Let $R$ be a ring. Let $Mod(R)$ be the category of left $R$-modules, and let $Proj(R) \subseteq Mod(R)$ be the full subcategory of projective $R$-modules. Let's say that $R$-projectives are closed under highly-filtered colimits if there exists a cardinal $\kappa$ such that $Proj(R)$ is closed under $\kappa$-filtered colimits in $Mod(R)$. For example,

  • When $R$ is a field (or division ring), we have $Proj(R) = Mod(R)$, and so vacuously we have that $R$-projectives are closed under highly-filtered colimits.

  • When $R = \mathbb Z$, the question (perhaps surprisingly) depends on set theory

    • Under the anti-large cardinal hypothesis $V=L$, Eklof and Mekler showed that there exist aribtrarily large non-free abelian groups all of whose smaller subgroups are free and it follows that $\mathbb Z$-projectives are not closed under highly-filtered colimits.
    • Whereas if there exists a strongly compact cardinal, then it follows from the powerful image theorem of Makkai and Pare that free abelian groups are an accessible, accessibly embedded subcategory of all abelian groups, and in particular $\mathbb Z$-projectives are closed under highly-filtered colimits.

The positive result of Makkai and Pare generalizes immediately to show that if $R$ is a PID (in the noncommutative case I'm not sure of the terminology, but the hypothesis is that every submodule of a projective $R$-module is free), then assuming the existence of a sufficiently-large strongly compact cardinal (I think the cardinal just has to be bigger than the cardinality of $R$) we have that $R$-projectives are closed under highly-filtered colimits.

In fact, Rosicky and Brooke-Taylor's "$\lambda$-pure" version of the powerful image theorem shows that if there is a sufficiently large strongly compact cardinal, then projective $R$-modules are closed under highly-filtered colimits provided that $R$ is of global projective dimension $\leq 1$ (every submodule of a projective module is projective rather than free), so this result holds even for some $R$ which are not domains.


  1. What are some other rings $R$ for which we can say (perhaps under set-theoretical hypotheses) whether $R$-projectives are closed under highly-filtered colimits?

  2. Are there any rings $R$ other than division rings for which the question "are $R$-projectives closed under highly-filtered colimits?" is decidable in ZFC?

  3. Is the precise large-cardinal strength of "$\mathbb Z$-projectives are closed under highly-filtered colimits" -- or (equivalently, I think) "any sufficiently-large abelian group all of whose smaller subgroups are free, is free" -- known?

  1. Let $R$ be a ring and $\kappa$ be a strongly compact cardinal such that $|R|<\kappa$. Then the class of all projective $R$-modules is closed under $\kappa$-filtered colimits. This is Theorem 3.3 in the recent preprint of J. Šaroch and J. Trlifaj "Test sets for factorization properties of modules", https://arxiv.org/abs/1912.03749

  2. For any left perfect ring $R$ (which includes both all left Artinian and all right Artinian rings), all flat left $R$-modules are projective, so the class of projective left $R$-modules is closed under $\aleph_0$-filtered colimits. For example, this applies to all the rings $\mathbb Z/n\mathbb Z$, $n\ge2$ (many of which are not division rings).

  • $\begingroup$ Thanks, this is fantastic! I will hold off for a bit on accepting in hopes of hearing about more ZFC examples and about negative results under anti-large-cardinal hypotheses, and about more precise consistency-strength calibrations, but I think this is primarily what I was looking for. $\endgroup$
    – Tim Campion
    Feb 24 '20 at 19:54
  • $\begingroup$ Actually, your (1) is stronger than I expected might be known. Do you know whether large cardinal hypotheses imply more generally that every small-projectivity class in a locally presentable category is accessibly embedded -- or perhaps even that the left half of an accessible wfs is accessibly embedded in the arrow category? $\endgroup$
    – Tim Campion
    Feb 24 '20 at 20:01
  • $\begingroup$ No, concerning (1), this is just a new (and very interesting) result which I happen to have heard about. I am not an expert on such things, have not studied the details, and do not know what the potential for generalizations of this theorem may be. $\endgroup$ Feb 24 '20 at 21:59

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