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\begin{document}
\normalsize
\section{Lecture 4: Specialization constructions}

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

\subsection{Introduction}
We recall that in order to prove Tate's Acyclicity Theorem (i.e., for an affinoid space $X$, the presheaf $\cO_X$ on the category of affinoid subdomains of $X$ -- or just rational subdomains -- with inclusions as morphisms satisfies the sheaf conditions with respect to {\em finite} coverings), one reduces to the case of Laurent coverings (in fact, Laurent coverings generated by a single function), on which the statement is checked ``by hand''. A proof of an analogue of Tate's Acyclicity for adic affinoids satisfying some mild finiteness
hypotheses will be absolutely essential, and in order to do so we need to get a handle on the topology of the valuation spectrum of commutative rings $A$, which will contain adic spectra $\Spa(A,A^+)$ (to be defined in a few weeks).

The reader may refer to  \cite[Thm. 2.5]{H2} for Huber's proof of the adic version of Tate's acyclicity, under strong finiteness assumptions on the rings involves. In \cite{BuVe}, K. Buzzard and A. Verberkmoes give a proof for very general adic spaces, and in both
cases the proof is based on a concrete understanding of specializations. We will address this in a later lecture.

We have already encountered a simple and concrete construction encoding specialization relations, which we shall call {\em vertical}. Namely, when $A=K$ a field, recall that $\Spv(K)$ is just the Riemann-Zariski space $\RZ(K,\bZ)$, and a point $v\in\Spv(K)$ is a specialization of $w\in\Spv(K)$, which is to say 
$$v\in\overline{\{w\}},$$
if and only if we have inclusions of their respective valuation rings:
$$R_v\subset R_w\subset K$$
where both $R_v$ and $R_w$ have field of fractions $K$. Given $w$, the specialization $v$ arises from a valuation on the residue field of $R_w$, as explained in the construction in the Appendix to Lecture 2. 
%The reader may want to review the aforementioned section, before going on.

The goal of this lecture is to describe yet another procedure providing specialization/generization relations, which we shall call {\em horizontal}, and explain how general specializations/generizations are decribed in terms of these two basic processes.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
\subsection{Review of vertical generizations}\ \\
We recall a few facts from the Appendix in Lecture 2.\\

Given an inclusion of valuation rings
$$R_v\subset R_w\subset K$$
whose fraction fields coincide with $K$, the value groups
$$\Gamma_v=v(K^{\times})=K^{\times}/R_v^{\times}\ \ \textup{and}\ \ \Gamma_w=w(K^{\times})=K^{\times}/R^{\times}_w$$
fit in the short exact sequence
$$1\to v(R_w^{\times})\to\Gamma_v\to\Gamma_w\to 1$$
where $H:=v(R_w^{\times})$ is a {\em convex} subgroup of $\Gamma_v$ (meaning that if $\gamma\in\Gamma_v$ 
satisfies $h\leq\gamma\leq h'$ for some $h,h'\in H$ then $\gamma\in H$). 

The value group
$\Gamma_w$ is endowed with the ``quotient ordering'' from $\Gamma_v$ in the following sense. If we are given a totally ordered abelian group $\Gamma$ and a convex subgroup $H\subset \Gamma$, then a total ordering on $\overline{\Gamma}:=\Gamma/H$ is induced by the assignment
$$\overline{\Gamma}_{\le 1}=(\Gamma_{\le 1}\cdot H)/H;$$
that is, for all $\overline{\gamma},\overline{\gamma}'\in\overline{\Gamma}$, $$\overline{\gamma}\le \overline{\gamma}'\ \ \textup{if and only if}\ \ \gamma/\gamma'\in\Gamma_{\le 1}\cdot H.$$
The above indeed induces a total ordering relation because $H$ is convex, which ensures antisimmetry
(i.e., if $\overline{\gamma} \le \overline{\gamma}'$ and $\overline{\gamma}' \le \overline{\gamma}$ then $\overline{\gamma} = \overline{\gamma}'$).

\begin{eg} \label{eg: later needed} Recall that in Example 2.3.4 in Lecture 2 we found a higher-rank point in 
$$\RZ(k(\!(x)\!)(\!(y)\!), k(\!(x)\!)[\![y]\!])$$ 
whose value group is $\bZ\times\bZ$ with the lexicographic ordering; i.e., 
$$(m,n)<(m',n')\ \ \textup{if and only if either} \ m<m'\ \textup{or}\ m=m'\ \textup{with }\ n<n'.$$
An example of a convex subgroup is $\{0\}\times\bZ$. On the other hand, $\bZ\times\{0\}$ is not convex
(e.g., $(0,0) < (1,-5) < (1,0)$ but $(1,-5) \not\in \bZ \times \{0\}$), 
nor is $\bZ$ diagonally embedded in $\bZ\times\bZ$ (check!). We'll get back to this example later.
\end{eg}

\begin{defin} \label{def: vertical H} Let $v:A\to\Gamma\cup\{0\}$ be a valuation on $A$, and call $\Gamma_v$ its value group; that is, 
$\Gamma_v$ is the subgroup of $\Gamma$ generated by $v(A-\fp_v)$. Let $H\subset \Gamma$ be a convex subgroup. We define
$$v_{/_H}:A\to(\Gamma/H)\cup\{0\}$$
by
$$v_{/_H}(a):=\left\{\begin{array}{cc}
v(a) \bmod H& \textup{if } v(a) \ne 0,\\
0 & \textup{if}\ v(a)=0.
\end{array}\right.$$
\end{defin}

\begin{rem} \label{rem: fiberwise} Note that $\textup{supp}(v_{/_H})=\textup{supp}(v)=\fp_v$.  We call $v_{/_H}$ a {\em vertical generaization}
because the generization procedure takes place in the fiber of $\Spv(A)\to\Spec(A)$ over $\fp_v$, which is just $\RZ(\kappa(\fp_v),\bZ)$. Basically, we return to the old picture ``$R_v\subset R_{v_{/H}}\subset\kappa(\fp_v)$'' fiberwise. 
\end{rem}

Remark \ref{rem: fiberwise} suggests an exhaustive description of such specializations/generizations:

 \begin{prop} \label{prop: exhaustive vertical} Let $v:A\to\Gamma\cup\{0\}$ be a valuation, and assume $\Gamma = \Gamma_v$; 
i.e., $\Gamma$ is generated by $v(A-\fp_v)$. Then we have a bijective correspondence
$$\{\textup{convex subgroups}\ H\subset\Gamma\}\leftrightarrow\{\textup{vertical generizations of}\ v\}$$
assigned by $H\mapsto v_{/_H}$.
\end{prop}

\begin{proof}[Proof of Proposition \ref{prop: exhaustive vertical}]
The proof amounts to the previous discussion and the Appendix to Lecture 2. See \cite[Prop. 2.14]{Wed} for further
details.  
\end{proof}


Notice that $H\subset H'$ if and only if $v_{/_{H'}}$ generizes $v_{/_H}$.
We shall write the specialization relation as
$$v_{/_{H'}}\rightsquigarrow v_{/_H}.$$

\begin{rem}\label{vertconvex}
Consider the special case when $A = K$ is a field, so $v$ is just a valuation ring $R$ with fraction field $K$.
In this case every prime ideal $\mathfrak{p}$ of $R$ gives rise to a valuation ring $R_{\mathfrak{p}}$ between $R$ and $K$,
and by \cite[Thm.\,10.1(i), (ii)]{Mat} this construction gives rise to {\em every} valuation ring between $R$ and $K$
without repetition.  So another way to describe Proposition \ref{prop: exhaustive vertical} is as a bijection between
$\Spec(R)$ and the set of convex subgroups of the value group $\Gamma_R = K^{\times}/R^{\times}$.
\end{rem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
\subsection{Horizontal specialization}

