Berkovich space

Remarks

Berkovich proved that $\mathcal{M}(A)$ is a nonempty compact Hausdorff space when $A \ne 0$.

Example: let $A = K\langle T \rangle$ be the Tate algebra of power series $\sum_n a_n T^n$ with $a_n \to 0$, with the Gauss norm $\norm{\sum a_n T^n} = \max_n \abs{a_n}$. Take $a \in K$ with $\abs{a} \le 1$ and a real number $0 \le r \le 1$. Expand $f \in A$ as $f = \sum_n c_n (T - a)^n$ and set $\abs{f(\zeta_{a,r})} = \max_n \abs{c_n} r^n$. This is a point $\zeta_{a,r}$ of $\mathcal{M}(A)$, the point of the disc of centre $a$ and radius $r$. Clause 3 holds because $\abs{c_n} \le \norm{f}$ when $\abs{a} \le 1$, and clause 2 is a form of Gauss's lemma. For $a = 0$ and $r = 1$ it is the Gauss norm itself. For $r = 0$ it is $f \mapsto \abs{f(a)}$, evaluation at $a$, a classical point.

Non-example: the supremum over the two-point set $\set{0, 1}$, namely $f \mapsto \max(\abs{f(0)}, \abs{f(1)})$, satisfies clauses 1 and 3 but not clause 2. For $f = T$ and $g = T - 1$ both values are $1$, while $fg$ vanishes at both points and has value $0$.

Let $X = \operatorname{Spec} B$ be an affine variety of finite type over $K$. Its analytification $X^{\mathrm{an}}$ is the set of multiplicative seminorms on $B$ that extend $\abs{\cdot}$ on $K$, with the same kind of topology.