The topology on Berkovich affine lines over complete valuation rings

TL;DR

This paper characterizes the topology of the Berkovich affine line over complete valuation rings with algebraically closed fraction fields, showing it is connected, locally path connected, and a completion of a specific metric space.

Contribution

It provides a comprehensive description of the topology of the Berkovich affine line over such rings, including its connectedness and local path connectedness, and relates it to a canonical uniform structure.

Findings

$ ext{A}_R^1$ is connected and locally path connected.

$ ext{A}_R^1$ is the completion of $K imes (1, ext{infinity})$ under a canonical uniform structure.

Description of the Berkovich spectrum of $ ext{Z}_p[G]$ for cyclic $p$-groups.

Abstract

In this article, we give a full description of the topology of the one dimensional affine analytic space $\mathbb{A}_R^1$ over a complete valuation ring $R$ (i.e. a valuation ring with "real valued valuation" which is complete under the induced metric), when its field of fractions $K$ is algebraically closed. In particular, we show that $\mathbb{A}_R^1$ is both connected and locally path connected. Furthermore, $\mathbb{A}_R^1$ is the completion of $K\times (1,\infty)$ under a canonical uniform structure. As an application, we describe the Berkovich spectrum $\mathfrak{M}(\mathbb{Z}_p[G])$ of the Banach group ring $\mathbb{Z}_p[G]$ of a cyclic $p$-group $G$ over the ring $\mathbb{Z}_p$ of $p$-adic integers.

Equations

$$E_{\epsilon}^{X}:=\big{\{}(\mu,\nu)\in\mathbb{A}_{S}^{n}\times\mathbb{A}_{S}^{n}:\big{|}|\mathbf{p}|_{\mu}-|\mathbf{p}|_{\nu}\big{|}<\epsilon,\text{ for any }\mathbf{p}\in X\big{\}},$$

$$S^{\star }:=S\setminus \{0_S\},$$

$$S\{n^{-1}\mathbf{t}\}:=\left\{\sum _{k=0}^{\infty }{a}_k{\mathbf{t}}^k:\underset{k}{\lim }\Vert a_k\Vert n^k=0\right\}$$

$${\mathbb{U}}_n^S:=\{\mu \in {\mathbb{A}}_S^1:|\mathbf{t}{|}_{\mu }<n\},$$

$$\tilde{Q}:R[\mathbf{t}]\to F[\mathbf{t}]$$

$$|\mathbf{p}{|}_{\lambda +s}:=|{\mathbf{p}}_s{|}_{\lambda }(\mathbf{p}\in K[\mathbf{t}])\text{and}\lambda -s:=\lambda +(-s).$$

$$D(s,\tau ):=\{t\in K:|t-s|\le \tau \},$$

$${\left|\sum _{k=0}^n{a}_k(\mathbf{t}-s{)}^k\right|}_{{\zeta }_{s,\tau }}=\underset{k=0,\dots ,n}{\max }|a_k|{\tau }^k(n\in \mathbb{N};a_0,\dots ,a_n\in K);$$

$$J_{\omega }^{\mathbb{A}}:{\mathbb{A}}_{K^{\omega }}^1\to {\mathbb{A}}_R^1.$$

$$Q^{\mathbb{A}}:{\mathbb{A}}_F^1\to {\mathbb{A}}_R^1.$$

$${\left|\sum _{l=0}^n{\mathbf{r}}_l{\mathbf{q}}^l\right|}_{{\gamma }_{\mathbf{q},\kappa }}:=\underset{{\mathbf{r}}_l\ne 0}{\max }{\kappa }^l$$

$${\gamma }_{\mathbf{t}-x,\kappa }={\zeta }_{x,\kappa }(x\in \mathfrak{k};\kappa \in {\mathbb{R}}_{+}).$$

$${\mathbb{A}}_{\mathfrak{k}}^1=\{{\gamma }_{\mathbf{q},\kappa }:\mathbf{q}\in \mathfrak{k}[\mathbf{t}{]}_{\mathrm{irr}}\setminus \{\mathbf{t}\};\kappa \in [0,1)\}\cup \{{\gamma }_{\mathbf{t},\tau }:\tau \in {\mathbb{R}}_{+}\}.$$

$$P^{\lambda }:=\{\mathbf{p}\in \mathfrak{k}[\mathbf{t}]:|\mathbf{p}{|}_{\lambda }<1\}.$$

$${\mathbb{A}}_{\mathfrak{k}}^1=\{{\zeta }_{s,\kappa }:s\in {\mathfrak{k}}^{\star };\kappa \in [0,1)\}\cup \{{\zeta }_{0_{\mathfrak{k}},\tau }:\tau \in {\mathbb{R}}_{+}\}.$$

$${\mathbb{A}}_R^1=Q^{\mathbb{A}}({\mathbb{A}}_F^1)\cup \bigcup _{\omega \in [1,\infty )}{J}_{\omega }^{\mathbb{A}}({\mathbb{A}}_{K^{\omega }}^1).$$

$$|a^{-1}\mathbf{p}{|}_{\bar{\lambda }}:=|a{|}_{\lambda }^{-1}|\mathbf{p}{|}_{\lambda }(a\in R^{\star };\mathbf{p}\in R[\mathbf{t}]),$$

$$|\mathbf{p}{|}_{{\lambda }^{\omega }}:=|\mathbf{p}{|}_{\lambda }^{\omega }(\mathbf{p}\in K[\mathbf{t}]).$$

$$\Lambda :{\mathbb{A}}_K^1\times [1,\infty )\to \bigcup _{\omega \in [1,\infty )}{J}_{\omega }^{\mathbb{A}}({\mathbb{A}}_{K^{\omega }}^1).$$

$$|a{|}^{{\omega }_{i_j}}=|a{|}_{{\lambda }_{i_j}^{{\omega }_{i_j}}}\to |Q(a){|}_{{\zeta }_{x,\tau }}=0,$$

$$|\mathbf{q}{|}_{{\lambda }_i}\le \underset{k=0,\dots ,n}{\max }|a_k||\mathbf{t}{|}_{{\lambda }_i}^k\le \underset{k=0,\dots ,n}{\max }|a_k|<1,$$

$$\mathbf{p}={\mathbf{t}}^k\cdot (\mathbf{t}-b_1)\cdots (\mathbf{t}-b_m)+{\mathbf{q}}_0.$$

$$\mathbb{A}_{F}^{1}=\big{\{}\zeta_{x,\tau}:(x,\tau)\in F\times[0,1)\cup\{0_{F}\}\times[1,\infty)\big{\}}.$$

$$|a_0{|}^{{\omega }_i/{\omega }_0}=|a_0{|}_{{\lambda }_i^{{\omega }_i}}^{1/{\omega }_0}\to |a_0|$$

$$|\mathbf{t}-c{|}_{{\lambda }_i}^{{\omega }_i}\to {\left|\mathbf{t}-Q(c)\right|}_{{\zeta }_{0_F,\kappa }}=\kappa (c\in R).$$

$${\mathrm{limsup}}_{i\in \mathfrak{I}}|\mathbf{t}{|}_{{\lambda }_i}\le 1,$$

$$|\mathbf{t}-c^{\prime }{|}_{{\lambda }_i}^{{\omega }_i}\to {\tau }_2(c^{\prime }\in R).$$

$$|\mathbf{q}{|}_{{\lambda }_i}\le \underset{k=0,\dots ,n}{\max }|a_k||\mathbf{t}{|}_{{\lambda }_i}^k<{\kappa }^{\frac{1}{n+1}}.$$

$$\mathbf{p}=(\mathbf{t}-c_1)\cdots (\mathbf{t}-c_m)+{\mathbf{q}}_0.$$

$$|\mathbf{t}-c^{\prime }{|}_{{\lambda }_i}=|\mathbf{t}-c{|}_{{\lambda }_i}(i\ge i_0).$$

Full text

1. Introduction and notation

Let $S$ be a commutative unital Banach ring with its norm being denoted by $\|\cdot\|$. The Berkovich spectrum $\mathcal{M}(S)$ as well as the $n$-dimensional affine analytic space $\mathbb{A}_{S}^{n}$ over $S$ was introduced by Vladimir Berkovich (see, e.g., [3]) and was further studied by Jérôme Poineau in [8]. More precisely, $\mathcal{M}(S)$ is the set of all non-zero contractive multiplicative semi-norms on $S$, while $\mathbb{A}_{S}^{n}$ is the set of all non-zero multiplicative seminorms on the $n$-variables polynomial ring $S[\mathbf{t}_{1},...,\mathbf{t}_{n}]$ whose restrictions on $S$ are contractive. The topologies on both $\mathcal{M}(S)$ and $\mathbb{A}_{S}^{n}$ are the ones given by pointwise convergence. The topology on $\mathbb{A}_{S}^{n}$ is also induced by the Berkovich uniform structure which is given by a fundamental system of entourages consisting of sets of the form

[TABLE]

where $\epsilon$ runs through all strictly positive real numbers and $X$ runs through all non-empty finite subsets of $S[\mathbf{t}_{1},...,\mathbf{t}_{n}]$. It is not hard to see that $\mathbb{A}_{S}^{n}$ is complete under this uniform structure.

In the case of a non-Archimedean field $L$, the properties of $\mathbb{A}_{L}^{n}$ plays an important role in the study of non-Archimedean geometry. For example, the Bruhat-Tits tree of $SL_{2}({\mathbb{Q}}_{p})$ can be realized as a subspace of the Berkovich projective line, which is a glue of two copies of $\mathbb{A}^{1}_{{\mathbb{Q}}_{p}}$ (see [10]).

When $L$ is an algebraically closed non-Archimedean complete valued field, Berkovich gave in [3] a full description of the space $\mathbb{A}_{L}^{1}$. This may then be used to describe the one-dimensional affine analytic spaces over not necessarily algebraically closed fields.

The aim of this article is to give a full description of the topology space $\mathbb{A}_{R}^{1}$ of a “complete valuation ring” $R$. Recall that an integral domain $R$ is a valuation ring if for every element $x$ in its field of fractions $K$, either $x\in R$ or $x^{-1}\in R$ (see e.g. [1, p.65] or Definition 2 of [4, §VI.1]). It is easy to see that if $R^{\times}$ and $K^{\times}$ are the sets of invertible elements in $R$ and $K$ respectively, then $K^{\times}/R^{\times}$ is a totally ordered abelian group and the canonical map $\nu_{R}:K^{\times}\to K^{\times}/R^{\times}$ is a “valuation” such that $R=\{x\in K^{\times}:\nu_{R}(x)\geq 0\}\cup\{0_{R}\}$.

Definition 1.1**.**

A valuation ring $R$ is called a complete valuation ring if $K^{\times}/R^{\times}$ is isomorphic to an ordered subgroup of $\mathbb{R}$ and $R$ is complete under the induced norm.

If $R$ is a complete valuation ring, then $K$ is a non-Archimedean complete valued field and $R$ coincides with the ring of integers, $\{a\in K:|a|\leq 1\}$, of $K$. Conversely, the ring of integers of a non-Archimedean complete valued field is always a complete valuation ring.

