Proof. We begin with two (well-known) observations on the affinoid adic space $\mathrm {Spa}(A,A^\circ )$; the first concerns finding suitable rings of definition, while the second concerns a description via formal models.

First, we claim that there exists an open and bounded tfp $\mathcal {O}_K$-subalgebra $A_0 \subset A^0$ such that $\mathrm {Spec}(\tilde {A}) \to \mathrm {Spec}(A_0/tA_0)$ is a universal homeomorphism and such that $A^\circ$ is the integral closure of $A_0$ in $A$. In fact, the second part follows from [Reference Bosch, Güntzer and RemmertBGR84, Remark following 6.3.4/1]. For the first, using Noether normalization, we can choose a surjective map $T \to A$, where $T = K \langle x_1,\ldots,x_n \rangle$ is a Tate algebra. By [Reference Bosch, Güntzer and RemmertBGR84, 6.3.4/2], the map $\tilde {T} \to \tilde {A}$ is module finite. Setting $A'_0 = \mathrm {im}(T^\circ \to A^\circ )$, we learn that $A_0'/(A^{\circ \circ } \cap A_0') \to \tilde {A}$ is module finite and injective, with $A_0'$ being open, bounded, and tfp over $\mathcal {O}_K$. Enlarging $A_0'$ inside $A^\circ$ by adding finitely topological generators of $\tilde {A}$ over $A_0'$, we find an open bounded tfp $\mathcal {O}_K$-subalgebra $A_0 \subset A^\circ$ such that $A_0/(A^{\circ \circ } \cap A_0) \to \tilde {A}$ is bijective. But we also know that $\sqrt {tA_0} = A^{\circ \circ } \cap A_0$: any element of the right side is topologically nilpotent and in $A_0$, and must thus have a large enough power inside $tA_0$. Thus, we have found the subalgebra $A_0$ indicated at the start of this paragraph.

Next, we also recall an alternative description of the locally ringed space $(\mathrm {Spa}(A,A^\circ ), \mathcal {O}^+)$. Consider the category of all proper maps $f_i:X_i \to \mathrm {Spec}(A_0)$ of schemes which are isomorphisms after inverting $t$. For each $f_i$, let $X_{i,t=0} \subset X_i$ be the special fibre (regarded merely as a closed subset). Set $Z = \lim _i X_{i,t=0}$, so $Z$ is a spectral space. Let $\pi _i:Z \to X_i$ be the structure map, and define the structure sheaf $\mathcal {O}_Z$ of $Z$ via $\mathcal {O}_Z := \mathop {\mathrm {colim}}_i \pi _i^{-1} \mathcal {O}_{X_i}$. Then it is a basic fact that $\mathrm {Spa}(A,A^\circ ) = Z$ as topological spaces, and $\mathcal {O}^+$ identifies with the $t$-adic completion of $\mathcal {O}_Z$. For future use, we remark that, by passing to a cofinal subsystem, we may (and do) assume that each $X_i$ is $\mathcal {O}_K$-flat, i.e. $\mathcal {O}_{X_i}$ is $t$-torsionfree. This condition implies that the generic fibre $X_i[1/t]$ is dense in each $X_i$, and thus all the transition maps $X_i \to X_j$ in the system are surjective: their image is a closed set containing a dense open. In particular, $\mathcal {O}_Z$ is $t$-torsionfree. By spectrality, the maps $\pi _i:Z \to X_{i,t=0}$ are also surjective for all $i$. Finally, we also remark that $\mathcal {O}_Z$ is integrally closed in $\mathcal {O}_Z[1/t]$ by generalities on blowups.

Using the preceding two observations, we prove the equivalences.

$(1) \Rightarrow (2)$: Fix $x \in \mathrm {Spa}(A)$ with $\mathop {\mathrm {sp}}(x) \in \mathop {\mathrm {Spec}} \tilde {A}$ being a generic point. Using a suitable Noether normalization, we then learn that $\tilde {A}_{\mathop {\mathrm {sp}}(x)}$ is a rank-$1$ valuation ring with pseudouniformizer $t$: this ring is the integral closure of a rank-$1$ $t$-complete valuation ring in a finite extension of its fraction field. Any point $y \in \mathop {\mathrm {sp}}^{-1}(\mathop {\mathrm {sp}}(x))$ is represented by an equivalence class of maps $A_0 \to V$ where $V$ is a $t$-complete $t$-torsionfree valuation ring with the property that the closed point of $\mathrm {Spec}(V)$ is carried to $\mathop {\mathrm {sp}}(x)$. But any such map factors uniquely as $A_0 \to \tilde {A}_{\mathop {\mathrm {sp}}(x)} \to V$. Thus, taking $V$ to be the $t$-completion of $\tilde {A}_{\mathop {\mathrm {sp}}(x)}$ gives the unique such map up to equivalence, showing that $y=x$.

