Functor $−\otimes_R L$ is right exact

Question

Consider

(*) A sequence of $R$-modules $M′ \to M \to M′′ \to 0$ is exact if and only if for all $R-$ modules $N$ the induced sequence

$0 \to Hom_R(M′′, N) \to Hom_R(M, N) \to Hom_R(M′, N)$ is exact

Now my problem : Assume $R$ is commutative. Let $L$ be an $R$-module. Use (*) plus Hom-$\otimes$-adjunction to prove that the functor $−\otimes_R L$ is right exact.

I have already proven (*), but I am clueless on how to do this. Please help me prove this. Thanks for your time

Answer 1

It boils down to writing out what the condition $(*)$ gives you in the case you're interested in, which is exactness of a sequence with tensor products. Let an exact sequence $M'\to M\to M''\to 0$ be given. You want to prove that $M'\otimes_R L\to M\otimes_R L\to M''\otimes_R L\to 0$ is also exact. By $(*)$, it suffices to take an arbitrary module $N$ and show that the sequence $0\to \mathrm{Hom}_R(M''\otimes_R L,N)\to\mathrm{Hom}_R(M\otimes_R L,N)\to\mathrm{Hom}_R(M'\otimes_R L,N)$ is exact. But by the tensor-hom adjunction, this sequence is isomorphic to the sequence $0\to \mathrm{Hom}_R(M'',\mathrm{Hom}_R(L,N))\to\mathrm{Hom}_R(M,\mathrm{Hom}_R(L,N))\to\mathrm{Hom}_R(M',\mathrm{Hom}_R(L,N))$, which is exact by $(*)$ because the sequence $M'\to M\to M''\to 0$ is exact. This is by the way how you can prove in general that left adjoint functors are right exact (and dually, that right adjoint functors are left exact).