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\begin{document}
\normalsize
\section{Lecture 9: Spectrality of $\Cont(A)$}


\subsection{Introduction}
This lecture is devoted to prove spectrality of $\Cont(A)$.  Next time this will be used to prove spectrality, and in
particular quasi-compactness, of affinoid adic spectra
(once these are defined).
The way to achieve spectrality of $\Cont(A)$ is by giving an ``algebraic'' description of the notion of continuous valuations on a Huber ring $A$, achieved in terms of cofinality of values of valuations.
The idea is to give an algebraic description of continuity in terms of $A$ and the ideal of $A$ generated by the set $A^{00}$ of topologically nilpotent elements of $A$.  Continuity is stable under horizontal specializations, and the idea is to exploit the algebraic input lying behind horizontal specializations. 


% HERE THERE WAS AN EXAMPLE OF Cont(A) in Spv(A) NOT OPEN BUT IT IS NOT USEFUL HERE I THINK


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\subsection{Towards an algebraic description of continuity}
%
%The reader is encouraged to step back to Example 8.2.2. 
%
%Our aim is to find a suitable algebraic counterpart of continuity for valuations: the strategy for proving spectrality of $\Cont(A)$ consists in essentially avoiding the topological input until the very end, while first performing certain algebraic construction we are about to discuss. Example 8.2.2 is already motivational: the first upshot is that horizontal generizations have no hope to capture well enough the topological properties of the continuous valuation spectrum, and the remaining specialization/generization constructions instead already begin to suggest a strategy to ``scan'' continuity of valuations in $\Spv(A)$ (we'll see this rough idea is due once again to the notion of convexity for subgroups of totally ordered abelian groups giving rise to horizontal specializations, and we will even see it in practice when we will picture the adic unit disc). However, continuity is stable under horizontal specializations, and the idea is to exploit the algebraic input lying behind horizontal specializations. 
%
%Before moving on, let us discuss a few more facts about analytic points. They will allow us to make more examples, along the way.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\paragraph{More properties of analytic points}\

Recall Proposition 8.3.2 and its Corollary 8.3.3, summarized in the following:

\begin{prop} \label{continuous val} Let $A$ be a Huber ring, and choose a couple of definition $(A_0,I)$. The point $v \in \Spv(A)$ is analytic if and only if $v(I) \neq 0$. In particular, if $A$ is Tate then
$$\Cont(A)=\Cont(A)_{\rm an}.$$
\end{prop}

The reader should keep in mind the following:

\begin{rem} \label{Tate incoming}
The statement $v(I)\neq 0$ is equivalent to saying that $v$ has non-open support. Without mentioning the couple $(A_0,I)$, one may simply rephrase this by saying that there exists a topologically nilpotent element $a\in A^{00}$ which is not killed by $v$.
In other words,
$$\{\fp\in\Spec(A)\mid\fp\ \textup{open}\}=V(A^{00} \cdot A)$$
and $v$ is in $\Cont(A)$ if and only if $v\notin V(A^{00} \cdot A)$. 
We stress once more the fact that $A^{00}$ is not in general an ideal of $A$, and that the ideal it generates in $A$ may well be $A$ itself: Tate rings contain topologically nilpotent units by definition. If $A$ is Tate then 
$$A^{00}\cdot A=A,$$
so $V(A^{00} \cdot A)=\varnothing$; this rephrases the last statement in Proposition \ref{continuous val}.
\end{rem}


Rank-1 valuations $v$ and $w$ on a field $K$ coincide as points in $\Spv(K)$
if (and only if) they induce the same valuation topology.  Indeed, in the rank-1 case the valuation ring is characterized
in terms of the topology on the fraction field as the set of power-bounded elements.   But for higher-rank
valuations on a field it can (and often does!) happen that its topology coincides with that induced by a rank-1 valuation.
We saw an instance of this in Example 6.2.1, and
the following characterization of such all such cases underlies the importance of rank-1 valuations (and hence
of rigid-analytic geometry!) in the general
theory of adic spaces:

\begin{prop} \label{enlargeable} Let $K$ be a field, and $v\in\Spv(K)$ nontrivial. Let $R\subset K$ be the valuation ring of $v$. The following conditions are equivalent:
\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(\arabic{enumi})}}
\item There exists a rank-$1$ valuation $w$ on $K$ defining the same topology on $K$ as $v$ does.
\item There exists a nonzero topologically nilpotent element in $K$.
\item $R$ has a prime ideal of height equal to $1$.
\end{enumerate}
\end{prop}

\begin{proof}
Assuming (1) holds, any $\varpi\neq 0$ in $K$ such that $w(\varpi)<1$ is topologically nilpotent because $w$ is of rank one, 
so (2) holds. Next assume (2) holds, so the valuation topology is non-trivial and makes $K$ a Huber ring by
Proposition 6.2.3.  More specifically, if $\varpi \in K^{\times}$ is topologically nilpotent
then we may replace it by a suitable power so that $\varpi$ lies in the open subring $R$ and then
$R$ is a ring of definition with $I = \varpi R$ as an ideal of definition.  Note that $I$ is a proper ideal (equivalently, 
$\varpi \not\in
R^{\times}$) since all elements of $I$ are topologically nilpotent.  Let $\mathfrak{p}$ be minimal among prime
ideals of $R$ over $I$ (i.e., it corresponds to a generic point of $\Spec(R/I)$).  This is a nonzero prime
(since $I \ne 0$, as $\varpi \ne 0$), and we claim that $\mathfrak{p}$
has height 1, which is to say that it is minimal as a nonzero prime ideal
of the domain $R$.  In other words, if $\mathfrak{q}$ is a prime ideal of $R$ strictly
contained in $\mathfrak{p}$ then we claim that $\mathfrak{q} = 0$.  The set of {\em all} ideals of a valuation ring is
totally ordered by inclusion.  (Indeed, if $J, J'$ are ideals in a valuation ring $A$ and there exists $a \in J$ and $a' \in J'$
with $a \not\in J'$ and $a' \not\in J$ then we reach a contradiction because either
$a \in a'A \subset J'$ or $a' \in aA \subset J$.)  Since $I \not\subseteq \mathfrak{q}$ (as otherwise
$\mathfrak{p}$ would not be minimal over $I$), it follows that $\mathfrak{q} \subsetneq I = \varpi R$.  In particular, $\varpi \not\in
\mathfrak{q}$.  Every
$x \in \mathfrak{q}$ can be written as $x = \varpi y$ for $y \in R$.  Rather generally, if
such an $x$ has the form $\varpi^n y_n$ for some $n \ge 1$ and $y_n \in R$
then the membership $\varpi^n y_n = x \in \mathfrak{q}$ forces
$y_n \in \mathfrak{q}$ since $\mathfrak{q}$ is prime and $\varpi \not\in \mathfrak{q}$.  But then $y_n \in I = \varpi R$, so
$y_n = \varpi y_{n+1}$ for some $y_{n+1} \in R$, so $x = \varpi^{n+1} y_{n+1}$.  This calculation shows
that $\mathfrak{q} \subseteq \cap_{n \ge 1} I^n$, and this intersection vanishes since the $I$-adic
topology of $R$ is the $v$-adic topology, which is separated.   This proves $\mathfrak{q} = 0$ as claimed,
so we have shown that (2) implies (3).


