$\DeclareMathOperator\THH{THH}\DeclareMathOperator\HH{HH}$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:

  1. In this setting, is it true that $\pi_*(\THH(A))$ is strictly graded commutative?
  2. 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 $p=2$.

  • 5
    $\begingroup$ "Of course, these questions are meaningful only when $p=2$" Given how vanishingly and hopelessly lost I feel even after reading only the first sentence of the post, I'm comforted to know that the questions are meaningful to someone in some situations :-) $\endgroup$ Jan 10 at 21:39
  • $\begingroup$ @Carl-FredrikNybergBrodda I wanted to say that, since "strictness" is automatic when $p\neq2$, these questions are meaningless except when $p=2$. I have modified that sentence. $\endgroup$
    – Z. M
    Jan 11 at 7:09
  • $\begingroup$ @ZM My comment was not a criticism, just a remark on how difficult this material appears to me. But the fact that the questions are now "trivial" in some cases makes me even more depressed... :-) $\endgroup$ Jan 11 at 11:25

1 Answer 1


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.

  • $\begingroup$ Thanks. I was too stupid to realize that it suffices to square the first degree elements. For that, one could simply invoke the connectivity of $\THH\to\HH$. $\endgroup$
    – Z. M
    Jan 11 at 7:43
  • $\begingroup$ Yes, so long as you are using Shukla homology / THH relative to Z. $\endgroup$ Jan 11 at 16:11

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