Patching over analytic fibers and the local–global principle

Abstract

As a starting point for higher-dimensional patching in the Berkovich setting, we show that this technique is applicable around certain fibers of a relative Berkovich analytic curve. As a consequence, we prove a local–global principle over the field of overconvergent meromorphic functions on said fibers. By showing that these germs of meromorphic functions are algebraic, we also obtain local–global principles over function fields of algebraic curves defined over a class of (not necessarily complete) ultrametric fields, thus generalizing the results of Mehmeti(Compos Math 155:2399–2438, 2019).

Appendices

Appendices

The author believes that most of the results in the Appendices are known to the mathematical community working with Berkovich spaces, but she did not manage to find references for them.

1.1 Appendix I: The sheaf of meromorphic functions

As in the complex setting, a sheaf of meromorphic functions can be defined satisfying similar properties. Moreover, its definition resembles heavily that of the sheaf of meromorphic functions for schemes (including the subtleties of the latter, see [24]). See [26, 7.1.1] for a treatment of meromorphic functions in the algebraic setting.

Let k denote a complete ultrametric field.

Definition 8.2

Let X be a good k-analytic space. Let ${\mathcal {S}}_X$ be the presheaf of functions on X,  which associates to any analytic domain U the set of analytic functions on U whose restriction to any affinoid domain in it is not a zero-divisor. Let ${\mathscr {M}}_{-}$ be the presheaf on X that associates to any analytic domain U the ring ${\mathcal {S}}_X(U)^{-1}{\mathcal {O}}_X(U).$ The sheafification ${\mathscr {M}}_X$ of the presheaf ${\mathscr {M}}_{-}$ is said to be the sheaf of meromorphic functions on X.

It is immediate from the definition that for any analytic domain U of X, ${\mathcal {S}}_X(U)$ contains no zero-divisors of ${\mathcal {O}}_X(U).$

Proposition 8.3

Let X be a good k-analytic space. Let U be an analytic domain of X.

• (1) ${\mathcal {S}}_X(U)=\{f \in {\mathcal {O}}_X(U): f \ \text {is a non-zero-divisor in} \ {\mathcal {O}}_{U,x} \ \text {for all} \ x\in U\}.$

• (2) ${\mathcal {S}}_X(U)=\{f \in {\mathcal {O}}_X(U): f \ \text {is a non-zero-divisor in} \ {\mathcal {O}}_U(G) \ \text {for any open subset} G \ \text {of} \ U\}.$

Proof

(1) By a direct application of the definition, the elements of ${\mathcal {S}}_X(U)$ are non-zero-divisors on ${\mathcal {O}}_{U,x}$ for all $x \in U.$

Let $f \in {\mathcal {O}}_X(U)$ be such that f is a non-zero-divisor in ${\mathcal {O}}_{U,x}$ for all $x\in U.$ This means that ${\mathcal {O}}_{U,x} \rightarrow {\mathcal {O}}_{U,x}, a \mapsto f \cdot a$, is an injective map for $x \in U.$

Let V be any affinoid domain in U. By [13, 4.1.11], for any $x \in V,$ the morphism ${{\mathcal {O}}_{U,x} \rightarrow {\mathcal {O}}_{V,x}}$ is flat. Consequently, the map ${\mathcal {O}}_{V,x} \rightarrow {\mathcal {O}}_{V,x}, b \mapsto f\cdot b,$ is injective, or equivalently, f is a non-zero-divisor in ${\mathcal {O}}_{V,x}.$ Suppose there exists $c \in {\mathcal {O}}_U(V)$ such that $f \cdot c=0.$ Then $c=0$ in ${\mathcal {O}}_{V,x}$ for all $x \in V,$ implying $c=0$ in ${\mathcal {O}}_U(V).$ As a consequence, f is a non-zero-divisor in ${\mathcal {O}}_U(V).$ We have shown that $f \in {\mathcal {S}}_X(U),$ concluding the proof of the first part of the statement.

Finally, (2) is a direct consequence of (1). $\square$

Lemma 8.4

Let X be a good k-analytic space. Let U be an affinoid domain in X. Then ${\mathcal {S}}_X(U)$ is the set of non-zero divisors of ${\mathcal {O}}_X(U).$

Proof

By definition, the elements of ${\mathcal {S}}_X(U)$ are not zero-divisors in ${\mathcal {O}}_X(U).$

Let f be an element of $A_U{:}{=}{\mathcal {O}}_X(U)$ that is a non-zero-divisor, i.e. such that the map $A_U \rightarrow A_U,$ $a \mapsto f \cdot a$, is injective. Let $V \subseteq U$ be any affinoid domain. Set $A_V{:}{=}{\mathcal {O}}_X(V).$ Then, by [1, Proposition 2.2.4(ii)], the restriction map $A_U \rightarrow A_V$ is flat. Consequently, the map $A_V \rightarrow A_V, b \mapsto f \cdot b,$ remains injective, meaning f is not a zero divisor in $A_V.$ This implies that $f \in {\mathcal {S}}_X(U),$ proving the statement. $\square$

The proof of the following statement resembles the proof of its algebraic analogue.

Corollary 8.5

Let X be a good k-analytic space. Then for any $x \in X,$ ${\mathcal {S}}_{X,x}$ is the set of elements of ${\mathcal {O}}_{X,x}$ that are non-zero-divisors.

Proof

Let $x \in X.$ Clearly, the elements of ${\mathcal {S}}_{X,x}$ are not zero divisors in ${\mathcal {O}}_{X,x}.$

Let $f \in {\mathcal {O}}_{X,x}$ be a non-zero-divisor. By restricting to an affinoid neighborhood of x if necessary, we may assume, without loss of generality, that X is an affinoid space and $f \in {\mathcal {O}}_X(X).$ Set $A={\mathcal {O}}_X(X).$ Set $I=\{a \in A: f \cdot a=0\}.$ This is an ideal of A, and gives rise to the following short exact sequence

$$\begin{aligned} 0 \rightarrow I \rightarrow A \rightarrow A, \end{aligned}$$

where $A \rightarrow A$ is given by $a \mapsto f \cdot a.$ Seeing as f is a non-zero-divisor in ${\mathcal {O}}_{X,x}$, we obtain that $I{\mathcal {O}}_{X,x}=0.$

The ring A is an affinoid algebra, and hence Noetherian (cf. [1, Proposition 2.1.3]). Consequently, I is finitely generated. Let $a_1, a_2, \dots , a_n \in A$ be such that $I=(a_1, a_2, \dots , a_n).$ By the above, the germs $a_{i,x} \in {\mathcal {O}}_{X,x}$ of $a_i$ at x are zero for all $i \in \{1,2,\dots , n\}.$ Consequently, there exists an affinoid neighborhood V of x in X such that $a_{i|V}=0$ for all i,  implying $I{\mathcal {O}}_X(V)=0.$