Throughout this article, for a commutative unital Banach ring $S$, we denote

[TABLE]

where $0_{S}$ is the zero element of $S$. The identity of $S$ will be denoted by $1_{S}$. For any $n\in\mathbb{N}$, we define

[TABLE]

and equip it with the norm $\left\|\sum_{k=0}^{\infty}a_{k}\mathbf{t}^{k}\right\|:=\max_{k}\|a_{k}\|n^{k}$. It is well-known that $S\{n^{-1}\mathbf{t}\}$ is a commutative unital Banach ring. For simplicity, we will use the notation $S\{\mathbf{t}\}$ for $S\{1^{-1}\mathbf{t}\}$.

It was shown in [3] that for each $n\in\mathbb{N}$, the compact space $\mathcal{M}(S\{n^{-1}\mathbf{t}\})$ can be identified with the subspace $\{\mu\in\mathbb{A}_{S}^{1}:|\mathbf{t}|_{\mu}\leq n\}$ of $\mathbb{A}_{S}^{1}$. If we set

[TABLE]

then $\mathbb{A}_{S}^{1}=\bigcup_{n\in\mathbb{N}}\mathbb{U}^{S}_{n}=\bigcup_{n\in\mathbb{N}}\mathcal{M}(S\{n^{-1}\mathbf{t}\})$ and hence $\mathbb{A}_{S}^{1}$ is both locally compact and $\sigma$-compact.

From now on, $R$ is a complete valuation ring and $K$ is its field of fractions. The absolute value on $K$ induced by $\nu_{R}$ will be denoted by $|\cdot|$. The residue field of $R$ is denoted by $F$ and $Q:R\to F$ is the quotient map. We set

[TABLE]

to be the map induced by $Q$.

Suppose that $s\in K$ and $\mathbf{p}\in K[\mathbf{t}]$. If $r_{0},\dots,r_{n}\in K$ are the unique elements with $\mathbf{p}=\sum_{k=0}^{n}r_{k}\mathbf{t}^{k}$, then we put $\mathbf{p}_{s}:=\sum_{k=0}^{n}r_{k}(\mathbf{t}+s)^{k}$. For any $\lambda\in\mathbb{A}_{K}^{1}$, we define

[TABLE]

It is easy to see that $\lambda+s\in\mathbb{A}_{K}^{1}$, and that $\lambda\mapsto\lambda+s$ is a bicontinuous bijection from $\mathbb{A}_{K}^{1}$ to itself.

For every $s\in K$ and $\tau\in\mathbb{R}_{+}$, we denote the closed ball with center $s$ and radius $\tau$ by $D(s,\tau)$, i.e.

[TABLE]

and define (as in [2]) $\zeta_{s,\tau}\in\mathbb{A}_{K}^{1}$ by $|\mathbf{p}|_{\zeta_{s,\tau}}:=\sup_{t\in D(s,\tau)}|\mathbf{p}(t)|$ ($\mathbf{p}\in K[\mathbf{t}]$). Because of the maximum modulus principle, one has

[TABLE]

here, we use the convention that $0^{0}:=1$.

Let us recall the following result from [3].

Theorem 1.2**.**

(Berkovich) Let $L$ be an algebraically closed non-Archimedean complete valued field with a non-trivial norm $|\cdot|$, and $|\cdot|_{\lambda}:L[\mathbf{t}]\to\mathbb{R}_{+}$ be a function. Then $\lambda\in\mathbb{A}_{L}^{1}$ if and only if there is a decreasing sequence $\{D(s_{n},\tau_{n})\}_{n\in\mathbb{N}}$ of closed balls in $L$ such that $|\mathbf{p}|_{\lambda}=\inf_{n\in\mathbb{N}}|\mathbf{p}|_{\zeta_{s_{n},\tau_{n}}}$.

The topology on $\mathbb{A}_{L}^{1}$ was also described in [3]. Moreover, as noted in [3], by using the fact that $D(t,\tau)=D(s,\tau)$ whenever $D(s,\tau)\subseteq D(t,\tau)$, one can show easily that $\mathbb{A}_{L}^{1}$ is path connected.

It is also well-known that $\mathbb{A}_{L}^{1}$ is locally path connected, in the sense that for every point in this topological space, there is a local neighborhood basis at that point consisting of open sets that are path connected under the induced topologies. Indeed, as noted in [2], any two points in $\mathbb{A}_{L}^{1}$ are joined by a unique path, and hence $\mathbb{A}_{L}^{1}$ is a $\mathbb{R}$-tree. The “weak topology” induced by this $\mathbb{R}$-tree structure coincides with the pointwise convergence topology on $\mathcal{M}(L\{\mathbf{t}\})$ (see e.g. [2, Proposition 1.13]). Since the canonical basic neighborhoods of the “weak topology” are path connected, we know that $\mathbb{U}^{L}_{1}\subseteq\mathcal{M}(L\{\mathbf{t}\})$ (see (2)) is locally path connected. Furthermore, as $\zeta_{s,0}=\zeta_{0_{S},0}+s\in\mathbb{U}^{L}_{1}+s$ (see (4)), the density of the image of $L$ in $\mathbb{A}_{L}^{1}$ implies $\mathbb{A}_{L}^{1}=\bigcup_{s\in L}\mathbb{U}^{L}_{1}+s$, and this gives the local path connectedness of $\mathbb{A}_{L}^{1}$.

Observe that if the complete valuation ring $R$ is actually a field, then the absolute value $|\cdot|$ on $K$ is trivial, and the structure of $\mathbb{A}_{R}^{1}$ is already given in [3, 1.4.4]. However, because we need a concrete presentation of this space for the general case, we will first have a closer look at this case in Proposition 2.1. As a sidetrack, we verify the fact that if two fields $\mathfrak{k}_{1}$ and $\mathfrak{k}_{2}$ are endowed with the trivial norm, then $\mathbb{A}_{\mathfrak{k}_{1}}^{1}\cong\mathbb{A}_{\mathfrak{k}_{2}}^{1}$ if and only if the cardinalities of the sets of monic irreducible polynomials over them are the same.

Suppose that $R$ is not a field, or equivalently, the absolute value $|\cdot|$ on $K$ is non-trivial. Let us pick an arbitrary number $\omega\in[1,\infty)$, and

set $K^{\omega}$ to be the field $K$ equipped with the equivalent norm $|\cdot|^{\omega}$.

As $|a|^{\omega}\leq|a|$ ($a\in R$), we know that every semi-norm $\lambda\in\mathbb{A}_{K^{\omega}}^{1}$ restricts to an element in $\mathbb{A}_{R}^{1}$ and this gives a map

[TABLE]

The map $J_{\omega}^{\mathbb{A}}$ is injective, because for any $\mathbf{p}\in K[\mathbf{t}]$, there exists $a\in R$ with $a\mathbf{p}\in R[\mathbf{t}]$. On the other hand, the surjection $\tilde{Q}$ as in (3) produces an injection

[TABLE]

It is not hard to see that one actually has $\mathbb{A}_{R}^{1}=Q^{\mathbb{A}}(\mathbb{A}_{F}^{1})\cup\bigcup_{\omega\in[1,\infty)}J^{\mathbb{A}}_{\omega}(\mathbb{A}_{K^{\omega}}^{1})$ (Proposition 2.4). Note, however, that in the case of a general Banach integral domain $S$, elements in $\mathbb{A}_{S}^{1}$ cannot be described in such an easy way; for example, if $S$ is the ring $\mathbb{Z}$ equipped with the trivial norm, the description of elements in $\mathbb{A}_{S}^{1}$ requires the knowledge of all multiplicative ultrametric norms on $\mathbb{Q}$, instead of just the trivial norm on $\mathbb{Q}$ (which is the one induced from $S$).

The topology on $\mathbb{A}_{R}^{1}$ is more difficult to describe, and we will give a full presentation of it in Theorem 2.6, in the case when $K$ is algebraically closed. Using this description, we obtain in Theorem 2.8(c) that $\mathbb{A}_{R}^{1}$ is first countable if and only if $F$ is countable and $\mathbb{A}_{K}^{1}$ is first countable. We will also verify, in Proposition 2.10, that $\mathbb{A}_{R}^{1}$ is second countable if and only if $R$ is separable as a metric space (or equivalently, $K$ is a separable metric space). Moreover, the Berkovich uniform space $\mathbb{A}_{R}^{1}$ is the completion of $K\times(1,\infty)$ under the induced uniform structure (see Remark 2.7(c)).

We also show that $\mathbb{A}_{R}^{1}$ is both connected and locally path connected (parts (a) and (b) of Theorem 2.8). Notice that, unlike the case of $\mathbb{A}_{K}^{1}$, any two points in a connected open subset of $\mathbb{A}_{R}^{1}$ are joined by infinitely many paths inside that subset (see Remark 2.9). Consequently, the topology on $\mathbb{A}_{R}^{1}$ cannot be described using the “weak topology” of a $\mathbb{R}$-tree structure.

Finally, we will apply our main result to give a description of the Berkovich spectrum of the Banach group ring $R[G]$ of a cyclic group $G$ over $R$ (Corollary 2.13). In the case when $K$ is not necessarily algebraically closed, one may obtain information about $\mathcal{M}(R[G])$ by looking at the corresponding spectrum over the completion of the algebraic closure of $K$. In particular, we will take a closer look at the case when $R=\mathbb{Z}_{p}$ and $G$ is a cyclic $p$-group, for a fixed prime number $p$ (Example 2.15).

2. The main results

Let us begin with a careful presentation of the content of the second line of [3, 1.4.4]. More precisely, we will give a concrete description of $\mathbb{A}_{\mathfrak{k}}^{1}$ when $\mathfrak{k}$ is a field (not necessarily algebraically closed) equipped with the trivial norm.

In the following, $\mathfrak{k}[\mathbf{t}]_{\rm irr}$ is the set of all monic irreducible polynomials in $\mathfrak{k}[\mathbf{t}]$. Consider $\mathbf{q},\mathbf{q}^{\prime}\in\mathfrak{k}[\mathbf{t}]_{\rm irr}$ as well as $\kappa,\kappa^{\prime}\in\mathbb{R}_{+}$. We define a semi-norm $\gamma_{\mathbf{q},\kappa}$ on $\mathfrak{k}[\mathbf{t}]$ by

[TABLE]

(again, $0^{0}:=1$), where $\mathbf{r}_{0},\dots,\mathbf{r}_{n-1}\in\mathfrak{k}[\mathbf{t}]$ and $\mathbf{r}_{n}\in\mathfrak{k}[\mathbf{t}]^{\star}$ are elements with degrees strictly less than $\deg\mathbf{q}$. Note that, because the absolute value on $\mathfrak{k}$ is trivial, one has (see (5))

[TABLE]

The semi-norm $\gamma_{\mathbf{q},1}$ is independent of $\mathbf{q}$ and equals the trivial norm on $\mathfrak{k}[\mathbf{t}]$. Furthermore, when $\gamma_{\mathbf{q},\kappa}=\gamma_{\mathbf{q}^{\prime},\kappa^{\prime}}$, we have $\kappa=\kappa^{\prime}$, and we will also have $\mathbf{q}=\mathbf{q}^{\prime}$ if, in addition, $\kappa=\kappa^{\prime}<1$. For $\kappa\in(1,\infty)$ and $x\in\mathfrak{k}$, one has $\gamma_{\mathbf{t}-x,\kappa}(\mathbf{p})=\gamma_{\mathbf{t},\kappa}(\mathbf{p})=\kappa^{\deg\mathbf{p}}$ ($\mathbf{p}\in\mathfrak{k}[\mathbf{t}]$).

