TITLE: If $M \otimes -$ is continuous, why is $M$ f.g. projective? Alternative proof
QUESTION [7 upvotes]: Let $R$ be a commutative ring and $M$ be some $R$-module such that $M \otimes -$ is continuous (i.e. preserves all limits). Then one can show that $M$ is f.g. projective.
One way to prove this is to use adjoint functor theorems (anyone you like): We obtain a left adjoint, which has the form $M^* \otimes -$ by Eilenberg-Watts, and then one obtains that $M$ is dualizable, etc.
Is there also a proof which is more direct or elementary? We can also use other characterizations, such as f.g. projective = f.p. flat. It is clear that $M$ is flat (since $M \otimes -$ preserves finite limits), but I do not know how to approach f.p.

REPLY [9 votes]: I am posting the comment above as one answer.  Kernels are examples of limits.  So if the functor is continuous, then $M$ is flat.  Next, consider the natural transformation $M\otimes_R \text{Hom}_R(-,R)\Rightarrow \text{Hom}_R(-,M)$.  Both functors convert colimits to limits, and the natural transformation is an isomorphism on finitely presented objects.  Every object is a colimit of finitely presented objects.  Thus, the natural transformation is an isomorphism.  In particular, consider $\text{Id}_M$ in $\text{Hom}_R(M,M)$.  That element is in the image of the natural transformation.  Thus, $M$ is finitely presented.