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?

One thought on “Distinguished affinoids”

  1. Hi David,

    Question 1) is true whenever A is absolutely reduced, i.e., for every complete field extension $K^\prime$ of $K$, the base change $A^\prime := A \widehat{\otimes}_K K^\prime$ is reduced. See Lemma 2.7 in https://link.springer.com/article/10.1007/BF02052727 . In fact, in this case, there exists a finite Galois extension $L/K$ such that $A \otimes_K L$ is distinguished. See Lemma 1.3 in https://arxiv.org/pdf/2309.14542.pdf. Every smooth affinoid algebra of pure dimension is absolutely reduced.

    I am not sure about the second question.

    Arun

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