Notice that $\gamma_{\mathbf{t},\kappa}\in\mathbb{A}_{\mathfrak{k}}^{1}$ for any $\kappa\in\mathbb{R}_{+}$, but $\gamma_{\mathbf{q},\kappa}$ is not submultiplicative when $\deg\mathbf{q}>1$ and $\kappa>1$. Nevertheless, if $\kappa\in[0,1)$, then $\gamma_{\mathbf{q},\kappa}\in\mathbb{A}_{\mathfrak{k}}^{1}$ (regardless of the degree of $\mathbf{q}$).

On the other hand, for any $\tau\in[-1,\infty)$ and $\mathbf{q}\in\mathfrak{k}[\mathbf{t}]_{\rm irr}$, we consider $\delta_{\mathbf{q},\tau}$ to be the function from $\mathfrak{k}[\mathbf{t}]_{\rm irr}$ to $[-1,\infty)$ that vanishes outside the point $\mathbf{q}$ and sends $\mathbf{q}$ to $\tau$. Observe that $\delta_{\mathbf{q},0}$ is the constant zero function for all $\mathbf{q}\in\mathfrak{k}[\mathbf{t}]_{\rm irr}$.

Proposition 2.1**.**

Let $\mathfrak{k}$ be a field endowed with the trivial norm. Then $\mathbb{A}_{\mathfrak{k}}^{1}$ is canonically homeomorphic to the subspace X:=\big{\{}\delta_{\mathbf{q},\tau}:\mathbf{q}\in\mathfrak{k}[\mathbf{t}]_{\rm irr}\setminus\{\mathbf{t}\};\tau\in[-1,0)\big{\}}\cup\big{\{}\delta_{\mathbf{t},\tau}:\tau\in[-1,\infty)\big{\}} of the product space $\prod_{\mathfrak{k}[\mathbf{t}]_{\rm irr}}[-1,\infty)$. Consequently, $\mathbb{A}_{\mathfrak{k}}^{1}$ is connected and locally path connected. Moreover, $\mathbb{A}_{\mathfrak{k}}^{1}$ is first countable if and only if $\mathfrak{k}$ is at most countable.

Proof.

Since the second and the third statements follow easily from the first one, we will only establish the first statement (observe that $\mathfrak{k}[\mathbf{t}]_{\rm irr}$ is countable if and only if $\mathfrak{k}$ is at most countable). Let us show that

[TABLE]

In fact, consider any $\lambda\in\mathbb{A}_{\mathfrak{k}}^{1}$ and set $\tau:=|\mathbf{t}|_{\lambda}$. If $\tau\in\mathbb{R}_{+}\setminus\{1\}$, one may deduce from the Isosceles Triangle Principle that $\lambda=\gamma_{\mathbf{t},\tau}$.

Suppose that $\tau=1$. Clearly, $|\mathbf{p}|_{\lambda}\leq|\mathbf{p}|_{\gamma_{\mathbf{t},1}}$ ($\mathbf{p}\in\mathfrak{k}[\mathbf{t}]$) and we consider the case when $\lambda\neq\gamma_{\mathbf{t},1}$. Let

[TABLE]

As $P^{\lambda}$ is a non-zero prime ideal of $\mathfrak{k}[\mathbf{t}]$, we know that $P^{\lambda}=\mathbf{q}\cdot\mathfrak{k}[\mathbf{t}]$ for a unique element $\mathbf{q}\in\mathfrak{k}[\mathbf{t}]_{\rm irr}\setminus\{\mathbf{t}\}$ (note that $|\mathbf{t}|_{\lambda}=\tau=1$), and we put $\kappa:=|\mathbf{q}|_{\lambda}\in[0,1)$. For any $n\in\mathbb{Z}_{+}$ and $\mathbf{r}_{0},\dots,\mathbf{r}_{n}\in\mathfrak{k}[\mathbf{t}]$ with $\mathbf{r}_{n}\neq 0$ and $\deg\mathbf{r}_{k}<\deg\mathbf{q}$ ($k=0,\dots,n$), we know from the Isosceles Triangle Principle that $\left|\sum_{k=0}^{n}\mathbf{r}_{k}\mathbf{q}^{k}\right|_{\lambda}=\left|\sum_{k=0}^{n}\mathbf{r}_{k}\mathbf{q}^{k}\right|_{\gamma_{\mathbf{q},\kappa}}$.

Next, we define a map $\Phi:\mathbb{A}_{\mathfrak{k}}^{1}\to X$ by $\Phi(\gamma_{\mathbf{q},\kappa}):=\delta_{\mathbf{q},\kappa-1}$ ($\gamma_{\mathbf{q},\kappa}\in\mathbb{A}_{\mathfrak{k}}^{1}$). Clearly, $\Phi$ is bijective, and is continuous on the two subsets $\{\gamma_{\mathbf{q},\kappa}:\mathbf{q}\in\mathfrak{k}[\mathbf{t}]_{\rm irr};\kappa\in[0,1)\}$ and $\{\gamma_{\mathbf{t},\tau}:\tau\in[1,\infty)\}$. As $\Phi$ restricts to a homeomorphism on the compact subset $\mathcal{M}(\mathfrak{k}\{\mathbf{t}\})$ of $\mathbb{A}_{\mathfrak{k}}^{1}$, it is not hard to verify that $\Phi$ is actually bicontinuous. ∎

If $\mathfrak{k}$ is algebraically closed, then we have, by Equalities (7) and (8),

[TABLE]

Now, consider $\mathfrak{k}_{1}$ and $\mathfrak{k}_{2}$ to be two (not necessarily algebraically closed) fields equipped with the trivial norms. If the cardinalities of $\mathfrak{k}_{1}[\mathbf{t}]_{\rm irr}$ and $\mathfrak{k}_{2}[\mathbf{t}]_{\rm irr}$ are the same, then Proposition 2.1 implies that $\mathbb{A}_{\mathfrak{k}_{1}}^{1}$ is homeomorphic to $\mathbb{A}_{\mathfrak{k}_{2}}^{1}$. Conversely, suppose that we have a homeomorphism $\Psi:\mathbb{A}_{\mathfrak{k}_{1}}^{1}\to\mathbb{A}_{\mathfrak{k}_{2}}^{1}$. For any $\mathbf{q}_{1}\in\mathfrak{k}_{1}[\mathbf{t}]_{\rm irr}$, it is easy to see that $\Psi$ will send $\gamma_{\mathbf{q}_{1},0}$ to $\gamma_{\mathbf{q}_{2},0}$ for some $\mathbf{q}_{2}\in\mathfrak{k}_{2}[\mathbf{t}]_{\rm irr}$ (and vice-versa), because elements of the form $\gamma_{\mathbf{q}_{i},0}$ are all the free end points of maximal line-segments of $\mathbb{A}_{\mathfrak{k}_{i}}^{1}$ ($i=1,2$). Hence, we have a bijection between $\mathfrak{k}_{1}[\mathbf{t}]_{\rm irr}$ and $\mathfrak{k}_{2}[\mathbf{t}]_{\rm irr}$. These produce part (a) of the following corollary. The other parts of this corollary follow from part (a) and some well-known facts.

Corollary 2.2**.**

Suppose that $\mathfrak{k}_{1}$ and $\mathfrak{k}_{2}$ are two fields equipped with the trivial norm.

(a) $\mathbb{A}_{\mathfrak{k}_{1}}^{1}$ and $\mathbb{A}_{\mathfrak{k}_{2}}^{1}$ are homeomorphic if and only if the cardinality of $\mathfrak{k}_{1}[\mathbf{t}]_{\rm irr}$ equals that of $\mathfrak{k}_{2}[\mathbf{t}]_{\rm irr}$.

(b) If both $\mathfrak{k}_{1}$ and $\mathfrak{k}_{2}$ are infinite (as sets), then $\mathbb{A}_{\mathfrak{k}_{1}}^{1}$ and $\mathbb{A}_{\mathfrak{k}_{2}}^{1}$ are homeomorphic if and only if $\mathfrak{k}_{1}$ and $\mathfrak{k}_{2}$ has the same cardinality.

(c) If $\check{\mathfrak{k}}_{1}$ is the algebraic closure of $\mathfrak{k}_{1}$ (again, endowed with the trivial norm), then $\mathbb{A}_{\mathfrak{k}_{1}}^{1}$ is homeomorphic to $\mathbb{A}_{\check{\mathfrak{k}}}^{1}$.

In the case when the complete valuation ring $R$ is a field, then $|\cdot|$ is a trivial norm, and Proposition 2.1 gives the full description of $\mathbb{A}_{R}^{1}$.

From now on, we consider the case when $R$ is not a field; or equivalently, $|\cdot|$ is non-trivial.

Let us start with the following simple fact.

Lemma 2.3**.**

(a) Suppose that $S$ is an integral domain. If $P\subseteq S[\mathbf{t}]$ is a prime ideal with $I:=P\cap S$ being non-zero, then one can find a prime ideal $J$ (not necessarily non-zero) of $(S/I)[\mathbf{t}]$ such that $P=\Upsilon^{-1}(J)$, where $\Upsilon:S[\mathbf{t}]\to(S/I)[\mathbf{t}]$ is the quotient map.

(b) Suppose that $P\subseteq R[\mathbf{t}]$ is a non-zero prime ideal. Then $P\cap R\neq\{0_{R}\}$ if and only if $P=\tilde{Q}^{-1}(J)$ for a (not necessarily non-zero) prime ideal $J$ of $F[\mathbf{t}]$.

In fact, as $\ker\Upsilon=I[\mathbf{t}]\subseteq P$, we know that $P=\Upsilon^{-1}(\Upsilon(P))$ and $\Upsilon(P)$ is a prime ideal of $(S/I)[\mathbf{t}]$. Furthermore, part (b) follows from part (a) and the fact that the only non-zero prime ideal of $R$ is its maximal ideal (see e.g. Propositions 6 and 7 of [4, §VI.4.5]).

Proposition 2.4**.**

Let $R$ be a complete valuation ring, which is not a field. Under the notations as in the Introduction, one has

[TABLE]

Proof.

We have already seen that $Q^{\mathbb{A}}(\mathbb{A}_{F}^{1})\cup\bigcup_{\omega\in[1,\infty)}J_{\omega}^{\mathbb{A}}(\mathbb{A}_{K^{\omega}}^{1})\subseteq\mathbb{A}_{R}^{1}$. For the other inclusion, let us pick an arbitrary element $\lambda\in\mathbb{A}_{R}^{1}$.

