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Let $A$ be a differential graded algebra over a commutative ring $R$. Suppose that $H_*(A)=0$, i.e. $A$ is acyclic.

Question: Does this imply that the Hochschild homology $HH_*(A)$ also vanishes identically? If not, what hypotheses does one need to impose on $R$ and $A$? (In the case I am mostly interested in, $R$ is a polynomial ring in one variable.)

Remark: if we assume that $R$ is a field, then it's a general fact that one can cook up an $A_{\infty}$-algebra structure on $H_*(A)$ such that $A \to H_*(A)$ is an $A_{\infty}$ quasi-isomorphism. Moreover, $A_{\infty}$-algebra quasi-isomorphisms are invertible and $HH_*(-)$ is functorial. So it follows that that $HH_*(A)=0$ is $H_*(A)=0$. So the result should be true in this case.

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  • $\begingroup$ I guess the answer depends on how precisely define the Hochschild homology over a general ring $R$ (i.e., are you using the derived tensor product or not?). $\endgroup$
    – Denis Nardin
    Commented Apr 19, 2020 at 8:05
  • $\begingroup$ I think I'd like to define HH_* just using the bar complex (i.e. the same way as I would defined it over a field). However, I'm happy to assume that A is flat over R, so I presume I don't have to derive anything? (In fact, in the case I'm interested in, A is a free associative algebra and R is a polynomial ring in one variable.) $\endgroup$
    – user155668
    Commented Apr 19, 2020 at 8:26
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    $\begingroup$ If you use the derived tensor product (which coincides with the classical one if $A$ is flat over $R$), then the definition of $HH_\ast(A)$ is manifestly invariant under quasi-isomorphism of algebras. In particular the map $A\to 0$ is a quasi-isomorphism... $\endgroup$
    – Denis Nardin
    Commented Apr 19, 2020 at 8:29
  • $\begingroup$ @DenisNardin Sorry, but while it's clear that a morphism of dg-algebras $A \to B$ induces a morphism $HH_*(A) \to HH_*(B)$, it's not clear why a quasi-isomorphism of dg-algebras induces an isomorphism on Hochschild homology. Maybe I'm using a bad definition for $HH_*(-)$? Could you say a few words about why this is clear? $\endgroup$
    – user155668
    Commented Apr 19, 2020 at 17:51
  • $\begingroup$ Both derived tensor products and geometric realizations of simplicial objects preserve quasi-isomorphisms, and the bar complex is obtained by composing the two operations. I'm not sure I can say much more than what is in Weibel's Homological algebra. $\endgroup$
    – Denis Nardin
    Commented Apr 19, 2020 at 18:58

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Hochschild homology is a derived invariant, and in particular a quasi-isomorphism invariant, since quasi-isomorphic algebras are derived invariant. Your algebra is non-unital, I assume, since $1$ is usually a non-trivial cycle in general.

In any case, however, you can consider the Hochschild cyclic complex of $A$, call it $C_*(A)$, which has an internal differential coming from $A$ and an external one coming from the usual Hochschild differential. This is a bicomplex and so there is a spectral sequence with second page $HH_*(H_*(A))$ that converges (perhaps you need to check this part carefully) to $HH_*(A)$. The Hochschild homology of $k$ is $k$, and hence so is that of $A$. If $A$ is non-unital and $H_*(A) = 0$, you would get that $HH_*(A) = 0$, too.

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  • $\begingroup$ K[x]/x^2 with dx=1 is a unital dg algebra that is acyclic $\endgroup$
    – Marco Farinati
    Commented Jul 12, 2020 at 22:39
  • $\begingroup$ @MarcoFarinati I guess I am too used to considering weight graded algebras. :) $\endgroup$
    – Pedro
    Commented Jul 13, 2020 at 9:59

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