## Distinguished affinoids

Fix a complete nonarchimedean field $K$ equipped with a fixed norm, with residue field $k$. Let $A$ be a $K$-affinoid algebra in the sense of classical rigid geometry. Here’s a funny definition I learned recently.

Definition. A surjection $\alpha : T_{n,K} \twoheadrightarrow A$ is distinguished if the associated residue norm $|\cdot|_\alpha$ equals the supremum seminorm $|\cdot|_{\mathrm{sup}}$. A $K$-affinoid algebra $A$ is distinguished if it admits a distinguished surjection from a Tate algebra.

Being distinguished imposes some obvious conditions on $A$: since the supremum seminorm is a norm iff $A$ is reduced, it certainly it implies
1) $A$ is reduced.
Since any residue norm takes values in $|K|$, it also implies
2) $|A|_{\mathrm{sup}} = |K|$.

If $K$ is stable (which holds if $K$ is discretely valued or algebraically closed), then the converse is true: 1) and 2) imply that $A$ is distinguished. Since 2) is automatic for $K$ algebraically closed, we see that any reduced $K$-affinoid is distinguished if $K$ is algebraically closed. It is also true that if $\alpha: T_{n,K} \to A$ is a distinguished surjection, then $\alpha^\circ: T_{n,K}^{\circ} \to A^\circ$ is surjective. Moreover, if $A$ satisfies 2), or $K$ is not discretely valued, then a surjection $\alpha: T_{n,K} \to A$ is distinguished iff $\alpha^\circ$ is surjective. Either way, if $A$ is distinguished then $A^\circ$ is a tft $K^\circ$-algebra.

All of this can be found in section 6.4.3 of BGR.

Question 1. If $A$ is reduced, is there a finite extension $L/K$ such that $A\otimes_K L$ is distinguished as an $L$-affinoid algebra?

This should be easy if it’s true. I didn’t think much about it.

Now suppose $A$ is distinguished, and let $\tilde{A} = A^{\circ} / A^{\circ \circ}$ be its reduction to a finite type $k$-algebra. As usual we have the specialization map $\mathrm{sp}: \mathrm{Sp}A \to \mathrm{Spec} \tilde{A}$. It is not hard to see that if $D(f) \subset \mathrm{Spec} \tilde{A}$ is a principal open, then $\mathrm{sp}^{-1}D(f)$ is a Laurent domain in $\mathrm{Sp}A$. Much less obvious is that for any open affine $U \subset \mathrm{Spec} \tilde{A}$, the preimage $\mathrm{sp}^{-1}U$ is an affinoid subdomain such that $A_U=\mathcal{O}(\mathrm{sp}^{-1}U)$ is distinguished and $\widetilde{A_U} = \mathcal{O}_{\mathrm{Spec} \tilde{A}}(U)$. This is buried in a paper of Bosch.

Loosely following Bosch, let us say an affinoid subdomain $V \subset \mathrm{Sp}A$ is formal if it can be realized as $\mathrm{sp}^{-1}U$ for some open affine $U \subset \mathrm{Spec} \tilde{A}$. Now let $X$ be a reduced quasicompact separated rigid space over $K$. Let us say a finite covering by open affinoids $U_1=\mathrm{Sp}A_1,\dots,U_n= \mathrm{Sp}A_n \subset X$ is a formal cover if
1) all $A_i$ are distinguished, and
2) for each $(i,j)$, the intersection $\mathrm{Sp}A_{ij}=U_{ij} := U_i \cap U_j$, which is automatically affinoid, is a formal affinoid subdomain in $U_i$ and in $U_j$.

This is a very clean kind of affinoid cover: we can immediately build a formal model for $X$ by gluing the tft formal affines $\mathrm{Spf}(A_i^\circ)$ along their common formal affine opens $\mathrm{Spf}(A_{ij}^\circ)$. Moreover, the special fiber of this formal model is just the gluing of the schemes $\mathrm{Spec}\widetilde{A_i}$ along the affine opens $\mathrm{Spec}\widetilde{A_{ij}}$.

Question 2. For $X$ a reduced qc separated rigid space over $K$, is there a finite extension $L/K$ such that $X_L$ admits a formal affinoid cover?