$(2) \Rightarrow (3)$: Write $x_i \in X_i$ for the image of $x$, so $\mathcal {O}^{+}_{\mathrm {Spa}(A,A^+),x}$ and $\mathop {\mathrm {colim}} \mathcal {O}_{X_i,x_i}$ identify after $t$-completion. Now if $x = \mathop {\mathrm {sp}}^{-1}(\mathop {\mathrm {sp}}(x))$, then $x_i = f_i^{-1}(\mathop {\mathrm {sp}}(x))$ is the unique preimage of $\mathop {\mathrm {sp}}(x)$. The map $A_{0,\mathop {\mathrm {sp}}(x)} \to \mathcal {O}_{X_i,x_i}$ is then integral (by properness of $X_i \to \mathrm {Spec}(A_0)$, base changed to $\mathrm {Spec}(A_{0,\mathop {\mathrm {sp}}(x)}$) and an isomorphism after inverting $t$. Taking a colimit, we learn that $A_{0,\mathop {\mathrm {sp}}(x)} \to \mathop {\mathrm {colim}}_i \mathcal {O}_{X_i,x_i}$ is integral and an isomorphism after inverting $t$. But the target is also integrally closed in its $t$-localization, so it must coincide with $A^{\circ }_{\mathop {\mathrm {sp}}(x)}$; here we implicitly use that $\mathrm {Spec}(\tilde {A}) \simeq \mathrm {Spec}(A_0/t)$ to identify $\mathop {\mathrm {sp}}(x)$ with a point of $\mathrm {Spec}(A^\circ )$, as well as the fact that $A^\circ$ is the integral closure of $A_0$. Thus, we have shown that the $t$-completion of $A^{\circ }_{\mathop {\mathrm {sp}}(x)} \to \mathcal {O}^{+}_{\mathrm {Spa}(A,A^+),x}$ is an isomorphism. As the $t$-completion of the target is $K_x^+$, the claim follows.

$(3) \Rightarrow (2)$: This is clear from the description of points of $\mathrm {Spa}(A,A^\circ )$ as equivalence classes of maps $A^\circ \to V$ to $t$-complete and $t$-torsionfree valuation rings.

$(2) \Rightarrow (1)$: Assume that $x$ is the unique preimage of $\mathop {\mathrm {sp}}(x)$ and yet that $\mathop {\mathrm {sp}}(x) \in \mathrm {Spec}(\tilde {A})$ is not a generic point; we shall obtain a contradiction. Let $\bar {y} \in \mathrm {Spec}(\tilde {A})$ be a generic point of an irreducible component $Y \subset \mathrm {Spec}(\tilde {A})$ containing $\mathop {\mathrm {sp}}(x)$. Applying the implications $(1) \Rightarrow (3)$ to a lift $y \in \mathrm {Spa}(A,A^+)$ of $\bar {y}$, we learn that $\bar {y}$ has a unique lift $y \in \mathrm {Spa}(A,A^+)$, and that $K_{y}^+$ is the $t$-completed local ring of $A^{\circ }_y$. In particular, the residue field of $K_{y}^+$ identifies with function field $K(Y)$ of the irreducible component $Y$. Choose a valuation subring $\bar {V} \subset K(Y)$ that is a $\tilde {A}$-algebra and has center $\mathop {\mathrm {sp}}(x) \in \mathrm {Spec}(\tilde {A})$; this valuation has rank $\geq 1$ as $\bar {y} \neq \mathop {\mathrm {sp}}(x)$. Let $V \subset K_{y}^+$ be the preimage of $\bar {V}$. Then $V$ is a $t$-complete valuation ring of rank $\geq 2$, is an $A^\circ$-algebra, and has center $\mathop {\mathrm {sp}}(x)$ on $\mathrm {Spec}(\tilde {A})$. The resulting map $A^\circ \to V$ then gives a point $x' \in \mathrm {Spa}(A,A^\circ )$ with rank $\geq 2$ and image $\mathop {\mathrm {sp}}(x)$ in $\mathrm {Spec}(\tilde {A})$. But $x$ is the unique preimage of $\mathop {\mathrm {sp}}(x)$, so $x = x'$. We now obtain a contradiction as $x$ had rank $1$ by assumption, while $x'$ has rank $\geq 2$ by construction.

$(1) \Longleftrightarrow (4)$: This is proven in [Reference BerkovichBer90, 2.4.4].

Proof. By definition, any open affinoid $U \subset X$ contains a weakly Shilov point.

Proof. $(1) \Longleftrightarrow (2)$: This follows by combining Lemme 4.4 and Corollaire 4.15 in [Reference PoineauPoi13].

$(3) \Rightarrow (1)$: If there exists a formal model $\mathfrak {X}$ as in (3), then taking $W \subset X$ be the preimage of any formal affine open $\mathfrak {X}$ containing the image of $x$ shows that $x$ is weakly Shilov.

$(1) \Rightarrow (3)$: Assume $x \in X$ is weakly Shilov. We can then find an affinoid open $\mathrm {Spa}(R) \subset X$ containing $x$ such that $x \in \mathrm {Spa}(R)$ is Shilov. By Raynaud, there is a formal model $\mathfrak {X}$ of $X$ such that $\mathrm {Spa}(R) \subset X$ is the preimage of a formal affine open $\mathrm {Spf}(R_0) \subset \mathfrak {X}$, cf. [Reference BoschBos14, Theorem 8.4.3]. By passing to a refinement, we can assume that the map $R_0/t \to \tilde {R}$ gives a homeomorphism of spectra; see first paragraph of the proof of Proposition 2.2. As $x$ is weakly Shilov, its image in $\mathrm {Spec}(\tilde {R})$ is a generic point. But then its image in $\mathrm {Spec}(R_0/t)$ is also a generic point since $\mathrm {Spec}(\tilde {R}) \to \mathrm {Spec}(R_0/t)$ is a homeomorphism by construction. As $\mathrm {Spf}(R_0) \subset \mathfrak {X}$ is a formal open immersion, it follows that $x$ gives a generic point of $\mathfrak {X}_s$ as well

Proof. As $Z \subset X$ is a nowhere dense Zariski closed set, we have $\dim _x(Z) < \dim _x(X)$ for all $x \in X$. The first statement now follows from the characterization of weak Shilov points in Proposition 2.9(2). The second statement follows from the first statement applied to the locus $Z \subset X$ of points that are not regular, which is a nowhere dense Zariski closed set by excellence considerations.