Finally, assume (3) and let $\mathfrak{p}$ be a height-1 prime ideal of $R$,so $R_{\fp}$ is a valuation ring of $K$ containing $R$.
The value group $K^{\times}/R_{\fp}^{\times}$ is a quotient of the value group 
$K^{\times}/R^{\times}$ of $v$.  Let $w:K \rightarrow (K^{\times}/R_{\fp}^{\times}) \cup \{0\}$ be the associated
valuation.  We claim that $w$ is rank-1, which is to say that its nontrivial value group is order-isomorphic to a subgroup of
$\mathbf{R}^{\times}_{>0}$ (or equivalently $\mathbf{R}$ in additive notation).  
We need to show that $\Gamma_w$ is order-isomorphic to a subgroup of
$\mathbf{R}^{\times}_{>0}$.  By \cite[Thm.\,10.6]{Mat} it is equivalent to show that $\Gamma_w$ satisfies the Archimedean
axiom.  This means that for {\em any} $\gamma \in \Gamma_w$ with $\gamma > 1$,
every $\gamma' \in \Gamma_w$ satisfies $\gamma' < \gamma^n$ for some $n > 0$
(equivalently, $1/\gamma$ is cofinal in $\Gamma_w$).   There is a very useful necessary and sufficient
condition for a totally ordered abelian group $G$ to satisfy the Archimedean axiom:
it should have no nontrivial proper convex subgroups.  The necessity of this condition
is clear, and for sufficiency suppose there are no such convex subgroups of $G$ and consider 
$g \in G$ with $g > 1$.  Define
$\Delta$ to be the set of elements $h \in G$ that satisfy $g^{-n} < h < g^n$ for some $n > 0$ (which may depend
on $h$).  Clearly $\Delta$ contains 1 and is stable under inversion and multiplication, so
it is a subgroup of $G$, and by design $\Delta$ is convex.  Also, $\Delta \ne 1$ since $g \in \Delta$
(as $g > 1$). The hypothesis on $G$ then forces $\Delta = G$, as desired.  
Thus, to prove that $w$ is a rank-1 valuation we just need to check that its nontrivial value group has
no nontrivial proper convex subgroups.
But in general there is a natural inclusion-preserving bijection between the set of prime ideals of a valuation ring
and the set of convex subgroups of its value group, explained by combining both parts of \cite[Prop.\,2.14]{Wed}.
Thus, since $R_{\fp}$ has no nonzero non-maximal prime ideals (as
$\fp$ has height 1), it follows that $\Gamma_w$ has no nonzero proper convex subgroups.

To conclude the proof that (3) implies (1), it suffices to show that the $w$-topology on $K$ coincides with the 
initial $v$-topology associated to $R$.  Rather generally, 
any pair of non-trivial valuations on a field whose valuation rings do {\em not} general the field as a ring
(e.g., this applies when one of the valuation rings contains the other) define the same
topology on the field \cite[Ch.\,VI, \S7.2, Prop.\,3]{bourbaki}. (This reference also proves the more difficult converse
that we do not need.)
\end{proof}

\begin{defin}\label{enlarge} We call a valuation $v$ on a field $K$ {\em microbial} 
if it satisfies the equivalent conditions in Proposition \ref{enlargeable}.
\end{defin}

We can now prove the following:

\begin{prop} \label{rank analytic} Let $A$ be a Huber ring and choose $v\in\Cont(A)_{\rm an}$. Then $v$ is microbial and 
has a unique rank-$1$ generization $w$ inside $\Cont(A)_{\rm{an}}$.
Moreover, $v$ has no proper generization inside $\Cont(A)_{\rm{an}}$ if and only if it is a rank-$1$ point.
\end{prop}

\begin{proof}  If $v$ were trivial then its support $\fp_v$ would be $\{v < 1\}$ and hence would be open by continuity of $v$,
yet $\fp_v$ is non-open by analyticity.  Thus, $v$ is a non-trivial valuation.  
By Proposition \ref{continuous val} the non-open prime $\fp_v$ is not contained in $V(A^{00} \cdot A)$. Hence,
there exists some nonzero topologically nilpotent element $a\in A^{00}$ such that $v(a)\neq 0$. In particular the image of $a$ in $A/\fp_v$ is nonzero and topologically nilpotent for the quotient topology
and thus for the $v$-topology since $v$ is {\em continuous} on $A$.  It follows that the residue field $\kappa(v)$ 
equipped with its non-trivial valuation arising from $v$ satisfies $(2)$ in Proposition \ref{enlargeable}, so $v$ is microbial.
In particular, $v$ admits a rank-1 vertical generization $w$.  Since $w$ is non-trivial it inherits
continuity from $v$, and its support is equal to $\fp_v$ so it is also analytic.   So we have produced a rank-1 generization
of $v$ inside $\Cont(A)_{\rm{an}}$.

By Proposition 8.3.10, all generizations inside $\Cont(A)_{\rm{an}}$ are vertical, so to prove uniqueness of $w$
we just need to check that $v$ has no other rank-1 {\em vertical} generization inside $\Spv(A)$.
If $v'$ is a rank-1 vertical generization of $v$ then it gives rise to a rank-1 valuation ring inside
$\kappa(v)$ that contains the valuation ring of $v$, and the two resulting valuation topologies on $\kappa(v)$
must coincide (as we saw at the end of the proof of Proposition \ref{enlargeable}). 
But $w$ also gives the same topology, and we have seen earlier the elementary fact that 
rank-1 points in $\Spv(\kappa(v))$ coincide if and only if they induce the same topology on $\kappa(v)$.
Thus, $v' = w$ as desired.  

If $v$ does not have rank 1 then we have produced a proper generization of it inside $\Cont(A)_{\rm{an}}$,
and if $v$ is rank 1 then any point $v' \in \Cont(A)_{\rm{an}}$ generizing $v$ must be vertical and so
corresponds to a rank-1 valuation ring of $\kappa(v)$ that contains the rank-1 valuation ring associated to $v$.
But it is elementary that there are no proper subrings of a field that strictly contain a rank-1 valuation ring,
so the valuation rings for $v$ and $v'$ inside $\kappa(v)$ are equal, forcing $v' = v$ as points of $\Spv(A)$.
\end{proof}

\begin{eg} \label{val on field} Let $K$ be a field, and $v$ be a microbial valuation on $K$. Endow $K$ with the valuation topology induced by $v$, thus making it into a non-archimedean ring (in fact a topological field). We claim $\Cont(K)$ is the 
subspace of $\Spv(K)$ consisting of those valuations $w$ on $K$ whose valuation topology on $K$ coincides with 
that induced by $v$. 

Since every valuation on a field has support $(0)$, and this is not $v$-open since $v$ is nontrivial, it follows that
$$\Cont(K)=\Cont(K)_{\rm an}.$$
The microbial hypothesis on $v$ implies that $K$ contains a topologically nilpotent unit, so by continuity 
every $w \in \Cont(K)$ admits a topologically nilpotent unit in $K$ for the $w$-topology
and thus is itself microbial.  By Proposition \ref{enlargeable}, $w$ therefore
admits a rank-1 generization that defines the same topology as $w$.
Thus, to prove that all such $w$ induces the same topology as $v$ we may assume $v$ has rank 1 and just need to check
that (i) $\Cont(K)$ has no rank-1 points aside from $v$ and (ii) any rank-1 point of $\Spv(K)$ whose 
topology coincides with that of $v$ is equal to $v$.  

Assertion (ii) is immediate from the fact that a rank-1 point is recovered from its
associated topology on $K$ (in such cases the valuation ring is precisely the set of power-bounded elements).
As for (i), if a rank-1 valuation $v'$ on $K$ is continuous for the $v$-topology then any topologically nilpotent $a \in K$
for the $v$-topology must satisfy $v'(a) < 1$ since $v'(a^n) \rightarrow 0$ for the order topology on 
$\Gamma_{v'} \cup \{0\}$.  But all elements of $\mathfrak{m}_v$ are topologically nilpotent
for the $v$-topology since $v$ has rank 1, so $\mathfrak{m}_v \subseteq \mathfrak{m}_{v'}$.
Thus, 
$$K - R_v = (\mathfrak{m}_v - \{0\})^{-1} \subseteq (\mathfrak{m}_{v'} - \{0\})^{-1} = K - R_{v'},$$
giving that $R_{v'} \subseteq R_v$.  But $R_v \ne K$ and $R_{v'}$ is a {\em rank-$1$} valuation ring,
forcing $R_{v'} = R_v$ since there are no rings strictly between
a rank-1 valuation ring and its fraction field (always false in higher rank!).
\end{eg}
%
%\begin{rem} Example \ref{val on field} also shows once again that the topology on $\Cont(A)$ is much finer than the topology given by the product of the topologies of the fibers of $\Cont(A)\hookrightarrow\Spv(A)\to\Spec(A)$. We will revisit this example soon.
%%Yes but careful correct later
%\end{rem}
%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

For a Huber ring $A$ and $(A_0,I)$ a couple of definition for $A$,  Remark 8.4.3 gave
$$\rad(I\cdot A)=\rad(A^{00}\cdot A).$$
Since $I\cdot A$ is a finitely generated ideal of $A$, this equality of radicals has geometric meaning:
recall from Proposition 3.2.1 that for any commutative ring $A$ an open subset $U \subset\Spec(A)$ is quasi-compact
 if and only if $\Spec(A) - U$ has the form $\Spec(A/J)$ for some finitely generated ideal $J$ of $A$, which is to say that
 the unique radical ideal defining the closed complement of $U$ is also the radical of a finitely generated
 ideal.  (In particular, $\Spec(A) - V(A^{00}\cdot A)$ is quasi-compact.) 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\paragraph{The setup for ``algebraic'' description of continuity of valuations on Huber rings}\ \\


\begin{defin} \label{cofinal} If $\Gamma$ is a totally ordered abelian group and $H\subset \Gamma$ is a subgroup, we say that $\gamma\in\Gamma\cup\{0\}$ is {\em cofinal for} $H$ if for all $h\in H$ there exists a sufficiently large integer $n\ge 0$ such that $\gamma^n<h$.
\end{defin}

Note that $0$ is always cofinal for {\em any} subgroup $H\subseteq \Gamma$.