Suppose that $\ker|\cdot|_{\lambda}\cap R\neq\{0_{R}\}$. Then Lemma 2.3(b) gives a unique prime ideal $J\subseteq F[\mathbf{t}]$ with $\ker|\cdot|_{\lambda}=\tilde{Q}^{-1}(J)$. From this, one concludes that $\lambda=\mu\circ\tilde{Q}$ for an element $\mu\in\mathbb{A}_{F}^{1}$, i.e. $\lambda\in Q^{\mathbb{A}}(\mathbb{A}_{F}^{1})$.

Suppose that $\ker|\cdot|_{\lambda}\cap R=\{0_{R}\}$. Let us extend $\lambda|_{R}$ to a function $|\cdot|_{\tilde{\lambda}}:K\to\mathbb{R}_{+}$ by setting $|r|_{\tilde{\lambda}}:=|r^{-1}|_{\lambda}^{-1}$ whenever $r\in K\setminus R$. It is not hard to check that $|\cdot|_{\tilde{\lambda}}$ is a multiplicative norm on $K$. For any $r\in K$, one has $|r|\leq 1$ if and only if $|r|_{\tilde{\lambda}}\leq 1$. Thus, $|\cdot|_{\tilde{\lambda}}$ is equivalent to $|\cdot|$, and one can find $\omega\in\mathbb{R}_{+}$ satisfying $|\cdot|_{\tilde{\lambda}}=|\cdot|^{\omega}$ (see e.g. Proposition 3 of [4, §VI.3.2]). Moreover, as $|a|_{\tilde{\lambda}}\leq|a|$ ($a\in R$), we know that $\omega\geq 1$. Finally, if we put

[TABLE]

then $|\cdot|_{\bar{\lambda}}$ is a well-defined multiplicative seminorm on $K[\mathbf{t}]$ with $|r|_{\bar{\lambda}}=|r|_{\tilde{\lambda}}$ ($r\in K$). Thus, $\bar{\lambda}\in\mathbb{A}_{K^{\omega}}^{1}$ and $\lambda=J_{\omega}^{\mathbb{A}}(\bar{\lambda})$. ∎

It is clear that both $Q^{\mathbb{A}}:\mathbb{A}_{F}^{1}\to\mathbb{A}_{R}^{1}$ and $J_{\omega}^{\mathbb{A}}:\mathbb{A}_{K^{\omega}}^{1}\to\mathbb{A}_{R}^{1}$ ($\omega\in[1,\infty)$) are homeomorphisms onto their images (by considering the compact spaces $\mathcal{M}(F\{n^{-1}\mathbf{t}\})$ and $\mathcal{M}(K^{\omega}\{n^{-1}\mathbf{t}\})$ for all $n\in\mathbb{N}$). If $\lambda\in\mathbb{A}_{K}^{1}$ and $\omega\in[1,\infty)$, let us define $\lambda^{\omega}\in\mathbb{A}_{K^{\omega}}^{1}$ by

[TABLE]

Obviously, $(\lambda,\omega)\mapsto J_{\omega}^{\mathbb{A}}(\lambda^{\omega})$ induces a continuous bijection

[TABLE]

Later on, we will verify that $\Lambda$ is actually a homeomorphism. We will also describe how the subspaces $Q^{\mathbb{A}}(\mathbb{A}_{K^{\omega}}^{1})$ and $\bigcup_{\omega\in[1,\infty)}J_{\omega}^{\mathbb{A}}(\mathbb{A}_{K^{\omega}}^{1})$ sit together in $\mathbb{A}_{R}^{1}$.

From now on, we will identify $\mathbb{A}_{F}^{1}$ as subspaces of $\mathbb{A}_{R}^{1}$, and may sometimes ignore the map $Q^{\mathbb{A}}$ if no confusion arises.

Lemma 2.5**.**

Suppose that $F$ is algebraically closed. Consider a net $\{(\lambda_{i},\omega_{i})\}_{i\in\mathfrak{I}}$ in $\mathbb{A}_{K}^{1}\times[1,\infty)$.

(a) If $\{\lambda_{i}^{\omega_{i}}\}_{i\in\mathfrak{I}}$ converges to an element in $\mathbb{A}_{F}^{1}$, then $\lim_{i}\omega_{i}=\infty$ and $\limsup_{i\in\mathfrak{I}}|\mathbf{t}|_{\lambda_{i}}\leq 1$.

(b) If $\lambda_{i}\in\mathbb{U}^{K}_{1}$ for all $i\in\mathfrak{I}$ (see (2)), $\lim_{i}\omega_{i}=\infty$ and $\lim_{i}|\mathbf{t}|_{\lambda_{i}}^{\omega_{i}}=\tau\in[0,1]$, then $\lambda_{i}^{\omega_{i}}\to\zeta_{0_{F},\tau}$.

Proof.

(a) By (9) and the assumption, one can find $(x,\tau)\in F^{\star}\times[0,1)\cup\{0_{F}\}\times\mathbb{R}_{+}$ such that $\lambda_{i}^{\omega_{i}}\to\zeta_{x,\tau}$. Assume on the contrary that $\{\omega_{i}\}_{i\in\mathfrak{I}}$ has a bounded subnet. Then there is a subnet $\{\omega_{i_{j}}\}_{j\in J}$ of $\{\omega_{i}\}_{i\in\mathfrak{I}}$ converging to a number $\omega_{0}\in[1,\infty)$. For any $a\in R$ with $|a|<1$, one gets from

[TABLE]

that $|a|^{\omega_{0}}=0$, and this contradicts the non-trivial assumption of $|\cdot|$. Hence, $\omega_{i}\to\infty$. On the other hand, since $\{|\mathbf{t}|_{\lambda_{i}}^{\omega_{i}}\}_{i\in\mathfrak{I}}$ converges to \big{|}\tilde{Q}(\mathbf{t})\big{|}_{\zeta_{x,\tau}}, one concludes that $\limsup_{i\in\mathfrak{I}}|\mathbf{t}|_{\lambda_{i}}\leq 1$, because otherwise, there exist $r>1$ and a subnet $\{\lambda_{i_{j}}\}_{j\in\mathfrak{J}}$ with $|\mathbf{t}|_{\lambda_{i_{j}}}\geq r$, which produces the contradiction that $|\mathbf{t}|_{\lambda_{i_{j}}}^{\omega_{i_{j}}}\geq r^{\omega_{i_{j}}}\to\infty$.

(b) If $b\in R$ with $|b|=1$, then $|\mathbf{t}-b|_{\lambda_{i}^{\omega_{i}}}=1$ (because $|\mathbf{t}|_{\lambda_{i}}<1$) for all $i\in\mathfrak{I}$. On the other hand, consider a polynomial $\mathbf{q}\in\ker\tilde{Q}$. There exist $a_{0},\dots,a_{n}\in\ker Q$ with $\mathbf{q}=\sum_{k=0}^{n}a_{k}\mathbf{t}^{k}$. Hence,

[TABLE]

and one has $|\mathbf{q}|_{\lambda_{i}^{\omega_{i}}}\leq\max_{k=0,\dots,n}|a_{k}|^{\omega_{i}}\to 0$ (along $i$).

Now, let $\mathbf{p}\in R[\mathbf{t}]^{\star}$ and $\{x_{1},\dots,x_{m}\}$ be all the non-zero roots of $\tilde{Q}(\mathbf{p})$ in $F$ (counting multiplicity). Pick $b_{1},\dots,b_{m}\in R$ with $Q(b_{l})=x_{l}$ ($l=1,\dots,m$). Then $|b_{l}|=1$ for all $l\in\{1,\dots,m\}$, and one can find $k\in\mathbb{Z}_{+}$ as well as $\mathbf{q}_{0}\in\ker\tilde{Q}$ satisfying

[TABLE]

The above and the hypothesis will then tell us that $|\mathbf{t}^{k}\cdot(\mathbf{t}-b_{1})\cdots(\mathbf{t}-b_{m})|_{\lambda_{i}^{\omega_{i}}}\to\tau^{k}$ and $|\mathbf{q}_{0}|_{\lambda_{i}^{\omega_{i}}}\to 0$. Consequently, $|\mathbf{p}|_{\lambda_{i}^{\omega_{i}}}\to\tau^{k}=|\tilde{Q}(\mathbf{p})|_{\zeta_{0_{F},\tau}}$ as is required (observe that $0\leq\tau\leq 1$). ∎

The following is our first main theorem that gives a full description of the topological space $\mathbb{A}_{R}^{1}$.

Theorem 2.6**.**

Let $R$ be a complete valuation ring which is not a field, $F$ be the residue field of $R$, and $K$ be the field of fractions of $R$ equipped with the induced absolute value $|\cdot|$. Suppose that $K$ is algebraically closed. We fix a cross section $\tilde{F}\subseteq R$ of $Q:R\to F$ that contains $0_{R}$.

(a) $\mathbb{A}_{F}^{1}$ is closed in $\mathbb{A}_{R}^{1}$.

(b) The map $\Lambda:\mathbb{A}_{K}^{1}\times[1,\infty)\to\bigcup_{\omega\in[1,\infty)}J_{\omega}^{\mathbb{A}}(\mathbb{A}_{K^{\omega}}^{1})$ in (10) is a homeomorphism.

(c) Suppose that $\{(\lambda_{i},\omega_{i})\}_{i\in\mathfrak{I}}$ is a net in $\mathbb{A}_{K}^{1}\times[1,\infty)$. Then $\{\lambda_{i}^{\omega_{i}}\}_{i\in\mathfrak{I}}$ converges to an element $\lambda_{0}\in\mathbb{A}_{F}^{1}$ if and only if $\omega_{i}\to\infty$ and either one of the following holds:

1. C1).

there exist $\tau_{1}\in[0,1)$ and $b\in R$ such that $|\mathbf{t}-b|_{\lambda_{i}}^{\omega_{i}}\to\tau_{1}$ (in this case, $\lambda_{0}=\zeta_{Q(b),\tau_{1}}$); 2. C2).

one can find $\tau_{2}\in(1,\infty)$ such that $|\mathbf{t}|_{\lambda_{i}}^{\omega_{i}}\to\tau_{2}$ (in this case, $\lambda_{0}=\zeta_{0_{F},\tau_{2}}$); 3. C3).

$|\mathbf{t}-c|_{\lambda_{i}}^{\omega_{i}}\to 1$* for any $c\in\tilde{F}$ (in this case, $\lambda_{0}=\zeta_{0_{F},1}$).*

(d) Under the homeomorphism in part (b), one may regard $\mathbb{A}_{K}^{1}\times[1,\infty)$ as an open dense subset of $\mathbb{A}_{R}^{1}$. Consequently, the image of $K\times(1,\infty)$ is dense in $\mathbb{A}_{R}^{1}$.

Proof.