Notation 2.11 Fix a complete non-archimedean field $K$ with valuation ring $V \subset K$ and a pseudouniformizer $t \in V$. For any map $A \to B$ of $V$-algebras, write $L^{\rm an}_{B/A}$ for the derived $t$-completion of $L_{B/A}$. A tft (or topologically finite type) $V$-algebra is a $V$-algebra $A$ of the form $V[x_1,\ldots,x_n]^{\wedge }/I$ (where the completion is $t$-adic). If moreover $I$ is finitely generated, we say that $A$ is tfp. The class of tfp $V$-algebras has good properties such as coherence and classical $t$-adic completeness, see [Reference Gabber and RameroGR03, Proposition 7.1.1]. Moreover, any tft $V$-algebra which is $t$-torsion-free is in fact tfp, see [Reference Fujiwara, Gabber and KatoFGK11, Corollary 7.3.6].

Proof. Part (1) is [Reference Gabber and RameroGR03, Theorem 7.1.31]. Part (2) is the key assertion checked in the proof of [Reference Gabber and RameroGR03, Theorem 7.2.39]. Part (3) is [Reference Gabber and RameroGR03, Theorem 7.1.29].

Proof. For (1), by noetherian approximation, we can write $V \to R$ as the base change of a finitely presented flat map $V_0 \to R_0$ of finitely generated $\mathbf {Z}$-algebras along some map $V_0 \to V$. Then $L_{R_0/V_0}$ is pseudocoherent, and $L_{R_0/V_0} \otimes _{R_0}^L R \simeq L_{R/V}$ by Tor independent base change, so $L_{R/V}$ is also pseudocoherent.

For (2), observe that if $R$ is $V$-finite, then $R$ is a finite free $V$-module (as any finitely presented flat $V$-module is finite free). In particular, a pseudocoherent $R$-complex is also pseudocoherent as a $V$-complex. The claim now follows as any pseudocoherent $V$-complex is derived $t$-complete (since $V$ itself is so).

Proof. For (1), we argue as follows. Since $W_0 \subset E^\circ$, choosing monic equations defining generators of $W_0$ shows that any such $W_0$ is a finitely presented flat $V$-algebra. Lemma 2.13 then implies that $L_{W_0/V}$ is pseudocoherent and thus already derived $t$-complete, so $L_{W_0/V} \simeq L^{\rm an}_{W_0/V}$. The last part of (1) follows by inverting $t$ and noting that formation of cotangent complexes commutes with localization.

Part (2) follows from (1) and a general fact: the cotangent complex of any extension of fields only has homology in degrees $1$ and $0$. Indeed, this follows from transitivity triangles and the fact that any field is ind-smooth over its prime subfield (which is perfect) by generic smoothness in algebraic geometry.

Proof. [Reference QuillenQui, Corollary 10.5] shows that $\mathfrak {m}$ is generated by a regular sequence if $H_2(L_{k/A}) = 0$. As the maximal ideal of a noetherian local ring is generated by a regular sequence exactly when the ring is regular, the claim follows.

Proof. The question is local, so we can assume $M = A^{\oplus n}$ is a free module. Let $x = (x_1,\ldots,x_n) \in M$ lie in the kernel of $M \to N$. The hypothesis implies that $x_i \in \mathfrak {m}$ for every maximal ideal $\mathfrak {m} \subset A$. But the intersection of all maximal ideals of $A$ is its nilradical (as $A$ is Jacobson) which is $0$ (as $A$ is reduced). So $x = 0$, as wanted.

Notation 2.17 Fix a complete non-archimedean field $C$ with valuation ring $\mathcal {O}_C \subset C$ and a pseudouniformizer $t \in \mathcal {O}_C$.Footnote ³

Proof. Our strategy is to first simplify $T$, and then argue that $R_V[1/t]$ is regular using the cotangent complex.

First, observe that the statement of the theorem is Zariski local around $\eta \in \mathrm {Spf}(T)$, so we may shrink $\mathrm {Spf}(T)$ around $\eta$ to assume $\mathrm {Spec}(T/t)$ is irreducible with generic point $\eta$. If $T' = \mathcal {O}_C \langle x_1,\ldots,x_n \rangle \to T$ is a finite injective map, then $\mathrm {Spec}(T/t) \to \mathrm {Spec}(T'/t)$ is finite and surjective. As both these schemes are irreducible, the image of the generic point $\eta \in \mathrm {Spec}(T/t)$ is the generic point $\eta ' \in \mathrm {Spec}(T'/t)$, and $\eta$ is the unique preimage of $\eta '$. In particular, we must have $T \widehat {\otimes }_{T'} V' \simeq V$ by base change. But then $R \widehat {\otimes }_{T'} V' \simeq R_V$. Thus, we may replace $T$ with $T'$ to assume that $T$ is a standard Tate algebra over $\mathcal {O}_C$. In particular, $T$ is formally smooth over $\mathcal {O}_C$. In this case, we have $T = T[1/t]^{\circ }$, so $V$ is actually a $t$-complete and $t$-torsionfree rank-$1$ valuation ring (either by explicit calculation, or as explained in Proposition 2.2).

Next, we describe $L^{\rm an}_{V/\mathcal {O}_C}$. By definition, the ring $V$ is a $t$-completed ind-Zariski localization of $T$. As the formation of the $t$-completed cotangent complex is compatible with $t$-completed localizations and $t$-completed filtered colimits, we learn that $L^{\rm an}_{V/\mathcal {O}_C} \simeq L^{\rm an}_{T/\mathcal {O}_C} \widehat {\otimes }_T^L V$. But $T/\mathcal {O}_C$ is formally smooth, so $L^{\rm an}_{T/\mathcal {O}_C}$ is a finite projective $T$-module placed in degree $0$, whence $L^{\rm an}_{V/\mathcal {O}_C}$ is a finite free $V$-module placed in degree $0$.

