TITLE: What is the topological Hochschild cohomology of $\mathbb{F}_p$?
QUESTION [6 upvotes]: 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.

REPLY [8 votes]: 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.