We want to visualize the general process by which a point $v$ in $\Spv(A)$ can specialize to another point $w$. 
The picture of $\Spv(A)$ lying over $\Spec(A)$ will guide our visualization.  The case when $v$ and $w$ lie in a common fiber has already been discussed, and now we want to explore cases when the supports are distinct.  
Adapting Definition \ref{def: vertical H}, we give the following:

\begin{defin} \label{def: temporary} Let $H\subset\Gamma$ be a subgroup, and let $v:A\to\Gamma\cup\{0\}$ be a valuation on $A$. We 
define the function 
$$v_{|_H}:A\to\Gamma\cup\{0\}$$
by
$$v_{|_H}(a):=\left\{\begin{array}{cc}
v(a) & \textup{if}\ v(a)\in H\\
0 & \textup{otherwise}
\end{array}\right.$$
We will later call $v_{|_H}$ a {\em horizontal} specialization of $v$ (once we figure out under what conditions on $H$ this 
is a valuation, and when so then when it is a specialization of $v$). We shall often use the notation 
$$v':=v_{|_H}.$$
\end{defin}

Later we will see that this construction produces unique generic points in (the non-empty) fibers of
$\overline{\{v\}}$ over $\Spec(A)$ when $v(A) \not\subset \Gamma_{\le 1}$ (see Corollary \ref{gencor}).  

To give some motivation for this construction, consider the following example.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  EXAMPLE
\begin{eg} \label{eg: Z} Let $A=\bZ$. We study the fiber of $\Spv(\bZ)\to\Spec(\bZ)$ over $(0)$. The fiber over $(0)\in\Spec(\bZ)$ is $\Spv(\bQ)=\RZ(\bQ,\bZ)$, which consists of the trivial valuation
and the (normalized, say) $p$-adic valuations for primes $p$. Let $v_p\in\Spv(\bQ)$ be the $p$-adic valuation on $\bQ$, $p$ prime. We have $\Gamma_{v_p}=p^{\bZ}\simeq\bZ$, and the only convex subgroups of $\bZ$ are $\{0\}$ and $\bZ$ itself. If $H=(0)\subset\bZ\simeq\Gamma_{v_p}$, then we have that $v_{/_H}$ is $v_p$ itself. Instead, if $H=\Gamma$, then $v_{/_{H}}$ is the trivial valuation
$v_{\rm triv}$, and we see that
$$v_{\rm triv}\rightsquigarrow v_p$$
for all primes $p$. The trivial valuation is the  unique generic point in $\Spv(\bQ)$. (Topologically, $\Spv(\bQ) = \Spec(\bZ)$.)

Explicitly, we have  $\Gamma_{v_p}=p^{\bZ}$ and $\Gamma_{v_{\rm triv}}=(1)$ fit into the following short exact sequence:
\begin{displaymath}
\xymatrix{1\ar[r] & v_p(R_{v_{\rm triv}}^{\times})\ar@{=}[d]\ar[r] & \Gamma_{v_p}\ar@{=}[d]\ar[r] & \Gamma_{v_{\rm triv}}\ar@{=}[d]\ar[r] & 1\\
		1 \ar[r] & p^{\bZ} \ar[r] & p^{\bZ}\ar[r] & 1\ar[r] & 1}
\end{displaymath}
\\

We consider now the remaining valuations on $\bZ$. Recall that by Definition 2.4.3 in Lecture 2, to specify a point $v$ 
in $\Spv(\bZ)$ we assign a support $\fp_v$ and a valuation on its residue field. The only possible supports are $(0)$, which we have just studied, and $(p)$ for $p$ prime. The residue field of this latter is $\bF_p$, and the only possible valuation on $\bF_p$ is the trivial valuation. We call $\overline{v}_p$ the corresponding valuation on $\bZ$, and get that $\Spv(\bF_p)=\{\overline{v}_p\}$.\\

Let us step back to the $p$-adic valuation $v_p$. If $x\notin (p)$, then $v_p(x)=1$, and it is not $1$ otherwise. If we choose $H=(1)=p^{(0)}\subset p^{\bZ}$ then $$\overline{v}_p=(v_p)_{|_H};$$
in this particular case, Definition \ref{def: temporary} indeed yelds a valuation on $\bZ$.

It is also easy to check that the $p$-adic valuation specializes to $\overline{v}_p$, and this is the case for each prime $p$. In fact, 
we have the following commutative square encoding the specialization relations encoutered so far:
\begin{displaymath}
\xymatrix{\Spv(\bZ)\ar[ddd] & \Spv(\bQ) & \Spv(\bF_p)\\
			&  v_{\rm triv} \ar@{~>}[d]\ar@{~>}[r] & \overline{v}_p\ar@{=}[d]\\
		 &	v_p\ar@{~>}[r] & \overline{v}_p\\
		\Spec(\bZ)		& (0) & (p)}
\end{displaymath}
The very same picture, notationally updated, can be obtained for any Dedekind domain whose residue fields at maximal ideals are finite.
\end{eg}
 We now discuss a less trivial example.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   EXAMPLE 
\begin{eg} \label{eg: power} Recall Example 2.3.4 in Lecture 2. We are going to build a new example on it. Let $K=k(\!(y)\!)(\!(x)\!) = {\rm{Frac}}(k(\!(y)\!)[\![x]\!])$ for a field $k$.

Consider $\Spv(k(\!(y)\!)[\![x]\!])$. This is actually a huge space since it
includes lots of valuations that are nontrivial on $k$ (when $k$ isn't algebraic over a finite field), 
but we'll study some valuations that are trivial on $k$.
Note that $k(\!(y)\!)[\![x]\!]$ is a discrete valuation ring, and hence $\Spec(k(\!(y)\!)[\![x]\!])$ is just $\{(0), (x)\}$. We first consider the fiber of $\Spv(k(\!(y)\!)[\![x]\!])\to \Spec(k(\!(y)\!)[\![x]\!])$ over the generic point $(0)$. From Example 2.3.4 in Lecture 2 
we know three points in $\Spv(K)$: the trivial valuation $\eta$, the $x$-adic valuation $v$, and the valuation $w$ arising from 
a procedure we briefly recall now.

The valuation ring $R_v$ of the $x$-adic valuation is $k(\!(y)\!)[\![x]\!]$, and we generate a specialization $w$ of $v$ with the same support
via a nontrivial valuation on the residue field of $R_v$ 
as follows. The maximal ideal $\fm_v\subset R_v$ is $xR_v$. We equip $R_v/\fm_v\simeq k(\!(y)\!)$ with the 
evident $y$-adic valuation $\overline{w}$ (trivial on $k$), and $w$ corresponds to the valuation ring preimage
$R_w = k[\![y]\!] + xk(\!(y)\!)[\![x]\!] \subset K$, so 
$\Gamma_v = x^{\bZ}$, $\Gamma_{\overline{w}}= y^{\bZ}$, and there is a natural exact sequence 
of abelian groups 
$$1 \rightarrow \Gamma_{\overline{w}} \rightarrow \Gamma_{w} \rightarrow \Gamma_v \rightarrow 1$$
which splits to give $$\Gamma_{w}\simeq \Gamma_v\times\Gamma_{\overline{w}}\simeq x^{\bZ}\times y^{\bZ}\simeq\bZ\times\bZ$$
in which $\bZ\times\bZ$ is totally ordered with the {\em lexicographical ordering} (left factor comes first!).\\

Convex subgroups $H$ of $\Gamma_w$ include $\Gamma_w$ (corresponding to the trivial valuation $\eta$), $1\times y^{\bZ}$
corresponding to $v=\eta_{/_{1\times y^{\bZ}}}$
 (see Example \ref{eg: later needed}), and $1\times 1$ corresponding to $w=v_{/_{1\times 1}}$.

