# Scholze and Weinstein's $\operatorname{Spa}\mathbf{Z}_p\times \operatorname{Spa}\mathbf{Z}_p$

In their Berkeley Lectures, to motivate the introduction of Diamonds, Scholze and Weinstein discuss what should be the definition of $$\operatorname{Spa}\mathbf{Z}_p\times \operatorname{Spa}\mathbf{Z}_p$$. The analogy is made with the "equal characteristic" setting, where the counterpart of $$\mathbf{Z}_p$$ is $$\mathbf{F}[\![t]\!]$$ for a certain finite field $$\mathbf{F}$$ and an indeterminate $$t$$. Then $$\operatorname{Spa} \mathbf{F}[\![t]\!]\times_{\operatorname{Spa} \mathbf{F}}\operatorname{Spa} \mathbf{F}[\![t]\!]$$ equals $$\operatorname{Spa} \mathbf{F}[\![t,u]\!]$$ where we named $$u$$ the right-hand indeterminate.

In that regard, why isn't $$\operatorname{Spa}\mathbf{Z}_p\times \operatorname{Spa}\mathbf{Z}_p$$ something as simple as $$\operatorname{Spa} W_p(\mathbf{Z}_p)$$ ? (Sorry if this is a naive question).

EDIT: I shall add more precisions to my question. As quoted from the Berkeley Lectures, $$\operatorname{Spa}\mathbf{Z}_p\times \operatorname{Spa}\mathbf{Z}_p$$ should contain $$\operatorname{Spa}\mathbf{Q}_p\times \operatorname{Spa}\mathbf{Q}_p$$ as a dense subset, so Scholze and Weinstein first construct the latter (on page 5). By analogy with the function fields side where $$\operatorname{Spa} \mathbf{F}(\!(t)\!)\times_{\operatorname{Spa} \mathbf{F}}\operatorname{Spa} \mathbf{F}(\!(t)\!)$$ is the punctured open unit disc over $$\mathbf{F}(\!(t)\!)$$, they consider the punctured open unit disc $$D_{\mathbf{Q}_p}^*$$ over $$\mathbf{Q}_p$$. Next comes two steps that I do not understand:

1. They consider the limit $$\tilde{D}_{\mathbf{Q}_p}=\varprojlim D_{\mathbf{Q}_p}$$ where the transition maps are $$x\mapsto (x+1)^p-1$$.

2. They consider the quotient $$\tilde{D}_{\mathbf{Q}_p}^*/\mathbf{Z}_p^{\times}$$.

In definition 1.2.1, they define $$\operatorname{Spa}\mathbf{Q}_p\times \operatorname{Spa}\mathbf{Q}_p$$ as $$\tilde{D}_{\mathbf{Q}_p}^*/\mathbf{Z}_p^{\times}$$. But why to not stop at $$D_{\mathbf{Q}_p}^*$$?

• what would be the symmetry corresponding to swapping the two factors? Sep 5 at 11:02
• Thank @DylanWilson for your comment, that is a good point (and sorry for the late reply). If I dig further in the text, the analogy is made with shtukas where, in the equal char. case, the two copies of $\mathbf{F}[\![t]\!]$ do not seem to play the same role (one is the coefficients, the other is the base). Shtukas involve this map $\varphi$ which acts as the Frobenius morphism on one side and as the identity on the other. So I wouldn't have guess that total symmetry would have been required somehow. Sep 9 at 7:43