't see how to do so. Would you please help me more? thanks in advance

February 7th, 2018, 09:49 AM  #4 
Member Joined: Jan 2016 From: Athens, OH Posts: 92 Thanks: 47 
Mona, Since you never answer my hints or proof outlines, here's a complete proof: 
February 7th, 2018, 12:03 PM  #5 
Senior Member Joined: Jan 2015 From: usa Posts: 104 Thanks: 1 
at the end of your proof the sum + are they simple sum or direct sum $\oplus$? and why $T\cong (M_{n1}\oplus T)/M_{n1}$?
Last edited by mona123; February 7th, 2018 at 12:14 PM. 
February 9th, 2018, 06:28 PM  #6 
Member Joined: Jan 2016 From: Athens, OH Posts: 92 Thanks: 47 
The sum is direct, but one doesn't need this. For any submodules A and B, $${A+B\over A}\simeq {B\over A\cap B}$$ In particular, for $A\cap B=0$, (here $0$ means the zero submodule), $${A+B\over A}\simeq {B\over 0}\simeq B$$ 

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