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Let $0 \to A \to B \to C \to 0$ be a short exact sequence of left $R$-modules. If $M$ is any left $R$-module, prove that there are exact sequences $$ 0 \to A \oplus M \to B \oplus M \to C \to 0 $$ and $$ 0 \to A \to B \oplus M \to C \oplus M \to 0. $$

(Original picture of the problem here.)

Proof Attempt: Let $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ be a short exact sequence of left $R$-modules. Let $M$ be any left $R$-Module. We want to adjoin $M$ to the corresponding exact sequence $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$.

At this point I'm noticing that $A \cong A \oplus M$ and $A \oplus M$ and $B \cong B \oplus M$ would need to true for the corresponding exact sequences $0 \rightarrow A \oplus M \rightarrow B \oplus M \rightarrow C \rightarrow0$, $0 \rightarrow A \rightarrow B \oplus M \rightarrow C \oplus M \rightarrow 0$.

However, my problem is trying to figure out how to implement the (internal) direct sums in the first exact sequence to get the two new exact sequences.

I would appreciate only advice and hints.

Jendrik Stelzner
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Alexander King
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    What you claim to notice, that $A\cong A\oplus M$ and $B\cong B\oplus M$ would need to be true, is incorrect. And what do you mean by "the (internal) direct sums in the first sequence? In the first sequence, $0\to A\to B\to C\to 0$, I don't see any direct sums. – Andreas Blass May 14 '18 at 21:44
  • I think I was trying to state how to get from the first exact sequence the the two new ones with the direct sum. Note also that I didn't think that A being isomorphic to direct sum A and B would be true. – Alexander King May 14 '18 at 21:48
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    There is a more general version which is not more difficult : if $$0\to A\to B\to C\to 0$$ $$0\to A'\to B'\to C'\to 0$$ are two short exact sequence, then $$0\to A\oplus A'\to B\oplus B'\to C\oplus C'\to 0$$ is also exact. Can you show it ? Can you see how this implies your claim ? – Roland May 15 '18 at 08:07

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