The non-triviality of $|\cdot|$ produces $a_{0}\in R$ with $|a_{0}|\in(0,1)$. Moreover, since $K$ is algebraically closed, so is $F$. As in (9), one has

[TABLE]

(a) Suppose on the contrary that there is a net $\{(x_{i},\tau_{i})\}_{i\in\mathfrak{I}}$ in $F\times[0,1)\cup\{0_{F}\}\times[1,\infty)$ with $\zeta_{x_{i},\tau_{i}}\to\lambda^{\omega}$ for some $(\lambda,\omega)\in\mathbb{A}_{K}^{1}\times[1,\infty)$. Then, $0=|Q(a_{0})|_{\zeta_{x_{i},\tau_{i}}}\to|a_{0}|^{\omega}$, which is absurd.

(b) As said in the paragraph preceding Lemma 2.5, the map $\Lambda$ is a continuous bijection. If $\{(\lambda_{i},\omega_{i})\}_{i\in\mathfrak{I}}$ is a net in $\mathbb{A}_{K}^{1}\times[1,\infty)$ satisfying $\lambda_{i}^{\omega_{i}}\to\lambda_{0}^{\omega_{0}}$ for some $(\lambda_{0},\omega_{0})\in\mathbb{A}_{K}^{1}\times[1,\infty)$, then

[TABLE]

and so $\omega_{i}\to\omega_{0}$. From this, one can also deduce that $|\mathbf{p}|_{\lambda_{i}}\to|\mathbf{p}|_{\lambda_{0}}$ for every $\mathbf{p}\in R[\mathbf{t}]$. Consequently, the inverse of $\Lambda$ is continuous.

(c) $\Rightarrow)$. Suppose that $\{\lambda_{i}^{\omega_{i}}\}_{i\in\mathfrak{I}}$ converges to $\zeta_{Q(b),\tau_{1}}$ for some $b\in R$ and $\tau_{1}\in[0,1)$. Then we have $\omega_{i}\to\infty$ (by Lemma 2.5(a)) and $|\mathbf{t}-b|_{\lambda_{i}}^{\omega_{i}}\to|\mathbf{t}-Q(b)|_{\zeta_{Q(b),\tau_{1}}}=\tau_{1}$. This verifies Condition (C1).

On the other hand, suppose that $\{\lambda_{i}^{\omega_{i}}\}_{i\in\mathfrak{I}}$ converges to $\zeta_{0_{F},\kappa}$ for some $\kappa\in[1,\infty)$. Again, one has $\omega_{i}\to\infty$ because of Lemma 2.5(a). Moreover, as $\kappa\geq 1$, one knows that

[TABLE]

Hence, either Conditions (C2) or (C3) holds (depending on whether $\kappa=1$).

$\Leftarrow)$. Suppose that $\omega_{i}\to\infty$ and Condition (C1) holds. Then $|\mathbf{t}|_{\lambda_{i}-b}^{\omega_{i}}=|\mathbf{t}-b|_{\lambda_{i}}^{\omega_{i}}\to\tau_{1}<1$ (see (4)), which implies that $|\mathbf{t}|_{\lambda_{i}-b}<1$ eventually. By Lemma 2.5(b), we know that $(\lambda_{i}-b)^{\omega_{i}}\to\zeta_{0_{F},\tau_{1}}$ and hence $\lambda_{i}^{\omega_{i}}\to\zeta_{Q(b),\tau_{1}}$.

Secondly, suppose that $\omega_{i}\to\infty$ and Condition (C2) holds. We first note that

[TABLE]

because otherwise, one can find a subnet such that $|\mathbf{t}|_{\lambda_{i_{j}}}^{\omega_{i_{j}}}\to\infty$. Let us also show that

[TABLE]

In fact, as $\tau_{2}>1$, when $i$ is large, one has $|\mathbf{t}|_{\lambda_{i}}>1$, and hence $|\mathbf{t}-c^{\prime}|_{\lambda_{i}}=|\mathbf{t}|_{\lambda_{i}}$ ($c^{\prime}\in R$) and Relation (12) follows.

Now, for any $\mathbf{q}\in\ker\tilde{Q}$, we let $a_{0},\dots,a_{n}\in\ker Q$ be the elements with $\mathbf{q}=\sum_{k=0}^{n}a_{k}\mathbf{t}^{k}$. As $\kappa:=\max\{|a_{0}|,\dots,|a_{n}|\}<1$, we obtain from (11) an element $i_{0}\in\mathfrak{I}$ satisfying $\sup_{i\geq i_{0}}|\mathbf{t}|_{\lambda_{i}}<1/\kappa^{\frac{1}{n+1}}$ and hence for $i\geq i_{0}$,

[TABLE]

This means that $|\mathbf{q}|_{\lambda_{i}^{\omega_{i}}}\to 0$. Consider now a polynomial $\mathbf{p}\in R[\mathbf{t}]$ of degree $m$. Let $\{y_{1},\dots,y_{m}\}$ be all the roots of $\tilde{Q}(\mathbf{p})$ in $F$ (counting multiplicity) and let $c_{1},\dots,c_{m}$ be elements in $R$ satisfying $Q(c_{l})=y_{l}$ ($l=1,\dots,m$). Then one can find $\mathbf{q}_{0}\in\ker\tilde{Q}$ with

[TABLE]

Thus, (12) and the above imply that $|\mathbf{p}|_{{\lambda_{i}^{\omega_{i}}}}\to\tau_{2}^{m}=\zeta_{0_{F},\tau_{2}}(\tilde{Q}(\mathbf{p}))$ (notice that $\tau_{2}\geq 1$).

Finally, we consider the case when $\omega_{i}\to\infty$ and Condition (C3) holds. As $0_{R}\in\tilde{F}$, we know that $|\mathbf{t}|_{\lambda_{i}}^{\omega_{i}}\to 1$ and Relation (11) is satisfied. Moreover, we have $|\mathbf{t}-c^{\prime}|_{\lambda_{i}}^{\omega_{i}}\to 1$ $(c^{\prime}\in R)$. Indeed, if $c^{\prime}\in R$, one can find $c\in\tilde{F}$ satisfying $|c^{\prime}-c|<1$. As $\omega_{i}\to\infty$, there exists $i_{0}\in\mathfrak{I}$ such that for any $i\geq i_{0}$, one has both $|c^{\prime}-c|^{\omega_{i}}<1/2$ as well as $|\mathbf{t}-c|_{\lambda_{i}}^{\omega_{i}}>1/2$, and hence

[TABLE]

This implies that $|\mathbf{t}-c^{\prime}|_{\lambda_{i}}^{\omega_{i}}\to 1$ as required. Now, it follows from the same argument as that for Condition (C2) that $|\mathbf{p}|_{{\lambda_{i}^{\omega_{i}}}}\to 1=\zeta_{0_{F},1}(\tilde{Q}(\mathbf{p}))$, for any $\mathbf{p}\in R[\mathbf{t}]$.

(d) If $\tau_{1}<1$, then it follows easily from part (c) that $\zeta_{b,\tau_{1}^{1/n}}^{n}\to\zeta_{Q(b),\tau_{1}}$ for every $b\in R$. If $\tau_{2}>1$, it is not hard to see from part (c) that $\zeta_{0_{K},\tau_{2}^{1/n}}^{n}\to\zeta_{0_{F},\tau_{2}}$. Furthermore, since $\zeta_{0_{K},1}(\mathbf{t}-c)=1$ for all $c\in R$, we know from part (c) that $\zeta_{0_{K},1}^{n}\to\zeta_{0_{F},1}$. These establish the first statement. The second statement follows from the first one, part (b) as well as the fact that $K$ is dense in $\mathbb{A}_{K}^{1}$. ∎

From now on, we will also identify $\mathbb{A}_{K}^{1}\times[1,\infty)$ with $\bigcup_{\omega\in[1,\infty)}J_{\omega}^{\mathbb{A}}(\mathbb{A}_{K^{\omega}}^{1})$ (i.e. $(\lambda,\omega)$ is identified with $\lambda^{\omega}$) as well as $K\times[1,\infty)$ with its image in $\bigcup_{\omega\in[1,\infty)}J_{\omega}^{\mathbb{A}}(\mathbb{A}_{K^{\omega}}^{1})$ (i.e. $(r,\omega)$ is identified with $\zeta_{r,0}^{\omega}$).

Remark 2.7**.**

(a) In the proof of Theorem 2.6(c), we see that $\lambda_{i}^{\omega_{i}}\to\zeta_{0_{F},\kappa}$ for some $\kappa\in[1,\infty)$ if and only if $\omega_{i}\to\infty$ and $|\mathbf{t}-c^{\prime}|_{\lambda_{i}}^{\omega_{i}}\to\kappa$, for all $c^{\prime}\in R$.

(b) Unlike the case of $\tau_{2}\in(1,\infty)$, the requirement $|\mathbf{t}|_{\lambda_{i}}^{\omega_{i}}\to 1$ does not imply $\lambda_{i}^{\omega_{i}}\to\zeta_{0_{F},1}$. For example, if we set $\lambda_{i}:=\zeta_{1_{K},0}$ ($i\in\mathfrak{I}$), then $|\mathbf{t}|_{\lambda_{i}}^{\omega_{i}}=1$ but $|\mathbf{t}-1|_{\lambda_{i}}^{\omega_{i}}=0$ for all $i\in\mathfrak{I}$.

(c) Consider the uniform structure on $K\times(1,\infty)$ that is induced by the fundamental system of entourages of the form:

[TABLE]

where $\epsilon$ runs through all positive numbers and $X$ runs through all non-empty finite subsets of $R[\mathbf{t}]$. As a uniform space, $\mathbb{A}_{R}^{1}$ is the completion of $K\times(1,\infty)$ under this structure (because of Theorem 2.6(d)).

Before continuing our discussion, let us set some more notations. For any $\mathbf{p}\in R[\mathbf{t}]$ as well as $\tau,\epsilon\in\mathbb{R}_{+}$, we denote

[TABLE]

It follows from the definition that $U^{\mathbf{p}}_{\tau,\epsilon}$ an open subset of $\mathbb{A}_{R}^{1}$.

Proposition 2.4 as well as parts (a) and (b) of Theorem 2.6 tell us that $\mathbb{A}_{R}^{1}$ contains the product space $\mathbb{A}_{K}^{1}\times[1,\infty)$ as an open subset with its complement being $\mathbb{A}_{F}^{1}$. Moreover, by Lemma 2.5(a), we know that for every $\kappa\geq 1$, the subset $\mathbb{A}_{K}^{1}\times[1,\kappa]$ is closed in $\mathbb{A}_{R}^{1}$.

Since the topology on the product space $\mathbb{A}_{K}^{1}\times[1,\infty)$ is well-known and Proposition 2.1 describes the topological space $\mathbb{A}_{F}^{1}$, the topology of $\mathbb{A}_{R}^{1}$ can be determined if one knows the description of neighborhood bases over elements in $\mathbb{A}_{F}^{1}$. Through Theorem 2.6, these bases will be described as follows:

1. N1)

Consider $\tau\in[0,1)$ and $b\in\tilde{F}$ (see Theorem 2.6). If $\kappa\geq 1$ and $0<\epsilon<1-\tau$, one can check easily, using Relation (9), that

[TABLE]

Thus, by Theorem 2.6(c) and Proposition 2.1, the countable collection:

[TABLE]

constitutes an open neighborhood basis of $\zeta_{Q(b),\tau}$. 2. N2)

Consider $\tau>1$. If $\kappa\geq 1$ and $0<\epsilon<\tau-1$, one may verify that

[TABLE]

Again, Theorem 2.6(c) and Proposition 2.1 imply that the countable collection:

[TABLE]

constitutes an open neighborhood basis of $\zeta_{0_{F},\tau}$. 3. N3)

Let $\mathcal{F}$ be the collection of all finite subsets of $\tilde{F}$ such that each of them contains $0_{R}$. If $X\in\mathcal{F}$, $\kappa\geq 1$ and $\epsilon\in(0,1)$, it is not hard to see that

[TABLE]

It now follows from Theorem 2.6(c) and Proposition 2.1 that the collection:

[TABLE]

constitutes an open neighborhood basis of $\zeta_{0_{F},1}$.

