
Citation 
 Permanent Link:
 http://digital.auraria.edu/AA00003138/00001
Material Information
 Title:
 Abelian extensions of Q
 Creator:
 Elliot, Jason W
 Publication Date:
 2005
 Language:
 English
 Physical Description:
 xi, 77 leaves : ; 28 cm
Thesis/Dissertation Information
 Degree:
 Master's ( Master of Science)
 Degree Grantor:
 University of Colorado Denver
 Degree Divisions:
 Department of Mathematical and Statistical Sciences, CU Denver
 Degree Disciplines:
 Applied Mathematics
 Committee Chair:
 Payne, Stan
 Committee Members:
 Cherowitzo, Bill
Lundgren, Rich
Subjects
 Subjects / Keywords:
 Functions, Abelian ( lcsh )
Dedekind sums ( lcsh ) Galois theory ( lcsh ) Dedekind sums ( fast ) Functions, Abelian ( fast ) Galois theory ( fast )
 Genre:
 bibliography ( marcgt )
theses ( marcgt ) nonfiction ( marcgt )
Notes
 Bibliography:
 Includes bibliographical references (leaves 7677).
 General Note:
 Department of Mathematical and Statistical Sciences
 Statement of Responsibility:
 by Jason W. Elliot.
Record Information
 Source Institution:
 University of Colorado Denver
 Holding Location:
 Auraria Library
 Rights Management:
 All applicable rights reserved by the source institution and holding location.
 Resource Identifier:
 62768584 ( OCLC )
ocm62768584
 Classification:
 LD1193.L622 2005m E54 ( lcc )

Full Text 
ABELIAN EXTENSIONS OF Q
by
Jason W. Elliot
Bachelor of Science, University of Colorado at Denver, 2002
A thesis submitted to the
University of Colorado at Denver
in partial fulfillment
of the requirements for the degree of
Master of Science
Applied Mathematics
2005
This thesis for the Master of Science
degree by
Jason W. Elliot
has been approved
by
3 C? 1 $00.5^
Date
Elliot, Jason W. (M.S., Applied Mathematics)
Abelian Extensions of Q
Thesis directed by Professor Stan Payne
ABSTRACT
Our main aim in this paper is to prove the theorem of Kronecker and Weber,
which states the following:
If K is a finite abelian extension of Q then K is contained in a
cyclotomic extension of Q.
In our proof of this theorem we avoid the theory of localization, which is used
in all proofs that this author has read.
Our proof requires theory of integral extensions of Dedekind domains, which
we develop to some extent here. Some familiarity of this subject is assumed of
the reader. Likewise, we develop some theory of algebraic numbers in Galois
extensions.
In Chapter 3, we state many results from modern algebra that find an
application in our investigations. This is meant as a reminder, or might serve
as reference. The reader is assumed to have an aquaintence with basic modern
algebra, including Galois theory. We state some relevant theorems from group
theory and theory of Cyclotomic fields in Chapters 4 and 5, respectively.
This abstract accurately represents the content of the candidates thesis. I
recommend its publication.
DEDICATION
I dedicate this thesis to Jennifer Hiromi Nakasone, my fiance, soon to be my
wife.
ACKNOWLEDGMENT
I would like to thank Bill Cherowitzo, Rich Lundgren, and Stan Payne, for
teaching me most of the mathematics upon which this thesis is based and for
encouraging me to write this thesis.
CONTENTS
Figures ................................................................ x
Tables............................................................... xi
Chapter
1. Introduction......................................................... 1
2. Conventions.......................................................... 2
3. Rings and Fields ............................................. 3
3.1 Quotient Fields.................................................... 3
3.2 Ideals ............................................................ 4
3.3 Modules............................................................ 6
3.4 Primes and Maximal Ideals ......................................... 7
3.5 Integral Extensions................................................ 8
3.6 Residue Class Fields.............................................. 10
3.6.1 Residue Class Vector Space...................................... 10
3.7 Fractional Ideals ................................................ 10
3.8 Invertible Ideals................................................. 11
3.9 Field Extensions ................................................. 13
3.10 Number Fields and Number Rings ................................... 15
4. Some Theorems on Abelian Groups..................................... 17
4.0.1 pgroups......................................................... 17
4.1 Abelian Field Extensions......................................... 19
5. Cyclotomic Extensions............................................... 21
5.1 Roots of Unity ................................................... 22
5.2 Cyclotomic Field Extensions....................................... 23
6. Trace, Norm, and Discriminant .................................... 25
6.1 The Discriminant.................................................. 26
6.2 Quadratic Extensions of Q......................................... 27
6.3 Cyclotomic Extensions of Q........................................ 28
7. Dedekind Domains.................................................... 29
7.1 Noethers Classification.......................................... 30
7.2 Ideals ........................................................... 33
7.3 Ramification Index................................................ 35
7.4 Residue Class Fields and Inertial Degree.......................... 36
7.4.1 Norm for Ideals................................................. 38
7.5 Cyclotomic Extensions of Q........................................ 39
8. The Different ...................................................... 40
8.1 Dual Basis....................................................... 41
9. Galois Extensions................................................... 43
9.1 Subgroups of the Galois Group .................................... 44
9.1.1 The Decomposition Group................... 45
9.1.2 The Inertia Group............................................... 46
9.1.3 The Ramification Groups ........................................ 48
9.2 The Frobenius Automorphism in the Classical Case.................. 50
9.3 Embedding Â£/Vi (t)/p)x ........................................... 51
9.3.1 A Better Result for Abelian Extensions in the Classical Case ... 53
9.4 Embedding Vm_1/'Vm S)/^p........................................... 56
9.5 Hilberts Formula for Totally Ramified Primes in the Classical Case 58
10. Abelian Extensions and the KroneckerWeber Theorem............... 60
10.1 Reduction to Prime Power Degree................................. 60
10.2 Reduction to One (Particular) Ramified Prime.................... 61
10.3 The Case for 2 .................................................... 64
10.4 The Case When p is Odd and m = 1................................ 67
10.4.1 The Case p is odd and m = 2................................... 69
10.4.2 Back to m = 1 ................................................ 70
10.5 The Case p is odd and m > 1..................................... 71
Appendix
A. Notation............................................................ 72
B. The Minkowski Bound................................................. 74
References .
