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Let $R$ be a commutative ring with unity. The Bass-Papp theorem states that any countable direct sum of injective $R$-modules is injective iff $R$ is Noetherian . Chase's theorem states that any direct product of projective $R$-modules is projective iff $R$ is Artinian . My question is : Is any characterization for commutative rings (with unity) are known such that any countable direct-product of projective modules over the ring is again projective ?

Also asked on MSE : https://math.stackexchange.com/questions/2501945/when-is-countable-direct-product-of-projective-modules-again-projective

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    $\begingroup$ If $R$ is countable and $R^{\aleph_0}$ is projective, then $R$ is Artinian by Theorem 3.1 of ams.org/journals/tran/1960-097-03/S0002-9947-1960-0120260-3/…. If $R^{\aleph_0}$ is free, then $R$ is a field by a theorem of Krull (I couldn't find a reference). If $R = \mathbb{Z}$ then $R^{\aleph_0}$ is not projective by Baer's theorem. Some sources claim that this also holds for any PID. $\endgroup$ – Luc Guyot Nov 8 '17 at 22:10
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    $\begingroup$ Have you looked into the book "Almost free modules" by Eklof and Mekler? I don't have my copy handy, but it seems to me there's probably some relevant information there, perhaps involving the notion of perfect rings. $\endgroup$ – Andreas Blass Nov 9 '17 at 1:31
  • $\begingroup$ @LucGuyot : I knew about Baer's result , I don't know whether it extends to any PID or not ... could you please present a proof of the theorem of Krull you mention that if $R^{\aleph_0}$ is free then $R$ is a field ? $\endgroup$ – user111524 Nov 9 '17 at 9:18
  • $\begingroup$ @AndreasBlass : Ah no I haven't .. I will check it then as you say .. $\endgroup$ – user111524 Nov 9 '17 at 9:19
  • $\begingroup$ @LucGuyot : A reference (no. 5) to this paper cyber.sci-hub.cc/MTAuMTA4MC8wMDkyNzg3OTMwODgyNDc2Nw==/… seems to deal with the question that when an infinite direct product of copies of $R$ is free , but I couldn't find this paper . If you have time , you may look for it and let me know if you find it. $\endgroup$ – user111524 Nov 9 '17 at 9:23
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Under the assumption that a countable direct product of modules over $R$ means a direct product of countably many modules over $R$, I answer OP's question when $R$ is Noetherian. In addition, I outline a remark for $R$ an arbitrary countable ring with identity.

Claim 1. Let $R$ be a commutative ring with identity. If $R$ is Noetherian then the following are equivalent:

$(i)$ Every direct product of countably many projective modules over $R$ is projective.

$(ii)$ $R$ is Artinian.

Proof. The implication $(ii) \Rightarrow (i)$ is given by Stephen Chase's Theorem [1, Theorems 3.3 and 3.4]. Assume that $(i)$ holds. Then $M \Doteq R^{\aleph_0}$ is a projective module. If $R$ is moreover connected, then $M$ is free over $R$ by [2, Corollary 4.5]. $R$. By John O'Neill's result [3, Lemma 1.1], the ring $R$ is then Artinian. If $R$ is not connected, it is the direct product of finitely many non-trivial connected rings $R_i$. By hypothesis, the module $M_i \Doteq R_i^{\aleph_0}$ is projective over $R$, and hence over $R_i$, for every $i$. As $R$ is a product of finitely many Artinian rings, it is Artinian.

For an arbitrary countable ring, the following weaker result holds.

Claim 2. If $R$ is a countable ring with identity such that $R^{\aleph_0}$ is projective over $R$, then $R$ is left perfect.

Proof. Apply [1, Theorem 3.1] and Bass's characterization of left perfect rings.

Coming back to the commutative setting, we get

Corollary. Let $R$ be an integral domain. If $R$ is countable then the following are equivalent:

$(i)$ Every direct product of countably many projective modules over $R$ is projective.

$(ii)$ $R$ is a field.

Proof. Any element of a perfect commutative ring is either a unit or a zero divisor. The result is then a direct consequence of Claim 2.


[1] S. Chase, "Direct product of modules", 1960.
[2] H. Bass, "Big projective modules are free", 1962.
[3] J. O'Neill, "When a ring is an $F$-ring", 1993.

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  • $\begingroup$ How do you apply Lemma 1.1 of J.O'Neil's paper ? How do you find a non-measurable cardinal $\alpha$ such that $R^\alpha$ is free or how do you find a cardinal number $\alpha \ge |R|$ such that $R^\alpha$ is free ? May be you should apply Corollary 3.2 ? $\endgroup$ – user111524 Nov 11 '17 at 6:51
  • $\begingroup$ $\aleph_0$ is non-measurable (en.wikipedia.org/wiki/Measurable_cardinal). $\endgroup$ – Luc Guyot Nov 11 '17 at 7:52
  • $\begingroup$ Ah right I forgot ... measurable cardinals had to be uncountable ... thanks a lot for your contribution $\endgroup$ – user111524 Nov 11 '17 at 13:59

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