# Finite Residue Field Case

Let be a field with a non-archimedean valuation . Recall that the set of with forms a ring , the ring of integers for . The set of with are a group under multiplication, the group of units for . Finally, the set of with is a maximal ideal , so the quotient ring is a field. In this section we consider the case when is a finite field of order a prime power . For example, could be and could be a -adic valuation, or could be a number field and could be the valuation corresponding to a maximal ideal of the ring of integers. Among other things, we will discuss in more depth the topological and measure-theoretic nature of the completion of at .

Suppose further for the rest of this section that is discrete. Then by Lemma 13.2.8, the ideal is a principal ideal , say, and every is of the form , where and is a unit. We call

the ord of  at . (Some authors, including me (!) also call this integer the valuation of  with respect to .) If , then is a unit, and conversely, so is independent of the choice of .

Let and be defined with respect to the completion of at .

Lemma 15.1.1   There is a natural isomorphism

and as an -ideal.

Proof. We may view as the set of equivalence classes of Cauchy sequences in such that for sufficiently large. For any , given such a sequence , there is such that for , we have . In particular, we can choose such that implies that . Let , which is well-defined. The map is surjective because the constant sequences are in . Its kernel is the set of Cauchy sequences whose elements are eventually all in , which is exactly . This proves the first part of the lemma. The second part is true because any element of is a sequence all of whose terms are eventually in , hence all a multiple of (we can set to 0 a finite number of terms of the sequence without changing the equivalence class of the sequence).

Assume for the rest of this section that is complete with respect to .

Lemma 15.1.2   Then ring is precisely the set of infinite sums

 (15.1)

where the run independently through some set of representatives of in .

By (15.1.1) is meant the limit of the Cauchy sequence as .

Proof. There is a uniquely defined such that . Then . Now define by . And so on.

Example 15.1.3   Suppose and is the -adic valuation, for some prime . We can take . The lemma asserts that

Notice that is uncountable since there are choices for each -adic digit''. We can do arithmetic with elements of , which can be thought of backwards'' as numbers in base . For example, with we have

 not in canonical form still not canonical

Here is an example of doing basic arithmetic with -adic numbers in Sage:

Theorem 15.1.4   Under the conditions of the preceding lemma, is compact with respect to the -topology.

Proof. Let , for running through some index set , be some family of open sets that cover . We must show that there is a finite subcover. We suppose not.

Let be a set of representatives for . Then is the union of the finite number of cosets , for . Hence for at lest one the set is not covered by finitely many of the . Then similarly there is an such that is not finitely covered. And so on. Let

Then for some . Since is an open set, for some  (since those are exactly the open balls that form a basis for the topology). This is a contradiction because we constructed so that none of the sets , for each , are not covered by any finite subset of the .

Definition 15.1.5 (Locally compact)   A topological space is locally compact at a point if there is some compact subset of that contains a neighborhood of . The space is locally compact if it is locally compact at each point in .

Corollary 15.1.6   The complete local field is locally compact.

Proof. If , then , and is a compact subset of by Theorem 15.1.4. Also contains the neighborhood of . Thus is locally compact at .

Remark 15.1.7   The converse is also true. If is locally compact with respect to a non-archimedean valuation , then
1. is complete,
2. the residue field is finite, and
3. the valuation is discrete.
For there is a compact neighbourhood of 0. Let be any nonzero with . Then for sufficiently large , so is compact, being closed. Hence is compact. Since is a metric, is sequentially compact, i.e., every fundamental sequence in has a limit, which implies (1). Let (for ) be a set of representatives in of . Then is an open covering of . Thus (2) holds since is compact. Finally, is compact, being a closed subset of . Let be the set of with Then (for ) is an open covering of , so for some , i.e., (3) is true.

If we allow to be archimedean the only further possibilities are and with equivalent to the usual absolute value.

We denote by the commutative topological group whose points are the elements of , whose group law is addition and whose topology is that induced by . General theory tells us that there is an invariant Haar measure defined on and that this measure is unique up to a multiplicative constant.

Definition 15.1.8 (Haar Measure)   A Haar measure on a locally compact topological group is a translation invariant measure such that every open set can be covered by open sets with finite measure.

Lemma 15.1.9   Haar measure of any compact subset of is finite.

