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.

  • $\begingroup$ For your last sentence, can you use the Specker phenomenon? $\endgroup$ 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. $\endgroup$ Oct 23, 2016 at 5:52

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Aug 11, 2020 at 16:49
  • 1
    $\begingroup$ 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. $\endgroup$ 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. $\endgroup$ Aug 12, 2020 at 8:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.