Set $A_V{:}{=}{\mathcal {O}}_X(V).$ By [1, Proposition 2.2.4(ii)], the restriction morphism $A \rightarrow A_V$ is flat, so the short exact sequence above induces the following short exact sequence:

$$\begin{aligned} 0 \rightarrow I \otimes _A A_V \rightarrow A_V \rightarrow A_V, \end{aligned}$$

where $A_V \rightarrow A_V$ is given by $b \mapsto f_{|V} \cdot b.$ Seeing as $A_V$ is a flat A-module, $I \otimes _A A_V$ is isomorphic to $IA_V=0.$ Consequently, multiplication by $f_{|V}$ is injective in $A_V,$ or equivalently, $f_{|V}$ is a non-zero-divisor in $A_V.$ By Lemma 8.4, this implies that ${f_{|V} \in {\mathcal {S}}_X(V)},$ and finally that $f \in {\mathcal {S}}_{X,x}.$ $\square$

By Corollary 8.5, if X is a good k-analytic space, then for any $x \in X,$ ${\mathscr {M}}_{X,x}$ is the total ring of fractions of ${\mathcal {O}}_{X,x}.$ In particular, if ${\mathcal {O}}_{X,x}$ is a domain, then ${\mathscr {M}}_{X,x}=\text {Frac} \ {\mathcal {O}}_{X,x}.$ When there is no risk of confusion, we will simply denote ${\mathcal {O}}$, resp. ${\mathscr {M}}$, for the sheaf of analytic, resp. meromorphic functions on X. We recall (see [27, Lemma 1.2] for a proof):

Lemma 8.6

Let X be an integral k-affinoid space. Then ${\mathscr {M}}(X)=\mathrm {Frac} \ {\mathcal {O}}(X).$

We now show that the meromorphic functions of the analytification of a proper scheme defined over an affinoid algebra are algebraic. It is a non-trivial result for which GAGA-type theorems (cf. [25, 32, Annexe A]) are crucial. The arguments to prove the following result were given in a Mathoverflow thread (see [30]). In the case of curves, this is shown in [1, Prop. 3.6.2].

Let us first mention some brief reminders on the notion of depth. Let R be a ring, I an ideal of R,  and M a finitely generated R-module. An M-regular sequence of length d over  I is a sequence $r_1, r_2, \dots , r_d \in I$ such that $r_i$ is not a zero divisor in $M/(r_1, \dots , r_{i-1})M$ for $i=1,2,\dots , d.$ The depth of M over I, denoted $\text {depth}_R(I,M)$ in [6, Section 1], is

• $\infty$ if $IM=M,$

• the supremum of the length of M-regular sequences over I,  otherwise.

In what follows, when $M=R,$ we will denote $\text {depth}_R(I,R)$ by $\text {depth}_{I} R.$ Remark that $\text {depth}_{I} R>0$ if and only if I contains a non-zero divisor of R.

Theorem 8.7

Let k be a complete ultrametric field. Let A be a k-affinoid algebra. Let X be a proper scheme over $\text {Spec} \ A.$ Let $X^{\mathrm {an}}/{\mathcal {M}}(A)$ denote the Berkovich analytification of X. Then ${\mathscr {M}}_{X^{\mathrm {an}}}(X^{an})={\mathscr {M}}_X(X)$, where ${\mathscr {M}}_{X^{\mathrm {an}}}$ (resp. ${\mathscr {M}}_X$) denotes the sheaf of meromorphic functions on $X^{\mathrm {an}}$ (resp. X).

When there is no risk of ambiguity and the ambient space is clear from context, we will simply write ${\mathscr {M}}$ for the sheaf of meromorphic functions.

Proof

As in Definition 8.2, let ${\mathcal {S}}_{X^{\mathrm {an}}}$ denote the presheaf of analytic functions on $X^{\mathrm {an}}$, which associates to any analytic domain U the set of analytic functions on U whose restriction to any affinoid domain in it is not a zero divisor. By Corollary 8.5, for any $x \in X^{\mathrm {an}},$ ${\mathcal {S}}_{X^{\mathrm {an}},x}$ is the set of non-zero-divisors of ${\mathcal {O}}_{X^{\mathrm {an}},x}.$

Let ${\mathcal {I}}$ be a coherent ideal sheaf on $X^{\mathrm {an}}$ that locally on $X^{\mathrm {an}}$ contains a section of ${\mathcal {S}} _{X^{\mathrm {an}}}.$ This means that for any $x \in X^{\mathrm {an}},$ ${\mathcal {S}} _{X^{\mathrm {an}},x} \cap {\mathcal {I}}_{x} \ne \emptyset .$ Let $s \in {\mathcal {S}} _{X^{\mathrm {an}},x} \cap {\mathcal {I}}_{x}.$ Then s is a non-zero divisor in ${\mathcal {O}} _{X^{\mathrm {an}},x},$ which implies $\mathrm {depth}_{{\mathcal {I}}_{x}} {\mathcal {O}} _{X^{\mathrm {an}},x}>0$. Suppose, on the other hand, that ${\mathcal {I}}$ is a coherent ideal sheaf on $X^{\mathrm {an}}$ such that $\mathrm {depth} _{{\mathcal {I}}_{x}} {\mathcal {O}}_{X^{\mathrm {an}},x} >0$ for all $x \in X^{\mathrm {an}}.$ Then there exists at least one element $s \in {\mathcal {I}}_{x}$ which is a non-zero-divisor in ${\mathcal {O}} _{X^{\mathrm {an}},x},$ implying $s \in {\mathcal {S}} _{X^{\mathrm {an}},x}.$ To summarize, a coherent ideal sheaf ${\mathcal {I}}$ on $X^{\mathrm {an}}$ contains locally on $X^{\mathrm {an}}$ a section of ${\mathcal {S}}_{X^{\mathrm {an}}}$ if and only if $\mathrm {depth}_{{\mathcal {I}}_{x}}({\mathcal {O}} _{X^{\mathrm {an}},x})>0$ for all $x \in X^{\mathrm {an}}.$

Let us show that for any coherent ideal sheaf ${\mathcal {I}}$ on $X^{\mathrm {an}}$ containing locally on $X^{\mathrm {an}}$ a section of ${\mathcal {S}}_{X^{\mathrm {an}}},$ there is an embedding $\text {Hom}_{X^{\mathrm {an}}}({\mathcal {I}}, {\mathcal {O}}_{X^{\mathrm {an}}}) \subseteq {\mathscr {M}}_{X^{\mathrm {an}}}(X^{\mathrm {an}}),$ where $\text {Hom}_{X^{\mathrm {an}}}({\mathcal {I}}, {\mathcal {O}}_{X^{\mathrm {an}}})$ denotes the global sections on $X^{\mathrm {an}}$ of the hom sheaf ${\mathscr {H}}om({\mathcal {I}}, {\mathcal {O}}_{X^{\mathrm {an}}}).$ Let $\varphi \in \text {Hom}_{X^{\mathrm {an}}}({\mathcal {I}}, {\mathcal {O}}_{X^{\mathrm {an}}}).$ For any $x \in X^{\mathrm {an}},$ $\varphi$ induces a morphism $\varphi _x: {\mathcal {I}}_{x} \rightarrow {\mathcal {O}}_{X^{\mathrm {an}},x}.$ Let $s_x \in {\mathcal {S}}_{X^{\mathrm {an}},x} \cap {\mathcal {I}}_{x},$ and set ${a_x=\varphi _x(s_x)}.$ There exists a neighborhood $U_x$ of x,  such that $s_x \in {\mathcal {I}}(U_x) \cap {\mathcal {S}}_{X^{\mathrm {an}}}(U_x), {a_x \in {\mathcal {O}}_{X^{\mathrm {an}}}(U_x)}$, and $\varphi (U_x)(s_x)=a_x.$ Set $f_x=\frac{a_x}{s_x} \in {\mathcal {S}}_{X^{\mathrm {an}}}(U_x)^{-1} {\mathcal {O}}_{X^{\mathrm {an}}}(U_x) \subseteq {\mathscr {M}}_{X^{\mathrm {an}}}(U_x)$ (the presheaf ${\mathcal {S}}_{X^{\mathrm {an}}}^{-1} {\mathcal {O}}_{X^{\mathrm {an}}}$ is separated, so ${\mathcal {S}}_{X^{\mathrm {an}}}^{-1}{\mathcal {O}}_{X^{\mathrm {an}}} \subseteq {\mathscr {M}}_{X^{\mathrm {an}}}$).