Proof. The whole group is open, so there is a covering of by open sets each of which has finite measure. Since is compact, there is a finite subset of the that covers . The measure of is at most the sum of the measures of these finitely many , hence finite.

Remark 15.1.10   Usually one defined Haar measure to be a translation invariant measure such that the measure of compact sets is finite. Because of local compactness, this definition is equivalent to Definition 15.1.8. We take this alternative viewpoint because Haar measure is constructed naturally on the topological groups we will consider by defining the measure on each member of a basis of open sets for the topology.

We now deduce what any such measure on must be. Since is compact (Theorem 15.1.4), the measure of is finite. Since is translation invariant,

is independent of . Further,

(disjoint union)

where (for ) is a set of representatives of . Hence

If we normalize by putting

we have , hence , and in general

Conversely, without the theory of Haar measure, we could define to be the necessarily unique measure on such that that is translation invariant. This would have to be the we just found above.

Everything so far in this section has depended not on the valuation but only on its equivalence class. The above considerations now single out one valuation in the equivalence class as particularly important.

Definition 15.1.11 (Normalized valuation)   Let be a field equipped with a discrete valuation and residue class field with elements. We say that is normalized if

where is the maximal ideal of .

Example 15.1.12   The normalized valuation on the -adic numbers is , where is a rational number whose numerator and denominator are coprime to .

Next suppose . Then the -adic valuation on extends uniquely to one on such that . Since for , this valuation is not normalized. (Note that the ord of is .) The normalized valuation is . Note that , or instead of .

Finally suppose that where has not root mod . Then the residue class field degree is , and the normalized valuation must satisfy .

The following proposition makes clear why this is the best choice of normalization.

Theorem 15.1.13   Suppose further that is complete with respect to the normalized valuation . Then

where is the Haar measure on normalized so that .

Proof. Since is translation invariant, . Write , where is a unit. Then since , we have

Here we have by the discussion before Definition 15.1.11.

We can express the result of the theorem in a more suggestive way. Let with , and let be a Haar measure on (not necessarily normalized as in the theorem). Then we can define a new Haar measure on by putting for . But Haar measure is unique up to a multiplicative constant and so for all measurable sets , where the factor  depends only on . Putting , shows that the theorem implies that  is just , when is the normalized valuation.

Remark 15.1.14   The theory of locally compact topological groups leads to the consideration of the dual (character) group of . It turns out that it is isomorphic to . We do not need this fact for class field theory, so do not prove it here. For a proof and applications see Tate's thesis or Lang's Algebraic Numbers, and for generalizations see Weil's Adeles and Algebraic Groups and Godement's Bourbaki seminars 171 and 176. The determination of the character group of is local class field theory.

The set of nonzero elements of  is a group under multiplication. Multiplication and inverses are continuous with respect to the topology induced on as a subset of , so is a topological group with this topology. We have

where is the group of units of and is the group of -units, i.e., those units with , so

The set is the open ball about 0 of radius , so is open, and because the metric is nonarchimedean is also closed. Likewise, is both open and closed.

The quotient is isomorphic to the additive group of integers with the discrete topology, where the map is

for

The quotient is isomorphic to the multiplicative group of the nonzero elements of the residue class field, where the finite gorup has the discrete topology. Note that is cyclic of order , and Hensel's lemma implies that contains a primitive th root of unity . Thus has the following structure:

(How to apply Hensel's lemma: Let and let be such that generates . Then and . By Hensel's lemma there is a such that and .)

Since is compact and the cosets of cover , we see that is locally compact.

Lemma 15.1.15   The additive Haar measure on , when restricted to gives a measure on that is also invariant under multiplication, so gives a Haar measure on .

Proof. It suffices to show that

for any and . Write . We have

which is an additive translate of , hence has the same measure.

Thus gives a Haar measure on by translating around to cover .

Lemma 15.1.16   The topological spaces and are totally disconnected (the only connected sets are points).

Proof. The proof is the same as that of Proposition 14.2.13. The point is that the non-archimedean triangle inequality forces the complement an open disc to be open, hence any set with at least two distinct elements falls apart'' into a disjoint union of two disjoint open subsets.

Remark 15.1.17   Note that and are locally isomorphic if has characteristic 0. We have the exponential map

defined for all sufficiently small with its inverse

which is defined for all sufficiently close to .

William Stein 2012-09-24