The reader may wish to have a look at the picture at the end of the example, before going on.

Now we discuss valuations on $k(\!(y)\!)[\![x]\!]$ whose support is $(x)$, using the construction in
Definition \ref{def: temporary}. We certainly have the trivial valuation $\overline{\eta}$ on $k(\!(y)\!)$, which arises from $\eta$ by ``horizontal specialization'' with respect to the trivial group $H=(1)$. On the other hand, if we consider the $x$-adic valuation $v$ on $k(\!(y)\!)[\![x]\!]$
then we may choose $H=1\times 1\subset \Gamma_v=x^{\bZ}\times 1$ and (according to Definition \ref{def: temporary})
get again the 
trivial valuation $\overline{\eta}=v_{|_{1\times 1}}$ on $k(\!(y)\!)$. On the other hand, we may also 
choose $H=1\times y^{\bZ}\subset x^{\bZ}\times y^{\bZ}=\Gamma_w$, and $w_{|_{1\times y^{\bZ}}}$
 is the $y$-adic valuation on $k(\!(y)\!)$ that we previously called $\overline{w}$. The situation is now like in the picture below.
\begin{displaymath}
\xymatrix{\Spv(k(\!(y)\!)[\![x]\!])\ar[dddd] & \Spv(K) & \Spv(k(\!(y)\!))\\
			&  \eta \ar@{~>}[d]\ar@{~>}[r] & \overline{\eta}=\eta_{|_{(1)}}\ar@{=}[d]\\
		 &	v\ar@{~>}[d]\ar@{~>}[r] & \overline{\eta}=v_{|_{1\times 1}}\ar@{~>}[d]\\
			& w \ar@{~>}[r] & \overline{w}=w_{|_{1\times y^{\bZ}}}\\
		\Spec(k(\!(y)\!)[\![x]\!])		& (0) & (x)}
\end{displaymath}
We remark that in specializing $v$, a choice like $x^{\bZ}\times 1=\Gamma_v$ was not allowed because we wanted any such specialization of $v$ to have support $(x)$. For the same reason, in specializing $w$ we could not choose $H$ to be $x^{\bZ}\times y^{\bZ}=\Gamma_w$ or $x^{\bZ}\times 1$. We observe that the latter is not a convex subgroup of $\Gamma_w$.
\end{eg}

For a valuation $v:A\to\Gamma\cup\{0\}$ and subgroup $H\subset\Gamma$ we now address the issue of finding suitable {\em general} conditions on $H$ so that $v_{|_H}$, defined in Definition \ref{def: temporary} is indeed a valuation on $A$ that moreover
lies in the closure of $\{v\}$.

\begin{prop} \label{prop: convex hull} A necessary condition for $v' := v_{|_H}$ to be a valuation is
$$\Gamma_{\geq 1}\cap v(A)\subseteq H;$$
that is, for all $a\in A$, if $v(a)\geq 1$ then $v(a)\in H$.
\end{prop}
\begin{proof} Suppose not. We have $1<v(a)$ for some $a\in A$ with $v(a)\notin H$. 
 It follows $v'(a)=0$. However, $v(a+1)>1$, and so $v'(a+1)=0$. Hence, 
 $a$ and $a+1$ are in the support of $v'$, so this support contains $1$. Contradiction.
\end{proof}

%Now consider the property that $v'$ (which we assume to be a valuation) lies in the closure of $\{v\}$.
%By continuity of the ``support'' map $\Spv(A) \rightarrow \Spec(A)$, the support
%$\mathfrak{p}_{v'}$ of $v'$ has to lie (as a point of $\Spec A$) in the closure of the $\mathfrak{p}_v$,
%which is to say $\mathfrak{p}_{v'} \supset \mathfrak{p}_v$.  This latter property automatically holds by the definition of $v'$:
%we have 
%$$\fp_{v'}=(v')^{-1}(0)=\{a\in A\mid v(a)\in(\Gamma-H)\cup\{0\}\}=v^{-1}((\Gamma-H)\cup\{0\})\supset v^{-1}(0)=\fp_v.$$
%

\begin{eg}\label{Rex}
Let $R'\subset R$ be an inclusion of valuation rings with the same fraction field $K$, with $v$ the valuation attached to $R$, and $v'$ the one attached to $R'$. Recall that the couple $(R', v')$ is induced by a valuation $\overline{v}'$ on the residue field $k=R/\fm_R$ of $R$, having valuation ring denoted $\overline{R}'\subset k$. Thus, we have the following diagram:
\begin{displaymath}
\xymatrix{\fm_R\ \ar@{^(->}[d]\ar@{^(->}[r] & R\ar@{=}[d] & \\
		R'\ \ar@{^(->}[r]\ar@{->>}[d] & R\ \ar@{^(->}[r]\ar@{->>}[d] & K\\
		\overline{R}'\ \ar@{^(->}[r] & k & }
\end{displaymath}
The above diagram induces, in turn, the following:
\begin{displaymath}
\xymatrix{{R'}^{\times}\ \ar@{^(->}[r]\ar@{->>}[d] & R^{\times}\ \ar@{^(->}[r]\ar@{->>}[d] & K^{\times}\\
		{\overline{R}'}^{\times}\ \ar@{^(->}[r] & k^{\times} & }
\end{displaymath}
The top row yelds the inclusion
$$(*)\ \ R^{\times}/{R'}^{\times}\subset K^{\times}/{R'}^{\times}=\Gamma_{v'},$$
as well as a surjection $$(**)\ \ R^{\times}/{R'}^{\times}\twoheadrightarrow k^{\times}/\overline{R'}^{\times}=\Gamma_{\overline{v}'},$$
which is seen to be an isomorphism by inspection.
By $(*)$, we can view $v'$ as a valuation on $R$ (after restriction from $K$) in the sense of Definition from Lecture 2. By $(**)$, we can do the same with $\overline{v}'$. Observe that $\fp_{v'} = (0)$ and $\fp_{\overline{v}'} = \fm_R$.

(The reader may wish to turn back to Example \ref{eg: Z} and consider respectively
 $$R=\bZ_{(p)}\subset\bQ=K,\ \  R'=R$$ and $v'=v_p$, $\overline{v}'=\overline{v}_p$, 
 as well as turn back to Example \ref{eg: power} and 
 consider $$R=k(\!(y)\!)[\![x]\!]\subset k(\!(y)\!)(\!(x)\!)=K,\ \  R'=\{f(x,y)\in R\mid f(0,y)\in k[\![y]\!]\},$$ $v'=w$, and $\overline{v}'=y$-adic valuation on $k(\!(y)\!)$. In those examples, we saw $\overline{v}'$ is a specialization of $v'$.)
 