We now begin proving the theorem. Since $V$ is a $t$-complete and $t$-torsionfree rank-$1$ valuation ring, the ring $K := V[1/t]$ is a nonarchimedean field extension of $C$ with $K^\circ = V$. As $R_V$ is a tfp $V$-algebra, we can (and will) regard $R_V[1/t]$ as an affinoid $K$-algebra. To show regularity of $R_V[1/t]$, it is enough to show that the local rings of $R_V[1/t]$ at all closed points are regular: the regular locus in $R_V[1/t]$ is open (as affinoid $K$-algebras are noetherian and excellent) and the maximal ideals are dense (as affinoid $K$-algebras are Jacobson). A closed point is given by a quotient $R_V[1/t] \to E$ where $E/K$ is a finite extension; fix such a point. By Theorem 2.15, it suffices to show that $L_{E/R_V[1/t]}$ has homology only in degrees $0,1$. (In fact, there is no $H_0$ as $R_V[1/t] \to E$ is surjective, but it will be convenient to formulate things this way.) Let $W_0 \subset E^\circ$ be the image of $R_V$ under this map. As $E^\circ$ is the integral closure of $V$ in $E$, the ring $W_0$ is a finitely presented finite flat $V$-algebra. Since the map $R_V \to W_0$ is a surjection of tfp $V$-algebras, we have $L_{W_0/R_V} \simeq L^{\rm an}_{W_0/R_V}$ by Theorem 2.12(3), and hence $L_{E/R_V[1/t]} \simeq L^{\rm an}_{W_0/R_V}[1/t]$ by inverting $t$, so it is enough to show that $L^{\rm an}_{W_0/R_V}[1/t]$ has homology only in degrees $0,1$. Consider the transitivity triangle for $\mathcal {O}_C \to R_V \to W_0$:

(2.1)\begin{equation} L^{\rm an}_{R_V/\mathcal{O}_C} \widehat{\otimes}_{R_V}^L W_0 \to L^{\rm an}_{W_0/\mathcal{O}_C} \to L^{\rm an}_{W_0/R_V}. \end{equation}

As $R \to R_V$ is a $t$-completed Zariski localization, we have

\[ L^{\rm an}_{R/\mathcal{O}_C} {\otimes}_R^L R_V \simeq L^{\rm an}_{R/\mathcal{O}_C} \widehat{\otimes}_R^L R_V \simeq L^{\rm an}_{R_V/\mathcal{O}_C}, \]

where the first isomorphism is from the pseudocoherence of $L^{\rm an}_{R/\mathcal {O}_C}$ coming from Theorem 2.12(1). The same reasoning also allows us to drop the completion on the leftmost term in (2.1). Inverting $t$ in (2.1) then gives a triangle

(2.2)\begin{equation} L^{\rm an}_{R_V/\mathcal{O}_C}[1/t] {\otimes}_{R_V[1/t]}^L E \to L^{\rm an}_{W_0/\mathcal{O}_C}[1/t] \to L^{\rm an}_{W_0/R_V}[1/t]. \end{equation}

Our task was to show that the term on the right has homology only in degrees $0,1$. The term on the left is a finite projective $E$-module in degree $0$: indeed, $L^{\rm an}_{R/\mathcal {O}_C}[1/t]$ is a finite projective $R[1/t]$-module by Theorem 2.12(2) and the smoothness assumption on $\mathrm {Spa}(R[1/t])$, and we have $L^{\rm an}_{R/\mathcal {O}_C} \otimes _{R[1/t]}^L R_V[1/t] \simeq L^{\rm an}_{R_V/\mathcal {O}_C}[1/t]$ by the reasoning explained above. By the long exact sequence, it thus suffices to check that $L^{\rm an}_{W_0/\mathcal {O}_C}$ has homology only in degrees $0,1$. For this, we consider transitivity triangle for $\mathcal {O}_C \to V \to W_0$:

\[ L^{\rm an}_{V/\mathcal{O}_C} \widehat{\otimes}_V^L E \to L^{\rm an}_{W_0/\mathcal{O}_C}[1/t] \to L^{\rm an}_{W_0/V}[1/t]. \]

Now $L^{\rm an}_{V/\mathcal {O}_C}[1/t]$ is a finite free $K$-module as explained previously, so the term on the left is a finite free $E$-module. It remains to observe that $L^{\rm an}_{W_0/V}[1/t]$ has homology in degrees $0,1$ by Corollary 2.14.

Proof. ‘Only if’ is clear. Conversely, suppose $\bar {f}$ is flat. If $J$ is nilpotent (e.g. if $A$ is Artinian), then flatness of $f$ follows from Proposition 0.8.3.7 in [Reference Fujiwara and KatoFK18].

In the general case, write $A$ as a filtered colimit $A \simeq \mathop {\mathrm {colim}}_{i \in I} A_i$ where $A_i \subset A$ is a filtered system of $\mathbf {Z}$-algebras of finite type. Set $J_i = J \cap A_i$, so $J_i \subset A_i$ is a nilpotent ideal. By standard approximation arguments, there is an index $i_0 \in I$ such that $f$ is the base change along $A_{i_0} \to A$ of a map $f_{i_0} : R_{i_0} \to S_{i_0}$ of finitely presented $A_{i_0}$-algebras. For all $i \geq i_0$, write $f_i : R_i \to S_i$ for the evident base change of $f_{i_0}$. By two applications of [Reference GrothendieckGro66, Théorème 11.2.6], $R_i$ and $S_i$ are flat $A_i$-algebras for all sufficiently large $i$.

