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Let $k$ be a commutative ring, and consider the category of associative non-positive DG-algebras over $k$ (thus, $A^i = 0$ for $i>0$, and the differential has degree $+1$).

Is there a closed model structure on this category with functorial cofibrant replacement, such that the weak equivalences are the quasi-isomorphisms and such that every cofibrant object is flat over $k$?

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  • $\begingroup$ If so, this model structure would be built via Cotorsion Pairs. Since you know what the weak equivalences and cofibrant objects are, you actually already know two of the three classes of objects (W, C, F). To learn about F, think about what the fibrant objects have to be. It cannot be that fibrations are surjections, because then your model category would agree with the usual one on chain complexes. However, it could still be that F = all complexes. Just check via lifting. Then check that (W,C,F) satisfies the conditions of Mark Hovey and Jim Gillespie so that it defines a model structure. $\endgroup$ May 5, 2016 at 12:56
  • $\begingroup$ @DavidWhite, I think I don't know what are the cofibrant objects. I asked that all of them will be flat, but did not ask that every flat is cofibrant. $\endgroup$ May 5, 2016 at 13:00
  • $\begingroup$ The paper arxiv.org/abs/1412.4085 shows that if $k$ is coherent then there is a model structure on $k$-Mod where $M$ is cofibrant if and only if $M$ is Gorenstein flat. Perhaps you can generalize this to chain complexes. This paper discusses when Gorenstein flat is the same as flat: researchgate.net/publication/… $\endgroup$ May 5, 2016 at 16:41
  • $\begingroup$ There is a model structure on $k$-DGAs transferred from the standard projective model structure on non-positive $k$-chain complexes, with quasi-isomorphisms as weak equivalences and surjections in negative degree as fibrations. Since this is cofibrantly generated, it's easy to check and it looks like every cofibrant DGA is degreewise projective as a $k$-module, which would imply degreewise flat (if degreewise flat is what you mean). $\endgroup$ Sep 30, 2016 at 11:59

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