Continuing with our example, we claim that 
$$\overline{v}'\in\overline{\{v'\}}$$ in $X:=\Spv(R)$. 
We have to show that there is no open neighborhood $U$ of $\overline{v}'$ which does not contain $v'$. 
It is sufficient to show this for all {\em rational} open neighbourhoods of $\overline{v}'$; that is, for all
$$X\left(\frac{f}{s}\right) = \{ w \in X \mid w(f) \leq w(s) \neq 0 \},$$ where $f,s\in R$, we must prove that if $\overline{v}' \in X(f/s)$
 then $v \in X(f/s)$. We define 
$$\overline{f}:=f\ \textup{mod}\ \fm_R,\ \ \textup{and}\ \ \overline{s}:= s\ \textup{mod}\ \fm_R.$$

We have to show that for all $f,s\in R$ such that
$$\overline{v}'(\overline{f}) \leq \overline{v}'(\overline{s}) \neq 0,$$
necessarily $$v'(f) \leq v'(s) \neq 0.$$
Since $\overline{v}'(\overline{s})\neq 0$, $s$ is a unit in $R$. We are reduced to show that
as soon as $\overline{f}/\overline{s} \in \overline{R}'$ then $f/s \in R'$. This is equivalent to show that if $\overline{f}\in\overline{s}\overline{R}'$, then $f\in sR'$. But the former assumption implies $$f\in sR' + \fm_R = sR'+s\cdot\fm_R\subset sR'$$ 
(since $\mathfrak{m}_R = s \cdot \mathfrak{m}_R$ because $s\in R^{\times}$). It follows $f\in sR'$, as required.
\end{eg}

Now we return to a valuation $v$ on a general ring $A$ and a subgroup $H \subset \Gamma_v$.
Example \ref{Rex} motivates interest in the case when $H$ is convex.  For such $H$,
if $v' := v_{|_H}$ is to be a valuation then $H$ has to contain
the minimal convex subgroup
$c\Gamma_v$ of $\Gamma_v$ containing $\Gamma_{\ge 1}\cap v(A)$ (as this is 
the intersection of all convex subgroups of $\Gamma_v$ containing $\Gamma_{\ge 1} \cap v(A)$); we call $c\Gamma_v$ 
the {\em characteristic subgroup} of $v$.

\begin{prop}  Let $H$ be a convex subgroup of $\Gamma_v$.  Then 
$v' := v_{|_H}$ is a valuation on $A$ 
if and only if $c\Gamma_v \subset H$, in which case it is a specialization of $v$.
\end{prop}
\begin{proof} The necessity has already been shown.
For the sufficiency (which is stated without proof in \cite[Lemma 1.2.1 (2)]{HuKne}, where the terminology ``primary'' is used for 
``horizontal'' and
likewise ``secondary'' is used for ``vertical''), suppose $H$ contains $c\Gamma_v$.   To show that $v'$ is multiplicative,
we have to prove that if $v(a), v(b) \not\in H$ then $v(ab) \not\in H$.  Necessarily $0 \le v(a), v(b) < 1$ since $\Gamma_{\ge 1} \cap v(A)
\subset H$, and by swapping
the labels if necessary we can assume $v(a) \le v(b)$.  If $v(a) = 0$ then all is clear, so we can assume $0 < v(a) \le v(b) < 1$.
 Since $v(ab) < v(b) < 1$ and $v(b) \not\in H$ but $1 \in H$, convexity of $H$ forces $v(ab) \not\in H$ as desired.
 Some easy case-checking shows that $v'(a+b) \le \max(v'(a), v'(b))$, so $v'$ is a valuation.
 
 To see that $v'$ is a specialization of $v$, we have to check that any open set around $v'$ in $X := \Spv(A)$ must contain $v$.
 In view of the definition of the topology, it is equivalent to show that if $a, b \in A$ with $v' \in X(a/b)$ then $v \in X(a/b)$,
 which is to say that if $v'(a) \le v'(b) \ne 0$ then $v(a) \le v(b) \ne 0$.  If $v'(a) \ne 0$ then necessarily $v(a), v(b) \in H$ and all is clear
 (by definition of $v'$).  If $v'(a) = 0$ then $v(a) \not\in H$ and $v(b) \in H$, so we just need to check that $v(a) \le v(b)$.
 Suppose to the contrary that $v(b) < v(a)$. But $\Gamma_{\ge 1} \cap v(A) \subset H$, so the condition $v(a) \not\in H$ forces
 $v(a) < 1$.  Hence, $v(a)$ lies between the elements $v(b), 1 \in H$ yet $v(a) \not\in H$, contradicting the convexity of $H$.
 Thus, indeed, $v'$ is a specialization of $v$.
\end{proof}

For a convex subgroup $H \subset \Gamma$
satisfying $c\Gamma_v \subset H$ we can reconstruct the specialization $v_{|_H}$ of $v$ from knowledge of its support.
 To express this efficiently, it is convenient to introduce some terminology.

\begin{defin}\label{def: v convex} We say that a subset $\Sigma\subset A$ is $v$-{\em convex} if 
any $a \in A$ satisfying $v(s) \le v(a) \le v(s')$ for $s, s' \in \Sigma$ necessarily lies in $\Sigma$.
\end{defin}

Note that a $v$-convex set is the preimage of its image under $v$, and a subset $\Sigma$ of $A$ satisfying 
$\Sigma = v^{-1}(v(\Sigma))$ is $v$-convex precisely when $v(\Sigma)$ is convex in $v(A)$;
if $0\in v(\Sigma)$ then this says exactly that any $\gamma \in v(A)$ satisfying 
$\gamma\le v(s)$ for some $s \in \Sigma$ necessarily lies in $v(\Sigma)$.

Note that any $v$-convex subset containing 0 must contain $v^{-1}(0) = \mathfrak{p}_v$; in particular, $v$-convex primes
contain $\mathfrak{p}_v$ and so can be viewed as prime ideals of $A/\mathfrak{p}_v$.

\begin{prop} Let $H \subset \Gamma_v$ be a convex
subgroup containing $c\Gamma_v$.  The prime ideal ${\rm{supp}}(v_{|_H})=v^{-1}((\Gamma_v \cup \{0\})-H)$ is $v$-convex.
\end{prop}

\begin{proof} Clearly ${\rm{supp}}(v_{|_H})$ is the $v$-preimage of its image under $v$, and it contains 0,
so we just have to check that if $a \in A$ and $v(a) \le v(s)$ for some $s \in A$ satisfying
$v(s) \not\in H$ then $v(a) \not\in H$. Note that $v(s) < 1$ since $v(A)_{\ge 1} \subset H$.
Hence, $v(a) \le v(s) < 1$, so by convexity of $H$ the hypothesis $v(s) \not\in H$ rules out the possibility that $v(a) \in H$.
\end{proof}

We finally give the official definition of horizontal specialization.

\begin{defin} \label{def: horizontal specialization} Let $v:A\to\Gamma\cup\{0\}$ be a valuation on $A$. A {\em horizontal specialization} of $v$ is a valuation on $A$ of the form $v_{|_H}$ for a convex subgroup $H\subset \Gamma_v$ containing $c\Gamma_v$.
\end{defin}