Now let us consider a commutative ring $A$, and an ideal $J\subset A$ such that its radical is  equal to the radical of a
{\em finitely generated} ideal.  We have seen that an example of such an ideal is $A^{00} \cdot A$ in a Huber ring $A$. 
For Huber rings $A$, the points of $\Spv(A)$ over $\Spec(A) - V(A^{00} \cdot A)$ are precisely the analytic points, so
for general commutative rings $A$ and $J$ as above we will imitate this and focus on the preimage under the natural continuous map 
$\Spv(A)\to\Spec(A)$ of the quasi-compact open set $\Spec(A)-V(J)$;
these are the valuations $v$ on $A$ such that $v(J)\neq 0$. 

The picture is a Cartesian diagram of topological spaces: 

\begin{displaymath}
\xymatrix{\Spv(A)\ar[d] &\ \{v\in\Spv(A)\mid v(J)\neq 0\} \ar@{_(->}[l] \ar[d]\\
		\Spec(A) & \ \Spec(A)-V(J) \ar@{_(->}[l]}
\end{displaymath}

For $v \in \Spv(A)$ we shall study the property of $v(a)$ of being cofinal in $\Gamma_v$
for $a\in J$, and in order to do so we shall separately consider
two cases: $v(J)\cap c\Gamma_v=\varnothing$, and $v(J)\cap c\Gamma_v\neq\varnothing$.
To that end, we first recall some notions related to horizontal specialization and give geometric
intepretations of when $v(J) \cap c\Gamma_v$ is empty or non-empty.

For $v\in\Spv(A)$ the horizontal specializations of $v$ in $\Spv(A)$  are given by 
$$
w_{\vert_H} : a \mapsto \left\{\begin{array}{cc} v(a) & \textup{if}\ a \in H\\ 
					0 & \textup{if}\ a \notin H
					 \end{array}\right.
$$
for $H$ a convex subgroup of $\Gamma_v$ containing the  characteristic subgroup $c\Gamma_v$, and 
we had the following:

\begin{thm} \label{horizontal Spec}
There is a bijection 
$$
\{ \textup{horizontal specializations of $v$} \} \leftrightarrow \{ v\textup{-convex primes}\},$$
via the formation of supports. 
Moreover, for two such specializations
$w$ and $w'$ of $v$, $w$ specializes to $w'$ if and only if $\fp_w \subset \fp_{w'}$ 
$($i.e., $\fp_w$ specializes to $\fp_{w'}$ in $\Spec(A)$$)$.
An inverse is given by assigning to any $v$-convex prime $\fq$ of $A$ the valuation 
$$
w_{\fq} : a \mapsto \left\{\begin{array}{cc} v(a) & \textup{if}\ a \notin \fq,\\ 
					0 & \textup{if}\ a \in \fq.
					 \end{array}\right.
$$
\end{thm}

In geometric terms, $v_{|_{c\Gamma_v}}$ is to be viewed as the ``most special'' among the horizontal
specializations of $v$.  Also, keep in mind that 
the above bijection is {\em not} between the set of $v$-convex primes of $A$ and 
the set of convex subgroups of $\Gamma_v$ containing $c\Gamma_v$ (see Example 4.4.1!).


\begin{rem} \label{equality} Let $v\in\Spv(A)$. For any convex subgroup $H$ of $\Gamma_v$, the containment
$$v(J)\cap H\subseteq v(\rad(J))\cap H$$
is an equality, because $\Gamma_v/H$ is torsion-free (in fact even totally ordered).
In particular, the condition of whether $v(J) \cap H$ is empty or not only depends on $J$ through its radical.
\end{rem}

\begin{rem} \label{nonvanishing loci} Let $v\in\Spv(A)$, and let $H$ be a convex subgroup of $\Gamma_v$ containing $c\Gamma_v$. Then we have that $v(J)\cap H$ is nonempty if and only if $v_{\vert_H}$ does {\em not} kill $J$, which is in turn equivalent to saying that $v_{\vert_H}$ lies over $\Spec(A)-V(J)$. Note that in the case $v(J)\cap H\neq\varnothing$, necessarily $v(J)\neq 0$!

When $v(J)\neq 0$, if $v(J)\cap H$ is empty then $v_{\vert_H}$ kills $J$ and hence lies over the vanishing locus of $J$.
$$[\textup{making picture!}]$$
\end{rem}

The upshot of Remark \ref{nonvanishing loci} is that the condition 
$$v(J)\cap c\Gamma_v\neq\varnothing$$
means exactly that {\em all} horizontal specializations of $v$ lie over $\Spec(A)-V(J)$.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%When describing horizontal specializations as being of the form $v_{\vert_H}$ for some convex subgroup $H$ of $\Gamma_v$
%containing $c\Gamma_v$, we need to rule out those $H$ such that there is some $a\in A$ with the property $v(a)>1$, but such that $v(a)$ is not in $H$. Otherwise, the following phenomenon would be allowed to happen. Since $v$ is a valuation, by the ultrametric inequality $v(a+1)=\max\{v(a), 1\}=v(a)$, which is not in $H$ by assumption. On the other hand, by definition of $v_{\vert_H}$, $$v_{\vert_H}(a+1)=0$$
%and if $v_{\vert_H}$ wants to be a valuation we should better have
%$$v_{\vert_H}(a+1)=\max\{v_{\vert_H}(a),1\}=1$$
%a contradiction. Essentially, assuming that $H$ contains $c\Gamma_v$ is necessary for the ultrametric inequality to hold for $v_{\vert_H}$. By Proposition, it is also sufficient. Finally, $c\Gamma_v$ is required to be a subgroup in order to ensure that the ultrametric inequality for $v_{\vert_H}$ is compatible with multiplicativity of $v_{\vert_H}$.
%
%$c\Gamma_v$ measures to what extent $v$ can be horizontally specialized. By Theorem \ref{horizontal Spec}, we can encode the containment $c\Gamma_v\subset H$ in terms of the support of $v_{\vert_{c\Gamma_v}}$. Such support is the preimage under $A\to \kappa(v)$ of the prime ideal of the valuation ring $R_v$ of $\kappa(v)$:
%$$\fp:=\{x\in R_v\subset \kappa(v)\mid\ \textup{for all}\ \gamma\in c\Gamma_v, v(x)<\gamma\}.$$
%Calling $\fm_v$ the maximal ideal of $R_v$, we have $\fm_v\subset\fp$ by minimality of $c\Gamma_v$ among the convex subgroups of $\Gamma_v$ containing $v(A)_{\ge 1}=v(\kappa(v))_{\ge 1}$, and being $\fm_v$ maximal we have equality.
%
%More in general, if we are given a convex subgroup $H$ of $\Gamma_v$ containing $c\Gamma_v$, then the support of $v_{\vert_H}$ is the preimage in $A$ of the prime ideal of the valuation ring $R_v$ of $v$ in $\kappa(v)$:
%$$\fp:=\{x\in R_v\mid \textup{for all}\ h\in H, v(x)<h\},$$
%under the canonical map $A\to\kappa(v)$.
%
%Let us put this observation into a more general context:

%\begin{eg} \label{what is this} Let us consider again our favourite example $k(\!(u)\!)(\!(t)\!)$. Call $A:=k(\!(u)\!)[\![t]\!]$ and call $v$ the rank two valuation on $A$ whose valuation ring is $k[\![u]\!]+tA=:R$, and value group $\Gamma_v:=a^{\bZ}\times b^{\bZ}$ ordered lexicographically, for two reals $0<a,b<1$.
%
%We compute $\fa$  for $H = b^{\bZ}$. Observe that on 
%the subfield $\kappa(v)=k(\!(u)\!)$, $v$ restricts to the $u$-adic valuation $\overline{v}$. Now given the definition of the total order on $\Gamma_v$, the condition $\overline{v}(x)<b^n$ for all $n\in\bZ$ means $\overline{v}(x)=0$, whence $\fa=(0)$, and the support of $v_{\vert_{c\Gamma_v}}$ is exactly $\fq:=tA$.
%
%At this point Theorem \ref{horizontal Spec} comes into play and allows us to reconstruct $c\Gamma_v$ as the minimal convex subgroup containing $v(A-\fq)$, which is $1\times b^{\bZ}$.
%
%We remark that the support of $v$ is $(0)$, and however $t\in A$ is such that $v(t)=(1,b)$ is cofinal for $H=\Gamma_v$. Therefore, notation being as in Lemma \ref{cofinality ideal}, the support of the horizontal specialization $v_{\vert_H}$ is {\em strictly} contained in $\fa$: these two ideals of $A$ can be quite different.
%\end{eg}
%
%\begin{rem} \label{understanding}
%The notation being that of the above Example \ref{what is this}, note that $\fq$ is manifestly maximal among the $v$-convex primes in $\Spec(A)$. Most likely the best way to understand $c\Gamma_v$ geometrically is to remember it is the smallest convex subgroup of $\Gamma_v$ containing $v(A-\fq)$, where $\fq$ is the maximal $v$-convex prime in $\Spec(A)$. This makes the picture quite explicit in the sense that, given Theorem \ref{horizontal Spec}, it is now very clear that $v_{\vert_{c\Gamma_v}}$ is going to be the most special among the horizontal specializations of $v$:
%\begin{displaymath}
%\xymatrix{\Spv(A)\ar[dd] & \Spv(\kappa(v)) & \Spv(\kappa(\fq)) \\
%			&  v \ar@{~>}[r] & v_{\vert_{c\Gamma_v}} \\
%		\Spec(A)	& \fp_v\ar@{~>}[r] & \fq =\textup{supp}(v_{\vert_{c\Gamma_v}})}
%\end{displaymath}
%The reader is encouraged to refer to Corollary 4.5.5 before going on.
%\end{rem}
%



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\paragraph{Understanding the condition $v(J)\cap c\Gamma_v\neq\varnothing$}\ \\


The next lemma says that checking whether $v(J)\cap c\Gamma_v$ is empty or not is equivalent to checking whether $v(J)\cap v(A)_{\ge 1}$ is empty or not.

\begin{lemma} \label{simpler cGamma} Let $A$ be a ring, $J$ an ideal of $A$ and $v\in\Spv(A)$. Then $v(J)\cap c\Gamma_v\neq\varnothing$ if and only if $v(J)\cap v(A)_{\ge 1}\neq\varnothing$.
\end{lemma}

\begin{proof}
Certainly if $v(J)\cap v(A)_{\ge 1}$ is nonempty, so is $v(J)\cap  c\Gamma_v$. Conversely,
suppose there exists $a\in J$ such that $v(a)\in c\Gamma_v$. If $v(a)\ge 1$ we are done. Let's assume $v(a)<1$.  By definition of
$c\Gamma_v$ there exist $b,b'\in A$ such that $v(b),v(b')\ge 1$ and
$$\frac{v(b')}{v(b)}\le v(a)<1.$$
It follows
$$1\le v(b')\le v(b)v(a)=v(ab),$$
and since $ab\in J$ we conclude.
\end{proof}

\begin{rem} \label{nonempty}
We can draw two consequences from the assumption $v(J)\cap c\Gamma_v\neq\varnothing$:
\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(\arabic{enumi})}}
\item For {\em any} convex subgroup $H\subset\Gamma_v$ containing $c\Gamma_v$, 
some element of $v(J)$ (e.g., anything in $v(J) \cap v(A)_{\ge 1} \ne \varnothing$) is not cofinal for $H$.
\item Every horizontal specialization of $v$ lies over $\Spec(A)-V(J)$ (since if $v(a) \in c\Gamma_v$
then for any convex subgroup $H$ of $\Gamma_v$ containing $c\Gamma_v$ clearly 
$(v_{\vert_H})(a) = v(a) \ne 0$).  Conversely, if this property holds for all horizontal specializations
then by considering the most special case $H = c\Gamma_v$ we see that $v(J) \cap c\Gamma_v$ is non-empty.
So this gives a precise geometric visualization of the condition $v(J) \cap c\Gamma_v \neq \varnothing$.
\end{enumerate}
\end{rem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\paragraph{Understanding the condition $v(J)\cap c\Gamma_v=\varnothing$}\ \\

The following result will be given a nice geometric interpretation via horizontal specialization in \S\ref{key},
and it is a striking contrast with both parts of Remark \ref{nonempty}.

\begin{prop} \label{minmax} Let $A$ be a ring, and $v\in\Spv(A)$ with value group $\Gamma_v$. Let $J$ be an ideal of $A$ such that its radical is equal to the radical of a finitely generated ideal, and assume $v(J)\cap c\Gamma_v=\varnothing$
$($i.e., the most special horizontal specialization $v_{|_{c\Gamma_c}}$ lies over $V(J)$$)$. 

\begin{enumerate}
\item[$(1)$] There exist
convex subgroups $H \subseteq \Gamma_v$ containing $c\Gamma_v$ with the property that all 
elements of $v(J)$ are cofinal for $H$, and among all such $H$ there is one $H_J$ that contains
all others. 
\item[$(2)$] If moreover $v(J)\neq 0$ $($so $c\Gamma_v \ne \Gamma_v$$)$
 then necessarily $H_J \ne c\Gamma_v$, $v(J) \cap H_J \ne \varnothing$, and $H_J$ is contained in every convex 
 subgroup $H$ of $\Gamma_v$ satisfying $v(J)\cap H\neq \varnothing$.
Geometrically, if $v$ lies over $\Spec(A) - V(J)$ then so does
the horizontal specialization $v_{|_{H_J}}$ and it is the ``unique most special'' such specialization
in the sense that every horizontal specializations of $v$
over $\Spec(A) - V(J)$ is a horizontal generization of $v_{|_{H_J}}$.
\end{enumerate}
\end{prop}

\begin{proof} If $v(J)=0$ then certainly $v(J)\cap c\Gamma_v=\varnothing$. In this case $H_J=\Gamma_v$ does the job.

We now assume $v(J)\neq 0$. Let $H$ be a convex subgroup of $\Gamma_v$ containing $c\Gamma_v$ and disjoint from  $v(J)$ (e.g., $c\Gamma_v$ is one such $H$).
Note that $v(J)\cap H=\varnothing$ if and only if we have
$$v(\rad(J))\cap H=\varnothing.$$
Indeed, if $a \in \rad(J)$ and $v(a) \in H$ then $v(a^n) \in H$ for any $n > 0$, so take $n$ large enough to ensure $a^n \in J$.
It follows that our problem only depends on $J$ through its radical, so we may
assume $J$ itself is finitely generated: say
$$J=(a_1,\dots,a_n).$$

Since $v(J) \ne 0$, some $v(a_i)$ is non-zero.  
The valuation of the generators $a_1,\dots,a_n$ of $J$ attains a maximum, which we call $h$, so $h \ne 0$. 
Note that $h \not\in H$ since $h \in v(J)$.  In particular, $h < 1$ since $v(A)_{\ge 1} \subseteq c\Gamma_v \subseteq H$
and likewise $h \not\in c\Gamma_v$. 
We call $$H_J:=\{\gamma\in\Gamma_v\mid h^n\le\gamma\le h^{-n}\ \textup{for some}\ n\ge 0\};$$
this is the convex subgroup ``generated'' by $h$ inside $\Gamma_v$, and $h \in H_J$ since $h < 1$;
in particular, $v(J) \cap H_J \ne \varnothing$.