Let $U_y, U_z$ be any non-disjoint elements of the cover $(U_x)_{x \in X^{\mathrm {an}}}$ of $X^{\mathrm {an}}.$ Then considering $\varphi$ is a morphism of sheaves of ${\mathcal {O}}_{X^{\mathrm {an}}}$-modules, $\varphi (U_y \cap U_z)(s_y \cdot s_z)=s_y \cdot a_z=a_y \cdot s_z$ in ${\mathcal {O}}_{X^{\mathrm {an}}}(U_y \cap U_z)$. Consequently, $f_{y|U_y \cap U_z}=f_{z|U_y \cap U_z}$ in ${\mathscr {M}}_{X^{\mathrm {an}}}(U_y \cap U_z),$ implying there exists $f \in {\mathscr {M}}_{X^{\mathrm {an}}}(X^{\mathrm {an}})$ such that $f_{|U_x}=f_x$ in ${\mathscr {M}}_{X^{\mathrm {an}}}(U_x)$ for all $x \in X^{\mathrm {an}}.$

We associate to $\varphi$ the meromorphic function f. Remark that if $f=0,$ then $a_x=0$ for all  x. This implies that for any $\alpha \in {\mathcal {I}}_x,$ $\varphi _x(s_x \cdot \alpha )=s_x \cdot \varphi _x(\alpha )=a_x \cdot \varphi _x(\alpha )=0,$ which, taking into account $s_x \in {\mathcal {S}}_{X^{\mathrm {an}},x}$, means that $\varphi _x(\alpha )=0.$ Consequently, $\varphi _x=0$ for all $x \in X^{\mathrm {an}},$ so $\varphi =0.$ Thus, the map $\psi _{{\mathcal {I}}}:\text {Hom}_{X^{\mathrm {an}}}({\mathcal {I}}, {\mathcal {O}}_{X^{\mathrm {an}}}) \rightarrow {\mathscr {M}}_{X^{\mathrm {an}}}(X^{\mathrm {an}})$ we have constructed is an embedding.

Remark that the set of coherent ideal sheaves on $X^{\mathrm {an}}$ containing locally on $X^{\mathrm {an}}$ a section of ${\mathcal {S}}_{X^{\mathrm {an}}}$ forms a directed set with respect to reverse inclusion (i.e. if ${\mathcal {I}}, {\mathcal {J}}$ satisfy these properties, then so does ${\mathcal {I}} \cdot {\mathcal {J}} \subseteq {\mathcal {I}}, {\mathcal {J}}$). Thus, by the paragraph above, there is an embedding $\varinjlim _{{\mathcal {I}}} \text {Hom}_{X^{\mathrm {an}}}({\mathcal {I}}, {\mathcal {O}}_{X^{\mathrm {an}}}) \hookrightarrow {\mathscr {M}}_{X^{\mathrm {an}}}(X^{\mathrm {an}}),$ where the direct limit is taken with respect to the same kind of coherent ideal sheaves ${\mathcal {I}}$ as above. Let us show that this embedding is an isomorphism.

For any $f \in {\mathscr {M}}_{X^{\mathrm {an}}}(X^{\mathrm {an}}),$ define the ideal sheaf $D_f$ as follows: for any analytic domain U of $X^{\mathrm {an}},$ set $D_f(U)=\{s \in {\mathcal {O}}(U): s \cdot f \in {\mathcal {O}}_{X^{\mathrm {an}}}(U) \subseteq {\mathscr {M}}_{X^{\mathrm {an}}}(U)\}.$ This is a coherent ideal sheaf on $X^{\mathrm {an}}$. Since ${\mathscr {M}}_{X^{\mathrm {an}},x}={\mathcal {S}}_{X^{\mathrm {an}},x}^{-1} {\mathcal {O}}_{X^{\mathrm {an}},x}$ for any $x \in X^{\mathrm {an}},$ there exist $s_x \in {\mathcal {S}}_{X^{\mathrm {an}},x}$ and $a_x \in {\mathcal {O}}_{X^{\mathrm {an}},x}$ such that $f_x=\frac{a_x}{s_x}$ in ${\mathscr {M}}_{X^{\mathrm {an}},x}.$ Considering $D_{f,x}=\{s \in {\mathcal {O}}_{X^{\mathrm {an}},x}: s \cdot f_x \in {\mathcal {O}}_{X^{\mathrm {an}},x}\},$ we obtain that $s_x \in D_{f,x},$ so $D_f$ contains locally on $X^{\mathrm {an}}$ a section of ${\mathcal {S}}_{X^{\mathrm {an}}}.$ To $f \in {\mathscr {M}}_{X^{\mathrm {an}}}(X^{\mathrm {an}})$ we associate the morphism $\varphi _f: D_f \rightarrow {\mathcal {O}}_{X^{\mathrm {an}}}$ which corresponds to multiplication by f (i.e. for any open subset U of $X^{\mathrm {an}}$, $D_f(U) \rightarrow {\mathcal {O}}_{X^{\mathrm {an}}}(U), s \mapsto f \cdot s$). Clearly, $\psi _{D_f}(\varphi _f)=f,$ implying the embedding $\varinjlim _{{\mathcal {I}}} \text {Hom}_{X^{\mathrm {an}}}({\mathcal {I}}, {\mathcal {O}}_{X^{\mathrm {an}}}) \hookrightarrow {\mathscr {M}}_{X^{\mathrm {an}}}(X^{\mathrm {an}})$ is surjective, so an isomorphism.

