Showing that if $0\to A\to B\to C\to 0$ is exact with $C$ projective, then $0\to A\otimes M\to B\otimes M\to C\otimes M\to 0$ is exact.

Question

I had a question on a past qual that asked just that, where $M$ was an arbitrary module. The question also said not to use the theory of Tor groups. I was able to show that it using Ext groups, but I don't feel that is in the spirit of what was meant by not using Tor groups. Could anyone help prove that $$0\to A\otimes M\to B\otimes M\to C\otimes M\to 0$$ is exact provided $0\to A\to B\to C\to 0$ is exact and $C$ projective? Thank you.

For how I did it with Ext groups, here's a proof sketch:

$$0\to \text{Hom}(C\otimes M,N)\to\text{Hom}(B\otimes M,N)\to \text{Hom}(A\otimes M,N)\to 0$$ is exact by the hom-tensor adjoint theorem iff $$0\to\text{Hom}(C,\text{Hom}(M,N))\to\text{Hom}(B,\text{Hom}(M,N))\to\text{Hom}(A,\text{Hom}(M,N))\to0$$ is exact. Well Since $C$ is projective the above is exact by the long exact sequence for $\text{Ext}$. Therefore for all $N$ $$0\to \text{Hom}(C\otimes M,N)\to\text{Hom}(B\otimes M,N)\to \text{Hom}(A\otimes M,N)\to 0$$ is exact, hence $$0\to A\otimes M\to B\otimes M\to C\otimes M\to 0$$ is exact.

Answer 1

If $C$ is projective, the sequence $0\to A\to B\to C\to 0$ splits. Therefore so does $0\to A\otimes M\to B\otimes M\to C\otimes M\to 0$ which implies it is also exact.

Answer 2

First, some notation: let's call these maps $i : A\to B$ and $p: B\to C$.

Next, a lemma: A module $P$ is projective if and only if every short exact sequence $$ 0 \to M \to N \to P \to 0$$ splits.

Since $C$ is projective, we know the sequence $$0\to A\to B\to C\to 0$$ splits, i.e. we have a map $q: B\to A$ such that $qi = 1_A$. This tells us that $i \otimes M : A \otimes M \to B\otimes M$ admits a retraction: $q\otimes M$. In particular, we must have that $i\otimes M$ is injective.