We now show that $H_J$ satisfies the required properties.
\paragraph{Step 1: $H_J$ strictly contains $c\Gamma_v$} We saw above that necessarily $h<1$.
Since $h$ does not lie in the convex subgroup $c\Gamma_v$, it follows that $h < \gamma$ for all $\gamma \in c\Gamma_v$.
(Indeed, if for some such $\gamma$ we have $\gamma \le h$ then the $h$ is sandwiches between $\gamma$ and $1$,
forcing $h \in c\Gamma_v$ by convexity.) 
 Applying inversion to the relation $h<c\Gamma_v$, we obtain
$$h<c\Gamma_v<h^{-1},$$
and being $H_J$ a convex subgroup of $\Gamma_v$ by construction, necessarily $c\Gamma_v\subseteq H_J$. The inclusion is strict, since $h\notin c\Gamma_v$.

\paragraph{Step 2: $v(J)$ is cofinal for $H_J$} 
By design, for any $\gamma \in H_J$ there exists $n > 0$ such that $h^n < \gamma$, so $v(a_i)^n \le h^n < \gamma$ for all
$i$.  Hence, the set
$$\{a\in A\mid v(a)\ \textup{cofinal for}\ H_J\}$$
contains $a_1, \dots, a_n$.  We want this set to contain the ideal $J$ generated by the $a_i$'s, so it suffices
for this set to be an ideal itself.   We cannot merely argue by ``$A$-linearity'' since $v(A)$ might not be $\le 1$ inside
$\Gamma_v$.  Nonetheless, since $c\Gamma_v$ is strictly contained in $H_J$ the hypotheses of the following surprising lemma 
are satisfied for $H_J$ and thereby gives the required ideal property:

\begin{lemma} \label{cofinality ideal} Let $A$ be a ring, and $v\in\Spv(A)$. Let $H$ be a 
subgroup of $\Gamma_v$ which strictly contains  $c\Gamma_v$. Then $$\fa:=\{a\in A\mid v(a)\ \textup{is cofinal for}\ H\}$$
is a radical ideal of $A$.
\end{lemma}

The surprise is that cofinality relative to $H$ defines an ideal even when $v(a) > 1$ for some $a \in A$.

\begin{proof}
Let $a,b\in \fa$. We want to show $v(a+b)$ is again cofinal for $H$. By the ultrametric inequality,
$$v(a+b)\le\max\{v(a),v(b)\}$$
and hence, since both $v(a)$ and $v(b)$ are cofinal for $H$, so is $v(a+b)$. Now let $a\in \fa$, and $b\in A$. If $v(b)\le 1$, then $v(ab)\le v(a)$, and we are done. Suppose $v(b)>1$. Then $v(b)$ is contained in $c\Gamma_v\subseteq H$. 
We now use the assumption that $c\Gamma_v$ is strictly contained in $H$, and that $v(b)>1$. 
There exists $h_0\in H$ not in $c\Gamma_v$, and by convexity of $c\Gamma_v$ we know that $h_0$
is either strictly larger or strictly smaller than all elements of $c\Gamma_v$.  Passing to its inverse if necessary,
we can arrange that 
$$h_0<c\Gamma_v.$$
It follows $h_0^{-1}>c\Gamma_v$, and we have $v(a)^nh_0^{-1}< 1$ for sufficiently large $n\ge 0$. Finally, 
$$v(ab)^{n+N}=v(a)^{n+N}v(b)^{n+N}<v(a)^{n+N}h_0^{-1}<v(a)^N$$
for $N\ge 1$, and we conclude that $v(ab)$ is cofinal for $H$ since $v(a)$ is cofinal for $H$. 
Thus, $\fa$ is indeed an ideal of $A$. 

If $a^n \in \fa$ then for any $h \in H$ we have $v(a^n)^m < h^n$ for all large $m$, so $v(a)^m < h$ for all large $m$,
so $a \in \fa$.  That is, $\fa$ is radical.
\end{proof}

\paragraph{Step 3: $H_J$ is maximal} Indeed, if $H$ is any convex subgroup of $\Gamma_v$ containing $c\Gamma_v$ and such that $v(J)$ is cofinal for $H$, then in particular $h$ is cofinal for $H$. It follows $H_{\le 1}\subseteq H_J$ by convexity of $H_J$, and then $H\subseteq H_J$.

It remains to address the minimality property for $H_J$.  Clearly
$v(J)\cap H_J\neq \varnothing$, as $h$ is in the intersection.  We want to prove that $H_J$ is contained in every
convex subgroup $H\subseteq\Gamma_v$ satisfying
$$v(J)\cap H\neq \varnothing.$$
We instead briefly sketch the idea, and refer the reader to \cite[Lemma 7.2]{Wed}. The assumption $v(J)\cap c\Gamma_v=\varnothing$ is used again to deduce that any such $H$ strictly contains $c\Gamma_v$, and convexity of $H$ and the choice of $h$ do the job to show $H_J$ is contained in $H$.
\end{proof}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The key construction}\label{key}

Using Remark \ref{nonempty} and Proposition \ref{minmax},
we are led to make the following definition.

\begin{defin} \label{key def} Let $A$ be a ring, and $v\in\Spv(A)$, with value group $\Gamma_v$. Let $J$ be an ideal of $A$ 
whose radical coincides with the radical of a finitely generated radical. We set:
$$c\Gamma_v(J):=\begin{cases} H_J & \textup{if}\ v(J)\cap c\Gamma_v=\varnothing\\
						c\Gamma_v & \textup{if}\ v(J)\cap c\Gamma_v\neq \varnothing.
\end{cases}
$$
\end{defin}

The ``reasonableness'' of the definition when $v(J) \cap c\Gamma_v$ is non-empty (which can only happen
when $v(J) \ne 0$) is explained by the minimality aspect of Proposition \ref{minmax}(3), even though that proposition
only concerns cases with $v(J)$ disjoint from $c\Gamma_v$).
Note also that always $c\Gamma_v(A) = c\Gamma_v$ (since $1 \in v(A)$).

The geometric idea underlying the definition of $c\Gamma_v(J)$ is that the associated horizontal
specialization $v_{|_{c\Gamma_c(J)}}$
 of $v$ is the one that is ``just barely'' still over $\Spec(A) - V(J)$
(with the understanding that it is equal to $v$ when $v$ itself lies over $V(J)$).
We summarize the basic properties of $c\Gamma_v(J)$, unifying the two cases of its dichotomous definition, in the following proposition (see especially part  (3)).

\begin{prop} \label{key duda}
For $A$ and $J$ as above and $v \in \Spv(A)$, the subgroup
$c\Gamma_v(J)$ of $\Gamma_v$ satisfies the following properties:
\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(\arabic{enumi})}}
\item $c\Gamma_v(J)$ is convex, and $c\Gamma_v\subseteq c\Gamma_v(J)$.
\item $c\Gamma_v(J)=\Gamma_v$ if and only if every proper horizontal specialization of $v$ lies over $V(J)\subset\Spec(A)$.
\item $c\Gamma_v(J)$ is minimal with respect to the property of being a convex subgroup of $\Gamma_v$ 
that contains $c\Gamma_v$ and meets $v(J)$.
\item If $v(J)\cap c\Gamma_v=\varnothing$ then $c\Gamma_v(J)$ is maximal among the convex subgroups of $\Gamma_v$ that contain $c\Gamma_v$ and relative to which all elements of $v(J)$ are cofinal.
\end{enumerate}
\end{prop}

There is no ambiguity about the use of ``minimal'' and ``maximal'' above since the collection of convex subgroups of
a totally ordered abelian group is itself totally ordered under inclusion (see \cite[Rem.\,1.10]{Wed}, an elementary proof
by contradiction).

\begin{proof}
$(1)$ is a consequence of Proposition \ref{minmax} and the definition of $c\Gamma_v$, 
as are $(3)$ and $(4)$. But $(2)$ requires an argument (using (3) crucially), as follows. 