%In contrast with vertical specialization, which is described by closure relations in a fiber
%of $\Spv(A) \rightarrow \Spec(A)$ and hence
%is trivially transitive, it is not immediately 
%obvious from the definitions if horizontal specialization enjoys the transitivity property.  The problem
%comes down to the difficulty of computing the value group of a horizontal specialization.  Let's explain the issue.
%
%Observe that in the definition of horizontal
%specialization we require $H \subset \Gamma_v$, and convexity is imposed relative to $\Gamma_v$, thereby keeping
%matters ``intrinsic'' to $v$ as a point in $\Spv(A)$ (i.e., we avoid any distraction with ``equivalence classes'' of valuations).
%It is {\em hard} to compute the value group of $v_{|_H}$ because
%$\Gamma_v$ is merely generated by $v(A - \mathfrak{p}_v)$.
%
%In particular, if $v' = v_{|_H}$ is a horizontal specialization of $v$ and $v'' = v'_{|_{H'}}$ is a horizontal
%specialization of $v'$ (with $H'$ a convex subgroup of $\Gamma_{v'}$ containing $c\Gamma_{v'}$)
%then it is {\em not clear if $H'$ is convex in $H$} (as $\Gamma_{v'}$ is a mysterious subgroup of $H$)
%and hence we cannot easily see if $H'$ is convex in $\Gamma_v$.  Thus, although for the subgroup
%$H' \subset H \subset \Gamma_v$ by definition the valuation $v'_{|_{H'}}$ 
%literally coincides with the function $v_{|_{H'}}$ (by some case-checking), so it is a specialization of $v$
%(as it lies in the closure of the point $v'$ that is a specialization of $v$), it isn't obvious if
%this is a {\em horizontal} specialization of $v$.   Fortunately, by replacing $H'$ with a convex hull
%we can get what we need:


\subsection{Comparison of several horizontal specializations}

It is generally difficult to compute the exact value group of a horizontal specialization, essentially because
(i) the value group $\Gamma_v$ of a valuation $v$ on $A$ is merely generated by $v(A - \mathfrak{p}_v)$ (so elements of $\Gamma_v$
need not have the form $v(a)$ for $a \in A$, in contrast with the case when $A$ is a field) 
and (ii) convexity is sensitive to the ambient ordered group.  Thus, it is not clear from the definitions
if $v_{|_H}$ determines $H$ for convex subgroups $H \subset \Gamma_v$ containing $c\Gamma_v$, and in fact this can fail
(so for horizontal specialization we don't have an analogue of Proposition \ref{prop: exhaustive vertical} or Remark \ref{vertconvex}).
The following example was suggested by John Pardon.

\begin{eg}
Let $A' = k[x,y]$ for a field $k$ and let $\Gamma = c^{\bZ} \times c^{\bZ}$ for your favorite $0 < c < 1$,
where $\Gamma$ is given the lexicographical ordering
(left factor comes first); we write $\Gamma$ in multiplicative notation since our axioms
for valuations are expressed in multiplication form (e.g., $\Gamma \cup \{0\}$ might lead to confusion
if we write $\Gamma$ additively).  Note that $1 \times c^{\bZ}$ is a convex subgroup of $\Gamma$
(see Example \ref{eg: later needed}).

Define the rank-2 valuation $v':A' \rightarrow \Gamma \cup \{0\}$ by
the requirement of triviality on $k$ and $v'(x^i y^j) = (c^i, c^j)$, so $v'(A') = \Gamma_{\ge 1} \cup \{0\}$
and hence $v'(A')_{\le 1} = 1$.  Thus, for any convex subgroup $H \subset \Gamma$
we may form the horizontal specialization $v'_{|_H}$.  Clearly
$v'_{|_1}$ is the trivial valuation with support at the origin
and for $H = 1 \times c^{\bZ}$ (convexity seen in Example \ref{eg: later needed})
we have that $v'_{|_H}$ has support equal to $xA'$ (i.e., it is the line $x=0$, which is to say the $y$-axis).

Now comes the geometric idea:  let $A = k[x,z]$ for $z = xy$, so $\Spec(A') \rightarrow \Spec(A)$
is the ``$x$ is a generator'' chart of the blow-up of the affine plane $\Spec(A)$ at the origin $(x,z) = (0,0)$.
In particular, $A$ and $A'$ have the same fraction field.
The line $x=0$ in $\Spec(A')$ is the exceptional divisor, so it is crushed onto the origin in $\Spec(A)$
(in algebraic terms, $A \rightarrow A'/xA'$ kills $(x,z)$), as is the origin of $\Spec(A')$, so the
restriction $v:A \rightarrow \Gamma \cup \{0\}$ of $v'$ to the subring $A \subset A'$ has the same value
group as $v'$ (via the equality of fraction fields) and 
satisfies ${\rm{supp}}(v_{|_1}) = {\rm{supp}}(v_{|_H})$.  By inspection, we see that
$v_{|_1}$ and $v_{|_H}$ each coincide with the trivial valuation on $A$ supported at the origin.

(To give some geometric motivation, in Theorem \ref{thm: horizontal Spec} we will record the general fact that a horizontal specialization is
{\em determined} by its support.  Hence, the equality of supports implies that $v_{|_1} = v_{|_H}$ in
$\Spv(A)$, which we saw directly.)
\end{eg}

\medskip

\begin{prop}\label{transprop} If $v'$ is a horizontal specialization of $v$ in $\Spv(A)$ and $w$ is a horizontal specialization of $v'$
then $w$ is a horizontal specialization of $v$.
\end{prop}

\begin{proof}
I am grateful to K\k{e}stutis \v{C}esnavi\v{c}ius and John Pardon for (independently) suggesting the following simple argument.
The main point is to reformulate horizontal specialization essentially without reference to the value
group.  Consider a valuation $$v:A \rightarrow \Gamma \cup \{0\}$$ (with $\Gamma_v$ possibly a proper subgroup of $\Gamma$),
and for any subgroup $H \subset \Gamma$ clearly $v_{|_H} = v_{|_{H \cap \Gamma_v}}$
as functions from $A$ into $H \cup \{0\}$.  More importantly,
if $H$ is convex in $\Gamma_v$ and if $G$ denotes the order-theoretic convex hull
$$G = \{\gamma \in \Gamma\,|\,h_1 \le \gamma \le h_2 \mbox{ for some } h_1, h_2 \in H\}$$
of $H$ in $\Gamma$ (so $G$ is obviously a subgroup of $\Gamma$) then $G \cap \Gamma_v = H$,
so $v_{|_H} = v_{|_G}$.  Note also that $v(A)_{\ge 1}$ lies in $H$ if and only if it lies in $G$,
and that in such cases $(v_{|_H})(A)_{\ge 1} = v(A)_{\ge 1}$.

Now we may write $v' = v_{|_G}$ for a convex $G \subset \Gamma$ containing $v(A)_{\ge 1}$, and
likewise we may write $v'' = v'_{|_{G'}}$ for the convex hull $G'$ in $G$ of some subgroup of
$\Gamma_{v'}$ containing $v'(A)_{\ge 1} = v(A)_{\ge 1}$.  But then $G'$ is convex in $\Gamma$ 
since it is convex in the convex subgroup $G$ of $\Gamma$, so 
$v_{|_{G'}}$ makes sense as a horizontal specialization of $v$ and it clearly coincides with $v''$.
\end{proof}

