In this post, we will discuss valuations.
We suppose that is a ring and is a totally ordered group, then a valuation is a map such that (1) for any ;(2) for any ; (3).
Note that this definition excludes immediately all the Euclidean metrics. A little remark on the morphism between : we say that is a morphism if , preserves the orders, and for any .
We should have heard valuation rings in commutative algebra. For example, is a field, and is a valuation, then we say that is the valuation ring of (associated to this valuation), and this valuation ring is a local ring, with the maximal ideal . Note that, for any , either or lies in . This is the most important property of a valuation ring. In fact, we can use this as the definition for valuation rings, that is:
A subring of a field is a valuation ring if and only if for any non-zero element , either or lies in .
Proof: If satisfies the condition above, then in fact we can define a totally ordered group using : we consider the group , and we say that if where . The condition above implies that this order is a total order, and we define the valuation as the projection . It is easy to show that the valuation ring of associated to this valuation is exactly , thus finishing the proof.
There is an important notion concerning valuations of a ring. For a valuation , , it is called the support of this valuation. This name is perhaps a little confusing at first, since it reminds us the support of a function, for example , then is just the closure of the set of the points such that their image is not zero. Still, we will use this name for a valuation as defined above.
This notion give some immediate consequences. In fact, we have that is a prime ideal of . It is rather easy. Furthermore, this valuation factors by . And now is an integral domain, and thus we can extend this valuation to its fraction field by for .
And then we can consider the valuation ring of this valuation on the field . To simplify notions, we will write , and is the valuation ring of associated with , is the maximal ideal of , and is the residue field of , that is .
A last notion is an equivalence relation on the set of valuations on a fixed ring. We say two valuations , then are called equivalent if there is an isomorphism between and . Equivalently, we can say that for any two , if and only if . Or and .
Using this last characterization, we get a bijection between and the set where is the equivalence relation defined above. If we have that is a valuation field with valuation and its valuation ring, then there is also a bijection between and , defined by sending to where is the canonical projection.
This last bijection is at first sight a little mysterious. For any valuation of contained in , it is clear that is again a valuation ring in , so defines a valuation . Now this valuation can be pulled back to , and then to , defining a valuation such that . Note that the valuation ring of is just according to the definition of , and the valuation ring of is which obviously contains . We have to show that . Suppose that , then since , this means that, according to the equivalent definition of valuation rings, . This shows that is a unit, and since , thus , this implies that , thus finishing the proof.
Note that a valuation on a field always has trivial support, that is or . In the last bijection, the essential assumption is that is a valuation ring.
In the above, we considered valuation rings contained in a valuation , next we will consider the valuation rings containing the valuation ring , that is(in the following the always assume that for a valuation , is surjective, this can always be possible since we can just take ):
Let be a valuation field with valuation , and its valuation group. (1)If we set , and , and we define two maps , and ;(2) Let be the set of prime ideals of , then we define another two maps, and where is the maximal ideal of . Then define two bijections among these sets.
Proof: (1) First we should show that is well-defined. For any , if and , then we should show that . But this is automatic since . First we show that for any . Indeed, for any , either , thus , thus , or , which means that , this shows that , and thus again , so this means that :Conversely, for any , this means that . Then either , this shows that , or , which implies that , and thus , so finally we get that . Then we show that for any . But this is rather straightforward.(2)These two maps are indeed well-defined. First we show that for any . But , the localization of with respect to , and the maximal ideal of . Then we have that . Yet, in general, for any and any , then , thus , so , this shows that . So, is a prime ideal of , so ;Conversely, we should show that . Note that for any , we have that , since is a local ring, thus , so , and thus . On the other hand, for any , if , so , so , thus , finishing the whole proof.
So, in this way, we get a one-to-one correspondence between the set of convex subgroups of and the set of valuation rings of containing . More specifically, for any a convex subgroup, we send it to the maximal ideal of the valuation ring , that is the set which can be shown to be a prime ideal, and conversely, for any prime ideal of , we send it to the image , that is the greatest convex subgroup of such that . Note that this subgroup indeed exists.
This last description clarifies the relation between prime ideals and convex subgroups.
In the next post, we will talk about the all these two types of valuations together.