76
FIGURES
Figure
4.1 The prime power decomposition of abelian extensions............ 20
10.1 The behavior of q ^ p.......................................... 62
x
TABLES
Table
9.1 Decomposition and Inertia Fields: Ramification and Inertial Degrees 48
1. Introduction
In 1853, Kronecker stated the theorem now known as the KroneckerWeber
Theorem, but his proof was incomplete there were difficulties with extensions
of degree a power of 2. Weber gave a proof in 1886, but even his proof contained
a gap. Both of these early proofs used theory of Lagrange resolvents. In 1896,
Hilbert gave a correct proof using analysis of ramification groups. [20]
Most modern proofs of the KroneckerWeber Theorem use local techniques,
theory of padic numbers, and theory of valuations. We avoid all of these subjects
entirely, and, following exercises in [10], we give a proof that, according to Milne
[12], is likely similar to Hilberts.
According to the theorem, finite abelian extensions of Q are, in a sense,
generated by the function e2mX at rational values of X. Kroneckers hope was
that finite abelian extensions of an arbitrary number field could be generated
in the same sense by a set of analytic functions at algebraic values. It turns
out that the abelian extensions of imaginary quadratic fields are generated by
certain values of elliptic functions. [14] The area that studies abelian extensions
of general number fields is known as class field theory. Kroneckers hope has
inspired much research in this area. Natural^, the first questions in class field
theory concern how to describe the abelian extensions of (Q>. Our investigation
therefore lies at the beginning of class field theory.
1
2. Conventions
As usual, N, Z, Q, R, C respectively denote nonnegative integers, integers,
rationals, reals, and complex numbers.
The following few paragraphs are an attempt to prepare the reader, but
they are nothing necessary. In this thesis, we make systematic use of fonts to
help remind the reader what symbols represent.
We reserve German letters for modules. Of course, rings and ideals are
modules, and we use the German letters for them, too. Fields are denoted with
letters such as k, F, K, L, M, N, T. A lower case letter in ordinary font indicates
a rational integer: p,q,r,e, f,g, etc. Groups are denoted with letters such as
T,Â£,S,1K,V. For polynomials we use script except in the case of the
minimal polynomial Ma k(W). Normal text font is generally reserved for certain
special objects, such as the trace Tr, norm N, discriminant Disc, different Diff,
and minimal polynomial M. These are all listed in Appendix A. Lowercase
Greek letters are generally used for elements of some algebraic structure more
general than Z, for example a group, ring, or field. Special symbols are listed
in the first appendix under the title Notation. We reserve the slash / for
quotients and use the vertical bar  for extensions.
2
3. Rings and Fields
In general, a ring need not have commutative multiplication nor have mul
tiplicative identity. In this thesis, however, we assume tacitly that all rings are
commutative with nonzero identity. In addition, we assume that any ring ho
momorphism preserves the identity. In particular, if one ring contains another,
then their identities coincide. We refer to a ring having no zero divisors as a
domain, and a ring with the unique factorization property as factorial.
3.1 Quotient Fields
Proposition 3.1 If r is a domain, then r can be embedded into a field.
Proof: We consider ordered pairs (a, (5) Â£ r x r such that (3 0, under the
relation
(a,0)~(7,5) a5 = 07.
This defines an equivalence relation. We use (a : 0) to indicate the equivalence
class containing (a,0). We define
(a : 0) (7 : 5) = (a/? : 75)
and
(a : 0) + (7 : 5) = (aS + /Ty : 05)
It is not hard to show that these definitions are natural, and that under these
two operations our set is a field. By identifying a e r with (a : 1) Â£ k, we see
3
that t is contained in this field. For more details, we refer the reader to [18, p.
41].
It is not hard to see that any field containing t must contain a field isomor
phic to the field constructed in the proof of Proposition 3.1. In this sense, this
field is the smallest field containing t, and we refer to it as the quotient field
of r.
Proposition 3.2 For each a Â£ k there is nonzero p Â£ r such that pa Â£ t.
Proof: Assume a = (an : 02) with 011,02 Â£ t, 02 0. Then 020 = (0201 :
o2) = (01 : 1) Â£ r.
An application of this result is the following:
Corollary 3.3 If D is a domain with quotient field k, then given a polynomial
over k there is a polynomial over 0 of the same degree with precisely the same
roots.
3.2 Ideals
An ideal is a subring that is stable under the action of multiplication. If a
and b are two ideals, then their product ab is defined to be the smallest ideal
containing all products o/3 with o Â£ a and (3 Â£ b. Equivalently, this is the set
of all finite sums Oj/^ such that ax Â£ a, Â£ b, and k Â£ N. Commutativity
for ideals is inherited from commutativity in the ring.
Let 3 be an ideal in r. If there is a set S such that every element of 3 can
be expressed as a finite sum Oip% with az Â£ S and px Â£ r, then 3 is said to
be generated by S. In this case we write 3 = (S). If S' is any set of elements
4
from the ring r, then there is a smallest ideal containing S, which is necessarily
generated by S. If the cardinality of S is finite, then (S) is said to be finitely
generated. If r is a subring of fH and S' is a set of elements of r, then the notation
(S) can be ambiguous, and we use the notation (S)x and (S)
ideal generated by S in r, and the ideal generated by S in fH, respectively.
A principal ideal is an ideal generated by a single element. In a commu
tative ring with identity, a principal ideal is the set of multiples of its generator.
That is, (a) = va. A ring in which all ideals are principal is itself called prin
cipal.
Following the categorical meaning, an ideal a is said to divide another ideal
b provided that there is an ideal t) such that ba = b. It should be clear that if a
and b are two ideals with a dividing b, then a D b. We shall see later when the
converse is true.
Assume that a and b are ideals of the domain 0. As with rational integers,
we say that ax is an exact divisor of b if axb, but ax+1 { b. Of course, if
ax+1 \ b, then neither does ax+2, etc. As with rational integers, we express this
as axJ b.
If a and b are two ideals in the same ring t, then we define a + b to be the
set of sums a + (3 with a a and 0 Â£ b.
Proposition 3.4 The ideal a + b is the smallest ideal containing a and b. In
symbols, a + b = (a, b).
Proof: If 3 is an ideal containing a and b, then 3 must contain sums a + 0
with a E a and 0 G b. Moreover, a + b clearly contains a and b.
5
3.3 Modules
If r is a ring, then an xmodule is an abelian group 971 together with an
action on it by r, that is, a ring homomorphism into the ring End97T of endo
morphisms of 971. We refer the reader to [5, p. 169] for more details. If 971 is an
rmodule, then we say 971 is a module over x.
If r is a subring of the ring 91, then 91 is automatically an rmodule; hence,
the quotient field of r is an rmodule. If m is a subring of r, we see that m is an
rmodule iff rm = m. Thus, we notice that ideals are precisely modules that are
subrings.
