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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.

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    $\begingroup$ I think the argument you outlined makes it look more complicated than it really is. I'd prefer this arrangement: first show any continuous functor on $Mod_R$ is of the form $\hom(N, -)$. If this is cocontinuous (as $M \otimes -$ is), then obviously $N$ is projective, and it's also finitely generated by considering $N$ as a directed colimit of its finitely generated submodules, and using preservation of this colimit by $\hom(N,-)$. But then $M \cong M \otimes R \cong \hom(N, R)$ is also f.g. projective (write $N$ as a direct summand of $R^m$). $\endgroup$
    – Todd Trimble ♦
    Mar 9, 2018 at 10:52
  • $\begingroup$ Thank you. I know this approach, but how do you show that every continuous functor has this form, again without using adjoint functor theorems? Also, why do you post as an answer in the comment section? $\endgroup$
    – HeinrichD
    Mar 9, 2018 at 11:00
  • $\begingroup$ Because it's not a complete answer. :-) Personally I'd use the SAFT, and not think of this as a big deal (alternatively, "beta reduce" the SAFT for this special case: it really just uses the fact that $Mod_R$ has a cogenerator). $\endgroup$
    – Todd Trimble ♦
    Mar 9, 2018 at 11:03
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    $\begingroup$ 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. $\endgroup$
    – Jason Starr
    Mar 9, 2018 at 11:06
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    $\begingroup$ @JasonStarr Now that's an answer! Post as such? $\endgroup$
    – Todd Trimble ♦
    Mar 9, 2018 at 11:10

1 Answer 1

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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.

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    $\begingroup$ Great answer! Actually, a preimage of $id_M$ is the same as writing $M$ as a direct summand of some $R^n$. So we get f.g. projective right away. $\endgroup$
    – HeinrichD
    Mar 9, 2018 at 11:16
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    $\begingroup$ This was a homework exercise in a course I took as a student. This is not my argument. $\endgroup$
    – Jason Starr
    Mar 9, 2018 at 11:24
  • $\begingroup$ The part about finite generation is also given in the Stacks Project. $\endgroup$
    – Fred Rohrer
    Mar 9, 2018 at 21:05

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