Next, note that $\bar {f}$ is the colimit of the diagrams $\overline {f_i}: R_i / J_i R_i \to S_i / J_i S_i$. Since $\bar {f}$ is flat by assumption, $\overline {f_i}$ is flat for all sufficiently large $i$ by another application of [Reference GrothendieckGro66, Théorème 11.2.6]. But $J_i$ is nilpotent, so flatness of $\overline {f_i}$ implies flatness of $f_i$ by the special case of the lemma treated in the first paragraph of the proof. Therefore $f_i$ is flat for all sufficiently large $i$, so $f$ is flat.

Proof. ‘Only if’ is clear. For the converse, choose a finitely generated ideal of definition $I \subset A$ contained in $J$; let $\overline {J}_n \subset A/I^n$ be the image of $J$, so $\overline {J}_n$ is locally nilpotent. Since $R/JR \to S/JR$ is flat, the previous lemma implies that $R/I^nR \to S/I^n S$ is flat for all $n\geq 1$. By Corollary 0.8.3.9 in [Reference Fujiwara and KatoFK18], we then deduce that $R \to S$ is flat.

Proof. By (the proof of) Theorem 1.3 in [Reference Bosch, Lütkebohmert and RaynaudBLR95], we can find a finite étale Galois extension $K'/K$ such that the unit ball $A_{K'}^{\circ } \subset A_{K'} \overset {\rm def}{=} A \otimes _K K'$ is topologically finitely presented over $K'^\circ$ and the special fiber $A_{K'}^{\circ } / K'^{\circ \circ } A_{K'}^{\circ }$ is (geometrically) reduced. Choose an open tfp $K'^\circ$-algebra $B_0 \subset B \otimes _K K'$ such that $(f \otimes _K K')(A_{K'} ^\circ ) \subset B_0$. Let $k'$ be the residue field of $K'^\circ$. Then $A_{K'}^{\circ } \otimes _{K'^\circ } k' \to B_0 \otimes _{K'^\circ } k'$ is a map of finite-type $k'$-algebras with reduced source, so there exists a non-zero-divisor $f \in A_{K'}^{\circ } \otimes _{K' \circ } k'$ such that $(A_{K'}^{\circ } \otimes _{K'^\circ } k')[1/f] \to (B_0 \otimes _{K'^\circ } k')[1/f]$ is flat. Choose any lift $\tilde {f} \in A_{K'}^{\circ }$, and let $C$ be the $\pi$-adic completion of $A_{K'}^{\circ }[1/\tilde {f}]$; similarly, let $D$ be the $\pi$-adic completion of $B_0[1/\tilde {f}]$. Applying the previous proposition with $A=K'^\circ$, $J = K'^{\circ \circ }$, $R=C$, and $S=D$, we deduce that the map $C \to D$ is flat. Then $\mathop {\mathrm {Spa}} C[1/\pi ] \to \mathop {\mathrm {Spa}} A_{K'}$ is the inclusion of the Laurent domain $U({1}/{\tilde {f}})$, and $\mathop {\mathrm {Spa}} B_{K'} \times _{\mathop {\mathrm {Spa}} A_{K'}} U({1}/{\tilde {f}}) \cong \mathop {\mathrm {Spa}} D[1/\pi ]$ by design, so $\mathop {\mathrm {Spa}} B_{K'} \times _{\mathop {\mathrm {Spa}} A_{K'}} U({1}/{\tilde {f}}) \to U({1}/{\tilde {f}})$ is flat. Moreover, $U({1}/{\tilde {f}})$ contains all the Shilov points of $\mathop {\mathrm {Spa}} A_{K'}$ by construction.

It remains to undo the base change from $K$ to $K'$. For this, let $h \in A^\circ$ be the image of $\tilde {f}$ under the norm map $A_{K'} \to A$, and let $U=U({1}/{h}) \subset \mathop {\mathrm {Spa}} A$ be the associated Laurent domain. We claim that $U$ satisfies the conclusions of the theorem. Indeed, writing $\pi : \mathop {\mathrm {Spa}} A_{K'} \to \mathop {\mathrm {Spa}} A$ for the evident finite étale map, it is clear from the definitions that $\pi ^{-1}(U) = \cap _{g \in \mathrm {Gal}(K'/K)} U({1}/{\tilde {f}})g$ as subsets of $\mathop {\mathrm {Spa}} A_{K'}$. Since each $g$-translate of $U({1}/{\tilde {f}})$ still contains all Shilov points of $\mathop {\mathrm {Spa}} A_{K'}$, we see that $\pi ^{-1}(U)$ contains all Shilov points of $\mathop {\mathrm {Spa}} A_{K'}$, and so $U$ contains all Shilov points of $\mathop {\mathrm {Spa}} A$. Moreover, the results in the previous paragraph show that $f_U: \mathop {\mathrm {Spa}} B \times _{\mathop {\mathrm {Spa}} A} U \to U$ becomes flat after base change along the surjective finite étale map $\pi ^{-1}(U) \to U$, so $f_U$ is flat by [Reference WarnerWar17, Proposition 3.1.12]. This concludes the proof.

Proof. The first part follows immediately from Theorem 2.18 thanks to the characterization of Shilov points in Proposition 2.2 once one observes that the local rings of $\mathrm {Spa}(R)$ for a regular tft $K$-algebra $R$ are regular. The last statement follows as a rigid analytic space over a characteristic zero non-archimedean field is smooth exactly when all of its local rings are regular.

Proof. This follows from (the proof of) Lemma 2.9.2 in [Reference WarnerWar17].

