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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 $D_{\mathbf{Q}_p}^*$?

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    $\begingroup$ what would be the symmetry corresponding to swapping the two factors? $\endgroup$ Sep 5 at 11:02
  • $\begingroup$ 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. $\endgroup$
    – Stabilo
    Sep 9 at 7:43

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