# I.5: Analytic adic spaces

Today I’m writing about an important class of adic spaces, namely analytic adic spaces. An important property is that all specializations on maps between analytic adic spaces are vertical. Analytic adic spaces should provide the correct level of generality for arithmetic considerations. Most of this part is taken from Conrad’s seminar (week 7 and week 6).

Definition 17 Let ${A}$ be a Huber ring. A point ${v \in \mathrm{Cont} (A)}$ is called analytic if its support ${\mathrm{supp}(v) \in \mathrm{Spec}(A)}$ is not an open ideal of ${A}$.

Lemma 18 Let ${A}$ be a Huber ring with a couple of definition ${(A_0,I)}$. A point ${v \in \mathrm{Spv}(A)}$ is non-analytic if and only if ${v(I)=0}$.

Proof: The support ${\mathrm{supp}(v)}$ is open in ${A}$ if and only if it contains ${I^n}$ for sufficiently large ${n}$. But ${\mathrm{supp}(v)}$ is radical so this is equivalent to ${\mathrm{supp}(v)}$ containing ${I}$, that is exactly ${v(I)=0}$. $\Box$

Let us denote by ${\mathrm{Cont}(A)_{\mathrm{an}}}$ the subspace of analytic points in ${\mathrm{Cont}(A)}$ endowed with the subspace topology.

Corollary 19 Let ${A}$ be a Tate ring. Then ${\mathrm{Cont}(A) = \mathrm{Cont}(A)_{\mathrm{an}}}$.

Proof: Let ${u}$ be the topologically nilpotent unit in ${A}$. Then ${v \in \mathrm{Spv}(A)}$ cannot kill ${u}$, even though ${u^n \in I}$ for all sufficiently large ${n}$. This implies ${v(I)\not = 0}$ and hence the result by the previous lemma. $\Box$

Let us now describe the structure of ${\mathrm{Cont}(A)_{\mathrm{an}}}$. We still denote by ${A}$ a Huber ring. Let ${(A_0, I)}$ be its couple of definitions and let ${\{f_1,...,f_n\}}$ be a set of generators of ${I}$. Let ${v \in \mathrm{Cont}(A) _{\mathrm{an}}}$. Then

$\displaystyle \sum ^n _{i=1} A_0 f_i = I \not \subset \mathrm{supp}(v)$

Hence for some ${r}$ we must have ${v(f_{r}) \not = 0}$. We choose ${r}$ for which ${v(f_r)}$ is maximal and hence ${v(f_j) \leq v(f_r) \not = 0}$ for all ${j}$. We can then form ${A\left( \frac{f_1}{f_r},...,\frac{f_n}{f_r}\right)}$. There is a natural map ${A \rightarrow A\left( \frac{f_1}{f_r},...,\frac{f_n}{f_r}\right)}$ which induces the map on ${\mathrm{Cont}(-)}$:

Proposition 20 The image of the natural continuous map

$\displaystyle \mathrm{Cont} \left(A\left( \frac{f_1}{f_r},...,\frac{f_n}{f_r}\right)\right) \rightarrow \mathrm{Cont}(A)$

is the subspace of ${\mathrm{Cont}(A)_{\mathrm{an}}}$ defined by ${\{v \in \mathrm{Cont}(A)_{\mathrm{an}} | \ \ v(f_j) \leq v(f_r) \textrm{ for all j}\}}$.

Using this result we get

Proposition 21 Let ${A}$ be a Huber ring with ${f_1,...,f_n}$ as before. We have

$\displaystyle \mathrm{Cont}(A)_{\mathrm{an}} = \bigcup ^n _{i=1} \mathrm{image} \left(\mathrm{Cont} \left(A\left( \frac{f_1}{f_r},...,\frac{f_n}{f_r}\right) \right) \right)$

where

$\displaystyle \mathrm{image} \left(\mathrm{Cont} \left(A\left( \frac{f_1}{f_r},...,\frac{f_n}{f_r}\right) \right) \right) = \mathrm{Spv}(A) \cap \mathrm{Spv} \left(A\left( \frac{f_1}{f_r},...,\frac{f_n}{f_r}\right)\right)$

I finish by stating a theorem on specializations

Theorem 22 Let ${A}$ be a Huber ring. Then all specializations inside ${\mathrm{Cont}(A)_{\mathrm{an}}}$ are vertical. In particular, if ${A}$ is Tate, then all specializations in ${\mathrm{Cont}(A)}$ are vertical.

Proof: One recalls a structure theorem for specializations which says that we can write each specialization as a vertical specialization of a horizontal specialization. Then it is enough to prove that there are no proper horizontal specializations. This is done for ${v\in \mathrm{Cont}(A)_{\mathrm{an}}}$ by choosing ${H \subset \Gamma _v}$ which contains ${v(A)_{\geq 1}}$. Hence ${v_{| H}}$ is horizontal specialization of ${v}$ and it lies in ${\mathrm{Cont}(A)}$. Assume that ${H\not =\Gamma _v}$ and derive a contradiction by showing that ${\mathrm{supp}(v_{|H})}$ is open, hence ${v}$ could not be analytic. $\Box$