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 , and we set . It is called the valuation spectrum of . Then what topology should we consider on ? This is a fundamental question. Recall what topology we pose on . The open sets are generated by sets of the form . If for a prime ideal , there is a valuation on such that , then means that . With this, perhaps we can try to give the topology on , it is generated by subsets of the form where .

First we examine some examples to get a feeling of this topology.(1)Take , the field of rational numbers. What are the possible valuations on ? Note that any valuation on gives a valuation ring . Yet, any valuation ring contains , thus any of its multiples, that is . We can show that any ring of containing is of the form where are primes. Yet a valuation ring is a local ring, so this means that for some prime. This gives all the possible valuation rings of , so we get that of which the elements are indexed by the primes( where integers coprime one to another and both are prime to , and is the trivial valuation). Moreover, the basic open sets are of the form . The condition is almost trivial as long as we take . means that . If we write , then . Yet the second set is always a finite set, and can be any finite subset of excluding . So, this means that the topology on is the same as .(2) Next we take , and consider . If has trivial support , then . And , thus or . If , then we can show that this valuation is just . If , then . If , we can consider the pair in place of , and continue the above argument, finally we can get all the valuations including the trivial valuation . So, if has trivial support, then is the usual valuation. If , a prime ideal generated by a prime , then factors by . Yet is a finite group, thus the only valuation on is the trivial valuation, which induces a valuation on , we denote this valuation as (more specifically, for any , if , then , otherwise it is ). So, all in all, we get that . Now consider the topology on . The basic open subsets are of the form . First if , then it is the whole set. If , then , writing , then . If , then write , and means that . Note that any open subset containing contains also , this shows that .

Now a little remark about morphism between valuation spectrums. Suppose that , then we have that induced by composition. Note that we also have that . Moreover, we have a map . It is worthwhile pointing out that is not always surjective. If is a valuation ring, then it is easy to see that is surjective. The converse is not true in general, as the above example shows. There are many interesting remarks to make concerning these maps .

It is easy to see that are always continuous. (1)If we take a multiplicative set, and set , then for the canonical map , we have that . Since , this shows that , that is . In fact, we can show that , and is a homeomorphism of onto its image. (2)If we take an ideal of , and set , then for the canonical map , this shows that for any , , and similarly, we can show that is a homeomorphism of to the image .(3) For the map , a basic open subset of is . Then since means that , so we get that , which shows that is continuous.(4) If is a field, then admits the trivial valuation as a generic point. Indeed, any basic open set containing a valuation simply means that , this in turn shows that , and thus , thus this basic open set contains , too. So the trivial valuation is a generic point in . This means that is an irreducible topological space.(5) If is a valuation on , then for the canonical map , we have that . It can be shown that is a homeomorphism, and thus is an irreducible subspace of .

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

Given two valuations , is said to be a **specialization** of if , and is said to be a **generization** of . is a specialization of means that any basic open subset containing contains , too. That is to say, implies that . If we take , then this is: implies that , which shows that . to memorize the relation, it is better to write this as if is a generization of .

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

**Lemma:Let be a field, and two valuations on , then is a specialization of if and only if **

This lemma is easy.

Then it is easy to see that, fixing a valuation on , , we have that

By the same reasoning, we see that

.

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 , and is a convex subgroup of , then we define

to be the composition,

where is the characteristic function of . It is easy to see that is again valuations on , as long as is a convex subgroup. As for , more condition should be posed. Before this, we should introduce an important convex subgroup of :

is the convex subgroup of generated by all the elements such that . It is called the **characteristic group** of .

Then we can state the condition:

** is a valuation on if and only if is a convex subgroup and contains .**

These two conditions are not trivial at first sight. Suppose that is a valuation, then if and , writing , we have that , thus , that is to say, , which implies that is convex; note that for any , this means that either or . Yet if and , we should have , thus . But (it is easy to show that if , then ), so , this is a contradiction, which implies that contains all the elements of the form and thus contains the subgroup . Conversely, if is a convex subgroup and contains , then is a valuation. Indeed, it is clear that . And since is a convex subgroup, if and only if , thus for any . Further more, for any , if , then either , thus , or , thus . If , then either , thus , which implies , or , which is impossible since , and implies that , thus . If , then , thus , showing that . All in all, we showed that .

The above proof is a bit long and not very illuminating. The essential point is that, to ensure that is multiplicative, should be a convex subgroup(more precisely, is a subgroup guarantees that the multiplication of two elements of is still in , and is a convex guarantees that if two elements do not lie in at the same time, then their product is not in , this leads to the facts that in ), and to ensure that satisfies the triangle inequality, should contain .

We call a **horizontal specialization** of and a **horizontal generization **of . It is easy to see that is indeed a specialization of , is indeed a generization of .

So, **vertical generizations of correspond to valuations of the form where is a convex subgroup, and horizontal generizations of correspond to valuations of the form where is a convex subgroup containing the characteristic subgroup.**

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