The rank of a module is the maximum number of linearly independent
elements and a basis is a maximal set of linearly independent elements. An
rmodule is said to be free if it is isomorphic to 0tg/ r for some set I. The free
module 0tg/ r has rank ffl, and consists of a basis of generators with the same
cardinality as I. See [3, pp. 439, 448], [9, p. 135], or [5, p. 181].
If k is a field, then a kmodule is a kvector space or a vector space over
k, and its rank is referred to as its dimension. For a kvector space y, the
collection of all klinear maps from V to k is itself a kvector space, which is
denoted by V' and called the dual space of y. If y is finite dimensional, then
dimlC = dim^'. See [3, p. 412] or [9, p. 142],
Proposition 3.5 Let x be a domain, k its quotient field. If DJI is a free xmodule
of rank n, then there is a kvector space y of dimension n containing DJI.
Proof: Consider the kspan of 971. See [15, pp. 107, 108] for the details.
Theorem 3.6 Let x be a domain and suppose that DJI is a free xmodule of rank
6
n. Then any two bases of VJl have n elements.
Proof: See [15, pp. 107, 108].
A module is said to be torsion free provided that there are no torsion
elements, that is, if pp = 0 then either p = 0 or p = 0.
Proposition 3.7 A finitely generated module over a principal domain is torsion
free iff it is free.
Proof: See [3, p. 442].
3.4 Primes and Maximal Ideals
An ideal is maximal iff it is not properly contained in any proper ideal.
Equivalently, the factor ring is a field; see [9, p. 93].
An ideal ip is a prime ideal provided that whenever the product af3 is in
ip either a G ip or (3 iP; equivalently, 93/ip is a domain. We shall refer to a
prime ideal of r as a prime of r. When primes in Z enter into our discussions,
we refer to them as rational primes to distinguish them. Prime ideals in Z
are precisely those generated by rational primes, and it is often easier to refer to
a rational prime than the prime ideal that it generates. Since fields are domains,
Proposition 3.8 every maximal ideal is prime.
Proof: For a direct proof, see [9, p. 92].
Suppose 93 is a ring with subring r. Let p be a prime ideal of t. If ip is a
prime ideal of 93 such that ip fl r = p, then we say ip lies above p or p lies
below ip. We write ipp to refer to this relationship. When this is the case, the
injection
induces an injection
t/p 9ty
of the factor rings, and gives the commutative diagram
r  r/p
iK > iH/qj
with all maps being the obvious ones. See also [8, p. 8].
3.5 Integral Extensions
Good treatments of this subject are given in [8, p. 4], [9, p. 333], and [11,
p. 64],
If 21 is a ring contained in a ring 23, then an element (3 G 23 is said to be
integral over 21 if (3 is a root of a monic polynomial with coefficients in 21. In
other words, there is a relation of the form f3n + an_1/3n_1 f + a0 = 0 with
OLi 21. A relation such as this makes the 21module 2t[/5] finitely generated.
Since the powers of (3 form a basis of 2l[/3] over 21, the converse is also true:
The element (3 G IB is integral over 21 iff 2l[/3] is finitely generated
over 21.
If every element of 23 is integral over 21 then we say that 23 is integral over 21.
The property of integrality is transitive:
Proposition 3.9 Assume 21 C 23 C Â£ is a tower of rings. If Â£ is integral over
23 and 23 is integral over 21, then . is integral over 21.
Proof: See [8, p. 5].
Also, integrality is invariant under homomorphism:
8
Proposition 3.10 If 91 is a ring that is integral over a subring r, and a is a
homomorphism of 91, then cr91 is integral over ax.
Proof: Apply a to any integral equation over x and it will be an integral
equation over ax. See [8, p. 5].
Consequently, if 91 is integral over x, and ip is a prime of 91 lying above the
prime p of r, then 91/fp is integral over r/p.
If x is a subring of 91, then the set of all elements of 91 that are integral over r
is a subring of 91, and is called the integral closure of r in 91. If every element
in 91 that is integral over r is actually in r, then r is said to be integrally closed
in 91, or in some literature normal in 91. If x is integrally closed in its field of
fractions then we merely say that r is integrally closed.
Proposition 3.11 Assume x is integrally closed m its quotient field k, and let
E  k be a separable extension of degree n. Let 91 be the integral closure of x m E.
Then there exist free xmodules 971 and 9Tt1 of rank n such that 971' C 91 C 971.
Proof: See [15, p. 108].
Consequently, in the setting of Proposition 3.11, 91 is an rmodule of rank
n, generated by n elements. Now, any field containing 91 must contain 971' and
k. Using the construction in Proposition 3.5, dimension considerations (over k)
yield that the quotient field of 91 is E.
Proposition 3.12 Assume x is a ring and 91 is a ring integral overx. Let ip be
a prime ideal of 91 lying above the prime p of x. Then ip is maximal if and only
if p is maximal.
9
Proof: See [8, p. 10].
3.6 Residue Class Fields
Assume that 0 is a domain and that D is a domain integral over D. If fp is
maximal, then of course fp is prime. Assume further that *p lies above the prime
p of D. Then, by Proposition 3.12, p is maximal, and both 2)/fp and D/p are
fields. In this case, the restriction of the (canonical) epimorphism 3) > 3D/*P to
the subring D induces an homomorphism D > 3D/fp whose kernel is 0 D
assumption, this intersection is p, so that we have an embedding D/p 5)/*p.
Thus, the field 3)/ip may be regarded as an extension of D/p.
3.6.1 Residue Class Vector Space
The natural map D > X>/(p)x> induces a map D 3}/(p)j) having kernel
(p)xi n D = p. Hence, there is an embedding D/p c> 5)/(p)j). Up to isomorphism,
therefore, we may regard D/p as a subring of 3D/(p)x As such, /(p)s is a
module over D/p, and since D/p is a field, this module is a vector space. A set
S of elements from 3D is said to be independent modulo p (or mod p) if the
set of residue classes of the elements of S is linearly independent in this vector
space.
3.7 Fractional Ideals
Much of this section and the next follows [21, V, Â§6].
If D is a domain, then we consider a submodule f of the quotient field of D a
fractional ideal provided that there is 7 6 D such that yf C D. In this case, it
is easy to verify that yf is an ideal of D, and hence fractional ideals are precisely
Dmodules of the form y^d where 3 is an ideal of D and y 6 D. In this case,
10
notice that if f is a fractional ideal of 0, then fO = 71jR) = 713 = f. With
7 = 1, we see that ideals are fractional ideals.
