# Huber space-8

In this post, we will consider the space of all the valuations on a ring. We have seen that this set is, intuitively a little larger than the spectrum of the ring(the set of prime ideals). Still we shall give this space a topology, to mimic the spectrum space.

We fix a ring $A$, and we set $Spv(A)=\{v|v\text{ valuation on } A\}/\approx$. It is called the valuation spectrum of $A$. Then what topology should we consider on $A$? This is a fundamental question. Recall what topology we pose on $Spec(A)$. The open sets are generated by sets of the form $Spec(A)(f)=\{p|p\text{ prime ideal of }A, f\not\in p\}$. If for a prime ideal $p$, there is a valuation $v$ on $A$ such that $supp(v)=p$, then $f\not\in p$ means that $v(f)\neq 0$. With this, perhaps we can try to give the topology on $Spv(A)$, it is generated by subsets of the form $Spv(A)(f/s)=\{v\in Spv(A)|v(f)\leq v(s)\neq 0\}$ where $f,s\in A$.

First we examine some examples to get a feeling of this topology.(1)Take $K=\mathbb{Q}$, the field of rational numbers. What are the possible valuations on $K$? Note that any valuation $v$ on $K$ gives a valuation ring $A(v)$. Yet, any valuation ring contains $1$, thus any of its multiples, that is $\mathbb{Z}\subset A(v)$. We can show that any ring $A$ of $K$ containing $\mathbb{Z}$ is of the form $A=\mathbb{Z}[(1/n_i)_{i\in I}]$ where $p_i\in\mathbb{N}-0$ are primes. Yet a valuation ring is a local ring, so this means that $A=\mathbb{Z}[1/p]$ for some prime. This gives all the possible valuation rings of $K$, so we get that $Spv(\mathbb{Q})=\{v_0,v_2,v_3,v_5,...\}$ of which the elements are indexed by the primes($v_p(p^k a/b)=p^{-k}$ where $a,b$ integers coprime one to another and both are prime to $p$, and $v_0$ is the trivial valuation). Moreover, the basic open sets are of the form $Spv(K)(n/m)=\{v_p|v_p(n)\leq v_p(m)\neq 0\}$. The condition $v_p(m)\neq0$ is almost trivial as long as we take $m\neq 0$. $v_p(n)\leq v_p(m)$ means that $v_p(n/m)\leq 1$. If we write $n/m=\pm\prod_k p_k^{e_k}$, then $Spv(K)(n/m)=\{p_k|e_k\geq0\}=Spec(\mathbb{Z})-\{p_k|e_k<0\}$. Yet the second set $\{p_k|e_k<0\}$ is always a finite set, and can be any finite subset of $Spec(\mathbb{Z})$ excluding $0$. So, this means that the topology on $Spv(K)$ is the same as $Spec(\mathbb{Z})$.(2) Next we take $A=\mathbb{Z}$, and consider $Spv(A)$. If $v\in Spv(A)$ has trivial support $supp(v)=0$, then $v(2)=v(1+1)\leq v(1)=1, ...,v(n)\leq 1$. And $v(1)=v(3-2)\leq \max(v(3),v(2))$, thus $v(2)=1$ or $v(3)=1$. If $v(2)<1$, then we can show that this valuation is just $v_2$. If $v(3)<1$, then $v=v_3$. If $v(2)=v(3)=1$, we can consider the pair $(3,4)$ in place of $(2,3)$, and continue the above argument, finally we can get all the valuations $v_p$ including the trivial valuation $v_0$. So, if $v$ has trivial support, then $v$ is the usual valuation. If $supp(v)=pA$, a prime ideal generated by a prime $p$, then $v$ factors by $v':A/pA=\mathbb{F}_p\rightarrow\Gamma\bigcup\{0\}$. Yet $\mathbb{F}_p^*$ is a finite group, thus the only valuation on $\mathbb{F}_p$ is the trivial valuation, which induces a valuation on $A$, we denote this valuation as $v_{p,0}$(more specifically, for any $a\in A$, if $p|a$, then $v_{p,0}(a)=0$, otherwise it is $1$). So, all in all, we get that $Spv(\mathbb{Z})=\{v_0,v_2,v_{2,0},...\}$. Now consider the topology on $Spv(A)$. The basic open subsets are of the form $Spv(A)(n/m)=\{v\in Spv(A)|v(n)\leq v(m)\neq 0\}$. First if $m=0,n=0$, then it is the whole set. If $m=0,n\neq 0$, then $v(n)=0$, writing $n=\pm\prod_k p_k^{e_k}$, then $Spv(A)(n/0)=\{v_{p_k,0}|e_k\neq 0\}\bigcup\{v_0\}$. If $m\neq 0$, then write $n/m=\prod_k p_k^{e_k}$, and $v(n)\leq v(m)$ means that $Spv(A)(n/m)=\{v_0\}\bigcup\{v_{p_k},v_{p_k,0}|e_k\geq 0\}$. Note that any open subset containing $v_{p,0}$ contains also $v_0,v_p$, this shows that $\overline{\{v_0\}}=Spv(A),\overline{\{v_p\}}=\{v_p,v_{p,0}\}$.