Let ${\mathcal {S}}_X$ denote the presheaf on X through which ${\mathscr {M}}_X$ is defined (see [26, Section 7.1.1]). Remark that since A is Noetherian ([1, Proposition 2.1.3]), the scheme X is locally Noetherian. Under this assumption, for any $x \in X,$ ${\mathcal {S}}_{X,x}$ is the set of all non-zero-divisors of ${\mathcal {O}}_{X,x}$ (see [26, 7.1.1, Lemma 1.12(c)]). Taking this into account, all the reasoning above does not make use of the fact that $X^{\mathrm {an}}$ is an analytic space, and can be applied mutatis mutandis to the scheme X and its sheaf of meromorphic functions ${\mathscr {M}}_X$. Thus, ${\mathscr {M}}_X(X) \cong \varinjlim _{{\mathcal {J}}} \text {Hom}_X({\mathcal {J}}, {\mathcal {O}}_X),$ where the direct limit is taken with respect to coherent ideal sheaves ${\mathcal {J}}$ on X,  for which $\text {depth}_{{\mathcal {J}}_{X,x}} {\mathcal {O}}_{X,x}>0$ for all $x \in X.$

Consequently, to show the statement, we need to show that $\varinjlim _{{\mathcal {J}}} \text {Hom}_X({\mathcal {J}}, {\mathcal {O}}_X)=\varinjlim _{{\mathcal {I}}} \text {Hom}_{X^{\mathrm {an}}}({\mathcal {I}}, {\mathcal {O}}_{X^{\mathrm {an}}}),$ where the direct limits are taken as above.

By [32, Annexe A] (which was proven in [25] in the case of rigid geometry), there is an equivalence of categories between the coherent sheaves on X and those on $X^{\mathrm {an}}$. Let us show that this induces an equivalence of categories between the coherent ideal sheaves on X and those on $X^{\mathrm {an}}.$ To see this, we only need to show that if ${\mathcal {F}}$ is a coherent sheaf on X such that ${\mathcal {F}}^{\mathrm {an}}$ is an ideal sheaf on $X^{\mathrm {an}},$ then ${\mathcal {F}}$ is an ideal sheaf on X. By [32, A.1.3], we have a sheaf isomorphism ${\mathscr {H}}om({\mathcal {F}}, {\mathcal {O}})^{\mathrm {an}} \cong {\mathscr {H}}om({\mathcal {F}}^{\mathrm {an}}, {\mathcal {O}}_{X^{\mathrm {an}}}),$ so ${\mathscr {H}}om({\mathcal {F}}, {\mathcal {O}})^{\mathrm {an}}$ has a non-zero global section $\iota$ corresponding to the injection ${\mathcal {F}}^{\mathrm {an}} \subseteq {\mathcal {O}}_{X^{\mathrm {an}}}.$ By [32, Théorème A.1(i)], ${\mathscr {H}}om({\mathcal {F}}, {\mathcal {O}})^{\mathrm {an}}(X^{\mathrm {an}}) \cong {\mathscr {H}}om({\mathcal {F}}, {\mathcal {O}})(X).$ Let $\iota ' \in {\mathscr {H}}om({\mathcal {F}}, {\mathcal {O}})(X)$ denote the element corresponding to $\iota .$ Then the analytification of $\iota ': {\mathcal {F}} \rightarrow {\mathcal {O}}_X$ is the morphism $\iota : {\mathcal {F}}^{\mathrm {an}} \hookrightarrow {\mathcal {O}}_{X^{\mathrm {an}}}.$ By flatness of $X^{\mathrm {an}} \rightarrow X,$ we obtain that $(\mathrm {ker} \ \iota ')^{\mathrm {an}}=\mathrm {ker} \ \iota '^{\mathrm {an}}=\mathrm {ker} \ \iota ,$ so $(\mathrm {ker} \ \iota ')^{\mathrm {an}}=0,$ implying $\mathrm {ker} \ \iota '=0.$ Consequently, there exists an embedding ${\mathcal {F}} \hookrightarrow {\mathcal {O}}_X,$ implying ${\mathcal {F}}$ is an ideal sheaf on X.

If to a coherent ideal sheaf ${\mathcal {J}}$ on X we associate the coherent ideal sheaf ${\mathcal {J}}^{\mathrm {an}}$ on $X^{\mathrm {an}},$ then as seen above: $\text {Hom}_{X}({\mathcal {J}}, {\mathcal {O}}_X)\cong \text {Hom}_{X^{\mathrm {an}}}({\mathcal {J}}^{\mathrm {an}}, {\mathcal {O}}_{X^{\mathrm {an}}})$.

Let us also show that a coherent ideal sheaf ${\mathcal {J}}$ on X satisfies ${\text {depth}_{{\mathcal {J}}_{x}} {\mathcal {O}}_{X,x}>0}$ for all $x \in X$ if and only if $\text {depth}_{{\mathcal {J}}^{\mathrm {an}}_y} {\mathcal {O}}_{X^{\mathrm {an}}, y}>0$ for all $y \in X^{\mathrm {an}}.$ To see this, recall that by [2, Proposition 2.6.2], the morphism $\phi : X^{\mathrm {an}} \rightarrow X$ is surjective and for any $y \in X^{\mathrm {an}},$ the induced morphism of local rings ${\mathcal {O}}_{X, x} \!\rightarrow \! {\mathcal {O}}_{X^{\mathrm {an}},y}$ is faithfully flat, where $x{:}{=}\phi (y).$ By [6, 1.3, Proposition 6], ${\text {depth}_{{\mathcal {J}}_x} {\mathcal {O}}_{X,x}=\text {depth}_{{\mathcal {J}}_x {\mathcal {O}}_{X^\mathrm {an},y}} {\mathcal {O}}_{X^{\mathrm {an}},y} \otimes _{{\mathcal {O}}_{X,x}} {\mathcal {O}}_{X,x}}.$ At the same time, seeing as the morphism ${\mathcal {O}}_{X, x} \rightarrow {\mathcal {O}}_{X^{\mathrm {an}},y}$ is flat, ${{\mathcal {J}}^{\mathrm {an}}_y={\mathcal {J}}_x \otimes _{{\mathcal {O}}_{X,x}} {\mathcal {O}}_{X^\mathrm {an},y}={\mathcal {J}}_x{\mathcal {O}}_{X^{\mathrm {an}},y}},$ so ${\text {depth}_{{\mathcal {J}}_x} {\mathcal {O}}_{X,x}=\text {depth}_{{\mathcal {J}}_y^{\mathrm {an}}} {\mathcal {O}}_{X^{\mathrm {an}}, y}}.$

From the above, $\varinjlim _{{\mathcal {J}}} \text {Hom}_X({\mathcal {J}}, {\mathcal {O}}_X)=\varinjlim _{{\mathcal {I}}} \text {Hom}_{X^{\mathrm {an}}}({\mathcal {I}}, {\mathcal {O}}_{X^{\mathrm {an}}}),$ where the direct limits are taken with respect to coherent ideal sheaves ${\mathcal {J}}$ on X (resp. ${\mathcal {I}}$ on $X^{\mathrm {an}}$), for which $\text {depth}_{{\mathcal {J}}_x} {\mathcal {O}}_{X, x}>0$ for all $x \in X$ (resp. $\text {depth}_{{\mathcal {I}}_x} {\mathcal {O}}_{X^{\mathrm {an}}, x}>0$ for all $x \in X^{\mathrm {an}}$). Finally, this implies that ${\mathscr {M}}_{X}(X)={\mathscr {M}}_{X^{\mathrm {an}}}(X^{\mathrm {an}}).$ $\square$

As an immediate consequence of the theorem above, we obtain that for any integral k-affinoid space Z,  ${\mathscr {M}}({\mathbb {P}}_Z^{1, \mathrm {an}})={\mathscr {M}}(Z)(T).$

1.2 Appendix II: Some results on analytic curves

Lemma 8.8

Let C be a normal irreducible projective k-analytic curve. Let U be a connected affinoid domain of C such that its boundary contains only type 3 points. Then for any $S \subseteq \partial {U},$ $U \backslash S$ is connected.

