Let $HH^{\cdot}(\mathbb{F}_p)$ be the Hochschild cohomology of $\mathbb{F}_p$ over $\mathbb{Z}$, which as a $E_1$ ring is simply $\mathbb{F}_p[x]$ with $x$ in cohomological degree $2$. Then it's clear that $$ \mathbb{F}_p \otimes_{HH^{\cdot}(\mathbb{F}_p)} \mathbb{F}_p$$ has homotopy groups $\mathbb{F}_p$ in cohomological degree $0$ and $1$. However I would like to know the algebra structure on it. It has an algebra structure because $HH^{\cdot}(\mathbb{F}_p)$ has an $E_2$ ring structure. Now from https://arxiv.org/pdf/1207.3461.pdf it seems that having homotopy groups as above leaves open many possibilities for what the algebra could be. For example it could be $Hom_{\mathbb{F}_p[[t]]}(\mathbb{F}_p,\mathbb{F}_p)$ (which I consider unlikely) or $Hom_{A}(\mathbb{F}_p,\mathbb{F}_p)$ for $A$ a totally ramified extensions of $\mathbb{Z}_p$ (giving a different algebra for each $A$). However, I am not sure how I can tell which algebra structure is on it.
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$\begingroup$ $HC$ might be dangerous notation, as it collides with cyclic homology. $\endgroup$– Achim KrauseCommented Mar 27, 2021 at 10:56
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1$\begingroup$ I think it's $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{F}_p, \mathbb{F}_p)$, which ought to follow from the universal property of Hochschild cohomology giving you an action of HH^*(F_p/Z) on F_p through Z-algebra maps (and then 'barring' it up). $\endgroup$– Dylan WilsonCommented Aug 28, 2021 at 11:52
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