Now a little remark about morphism between valuation spectrums. Suppose that $f:A\rightarrow B$, then we have that $Spv(f):Spv(B)\rightarrow Spv(A)$ induced by composition. Note that we also have that $Spec(f):Spec(B)\rightarrow Spec(A)$. Moreover, we have a map $\pi_A: Spv(A)\rightarrow Spec(A), v\mapsto supp(v)$. It is worthwhile pointing out that $\pi_A$ is not always surjective. If $A$ is a valuation ring, then it is easy to see that $\pi_A$ is surjective. The converse is not true in general, as the above example $Spv(\mathbb{Z}), Spec(\mathbb{Z})$ shows. There are many interesting remarks to make concerning these maps $Spv(f),Spec(f),\pi_A$.

It is easy to see that $Spv(f), Spec(f)$ are always continuous. (1)If we take $S\subset A$ a multiplicative set, and set $B=S^{-1}A$, then for the canonical map $f:A\rightarrow B$, we have that $Spv(f)(v)=v\circ f=v'$. Since $v(1/s)v(s)=v(1)=1$, this shows that $v(s)\neq 0$, that is $supp(v')\bigcap S=\emptyset$. In fact, we can show that $Im(Spv(f))=\{v'\in Spv(A)|supp(v')\bigcap S=\emptyset\}$, and $Spv(f)$ is a homeomorphism of $Spv(B)$ onto its image. (2)If we take $I$ an ideal of $A$, and set $B=A/I$, then for the canonical map $f:A\rightarrow A/I$, this shows that for any $v\in Spv(B)$, $supp(Spv(f)(v))\supset I$, and similarly, we can show that $Spv(f)$ is a homeomorphism of $Spv(B)$ to the image $Im(Spv(f))=\{v\in Spv(A)|supp(v)\supset I\}$.(3) For the map $\pi_A:Spv(A)\rightarrow Spec(A)$, a basic open subset of $Spec(A)$ is $Spec(A)(f)=\{p\in Spec(A)|f\not\in p\}$. Then since $\pi_A(v)=supp(v)\in Spec(A)(f)$ means that $v(f)\neq 0$, so we get that $\pi_A^{-1}(Spec(A)(f))=Spv(A)(0/f)$, which shows that $\pi_A$ is continuous.(4) If $K$ is a field, then $Spv(K)$ admits the trivial valuation $v_0$ as a generic point. Indeed, any basic open set $Spv(K)(f/s)$ containing a valuation $v$ simply means that $v(f)\leq v(s)\neq 0$, this in turn shows that $s\neq 0$, and thus $v_0(f)\leq v_0(s)=1$, thus this basic open set contains $v_0$, too. So the trivial valuation is a generic point in $Spv(K)$. This means that $Spv(K)$ is an irreducible topological space.(5) If $v$ is a valuation on $A$, then for the canonical map $f:A\rightarrow K(v)$, we have that $Im(Spv(f))=\{v'\in Spv(A)|supp(v')=supp(v)\}$. It can be shown that $Spv(f)$ is a homeomorphism, and thus $Im(Spv(f))$ is an irreducible subspace of $Spv(A)$.

We have seen in the above that in $Spv(A)$ some points are more special than other points, just like in $Spec(A)$, that is the generic points. We give definitions concerning this situation:

Given two valuations $v,v'\in Spv(A)$, $v$ is said to be a specialization of $v'$ if $v\in \overline{\{v\}}$, and $v'$ is said to be a generization of $v$. $v$ is a specialization of $v'$ means that any basic open subset $Spv(A)(f/s)$ containing $v$ contains $v'$, too. That is to say, $v(f)\leq v(s)\neq 0$ implies that $v'(f)\leq v'(s)\neq 0$. If we take $f=0$, then this is: $v'(s)=0$ implies that $v(s)=0$, which shows that $supp(v')\subset supp(v)$. to memorize the relation, it is better to write this as $A-supp(v')\supset A-supp(v)$ if $v'$ is a generization of $v$.

Here there are two cases to consider, the first is that $v$ is a specialization of $v'$ with $supp(v)=supp(v')$, in this case we will call $v$ to be a vertical specialization of $v'$(and of course, $v'$ is a vertical specialization of $v$), and the second is that $v$ is a specialization of $v'$ with $supp(v)\supset supp(v')$. To see another characterization of vertical specialization, we need:

Lemma:Let $K$ be a field, and $v,v'$ two valuations on $K$, then $v$ is a specialization of $v'$ if and only if $A(v)\subset A(v')$

This lemma is easy.

Then it is easy to see that, fixing a valuation on $A$, $v:A\rightarrow\Gamma\bigcup\{0\}$, we have that

$\{v'\in Spv(A)|v'\text{ vertical generization of }v\}=\{v'\in Spv(K(v))|v'\text{ generization of }v\}=\{v'\in Spv(K(v))|A(v')\supset A(v)\}=\{H|H\text{ convex subgroup of }\Gamma\}$

By the same reasoning, we see that

$\{v'\in Spv(A)|v'\text{ vertical specialization of }v\}=\{v'\in Spv(K(v))|A(v')\subset A(v)\}=Spv(\kappa(v))$.

