Huber space-9

In this post, we will study more closely the generizations of a valuation.

Before all, we should make an important remark: suppose that $v:A\rightarrow\Gamma\bigcup\{0\}$ is a valuation, $\pi:A\rightarrow K(v)$ is the canonical morphism, $H$ is a convex subgroup of $\Gamma$ and $p\subset A(v)$ the prime ideal $p=\{a\in A(v)|v(a). Then

(1)$c\Gamma_v\subset H$ if and only if $\pi(A)\subset A(v)_p$.

Proof: if $c\Gamma_v\subset H$, then for any $a\in \pi(A)-A(v)$, we have that $1/a\in A(v)$. This shows that $v(a)\geq 1$, thus $v(a)\in c\Gamma_v\subset H$. It is worth pointing out that in the case $c\Gamma_v\subset H$, we have that for any $a\in A$, either $v(a)\in H$, either $v(a)(just consider the two cases: (1)$v(a)\geq1$, then $v(a)\in c\Gamma_v\in H$;(2)$v(a)<1$, then if $v(a)\not\leq H$, say $h\leq v(a)<1$ for some $h\in H$, since $H$ is convex, thus $v(a)\in H$). Thus $v^{-1}(H)=A-\pi^{-1}(p)$, so $\pi(a)\in A(v)_p$. This shows that $\pi(A)\subset A(v)_p$. Conversely, if $\pi(A)\subset A(v)_p$, then for any $v(a)>1$, since $\pi(a)\in A(v)_p$, thus $\pi(a)=a'/a''$ with $a'\in A(v),a''\in A(v)-p$. So we have that $v(a)v(a'')=v(a')$(we use the same letter for the valuations on $A$ and the induced valuation on $K(v)$). And since $v(a'')\geq h$ for some $h\in H$ and $v(a)>1$, so $v(a')>h$. So we have that $a',a''\not\in p$, which implies that $\pi(a)\not\in p$ since $p$ is a prime ideal. So, $v(a)>v(p)$, and use the correspondence of the set of prime ideals of $A(v)$ and the set of convex subgroups of $\Gamma$, we have that $v(a)\in H$, which shows that $c\Gamma_v\subset H$.

Again this proof is not illuminating. One way to understand this proof is to note that, $c\Gamma_v\subset H$ means that $H$ contains all the elements of $\Gamma$ that are $\geq1$. Thus for any $a\in A$, if $v(a)\leq 1$, then $v(a)\in A(v)$ automatically, if $v(a)>1$, then $v(a)\in H$, and use the fact that $A(v)$ is a valuation ring, $1/\pi(a)\in A(v)$, so we must have that $\pi(a)\in(v^{-1}(H))^{-1}A(v)$. Yet this multiplicative set is just the complement of $p$ in $A(v)$ provided that $c\Gamma_v\subset H$.

Note that we have transformed problems on $A$ to problems on $A(v)$, latter of which is a subring of $K(v)=Frac(A/supp(v))$, this is because we want to use the correspondence of prime ideals of a ring and the convex subgroups of a totally ordered group, and this fact is established only for valuation rings. That is why we do this in the above.

(2)We assume that $c\Gamma_v\subset H$. Then it is easy to see that $supp(v|_H)=\pi^{-1}(pA(v)_p)$. This is just the definition. Moreover, $supp(v)\subset supp(v|_H)$. And it is clear that $supp(v)=supp(v|_H)$ if and only if $H=\Gamma$, if and only if $p=0$.

(3)We assume that $c\Gamma_v\subset H$. Then consider the local ring $A(v)_p$. Its residue field is $A(v)_p/p$, with a valuation ring $A(v)/p$. Note that we can not extend directly the valuation $v$ to $A(v)_p/p$, we have to kill all the elements $v(p)$. This reminds us to use $v|_H$ to do this extension. Indeed, $v|_H:A(v)_p/p\rightarrow H\bigcup\{0\}$ defines well a valuation. And it is easy to see that $A(v|_H)=A(v)/p$. So, here we get an explicit description of $v|_H$, horizontal generizations of $v$.

In the following, we shall prove a series of proposition to get the result that any generization of $v$ is the composite of vertical generization and horizontal generization(similar results for specializations), as promised at the end of the precedent post.

The first proposition below gives a description of horizontal specializations of $v$. Firstly, we need a concept:

Definition: if $v:A\rightarrow\Gamma\bigcup\{0\}$ is a valuation, a subset $T\subset A$ is call $v$-convex if for any $v(t)\leq v(a)\leq v(t')$ in $A$ with $t,t\in T$ implies that $a\in T$.

Now another description of horizontal specializations of a given valuation:

Suppose $v:A\rightarrow\Gamma\bigcup\{0\}$ is a valuation, and $C_v$ is the set of $v$-convex prime ideals, equipped with an order of inclusion, and $S_h$ is the set of horizontal specializations of $v$, equipped with an order of being specialization($v'\leq v''$ if $v'$ is a specialization of $v''$), then $C_v$ and $S_h$ are totally ordered sets and the map $F:S_h\rightarrow C_v,v\mapsto supp(v)$ is an order-preserving bijection.

