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. 



Hochschild homology of acyclic complex
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.


