Does localization functor have both sides adjoint functors? Let $A$ be commutative ring, and $S$ a multiplicative set. The localization $S^{-1}$: $A$-module $\rightarrow$ $S^{-1}A$-module. Functor $F$: $S^{-1}A$-module $\rightarrow$ $A$-module, regard $A$ as subring of $S^{-1}A$ and acts on an $S^{-1}A$-module gives one side. Is there another functor that gives the other side? So I can prove the exactness of localization.
 A: No. Here is a general statement.
Theorem: Let $A, B$ be rings and let $_A M_B$ be an $(A, B)$-bimodule. Then tensoring $(-) \otimes_A M$ defines a functor $\text{Mod-}A \to \text{Mod-}B$. This functor always has a right adjoint given by $\text{Hom}_B(M, -)$, but it has a left adjoint iff $M$ is finitely generated projective as a left $A$-module. In this case, the left adjoint is given by $(-) \otimes_B M^{\ast}$, where $M^{\ast} \cong \text{Hom}_A(M, A)$. 
Localization is the case that $M$ is $S^{-1}(A)$, regarded as an $(A, S^{-1}(A))$-bimodule, and $S^{-1}(A)$ is almost never even finitely generated as an $A$-module in general. Indeed, suppose $A$ is an integral domain, $s \in A$ is nonzero and not already invertible, and we localize by inverting $s$. I claim that the associated functor $M \mapsto M \otimes_A A[s^{-1}]$ fails to preserve infinite products, and so cannot have a left adjoint: in particular, the natural map
$$(\prod_{i \in \mathbb{N}} A) \otimes A[s^{-1}] \to \prod_{i \in \mathbb{N}} (A[s^{-1}])$$
fails to be an isomorphism because its image does not contain, for example, the element $(1, s^{-1}, s^{-2}, s^{-3}, ...)$. 

Note that if you only want $(-) \otimes_A M$ to be exact then this is equivalent to saying that $M$ is flat as a left $A$-module, and this is a much weaker condition than being finitely generated projective. In particular, any filtered colimit of flat modules is flat, and $S^{-1}(A)$ is naturally a filtered colimit of free modules. (I'm not saying this is how you should prove that localization is exact. I think you can just verify this directly.) 