Proof. Let $y \in Y$ be any weakly Shilov point. By Theorem 2.21, we can choose some open subset $U(y) \subset Y$ containing $y$ such that $X \times _Y U(y) \to U(y)$ is flat. Moreover, by Theorem 2.25, the entire fiber $X_y$ is smooth over $y$. Applying Lemma 2.26 and the openness of the smooth locus in the source, we deduce that every point $x \in f^{-1}(y)$ admits a quasicompact open neighborhood $W_x$ in $X$ such that $W_x \to Y$ is smooth. Forming a suitable union of the $W_x$ and using the quasicompacity of $f$, we deduce that there is a quasicompact open neighborhood $W \subset X$ of the fiber $f^{-1}(y)$ such that $W \to Y$ is smooth. Shrinking $U(y)$ further, we may assume by an easy quasicompacity argument that $f^{-1}(U(y)) \subset W$. In particular, $X \times _Y U(y) \to U(y)$ is smooth.

Since weakly Shilov points are dense in $Y$, setting $U = \cup _{y\ \mathrm {weakly\ Shilov}} U(y)$ concludes the proof.

Proof. Let $W \subset X$ be the Zariski-open subset where $X \to Y$ is smooth, and let $Z = X-W$ be the complement regarded as a rigid space with its induced reduced structure. The composite morphism $g:Z \to Y$ is proper, so by Kiehl's results $g_{\ast } \mathcal {O}_Z$ is a coherent sheaf on $Y$. Since $\mathrm {Supp}(g_{\ast } \mathcal {O}_Z)=f(Z)$, we deduce that $f(Z) \subset Y$ is Zariski-closed, so then $S_f=Y - f(Z)$ is Zariski-open, and density follows from the previous theorem.

Warning 3.2 Let us record some subtleties concerning this notion.

(1) Given a Zariski-open immersion $j:U \to X$ and a Zariski-constructible sheaf $\mathscr {F}$ on $U$, the extension $j_! \mathscr {F}$ can fail to be Zariski-constructible $X$, unlike the situation in algebraic geometry (see next example). The main problem is that the operation of taking Zariski-closures in $X$ of Zariski-closed subsets of $U$ is poorly behaved in general (e.g. it does something non-trivial over $U$); this issue does not arise if $\mathscr {F}$ is itself locally constant.

Proof. Parts (1)–(3) are proved in [Reference HansenHan20]. For (4), one direction is clear. For the other direction, let $f_0: \tilde {X} \to X$ be the normalization of $X$, and let $f_1: \tilde {X} \times _X \tilde {X} \to X$ be the evident map. The hypothesis guarantees that $f_0^\ast \mathscr {F}$ (and then also $f_1^{\ast } \mathscr {F}$) is Zariski-constructible, since $\tilde {X} = \coprod _i \tilde {X}_i$ is the disjoint union of normalizations of the irreducible components of $X$. Since $f_0$ and $f_1$ are both finite, pushforward along these maps preserves Zariski-constructibility by (3). The exact sequence $0 \to \mathscr {F} \to f_{0 \ast } f_0^{\ast } \mathscr {F} \to f_{1 \ast } f_1^{\ast } \mathscr {F}$ now exhibits $\mathscr {F}$ as the kernel of a map between Zariski-constructible sheaves, so we conclude by (1).

Proof. By induction on $\dim X$ and devissage, it is enough to show that if $j:U \to X$ is a dense Zariski-open and $\mathscr {F}$ is lisse and killed by a prime $\ell$, then $j_!{\mathscr {F}}$ lies in the claimed subcategory. For this, choose (as in [Sta18, Tag 0A3R]) a finite étale cover $g:U'\to U$ of prime-to-$\ell$ degree such that $g^\ast \mathscr {F}$ is an iterated extension of copies of $\mathbf {F}_\ell$. Then $\mathscr {F}$ is a summand of $g_\ast g^\ast \mathscr {F}$, so $\mathscr {F}$ is a summand of an iterated extension of copies of $g_{\ast }\mathbf {F}_\ell$. Now extend $g$ to a finite cover $f:X' \to X$ as in [Reference HansenHan20], so $j_!\mathscr {F}$ is a summand of an iterated extension of copies of $\mathscr {G} = j_!g_{\ast }\mathbf {F}_\ell = f_{\ast } j'_!\mathbf {F}_\ell$, where $j':U' \to X'$ is the evident map. Letting $i:Z\to X'$ be the complement of $j'$, the exact sequence $0 \to f_{\ast } j'_!\mathbf {F}_\ell \to f_{\ast } \mathbf {F}_\ell \to (f \circ i)_\ast \mathbf {F}_\ell \to 0$ shows that $\mathscr {G}$ lies in the desired subcategory.

Proof. (1) This is exactly [Reference HansenHan20, Theorem 1.8].

(2) The full faithfulness is a special case of [Reference HansenHan20, Theorem 1.10.ii]. Essential surjectivity in the case where $\mathcal {X} / \mathcal {S}$ is proper can be checked on hearts. We can also assume that $\mathcal {X}$ and $X$ are reduced. Arguing by induction on $\dim \mathcal {X}$ as in the proof of [Reference HansenHan20, Theorem 1.7], one reduces to checking that any sheaf on $X$ of the form $j_! \mathscr {F}$ is in the essential image of $(-)^{\rm an}$; here $j:U \subset X$ is the inclusion of any normal Zariski-open subset and $\mathscr {F}$ is any lisse sheaf. To do this, first note that the complement $Z = X-U$ algebraizes to a closed subscheme $\mathcal {Z} \subset \mathcal {X}$ by relative rigid GAGA, so then $\mathcal {U} = \mathcal {X} - \mathcal {Z}$ is an algebraization of $U$. We're now reduced to proving that the analytification functor $\mathrm {F\acute {E}t}(\mathcal {U}) \to \mathrm {F\acute {E}t}(U)$ is an equivalence of categories. We explain the construction of an essential inverse. Suppose $V \to U$ is any finite étale map. By [Reference HansenHan20, Theorem 1.6], this extends uniquely to a branched covering $V' \to X^n$, where $X^n$ is the normalization of $X$. By relative rigid GAGA again, this algebraizes to a branched covering $\mathcal {V}' \to \mathcal {X}^n$, and then $\mathcal {V}:= \mathcal {V}' \times _{\mathcal {X}^n} \mathcal {U} \to \mathcal {U}$ is the desired algebraization of $V$.

