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 February 5th, 2018, 04:20 PM #1 Senior Member   Joined: Jan 2015 From: usa Posts: 103 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: 628 Thanks: 190 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: 103 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: Thanks from Joppy
 February 7th, 2018, 12:03 PM #5 Senior Member   Joined: Jan 2015 From: usa Posts: 103 Thanks: 1 at the end of your proof the sum + are they simple sum or direct sum $\oplus$? and why $T\cong (M_{n-1}\oplus T)/M_{n-1}$? 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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