CHAPTER 1
The Berkovich unit disc
In this chapter, we recall a theorem of Berkovich which states that points
of the Berkovich unit disc D(0, 1) over K can be identified with equivalence
classes of nested sequences of closed discs {D(ai,ri)}i=1,2,... contained in the
closed unit disc D(0, 1) of K. This leads to an explicit description of the
Berkovich unit disc as an “infinitely branched tree”; more precisely, we show
that D(0, 1) is an inverse limit of finite R-trees.
1.1. Definition of D(0, 1)
Let A = K T be the ring of all formal power series with coefficients
in K, converging on D(0, 1). That is, A is the ring of all power series
f(T) =
∑∞
i=0
aiT
i
K[[T]] such that limi→∞ |ai| = 0. Equipped with the
Gauss norm defined by f = maxi(|ai|), A becomes a Banach algebra
over K.
A multiplicative seminorm on A is a function [ ]x : A R≥0 such that
[0]x = 0, [1]x = 1, [f · g]x = [f]x · [g]x, and [f + g]x [f]x + [g]x for all
f, g A. It is a norm provided that [f]x = 0 if and only if f = 0.
A multiplicative seminorm [ ]x is called bounded if there is a constant Cx
such that [f]x Cx f for all f A. It is well known (see [49, Proposition
5.2]) that boundedness is equivalent to continuity relative to the Banach
norm topology on A. The reason for writing the x in [ ]x is that we will be
considering the space of all bounded multiplicative seminorms on A, and we
will identify the seminorm [ ]x with a point x in this space.
It can be deduced from the definition that a bounded multiplicative
seminorm [ ]x on A behaves just like a non-Archimedean absolute value,
except that its kernel may be nontrivial. For example, [ ]x satisfies the
following properties:
Lemma 1.1. Let [ ]x be a bounded multiplicative seminorm on A. Then
for all f, g A,
(A) [f]x f .
(B) [c]x = |c| for all c K.
(C) [f + g]x max([f]x, [g]x), with equality if [f]x = [g]x.
Proof. (A) For each n,
([f]x)n
= [f
n]x
Cx f
n
= Cx f
n,
so [f]x
Cx/n 1
f , and letting n gives the desired inequality.
(B) By the definition of the Gauss norm, c = |c|. If c = 0, then
trivially [c]x = 0; otherwise, [c]x c = |c| and
[c−1]x

c−1
=
|c−1|,
1
http://dx.doi.org/10.1090/surv/159/01
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