In this post, we will study more closely the generizations of a valuation.
Before all, we should make an important remark: suppose that is a valuation, is the canonical morphism, is a convex subgroup of and the prime ideal . Then
(1) if and only if .
Proof: if , then for any , we have that . This shows that , thus . It is worth pointing out that in the case , we have that for any , either , either (just consider the two cases: (1), then ;(2), then if , say for some , since is convex, thus ). Thus , so . This shows that . Conversely, if , then for any , since , thus with . So we have that (we use the same letter for the valuations on and the induced valuation on ). And since for some and , so . So we have that , which implies that since is a prime ideal. So, , and use the correspondence of the set of prime ideals of and the set of convex subgroups of , we have that , which shows that .
Again this proof is not illuminating. One way to understand this proof is to note that, means that contains all the elements of that are . Thus for any , if , then automatically, if , then , and use the fact that is a valuation ring, , so we must have that . Yet this multiplicative set is just the complement of in provided that .
Note that we have transformed problems on to problems on , latter of which is a subring of , 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 . Then it is easy to see that . This is just the definition. Moreover, . And it is clear that if and only if , if and only if .
(3)We assume that . Then consider the local ring . Its residue field is , with a valuation ring . Note that we can not extend directly the valuation to , we have to kill all the elements . This reminds us to use to do this extension. Indeed, defines well a valuation. And it is easy to see that . So, here we get an explicit description of , horizontal generizations of .
In the following, we shall prove a series of proposition to get the result that any generization of 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 . Firstly, we need a concept:
Definition: if is a valuation, a subset is call -convex if for any in with implies that .
Now another description of horizontal specializations of a given valuation:
Suppose is a valuation, and is the set of -convex prime ideals, equipped with an order of inclusion, and is the set of horizontal specializations of , equipped with an order of being specialization( if is a specialization of ), then and are totally ordered sets and the map is an order-preserving bijection.
Proof: (1)Note that for a -convex subset of and an element , either , or , or . Moreover, if , then with , we have . So, for two -convex prime ideal of , if , then we should have that . This means that . So, any two elements in are comparable, thus is a totally ordered set. For the present we shall not show directly that is a totally ordered set, we will show that is an order-preserving bijection, which implies that is totally ordered.
(2)It is easy to show that is well-defined. Indeed, for any horizontal specialization of , , thus showing that is indeed a -convex prime ideal. Moreover, if two convex subgroups of , then is a horizontal specialization of , that is . It is also easy to see that where are the corresponding prime ideals for . Thus preserves the order.
(3)It remains to see that is a bijection. It suffices to construct the inverse of . For any , we define . Then is a convex subgroup. Indeed, if , that is , then , thus , that is . Moreover, note that , that is . So, if , and if , that is for some , then , thus , contradiction, showing that is indeed a subgroup. In addition, if in with , then , otherwise . So, this shows that is a convex subset. Now we can define , and it is not difficult to show that are inverse one to the other.
The next proposition is very important for later use:
Suppose is a horizontal specialization of , that is for some , then(1) for any a vertical specialization of (), there exists a unique vertical specialization of () such that is a horizontal specialization of ();(2) if is a vertical specialization of (), then there exists a vertical specialization of () such that is a horizontal specialization of ().
In other words, for (1), if and , then there exists a unique such that and ; for (2), if , then .
The essential point for this proposition is a simple commutative diagram, for , we have that . The uniqueness of the first point also follows. And the non-uniqueness in the second point comes from the fact that can be a trivial valuation, that is in the commutative diagram. If it is not trivial, then the uniqueness is guaranteed.
A corollary of this proposition is
If is a prime ideal such that , then there exists a horizontal generization of () such that .
Proof: the proof is a bit tricky. Let be the canonical projection, , then is a local ring, a subring of . Thus there exists a valuation ring of dominating . Suppose then is the valuation on , thus on , with . Then it is easy to find a convex subgroup such that .
Now we give the main result about specializations of valuations:
Let be a valuation, be a specialization of , (1)then is a horizontal specialization of a vertical specialization of ;(2) is also a vertical specialization of which is in turn a horizontal specialization of or and is a trivial valuation whose support contains .
In other words, if is a specialization of , then for some convex subgroup and some convex subgroup such that . It is easier to first consider the second point. We shall give only a sketch of the proof. We first show that is -convex, and thus according to a precedent result, there exists a horizontal specialization of with . Then it remains to see that is a specialization of .
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.