Let $R$ be a commutative ring and $M$ be an $R$ module. The only proof that I know of the right exactness of $ -\otimes_R M$ uses the left exactness of $Hom_R(-,M)$. I wondered if there was a way of deriving the long exact sequence of tensor without using the right exactness of $ -\otimes_R M$. This would allow me to deduce the right exactness of tensor.
Let $0 \to M \to N \to K \to 0$ be an exact sequence of $R$ modules. Let $\mathfrak M$ be a free resolution of $M$, $\mathfrak N$ a free resolution of $N$, $\mathfrak K$ a free resolution of $K$. We get as usual that there is a short exact sequence of chain complexes $0 \to \mathfrak M \to \mathfrak N \to \mathfrak K \to 0$ which stays exact after tensoring with an $R$ module $L$.
What I would appreciate assistance in is showing that the cohomology of $\mathfrak M \otimes_R L$ at the first entry is $M \otimes_R L$. This is where I see the argument would ordinarily need the right exactness of tensor.