Let $R$ be a cofibrant commutative $S$-algebra (in the sense of Elmendorf-Kriz-Mandell-May; they call them "$q$-cofibrant") and $A$ be a cofibrant commutative $R$-algebra.

Does $A\wedge_R-:RMod→RMod$ preserve all weak equivalences?

This could be rephrased as "$A$ is a flat $R$-module". It is true that cofibrant $R$-modules are flat (EKMM III.3.8), however, the underlying $R$-modules of cofibrant commutative $R$-algebras are not generally cofibrant. Another positive result is that in the above hypotheses, $A\wedge_R-$ preserves weak equivalences of cofibrant commutative $R$-algebras (EKMM VII.7.4).

  • $\begingroup$ You wrote that "the underlying R-modules of cofibrant commutative R-algebras are not generally cofibrant" - do you have an example in mind? This is why in my answer below I suggested Shipley's approach of working with positive cofibrations (which is also what I did in my thesis), but I thought that for EKMM S-modules one could often avoid the shift to positive cofibrancy, since the unit is already not cofibrant (so Lewis's obstruction doesn't apply) $\endgroup$ Nov 25 '17 at 21:15
  • $\begingroup$ @DavidWhite $\mathbb S$ is a cofibrant commutative $\mathbb S$-algebra (it is cell), but $\mathbb S$ is not a cofibrant $\mathbb S$-module. $\endgroup$ Nov 26 '17 at 7:14
  • $\begingroup$ Thanks for the response. What I meant to ask was: can it be cofibrant as a commutative R-algebra but not as an R-algebra? When you forget all the way to just R-modules, you can lose cofibrancy just as you point out, but what's usually true is that a cofibration with cofibrant source forgets to a cofibration. So you have a kind of relative cofibrancy, and that's enough for the flatness property you asked about. $\endgroup$ Nov 26 '17 at 12:00
  • $\begingroup$ @DavidWhite Two remarks: the first one is that the unit $R\to A$ of a cofibrant $R$-algebra or commutative $R$-algebra is a cofibration of underlying $R$-modules (EKMM, right after VII.4.14). $\endgroup$ Nov 26 '17 at 14:10
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    $\begingroup$ The second is a quote of the paper "Topological Hochschild Homology" by Schwänzl-Vogt-Waldhausen: "We have to distinguish between the associative and commutative case, because the forgetful functor $RCAlg\to RAlg$ does not preserve q-cofibrant objects. This is a well-known phenomenon: in ordinary algebra free associative resolutions use tensor algebras, while free associative and commutative resolutions use symmetric algebras". (Also, thank you for the interest.) $\endgroup$ Nov 26 '17 at 14:10

I'm a bit out of practice with these types of questions, and I've never really worked with $S$-modules, but I think the answer is yes. Here's why I think so. First, it's sufficient to check it for the case when A is a (co)domain of a generating cofibration in CAlg(R), by a standard cellular induction (for details, see Theorem A.2 in Hovey's paper on Smith ideals). Thus, we can assume A has the form $Sym(B) \wedge R$ where $B$ is a (co)domain of a generating cofibration of $S$-algebras. Thus, $A\wedge_R -$ is a weak equivalence of $R$-modules if and only if $Sym(B) \wedge -$ is a weak equivalence (in the underlying category of $S$-modules). Since (co)domains of the generating cofibrations of $S$-algebras are cofibrant (see MMSS), $B$ is cofibrant. Thus, $Sym(B)$ is cofibrant as a commutative $S$-algebra. Next, Lemma 3.7 of my paper on commutative monoids (accepted to JPAA), proves that $Sym(B)$ is cofibrant as an $S$-algebra (so the EKMM result you cite finishes the proof). I hasten to note that it's not recorded in print anywhere that the category of $S$-algebras satisfies the strong commutative monoid axiom, but for the crucial place in the proof of 3.7 where this is needed, you can use Proposition 4.2 of Shipley's A convenient model category for commutative ring spectra instead. Basically, the idea is that it's good enough to work with positive cofibrations to get the result you want. The last section of chapter VII of EKMM is also relevant, and could perhaps avoid the shift into positive cofibrations.

  • $\begingroup$ I'm starting a bounty on the question -- more details on your answer would be very welcome, thanks. $\endgroup$ Nov 28 '17 at 2:56
  • $\begingroup$ If I understand correctly (which I probably don't), Shipley's result says that if you have some flavor of positively cofibrant commutative algebra, then the underlying module is cofibrant (but not necessarily positively cofibrant). But the OP is precisely not asking that the algebra A be some type of positively cofibrant. Also, I think I'd be surprised if S-modules satisfies the 'strong commutative monoid axiom' since it looks to me like $S^{\wedge 2}_{\Sigma_2}$ is just $S$, which isn't homotopically correct. (But again, I'm likely wrong since I'm not used to these things.) $\endgroup$ Nov 28 '17 at 3:59
  • $\begingroup$ @DylanWilson: using different terminology, Proposition 4.2 says: if $f$ is a map in $CAlg(R)$ which is a projective cofibration in $CAlg(R)$, then it is a positive flat cofibration in $CAlg(R)$. This implies that that it is a positive flat cofibration in $Mod(R)$ (so in particular, a flat cofibration in $Mod(R)$, but not necessarily a projective cofibration in $Mod(R)$). As for your last statement: Shipley's Proposition 3.3 proves that an $R$-module which is positive flat, then the standard map $\pi:X^{\wedge_R n}_{h\Sigma_n}\to X^{\wedge_R n}_{\Sigma_n}$ is a stable equivalence. $\endgroup$ Nov 28 '17 at 8:49
  • $\begingroup$ (Above, I meant: if X is an R-module which is positively flat. Couldn't edit it on time). An analogous result is true in EKMM (III.5.1): if R is a cofibrant commutative S-algebra and $X$ is a cell $R$-module, then $\pi$ is a homotopy equivalence of spectra. $\endgroup$ Nov 28 '17 at 8:51
  • $\begingroup$ @BrunoStonek at the moment I'm in Patagonia, about to start the W trek. So, I don't think I'll have any internet for the next 5 days. Towards Dylan's point, because I never verified the axiom for S-modules, I sketched how to get around it using Shipley's work. Basically, cofibrant commutative monoids don't forget to honestly cofibrant objects, but should forget to a class of objects contained in the flat objects. I agree with your comment answering Dylan. Good luck! $\endgroup$ Nov 29 '17 at 2:43

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