As the title suggests, i'm trying to understand this calculation of the algebra structure on $HH(\mathbb{F}_p)$, which I will outline below:

We can calculate $HH(\mathbb{F}_p)$ as $$\mathbb{F}_p \otimes^{L}_{\mathbb{F}_p \otimes^L_{\mathbb{Z}} \mathbb{F}_p} \mathbb{F}_p,$$

So first we need to understand $\mathbb{F}_p \otimes^L_{\mathbb{Z}} \mathbb{F}_p$, which is easy since tensoring the multiplication by $p$ resolution $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{F}_p \to 0$ lets us see $A := \mathbb{F}_p \otimes^L_{\mathbb{Z}} \mathbb{F}_p = \mathbb{F}_p[\varepsilon]/\varepsilon^2$.

Here's the part I feel like I really don't understand:

Then the claim is that we can resolve $\mathbb{F}_p$ as an $A$-algebra by:

$$A\langle x \rangle = A[x_1, x_2, ....]/(x_ix_j = \binom{i + j}{i}x_{i + j}),$$ where $dx_i = \varepsilon x_{i - 1}$, and $|x_i| = 2i$.

As a chain complex I can see that this is just the multiplication by $\varepsilon$ resolution:

$$... \rightarrow \mathbb{F}_p[\varepsilon]/\varepsilon^2 \xrightarrow{\varepsilon} \mathbb{F}_p[\varepsilon]/\varepsilon^2 \xrightarrow{\varepsilon} \mathbb{F}_p[\varepsilon]/\varepsilon^2 \xrightarrow{\varepsilon \mapsto 0} \mathbb{F}_p,$$

but I don't understand why this is the correct algebra structure on this resolution. For instance, if I didn't quotient by the PD relations what would prevent me from concluding that $HH(\mathbb{F}_p) = \mathbb{F}_p[x]$ as opposed to the correct $\mathbb{F}_p\langle x \rangle$?


Well, I think i've figured it out and i'll leave an answer in case anyone is curious:

According to the theorem stated in this stackexchange post we just need to show that the dga above is a resolution and then it will give us the algebra structure.

First, from staring at the resolution at the level of modules $$... \to \mathbb{F}_p[\varepsilon]/\varepsilon^2 \to \mathbb{F}_p[\varepsilon]/\varepsilon^2 \to \mathbb{F}_p[\varepsilon]/\varepsilon^2 \to \mathbb{F}_p$$

we can see that in each degree we have $\mathbb{F}_p \cdot y$ for some generator $y$; this is where the most of my confusion was I think since the above chain complex is slightly misleading! The term $\mathbb{F}_p[\varepsilon]/\varepsilon^2$ is not homogenous, since $\varepsilon$ has degree 2! Really the picture is:

$$.... \to \mathbb{F}_p \cdot x_2 \to \mathbb{F}_p \cdot x_1\varepsilon \to \mathbb{F}_p \cdot x_1 \to \mathbb{F}_p \cdot \varepsilon \to \mathbb{F}_p \to \mathbb{F}_p$$

To see that it's a resolution, let's go degree by degree and see what we need the differential to do in order for it to be exact.

0, 1: in these degrees the map is forced on us by requiring it to be a map of $\mathbb{F}_p[\varepsilon]/\varepsilon^2$ algebras; $1 \mapsto 1$ and $\varepsilon \mapsto 0$.

2: In degree 2 we have a generator $x_1$, and the kernel of $d_1$ is generated by $\varepsilon$, so for this to be exact we must have $x_1 \mapsto \varepsilon$.

3: In degree 3 the generator is $x_1\varepsilon$, and the Leibniz rule forces $d(x_1\varepsilon) = x_1d(\varepsilon) + \varepsilon d(x_1) = 0 + 0$.

4: In degree 4 we have the generator $x_2$ and since the kernel of the degree 3 stuff is generated by $x_1\varepsilon$ we must have $d x_2 = \varepsilon x_1$.

This is the first degree where something interesting happens, namely $x_1^2$ also has degree 4, and since the degree 4 piece is generated by $x_2$ this means $x_1^2 = ax_2$ for some $a \in \mathbb{F}_p$. To calculate what $a$ is we can use the $\mathbb{F}_p$ linearity of $d$, which tells us that $d(x^2_1) = 2x_1d(x_1) = 2x_1\varepsilon$, and $d(ax_2) = ad(x_2) = ax_1\varepsilon$; so $a = 2 = \binom{2}{1}$.

This is where the divided powers come from!

Then one can keep going and prove in general that all of the variables have divided power relations in a similar way.


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