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Let $0 → A → B → C → 0$ be a short exact sequence of $R$-modules. Prove that for any $R$-module $M$ , there is a short exact sequence $0 → A \oplus M → B \oplus M → C → 0$.

Can anyone please help me with this? I don't even have a clue how to start. Have been staring books for hours and I still don't have a clue.

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    $\begingroup$ If you have a map $f:A\to B$, can you construct a map $g:A\oplus M\to B\oplus M$? $\endgroup$ – Quimey May 15 '14 at 17:07
  • $\begingroup$ cant you define g:A⊕M → B⊕M by g(a+m) = (b+m) for some b in B? $\endgroup$ – Lucy May 15 '14 at 17:19
  • $\begingroup$ Yes, but you want $g$ to be related to $f$ in some way. for instance you may take $b=f(a)$. Using this particular $g$ you can get the desired short exact sequence. $\endgroup$ – Quimey May 15 '14 at 17:20
  • $\begingroup$ Sorry to bother but I am still not seeing it. $\endgroup$ – Lucy May 15 '14 at 17:24
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Let $0\to A \stackrel{f}{\to} B \stackrel{g}{\to} C \to 0$ be an exact sequence of $R$-modules and $M$ any $R$-module. You can construct a new short exact sequence $0\to A \oplus M \stackrel{\overline f}{\to} B \oplus M \stackrel{\overline g}{\to} C \to 0$ where $\overline f (a,m)=(f(a),m)$ and $\overline g (b,m)=g(b)$.

Things you need to check:

  • $\overline f$ and $\overline g$ are morphisms of $R$-modules,
  • $\overline f$ is injective,
  • $\overline g$ is surjective,
  • $\ker \overline g = \textrm{im} \overline f$
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