Proof

Suppose that C is generically quasi-smooth. Since $\partial {U}$ contains only type 3 points, all of the points of S are quasi-smooth in C.

Let $x,y \in \text {Int} \ U.$ Since U is connected, there exists an arc $[x, y] \subseteq U$ connecting x and y. Let $z \in S.$ We aim to show that $z \not \in [x,y],$ implying $[x,y] \subseteq U \backslash S,$ and thus the connectedness of $U \backslash S.$

By [10, Théorème 4.5.4], there exists an affinoid neighborhood V of z in U such that it is a closed virtual annulus, and its Berkovich boundary is $\partial _B(V)=\{z,u\}$ for some $u \in U.$ We may assume that $x,y \not \in V.$ Since V is an affinoid domain in U,  by [2, Proposition  1.5.5], the topological boundary $\partial _U{V}$ of V in U is a subset of $\partial _B(V)=\{z,u\}.$ Since V is a neighborhood of z,  $\partial _U{V}=\{u\}.$

Suppose $z \in [x,y].$ Then we could decompose $[x,y]=[x,z] \cup [z,y].$ Since $x,y \not \in V$, and $z \in V,$ the sets $[x,z] \cap \partial _U{V},$ $[z,y] \cap \partial _U{V}$ are non-empty, thus implying u is contained in both [x, z] and [z, y],  which contradicts the injectivity of [x, y]. Consequently, $U \backslash S$ is connected.

Let us get back to the general case. Let $C^{\mathrm {alg}}$ denote the algebraization of C (i.e. the normal irreducible projective algebraic curve over k whose analytification is C). Since it is normal, there exists a finite surjective morphism $C^{\mathrm {alg}} \rightarrow {\mathbb {P}}_k^{1}.$ This induces a finite field extension $k(T) \hookrightarrow k(C^{\mathrm {an}})={\mathscr {M}}(C)$ of their function fields. Let F denote the separable closure of k(T) in k(C). Then there exists an irreducible normal algebraic curve X over k such that $k(X)=F.$ Seeing as $k(T) \hookrightarrow k(C)$ is separable, the induced morphism $X \rightarrow {\mathbb {P}}_k^1$ is generically étale, so X is generically smooth. On the other hand, the finite field extension k(C)/F is purely inseparable, implying the corresponding finite morphism $C^{\mathrm {alg}} \rightarrow X$ is a homeomorphism.

Finally, the analytification $X^{\mathrm {an}}$ is a normal irreducible projective k-analytic curve that is generically quasi-smooth, and there is a finite morphism $f: C \rightarrow X^{\mathrm {an}}$ that is a homeomorphism. By [10, Proposition 4.2.14], f(U) is a connected proper closed analytic domain of $X^{\mathrm {an}}$. By [10, Théorème 6.1.3], f(U) is an affinoid domain of $X^{\mathrm {an}}.$ Clearly, $\partial {f(U)}=f(\partial {U}).$ Let $S \subseteq \partial {U},$ and set $S'=f(S).$ As shown above, $f(U) \backslash S'$ is connected. Consequently, $U \backslash S$ is connected. $\square$

Corollary 8.9

Let C be a normal irreducible k-analytic curve. Let U be an affinoid domain in C containing only type 3 points in its boundary. If $\text {Int}(U) \ne \emptyset$, then $(\text {Int}\ U)^c$ is an affinoid domain in C containing only type 3 points in its boundary.

Proof

Seeing as U is an affinoid domain, it has a finite number of connected components, and by [1, Corollary 2.2.7(i)], they are all affinoid domains in C. Furthermore, each of the connected components of U contains only type 3 points in its boundary. Consequently, by Lemma 8.8, $\text {Int}(U)$ has only finitely many connected components. Thus, by [10, Proposition  4.2.14], $(\text {Int} \ U)^c$ is a closed proper analytic domain of C. By [10, Théorème 6.1.3], it is an affinoid domain in C. $\square$

Proposition 8.10

Let C be a compact k-analytic curve. For any $x, y \in C,$ there exist only finitely many arcs in C connecting x and y.

Proof

By [10, Théorème 3.5.1], C is a real graph. By [10, 1.3.13], for any $z \in C,$ there exists an open neighborhood $U_z$ of z such that: (1) $U_z$ is uniquely arcwise-connected; (2)  the closure $\overline{U_z}$ of $U_z$ in C is uniquely arcwise-connected; (3) the boundary $\partial {U_z}$ is finite, implying in particular $\partial {U_z}=\partial {\overline{U_z}}.$ Seeing as C is compact, the finite open cover $\{U_z\}_{z \in C}$ admits a finite subcover ${\mathcal {U}}{:}{=}\{U_1, U_2, \dots , U_n\}.$ Set $S{:}{=}\bigcup _{i=1}^n \partial {U_i}.$ This is a finite subset of C.

Let x, y be any two points of C. Let $\gamma : [0,1] \rightarrow C$ be any arc in C connecting x and y. Set $S_{\gamma }{:}{=}S \cap \gamma ([0,1]) \backslash \{x,y\}.$ It is a finite (possibly empty) subset of C. For any $\alpha \in S_{\gamma },$ there exists a unique $a \in [0,1]$ such that $\gamma (a)=\alpha .$ This gives rise to an ordering of the points of $S_{\gamma }.$ Set $S_{\gamma }=\{\alpha _1, \alpha _2, \dots , \alpha _m\}$ such that the order of the points is the following: ${\alpha _1< \alpha _2< \cdots < \alpha _{m}}$ (meaning $\gamma ^{-1}(\alpha _1)< \gamma ^{-1}(\alpha _2)< \cdots < \gamma ^{-1}(\alpha _m)$). To the arc $\gamma$ we associate the finite sequence ${\overline{\gamma }}{:}{=}(\alpha _1, \alpha _2, \dots , \alpha _m)$ of points of $S_{\gamma }.$ Set $\alpha _0=x$ and $\alpha _{m+1}=y.$

For any $i \in \{0,1,\dots , m+1\},$ set $\gamma _i{:}{=}\gamma ([\gamma ^{-1}(\alpha _i), \gamma ^{-1}(\alpha _{i+1})])$. This is an arc in C connecting $\alpha _i$ and $\alpha _{i+1}.$ By construction, for any i,  $\gamma _i \cap S \subseteq \{\alpha _i, \alpha _{i+1}\}.$ Remark that $\gamma ([0,1])=\bigcup _{i=0}^{m+1} \gamma _i.$

