# Short Exact Sequences and R modules

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.

• If you have a map $f:A\to B$, can you construct a map $g:A\oplus M\to B\oplus M$? – Quimey May 15 '14 at 17:07
• cant you define g:A⊕M → B⊕M by g(a+m) = (b+m) for some b in B? – Lucy May 15 '14 at 17:19
• 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. – Quimey May 15 '14 at 17:20
• Sorry to bother but I am still not seeing it. – Lucy May 15 '14 at 17:24

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