One may use the above information to obtain certain topological properties of $\mathbb{A}_{R}^{1}$. The following is an illustration.

Theorem 2.8**.**

Let $R$ be a complete valuation ring which is not a field, with its field of fractions being algebraically closed. The following properties hold.

(a) $\mathbb{A}_{R}^{1}$ is path connected.

(b) $\mathbb{A}_{R}^{1}$ is locally path connected.

(c) $\mathbb{A}_{R}^{1}$ is first countable if and only if $F$ is countable and $\mathbb{A}_{K}^{1}$ is first countable.

Proof.

(a) By Proposition 2.1, $\mathbb{A}_{F}^{1}$ is path connected. Moreover, as said in the Introduction, $\mathbb{A}_{K}^{1}$ is path connected, and this gives the path connectedness of $\mathbb{A}_{K}^{1}\times[1,\infty)$. Consider $(x_{0},\tau_{0})\in F\times[0,1)\cup\{0_{F}\}\times[1,\infty)$ and $(\lambda_{0},\omega_{0})\in\mathbb{A}_{K}^{1}\times[1,\infty)$. As in the proof of Theorem 2.6(d), one has $\zeta_{0_{K},0}^{\omega}\to\zeta_{0_{F},0}$ (when $\omega\to\infty$), and this produces a path joining $\zeta_{0_{K},0}$ to $\zeta_{0_{F},0}$. By considering a path in $\mathbb{A}_{F}^{1}$ (respectively, $\mathbb{A}_{K}^{1}\times[1,\infty)$) joining $\zeta_{x_{0},\tau_{0}}$ to $\zeta_{0_{F},0}$ (respectively, joining $\lambda_{0}^{\omega_{0}}$ to $\zeta_{0_{K},0}$), one obtains a path that joins $\zeta_{x_{0},\tau_{0}}$ to $\lambda_{0}^{\omega_{0}}$.

(b) As recalled in the Introduction, $\mathbb{A}_{K}^{1}$ is locally path connected, and hence so is the open subset $\mathbb{A}_{K}^{1}\times[1,\infty)$ of $\mathbb{A}_{R}^{1}$.

Let us now verify the path connectedness of open sets of the form as in (N1). Let $\tau\in[0,1)$ and $b\in\tilde{F}$. Fix $\kappa\geq 1$ as well as $\epsilon\in(0,1-\tau)$. For simplicity, we denote

[TABLE]

It is easy to see that \big{\{}\zeta_{Q(b),\upsilon}:\upsilon\in[0,1);|\upsilon-\tau|<\epsilon\big{\}} is path connected (see Proposition 2.1). Thus, in order to verify the path connectedness of $U^{\mathbf{t}-b}_{\tau,\epsilon}\setminus\mathbb{A}_{K}^{1}\times[1,\kappa]$, it suffices to show that an arbitrary fixed element $\lambda_{0}^{\omega_{0}}\in U^{\mathbf{t}-b}_{\tau,\epsilon}\cap V^{\kappa}$ can be joined to $\zeta_{Q(b),\tau}$ through a path inside $U^{\mathbf{t}-b}_{\tau,\epsilon}\setminus\mathbb{A}_{K}^{1}\times[1,\kappa]$.

Notice that the subset

[TABLE]

is open in $\mathbb{A}_{K}^{1}$ and contains $\lambda_{0}^{\omega_{0}}$. Hence, by the locally path connectedness of $\mathbb{A}_{K}^{1}$ and the density of the image of $K$ in $\mathbb{A}_{K}^{1}$, one may assume that $\lambda_{0}=\zeta_{s_{0},0}$ for an element $s_{0}\in K$. The requirement of $\zeta_{s_{0},0}^{\omega_{0}}\in U^{\mathbf{t}-b}_{\tau,\epsilon}$ implies that

[TABLE]

For every \upsilon\in\big{[}0,|s_{0}-b|\big{]}, the relation

[TABLE]

tells us that $\zeta_{s_{0},\upsilon}^{\omega_{0}}\in U^{\mathbf{t}-b}_{\tau,\epsilon}$. Thus, \big{\{}\zeta_{s_{0},\upsilon}^{\omega_{0}}:\upsilon\in[0,|s_{0}-b|]\big{\}} is a path in $U^{\mathbf{t}-b}_{\tau,\epsilon}$ joining $\zeta_{s_{0},0}^{\omega_{0}}$ to $\zeta_{s_{0},|s_{0}-b|}^{\omega_{0}}=\zeta_{b,|s_{0}-b|}^{\omega_{0}}$.

Similarly, the path \big{\{}\zeta_{b,\upsilon}^{\omega_{0}}:\upsilon\text{ is in between }|s_{0}-b|\text{ and }\tau^{1/\omega_{0}}\big{\}} that joins $\zeta_{b,|s_{0}-b|}^{\omega_{0}}$ to $\zeta_{b,\tau^{1/\omega_{0}}}^{\omega_{0}}$ also lies inside $U^{\mathbf{t}-b}_{\tau,\epsilon}$. Moreover, it follows from

[TABLE]

and Theorem 2.6(c) that the net $\{\zeta_{b,\tau^{1/\omega}}^{\omega}\}_{\omega\geq\omega_{0}}$ converges to $\zeta_{Q(b),\tau}$ when $\omega\to\infty$. Therefore,

[TABLE]

is a path in $\mathbb{A}_{K}^{1}$ joining $\zeta_{b,\tau^{1/\omega_{0}}}^{\omega_{0}}$ to $\zeta_{Q(b),\tau}$. Furthermore, (13) also ensures that this path lies inside $U^{\mathbf{t}-b}_{\tau,\epsilon}$. Consequently, $U^{\mathbf{t}-b}_{\tau,\epsilon}\setminus\mathbb{A}_{K}^{1}\times[1,\kappa]$ is path connected (note that $\omega_{0}$ as well as all the $\omega$ in the above are strictly bigger than $\kappa$).

In the same way, for any fixed $\tau\in(1,\infty)$, one can establish the path connectedness of open sets as in (N2), i.e., $U^{\mathbf{t}}_{\tau,\epsilon}\setminus\mathbb{A}_{K}^{1}\times[1,\kappa]$ for $\kappa\geq 1$ and $\epsilon\in(0,\tau-1)$.

It remains to show that open sets of the form as in (N3), namely, the set $\bigcap_{c\in X}U^{\mathbf{t}-c}_{1,\epsilon}\setminus\mathbb{A}_{K}^{1}\times[1,\kappa]$ (for $X\in\mathcal{F}$, $\kappa\geq 1$ and $\epsilon\in(0,1)$) is path connected. As in the above, we need to find a path in $\bigcap_{c\in X}U^{\mathbf{t}-c}_{1,\epsilon}\setminus\mathbb{A}_{K}^{1}\times[1,\kappa]$ that joins an arbitrary chosen element $\lambda_{0}^{\omega_{0}}\in\bigcap_{c\in X}U^{\mathbf{t}-c}_{1,\epsilon}\cap V^{\kappa}$ to $\zeta_{0_{F},1}$. Again, we may assume that $\lambda_{0}=\zeta_{s_{0},0}$ for an element $s_{0}\in K$. The condition $\zeta_{s_{0},0}^{\omega_{0}}\in\bigcap_{c\in X}U^{\mathbf{t}-c}_{1,\epsilon}$ implies that

[TABLE]

and, in particular, $|s_{0}|^{\omega_{0}}\in(1-\epsilon,1+\epsilon)$ (because $0_{R}\in X$ by the assumption of $\mathcal{F}$). For every $\upsilon\in[0,|s_{0}|]$, the equality

[TABLE]

will then ensure that $\zeta_{s_{0},\upsilon}^{\omega_{0}}\in\bigcap_{c\in X}U^{\mathbf{t}-c}_{1,\epsilon}$. In addition, a similar relation as (14) tells us that the path \big{\{}\zeta_{0_{K},\upsilon}^{\omega_{0}}:\upsilon\text{ is in between }|s_{0}|\text{ and }1\big{\}} lies inside $\bigcap_{c\in X}U^{\mathbf{t}-c}_{1,\epsilon}$. On the other hand, using Theorem 2.6(c), we see that $\{\zeta_{0_{K},1}^{\omega}\}_{\omega\geq\omega_{0}}$ converges to $\zeta_{0_{F},1}$ when $\omega\to\infty$, and this produces a path joining $\zeta_{0_{K},1}^{\omega_{0}}$ to $\zeta_{0_{F},1}$. Moreover, the equalities $\zeta_{0_{K},1}(\mathbf{t}-c)=1$ ($c\in X$) gives $\zeta_{0_{K},1}^{\omega}\in\bigcap_{c\in X}U^{\mathbf{t}-c}_{1,\epsilon}$ ($\omega\geq\omega_{0}$). Consequently, we obtain a path in $\bigcap_{c\in X}U^{\mathbf{t}-c}_{1,\epsilon}\setminus\mathbb{A}_{K}^{1}\times[1,\kappa]$ joining $\lambda_{0}^{\omega_{0}}$ to $\zeta_{0_{F},1}$.

(c) Clearly, the countability of $F$ and the first countability of $\mathbb{A}_{K}^{1}$ will imply the first countability of $\mathbb{A}_{R}^{1}$. Conversely, suppose that $\mathbb{A}_{R}^{1}$ is first countable. Then $\mathbb{A}_{K}^{1}\times[1,\infty)$ is first countable and so is $\mathbb{A}_{K}^{1}$. Moreover, there is a countable subcollection

[TABLE]

that form a neighborhood basis for $\zeta_{0_{F},1}$ in $\mathbb{A}_{R}^{1}$. Let $C:=\bigcup_{k\in\mathbb{N}}X_{k}$. We know that whenever a net $\{(\lambda_{i},\omega_{i})\}_{i\in\mathfrak{I}}$ in $\mathbb{A}_{K}^{1}\times[1,\infty)$ satisfying $\omega_{i}\to\infty$ as well as $|\mathbf{t}-b|_{\lambda_{i}}^{\omega_{i}}\to 1$ ($b\in C$), we have $\lambda_{i}^{\omega_{i}}\to\zeta_{0_{F},1}$. Assume on the contrary that $\tilde{F}$ is uncountable, then there exists $c_{0}\in\tilde{F}\setminus C$ (which implies $|c_{0}-b|=1$ for all $b\in C$). If we set $\omega_{n}:=n$ and $\lambda_{n}:=\zeta_{c_{0},0}$ ($n\in\mathbb{N}$), then $|\mathbf{t}-b|_{\lambda_{n}}^{\omega_{n}}=1$ ($n\in\mathbb{N}$), for each $b\in C$, but $|\mathbf{t}-c_{0}|_{\lambda_{n}}^{\omega_{n}}=0$ ($n\in\mathbb{N}$), which contradicts $\lambda_{n}^{\omega_{n}}\to\zeta_{0_{F},1}$. ∎

Remark 2.9**.**

Unlike $\mathbb{A}_{K}^{1}$, the topological space $\mathbb{A}_{R}^{1}$ is far from having a $\mathbb{R}$-tree structure. Actually, for any connected open subset $V\subseteq\mathbb{A}_{R}^{1}$ and any $\lambda_{1},\lambda_{2}\in V$, there exist infinitely many paths in $V$ joining $\lambda_{1}$ and $\lambda_{2}$.