Let us show that for any $i \in \{0,1,\dots , m\}$, there exists a unique arc $[\alpha _i, \alpha _{i+1}]_0$ in C connecting $\alpha _i$ and $\alpha _{i+1}$ such that $[\alpha _i, \alpha _{i+1}]_0 \cap S \subseteq \{\alpha _i, \alpha _{i+1}\}.$ Let $[\alpha _i, \alpha _{i+1}]$ be any such arc (the existence is guaranteed by the paragraphs above). Let $j \in \{1,2,\dots , n\}$ be such that $[\alpha _i, \alpha _{i+1}] \cap U_j \ne \emptyset .$ Let $z \in [\alpha _i, \alpha _{i+1}] \cap U_j;$ since $[\alpha _i, \alpha _{i+1}] \cap U_j$ is open in $[\alpha _i, \alpha _{i+1}],$ we may choose z such that $z \not \in \{\alpha _i, \alpha _{i+1}\}$. Let us denote by $[\alpha _i,z]$, resp. $[z, \alpha _{i+1}]$ the arc in C induced by $[\alpha _i, \alpha _{i+1}]$ connecting $\alpha _i$ and z, resp. z and $\alpha _{i+1}.$ Clearly, $[\alpha _i, \alpha _{i+1}]=[\alpha _i, z] \cup [z, \alpha _{i+1}].$

Suppose there exists $u \in [\alpha _i, \alpha _{i+1}] \backslash \overline{U_j}.$ Again, as $[\alpha _i, \alpha _{i+1}] \backslash \overline{U_j}$ is open in $[\alpha _i, \alpha _{i+1}],$ we may assume that $u \not \in \{\alpha _i, \alpha _{i+1}\}.$ Without loss of generality, let us suppose that $u \in [\alpha _i, z]$. Let $[\alpha _i, u]$, resp. [u, z], be the induced arcs connecting $\alpha _i$ and u, resp. u and z. Seeing as $z \in U_j$ and $u \not \in U_j,$ $[z,u] \cap \partial {U_j} \ne \emptyset .$ At the same time, $\emptyset \ne [z, u] \cap \partial {U_j} \subseteq [\alpha _i, \alpha _{i+1}] \cap \partial {U_j} \subseteq [\alpha _i, \alpha _{i+1}] \cap S \subseteq \{\alpha _i, \alpha _{i+1}\},$ which contradicts the injectivity of $[\alpha _i, \alpha _{i+1}].$

Consequently, $[\alpha _i, \alpha _{i+1}] \subseteq \overline{U_j}.$ Seeing as $\overline{U_j}$ is uniquely arcwise-connected, we obtain that the arc $[\alpha _i, \alpha _{i+1}]$ in C connecting $\alpha _i$ and $\alpha _{i+1},$ and satisfying the property $[\alpha _i, \alpha _{i+1}] \cap S \subseteq \{\alpha _i, \alpha _{i+1}\},$ is unique. Thus, $\gamma _i=[\alpha _i, \alpha _{i+1}],$ and the arc $\gamma$ is uniquely determined by its associated ordered sequence ${\overline{\gamma }}.$

Seeing as S is finite, the set of all finite sequences $(\beta _l)_l$ over S such that $\beta _{l'} \ne \beta _{l''}$ whenever $l' \ne l'',$ is also finite. Consequently, the set of arcs in C connecting x and y is finite. $\square$

1.3 Appendix III: Some results on the analytic projective line

We show here some auxiliary results on the analytic projective line. We recall that there is a classification of points of this analytic curve (see e.g. [1, 1.4.4] or [31, 1.1.2.3]). See also [27, Definition  2.2, Proposition 2.3] for an exposition on the nature of points of ${\mathbb {P}}^{1, \mathrm {an}}.$

Recall that for a complete ultrametric field K, ${\mathbb {P}}_K^{1, \mathrm {an}}$ is uniquely arcwise-connected. For any $x,y \in {\mathbb {P}}_K^{1,\mathrm {an}},$ we denote by [x, y] the unique arc in ${\mathbb {P}}_K^{1, \mathrm {an}}$ connecting x and y.

Finally, for $a \in K$ and $r \in {\mathbb {R}}_{>0},$ recall the notation $\eta _{a,r}$ for a point of ${\mathbb {P}}_K^{1, \mathrm {an}}$ in [31, 1.1.2.3].

Proposition 8.11

Let K be a complete ultrametric field. Let U be a connected affinoid domain of ${\mathbb {P}}_K^{1, \mathrm {an}}$ with only type 3 points in its boundary. Suppose U is not a point. Let us fix a copy of ${\mathbb {A}}_K^{1, \mathrm {an}}$ and a coordinate T on it. Let $\partial {U}=\{\eta _{R_i, r_i}:i=1,2,\dots , n\},$ where ${R_i \in K[T]}$ are irreducible polynomials and $r_i \in {\mathbb {R}}_{>0} \backslash \sqrt{|K^\times |}.$ Then ${U=\bigcap _i \{x : |R_i|_x \bowtie _i r_i\}},$ where $\bowtie _i \in \{\leqslant , \geqslant \}, i=1,2,\dots , n.$

Proof

We need the following two auxiliary results:

Lemma 8.12

For any $i \in \{1, 2, \dots , n\},$ either $U \subseteq \{x: |R_i|_x \leqslant r_i\}$ or ${U \subseteq \{x: |R_i|_x \geqslant r_i \}}.$

Proof

To see this, assume that the open subsets $V_1{:}{=}U \cap \{x: |R_i|_x < r_i\}$ and ${V_2{:}{=}U \cap \{x: |R_i|_x >r_i\}}$ of U are non-empty. As intersections of two connected subsets of ${\mathbb {P}}_K^{1, \mathrm {an}}$, both $V_1$ and $V_2$ are connected. Assume $V_j \cap \text {Int}(U) =\emptyset , j=1,2.$ This implies $V_j \subseteq \partial {U},$ and since $V_j$ is connected, it is a single type 3 point $\{\eta _j\}.$ But then, this would be an isolated point of U,  which is in contradiction with the connectedness of U. Consequently, there exist $x_j \in V_j \cap \text {Int}(U), j=1,2.$ By Lemma 8.8, $\text {Int}(U)$ is a connected set, so there exists a unique arc $[x_1, x_2]$ connecting $x_1, x_2$ that is entirely contained in $\text {Int}(U).$ Since $|R_i|_{x_1}<r_i$, $|R_i|_{x_2}>r_i,$ there exists $x_0 \in [x_1, x_2]$ such that $|R_i|_{x_0}=r_i.$ Since there is a unique point satisfying this condition ([27, Proposition 2.3(2)]), and it is $\eta _{R_i, r_i},$ we obtain that $\eta _{R_i, r_i} \in [x_1, x_2] \subseteq \text {Int}(U),$ which is in contradiction with the fact that $\eta _{R_i, r_i} \in \partial {U}.$ Thus, there exists $j \in \{1,2\}$ such that $V_j =\emptyset ,$ implying the statement. $\square$

Lemma 8.13

For $n \in {\mathbb {N}},$ let $W_i{:}{=}\{x \in {\mathbb {P}}_K^{1, \mathrm {an}}: |P_i| \bowtie _i r_i\}$, where $P_i \in K[T]$ is irreducible, $r_i \in {\mathbb {R}}_{>0} \backslash \sqrt{|k^\times |},$ $\bowtie _i \in \{\leqslant , \geqslant \},$ $i \in \{1,2,\dots , n\}.$ Suppose for all $i \ne j,$ $W_i \not \subseteq \mathrm {Int}(W_j).$ Then, for $V{:}{=}\bigcap _{i=1}^n W_i,$ $\partial {V}=\bigcup _{i=1}^n \partial {W_i}.$

Proof

Since $\text {Int}(V)=\bigcap _{j=1}^n \text {Int}(W_j),$ we obtain that

$$\partial {V}=\left( \bigcap _{j=1}^n W_j\right) \backslash \left( \bigcap _{i=1}^n \text {Int}(W_i)\right) = \bigcup _{i=1}^n \bigcap _{j=1}^n (W_i \backslash \text {Int}(W_j)).$$

Suppose there exist $i, j \in \{1,2,\dots , n\}$ such that $W_i \backslash \text {Int}(W_j)=\emptyset .$ Then $W_i \subseteq \text {Int}(W_j),$ contradicting the hypothesis of the statement.