If $v(J)=0$ then (2) is trivially true since $v$ lies already over $V(J)$. Suppose instead $v(J)\neq 0$.
First assume $c\Gamma_v(J) = \Gamma_v$.  
Consider a proper horizontal specialization $v_{|_H}$, so $H$ is a {\em proper} convex subgroup of $\Gamma_v$ containing
$c\Gamma_v$. By $(3)$, $v(J)\cap H$ must be empty, which implies exactly $v_{\vert_H}(J)=0$; i.e., every proper horizontal specialization of $v$ lies over $V(J)$. Conversely, suppose every proper horizontal specialization of $v$ lies over $V(J)$. Then for every proper convex subgroup $H\subsetneq\Gamma_v$ containing $c\Gamma_v$ the specialization
$v_{|_H}$ has support in $V(J)$, so $v(J)\cap H$ must be empty. This says that among all
{\em proper} convex subgroups of $\Gamma_v$ containing $c\Gamma_v$ none meet $v(J)$.
Hence, $\Gamma_v$ is the only convex subgroup of $\Gamma_v$ containing $c\Gamma_v$ and meeting $v(J) \ne 0$,
so by (3) it must equal $c\Gamma_v(J)$.
\end{proof}

Let us recast the geometric meaning of Proposition \ref{key duda} in terms that will be useful for what follows.
By part (1), $c\Gamma_v(J)$ is a convex subgroup containing $c\Gamma_v$, so $v_{|_{c\Gamma_v(J)}}$ makes
sense as a horizontal specialization of $v$. In particular, the assignment
$$r:v\longmapsto v_{\vert_{c\Gamma_v(J)}}$$
makes sense as a map of sets $\Spv(A)\to\Spv(A)$.
By part (2), $r(v)=v$ if and only if all {\em proper} horizontal specializations of $v$ lie 
over $V(J)$; this is 
a vacuous condition if there are no proper horizontal specializations, and it is a trivial condition if $v$ itself lies over $V(J)$.
Part (3) says that when $v$ lies over $\Spec(A) - V(J)$ then 
$r(v)$ is the ``most special'' among specializations of $v$ which lie over $\Spec(A) - V(J)$.
So the map $r$ should be viewed over $\Spec(A) - V(J)$ as carrying each $v$ to the ``last'' among its horizontal
specializations that do not lie over $V(J)$.
If $v(J)\cap c\Gamma_v$ is nonempty then even the most special among all horizontal specializations, namely
$v_{|_{c\Gamma_v}}$, does not lie over $V(J)$. But if $v(J)\cap c\Gamma_v$ is empty then all proper horizontal specializations of $r(v)$ lie over $V(J)$ but $r(v)$ does not. So $r$ acts as the identity on the part of $\Spv(A)$ over $V(J)$
and in general the geometric interpretation (and the {\em transitivity} of the relation
``horizontal specialization'') makes it clear that $r(r(v)) = r(v)$ for any $v$ whatsoever.

To summarize, $r$ is a retraction of the entire space $\Spv(A)$ onto the union of
the locus over $V(J)$ and the subset of the locus over $\Spec(A) - V(J)$ that is its ``interior edge'' relative
to horizontal specialization.

The preceding considerations motivate interest in the following definition:

\begin{defin} \label{Spv A J} For a ring $A$ and ideal $J$ such that $\rad(J)$ is the radical of a finitely generated ideal, define
$$\Spv(A,J):=\{v \in \Spv(A)\,|\, r(v)=v\} = \{v\in\Spv(A)\mid c\Gamma_v(J)=\Gamma_v\}.$$
\end{defin}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Thus, $r$ is a retraction of $\Spv(A)$ onto $\Spv(A,J)$. 
As we have seen above, 
$$\Spv(A,J) \supseteq \{v\in\Spv(A)\mid v(J)=0\}.$$
Since the case $J=A$ is relevant to the case of Tate rings later on, we note
that the space $\Spv(A,A) := \{v\,|\,c\Gamma_v = \Gamma_v\}$ can be described
geometrically as the locus of points $v \in \Spv(A)$ with no proper horizontal specializations (since if 
$H$ is any {\em proper} convex  subgroup of the group $\Gamma_v$ 
and $H \supseteq c\Gamma_v$
then necessarily $v_{|_H} \ne v$: some $a \in A - \mathfrak{p}_v$ must satisfy $v(a) \not\in H$ since
$\Gamma_v$ is generated by $v(A - \mathfrak{p}_v)$, 
so ${\rm{supp}}(v_{|_H})$ contains $a$ whereas ${\rm{supp}}(v)$ does not contain $a$).

Here is a concrete description of  $\Spv(A,J)$ in the spirit of continuity conditions for valuations on Huber rings (though presently
$A$ is just a commutative ring):

\begin{lemma} \label{rep} Let $a_1, \dots, a_n$ satisfy $\rad(a_1,\dots,a_n) = \rad(J)$.
For $v\in\Spv(A)$, the following are equivalent:
\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(\arabic{enumi})}}
\item $c\Gamma_v(J)=\Gamma_v$ $($i.e., $v \in \Spv(A,J)$$)$.
\item $\Gamma_v=c\Gamma_v$ or $v(a)$ is cofinal for $\Gamma_v$ for all $a\in J$.
\item $\Gamma_v=c\Gamma_v$ or $v(a_1),\dots , v(a_n)$ are all cofinal for $\Gamma_v$.
\end{enumerate}
\end{lemma}

\begin{proof} $(1)$ and $(2)$ are equivalent by Definition \ref{Spv A J} and Proposition \ref{minmax}. $(2)$ trivially implies $(3)$. On the other hand, by Lemma \ref{cofinality ideal} and Proposition \ref{key duda}, $(3)$ implies $(2)$ as well: indeed either $\Gamma_v=c\Gamma_v$, or if not then the assumptions of Lemma \ref{cofinality ideal} are satisfied.
\end{proof}

The technical importance of $\Spv(A,J)$ is due to:

\begin{prop} \label{fundamental} Let $A$ and $J$ be as above, and give $\Spv(A,J)$ the subspace topology from $\Spv(A)$.  Then: 
\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(\arabic{enumi})}}
\item $X:=\Spv(A,J)$ is a spectral space.
\item A base of quasi-compact open sets for the topology of $X$ is given by
$$X(T/s):=\{v\in X\mid v(g_1)\le v(s)\neq 0,\dots,v(g_n)\le v(s)\neq 0\}$$
for non-empty finite sets $T=\{g_1,\dots, g_n\}$ such that
$J\subseteq\rad(T\cdot A)$.
\item The retraction $r:\Spv(A)\to\Spv(A,J)$ is a spectral map $($in particular, continuous$)$.
\item If $v\in\Spv(A)$ lies over $\Spec(A)-V(J)$ then so does $r(v)$.
\end{enumerate}
\end{prop}

\begin{proof} The proof of (4) has been given in our discussion of the geometric meaning of $r$.  
We divide the proof of the rest into steps. As we have seen a few times already, we may and do assume $J$ is finitely generated.
\paragraph{Proof of (2) apart from quasi-compactness:} Choose any such $T$ and $s$, so 
$$X(T/s)=X\cap\Spv(A)(T/s)$$
 is open in $X$. To check that these open sets are stable under finite intersection, first note that 
$$X(T/s)=X\left(\frac{T\cup\{s\}}{s}\right),$$
so we are reduced to the case in which $T$ contains $s$. 
Pick two finite nonempty subsets $T$ and $T'$ in $A$, and $s \in T$ and $s' \in T'$ with the property
$$J\subseteq\rad(T\cdot A)\cap\rad(T'\cdot A).$$
If we define $T'':=\{t\cdot t', t\in T\ \textup{and}\ t'\in T'\}$ then ($ss' \in T''$ and) the ideal $\rad(T''\cdot A)$ contains $J$
since it  contains $J^2$.  Thus, $X(T''/ss')$ makes sense and clearly 
$$X(T/s)\cap X(T'/s')\subseteq X(T''/ss').$$
The reverse inclusion holds because if $v(tt') \le v(ss') \ne 0$ for some $t \in T$ and $t' \in T'$
then it is not possible that both $v(t) > v(s)$ and $v(t') > v(s')$.

For the topological base assertion in (2) it remains 
to show that given $v\in\Spv(A,J)$ and any open neighbourhood $U$ of $v$ in $\Spv(A)$, 
there exists an open neighbourhood of $v$ in $U \cap \Spv(A,J)$ given by such an $X(T/s)$.
We first assume $c\Gamma_v=\Gamma_v$. By definition of the topology on $\Spv(A)$,
we can choose $t_1,\dots,t_n,s\in A$ such that
$$v\in\Spv(A)\left(\frac{t_1,\dots,t_n}{s}\right)\subseteq U.$$
We claim we can arrange the choice of such $t_i$'s and $s$ such that one of the $t_i$'s is $1$
(so trivially $J \subseteq \rad(t_1,\dots,t_n)$).

