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Let $R$ be a ring and $M$ be an $R$-module. Say that $M$ is cocompact if the functor $\operatorname{Hom}(-,M)$ turns filtered limits into filtered colimits. (This is just saying that $M$ is compact as an object of $R\text{-Mod}^{op}$.) My question is:

Is it possible for a nonzero module $M$ to be cocompact?

I suspect the answer is no. For instance, by considering an infinite product as the inverse limit of finite subproducts, it is easy to show that if $R$ is a field then no nonzero vector space is cocompact.

More strongly, I can prove the answer is no assuming that there exists a proper class of measurable cardinals. Indeed, suppose $M$ is a nonzero module and $\kappa>|M|$ is a measurable cardinal. Let $U$ be a nonprincipal $\kappa$-complete ultrafilter on $\kappa$. There is then a homomorphism $L:M^\kappa\to M$ that sends a function $f:\kappa\to M$ to the unique element $m\in M$ such that $f^{-1}(\{m\})\in U$. Since $U$ is nonprincipal and $M$ is nonzero, $L$ does not factor through any finite subproduct of $M^\kappa$, and so $\operatorname{Hom}(-,M)$ does not preserve the limit of the inverse system consisting of finite subproducts of $M^\kappa$.

On the other hand, without assuming a measurable cardinal, I have not even been able to prove that $\mathbb{Z}$ is not cocompact as a $\mathbb{Z}$-module.

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  • $\begingroup$ For your last sentence, can you use the Specker phenomenon? $\endgroup$
    – Mariano Suárez-Álvarez
    Oct 23, 2016 at 5:35
  • $\begingroup$ Well, the Specker phenomenon says that $\operatorname{Hom}(-,\mathbb{Z})$ does preserve certain inverse limits, such as a countable product of copies of $\mathbb{Z}$ as a limit of finite products. It seems plausible that this might somehow generalize to show that $\mathbb{Z}$ is cocompact if no measurable cardinals exist, but I don't know how such an argument would go. $\endgroup$
    – Eric Wofsey
    Oct 23, 2016 at 5:52

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No nonzero module is cocompact. Indeed, suppose $M$ is a nonzero module, and let $X_n=\bigoplus_{i=n}^\infty M$. The modules $X_n$ form an inverse system using the obvious inclusion maps, and the inverse limit is $0$ since their intersection (as submodules of $X_0$) is $0$. On the other hand consider the homomorphism $\Delta:X_0\to M$ which is the identity on every coordinate. Then the restriction of $\Delta$ to each $X_n$ is nontrivial (since $M$ is nonzero), and so $\Delta$ represents a nonzero element of the colimit $\varinjlim \operatorname{Hom}(X_n,M)$. Thus $$\operatorname{Hom}(\varprojlim X_n,M)=0\neq \varinjlim \operatorname{Hom}(X_n,M),$$ so $M$ is not cocompact.

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  • $\begingroup$ The argument in your question shows that, if there are arbitrarily large measurable cardinals, then there is no nonzero module $M$ such that $\text{Hom}(-,M)$ turns products into coproducts. But I don't see how to adapt your answer to prove this without measurable cardinals. Have you thought about this? $\endgroup$
    – Jeremy Rickard
    Aug 11, 2020 at 16:49
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    $\begingroup$ I haven't thought about it in a while but I would not be surprised if the non-existence of measurable cardinals implies that $\operatorname{Hom}(-,\mathbb{Z})$ turns products into coproducts. $\endgroup$
    – Eric Wofsey
    Aug 11, 2020 at 17:10
  • $\begingroup$ This won't surprise you, then, but I think it's true. The details would make it a bit long for a comment, but I think it can be proved by restricting to products of cyclic subgroups, reducing it to the theorem of Zeeman that $\text{Hom}(\prod\mathbb{Z},\mathbb{Z})=\bigoplus\text{Hom}(\mathbb{Z},\mathbb{Z})$ for arbitrary index sets for the product, if there are no measurable cardinals. $\endgroup$
    – Jeremy Rickard
    Aug 12, 2020 at 8:32

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