Hence, for any i, j, $W_i \backslash \text {Int}(W_j) \ne \emptyset .$ In particular, this means that $W_i \cap \text {Int}(W_j)$ is a strict open subset of $W_i,$ so it is contained in $\text {Int}(W_i).$ Consequently, $\{\eta _{P_i, r_i}\}=W_i \backslash \text {Int}(W_i) \subseteq W_i \backslash (W_i \cap \text {Int}(W_j)) \subseteq W_i \backslash \text {Int}(W_j).$ This implies that for any i, ${\bigcap _{j=1}^n (W_i \backslash \text {Int}(W_j))=\{\eta _{P_i, r_i}\}}.$

Finally, $\partial {V}=\{\eta _{P_i, r_i}: i=1,2,\dots , n\},$ proving the statement. $\square$

If $U \subseteq \{x : |R_i|_x \leqslant r_i\}$ (resp. $U \subseteq \{x : |R_i|_x \geqslant r_i\}$), set $U_i=\{x : |R_i|_x \leqslant r_i\}$ (resp. $U_i{=}\{x : |R_i|_x {\geqslant } r_i\}$). Remark that for all i,  $U_i$ is connected and contains U. Set $V=\bigcap _{i=1}^n U_i.$ Let us show that $\partial {V}=\partial {U}.$ Assume there exist i, j such that $U_i \subseteq \text {Int}(U_j).$ Then $\eta _{R_j, r_j} \not \in U_i$, so $\eta _{R_j, r_j} \not \in U$, contradiction. Thus, Lemma 8.13 is applicable, and so $\partial {V}=\{\eta _{R_i, r_i}\}_{i=1}^n=\partial {U}.$

Remark that V is a connected affinoid domain (as an intersection of connected affinoid domains) of ${\mathbb {P}}_K^{1, \mathrm {an}}$. Also, $U \subseteq V$ and $\partial {U}=\partial {V}.$ Let us show that $U=V.$ Suppose there exists some $x \in V \backslash U.$ Then $x \in \text {Int}(V).$ Let $y \in \text {Int}(U) \subseteq \text {Int}(V).$ The unique arc [x, y] in ${\mathbb {P}}_K^{1, \mathrm {an}}$ connecting x and y is contained in $\text {Int}(V)$ (by connectedness of the latter, see Lemma 8.8). At the same time, since $x \not \in U$ and $y \in U,$ the arc [x, y] intersects $\partial {U}=\partial {V},$ contradiction. Thus, $U=V=\bigcap _{i=1}^n U_i,$ concluding the proof of Proposition 8.11. $\square$

In particular, the result above implies that every connected affinoid domain of ${\mathbb {P}}_K^{1, \mathrm {an}}$ with only type 3 points in its boundary is a rational domain.

Lemma 8.14

Let K be a complete ultrametric field. Let U, V be connected affinoid domains of ${\mathbb {P}}_K^{1, \mathrm {an}}$ containing only type 3 points in their boundaries, such that $U \cap V=\partial {U} \cap \partial {V}$ is a single type 3 point $\{\eta _{R,r}\}$ (i.e. R is an irreducible polynomial over K and $r \in {\mathbb {R}}_{>0} \backslash \sqrt{|K^\times |}$).

• (1) If $U \subseteq \{x \in {\mathbb {P}}_K^{1, \mathrm {an}}: |R|_x \leqslant r\}$ (resp. $U \subseteq \{x \in {\mathbb {P}}_K^{1, \mathrm {an}}: |R|_x \geqslant r\}$), then $V \subseteq \{x \in {\mathbb {P}}_K^{1, \mathrm {an}}: |R|_x \geqslant r\}$ (resp. $V \subseteq \{x \in {\mathbb {P}}_K^{1, \mathrm {an}}: |R|_x \leqslant r\}$).

• (2) Suppose $U \subseteq \{x \in {\mathbb {P}}_K^{1, \mathrm {an}}: |R|_x \leqslant r\}.$ Set $\partial {U}=\{\eta _{R,r}, \eta _{P_i, r_i}\}_{i=1}^n$ and ${\partial {V}=\{\eta _{R,r}, \eta _{P_j', r_j'}\}_{j=1}^m},$ so that $U=\{x \in {\mathbb {P}}_K^{1, \mathrm {an}}: |R|_x \leqslant r, |P_i|_x \bowtie _i r_i, i\}$ and $V=\{x \in {\mathbb {P}}_K^{1, \mathrm {an}}: |R|_x \geqslant r, |P_j'|_x \bowtie '_j r_j', j\}$, where $\bowtie _i, \bowtie _j' \in \{\leqslant , \geqslant \},$ $P_i, P_j' \in K[T]$ are irreducible, and $r_i, r_j' \in {\mathbb {R}}_{>0} \backslash \sqrt{|K^{\times }|}$ for all i, j.

  Then $U \cup V=\{x \in {\mathbb {P}}_K^{1, \mathrm {an}}:|P_i|_x \bowtie _i r_i, |P_j'|_x \bowtie _j' r_j', i=1,\dots , n, j=1,\dots , m\}.$ If $n=m=0,$ this means that $U \cup V={\mathbb {P}}_K^{1, \mathrm {an}}.$

Proof

(1) Remark that if $U \subseteq V,$ then $U=\{\eta _{R,r}\},$ so the statement is trivially satisfied. The same is true if $V \subseteq U.$ Let us suppose that neither of U, V is contained in the other.

Suppose $U \subseteq \{x \in {\mathbb {P}}_K^{1, \mathrm {an}}: |R|_x \!\leqslant \! r\}$ and $V \subseteq \{x \!\in \! {\mathbb {P}}_K^{1, \mathrm {an}}: |R|_x \leqslant r\}.$ Let ${u \in U \backslash V}$ and $v \in V \backslash U.$ Since $u, v \in \{x:|R|_x < r\}$—which is a connected set (Lemma 8.8), ${[u,v] \subseteq \{x: |R|_x<r\}}.$ At the same time, since $[u, \eta _{R,r}] \subseteq U$ and $[\eta _{R,r}, v] \subseteq V,$ ${[u, \eta _{R,r}] \cap [\eta _{R,r}, v]=\{\eta _{R,r}\}},$ so the arc $[u,v]=[u, \eta _{R,r}] \cup [\eta _{R,r}, v]$ contains the point $\eta _{R,r}.$ This is in contradiction with the fact that $[u,v] \subseteq \{x:|R|_x < r\}.$

The case $U, V \subseteq \{x \in {\mathbb {P}}_K^{1, \mathrm {an}}: |R|_x \geqslant r\}$ is shown to be impossible in the same way. (Property (1) is in fact true regardless of whether $\partial {U} \backslash \{\eta _{R,r}\}$ and $\partial {V} \backslash \{\eta _{R,r}\}$ contain only type 3 points or not.)

