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e, f, and the remainder theorem *September 12, 2009*

*Posted by Akhil Mathew in algebraic number theory, algebra, number theory, commutative algebra.*

Tags: Dedekind domains, ramification, Chinese remainder theorem

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Tags: Dedekind domains, ramification, Chinese remainder theorem

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So, now to the next topic in introductory algebraic number theory: ramification. This is a measure of how primes “split.” (*No, definitely wrong word there…)*

**e and f**

Fix a Dedekind domain with quotient field ; let be a finite separable extension of , and the integral closure of in . We know that is a Dedekind domain.

(By the way, I’m now assuming that readers have been following the past few posts or so on these topics.)

Given a prime , there is a prime lying above . I hinted at the proof in the previous post, but to save time and avoid too much redundancy I’ll refer interested readers to this post.

Now, we can do a prime factorization of say . The primes contain and consequently lie above . Conversely, any prime of containing must lie above , since if is an ideal in a Dedekind domain contained in a prime ideal , then occurs in the prime factorization of (to see this, localize and work in a DVR).

In particular, only finitely many primes of lie above a given prime of .

Definition 1If lies above , we write for the number of times occurs in the prime factorization of . We call this theramification index.We let be the degree of the field extension . This is called theresidue class degree.

The ramification index has an interpretation in terms of discrete valuations. Let be the absolute value on corresponding to the prime and, by abuse of notation, its extension to corresponding to . Then

This is because if is a uniformizer of (so that generates the cyclic group ), and at , then is a unit with respect to the absolute value , or in the discrete valuation ring .

A basic fact about and is that they are **multiplicative in towers**; that is, if is a finite separable extension, the integral closure in , a prime lying over which lies over , we have:

The assertion about follows from (1), and that about by the multiplicativity of degrees of field extensions. The degree is also multiplicative in towers for the same reason. There is a similarity.

Proposition 2For , we have

Indeed, we may replace with and with , where . Localization preserves integral closure, and the localization of a Dedekind domain is one too (unless it is a field). Finally, and are stable under localization, which follows from the definitions.

In this case, is assumed to be a DVR, hence a PID. Thus is a torsion-free, hence free, finitely generated -module. Since is free over of rank , the rank of over is too. Thus is a vector space over of rank . I claim that this rank is also .

Indeed, let the factorization be . Now we have for , , so taking high powers yields .

By the remainder theorem below, we have

as rings. We need to compute the dimension of each factor as an -vector space. Now we have

and since is principal (see below) all the successive quotients are isomorphic to , which has dimension . So counting dimensions gives the proof.

**The remainder theorem and a consequence **

The remainder theorem below was known for the integers for thousands of years, but its modern form is elegant.

Theorem 3 (Chinese Remainder Theorem)Let be a ring and be ideals with for . Then the homomorphismis surjective with kernel .

First we tackle surjectivity.

If the assertion is trivial. If , then say . We then have where . So, if we fix and want to choose with

the natural choice is .

For higher , it will be sufficient to prove that each vector with all zeros except for one 1 in the product occurs in the image. By symmetry, we need only show that there is such that while for .

Since for ,

i.e. . Now apply the assertion to to find satisfying the conditions.

Now for the kernel assertion: we must show . For two ideals it follows because one chooses as above, and notes that

Then one uses induction and the fact that as above, etc.

As a corollary, we get a criterion for when a Dedekind domain is principal:

Theorem 4A Dedekind domain with finitely many prime ideals is principal.

To do this, we need only show that each prime is principal. Let the (nonzero) primes be ; we show is principal. Choose a uniformizer at , i.e. . Now choose such that

Then and have equal orders at all primes, hence are equal.

We tacitly used this theorem above: there has only finitely many prime ideals, which are the localizations of , so is principal.

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