For every ses $A \to B \to C$ and module $M$, there exists a ses $A \oplus M \to B \oplus M \to C$

Question

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

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

Answer 1

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,
• $\ker \overline g = \textrm{im} \overline f$