It will sometimes be simpler to refer to fractional ideals simply as ideals. To
distinguish ideals from fractional ideals, we refer to them as integral ideals.
We extend the definition of divide to fractional ideals. An ideal f is said
to divide another ideal t provided that there is an integral ideal o such that
fa = t
We also notice that finitely generated submodules of the quotient field are
fractional ideals. If {/?,} are generators of the module Oil, then find {/r^} C 0
such that Hifii G 0. Then /q) 371 C 0.
For a pair of ideals b, b' we adopt the following notation:
(b : b') = {a 6 k : ab' C b}.
It is easily verified that (b : b') is a 0module. In fact,
Proposition 3.13 (b : b') is fractional ideal.
Proof: Assume 7b C 0 and 7V C 0. Then, if /i G 77'b', then /r(b : b') C 7'0 C
0. Moreover, 77'b' C 70 C 0.
3.8 Invertible Ideals
A fractional ideal f is said to be invertible provided that there is a fractional
ideal f1 such that ff1 = 0.
Lemma 3.14 Iff has an inverse, then it is unique. Particularly,
f = (0 : f)
11
Proof: If ff_1 = 0, then f1 C (D : f). On the other hand, f(h : f) C ?), so if f_1
is an inverse of f, then (D : f) = f1f(h : f) C f_1D = f1.
We note that there is a dependence of the inverse on the ring 5, which is not
represented in the notation. Usually such a representation will be unnecessary,
as the intended ring will be clear. When we must distinguish, we shall use f1^.
It is easy to see that the inverse operation is contravariant (i.e., inclusion
reversing).
We now prove a series of lemmas that will be useful to us later.
Lemma 3.15 If all nonzero integral ideals are invertible, then the collection of
fractional ideals is a group under multiplication.
Proof: Since multiplication of ideals is associative, and h acts as an identity, it
suffices to show that fractional ideals are invertible provided that integral ideals
are invertible. Indeed, if f is a fractional ideal then writing f = 7_1ff shows us
that f has inverse yj1.
Lemma 3.16 If a finite product of integral ideals is invertible, then the ideals
themselves are invertible.
Proof: If (r1 []i = 0, then a3 (V1 at) = 0.
Lemma 3.17 Invertible ideals are finitely generated.
Proof: If aa"1 = 91, then we may write 1 = with {c^} C a and
{a'i} C a1. For arbitrary p Â£ 91, we have p 'f2,po!iai, and moreover, the
coefficients pa[ Â£ 91. Thus, {;} generates a. [21, p. 272],
12
Lemma 3.18 If a is a product of invertible prime ideals, then the a factorization
of a into prime ideals is unique.
Proof: See [21, p. 272],
3.9 Field Extensions
We assume that the reader has a basic knowledge of the theory of field
extensions, including Galois theory. We review here only a few facts. Many of
the details can be found in [3, Chapter 13], If k is a subfield of E then we say
that E is an extension field of k, and we refer to the extension Ek. In such a
situation, E is a vector space over k, and we write [E : k] for its dimension. If
[E : k] < oo we say Ek is finite. If a Â£ E, then k(n) denotes the smallest subfield
of E that contains k and a, and is called the field generated by a and k.
Assume that Ek is an extension of fields. An element a G E is algebraic
over k iff a is a root of a polynomial in k[X].
If a is algebraic then we may as well take the polynomial a satisfies to be
monic, since we can divide by the leading coefficient. Indeed, every element of E
that is algebraic over k satisfies a monic polynomial, and it is not difficult to see
that there is a unique such polynomial of minimal degree, and that it must be
an irreducible element of k[AT], We denote this polynomial by MQ!k(x) and refer
to it as the minimal polynomial of a over k. When there can be no confusion,
we write MQ(X) for Ma>k(W). An equivalent condition for a to be algebraic over
k is that the extension k(a)k be finite, since
k(a) = k[W]/Ma(W),
which is spanned over k by the powers of a.
13
Proposition 3.19 For ^(X) G k[A] we have a) = 0 ijfMa>k\^ (in k[A]j.
Proof: Apply the division algorithm. For the details, see [3, p. 500].
If a is algebraic over the field k, then the degree of ct is the dimension of
k(ai)k, which is the degree of Maji<. We say that the extension Ek is a algebraic
provided that every element of E is algebraic over k.
Proposition 3.20 Finite extensions are algebraic.
Proof: If [E : k] < oo, then for a G E, [k(a) : k] < oo as k(ct) is a subspace of
E. Thus, a is algebraic over k.
A field extension Ek is said to be simple provided that E = k(a) for some
a G E. As we have seen, a basis for the simple extension k(a)k consists of the
powers of a.
If a G E is algebraic over k, then a is said to be separable over k if its
multiplicity as a root of MQ>k is 1. The extension Ek is called separable if every
a G E is separable over k.
Proposition 3.21 (The Primitive Element Theorem) Finite separable ex
tensions are simple.
Proof: See [5, p. 287].
If all algebraic extensions of a field k are separable, then k is called perfect.
See [5, p. 289].
Proposition 3.22 Q is perfect.
Proof: See [3, p. 531] or [5, p. 287].
14
Corollary 3.23 Thus, all finite extensions of Q are simple.
A field L is algebraically closed if the only algebraic extension of L is
trivial, or equivalently, any element of any extension that is algebraic over L
is in L. The algebraic closure of L is the smallest extension of L that is
algebraically closed. Every field has a unique algebraic closure; we denote the
algebraic closure of L by La/. The algebraic closure of a field is algebraically
closed. For a more thorough development, see [9, p. 231].
A splitting field for a polynomial TP (E k[AT] is an extension of k in which &
splits into linear factors. The extension Ej k is called normal if every embedding
of E into kal induces an automorphism of E. Equivalently, E is the splitting field
of a set of polynomials in k[W]. See [9, p. 237]. The smallest field that is normal
over k and contains E is called the normal closure of E over k. An extension
E]k is called Galois if it is normal, separable, and algebraic. We assume that
the reader is familiar with the Fundamental Theorem of Galois Theory, which
is treated in [3, p. 554], We denote the Galois group of the extension Ek by
Sek or S(Ek), whichever is more convenient.
