# Global Fields

Definition 18.1.1 (Global Field)   A global field is a number field or a finite separable extension of , where is a finite field, and is transcendental over .

In this chapter, we will focus attention on number fields, and leave the function field case to the reader.

The following lemma essentially says that the denominator of an element of a global field is only nontrivial'' at a finite number of valuations.

Lemma 18.1.2   Let be a nonzero element of a global field . Then there are only finitely many inequivalent valuations of for which

Proof. If or then the lemma follows by Ostrowski's classification of all the valuations on  (see Theorem 13.3.2). For example, when , with , then the valuations where we could have are the archimedean one, or the -adic valuations for which .

Suppose now that is a finite extension of , so satisfies a monic polynomial

for some and . If is a non-archimedean valuation on , we have

Dividing each side by , we have that

so in all cases we have

 (18.1)

We know the lemma for  , so there are only finitely many valuations  on  such that the right hand side of (18.1.1) is bigger than . Since each valuation of  has finitely many extensions to , and there are only finitely many archimedean valuations, it follows that there are only finitely many valuations on  such that .

Any valuation on a global field is either archimedean, or discrete non-archimedean with finite residue class field, since this is true of and and is a property preserved by extending a valuation to a finite extension of the base field. Hence it makes sense to talk of normalized valuations. Recall that the normalized -adic valuation on is , and if  is a valuation on a number field  equivalent to an extension of , then the normalization of is the composite of the sequence of maps

where is the completion of at .

Example 18.1.3   Let , and let . Because , there is exactly one extension of to , and it sends to

Thus the normalized valuation of is .

There are two extensions of to , since , as . The image of under each embedding into is a unit in , so the normalized valuation of is, in both cases, equal to . More generally, for any valuation of of characteristic an odd prime , the normalized valuation of is .

Since in two ways, there are exactly two normalized archimedean valuations on , and both of their values on equal . Notice that the product of the absolute values of with respect to all normalized valuations is

This product formula'' holds in much more generality, as we will now see.

Theorem 18.1.4 (Product Formula)   Let be a nonzero element of a global field . Let run through the normalized valuations of . Then for almost all , and

We will later give a more conceptual proof of this using Haar measure (see Remark 18.3.9).

Proof. By Lemma 18.1.2, we have for almost all . Likewise, for almost all , so for almost all .

Let run through all normalized valuations of (or of ), and write if the restriction of to is equivalent to . Then by Theorem 17.2.2,

so it suffices to prove the theorem for .

By multiplicativity of valuations, if the theorem is true for and then it is true for the product and quotient (when ). The theorem is clearly true for , which has valuation at all valuations. Thus to prove the theorem for it suffices to prove it when is a prime number. Then we have , , and for primes that . Thus

as claimed.

If is a valuation on a field , recall that we let denote the completion of with respect to . Also when is non-archimedean, let

be the ring of integers of the completion.

Definition 18.1.5 (Almost All)   We say a condition holds for almost all elements of a set if it holds for all but finitely many elements.

We will use the following lemma later (see Lemma 18.3.3) to prove that formation of the adeles of a global field is compatible with base change.

Lemma 18.1.6   Let be a basis for , where is a finite separable extension of the global field of degree . Then for almost all normalized non-archimedean valuations on we have

 (18.2)

where are the extensions of to . Here we have identified with its canonical image in , and the direct sum on the left is the sum taken inside the tensor product (so directness means that the intersections are trivial).

Proof. The proof proceeds in two steps. First we deduce easily from Lemma 18.1.2 that for almost all the left hand side of (18.1.2) is contained in the right hand side. Then we use a trick involving discriminants to show the opposite inclusion for all but finitely many primes.

Since for all , the left hand side of (18.1.2) is contained in the right hand side if for and . Thus by Lemma 18.1.2, for all but finitely many  the left hand side of (18.1.2) is contained in the right hand side. We have just eliminated the finitely many primes corresponding to denominators'' of some , and now only consider  such that for all .

For any elements , consider the discriminant

where the trace is induced from the trace. Since each is in each , for , the traces lie in , so

Also note that since each is in . Now suppose that

with . Then by properties of determinants for any with , we have

 (18.3)

The left hand side of (18.1.3) is in , so the right hand side is well, i.e.,

(for )

where . Since are a basis for over and the trace pairing is nondegenerate, we have , so by Theorem 18.1.4 we have for all but finitely many . Then for all but finitely many  we have that . For these , that implies since , i.e., is in the left hand side of (18.1.2).

Example 18.1.7   Let and . Let and . In the first stage of the above proof we would eliminate because is not integral at . The discriminant is

As explained in the second part of the proof, as long as , we have equality of the left and right hand sides in (18.1.2).

William Stein 2012-09-24