%
%\subsection{Transitivity of horizontal specialization}
%
%We need to overcome
%the problem highlighted in Remark \ref{transhor}.  As a preliminary step, consider
%a general pair $v, v' \in \Spv(A)$ with $v' = v_{|_H}$ a horizontal specialization of $v$
%(here $H$ is a convex subgroup of $\Gamma_v$ containing $c\Gamma_v$).
%Since $H$ is convex and contains $\Gamma_{\ge 1} \cap v(A)$,
%elements of $v(A)$ outside $H$ must be $< 1$ and hence cannot be bounded below
%by anything in $H$.  Thus, 
%$$\mathfrak{p}_{v'} = \{a \in A\,|\,v(a) < h \mbox{ for all } h \in H\}.$$
%This makes quite explicit that $\mathfrak{p}_{v'} \supset \mathfrak{p}_v$.
%
%For our initial purposes it will simplify the notation and cause no loss of generality to
%replace $A$ with $A/\mathfrak{p}_v$, so suppose $A$ is a domain with fraction field $K = K(v)$, and we write 
%$\mathfrak{p}$ to denote the support of $v'$.  By viewing $v$ as a valuation on $K$, convexity of $H$ implies that 
%$$\mathfrak{P} := \{x \in R_v\,|\,v(x) < h \mbox{ for all } h \in H\}$$
%is a prime ideal of $R_v$, with $\mathfrak{P} \cap A = \mathfrak{p}$.
%Thus, the localization $(R_v)_{\mathfrak{P}}$ consists of fractions
%$r/r'$ with $r, r' \in R_v$ such that $v(r') \in H$, and its units are exactly
%its elements whose valuation under $v$ on $K$ lands inside $H$ (check!).
%In particular, once again using convexity, the maximal ideal of
%$(R_v)_{\mathfrak{P}}$ consists of those $x \in (R_v)_{\mathfrak{P}}$ such that $v(x) < h$ for all $h \in H$.
%
%Although $A$ might not lie inside $R_v$, we claim (using that $c\Gamma_c \subset H$) that $A \subset (R_v)_{\mathfrak{P}}$.
%Consider $a \in A$ with $a \not\in R_v$, so $v(a) > 1$ and hence $v(a) \in H$.
%Hence, $r' := 1/a \in R_v$ with $v(r') = 1/v(a) \in H$, so $a = 1/r' \in (R_v)_{\mathfrak{P}}$, as desired.
%Via the containment 
%$A \subset (R_v)_{\mathfrak{P}}$, the contraction of the maximal ideal consists of exactly
%those $a \in A$ such that $v(a) < h$ for all $h \in H$, which is to say it is $\mathfrak{p}$.  Thus, we arrive
%at the {\em local} inclusion
%$$A_{\mathfrak{p}} \hookrightarrow (R_v)_{\mathfrak{P}}$$
%inside the fraction field $K$ of $A$.  Note also that any localization of a valuation ring is a valuation ring (exercise),
%so $(R_v)_{\mathfrak{P}}$ is a valuation ring of $K$ (with fraction field $K$) dominating $A_{\mathfrak{p}}$.
%Likewise, any quotient of a valuation ring modulo a prime ideal is again a valuation ring (exercise), so
%$R_v/\mathfrak{P}$ is a valuation ring with fraction field $\kappa(\mathfrak{P})$ equal to the residue field
%of $(R_v)_{\mathfrak{P}}$.  
%Consider the resulting map
%$$w:A \rightarrow A_{\mathfrak{p}} \rightarrow (R_v)_{\mathfrak{P}}/\mathfrak{P}(R_v)_{\mathfrak{P}} = \kappa(\mathfrak{P})$$
%where the target is viewed as a valued field via its valuation ring $R_v/\mathfrak{P}$ (which does {\em not} necessarily
%contain the subring $A/\mathfrak{p}$).
%
%\begin{lemma} The valuation of $R_v/\mathfrak{P}$ is identified with the map given away from $0$ by
%$x \bmod \mathfrak{P} \rightarrow v(x)$, and it 
%identifies the value group of $R_v/\mathfrak{P}$ with $H$.  In particular, $w = v_{|_H}$.
%\end{lemma}
%
%Note that we do {\em not} claim that $v'$ on $A$ has value group equal to $H$ (it may be a proper subgroup of $H$)
%because $\kappa(\mathfrak{p})$ is merely a subfield of $\kappa(\mathfrak{P})$.
%
%\begin{proof}
%The map $R_v/\mathfrak{P} \rightarrow H \cup \{0\}$ that carries $0$ to $0$
%and $x \bmod \mathfrak{P}$ to $v(x)$ for $x \not\in \mathfrak{P}$ is well-defined:
%if $x \not\in \mathfrak{P}$ and $y \in \mathfrak{P}$ then $v(x) \in H$ and $v(y) < h$ for all $h \in H$, so
%$v(y) < v(x)$ and hence $v(x+y) = v(x)$.   It is clear that this is multiplicative and extends
%to a valuation on the fraction field $\kappa(\mathfrak{P})$ with value group inside $H$.
%The elements of $R_v/\mathfrak{P}$ are carried into $H_{\le 1}$ and those with value 1 are precisely the image of $R_v^{\times}$.
%Hence, this valuation on $\kappa(\mathfrak{P})$ has $R_v/\mathfrak{P}$ as its valuation ring, so
%it identifies the value group with a subgroup of $H$.
%
%To prove that all elements of $H \subset \Gamma_v$ arises as values, we can restrict attention to $h \le 1$.
%Writing $h = v(a)/v(b)$ for $a, b \in A$ with $b \ne 0$, we have $v(a/b) \le 1$, so $a/b \in R_v$.
%The image of this class in $R_v/\mathfrak{P}$ is nonzero and has valuation $h$.
%\end{proof}
%
%Now we can run the process in reverse to give an alternative characterization of precisely the valuations on $A$ that are horizontal
%specializations of $v$ (where $A$ is a domain with $v$ a valuation on its fraction field,
%to avoid a profusion of unnecessary notation).   We use the following general fact:
%for any valuation ring $R$ with fraction field $K$
%and value group $\Gamma_R = K^{\times}/R^{\times}$, its set of prime ideals $\mathfrak{q}$ is in natural inclusion-preserving
%bijection with the set of convex subgroups $\Delta$ of its value group:  to each
%$\mathfrak{q}$ we associate the group $R_{\mathfrak{q}}^{\times}/R^{\times} \subset \Gamma_R$
%and to each convex $\Delta$ we associate the prime ideal of elements $r \in R$ such that $r R^{\times} \in (\Gamma_R)_{\le 1} \Delta$.
%(See \cite[Prop.\,2.14]{Wed} for a short proof.)
%
%For any prime ideal $\mathfrak{P}$ of the valuation ring $R_v \subset K$ and the corresponding
%convex subgroup $H = v((R_v)_{\mathfrak{P}}^{\times})$ the valuation ring $R_v/\mathfrak{P}$
%gives the residue field $\kappa(\mathfrak{P})$ a structure of valued field. 
%For any nonzero $x \in \mathfrak{P}$ we claim that $v(x) < h$ for all $h \in H$.
%Indeed, since $v(x) \le 1$ and $H$ is convex, if $v(x) \ge h$ for some $h \in H$ then
%$v(x) \in H$, so $v(x) = v(u)$ for some $u \in (R_v)_{\mathfrak{P}}^{\times}$.
%Hence, $x = u(x/u)$ with $x/u \in R_v^{\times}$, giving $x \in (R_v)_{\mathfrak{P}}^{\times}$, a contradiction.
%It is obvious that conversely any $x \in R_v$ satisfying $v(x) \not\in H$ must not be a unit in
%$(R_v)_{\mathfrak{P}}^{\times}$, so $x \in \mathfrak{P}$.  In other words,
%$$\mathfrak{P} = \{x \in R_v\,|\, v(x) < h \mbox{ for all } h \in H\}.$$
%
%We thereby get a {\em well-defined} map $$w:\kappa(\mathfrak{P}) \rightarrow H \cup \{0\}$$ by sending the nonzero
%class of any unit $u \in (R_v)_{\mathfrak{P}}^{\times}$ to $v(u) \in H$, and
%$w$ is easily seen to be a valuation relative to which $R_v/\mathfrak{P}$ lies
%in the valuation ring and nonzero elements of $\mathfrak{m}_v/\mathfrak{P}$ are non-units
%(as their reciprocal has valuation in $H_{>1}$).  Hence,
%the inclusion of the valuation ring $R_v/\mathfrak{P}$ into the valuation ring of $w$ (inducing equality of fraction
%fields) is a {\em local} inclusion and thus an equality of rings.
%The contraction $\mathfrak{p} = A \cap \mathfrak{P}$ coincides with the support
%$\{a \in A\,|\,v(a) \not\in H\}$ of $v_{|_H}$, and 
%the composition of $w$ with $A \rightarrow 
%\kappa(\mathfrak{p}) \hookrightarrow \kappa(\mathfrak{P})$ is readily seen to coincide with $v_{|_H}$.
%In view of the convexity of $H$, $v_{|_H}$ is a valuation precisely when
%$H$ contains $v(a)$ whenever $v(a) \ge 1$, so $A \subset (R_v)_{\mathfrak{P}}$. 
%
%We can summarize the above considerations as follows:
%
%\begin{prop}\label{genhor}  Let $v$ be a valuation on a ring $A$, and let $R_v$ be the associated
%valuation ring of $\kappa(\mathfrak{p}_v)$.  The horizontal
%specializations of $v$ correspond bijectively to prime ideals
%$\mathfrak{P}$ of $R_v$ such that $A \subset (R_v)_{\mathfrak{P}}$, in which case
%the corresponding horizontal valuation is the composite map
%$$A \rightarrow (R_v)_{\mathfrak{P}} \twoheadrightarrow \kappa(\mathfrak{P})$$
%to the fraction field $\kappa(\mathfrak{P})$ equipped with the valuation ring $R_v/\mathfrak{P}$.
%\end{prop}
%
%The prime ideals $\mathfrak{P}$ as in Proposition \ref{genhor} have a direct characterization:
%