Assume first $v(s)\ge 1$. Then we simply append an extra $t':=1$ to the $t_i$'s.

Let now $v(s)<1$.   It follows $v(s^{-1})=v(s)^{-1}\ge 1$, and $v(s)^{-1}\in \Gamma_v=c\Gamma_v$.
We claim that $v(s)^{-1}\le v(a)$ for some $a\in A$. In this case we will have
$$1\le v(as)$$
and therefore
$$v\in X\left(\frac{t_1a,\dots,t_na,1}{sa}\right)\subseteq U,$$
as required. Since $v(s) \in \Gamma_v = c\Gamma_v$, there exist $a', a \in A$ such that $v(a'), v(a) \ge 1$
and $v(a')/v(a) \le v(s)$.  If $v(as) \ge 1$ then we are done, so we may assume $v(as) < 1$, or equivalently
$v(s) < 1/v(a)$.  Thus,
$$v(s) \ge v(a')/v(a) > v(a')v(s),$$
so $v(a') < 1$, a contradiction.

Now assume $\Gamma_v\neq c\Gamma_v$. Let $\{a_1,\dots,a_m\}$ be generators for $J$. 
Since $v(s) \ne 0$, by Lemma \ref{rep} $(3)$ we know that for sufficiently large $k\ge 0$, $$v(a_i)^k\le v(s)\ \ \textup{for all}\ i=1,\dots, m.$$
This implies
$$v\in X\left(\frac{t_1,\dots,t_n, a_1^k,\dots, a_m^k}{s}\right)\subseteq U.$$
This new collection of ``numerators'' clearly does the job.

\paragraph{Proof of (3):} We let $s\in A$, $T\subset A$ be a finite nonempty subset of $A$, and $I\subset\rad(T\cdot A)$. Let $$U:=X(T/s)\ \ \ \textup{and}\ \ \ U':=\Spv(A)(T/s).$$
We know that $U'$ is quasi-compact from our work on the spectrality of $\Spv(A)$.
We claim $U'=r^{-1}(U)$, which will achieve continuity of $r$ due to the settled part of (2), and also the spectrality
once $U$ is known to be quasi-compact (the part of (2) not yet proved). 
We know $U\subseteq U'$, and every point of $r^{-1}(U)$ is a horizontal generization of a point of $U$. Since $U'$ is open, it is stable under generization and hence contains $r^{-1}(U)$.

Now choose $w\in U'$ and we have to show $r(w)\in U$. If $w$ lies over $V(J)$ then $r(w) = w \in U' \cap X = X(T/s) = U$.
 Assume now $w(J)\neq 0$. Thus, $w$ lies over $\Spec(A) - V(J)$, so the same holds for $r(w)$.  That is,
 $r(w)$ does not kill $J$.  But $J \subseteq \rad(T \cdot A)$ by hypothesis, so $r(w)$ cannot kill $T$.
 That is, $r(w)(t_0) \ne 0$ for some $t_0 \in T$.
   We have by assumption 
$$w(t)\le w(s)\neq 0$$
for all $t\in T$.

It is a general fact that if $v' = v_{|_H}$ is a horizontal specialization of $v \in \Spv(A)$
and $a, b \in A$ satisfy $v(a) \le v(b)$ then $v'(a) \le v'(b)$.  Indeed, this is obvious if $v'(b) \ne 0$ (i.e., $v(b) \in H$)
since in such cases $v'(b) = v(b)$ whereas always $v'(a)$ is equal either to $v(a)$ or to 0 (depending on whether or not
$v(a) \in H$).   So we just need to rule out the possibility $v(b) \not\in H$ yet $v(a) \in H$.   But 
$v(A)_{\ge 1} \subseteq c\Gamma_v \subseteq H$, so if $v(b) \not\in H$ then $v(b) < 1$, so convexity of $H$
and the inequalities $v(a) \le v(b) < 1$ would force $v(a) \not\in H$.
Applying this general conclusion to the horizontal specialization $v' = r(w)$, we have $r(w)(t)\le r(w)(s)$ for all $t\in T$.
Taking $t = t_0$ thereby prevents $r(w)(s)$ from vanishing, so $r(w) \in U' \cap X = U$.

\paragraph{End of the proof} 
Consider the Boolean algebra generated by the subsets $X(T/s)$ of $X$, and 
endow $X$ with the topology that this algebra generates (so each $X(T/s)$ is open and closed for this new topology). 
The key lies in $(3)$, where we proved that for any member $U$ of this algebra,
$r^{-1}(U)$ is {\em constructible} in the spectral space $\Spv(A)$. 
Using Theorem 3.3.9 (Hochster's criterion) one deduces that $X$ is spectral
{\em and} that every $X(T/s)$ is quasi-compact, so also $r$ is spectral.  See the last paragraph of the proof of
\cite[Prop.\,2.6]{H1} for the details.  (This argument uses the quasi-compactness of
the {\em constructible topology} of spectral spaces, applied to the spectral space $\Spv(A)$.)
\end{proof}




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\subsection{The Main Theorem}


Recall we started seeking a suitable algebraic description of continuity for valuations on $A$. Now $A$ shall be a Huber ring, and our constructions from the previous section do finally produce the desired description! It turns out that cofinality 
captures continuity when we choose the appropriate $J$:

\begin{thm} \label{continuity algebraic} Let $A$ be a Huber ring, and $(A_0, I)$ a couple of definition. Then we have:
$$\Cont(A)=\{v\in\Spv(A, A^{00}\cdot A)\mid v(a)<1\ \textup{for all}\ a\in A^{00}\}.$$
\end{thm}

Note first of all that $\rad(A^{00}\cdot A)=\rad(I\cdot A)$, so since $I\cdot A$ is a finitely generated ideal of $A$
it follows that the construction explained throughout the previous section is applicable. 

The Theorem implies that $v\in\Spv(A)$ is continuous if and only if for every $a \in I$
the value $v(a)$ is cofinal in $\Gamma_v$
(not just $v(a) < 1$; in Example 10.2.3 we will illustrate that this is weaker than
cofinality for higher-rank $v$).  In fact, we can push this a bit further in terms of 
a finite generating set of $I$ as an ideal of $A_0$, subject to (necessary) boundedness for $v$ on the entirety of $A_0$;
see Corollary \ref{vbound}.

Granting Theorem \ref{continuity algebraic} for a moment, we finally prove spectrality of $\Cont(A)$:

\begin{cor} \label{spectrality achieved !} Let $A$ be a Huber ring, and $(A_0, I)$ a couple of definition. Then $\Cont(A)$ is closed in $\Spv(A, I\cdot A)$, which is spectral. In particular, $\Cont(A)$ is spectral and thus quasi-compact.
\end{cor}

\begin{proof} Indeed,
$$\Spv(A, I\cdot A)-\Cont(A)=\bigcup_{a\in I}\Spv(A, I\cdot A)\left(\frac{1}{a}\right),$$
which is open. By $\S 3.3$ (Lecture 3), since $\Cont(A)$ is a closed subspace of a spectral space, it is spectral.
\end{proof}

We now prove Theorem \ref{continuity algebraic}. The reader may refer also to \cite[Thm. 7.10]{Wed}, and to \cite[Thm. 3.1]{H1}.

\begin{proof}[Proof of Theorem \ref{continuity algebraic}]\
\paragraph{Step 1: $\Cont(A)\subseteq$ right side.} Let $v\in\Cont(A)$. For all 
$a \in A^{00}$  the continuity of $v$ implies both
$v(a)<1$ and the cofinality of $v(a)$ for $\Gamma_v$. 
Now Lemma \ref{rep} yields $c\Gamma_v(A^{00}\cdot A)=\Gamma_v$, so
$v\in \Spv(A, A^{00}\cdot A)$. 

