What is the topological Hochschild cohomology of $\mathbb{F}_p$?

Question

Following the computation of the THH (topological Hochschild homology) of $\mathbb{F}_p$ as outlined in Krause-Nikolaus. We use the fact that $\mathbb{F}_p$ is initial $E_2$ ring with $0=p$ to compute

$$\mathbb{F}_p \otimes_{\mathbb{S}} \mathbb{F}_p \cong \mathbb{F}_p[{\Omega^2 S^3}]$$

Then,

$$THC(\mathbb{F}_p) \cong \mathrm{Hom}_{\mathbb{F}_p[{\Omega^2 S^3}]}(\mathbb{F}_p,\mathbb{F}_p)$$

Now using

$$\mathrm{colim}_{BG} {G} = \{ \star\}$$

We have that

$$\mathrm{colim}_{\Omega S^3} {\mathbb{F}_p[{\Omega^2 S^3}]} = \mathbb{F}_p$$

and hence,

$$THC(\mathbb{F}_p) \cong \lim_{\Omega S^3}{\mathbb{F}_p} \cong \mathbb{F}_p^{\Omega^2 S^3}$$

where the action is trivial because $\mathbb{F}_p$ is discrete. Assuming the above is correct, I now am not sure how to compute this.

Answer 1

Let me write $HH^S(B) = THH(B) = B \wedge_{B^e} B$ for topological Hochschild homology, and $HH_S(B) = F_{B^e}(B, B)$ for topological Hochschild cohomology, where $B^e = B \wedge_S B^{op}$. For $B$ commutative the $B^e$-module action on $B$ factors through $\mu : B^e \to B$, so by adjunction we have $F_{B^e}(B, B) \cong F_B(B \wedge_{B^e} B, B)$. Hence $HH_S(B) \cong F_B(HH^S(B), B)$. For $B = H\mathbb{F}_p$ we have $\pi_* HH^S(B) = HH^S_*(B) = \mathbb{F}_p[x]$ with $|x|=2$ a primitively generated Hopf algebra, so dually $HH_S^{-*}(B) = \pi_* HH_S(B) = Hom_{\mathbb{F}_p}(\mathbb{F}_p[x], \mathbb{F}_p) = \Gamma_{\mathbb{F}_p}(\xi)$ is a divided power algebra with $\xi$ dual to $x$.

Closer to the approach you outline: The final lim calculates the mod $p$ cohomology of $\Omega S^3$, which is the Hopf algebra dual to the mod $p$ homology of $\Omega S^3$. Either one can be calculated with the Wang sequence, or Serre spectral sequence, or by reference to the James construction.