Realizing the Berkovich affine line as a union of Berkovich spectrums

Question

I am trying to understand what is the relation of the affine Berkovich space to the Berkovich space on an appropriate polynomial ring. A more exact version of the question is as follows:

Let $(K,\Vert \cdot \Vert)$ be a complete valued field. The affine Berkovich space is defined as:

$ A_K^{n, \text{an}}:= \begin{Bmatrix} & \vert f+g\vert_x \leq \vert f\vert_x +\vert g\vert_x \\ \vert \cdot \vert _x:K[x_1,...,x_n]\rightarrow [0,\infty) \quad \Bigg \vert & \vert f\cdot g\vert_x= \vert f\vert_x \cdot \vert g\vert_x \\ & \vert a\vert_x= \Vert a\Vert \; \text{for all } \; a\in K \end{Bmatrix} $

i.e, non-trivial multiplicative semi-norms on $K[x_1,...,x_n]$ extending $\Vert \cdot \Vert$. Given a Banach ring $(A,\Vert \cdot \Vert)$, it's Berkovich spectrum is:

$ \mathcal{M}(A):= \begin{Bmatrix} \text{multiplicative bounded seminorms} \\ \vert \cdot \vert : A \rightarrow [0,\infty) \end{Bmatrix} $

They both have the weakest topology induced by evaluations.

I have been trying to understand what is the relation between $A_K^{n, \text{an}}$ and a Berkovich spectrum of $K[x_1,...,x_n]$. They would coincide were it not for the extra assumption of boundedness. I saw somewhere that the Berkovich affine space is not a Berkovich spectrum, but instead an increasing union of Berkovich spectrums of power rings.

I have not found thus far how these spectrums are defined, and would appreciate anyone explaining, or directing to a reference where it is indeed written.

Answer 1

Let me see if I can help a bit; please let me know if you need more clarification. First of all, I don't know of many basic references on Berkovich spaces. One you might consider is the first two chapters in Baker–Rumely (which works over an algebraically closed field). For another reference, you might find this set of notes (especially §5) and the citations therein helpful.

To make this discussion a bit simpler, we will work over a complete non-Archimedean valued field $(K,\lvert\cdot\rvert)$, and only consider the case when $n = 1$.

Let's first recall that Berkovich's notion of a spectrum $\mathcal{M}(A)$ needs $A$ to be a Banach ring. To make sense of the Berkovich spectrum of $K[x]$, then, one would want to put a norm on this ring that makes it complete. The usual choice (the "Gauss norm") $$\biggl\lVert \sum_{i=0}^\infty a_ix^i \biggr\rVert = \max_i \lvert a_i \rvert$$ would make it so that the polynomial ring is not complete unless the field $K$ is trivially valued. In general, we therefore consider the Banach ring (the "Tate algebra") $$K\{x\} := \biggl\{ f = \sum_{i=0}^\infty a_ix^i \biggm\vert a_i \in K,\ \lim_{i \to \infty} \lvert a_i \rvert = 0 \biggr\}$$ which can be thought of as the "ring of convergent power series on the closed disc of radius $1$." Note that in the trivially valued case, this ring is just the usual polynomial ring.

Now borrowing the usual intuition from algebraic geometry, the ring $A$ that we take the spectrum of should consist of all regular functions on the space $\mathcal{M}(A)$. In this case, $\mathcal{M}(K\{x\})$ should therefore be thought of as a closed disc of radius $1$; after all, if you take an element of $K\{x\}$ and plug an element of $K$ with norm $>1$, then the power series can diverge!

To get closed discs with larger radii, then, Berkovich changes the norm and considers the ring $$K\{r^{-1}x\} := \biggl\{ f = \sum_{i=0}^\infty a_ix^i \biggm\vert a_i \in K,\ \lim_{i \to \infty} \lvert a_i \rvert r^i = 0 \biggr\}$$ for each $r > 0$, which now can be thought of as the "ring of convergent power series on the closed disc of radius $r$." The closed disc of radius $r$ is $$E(r) := \mathcal{M}(K\{r^{-1}x\}),$$ and we can define the affine line as $$\mathbf{A}^1_{\mathrm{Berk}} := \bigcup_{r > 0} E(r).$$