3.10 Number Fields and Number Rings
An algebraic number is an element of Qa/. Equivalently, an algebraic
number is a complex number that satisfies a monic polynomial over Q. If a
is an algebraic number, then there is a unique irreducible polynomial in Z[xj
having a as a root, which we obtain from Mq q by multiplying by the led of its
coefficients. If a is an algebraic number such that this irreducible polynomial
over Z is monic, then we say that a is an algebraic integer. Put differently,
an algebraic integer is a complex number that is integral over Z. As we noted
15
1
I
i
i
earlier in greater generality, this is equivalent to the Zmodule Z[a] being finitely
generated.
A number field is a finite extension of the rational numbers. Such an
extension is necessarily algebraic, and since Q is perfect, the extension is also
simple. Thus, any number field has the form
Q(a) = {ao + + + an_ictn 1 ao,..., an_i G Q}
for some algebraic number a, which, in this case, has degree n over Q. A number
ring is the ring of algebraic integers in a number field, i.e., the integral closure
of Z in a number field.
Assume that k is a finite extension of Q. We denote the integral closure of
Z in k by Op. If E is a finite separable extension of k, then the integral closure
of Ok in E is the same as the integral closure of Z in E, by Proposition 3.9. In
other words,
Proposition 3.24 the integral closure of Ok in E is Oe
In particular,
Corollary 3.25 if F is a finite extension of Q, then Op is integrally closed.
Assume F is a finite extension of Q. Then by Proposition 3.11, Op is a finitely
generated Zmodule of rank n. Since Op is a domain it is torsion free over Z. Since
Z is a principal domain, by Proposition 3.7, Of is free over Z. By an integral
basis we mean a basis for Of over Z. Notice that any integral basis must be a
basis for F over Q.
16
4. Some Theorems on Abelian Groups
Here we state a few results concerning abelian groups that we will need later
in this thesis.
Theorem 4.1 For any group S, ifK< S, then S/K is abelian iff IK is contained
m the commutator subgroup of S.
Proof: See [16, p. 33].
Lemma 4.2 If S is a finite abelian group whose order is divisible by a prime p,
then S contains an element of order p.
In fact, this theorem is true more generally for any finite group, not necessarily
abelian. We only need the result for abelian groups. For the proof of both
results, see [16, pp. 73, 4].
Theorem 4.3 Every finite subgroup of the multiplicative group of a field is
cyclic.
Theorem 4.4 A cyclic group of order n has a unique subgroup of order d for
every d dividing n.
For the proofs of Theorems 4.3 and 4.4, See [16, p. 28].
4.0.1 pgroups
Assume that p is a rational prime, m > 1 is a rational integer, and S is a
group of order pm.
17
Lemma 4.5 Assume that S is abelian and that TC < S has order ph. Then for
every h! satisfying h < h! < m there is a subgroup TC of S having order ph> and
containing TC.
Proof: It suffices to argue for h! = h + 1. The quotient group S = S/TC is a
pgroup, so there is 7 G S of order p. Let Oi' be the group generated by TC and
7. Then TC TC (because 7 ^ IK). Since yp G K,
K' = KuK7u uK7p~1.
Thus K' has order ph+l. [15, p. 276]
Proposition 4.6 The group S is cyclic if S is abelian with a unique subgroup
of order pm~l.
Proof: Let Ji be the subgroup of order pm~l and let 7 G S \ TC. We show that
7 has order pm. Suppose not and let ph be the order of 7. By Lemma 4.5, (7)
is contained in a subgroup of order pm_1, which is necessarily TC by hypothesis.
Thus 7 G IK, which is a contradiction. [15, p. 276]
Proposition 4.7 A pgroup S is cyclic iff it is abelian and contains exactly one
subgroup of order d for each d dividing
Proof: See [16, p. 29].
Theorem 4.8 Any finite abelian group is isomorphic to a direct product of
groups of prime power order. Particularly, if {pt} are the distinct primes di
viding the order of the abelian group S, and
SPi = {7 G S : the order of 7 is a power of Pi},
18
then S = El Spi
Proof: See [15, p. 47].
Lemma 4.9 If S is a cyclic group of order pm, and S = Uf x X, then either
% = 1 or X = 1.
Proof: This follows from the uniqueness assertion in the Fundamental Theorem
of Finite Abelian Groups. See [16, p. 131].
4.1 Abelian Field Extensions
We now prove a theorem concerning finite abelian field extensions. The case
when the base field is Q will play a crutial role in proving our main theorem.
Theorem 4.10 Every finite abelian extension is the compositum of abelian ex
tensions of prime power degree.
Proof: Assume Ek is a finite abelian extension. Using Theorem 4.8, write
5
sEik
i=1
for some s with TCj = pl and [E : k] = Let Sj = fr each
j = 1,... ,s and let Kj denote the fixed field of Sj. Then 9e[Kj Sj, by the
Fundamental Theorem of Galois Theory. Now,
S S
SeK!Ks < P SeK, = P) 9j = 1,
j=1
since the order of Sj is not divisible by p0, so the intersection is not divisible by
any primes. Thus, E = K; Ks.
19
j
j
I
Figure 4.1: The prime power decomposition of abelian extensions
Theorem 4.11 Assume E and F are extensions of a common field k. Then there
is an embedding of Sefi< into the subgroup of 3Ek x SFk consisting of(a,r) such
that aEnF = TEnF.
Proof: Consider a Sefi< By the containment of the fields E and F in the
compositum EF we may consider the restrictions crE and af of a to E and F,
respectively. Consider the map
O (crE,crF).
The codomian of this map is the set of (a, r) such that aEnF = rEnF Moreover,
if crE = rE and crE = rF, then necessarily a = r, so the map is injective. See
also [15, p. 17].
Corollary 4.12 If Ek and Fk are abelian extensions, then so is EFk.
20
5. Cyclotomic Extensions
Given a complex number Â£ Â£ + iv, the modulus of Â£ is given by Â£ =
Im(Â£) = v. The plane spanned by 1 and i is the complex plane, which we
coordinatize with Re(Â£) and Im(Â£). If 9 represents the geometric angle of Â£ as
a point in the complex plane, then we can write Â£ \C,\el9 We have the famous
identity
Since sin(X) and cos(X) both have period 2ir, we can always take 6 G [0, 27t).
When this is done we call 6 the argument of Â£ and write 9 Arg(Â£). Via this
same identity we also have 1 = el2kn for any k G Z, but the argument of 1 is 0.
Multiplicativity of the modulus tells us that if Â£n = 1 then Â£ = 1. (z" =
\z\n) Thus, Â£ = eld, and n9 = 2kir for some k G Z. For each k, the number
is an nth root of unity. These numbers can be represented in the complex plane
as points on the unit circle.
