From tchow@math.lsa.umich.edu Thu Nov 28 00:45:14 EST 1996
Article: 163724 of sci.math
Path: news.lsa.umich.edu!artscare.math.lsa.umich.edu!tchow
From: tchow@math.lsa.umich.edu (Timothy Chow)
Newsgroups: sci.math
Subject: Class field theory (repost)
Date: 28 Nov 1996 05:42:49 GMT
Organization: UMich, but the views expressed below are mine not theirs.
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A little over a year ago I posted an article giving the absolute minimum
bare-bones description of class field theory and got a number of favorable
responses. I also got some particularly helpful email from Don Kersey and
Keith Conrad which filled in some gaps in my understanding. Therefore, I
have decided to repost the article, with a couple of corrections and
additions. For those who saw the original version, the major addition
is a rough explanation of why the Artin reciprocity law is called a
reciprocity law, since it's not obvious at first what is reciprocal
to what.
======================== Revised repost follows ==========================
Some years ago, I took a course where the fundamentals of class field theory
were presented, but it went way over my head. Over the years I've gone back
to my course notes from time to time, and as a result have started to get a
very rudimentary idea of what the subject is about. In this article I try
to explain what I understand of it (which isn't much), both because such an
article is something that I wish I had seen when I was taking the course for
the first time, and because I hope to get some number theorists to correct
my errors and improve my understanding.
Prerequisites include familiarity with the elementary concepts of algebra,
e.g., group, ring, and field. We will also be using the term "finite
algebraic extension" a lot; if F is a subfield of K, then K is a finite
algebraic extension of F if K is finite-dimensional when considered as a
vector space over F.
Number theory may be thought of as the study of the integers, but somewhat
more generally it may be thought of as the study of *algebraic* integers.
An algebraic integer is any number that is the root of some polynomial
equation with integer coefficients whose leading coefficient is 1. The
definition seems artificial, but it turns out to have many important
properties. For example, the set of all algebraic integers forms a ring
(with the usual addition and multiplication). Moreover, if we fix a
finite algebraic extension F of the rationals, the set of all algebraic
integers lying in F forms a subring of F called the "ring of integers" of F.
Rings of algebraic integers have many properties analogous to those of the
usual integers. For example, one can define the notion of a prime number
in an exactly analogous way. The big catch is that in general, one does
not have a unique factorization theorem: elements in a ring of integers may
have several essentially distinct factorizations into primes.
There is, however, a way of fixing up the situation to some extent. Roughly
speaking, we consider certain *subsets* of the ring---the "ideals"---which
we can multiply together in a natural way (more precisely, the product of
the ideal A with the ideal B is the set of all elements of the form
a_1 * b_1 + ... + a_n * b_n with all a_i in A and all b_i in B). Once we
have a multiplication, it makes sense to talk about factoring ideals into
irreducible or "prime" ideals. The amazing fact is that in any ring of
integers, every ideal can be uniquely factored into prime ideals. (The
major architects of this theory were Kummer, who was partially motivated
by trying to fix a certain false proof of Fermat's Last Theorem whose
key false step assumes unique factorization in the naive sense, and
Dedekind.)
Some rings of integers are "principal ideal domains," i.e., every ideal
just consists of the set of all multiples of x for some element x in the
ring. This is the case for the ordinary integers: for each integer n,
the set of all multiples of n is an ideal, and these are all the ideals.
In principal ideal domains, the unique factorization theorem for ideals
implies that all *numbers* in the ring have a unique factorization into
*prime numbers* in the ring. In an arbitrary ring of integers, however,
there may exist ideals that are not principal.
To get a handle on rings of integers where not every ideal is principal,
we would like to have some measure of how far off the ring of integers
is from being a principal ideal domain. If you've had a fair amount of
algebra you can probably guess that the way to do this is to "mod out"
the set of all ideals by the set of principal ideals. This is almost
right, except that it turns out to be better to introduce extra ideals
called "fractional ideals" in order to make the set of all ideals a
*group* under multiplication. Modding out the group of all fractional
ideals by the group of all principal ideals gives an entity called the
"ideal class group" or just the "class group" of the ring of integers
(or of the finite algebraic extension).
It is a fundamental theorem of algebraic number theory that for any
finite algebraic extension of the rationals, the class group is always
a *finite* group. This is proved using geometric (!) methods. The order
of the class group is called the *class number* of the algebraic extension.
