If $F$ is a free module then $-\otimes F$ is an exact functor

Question

For any module $M$, the functor $-\otimes M $ is right exact. I want to show that if $F$ is a free module then $-\otimes F$ is a exact.

I could not see how to adapt the proof for the case of $M$ not being free to this case.

Answer 1

Given an exact sequence of $R$-modules $$ 0\to M'\to M\to M''\to 0 $$ the general result states that $$ M'\otimes F\to M\otimes F\to M''\otimes F\to 0 $$ is exact. The only thing missing is to show that the first arrow here is injective.

Since $F$ is free, it's isomorphic to a direct sum $\bigoplus_I R$ over some index set $I$. And we have, for any $R$-module $N$, that $$ N\otimes_R\left(\bigoplus_I R\right)\cong \bigoplus_I N $$ So, what we are really after is this: If $M'\to M$ is an injective map, is the induced componentwise map $\bigoplus_I M'\to\bigoplus_I M$ injective? And it clearly is, giving us our result.