Over the algebraic closure Qal, any polynomial & has as many roots, count
ing multiplicity, as its degree deg &. If the polynomial is irreducible over some
subfield of Qal, then since algebraic extensions of Q are separable (Proposi
tion 3.22), 3? has precisely as many distinct roots as its degree, that is, all of
the multiplicities are 1.
\JÂ£2 + v'2, the real part of Â£ by Re(Â£) = Â£, and the imaginary part of Â£ by
el9 = cos 9 + zsin#.
21
The polynomial Pn(X) = Xn 1 is not irreducible over Q if n > 1 since
Pn(l) = 0 implies that (X 1) divides Pn(X) is [AT. Degree considerations
show that their quotient is nonconstant when n > 1. However, PrL does have n
distinct roots, as we shall see.
5.1 Roots of Unity
In a field k, an nth root of unity is an element Â£ such that Â£n = 1, that
is, Â£ is a root of the polynomial Xn 1. Denote Xn 1 by Pn(X). If p is
the characteristic of k, then, considering the Frobenius endomorphism, Pp has
only one root, namely 1. If n is not divisible the characteristic of k, then Pn is
separable because the derivative nXn~l is not identically zero. The only root of
nXn~l is 0, which is not a root of Pn. Thus, Pn shares no common root with its
derivative and in kal; hence, all of the roots of Pn are distinct.
Clearly if Â£ and Â£ are nth roots of unity then so is Â£Â£~\ so the nth roots
of unity form a group, which we denote pn. By the Fundamental Theorem of
Algebra, pn has order n. As a multiplicative subgroup of the field kal, pn must
be cyclic. A generator of pn is called a primitive nth root of unity.
Assume that Â£n is a primitive nth root of unity in kal. Since Â£Â£ has or
der n/ gcd(j, n) in the group pn, we have a onetoone correspondence between
primitive nth roots of unity and positive integers < n that are prime to n. Thus
the Euler totient function 4>{n) represents precisely the number of primitive nth
roots of unity. Notice also that if j = i (mod n) then the orders of Q and Â£* are
the same. Thus, the natural association of positive integers < n with elements of
the group Z/nZ preserves group structure and induces a canonical isomorphism
Pn = Z/nZ,
22
where the latter group is considered additive.
The elements of Z/nZ prime to n form a multiplicative group, usually de
noted (Z/nZ)x. They correspond to the primitive nth roots of unity. Thus, the
polynomial
*.k(v) = n o
has as roots precisely the primitive nth roots of unity, and clearly has degree
4>{n). We refer to d>n]i< as the nth cyclotomic polynomial over k.
A particularly interesting case is when k = Q. When this is so we simply
write for d^.
Proposition 5.1 d>n(x) Â£ Z[x] for every n.
Proof: See [6, p. 194].
Theorem 5.2 d> is irreducible over Q (hence overZ).
Proof: See [6, p. 195].
Since d>n is monic, d>n is the minimal polynomial of Â£n over Q.
5.2 Cyclotomic Field Extensions
As a corollary to the above discussion, we obtain
Proposition 5.3 [Q(Cn) : Q] =
If (n is a primitive nth root of unity over Q, then (n generates the group of nth
roots of unity, the nth cyclotomic field Q(Cn) contains all nth roots of unity.
Hence, d> splits in Q(Â£n). In fact, Q(Â£) is the splitting field of d>n over Q since
Q((n) = Q[x]/($(x)). The extension Q(C)Q is therefore Galois.
Notice that Q(Cm,Cn) Q Q(Cicm[m,n])j and this clearly extends by induction.
23
Theorem 5.4 We have
S(Q(C)IQ) = (Z/Z)*;
the isomorphism is given by the map a (mod n) i aa, where aa ((n) = Q.
1 Proof: [3, p. 577].
Thus, the extension Q(Cn)Q is abelian. In fact, in many cases, the Galois
group of this extension is cyclic.
Proposition 5.5 The group (X/nX)x is cyclic iff n is of the form 2, 4, pa, or
2pa for an odd prime p and positive integer a.
Proof: See [6, p. 44]
By Proposition 4.4 and the Fundamental Theorem of Galois Theory, we
obtain the following:
Corollary 5.6 If n is of one of the forms mentioned in Proposition 5.5, then
for every d dividing
The case when n = pa will be especially useful to us in proving our main
theorem. 1
1We notice an interesting relationship between the groups and Gal(Q(Â£n)Q).
24
6. Trace, Norm, and Discriminant
If E k is a finite separable extension of fields and Â£ G E, then we define the
trace of Â£ relative to the extension Ek to be
TVeh?=
where lHEk denotes the homomorphisms (necessarily monomorphisms) of E
fixing k pointwise. In the case where Ejk is normal, f>CEk is the Galois group
S E k
We note that the trace fixes k pointwise and is klinear. We extend this
definition to sets S' C E by defining
TrEkS = {WEkX : X Â£ S'}.
If S' is an additive subgroup of E, then Tr^kS1 is an additive subgroup of k.
With the same situation as above, the norm 1 of Â£ relative to the extension
E j k is defined to be
nEii<= n
eMejk
When the base field is Q, we write Â£ for NEqÂ£ and refer to the norm as the
absolute norm. We note that the norm fixes k pointwise and is multiplicative.
See [15, p. 20].
lfThe norm and trace can be defined more generally when the extensions are not necessarily
separable, see [5, p. 289). We only need the definitions given.
25
Proposition 6.1 The norm and trace are transitive, meaning that if k C E C F
is a tower of fields, then
TrEk(TrFE(0) = TrFk(Â£),
and
NEk(NFE(0) = NFk(0
for all Â£ Â£ F.
Proof: See [15, p. 20].
The trace is nondegenerate (in separable extensions), meaning that there is
always an element Â£ Â£ E such that TrEkÂ£ ^ 0. The general proof of this is non
trivial and uses Dedekinds theorem on the independence of kmonomorphisms
of E. See [15, p. 20].
6.1 The Discriminant
The facts in this section are developed more fully in [15, pp. 20, 117]. Given
a set of n elements {Â£*,..., Â£n} in E, we define the discriminant to be
Disc(Â£i, ...,Â£) = Det(TrEk(Â£iÂ£j)).
If (<7j) are the kmonomorphisms of E, then
DiscEk(G, = Det(al(^))2.
Another representation of the discriminant is given as follows. If Â£' =
Â£,n=i aufi, then
DiscEk(Â£i, ,C) = Det(a,J)2DiscEk(^i,..
A corollary to this is the following:
26
Proposition 6.2 //(Â£) and (Â£) are two integral bases for Ek, then their dis
criminants are equal.
