Suppose further for the rest of this section that is discrete. Then by Lemma 15.2.8, the ideal is a principal ideal , say, and every is of the form , where and is a unit. We call

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

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

not in canonical form | ||

still not canonical | ||

Basic arithmetic with the -adics in is really weird (even weirder than it was a year ago... There are presumably efficiency advantages to using the formalization, and it's supposed to be better for working with extension fields. But I can't get it to do even the calculation below in a way that is clear.) In PARI (gp) the -adics work as expected:

? a = 1 + 2*3 + 3^2 + O(3^3); ? b = 2 + 2*3 + 3^2 + O(3^3); ? a+b %3 = 2*3 + O(3^3) ? sqrt(1+2*3+O(3^20)) %5 = 1 + 3 + 3^2 + 2*3^4 + 2*3^7 + 3^8 + 3^9 + 2*3^10 + 2*3^12 + 2*3^13 + 2*3^14 + 3^15 + 2*3^17 + 3^18 + 2*3^19 + O(3^20) ? 1/sqrt(1+2*3+O(3^20)) %6 = 1 + 2*3 + 2*3^2 + 2*3^7 + 2*3^10 + 2*3^11 + 2*3^12 + 2*3^13 + 2*3^14 + 3^15 + 2*3^16 + 2*3^17 + 3^18 + 3^19 + O(3^20)

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

- is complete,
- the residue field is finite, and
- the valuation is discrete.

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.

We now deduce what any such measure on must be. Since is compact (Theorem 17.1.4), the measure of is finite. Since is translation invariant,

(disjoint union)

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

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.

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.

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

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:

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

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

William Stein 2004-05-06