Let $k$ be a finite field, $THH(k)$ its topological Hochschild homology spectrum. For essentially formal reasons, we know that it's an $E\_\infty$-algebra over the Eilenberg-Mac Lane spectrum $Hk$, and so can be modeled by an $E\_\infty$-dga over $k$.


Question: Is this $E\_\infty$-structure equivalent to a strictly commutative multiplication?


We know that the homotopy algebra $\pi\_\* THH(k)$ is a polynomial ring on a generator in degree $2$, so in particular it cannot be equivalent to a simplicial commutative $k$-algebra (for then it would support a divided-power structure relative to its ideal of positive-degree elements.) So in a sense a positive answer to the question above would say that $THH(k)$ is "as commutative as it could possibly be" given its homotopy algebra.



The multiplication on $THH$ of finite fields
This is not the case, and you can use Dyer-Lashof operations to see so. In the following I'll show this for $k = \Bbb F\_2$ because that's the easiest case to compute. Bokstedt proved that $THH\_\*(\Bbb F\_2) = \Bbb F\_2[\sigma]$ where $|\sigma| = 2$. If it came from a commutative DGA, then the only nonzero Dyer-Lashof operations would be the squaring operations (because the operad action would factor through the projection down to the bottom class). We'll show that $Q^6 \sigma = \sigma^4$.


The description of $THH(k)$ by the cyclic bar construction makes it the derived smash product $k \wedge\_{k \wedge^{\Bbb L} k^{op}} k$. The associated Kunneth spectral sequence comes from a simplicial commutative $k$-algebra, which gives the associated spectral sequence compatible Dyer-Lashof operations. It sits inside the Bokstedt spectral sequence (this is how Bokstedt computed the hidden extensions to show $THH\_\*(\Bbb F\_2) = \Bbb F\_2[\sigma]$). In particular, this spectral sequence takes the form of a Tor-spectral sequence
$$
Tor\_{A\_\*} (\Bbb F\_2, \Bbb F\_2) \Rightarrow THH\_\*(\Bbb F\_2)
$$
where $A\_\*$ is the dual Steenrod algebra $\Bbb F\_2[\xi\_i]$. The Tor-groups form an exterior algebra generated by $I/I^2$, where $I$ is the augmentation ideal of the dual Steenrod algebra. The exterior algebra is therefore generated by $[\xi\_i]$.


However, Steinberger's calculations from the $H\_\infty$ book tell us that
$$Q^6 \xi\_1 =\xi\_1^7 + \xi\_1^4 \xi\_2 + \xi\_1 \xi\_2^2 + \xi\_3,$$
and therefore in $THH(k)$ we have an identity
$$Q^6 [\xi\_1] = [\xi\_3]$$
because everything else is decomposable. The spectral sequence degenerates and so $Q^6$ on the generator in degree two hits the only nonzero thing in degree 8.