(3) Since we already have full faithfulness, it suffices to prove essential surjectivity on the hearts, i.e. we want to realize a lisse sheaf on $X$ as the analytification of a unique lisse sheaf on $\mathcal {X}$. By uniqueness and Zariski/analytic descent for lisse sheaves in algebraic/analytic geometry, we may assume that $\mathcal {X}$ is separated (or even affine). By finite descent for lisse sheaves in both algebraic and analytic geometry, we may also assume $\mathcal {X}$ is normal. In this case, we can realize $\mathcal {X}$ as an open subscheme of a normal proper $\mathcal {S}$-scheme $\bar {\mathcal {X}}$; the argument used in the proof of (2) now yields the desired algebraization.

Proof. As local constancy is a local property, the collection of all opens $V \subset X$ such that $\mathscr {F}|_V$ is locally constant is stable under taking unions. Taking the union of all such opens then gives the maximal open $U_{\mathscr {F}}$ such that $\mathscr {F}|_{U_{\mathscr {F}}}$ is open. It is also clear from this description that $U_{\mathscr {F}}$ agrees with the set in (2). The set in (2) is trivially contained in the set in (1). Conversely, as the functor sending a scheme $Y$ to its category of locally constant sheaves (respectively, constructible sheaves) is locally finitely presented (i.e. carries cofiltered limits of affine schemes to direct limits), the set in (1) is also contained in the set in (2), so the two sets coincide. The final statement is clear from the description of $U_{\mathscr {F}}$ given by the set in (1).

Proof. Let $f:Y \to X$ be a universally generalizing map of schemes. We have an obvious containment $f^{-1} (U_{\mathscr {F}}) \subset U_{f^* \mathscr {F}}$, and we must show it is an equality. Assume towards contradiction that there exists some $y \in U_{f^* \mathscr {F}} - f^{-1}(U_{\mathscr {F}})$. Thus, $(f^* \mathscr {F})|_{Y_y}$ is locally constant but $\mathscr {F}|_{X_{f(y)}}$ is not locally constant. There is then a specialization $x_1 \rightsquigarrow x_2$ of geometric points of $X_{f(y)}$ such that the corresponding cospecialization map $\mathscr {F}_{x_2} \to \mathscr {F}_{x_1}$ is not an isomorphism. Choose an absolutely integrally closed valuation ring $V$ and a map $\eta _X:\mathrm {Spec}(V) \to X_{f(y)}$ that witnesses the specialization $x_1 \rightsquigarrow x_2$, so $\eta _X^*(\mathscr {F}|_{X_{f(y)}})$ is not locally constant. Now that the map $Y_y \to X_{f(y)}$ is universally generalizing (it factors as $Y_y \to X_{f(y)} \times _X Y \to X_{f(y)}$, with both maps being universally generalizing) and surjective (all points of $X_{f(y)}$ specialize to $y$, so surjectivity follows from the universally generalizing property). By stability of universally generalizing surjective maps under base change, we can replace $V$ with an extension if necessary to lift $\eta _X$ to a map $\eta _Y:\mathrm {Spec}(V) \to Y_y$. But then we have $\eta _X^* (\mathscr {F}|_{X_{f(y)}}) = \eta _Y^* (f^* \mathscr {F}|_{Y_y})$; this is a contradiction as the left side is not locally constant by choice of $\eta _X$, while the right side is locally constant by choice of $y$.

Proof. By Theorem 3.5, we may assume $Y=\mathrm {Spa}(A)$ for an affinoid $K$-algebra $A$. In particular, $X$ and $Y$ are both qcqs. We may clearly also assume that $Y$ is reduced. Our task is to show that $Rf_*$ preserves $D^{\rm b}_{\rm zc}(-,\mathbf {Z}/n)$ in this situation. As $X$ is quasi-compact, we may apply Proposition 3.6, so it suffices to show that $Rf_* \mathbf {Z}/n \in D^{\rm b}_{\rm zc}$. In fact, as the fibres of $f$ have bounded dimension (e.g. by checking on formal models), we know by cohomological dimension estimates and proper base change that $Rf_*$ has finite cohomological dimension, so it is enough to show that $Rf_* \mathbf {Z}/n \in D_{\rm zc}$, i.e. that each $R^i f_*\mathbf {Z}/n$ is Zariski-constructible on $Y$. By proper base change and induction on $\dim (Y)$, it suffices to find a dense open $U \subset Y$ such that $(R^if_* \mathbf {Z}/n)|_U$ is locally constant. As $K$ has characteristic $0$, Temkin's [Reference TemkinTem18, Theorem 1.1.13(i)] gives a proper hypercover $\epsilon :X^\bullet \to X$ with each $X^i$ being $K$-smooth. Cohomological descent enables us to compute $Rf_* \mathbf {Z}/n$ as the totalization of $Rg_* \mathbf {Z}/n$, where $g = f \circ \epsilon$. As $\mathcal {H}^i$ of the totalization of a cosimplicial object $K(\bullet ):\Delta \to \mathcal {D}^{\geq 0}$ only depends on truncated cosimplicial object $K(\bullet )|_{\Delta _{\leq i+1}}$, we can replace $X^\bullet$ with the finite diagram $X^{\leq i+1}$ and then by each $X^j$ to assume that $X$ is smooth. As $Y$ is reduced, our generic smoothness result (Theorem 2.29) yields a Zariski-dense Zariski-open $U \subset X$ such that $f:X \to Y$ is smooth over $U$. It then suffices to show that $R^i g_* \mathbf {Z}/n$ is locally constant when $g$ is both proper and smooth. To check this, we may assume $n=\ell$ is a prime. Now if $\ell \neq p$, the claim reduces to [Reference HuberHub96, Corollary 6.2.3], while the claim for $\ell = p$ reduces to [Reference Scholze and WeinsteinSW20, Theorem 10.5.1].