Two more pieces of background are needed before we go on to class field
theory. Suppose F and K are finite algebraic extensions of the rationals,
and suppose F sits inside K. Let their respective rings of integers be
R_F and R_K. It turns out that there is a natural way to associate an
ideal of R_K with each ideal in R_F (the ideal in R_K is said to be
"generated" by the ideal in R_F). A natural question to ask is whether
the prime factorization of an ideal I of R_F necessarily "lifts" to a
prime factorization of the ideal of R_K generated by I. Unfortunately,
things are not that simple. Even if one takes a prime ideal P of R_F,
the ideal of R_K generated by P might not be prime; for example, it might
be a power of a prime (in which case P is said to "ramify" in K) or it
might be the product of distinct primes (in which case P is said to
"split" in K). If no primes ramify, K is said to be an unramified
extension of F. (Here I'm lying slightly, because I really should
include "infinite" or "archimedean" primes when I talk about unramified
extensions, but that would be too much of a digression.)
The other piece of background is Galois theory. If F and K are as above,
then it turns out that the set of all automorphisms of K that leave every
element of F fixed is an object of fundamental interest. This set of all
automorphisms actually forms a group under the operation of composition.
In certain special cases, where K is a "normal extension" of F (I won't
define this here), this group is called the "Galois group of K over F,"
written Gal(K/F). Why the Galois group is important is unfortunately
beyond the scope of this article; let's just take it on faith for now.
O.K., let's take stock. We have two kinds of groups floating around:
class groups and Galois groups. It is therefore a natural question to
ask if there is any relationship between them. Now, since they are
defined in such different ways, it is not at all obvious that there is
any relationship between them at all. In a nutshell, class field theory
is the amazing fact that there is actually a very close relationship
between the two: certain class groups are actually equal to certain
Galois groups.
Let's make this a little more precise. First of all, Galois groups
need not be abelian, whereas class groups are always abelian (since
multiplication of ideals is defined in terms of ordinary multiplication
of numbers, which is commutative). So if F is a finite algebraic
extension of the rationals, the only extensions K of F for which Gal(K/F)
could possibly be equal to the class group of F are those for which
Gal(K/F) is abelian (the "abelian extensions of F"). One of the basic
theorems of class field theory is the claim that given any F, there
exists a unique maximal unramified abelian extension K of F (the
"Hilbert class field of F") and that for this extension K, the class
group of F is isomorphic to Gal(K/F). Moreover, the isomorphism is
given explicitly by the so-called Artin map, which associates a
"Frobenius element" to each prime ideal in R_F (details omitted).
Class field theory also tells us about extensions that are not unramified.
Given any finite set S of prime ideals in R_F, there exists a unique maximal
abelian extension of F in which only prime ideals in S ramify. These
extensions are called "ray class fields," and their Galois groups are
isomorphic to certain modified class groups of F. Again, the isomorphism
is given explicitly by the Artin map. (I'm being vague here because the
exact definitions are rather technical.)
That the Artin map is an isomorphism is the famous "Artin reciprocity
theorem." (It is usually stated for arbitrary abelian extensions of F
rather than just the Hilbert class field and the ray class fields, but
these maximal class fields embody the essence of the theorem.)
The basic point of this is that by looking only at "internal" invariants
of F such as the class group of F, one gets a great deal of information
about (one might even say a classification of) the maximal abelian
extensions of F (and Galois theory tells you about the intermediate
abelian extensions), which are "external" to F.
The main limitation of class field theory is that it gives information
only about abelian extensions of F. The search for an analogue of class
field theory in the nonabelian case is one of the basic motivations of
the Langlands program, about which I know even less than I know about
class field theory, so I'll shut up at this point.
--
Tim Chow tycchow@math.mit.edu
Where a calculator on the ENIAC is equipped with 18,000 vacuum tubes and weighs
30 tons, computers in the future may have only 1,000 vacuum tubes and weigh
only 1 1/2 tons. ---Popular Mechanics, March 1949
========================= Additional comments ==============================
Don Kersey emailed me to point out that I neglected to mention the
fundamental example of this theory:
>I think you should mention the canonical example here - abelian extensions
>of the rationals are all cyclotomic, and the ray class field corresponding
>to the ideal generated by a rational integer m is just the field obtained
>by adjoining the mth roots of unity to Q.