\paragraph{Step 2: cofinality.}  To show that the right side is contained in the left side, choose
$v\in\Spv(A, I\cdot A)$ such that $v(a)<1$ for all $a\in I$.  We first check 
that $v(a)$ is cofinal for $\Gamma_v$ for all $a\in I$.
This is immediate if $c\Gamma_v\neq\Gamma_v$ by Lemma \ref{rep}$(2)$. 
Assume now $c\Gamma_v=\Gamma_v$. We have to show that for any $\gamma$ in $\Gamma_v$ and $a$ in $I$, if $n$ is large enough (depending on $a$ and $\gamma$) then $v(a)^n < \gamma$.  The case $\gamma \ge 1$ is trivial (as $v(a) < 1$ by hypothesis), so assume $\gamma < 1$.  But $\gamma \in c\Gamma_v$, so it is bounded below by a fraction $v(b')/v(b)$ with $b, b'$ in $A$ such that $v(b), v(b') \ge 1$.  Hence,

$$1/v(b) \le v(b')/v(b) \le \gamma.$$

Thus, we are reduced to show that for large enough $n$ (such $n$ depending on $a$) we have $v(a)^n < 1/v(b)$, or equivalently $v(a^n b) < 1$.  But $a^n b \to 0$ in $A$ as $n \to \infty$ since $a$ is in $I$, so for $n$ sufficiently large we have $a^n b \in I$ and hence (by our initial hypotheses on $v$!) $v(a^n b) < 1$, as required.

\paragraph{Step 3: continuity.}
We finally deduce continuity for $v$ as in Step 2. Let $T:=\{a_1,\dots,a_n\}$  be a set of generators for $I$ as an ideal of $A_0$. 
Since $\Gamma_v$ is totally ordered, upon reindexing we can assume
$$v(a_1)\ge v(a_2)\ge\dots\ge v(a_n).$$
For any $m$-fold product $a$ of the $a_i$'s we have $v(a) \le v(a_1)^m$, so
by the established cofinality of $v(a_1)$ in $\Gamma_v$ it follows that 
for all $\gamma\in\Gamma_v$ and sufficiently large $m\ge 0$ we have
$$v(a)<\gamma.$$
Since $v(b) < 1$ for {\em all} $b \in I$ by hypothesis (!), for such $m\ge 0$ we have 
$$v(T^m\cdot I)<\gamma.$$
But $T^m\cdot I=I^{m+1}$, so this displayed inequality expresses exactly that $v$ is continuous. 
\end{proof}

\begin{cor}\label{vbound} Let $A$ be a Huber ring with couple of definition $(A_0, I)$, and let
$\{a_1, \dots, a_n\}$ be a finite generating set of $I$ as an $A_0$-module.  For $v \in \Spv(A)$, the following are equivalent:
\begin{enumerate}
\item[{\rm{(1)}}] $v$ is continuous;
\item[{\rm{(2)}}] $v(a)$ is cofinal in $\Gamma_v$ for all $a \in I$,
\item[{\rm{(3)}}] $\gamma_i := v(a_i)$ is cofinal in $\Gamma_v$ for all $i$ and when $\gamma := \max_i \gamma_i \ne 0$
then $v(a) < 1/\gamma$ for all $a \in A_0$.
\end{enumerate}
\end{cor}

This is an ``algebraic'' description of continuity (up to the fact that the specification of $(A_0, I)$ is of topological nature). 
beware that it is {\em not} necessary for $v \in \Cont(A)$ that $v(a) \le 1$ for all $a \in A_0$; we will give pervasive 
counterexamples with higher-rank $v$ in Example 10.2.4.

\begin{proof}  Trivially (1) implies (2) due to topological nilpotence of elements of $I$.
To see that (2) implies (3) we first note that cofinal elements of $\Gamma_v$ must be $< 1$, so
since $aa_i \in I$ for all $a \in A_0$ we see that $v(aa_i) < 1$ for all $i$ and all $a \in A_0$.
This implies $v(a) < 1/\gamma_i$ whenever $\gamma_i \ne 0$.
Thus, if $\gamma := \max_i \gamma_i$ is nonzero then necessarily $v(a) < 1/\gamma$ for all $a \in A_0$.

Finally, assume $\gamma_i$ is cofinal in $\Gamma_v$ for all $i$ and moreover that if some $\gamma_i$ is nonzero (so $\gamma \ne 0$) 
then in fact $v(a) < 1/\gamma$ for all $a \in A_0$.  We want to show that $v$ is continuous.  If all $\gamma_i$ vanish
then certainly $v(I) = \{0\}$, so $v$ is trivially continuous (as it factors through the discrete quotient $A/I\cdot A$ of $A$).
Thus, now we assume some $\gamma_i$ is nonzero, so $\gamma \ne 0$ and by hypothesis
$v(a) < 1/\gamma$ for all $a \in A_0$.  Hence, for any $i$ we have
$$v(a a_i) = \gamma_i v(a) \le \gamma v(a) < 1,$$
so every $A_0$-linear combination $b$ of the $a_i$'s satisfies $v(b) < 1$. Such linear combinations
exhaust $I$, so $v(b) < 1$ for all $b \in I$, and hence $v(a) < 1$ for all $a \in A^{00}$ (as any such $a$
satisfies $a^m \in I$ for some $m > 0$, since $I$ is open around 0).   By Theorem \ref{continuity algebraic},
continuity of such $v$ is reduced to checking that $v \in \Spv(A, A^{00} \cdot A)$.
Since ${\rm{rad}}(\sum A a_i) = {\rm{rad}}(A^{00}\cdot A)$, by Lemma \ref{rep} we are done
because $v(a_i)$ is cofinal in $\Gamma_v$ for all $i$ by hypothesis. 
\end{proof}

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%\subsection{An example}
%
%We now revisit Example \ref{val on field}, to which the reader is encouraged to refer before going on.
%
%Let us show the following:
%
%\begin{prop} \label{dependent} Let $v$ be a rank one valuation on a field $K$, endowed with the valuation topology yeld by $v$. A valuation $w\in\Spv(K)$ is dependent on $v$ if and only if $\fm_v\subset\fm_w$, where $\fm_v$ and $\fm_w$ denote the maximal ideals of the respective valuation rings $R_v$ and $R_w$.
%\end{prop}
%
%\begin{proof}
%We first observe that since $v$ is of rank $1$, $K$ is a Huber ring with ring of definition given by the valuation ring $R_v$ of $v$, and set of topologically nilpotent elements given by its maximal ideal $\fm_v$.
%
%Suppose $\fm_v\subset\fm_w$. Then
%$$w(\fm_v)\cap w(K)_{\ge 1}=\varnothing$$
%which implies $w(\fm_v)$ is cofinal for $c\Gamma_w=\Gamma_w$. Then $w$ is continuous, and by Example \ref{val on field}, $w$ depends on $v$.
%%% I HAVE ALSO ANOTHER PROOF OF THIS: DIRECT BUT LONGER
%
%Conversely, suppose that $w$ depends on $v$. By Example \ref{val on field}, $w\in\Cont(K)$. Now suppose there exists some $t\in\fm_v-\fm_w$. It means that $w(t)\ge 1\in\Gamma_w$, and hence $w(t)$ is not cofinal for $\Gamma_w$, and $w$ is not continuous. Contradiction.
%\end{proof}
%
%
%
%We fix ideas on the following example:
%\begin{eg} \label{Cont K} Let $A=K$ be a field equipped with a rank one valuation $v$. 
%
%By Proposition \ref{dependent} and Example \ref{val on field}, $w\in\Spv(K)$ is actually in $\Cont(K)$ if and only if $\fm_v\subset\fm_w$. We can rephrase this condition saying that
%\begin{flalign*}
%\Cont(K)=&\ \{w\in\Spv(K)\mid \fm_v\subset\fm_w\}\\
%		=&\ \{w\in\Spv(K,K)\mid w(x)<1,\ \textup{for all}\ x\in\fm_v\}\\
%		=&\ \{w\in\Spv(K, K^{00}\cdot K)\mid w(x)<1,\ \textup{for all}\ x\in K^{00}\}
%\end{flalign*}
%which is exactly the statement of Theorem \ref{continuity algebraic}.
%\end{eg}


%ADD EXAMPLE OF SPV(A,J) IN SPV(A) NON SPECTRAL? LATER NEEDS CHECK

% NOTE INCLUSION OF VALUATION RINGS REVERSED WITH RESPECT TO YOUR HINT! DUE TO THE FACT THAT m_v IS THE SET OF TOP NILPOTENTS


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\small
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