Proof: (1)Note that for a $v$-convex subset $T$ of $A$ and an element $a\in A$, either $v(a), or $v(a)>v(T)$, or $a\in T$. Moreover, if $0\in T$, then $v(a)\leq v(t)$ with $t\in T$, we have $s\in T$. So, for two $v$-convex prime ideal $p,p'$ of $A$, if $a\in p-p'$, then we should have that $v(a)>v(p')$. This means that $p'\subset p$. So, any two elements in $C_v$ are comparable, thus $C_v$ is a totally ordered set. For the present we shall not show directly that $S_h$ is a totally ordered set, we will show that $F$ is an order-preserving bijection, which implies that $S_h$ is totally ordered.

(2)It is easy to show that $F$ is well-defined. Indeed, for any $v|_H$ horizontal specialization of $v$, $supp(v|_H)=Ker(A\rightarrow A(v)_p/p)$, thus showing that $supp(v|_H)$ is indeed a $v$-convex prime ideal. Moreover, if $H\subset H'$ two convex subgroups of $\Gamma$, then $v|_{H'}$ is a horizontal specialization of $v|_H$, that is $v|_{H'}\leq v|_H$. It is also easy to see that $Ker(A\rightarrow A(v)_p/p)\supset Ker(A\rightarrow A(v)_{p'}/p')$ where $p,p'$ are the corresponding prime ideals for $H,H'$. Thus $F$ preserves the order.

(3)It remains to see that $F$ is a bijection. It suffices to construct the inverse of $F$. For any $p\in C_v$, we define $H=\{h\in \Gamma|h>v(p)\}$. Then $H$ is a convex subgroup. Indeed, if $v(a),v(a')\in H$, that is $v(a),v(a')>v(p)$, then $a,a'\not\in p$, thus $a a'\not\in p$, that is $v(a a')>v(p)$. Moreover, note that $1\not\in p$, that is $1\in H$. So, if $v(a)\in H$, and if $v(a)^{-1}\in v(p)$, that is $v(a k)=1$ for some $k\in p$, then $a k\in p$, thus $1\in v(p)$, contradiction, showing that $H$ is indeed a subgroup. In addition, if $h in $\Gamma$ with $h,h'\in H$, then $s\not\in v(p)$, otherwise $h\in v(p)$. So, this shows that $H$ is a convex subset. Now we can define $G:C_v\rightarrow S_h,p\mapsto H$, and it is not difficult to show that $F,G$ are inverse one to the other.

The next proposition is very important for later use:

Suppose $w$ is a horizontal specialization of $v$, that is $w=v|_H$ for some $H$, then(1) for any $v'$ a vertical specialization of $v$($v'=v/H'$), there exists a unique vertical specialization $w'$ of $w$($w'=w/L'$) such that $w'$ is a horizontal specialization of $v'$($w'=v'|_L$);(2) if $w'$ is a vertical specialization of $w$($w'=w/L$), then there exists a vertical specialization $v'$ of $v$($v'=v/H'$) such that $v'$ is a horizontal specialization of $v'$($w'=v'|_{L'}$).

In other words, for (1), if $w=v|_H$ and $v'=v/H'$, then there exists a unique $w'$ such that $w'=w/L$ and $w'=v'|_{L'}$; for (2), if $w'=(v|_H)/L$, then $w'=(v/H')|_{L'}$.

The essential point for this proposition is a simple commutative diagram, for $\Gamma\rightarrow H,\Gamma\rightarrow \Gamma/H'$, we have that $H\rightarrow H/(H\bigcap H'),\Gamma/H'\rightarrow H/(H\bigcap H')$. The uniqueness of the first point also follows. And the non-uniqueness in the second point comes from the fact that $w'$ can be a trivial valuation, that is $H\subset H'$ in the commutative diagram. If it is not trivial, then the uniqueness is guaranteed.

A corollary of this proposition is

If $p$ is a prime ideal such that $p\subset supp(v)$, then there exists a horizontal generization $w$ of $v$($v=w|_H$) such that $p=supp(w)$.

Proof: the proof is a bit tricky. Let $\pi:A\rightarrow A/p$ be the canonical projection, $R=(A/p)_{\pi(supp(v))}$, then $R$ is a local ring, a subring of $Frac(A/p)$. Thus there exists a valuation ring $B$ of $Frac(A/p)$ dominating $R$. Suppose then $w$ is the valuation on $B$, thus on $A$, with $supp(w)=p$. Then it is easy to find a convex subgroup $H$ such that $w|_H=v$.

Now we give the main result about specializations of valuations:

Let $v:A\rightarrow\Gamma\bigcup\{0\}$ be a valuation, $w$ be a specialization of $v$, (1)then $w$ is a horizontal specialization of a vertical specialization of $v$;(2)$w$ is also a vertical specialization of $w'$ which is in turn a horizontal specialization of $v$ or $c\Gamma_v=1$ and $w'$ is a trivial valuation whose support contains $supp(v|_1)$.

In other words, if $w$ is a specialization of $v$, then $w=(v/L)|_H$ for some convex subgroup $L\subset \Gamma$ and some convex subgroup $H\subset \Gamma/L$ such that $H\supset c\Gamma_{v/L}$. It is easier to first consider the second point. We shall give only a sketch of the proof. We first show that $supp(w)$ is $v$-convex, and thus according to a precedent result, there exists a horizontal specialization $u$ of $v$ with $supp(u)=supp(w)$. Then it remains to see that $u$ is a specialization of $w$.

So, until now, we finished the part on valuation theory.

In the next several posts, we shall focus on spectral spaces, which are very important for proving some properties of affinoid spaces to be defined later.