Equivalence of topological Hochschild homology and Mac Lane homology via an equivalence $QA\simeq HA \wedge_{\mathbb{S}} H\mathbb{Z}$

Question

Mac Lane homology is a homology theory for (not necessarily commutative) rings. Given a ring $A$, Eilenberg and Mac Lane define its cubical construction $QA$ to be a certain connective chain complex, whose homology is isomorphic to the stable homology of Eilenberg-Mac Lane spaces: $$H_i(QA)\cong H_{i+j}(K(A,j)) \text{ for } j \gg i.$$In fact, $QA$ is a dg-ring, and it comes with a ring map $QA \to A$ inducing an isomorphism on $H_0$. You can find the construction and basic properties of the cubical construction in [1] or [2, Chapter 13]. One then defines the Mac Lane homology of $A$ to be the Hochschild homology $$HML_i(A):= HH_i(QA,A)$$and similarly the Mac Lane cohomology is $HML^i(A):=HH^i(QA,A)$.

In 1992, Pirashvili and Waldhausen [3] proved that $HML_i(A)\cong THH_i(A)$, where the right-hand side is topological Hochschild homology. The proof goes by identifying both of them with a functor homology group. Subsequently in 1995, Fiedorowicz-Pirashvili-Schwaenzl-Vogt-Waldhausen [1] outlined a `brave new algebra' proof of this fact. In modern terms, they noticed that the Pirashvili-Waldhausen result would follow if one has a stable equivalence of $H\mathbb{Z}$-algebra spectra $$HQA \simeq HA \wedge_{\mathbb{S}} H\mathbb{Z}.$$Here, $H$ denotes the Eilenberg-Mac Lane functor taking dgas to $H\mathbb{Z}$-algebra spectra, and $\mathbb{S}$ is the sphere spectrum. (For concreteness, my preferred model of spectra is symmetric spectra in simplicial sets.) Indeed, if one has such a stable equivalence, then it follows by base change results that $HML(A)\simeq THH(A)$, for both homology and cohomology.

My question: Is it known that there is a stable equivalence $HQA \simeq HA \wedge_{\mathbb{S}} H\mathbb{Z}$ of $H\mathbb{Z}$-algebra spectra?

This may be a fact known to experts in $THH$, but I was unable to find anything in the literature. One can almost get there: because the homology of $QA$ is the stable cohomology of Eilenberg-Mac Lane spaces, it follows that one has $$\pi_i(HQA)\cong \pi_i(HA \wedge_{\mathbb{S}} H\mathbb{Z}).$$Moreover, both are $H\mathbb{Z}$-module spectra, hence wedges of Eilenberg-Mac Lane spectra, and so an isomorphism on $\pi_*$ actually lifts to a stable equivalence $HQA \simeq HA \wedge_{\mathbb{S}} H\mathbb{Z}$ of $H\mathbb{Z}$-module spectra. This argument already appears in [1], but it wasn't clear to the authors then, and it's certainly not clear to me now, how to lift this to a stable equivalence of $H\mathbb{Z}$-algebra spectra. I guess because $QA$ is connective, in theory one could write down both sides of the equation and match up the algebra structure somehow, but I am not sure if this is very tractable.

Thanks for your time.

References:

[1] Fiedorowicz, Z.; Pirashvili, T.; Schwänzl, R.; Vogt, R.; Waldhausen, F. Mac Lane homology and topological Hochschild homology. Math. Ann. 303 (1995)

[2] Jean-Louis Loday, Cyclic homology, Springer 1998

[3] Pirashvili, Teimuraz; Waldhausen, Friedhelm. Mac Lane homology and topological Hochschild homology. J. Pure Appl. Algebra 82 (1992)

Answer 1

The answer to your question is yes, these two $H\mathbb{Z}$-algebras are equivalent (and if $A$ is commutative, they are equivalent as commutative $H\mathbb{Z}$-algebras). I am in fact in the process of writing this up with Maxime Ramzi. Hopefully the paper will be finished in a week or 2.

Update : The paper has now appeared on arXiv