From above we see that convex subgroups play an important role in the classification of valuations on a ring. Here we give some valuations induced from some convex subgroups:

Suppose that we have a valuation $v:A\rightarrow\Gamma\bigcup\{0\}$, and $H$ is a convex subgroup of $\Gamma$, then we define

$v/H=A\rightarrow\Gamma\bigcup\{0\}\rightarrow\Gamma/H\bigcup\{0\}$

to be the composition,

$v|_H=v*\chi_H$

where $\chi_H:\Gamma\bigcup\{0\}\rightarrow \{0,1\}$ is the characteristic function of $H$. It is easy to see that $v/H$ is again valuations on $A$, as long as $H$ is a convex subgroup. As for $v|_H$, more condition should be posed. Before this, we should introduce an important convex subgroup of $\Gamma$:

$c\Gamma_v$ is the convex subgroup of $\Gamma$ generated by all the elements $v(a)$ such that $v(a)\geq 1$. It is called the characteristic group of $v$.

Then we can state the condition:

$v|_H$ is a valuation on $A$ if and only if $H$ is a convex subgroup and contains $c\Gamma_v$.

These two conditions are not trivial at first sight. Suppose that $v'=v|_H$ is a valuation, then if $h,h'\leq 1$ and $h h'\in H$, writing $v(a)=h,v(a')=h'$, we have that $v'(a)v'(a')=v'(aa')=v(aa')=hh'\neq 0$, thus $v'(a),v'(a')\in H$, that is to say, $h,h'\in H$, which implies that $H$ is convex; note that $v'(1)\leq\max v'(a),v'(1-a)$ for any $a\in A$, this means that either $v(a)\in H$ or $v(1-a)\in H$. Yet if $v(a)>1$ and $v(a)\not\in H$, we should have $v'(a)=0$, thus $v'(1-a)\geq 1$. But $v(1-a)=\max{v(1),v(a)}=v(a)\not\in H$(it is easy to show that if $v(a)\neq v(b)$, then $v(a+b)=\max{v(a),v(b)}$), so $v'(1-a)=0$, this is a contradiction, which implies that $H$ contains all the elements of the form $v(a)\geq 1$ and thus contains the subgroup $c\Gamma_v$. Conversely, if $H$ is a convex subgroup and contains $c\Gamma_v$, then $v'=v|_H$ is a valuation. Indeed, it is clear that $v'(0)=0,v'(1)=1$. And since $H$ is a convex subgroup, $v(a),v(a')\in H$ if and only if $v(a a')\in H$, thus $v'(a)v'(a')=v'(a a')$ for any $a,a'\in A$. Further more, for any $a,a'\in A$, if $v(a),v(a')\in H$, then either $v(a+a')\in H$, thus $v'(a+a')=v(a+a')\leq \max{v(a),v(a')}=\max{v'(a),v'(a')}$, or $v(a+a')\not\in H$, thus $v'(a+a')=0\leq\max{v'(a),v'(a')}$. If $v(a)\not\in H,v(a')\in H$, then either $v(a), thus $v(a+a')=\max{v(a),v(a')}=v(a')\in H$, which implies  $v'(a+a')=v(a+a')=\max{v(a),v(a')}=\max{v'(a),v'(a')}$, or $v(a)>H$, which is impossible since $H\supset c\Gamma_v$, and $v(a)>H$ implies that $v(a)>1$, thus $v(a)\in c\Gamma_c\subset H$. If $v(a),v(a')\not\in H$, then $v(a),v(a'), thus $v(a+a'), showing that $v'(a+a')=0\leq \max{v'(a),v'(a')}$. All in all, we showed that $v'(a+a')\leq\max{v'(a),v'(a')}$.

The above proof is a bit long and not very illuminating. The essential point is that, to ensure that $v|_H$ is multiplicative, $H$ should be a convex subgroup(more precisely, $H$ is a subgroup guarantees that the multiplication of two elements of $H$ is still in $H$, and $H$ is a convex guarantees that if two elements do not lie in $H$ at the same time, then their product is not in $H$, this leads to the facts that $0h=h$ in $H\bigcup\{0\}$), and to ensure that $v|_H$ satisfies the triangle inequality, $H$ should contain $c\Gamma_v$.

We call $v|_H$ a horizontal specialization of $v$ and $v$horizontal generization of $v|_H$. It is easy to see that $v|_H$ is indeed a specialization of $v$, $v$ is indeed a generization of $v|_H$.

So, vertical generizations of $v$ correspond to valuations of the form $v/H$ where $H$ is a convex subgroup, and horizontal generizations of $v$ correspond to valuations of the form $v|_H$ where $H$ is a convex subgroup containing the characteristic subgroup.

In the next post, we shall see that all generization of $v$ is a composite of these two kind of generizations.