What does $\mathrm{Spét}(H\mathbb{Z})\times_{\mathrm{Spét}(\mathbb{S})}\mathrm{Spét}(H\mathbb{Z})$ look like?

Question

See also: What does $\mathrm{Spec}(\mathbb{Z})\times_{\mathrm{Spec}(\mathbb{F}_{1})}\mathrm{Spec}(\mathbb{Z})$ look like?.

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One of the most mysterious objects in mathematics is the elusive "field with one element", and coming with it is the arithmetic curve $\mathrm{Spec}(\mathbb{Z})\times_{\mathrm{Spec}(\mathbb{F}_{1})}\mathrm{Spec}(\mathbb{Z})\cong\mathrm{Spec}(\mathbb{Z}\otimes_{\mathbb{F}_{1}}\mathbb{Z})$. I've heard people compare algebraic geometry over the field with one element a bit with spectral algebraic geometry (e.g. here), and (from what I understand) the analogue of $\mathrm{Spec}(\mathbb{Z})\times_{\mathrm{Spec}(\mathbb{F}_{1})}\mathrm{Spec}(\mathbb{Z})$ there should be the spectral scheme $\mathrm{Spét}(H\mathbb{Z})\times_{\mathrm{Spét}(\mathbb{S})}\mathrm{Spét}(H\mathbb{Z})$.

What does the tensor product $H\mathbb{Z}\wedge_{\mathbb{S}}H\mathbb{Z}$ of the Eilenberg Mac Lane $E_\infty$-rings associated to $\mathbb{Z}$ look like?

And $\mathrm{Spét}(H\mathbb{Z})\times_{\mathrm{Spét}(\mathbb{S})}\mathrm{Spét}(H\mathbb{Z})$?

Answer 1

Just like in classical algebraic geometry, we have a natural equivalence of spectral Deligne--Mumford stacks $$ \DeclareMathOperator{\Spet}{Spét} \Spet (\mathbf{Z}\underset{\mathbf{S}}{\otimes}\mathbf{Z})\xrightarrow{\simeq} \Spet\mathbf{Z} \underset{\Spet \mathbf{S}}{\times} \Spet \mathbf{Z}, $$ which you can find in Proposition.1.4.11.1.(3) in Lurie's ''Spectral Algebraic Geometry''. The universal property of the Cartesian product on the right is perhaps not super exciting: mapping into this product from an $X$ is just a choice of two maps $f,g\colon X\to \Spet \mathbf{Z}$ (as there is a unique map $X\to \Spet \mathbf{S}$).

As far as I am aware, people think of the homotopy groups of $\mathbf{Z}\otimes_{\mathbf{S}}\mathbf{Z}$ as quite complicated. A hint towards this ''fact'' is that $\mathbf{F}_p\otimes_{\mathbf{S}}\mathbf{F}_p$ is the mod $p$ dual Steenrod algebra. Maybe one could use Bockstein spectral sequences to try and calculate the homotopy groups of $\mathbf{Z}\otimes_{\mathbf{S}}\mathbf{Z}$ from this fact, but even the integral cohomology of things like $K(\mathbf{Z}/2,n)$ is pretty complicated (see Theorem 10.4 of May's ``A general approach to Steenrod operations''). One could also think about using a Tor spectral sequence to calculate $\pi_\ast \mathbf{Z}\otimes_{\mathbf{S}}\mathbf{Z}$, but once one tries to write down an $E_2$-page, one realises this is unfeasible.

However, general stable homotopy theory tells us that $\pi_0 \mathbf{Z}\otimes_{\mathbf{S}}\mathbf{Z}\simeq \mathbf{Z}$, as more generally we have a natural equivalence of abelian groups $$\pi_0 M\underset{R}{\otimes} N\simeq \pi_0 M\underset{\pi_0 R}{\otimes} \pi_0 N$$ for an $E_1$-ring $R$ and left module $N$ and right module $M$, where $R,M$, and $N$ only have homotopy groups in nonnegative degree. We also know that $\Spet R$ has underlying DM-stack $\Spet \pi_0 R$. This says that the underlying classical DM-stack for $\Spet (\mathbf{Z}\underset{\mathbf{S}}{\otimes}\mathbf{Z})$ is just $\Spet\mathbf{Z}$; all the extra information is ``higher'' information.

(This is more of a comment than an answer...I'd also be interested if anyone else had more to add!)