$\DeclareMathOperator\THH{THH}\DeclareMathOperator\HH{HH}$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 $p=2$.