This is probably a good place for me to insert a couple of historical
remarks. That every abelian extension of the rationals can be embedded
in a cyclotomic extension (i.e., an extension obtained by adding roots
of unity) is known as the Kronecker-Weber theorem, which was proved in
the 19th century. The problem of understanding abelian extensions was
the twelfth in the list of Hilbert's famous 23 problems. The Hilbert
class field case was proved by Furtwangler (1907), the ray class field
case by Takagi (1920) and the description of the isomorphism as the
explicit Artin map by Artin (1927).
Keith Conrad, among others, has remarked to me that even though there
now exist several different proofs of the theorems of class field
theory, none of them fully "explains" the theorems in a psychologically
satisfactory sense. I would venture to say that even quadratic
reciprocity, despite its 150-odd published proofs, remains quite
an astonishing fact even after you know how to prove it.
Speaking of quadratic reciprocity, let me now try to explain in very
rough terms why the Artin reciprocity law is called a reciprocity law,
and in what sense it generalizes quadratic reciprocity. I am indebted
to the article "What is a reciprocity law?" by B. F. Wyman, Amer. Math.
Monthly 79 (1972), 571-586 for many helpful insights.
A reciprocity law is a theorem that is (approximately) of the form,
"the behavior of X modulo Y is determined by the behavior of Y modulo X."
For example, quadratic reciprocity tells us that if p and q are odd primes,
then whether q is a square modulo p is (more or less) determined by whether
p is a square modulo q.
To see why the Artin reciprocity law is a reciprocity law that generalizes
quadratic reciprocity, let us begin by restating quadratic reciprocity as
the question of how the polynomial x^2 - q behaves modulo p. This allows
us to formulate the following more general question: take a finite algebraic
extension F of the rationals, let f(x) be an arbitrary polynomial with
coefficients in the ring of integers R_F of F, let P be a nonzero prime
ideal in R_F, and ask about the behavior of f(x) modulo P. I now claim that
the Artin reciprocity law tells us that "the behavior of f(x) modulo P"
is essentially determined by "the behavior of P modulo f(x)."
If this claim is correct then it certainly shows that Artin reciprocity
generalizes quadratic reciprocity and that Artin reciprocity deserves the
name "reciprocity." But there are two questions we must answer. First,
what does "the behavior of P modulo f(x)" mean? Second, how does the
Artin reciprocity law as we've stated it (as an isomorphism of groups)
imply the above relationship between "f(x) modulo P" and "P modulo f(x)"?
The answer to the first question is that by "P modulo f(x)" I just
mean "the residue class of P in a certain modified class group that is
determined by f(x)." This is not standard terminology, but I justify
it by suggesting that the locution "P modulo something" ought to be
reminiscent of class groups. In the case of quadratic reciprocity,
f(x) = x^2 - q and the relevant class group is just Z/qZ.
Note that this partially answers the second question, too. Artin
reciprocity gives a relation between Galois groups and modified class
groups, and I've just said that "P modulo f(x)" has to do with class
groups. It remains to demonstrate that "the behavior of f(x) modulo P"
has to do with Galois groups. I won't go into full details here,
referring you instead to the abovementioned paper by Wyman, but I will
at least state the relevant fact. It turns out the factorization of
f(x) in R_F/P is (almost always) the same as the prime factorization
of the ideal in R_K generated by P, and the latter is intimately related
to the way the Galois group Gal(K/F) acts on prime ideals. (This fact
is not trivial, but its proof is considerably easier than the proofs of
class field theory.)
The one annoying detail in this picture is the assumption that Gal(K/F)
is abelian. Can we get rid of it? We can still ask vaguely if there
is a relationship between "f(x) modulo P" and "P modulo f(x)" when f(x)
is arbitrary. Conjecturally, the answer is yes, provided we interpret
"f(x) modulo P" and "P modulo f(x)" appropriately---namely, each half
of the equation can be represented by a certain kind of "L-function"
and the two L-functions are conjecturally equal. At this point I've
reached the limits of my competence so I'll stop.
--
Tim Chow tchow@umich.edu
Where a calculator on the ENIAC is equipped with 18,000 vacuum tubes and weighs
30 tons, computers in the future may have only 1,000 vacuum tubes and weigh
only 1 1/2 tons. ---Popular Mechanics, March 1949