Proof. The statement is local on the target by Theorem 3.5, so we may assume $Y=\mathrm {Spa}(A)$ is affinoid. By assumption, we can factor $f$ as $X \xrightarrow {j} \bar {X} \xrightarrow {g} Y$ with $j$ being a Zariski-open immersion and $g$ being proper. Using Theorem 3.10, we can reduce to the case $f=j$ is a Zariski-open immersion. The claim for $Rf_!$ is clear from the definition of Zariski-constructible sheaves (using $\{X,\bar {X}-X\}$ as the stratification on $\bar {X}$ witnessing Zariski-constructibility), so it remains to check the assertion for $Rf_*$. As $Y$ is affinoid, the Zariski-open immersion $f:X \hookrightarrow Y$ is the algebraization of a unique open immersion $g:\mathcal {X} \to \mathcal {Y} = \mathrm {Spec}(A)$. Moreover, any object of $D^{\rm b}_{\rm lis}(X,\mathbf {Z}/n)$ is the analytification of a unique object in $D^{\rm b}_{\rm lis}(\mathcal {X},\mathbf {Z}/n)$ by Proposition 3.7(3). Using the compatibility of pushforwards with analytification from Proposition 3.7(1), the claim follows from Gabber's constructibility theorem in [Reference Illusie, Laszlo and OrgogozoILO14, Expose XIII, Théorème 1.1.1].

Proof. By Theorem 3.5, the assertion is étale-local on $Y$, so we may assume both $X$ and $Y$ are affinoid, corresponding to a finite map $A \to B$ of affinoid $K$-algebras. Fix $F \in D^{\rm b}_{\rm zc}(Y,\mathbf {Z}/n)$. Let $S_F \subset Y$ be the smallest Zariski-closed subset of $Y$ containing the support of $F$. We shall prove the claim by induction on $d_F = \dim (S_F)$.

If $d_F = 0$, then $F$ is supported at finitely many points. As the claim is étale local on $Y$, we may then assume that $F$ is a finite direct sum of sheaves of the form $k_* \mathbf {Z}/n$, where $k:W \to Y$ is the inclusion of a Zariski-closed point. Now $f^!$ and $k_*$ commute: the corresponding statement for left adjoints is the proper base change theorem for $f$. We are thus reduced to checking the statement when $Y$ (and thus $X$) are $0$-dimensional. In this case, up to universal homeomorphisms, the map $f$ is finite étale, so $Rf^! = f^*$, so the claim is clear.

Now assume $d_F > 0$. We may then choose a Zariski open subset $j:U \hookrightarrow Y$ such that $U \cap S_F$ is dense is $S_F$, the restriction $L := F|_{U \cap S_F}$ is lisse, and $f$ is finite étale (up to universal homeomorphisms) over $U$: one can find such an open $U$ as the algebraization of an open $\mathcal {U} \subset \mathrm {Spec}(A)$ satisfying the analogous properties for the map $\mathrm {Spec}(B) \to \mathrm {Spec}(A)$ and the algebraization $\mathcal {F}$ of $F$ (in the sense of Proposition 3.7). Let $i:Z \hookrightarrow Y$ be the closed complement, so we have the standard exact triangle

\[ i_* Ri^! F \to F \to Rj_* (F|_U). \]

Now $Rj_*(F|_U) \simeq k_* Rj'_* L$, where $j':U \cap S_F \to S_F$ and $k:S_F \to X$ are the natural maps (and thus Zariski open and Zariski closed immersions respectively). By Corollary 3.11, the third term in the triangle above is then Zariski-constructible. The remaining term $i_* Ri^! F$ in the triangle is then also Zariski-constructible, so $Ri^! F$ is itself Zariski-constructible. Applying $Rf^!$ to the above triangle gives a triangle

\[ Rf^!i_* i^! F \to f^! F \to Rf^! k_* Rj'_* L. \]

By proper base change for $f$ as in the previous paragraph, the last term identifies with $k_{X,*} Rj'_{X,*} (f|_{U \cap S_F})^! L$, where $k_X$ and $j'_X$ are the base changes of $k$ and $j'$ along $f$. As $f$ is finite étale up to universal homeomorphisms over $U$, we have $(f|_{U \cap S_F})^! L \simeq (f|_{U \cap S_F})^* L$, so this object is lisse on $f^{-1}(U \cap S_F)$. Corollary 3.11 then implies that the third term in the triangle above is lisse. For the first term, using proper base change again lets us write it as $i_{X,*} f_Z^! i^! F$, where $i_X$ and $f_Z$ are the base changes of $i$ and $f$ against $f$ and $i$. As $i^! F$ is known to be Zariski constructible, the induction hypothesis then shows that the first term is also Zariski constructible.