In fact, consider $i\in\{1,2\}$. Let $U_{i}\subseteq V$ be a path connected open neighborhood of $\lambda_{i}$ (see Theorem 2.8(b)) and fix any $(\mu_{i},\kappa_{i})\in\mathbb{A}_{K}^{1}\times(1,\infty)\cap U_{i}$ (see Theorem 2.6). Thus, $\lambda_{i}$ is joined to $(\mu_{i},\kappa_{i})$ through a path inside $U_{i}$. Choose a connected open neighborhood $W_{i}$ of $\mu_{i}$ in $\mathbb{A}_{K}^{1}$ as well as a number $\epsilon\in(0,1-\kappa_{i})$ such that $W_{i}\times(\kappa_{i}-\epsilon,\kappa_{i}+\epsilon)\subseteq U_{i}$. Pick any $\nu_{i}\in W_{i}$. There exist infinitely many paths in $W_{i}\times(\kappa_{i}-\epsilon,\kappa_{i}+\epsilon)$ joining $(\mu_{i},\kappa_{i})$ to $(\nu_{i},\kappa_{i})$. As $V$ is path connected, there exists a path in $V$ joining $(\nu_{1},\kappa_{1})$ to $(\nu_{2},\kappa_{2})$. In this way, we obtain infinitely many paths joining $\lambda_{1}$ and $\lambda_{2}$.

Similar to Theorem 2.8(c), there is also a description of the second countability of $\mathbb{A}_{R}^{1}$. In fact, one has the following more general result. This result could be a known fact, but since we do not find an explicit reference for it, we present its simple argument here.

Proposition 2.10**.**

If $(S,\|\cdot\|)$ is a commutative unital Banach ring with multiplicative norm, the following statements are equivalent.

1. S1).

$\mathbb{A}_{S}^{1}$* is second countable.* 2. S2).

$S$* is separable as a metric space.* 3. S3).

$\mathcal{M}(S\{n^{-1}\mathbf{t}\})$* is metrizable for every $n\in\mathbb{N}$.*

Proof.

$(S1)\Rightarrow(S2)$. For any $a\in S$, we defined $\zeta_{a}\in\mathbb{A}_{S}^{1}$ by $\zeta_{a}(\mathbf{p}):=\|\mathbf{p}(a)\|$ ($\mathbf{p}\in S[\mathbf{t}]$). It is easy to see that $a\mapsto\zeta_{a}$ is a homeomorphism from $S$ onto its image in $\mathbb{A}_{S}^{1}$. Thus, $S$ is also second countable and hence is separable.

$(S2)\Rightarrow(S3)$. Suppose that $S$ contains a countable dense subset $S_{0}$. We denote by $\mathcal{F}_{0}$ the collection of non-empty finite subsets of $S[\mathbf{t}]$ consisting of polynomials with coefficients in $S_{0}$. For a fixed $n\in\mathbb{N}$, since $|\mathbf{t}|_{\mu}\leq n$ for any $\mu\in\mathcal{M}(S\{n^{-1}\mathbf{t}\})$, it is not hard to see that the countable family

[TABLE]

(see Relation (1) for the meaning of $E_{1/k}^{X}$) forms a fundamental system of entourages for the Berkovich uniform structure on $\mathcal{M}(S\{n^{-1}\mathbf{t}\})$. Consequently, the Berkovich uniform structure (and hence the topology defined by it) is pseudo-metrizable (see, e.g., Theorem 13 in chapter 6 of [6]). However, since the topology on $\mathcal{M}(S\{n^{-1}\mathbf{t}\})$ is Hausdorff, we conclude that this topology is indeed metrizable.

$(S3)\Rightarrow(S1)$. Since $\mathcal{M}(S\{n^{-1}\mathbf{t}\})$ is a compact metric space, it is second countable, and so, the open subset $\mathbb{U}^{S}_{n}\subseteq\mathcal{M}(S\{n^{-1}\mathbf{t}\})$ (see (2)) is also second countable. Consequently, $\mathbb{A}_{S}^{1}$ is second countable. ∎

Clearly, the Banach ring $R$ is separable (equivalently, second countable) if and only if $K$ is separable as a metric space. Note also that if a complete valued field is algebraically closed and spherically complete, then it is not separable (see e.g. [5, Remark 1.4]). However, the completion of the algebraic closure of a separable complete valued field is again separable (see e.g. [5, Remark 1.3]).

Example 2.11**.**

Let $\mathfrak{k}$ be a field equipped with the trivial norm. By Proposition 2.1, one can define a metric on $\mathbb{A}_{\mathfrak{k}}^{1}$ through the “geodesic distance” $d_{G}$ of two given points. If $\mathfrak{k}$ is uncountable, then Proposition 2.10 tells us that $\mathbb{A}_{\mathfrak{k}}^{1}$ is not metrizable, and hence the topology induced by $d_{G}$ is different from the pointwise convergence topology on $\mathbb{A}_{\mathfrak{k}}^{1}$.

On the other hand, suppose that $\mathfrak{k}$ is at most countable and $\mathbf{p}_{1},\mathbf{p}_{2},\dots$ are all the elements in $\mathfrak{k}[\mathbf{t}]_{\rm irr}$. If we rescale the “geodesic distance” so that $d_{G}(\gamma_{\mathbf{p}_{k},0},\gamma_{\mathbf{t},1})\to 0$ when $k\to\infty$, then it is not hard to see that this metric defines the topology of pointwise convergence on $\mathbb{A}_{\mathfrak{k}}^{1}$. In this case, $\mathbb{A}_{\mathfrak{k}}^{1}$ is the following closed subspace of $\mathbb{R}^{2}$:

[TABLE]

In the following, we will have a look at the set of “type I points” of $\mathbb{A}_{R}^{1}$, i.e. those multiplicative semi-norms $\lambda\in\mathbb{A}_{R}^{1}$ with $\ker|\cdot|_{\lambda}\neq\{0_{R}\}$.

Corollary 2.12**.**

Let $R$, $(K,|\cdot|)$, $F$ and $\tilde{F}$ be as in Theorem 2.6. Denote

[TABLE]

(a) $\mathbb{A}_{R,Z}^{1}=\mathbb{A}_{F}^{1}\cup K\times[1,\infty)$ (and hence, $\mathbb{A}_{R,Z}^{1}$ is dense in $\mathbb{A}_{R}^{1}$).

(b) Suppose that $\{(s_{i},\omega_{i})\}_{i\in\mathfrak{I}}$ is a net in $K\times[1,\infty)$.

• •

$\zeta_{s_{i},0}^{\omega_{i}}\to\zeta_{Q(b),\tau_{1}}$* for some $b\in R$ and $\tau_{1}\in[0,1)$ if and only if $\omega_{i}\to\infty$ and $|s_{i}-b|^{\omega_{i}}\to\tau_{1}$;*

• •

$\zeta_{s_{i},0}^{\omega_{i}}\to\zeta_{0_{F},\tau_{2}}$* for a number $\tau_{2}\in(1,\infty)$ if and only if $\omega_{i}\to\infty$ and $|s_{i}|^{\omega_{i}}\to\tau_{2}$;*

• •

$\zeta_{s_{i},0}^{\omega_{i}}\to\zeta_{0_{F},1}$* if and only if $\omega_{i}\to\infty$ and $|s_{i}-c|^{\omega_{i}}\to 1$, for any $c\in\tilde{F}$.*

(c) If $\{(s_{i},\omega_{i})\}_{i\in\mathfrak{I}}$ is a net in $K\times[1,\infty)$ such that $\zeta_{s_{i},0}^{\omega_{i}}\to\zeta_{Q(b),\tau_{1}}$ for some $b\in R$ and $\tau_{1}\in[0,1)$, then $s_{i}$ eventually belongs to the “open ball” of $K$ of radius $1$ and center $b$.

Proof.

Since part (b) follows from Theorem 2.6 and part (c) follows directly from part (b), we will only establish part (a). In fact, it is clear that $\mathbb{A}_{F}^{1}\cup K\times[1,\infty)\subseteq\mathbb{A}_{R,Z}^{1}$. Consider any element $\lambda\in\mathbb{A}_{R,Z}^{1}$. If $\ker|\cdot|_{\lambda}\cap R\neq\{0_{R}\}$, the argument of Proposition 2.4 tells us that $\lambda\in\mathbb{A}_{F}^{1}$. Suppose that $\ker|\cdot|_{\lambda}\cap R=\{0_{R}\}$. As in the proof of Proposition 2.4, there is a unique positive number $\omega\in[1,\infty)$ such that $\lambda$ extends to an element $\bar{\lambda}\in\mathbb{A}_{K^{\omega}}^{1}$. Since $\ker|\cdot|_{\bar{\lambda}}$ (which contains $\ker|\cdot|_{\lambda}$) is a non-zero prime ideal of $K[\mathbf{t}]$ and $K$ is algebraically closed, one can find a (unique) element $s\in K$ with $\ker|\cdot|_{\bar{\lambda}}=(\mathbf{t}-s)\cdot K[\mathbf{t}]$ and it is not hard to check that $\lambda=\zeta_{s,0}^{\omega}$. ∎

In the following, we will use Theorem 2.6 to obtain the Berkovich spectra of Banach group rings of finite cyclic groups over $R$. Let $G$ be a finite abelian group and $(S,\|\cdot\|)$ be a commutative unital Banach ring. We denote by $S[G]$ the group ring of $G$ over $S$, and endowed it with the norm $\|\sum_{g\in G}a_{g}g\|:=\max_{g\in G}\|a_{g}\|$. Clearly, $S[G]$ is a commutative unital Banach ring.

Corollary 2.13**.**

Let $R$, $F$, $Q$ and $(K,|\cdot|)$ be as in Theorem 2.6. Denote by $|\cdot|_{0}$ the trivial norm on $F$. Let $G$ be a cyclic group of order $M$ with $u$ being a generator of $G$. Suppose $b_{1},\dots,b_{n}$ are all the distinct $M$-th roots of unity in $K$.