Proof: For the details, see [15, p. 116].
We refer to the common discriminant of an integral basis as the discriminant
of E (over k), which we denote by DiscEkOE
Proposition 6.3 The norm, trace, and discriminant lie in the base field.
Proof: They are invariant under automorphisms of E fixing k.
For a (E E, we write DiscEk a to mean
DiscEk(l,o,a2,... ,ofE:k,1).
6.2 Quadratic Extensions of Q.
We now wish to compute the discriminant of any quadratic extension of Q.
It is not hard to see that every quadratic extension of Q is of the form Q(v/a)
for some squarefree rational integer a. See [6, p. 188] for details.
Proposition 6.4 The algebraic integers m Q(\/a) are
Z Lyfa, if a = 2 or 3 (mod 4);
Z Z ~ , if a = 1 (mod 4).
Proof: See [6, p. 189].
From this, it is a straightforward calculation to verify that
4a if a = 2 or 3 (mod 4);
Disc Q(\/a) =
a if a = 1 (mod 4).
For the detais, see [6, p. 189].
27
6.3 Cyclotomic Extensions of Q.
If Â£ is a primitive pth root of unity, then DiscQ(^)Q(oQ(<;)) = For a
proof of this result, see [15, p. 118].
28
7. Dedekind Domains
A Dedekind domain is a domain in which every proper nonzero ideal is
a product of finitely many prime ideals. We shall see that in fact this product
is necessarily unique up to permutations of the factors. We shall refer to this
as the prime decomposition of an ideal. The ideals of a Dedekind domain
provide a natural generalization of rational integers. In fact, there are many
equivalent characterizations of Dedekind domains. See [21, V, Â§6], [5, p. 406],
or [10, p. 59].
We first notice the following:
Proposition 7.1 Principal ideal domains are Dedekind domains.
Proof: In a principal ideal domain, prime ideals are generated by irreducible
elements. Moreover, principal ideal domains are factorial. For these facts, we
refer the reader to [21, p. 243]. The result follows.
Theorem 7.2 In a Dedekind domain, fractional ideals are invertible.
Proof: See [21, p. 274] or [4, p. 38].
By the Lemmas in Section 3.8, fractional ideals of a Dedekind domain are
finitely generated and factor uniquely as a product of prime ideals to integer
powers. See [21, p. 274] for more details.
Corollary 7.3 The fractional ideals under multiplication form a free group gen
erated by the prime ideals.
29
In fact, integral ideals can be written canonically (up to permutation of
factors) as a product of distinct prime ideals to positive powers. See [21, p. 274]
for more details.
Assume that tp is a prime ideal in a Dedekind domain. Then (ipm)=0
is an infinite descending tower of ideals. The setP=0fpm easily satisfies the
properties of ideals. By unique factorization, the sequence never terminates, and
hence this ideal is not expressible as a finite product of primes. Consequently,
(XUV"1 must be the zero ideal (clearly it is not (1)).
7.1 Noethers Classification
Let P be a set partially ordered by the relation <, i.e., < is reflexive and
transitive, and X X = Y.
Proposition 7.4 The following conditions are equivalent:
1. Every increasing sequence Xi < X
such that Xfc XkJr\ .
2. Every nonempty subset of S has a maximal element.
Proof: [2, p. 74]
If o is a domain and DJI is an omodule, then the osubmodules of DJI can be
partially ordered by C In this case, DJI is said to be Noetherian (a Noetherian
omodule) provided that the conditions in Proposition 7.4 hold.
Incidentally, if the submodules are ordered by 3>, then a module satisfy
ing the conditions in Proposition 7.4 is said to be Artinian. As seen above,
Dedekind domains are not Artinian.
30
Theorem 7.5 An omodule 2Jt is Noetherian iff all of its osubmodules are
finitely generated over o.
Proof: [2, p. 75]
The ring o is said to be Noetherian if it is Noetherian as an omodule, i.e.,
if Proposition 7.4 holds for its ideals. Equivalently, by Theorem 7.5, all of its
ideals are finitely generated.
Theorem 7.6 Dedekind domains are Noetherian.
Proof: It follows immediately from Theorem 7.2 and Proposition 3.17 that the
fractional ideals of a Dedekind domain are finitely generated. Thus a Dedekind
domain is Noetherian.
Theorem 7.7 In a Dedekind domain, every nonzero prime ideal is maximal.
Proof: See [15, p. 125].
Incidentally, this means that a Dedekind domain has Krull dimension 1.
Theorem 7.8 A Dedekind domain is integrally closed.
Proof: See [21, p. 275].
We have taken the historical approach in defining Dedekind domains as
Dedekind did, and developing their properties. In fact, we have an alternative
definition, which was discovered by Noether.
Theorem 7.9 (Noethers Classification of Dedekind Domains) A ringQ
is a dedekind domain iff 3D is
1. Noetherian,
31
2. integrally closed, and
3. every nonzero prime ideal is maximal.
Proof: In consideration of our development so far, it suffices to prove suffi
ciency. See [8, pp. 1820].
Proposition 7.10 Let 91 be an integrally closed Noetherian ring, E a finite
separable extension of its quotient field k. Then the integral closure of 01 in E is
a finitely generated 01module.
Proof: See [8, p. 6].
Proposition 7.11 Assume 91 is a principal domain and E is a finite separable
extension of degree n of the quotient field of 91. Then the integral closure of 91
in E is a free Timodule of rank n.
Proof: [9, p. 7]
Proposition 7.12 Assume D is a Dedekind domain and E a finite separable
extension of the quotient field ofi). LetD be the integral closure ofD in E. Then
D is a Dedekind domain.
Proof: It is simplest to use Noethers classification of Dedekind domains. By
construction, T) is integrally closed. Prime ideals are maximal by 3.12. For the
details on why 2) is a Noetherian domain, see [21, p. 281].
Applying Proposition 3.11 to the setting in Proposition 7.12, the quotient
field of D is E. Notice that, in particular, since Z is a Dedekind domain, if E
is a finite separable extension of Q, the number ring 0e is a Dedekind domain.
32
i
i
i
]
j
Indeed, this class of examples is our main motivation for developing the theory
of Dedekind domains in this thesis. We shall refer to this case as the classical
case.
7.2 Ideals
Proposition 7.13 If 3 and 3 are two fractional ideals in a Dedekind domain
D, then 33 iff JD 3.
Proof: See [15, p. 129]
In other words, for ideals, to contain is to divide.