\begin{thm} \label{thm: horizontal Spec}
Let $v:A \rightarrow \Gamma \cup \{0\}$ be a valuation. The support of a horizontal specialization
of $v$ is a $v$-convex prime ideal and there is a natural bijection
$$
\{ \textup{horizontal specializations of $v$} \} \leftrightarrow \{ v\textup{-convex primes}\},$$
defined by $w \mapsto \mathfrak{p}_w := \textup{supp}(w)$.
Moreover, for horizontal specializations $w$ and $w'$ of $v$ we have that
$w'$ is a specialization of $w$ in $\Spv(A)$ if and only if $\{\mathfrak{p}_{w'}\}$ is a specialization of
$\{\mathfrak{p}_w\}$ in $\Spec(A)$ $($equivalently, $\fp_w \subset \fp_{w'}$$)$.

An inverse map is given explicitly by assigning to any $v$-convex prime $\fq$ in $\Spec(A)$ the valuation 
$$
w_{\fq} : a \mapsto \left\{\begin{array}{cc} v(a) & \textup{if}\ a \notin \fq,\\ 
					0 & \textup{if}\ a \in \fq
					 \end{array}\right.
$$
which is equal to $v|_{H_{\fq}}$ for the minimal convex subgroup $H_{\fq}$ of $\Gamma_v$
containing $v(A - \mathfrak{q})$. 
\end{thm}

\begin{proof} See \cite[Prop. 4.18]{Wed} for a proof of the bijectivity, and see the {\em proof}
of \cite[Lemma 1.2.3]{HuKne} for the explicit description of the inverse (including the recipe for $H_{\fq}$).
\end{proof}

The meaning of Theorem \ref{thm: horizontal Spec} is conveyed by the following picture: 
$$\xymatrix{{\rm{horz}}(v) \ar[r] \ar[d]_-{\simeq} & \overline{\{v\}} \ar[r] \ar[d] & \Spv(A) \ar[d] \\
{\rm{conv}}(v) \ar[r] & \Spec(A/\mathfrak{p}_v) \ar[r] & \Spec(A)}$$
where ${\rm{horz}}(v)$ is the set of horizontal specializations of $v$ and ${\rm{conv}}(v)$ is the set
of $v$-convex prime ideals of $A$.

The following corollary is in the spirit of the going-up theorem (which holds for flat maps of affine schemes),
but concerns the map of topological spaces $\Spv(A)\to\Spec(A)$:

\begin{cor}{\textup{(``going-up'')}} Let $v$ be a valuation on $A$, and $\fp_v$ its support. Let $\fp\in\Spec(A)$ be a point which specializes to $\fp_v$ in $\Spec(A)$. Then there exists a point $w\in\Spv(A)$ such that $\fp_w=\fp$, and $w$ specializes to $v$ in $\Spv(A)$.
\end{cor}

\begin{proof} The proof can be found in \cite[Cor. 4.20]{Wed} (it rests on
Proposition \ref{prop: exchange lemma}(1) below, but we present it here first
purely for expository reasons, since we are not providing the proof of either anyway), and 
is also proved in \cite[Lemma 1.2.6]{HuKne}.
\end{proof}

We will see below that as long as 
$v(A) \not\subset \Gamma_{\le 1}$ (a very typical situation; see Remark \ref{remnever}),
${\rm{conv}}(v)$ is the {\em exact image} of $\overline{\{v\}}$ in $\Spec(A)$
and the horizontal specializations of $v$ are precisely the ``most generic'' points
in the fibers of $\overline{\{v\}}$ over its image in $\Spec(A)$ (see Corollary \ref{gencor});
this is generally false if $v(A) \subset \Gamma_{\le 1}$ (e.g., the trivial valuation in a domain
that is not a field).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
\subsection{Robustness of the two specialization constructions}

Before we state the general result (our main goal today) which expresses the ubiquity of horizontal and vertical specialization to describe
{\em arbitrary} specializations inside $\Spv(A)$, we first discuss how these two operations interact with each other.

\begin{prop}{\textup{(``Exchange Lemma'')}} \label{prop: exchange lemma}
\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(\arabic{enumi})}}
\item If $v \rightsquigarrow w$ is a horizontal specialization and $w \rightsquigarrow w'$ is a vertical specialization, then there exists a vertical specialization $v \rightsquigarrow v'$ that admits $w'$ as a horizontal specialization; that is, 
\begin{displaymath}
\xymatrix{v\ar@{~>}[d]_{\exists}\ar@{~>}[r] & w\ar@{~>}[d]\\
		 v' \ar@{~>}[r]_-{{\rm{such\,\,that}}} & w'}
\end{displaymath} 
\item If $v \rightsquigarrow w$ is a horizontal specialization and $v \rightsquigarrow v'$ is a vertical specialization, then there exists a \emph{unique} horizontal specialization $v' \rightsquigarrow w'$ such that $w'$ is a vertical
specialization of $w$; that is, 
\begin{displaymath}
\xymatrix{v\ar@{~>}[d]\ar@{~>}[r] & w\ar@{~>}^-{\rm{such\,\,that}}[d]\\
		 v' \ar@{~>}[r]_{\exists !} & w'}
\end{displaymath} 
That is, among all horizontal specializations of $v'$, there is exactly one which is a vertical specialization of $w$.
\end{enumerate}
\end{prop}

\begin{proof}
The proof consists of bare-hands constructions with valuations on fields. The reader is referred to \cite[Lemma 4.19 (2), (1)]{Wed} and \cite[Lemma 1.2.5 (ii), (iii)]{HuKne} for proofs. 
\end{proof}