(2) The statement is clearly true if $m=n=0,$ so we may assume that is not the case.

Remark that $\partial (U \cup V) \subseteq \partial {U} \cup \partial {V}.$ Let $\eta \in \partial {U} \backslash V.$ Let G be any neighborhood of $\eta$ in ${\mathbb {P}}_K^{1, \mathrm {an}}.$ Since V is closed, there exists a neighborhood $G' \subseteq G$ of $\eta$ such that $G' \cap V=\emptyset .$ Since $\eta \in \partial {U},$ $G'$ contains points of both U and $U^C.$ Consequently, $G'$, and thus G, contain points of both $U \cup V$ and $U^C \cap V^C=(U \cup V)^C.$ Seeing as this is true for any neighborhood G of $\eta ,$ we obtain that $\eta \in \partial (U \cup V),$ implying $\partial {U} \backslash V \subseteq \partial {(U \cup V)}.$ Similarly, $\partial {V} \backslash U \subseteq \partial (U \cup V).$ It only remains to check for the point $\eta _{R,r}.$

Let $x \in \text {Int}(U) \subseteq \text {Int}(U \cup V)$ and $y \in \text {Int}(V) \subseteq \text {Int}(U \cup V).$ Remark that $x \not \in V$ and $y \not \in U.$ Furthermore, $|R|_x<r$ and $|R|_y>r.$ Consequently, $\eta _{R,r} \in [x,y].$ Since $U \cup V$ is a connected affinoid domain containing only type 3 points in its boundary, its interior is connected (see Lemma 8.8). Consequently, $[x, y] \subseteq \text {Int}(U \cup V),$ and hence $\eta _{R,r} \in \text {Int}(U\cup V).$

We have shown that ${\partial (U \cup V)=\{\eta _{P_i, r_i}, \eta _{P_j', r_j'}: i,j\}}.$ Since ${U \subseteq \{x:|P_i|_x \bowtie _i r_i\}}$ and $V \subseteq \{x: |P_j'|_x \bowtie _j' r_j'\}$ for all i, j,  we obtain that

$$\begin{aligned} U \cup V=\{x: |P_i|_{x} \bowtie _i r_i, |P_j'|_x \bowtie _j' r_j', i,j\}. \end{aligned}$$

$\square$

The next result describes certain affinoid domains of the analytic projective line.

Lemma 8.15

Let K be a complete ultrametric field. Let R(T) be a split unitary polynomial over K. Let $r \in {\mathbb {R}}_{>0}.$ Then for any root $\alpha$ of R(T) there exists a unique positive real number $s_\alpha$ such that $\{y \in {\mathbb {P}}_K^{1,\mathrm {an}}: |R(T)|_y=r\}=\bigcup _{R(\alpha )=0}\{y \in {\mathbb {P}}_K^{1,\mathrm {an}}: |T-\alpha |_y=s_{\alpha }\}.$ The point $\eta _{\alpha , s_{\alpha }}$ is the only point y of the arc $[\eta _{\alpha , 0}, \infty ]$ in ${\mathbb {P}}_K^{1, \mathrm {an}}$ for which $|R(T)|_y=r.$ Furthermore, $r=s_{\alpha } \cdot \prod _{R(\beta ) = 0, \alpha \ne \beta } \max (s_{\alpha }, |\alpha -\beta |).$

Proof

Remark that if $y \in {\mathbb {P}}_K^{1,\mathrm {an}}$ is such that ${|R(T)|_y=0},$ then ${\prod _{R(\alpha )=0} |T-\alpha |_y=0}$, meaning there exists a root $\alpha _0$ of R(T) such that $|T-\alpha _0|_{y}=0$ (notice that we haven’t assumed R(T) to be separable, i.e. there could be roots with multiplicities). This determines the unique point $\eta _{{\alpha }_0, 0}$ in ${\mathbb {P}}_K^{1,\mathrm {an}}.$ Thus, the zeros of R(T) in ${\mathbb {P}}_K^{1,\mathrm {an}}$ are $\eta _{\alpha , 0}, R(\alpha )=0.$ Remark also that R has only one pole in ${\mathbb {P}}_K^{1, \mathrm {an}}$ and that is the point $\infty .$

By [10, 3.4.23.1], the analytic function R(T) on ${\mathbb {P}}_K^{1,\mathrm {an}}$ is locally constant everywhere outside of the finite graph $\Gamma {:}{=}\bigcup _{R(\alpha )=0} [\eta _{\alpha ,0}, \infty ].$ Furthermore, its variation is compatible with the canonical retraction $d:{\mathbb {P}}_K^{1, \mathrm {an}} \rightarrow \Gamma$ in the sense that $|R(T)|_y=|R(T)|_{d(y)}$ for any $y \in {\mathbb {P}}_K^{1, \mathrm {an}}$ (cf. [10, 3.4.23.8]). By [10, 3.4.24.3], R(T) is continuously strictly increasing in all the arcs $[\eta _{\alpha , 0}, \infty ], R(\alpha )=0,$ where $|R(T)|_{\eta _{\alpha , 0}}=0$ and $|R(T)|_{\infty }=+\infty .$ Consequently, |R(T)| attains the value r exactly one time on each arc $[\eta _{\alpha , 0}, \infty ].$ Suppose $s_{\alpha }$ is the unique positive real number for which $|R(T)|_{\eta _{\alpha , s_{\alpha }}}=r.$ Then $\prod _{R(\beta )=0}|T-\beta |_{\eta _{\alpha , s_{\alpha }}}=s_\alpha \cdot \prod _{R(\beta )=0, \alpha \ne \beta } \max (s_{\alpha }, |\alpha -\beta |)=r.$

We have shown that there exist positive real numbers $s_{\alpha }$ such that ${\{y \in \Gamma : |R|_y=r\}=}$ $\{\eta _{\alpha , s_{\alpha }}: R(\alpha )=0\}.$ As mentioned before, the variation of R is compatible with the canonical retraction d of ${\mathbb {P}}_K^{1, \mathrm {an}}$ to $\Gamma .$ Since $d^{-1}(\eta _{\alpha ,s_{\alpha }})=\{y \in {\mathbb {P}}_K^{1, \mathrm {an}}: |T-\alpha |_y=s_{\alpha }\},$ we finally obtain that $\{y \in {\mathbb {P}}_K^{1, \mathrm {an}}: |R|_y=r\}=\bigcup _{R(\alpha )=0} \{y \in {\mathbb {P}}_K^{1, \mathrm {an}}: |T-\alpha |_y=s_{\alpha }\}$ with $s_{\alpha }$ as above. $\square$