References:
Spectral Theory and Analytic Geometry over Non-Archimedean Fields by Vladimir Berkovich
Math 731 Topics in Algebraic Geometry I Berkovich Spaces by Mattias Jonsson
There are various ways to deal with non-archimedean geometry, including
Rigid Analytic Spaces by J.Tate, they use Grothendieck topology.
It is good for category of sheaves, but does not follow a direct intuitive comparison with classical complex field.
Berkovich Spaces, they use the theory of valuations.
They are elegant objects possessing topological properties similar to that of complex analytic spaces and are sufficient for homology theory in the usual sense.
Adic Spaces, they use the theory of Huber rings.
Similar to the theory of valuations, where the valuations in a field like \(\mathbb{Q}\) corresponds to points in \(\mathrm{Spec}(\mathbb{Z})\), Berkovich spaces generalize this idea,, to use seminorms on a valued field to define a space.
Think of each valuation corresponds to the valuation ring with a maximal ideal.
A norm on a field \(k\) is a function \(|\cdot| : k \to \mathbb{R}_{\geq 0}\) satisfying the following properties:
\(|x| = 0\) if and only if \(x = 0\),
\(|xy| = |x| \cdot |y|\),
\(|x + y| \leq |x| + |y|\).
A seminorm however, does not require the regularity property, i.e. you can have \(f\neq 0\) with \(|f| = 0\).
Let \(k\) be a complete valued field. The Berkovich affine space \(\mathbb{A}^{n,an}_k\) is defined as the set of multiplicative seminorms on \(k[T_1,\ldots,T_n]\) extending the norm on \(k\).
\[\mathbb{A}^{n,an}_k = \left\{ |\cdot| : k[T_1,\ldots,T_n] \to \mathbb{R}_{\geq 0} \mid |\cdot|_{k} = |\cdot| \text{ on } k \right\}\]
The topology on \(\mathbb{A}^{n,an}_k\) is the weakest topology making the evaluation maps \(\mathrm{ev}_f: |\cdot|\mapsto |f|\) continuous for all \(f\in k[T_1,\ldots,T_n]\).
\(\mathbb{A}^{n,an}_k\) is Hausdorff, locally compact, and path-connected.
\(\mathrm{mSpec}(k[T_1,\ldots,T_n]) \hookrightarrow \mathbb{A}^{n,an}_k\), because for any point \(a\in k^n\), the evaluation at \(a\) followed by the norm on \(k\) gives us a seminorm!
\[|f|_a := |f(a)|\]
and you see that it is a seminorm because you cannot reasonably require \(f(a) = 0\) to imply \(f = 0\).
If we take \(k=\mathbb{C}\) with the usual absolute value, then the Berkovich affine space is just the usual complex affine space.
For simplicity assume \(k\) is algebraically closed, this simplifies valuations on \(k[T]\) to evaluating norms on the linear factors. Simplify further we elt \((k,|\cdot|)\) be the trivial norm. For \(x\in \mathbb{A}^{1,an}_k\) be a point in our line, there are some cases:
\(|\cdot|_x\) is the trivial norm on \(k[T]\).
All \(|T-a|_x=1\).
\(|T|_x= r>1\), since \(|\cdot|\) is trivial, we have \(|T-a|_x=r>1\) for all \(a\in k\).
All \(|T-a|>1\).
There exists \(a\in k\) such that \(|T-a|_x = r < 1\), and then for any other \(b\) since \(|a-b|=1\) we have \(|T-b|=1\).
Only one \(|T-a|<1\), all others are \(1\).