Now comes the main result which provides exhaustiveness of the two specialization constructions:

\begin{thm}\label{main}
Suppose $v \rightsquigarrow w$ is an arbitrary specialization in $\Spv(A)$. Then always
there exists a vertical specialization $v \rightsquigarrow v'$ that admits $w$ as a horizontal specialization:
\begin{displaymath}
\xymatrix{v\ar@{~>}[d]_{\exists}\ar@{~>}[rd] & \\
		v'\ar@{~>}_-{\rm{such\,\,that}}[r] & w}
\end{displaymath}
Also, there exists a horizontal specialization $v \rightsquigarrow w'$ admitting $w$ as a vertical specialization,
which is to say 
\begin{displaymath}
\xymatrix{v\ar@{~>}[r]^{\exists}\ar@{~>}[rd] & w'\ar@{~>}^-{\rm{such\,\,that}}[d]\\
		  & w}
\end{displaymath} 
except for possibly when $v(A) \subset \Gamma_{\le 1}$ and $v(A - \mathfrak{p}_w) = 1$.
\end{thm}

\begin{proof}
We refer the reader to \cite[Prop. 4.21]{Wed}, or \cite[Prop. 1.2.4]{HuKne}.  Note that neither of these
formulates the exceptional case (i.e., $v(A) \subset \Gamma_{\le 1}$ and $v(A - \mathfrak{p}_w) = 1$) in the way that we do,  instead
speaking in terms of triviality of some valuations; it is equivalent to the version above upon unraveling some definitions.
Also, the proof of the second assertion produces $w'$ as a succession of two horizontal specializations
so it is essential to know via Proposition \ref{transprop} that horizontal specialization is a transitive relation.  
\end{proof}


\begin{rem}\label{remnever}
The situation $v(A) \subset \Gamma_{\le 1}$
never happens if $A$ contains a field $k$ with $v\vert_k$ a nontrivial valuation. 
\end{rem}



\begin{cor}\label{allver}
If $v(A) \not\subset \Gamma_{\le 1}$ then the horizontal specialization $v_{|_{c\Gamma_v}}$ has {\em only} vertical specializations. 
\end{cor}

\begin{proof}
%First assume $v(A) \not\subset \Gamma_{\ge 1}$.
Let $w$ be a specialization of $v_{|_{c\Gamma_v}}$.  By the second assertion in Theorem \ref{main}, 
there is a horizontal specialization $w'$ of $v_{|_{c\Gamma_v}}$ such that $w$ is a vertical specialization of $w'$.
But $v \rightsquigarrow v_{|_{c\Gamma_v}} \rightsquigarrow w'$ is a chain of horizontal specializations,
so it is also a horizontal specialization (see Proposition \ref{transprop}).  That is, $w' = v_{|_H}$ for a convex subgroup
$H \subset \Gamma_v$ containing $c\Gamma_v$.  But $w'$ is a specialization of 
$v_{|_{c\Gamma_v}}$, forcing $c\Gamma_v \supset H$, so $H = c\Gamma_v$.  Thus, $w$ is a vertical
specialization of $w' = v_{|_{c\Gamma_v}}$.
%
%Now assume $v(A) \subset \Gamma_{\ge 1}$, so $c\Gamma_v = \{1\}$.
%Thus, $v_{|_{c\Gamma_v}}$ is the trivial valuation on $A$ associated to the prime
%$\mathfrak{q} = \ker(A \rightarrow R_v \rightarrow R_v/\mathfrak{m}_v)$.  Since this
%valuation has trivial value group, it admits no horizontal specialization.
\end{proof}


Intuitively, Corollary \ref{allver} amounts to the fact 
that $v|_{c\Gamma_v}$ is the ``most generic'' horizontal specialization of $v$, so it cannot have 
further horizontal specializations of its own. 
(It is instructive to compute this for $v$ as in Example \ref{eg: power} subject to the hypothesis $v(A) \not\subset \Gamma_{\le 1}$.)  

The conclusion in this corollary is generally false when the hypothesis on $v(A)$ is not satisfied.
For example, consider the trivial valuation $v$ on the fraction field of a domain $A$ that is not a field.
Then $c\Gamma_v = 1$ and $v_{|_{c\Gamma_v}} = v$.  This admits as a vertical specialization
a valuation $v_{\mathfrak{p}}$ dominating the local ring $A_{\mathfrak{p}}$ for any
nonzero prime ideal $\mathfrak{p}$ of $A$.
Since $v_{\mathfrak{p}}(a) \le 1$ for all $a \in A$, we have $c\Gamma_{v_{\mathfrak{p}}} = \{1\}$.
The resulting horizontal specialization $(v_{\mathfrak{p}})_{|_{1}}$ is the trivial valuation associated
to the residue field at $\mathfrak{p}$, so $v$ admits specializations with support equal to any prime ideal of $A$.

\begin{cor}\label{gencor}
If $v(A) \not\subset \Gamma_{\le 1}$ then the image of $\overline{\{v\}}$ in $\Spec(A)$ coincides
with the set ${\rm{conv}}(v)$ of $v$-convex primes
and  the topological fiber of $\overline{\{v\}} \twoheadrightarrow {\rm{conv}}(v)$ over a $v$-convex prime $\fq$
contains a unique generic point, namely the corresponding horizontal specialization.
\end{cor}

\begin{proof}
Since $v(A) \not\subset \Gamma_{\le 1}$, the second option in Theorem \ref{main} applies:
every $w \in \overline{\{v\}}$ is the vertical specialization of a horizontal specialization
$w' = v|_{_H}$ for a convex subgroup $H \subset \Gamma_v$ containing $c\Gamma_v$.
Thus, ${\rm{supp}}(w) = {\rm{supp}}(w')$ is a $v$-convex prime.  Hence, ${\rm{conv}}(v)$ is the image of
$\overline{\{v\}}$ in $\Spec(A)$.

For a $v$-convex prime $\fq$, consider a specialization $w$ of $v$ with support $\fq$.  By the same reasoning,
$w$ is the vertical specialization of a horizontal specialization $w'$ of $v$, so $w'$ has to be the unique horizontal
specialization $v_{\fq}$ of $v$ with support equal to the $v$-convex prime ${\rm{supp}}(w) = \fq$.
Hence, we conclude that all points in the $\fq$-fiber of $\overline{\{v\}}$
are specializations of $v_{\fq}$, so $v_{\fq}$ is the unique generic point of
the topological fiber of $\overline{\{v\}}$ over $\fq$.
\end{proof}

The preceding corollary is generally false if we drop the hypothesis on $v(A)$.  Indeed, if $v$ is the trivial
valuation on a domain $A$ that is not a field then the only $v$-convex prime is $\{0\}$ but we have seen that 
$v$ admits specializations in $\Spv(A)$ with any desired support in $\Spec(A)$.

Despite the robustness of horizontal and vertical specialization for probing the topology of the valuation spectrum, 
when we restrict attention to distinguished subspaces of $\Spv(A)$ (such as an adic spectrum when
$A$ is in a reasonable class of topological rings) then we will need to revisit the above results
in order to study specialization relations with such subspaces since the ``intermediate'' valuations
produced by the Exchange Lemma and Theorem \ref{main} might not lie in the subspace of interest.



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


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\end{document}