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


to be the composition,


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)<H, 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')<H, thus v(a+a')<H, 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 vhorizontal 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.


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