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.



If $M \otimes -$ is continuous, why is $M$ f.g. projective? Alternative proof
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.


