Strict graded commutativity of $\pi_*(\operatorname{THH}(A))$? $\DeclareMathOperator\THH{THH}\DeclareMathOperator\HH{HH}$A version of the strict graded commutativity (i.e. graded commutativity & $x^2=0$ for every homogeneous element $x$ of odd degree) of $\pi_*(\THH(A))$ seems to be used in Construction 6.8 of Bhatt–Morrow–Scholze, Topological Hochschild homology and integral $p$-adic Hodge theory to establish a form of HKR theorem due to Hesselholt (as indicated). Let me briefly summarize what is happening.
Let $R$ be a perfectoid $\mathbb Z_p$-algebra (for example, a perfect field of characteristic $p$), and $A$ an $R$-algebra. The goal is to produce a map $(\Omega_{A/R}^*)_p^\wedge\to\pi_*(\THH(A;\mathbb Z_p))$ of graded $p$-complete $A$-module.
The key step is to produce a map $\Omega_{A/\mathbb Z}^*\to\pi_*(\THH(A))$ of graded commutative $A$-algebras. The map $\THH(A)\to\HH(A):=\HH(A/\mathbb Z)$ becomes an equivalence after truncation $\tau_{\le2}$, and in particular, $\pi_1(\THH(A))\cong\pi_1(\HH(A))\cong\Omega_{A/\mathbb Z}^1$. Now they claim that, since $\pi_*(\HH(A))$ is strictly graded commutative as $\HH(A)$ is an animated $A$-algebra, the map $\Omega_{A/\mathbb Z}^1\to\pi_1(\THH(A))$ extends to a map $\Omega_{A/\mathbb Z}^*\to\pi_*(\THH(A))$ by the universal property of the exterior product.
If I understand correctly, in order to construct such a map, one needs the strict commutativity of $\THH(A)$, not that of $\HH(A)$. However, $\THH(A)$ is no longer the underlying $\mathbb E_\infty$-ring of an animated ring in general.
Update to clarify: no, my understanding was incorrect. The strict commutativity of $\THH$ is not needed to produce the map. Only that of $\HH$ is needed. See Tyler Lawson's answer. In retrospect, the text is clear and it is I who am stupid.
In fact, by Bökstedt's periodicity, the graded ring $\pi_*(\THH(\mathbb F_p))$ is isomorphic to $\mathbb F_p[u]$ where $u$ is of degree $2$, but for animated rings, every element of $\pi_2$ has divided powers. In particular, if $\THH(\mathbb F_p)$ is the underlying $\mathbb E_\infty$-ring of an animated ring, this implies that $u^p=p!v$ for some $v\in\pi_{2p}(\THH(\mathbb F_p))$, which implies that $u^p=0$, contradiction (this argument also shows that $\THH(R;\mathbb Z_p)$ is not an animated ring, therefore neither is $\THH(R)$).
Here are my questions:

*

*In this setting, is it true that $\pi_*(\THH(A))$ is strictly graded commutative?

*More generally, let $A$ be an animated ring. Is it true that $\pi_*(\THH(A))$ is strictly graded commutative?

Of course, these questions are trivial except when $p=2$.
 A: To my knowledge, there is no such result for THH of a commutative ring. (Could be?)
However, we need less for this result; we only need degree 1 elements to square to zero. Every element in $\pi_1 THH(A)$ is a finite linear combination of elements $a \cdot \sigma(b)$ for $a, b$ in $A$, where $\sigma$ is the circle action. Therefore it suffices to show all elements in degree 1 square to zero in the case where $A$ is a finite polynomial algebra over $\Bbb Z$, or more generally a monoid algebra.
If $M$ is a commutative monoid, then there is an equivalence of ring spectra
$$
THH(\Bbb Z[M]) \simeq THH(\Bbb Z) \otimes Z(M)_+
$$
where $Z(M)$ is the cyclic bar construction, a simplicial commutative monoid. This agrees through degree 2 with
$$
\Bbb Z \otimes Z(M)_+ \simeq HH(\Bbb Z[M])
$$
because $THH(\Bbb Z) \to \Bbb Z$ is a split map of ring spectra and an equivalence through degree 2.
Therefore, it suffices to show that all elements in $HH_1$ square to zero, and in this case it is true due to coming from a simplicial commutative ring.