(a) If we set $\left|\sum_{l=0}^{M-1}a_{l}u^{l}\right|_{\alpha_{b_{k}}^{\omega}}:=\left|\sum_{l=0}^{M-1}a_{l}b_{k}^{l}\right|^{\omega}$ and $\left|\sum_{l=0}^{M-1}a_{l}u^{l}\right|_{\beta_{Q(b_{k})}}:=\left|\sum_{l=0}^{M-1}Q(a_{l}b_{k}^{l})\right|_{0}$, then $\mathcal{M}(R[G])=\{\alpha_{b_{k}}^{\omega}:\omega\in[1,\infty);k=1,\dots,n\}\cup\{\beta_{Q(b_{k})}:k=1,\dots,n\}$.

(b) As a topological space, $\mathcal{M}(R[G])$ consists of $n$ intervals of the form $[1,\infty]$ corresponding to the elements $b_{1},\dots,b_{n}$ such that the “$1$-ends” of all these intervals are free, while the “$\infty$-ends” of the two intervals corresponding to $b_{k}$ and $b_{l}$ are identified with each other if $Q(b_{k})=Q(b_{l})$.

Proof.

(a) There is a contractive and surjective ring homomorphism

[TABLE]

sending $\mathbf{t}$ to $u$, and it is not hard to check that $\ker q_{G}=(\mathbf{t}^{M}-1)\cdot R\{\mathbf{t}\}$. Hence, one may regard $\mathcal{M}(R[G])$ as a topological subspace of $\mathcal{M}(R\{\mathbf{t}\})$ through $q_{G}$ in the following way:

[TABLE]

It is obvious that $\alpha_{b_{k}}^{\omega}$ and $\beta_{Q(b_{k})}$ are well-defined elements in $\mathcal{M}(R[G])$, and they can be identified, respectively, with the elements $\zeta_{b_{k},0}^{\omega}$ and $\zeta_{Q(b_{k}),0}$ in $\mathbb{A}_{R,Z}^{1}$.

On the other hand, let us pick an arbitrary element $\lambda\in\mathcal{M}(R[G])$. By Corollary 2.12(a), either $\lambda=\zeta_{s,0}^{\omega}$ for a unique $(s,\omega)\in K\times[1,\infty)$ or $\lambda\in\mathbb{A}_{F}^{1}$. In the first case, the condition $|\mathbf{t}^{M}-1|_{\zeta_{s,0}^{\omega}}=0$ will force $s=b_{k}$ for some $k\in\{1,\dots,n\}$, which means that $\lambda=\alpha_{b_{k}}^{\omega}$. In the second case, there exist $x\in F$ and $\tau\in\mathbb{R}_{+}$ satisfying $\lambda=\zeta_{x,\tau}$, and the condition $|\mathbf{t}^{M}-1|_{\zeta_{x,\tau}}=0$ tells us that $\tau=0$ and $x^{M}=1$. Since $Q(b_{1}),\dots,Q(b_{n})$ are all the roots of $\mathbf{t}^{M}-1$ in $F$, we conclude that $\lambda=\beta_{Q(b_{k})}$ for some $k=1,\dots,n$.

(b) By Corollary 2.12(a), it is not hard to see that the subset $\{\zeta_{b_{k},0}^{\omega}:\omega\in[1,\infty);k=1,\dots,n\}$ of $\mathbb{A}_{R,Z}^{1}$ are $n$ disjoint intervals of the form $[1,\infty)$. Assume that $(s_{i},\omega_{i})\in\{b_{1},\dots,b_{n}\}\times[1,\infty)$ ($i\in\mathfrak{I}$) such that $\{\zeta_{s_{i},0}^{\omega_{i}}\}_{i\in\mathfrak{I}}$ converges to $\zeta_{Q(b_{k}),0}$ for some $k\in\{1,\dots,n\}$. Then Corollary 2.12(c) tells us that $|s_{i}-b_{k}|<1$ eventually. In other words, $Q(s_{i})=Q(b_{k})$ eventually. Conversely, it follows from Corollary 2.12(b) that the conditions $Q(s_{i})=Q(b_{k})$ for large $i$ and $\omega_{i}\to\infty$ will imply $\zeta_{s_{i},0}^{\omega_{i}}\to\zeta_{Q(b_{k}),0}$. This completes the proof. ∎

In the case when the field $K$ is not algebraically closed, one may use the above as well as [3, Corollary 1.3.6] to describe $\mathcal{M}(R[G])$. In the following, we will consider the case when $R$ is the ring $\mathbb{Z}_{p}$ of $p$-adic integers and $G$ is a cyclic $p$-group (for a fixed prime number $p$). Let us start with the following possibly known lemma.

Lemma 2.14**.**

For any $l\in\mathbb{Z}_{+}$, the polynomial $\mathbf{q}_{l+1}:=\mathbf{t}^{p^{l}(p-1)}+\mathbf{t}^{p^{l}(p-2)}+\cdots+\mathbf{t}^{p^{l}}+1$ is irreducible in $\mathbb{Z}_{p}[\mathbf{t}]$.

Proof.

We first consider the case when $l=0$. The equalities

[TABLE]

gives $\sum_{i=1}^{p}(\mathbf{t}+1)^{p-i}=\sum_{i=1}^{p}\binom{p}{i-1}\mathbf{t}^{p-i}$. Thus, it follows from the Eisenstein’s criterion that the polynomial $\sum_{i=1}^{p}(\mathbf{t}+1)^{p-i}$ is irreducible in $\mathbb{Z}_{p}[\mathbf{t}]$, and hence so is $\sum_{i=1}^{p}\mathbf{t}^{p-i}$.

In the case of $l\geq 1$, we set $\mathbf{s}:=(\mathbf{t}+1)^{p^{l}}-1$. Since $\sum_{i=1}^{p}(\mathbf{s}+1)^{p-i}=\sum_{i=1}^{p}\binom{p}{i-1}\mathbf{s}^{p-i}$, we have

[TABLE]

From the left hand side, we see that the coefficient of $\mathbf{t}^{p^{l}(p-1)}$ is $1$ and the constant coefficient is $p$. From the right hand side, we know that all the other coefficients of $\mathbf{t}^{k}$ are divisible by $p$. Thus, the Eisenstein’s criterion tells us that $\sum_{i=1}^{p}(\mathbf{t}+1)^{p^{l}(p-i)}$ is an irreducible polynomial in $\mathbb{Z}_{p}[\mathbf{t}]$ and hence so is $\sum_{i=1}^{p}\mathbf{t}^{p^{l}(p-i)}$. ∎

Example 2.15**.**

Let $G$ be the cyclic group of order $p^{N}$ for a positive integer $N$. We set $\mathbb{F}_{p}:=\mathbb{Z}/p\mathbb{Z}$ and consider $q_{G}:\mathbb{Z}_{p}\{\mathbf{t}\}\to\mathbb{Z}_{p}[G]$ to be the canonical quotient map. As in the argument of Corollary 2.13, we may identify, through $q_{G}$:

[TABLE]

(see Proposition 2.4). By Lemma 2.14, the prime factorization of $\mathbf{t}^{p^{N}}-1$ in $\mathbb{Z}_{p}[\mathbf{t}]$ is

[TABLE]

where $\mathbf{q}_{0}:=\mathbf{t}-1$. On the other hand, the following is the prime factorization of $\mathbf{t}^{p^{N}}-1$ in $\mathbb{F}_{p}[\mathbf{t}]$:

[TABLE]

Let $\mathbb{C}_{p}$ be the completion of the algebraic closure of $\mathbb{Q}_{p}$, let $R_{p}$ be the ring of integers of $\mathbb{C}_{p}$ and let $F_{p}$ be the residue field of $R_{p}$. For $1\leq k\leq N$, we consider $\{r_{k,1},\cdots,r_{k,n_{k}}\}$ to be the set of all distinct roots of $\mathbf{q}_{k}$ in $\mathbb{C}_{p}$. It follows from (16) that $\mathbf{t}^{p^{N}}-1=(\mathbf{t}-1)^{p^{N}}$ in $F_{p}[\mathbf{t}]$. Hence, $Q(r_{k,i})=1$ for all possible $k$ and $i$. Therefore, Corollary 2.13(a) tells us that

[TABLE]

As in Corollary 2.13(b), the topological space $\mathcal{M}(R_{p}[G])$ consists of $1+\sum_{k=1}^{N}n_{k}$ intervals of the form $[1,\infty]$ with all the “$1$-ends” being free but with all the “$\infty$-ends” being identified with one point, namely, $\beta_{1}$.

Again, the prime factorization as in (16) ensures that \mathcal{M}(\mathbb{Z}_{p}[G])\cap Q^{\mathbb{A}}\big{(}\mathbb{A}_{\mathbb{F}_{p}}^{1}\big{)}=\{Q^{\mathbb{A}}(\zeta_{1_{\mathbb{F}_{p}},0})\}, and the semi-norm $\bar{\beta}_{1}:=Q^{\mathbb{A}}(\zeta_{1_{\mathbb{F}_{p}},0})$ coincides with the one induced by $\beta_{1}\in\mathcal{M}(R_{p}[G])$ through restriction. On the other hand, as in [3, Corollary 1.3.6], any element $\lambda\in\mathcal{M}(\mathbb{Z}_{p}[G])\cap J_{\omega}^{\mathbb{A}}(\mathbb{A}_{\mathbb{Q}_{p}^{\omega}}^{1})$ can be extended to an element $\bar{\lambda}\in J_{\omega}^{\mathbb{A}}(\mathbb{A}_{\mathbb{C}_{p}^{\omega}}^{1})$. It follows from $|\mathbf{t}|_{\bar{\lambda}}\leq 1$ and $|\mathbf{t}^{p^{N}}-1|_{\bar{\lambda}}=0$ that $\bar{\lambda}$ is either $\alpha_{1}^{\omega}$ or $\alpha_{r_{k,i}}^{\omega}$ for suitable $k$ and $i$.

Let us set $\bar{\alpha}_{0}^{\omega}$ to be the element in $\mathcal{M}(\mathbb{Z}_{p}[G])$ induced by $\alpha_{1}^{\omega}$. On the other hand, for a fixed $k\in\{1,2,\dots,N\}$, by considering an automorphism in the Galois group of the splitting field of the irreducible polynomial $\mathbf{q}_{k}$ over $\mathbb{Q}_{p}$, we know that $\alpha_{r_{k,i}}^{\omega}$ and $\alpha_{r_{k,j}}^{\omega}$ restrict to the same element in $\mathcal{M}(\mathbb{Z}_{p}[G])$ for any $i,j\in\{1,\dots,n_{k}\}$. We denote the resulting element by $\bar{\alpha}_{k}^{\omega}$. As $\bar{\alpha}_{k}^{\omega}$ comes from different irreducible factors of $\mathbf{t}^{p^{N}}-1$ for different $k$, they are all distinct.

Consequently, as a quotient space of $\mathcal{M}(R_{p}[G])$, the topological space $\mathcal{M}(\mathbb{Z}_{p}[G])$ is the following subspace of $\mathbb{R}^{2}$:

[TABLE]

Acknowledgement

This work is supported by the National Natural Science Foundation of China (11471168).