Lemma 7.14 Assume d is a Dedekind domain and 3 is an ideal of 0. Let
{a;}^=0 C Â£) and assume for each i that 3X'\\ (a*). If the are all distinct,
then
jmin{a;i} a^.
Proof: Let x denote min{xj}. Since a* G 3X for each i, we see that Ylai Â£ 3X;
thus (J2 af) C 3X. Assume without loss of generality that x = xq. Then for i > 0
we have Xi > x, so a* G 3X+1 for i > 0; thus, a; G 3X+1. Assuming that
Yli=oai Â£ 3X+1 would imply that the difference a0 G 3x+l. Then (a0) C Jx+1,
contradicting that 3X (an) Thus, ^ 3X+1, and (J2i=oai) $= 3x+l.
If a and b are two ideals in a ring 91, then the ideal generated by a and
b, in view of Proposition 7.13, is the smallest ideal dividing them. We are
motivated to therefore say that a and b are coprime or relatively prime
provided that the ideal (3,3) is the ring 31) which, incidentally, is the ideal (1).
Distinct maximal ideals are clearly coprime, and thus, in a Dedekind domain,
distinct primes are coprime.
33
Proposition 7.15 If a and b are coprime, then an and b are coprime.
Proof: If 1 = a + (5, then
Proposition 7.16 If a and b are coprime, then aft b = ab.
Proof: See [6, p. 181]
Theorem 7.17 (The Chinese Remainder Theorem) If is a finite col
lection of pairwise coprime ideals in a ring 91, then
Proof: See [6, p. 181].
Theorem 7.18 Let d be a Dedekind domain, k its quotient field, E a finite
separable extension of k and D the integral closure of 0 in E. Assume that is
a prime of T) and p is a prime ofd. Then the following are equivalent:
1 *P(P)
2 (Ph C ^
3. pC<Â£
4 fp n 0 = p (ty lies above p)
5. ^3 D k = p.
ionli
34
Proof: Since t) is integrally closed, 29 fl k = h, and 4 <=> 5 follows. The
equivalence of 1 and 2 follows from Proposition 7.13. Since p C (p)s we have
2^3; the converse follows from (p)s C (ip)^. Obviously
4=^3. Conversely, if p C ip, then p C ip D 0. Since p is maximal, ip fl 0 = p or
h. Since ip ^ 3D, 1 ^ ip and D ^ ip, so ip fl c) h; hence 3=>4
Proposition 7.19 If IK is integral over t, then there is exactly one nonzero
prime of x lying under a given nonzero prime of 92.
Proof: Assume ip is a prime of 91. Since ip ^ 29, 1 ^ ip and D $Â£ ip. Thus,
*P fl 0 D. Now, the ideal ip D r is easily seen to be prime. By Proposition 3.12,
ip fl r is maximal, hence nonzero.
7.3 Ramification Index
Now, assume 3 is a Dedekind domain and 29 is the integral closure in a
finite separable extension of the quotient field of 0. If p is a prime of 0, then we
consider the prime decomposition of the integral ideal (p) in 29, to which we
shall refer as the lifting of p to 29. Assume
<(> = VT
with ei > 1 for each i. Then prime ip of 29 occurs in this decomposition iff ip lies
above p. Given this decomposition, we call e, the ramification index of ipi
over p and write el e(iPip). When the ring h is understood, then considering
Proposition 7.19 we write e
Proposition 7.20 The ramifictation index is multiplicative in towers.
35
Proof: See [15, p 191].
The prime p of fi is said to be ramified in 2), or to ramify in 2), provided
that p2) is not squarefree, that is, for some prime *P lying over p we have
eCPlp) > 1. In this case we might also say that tp is ramified over p, or over
D. It is also common to say that ip (or p) ramifies in the extension of the quotient
fields.
Theorem 7.21 Assume that 0 is a Dedekind domain with quotient field k, E a
finite separable extension of k, and 2) the integral closure of t) in E. A prime p
oft) ramifies in 2) if and only z/pDiscek(2)).
Proof: [15, p. 238]
7.4 Residue Class Fields and Inertial Degree
If p is a nonzero prime ideal in the Dedekind domain D, then p is maximal
and our previous discussion on residue class fields applies. If k is the quotient
field of 0, E a finite separable extension of k, 2) the integral closure of D in E,
and tp a prime ideal of 2) lying above p, then 2)/tp is an extension field of D/p.
The degree of the extension 2)/
inertial degree of fp over p and is denoted by /(tpp), or /
ambiguity.
Proposition 7.22 The intertial degree is multiplicative in towers.
Proof: [15, p 191].
Proposition 7.23 The residue class vector space 2)/(p)s has dimension [E : k]
over D/p.
36
Proof: See [15, p. 212],
If e
completely or totally decompose in 2). The prime p is said to be totally
ramified in 2) if (p) = for some n > 1 with fp prime, and /(fpjp) = 1. In
this case we might also say that fp is totally ramified over p, over 0, or over
k.
We point out that if ip is both totally ramified and unramified over p, then
Ek must be a trivial extension. This seems trivial, but we shall use this in the
proof of our main theorem.
Notice that if a prime p is totally ramified in an extension Ek, then p is
totally ramified in any intermediate extension. More precisely,
Proposition 7.24 if k C F C E is a tower of finite separable extensions of
fields, and p is totally ramified in Ek, then p is totally ramified in EF and m
Fk.
Proof: This follows from Proposition 7.19 and Proposition 7.22.
Proposition 7.25 For any prime p o/h,
[E : k] = ^ etp/
Proof: See [8, p. 25], [10, p. 68] and use Proposition 7.23. There is also a
proof in [21, p. 287], The most elementary proof, however, is in [15, p. 193],
but Ribenboim only proves this for number rings.
Corollary 7.26 The extension 2)/fpD/p is finite.
37
Corollary 7.27 Ifty is totally ramified over p, then (p)x> =
7.4.1 Norm for Ideals
Theorem 7.28 Assume that = pf. Then = pefi
Proof: See [6, p. 182].
For a prime ideal ^3, we define the norma (for ideals) relative to the extension
Ek by
NEk^P = P*v,
and we extend our map NEk to fractional ideals by multiplicativity, since the
fractional ideals form a free abelian group generated by the prime ideals.
In the case when c) = Z, so k = Q, T) = oE, and p = pZ for some rational
prime p, we see that oE/fp is a finite field of characteristic p. In this case, the
norm is the (principal) ideal of Z generated by the cardinality of oE/!)3, and this
fact extends to ideals that are not prime by Theorems 7.17 and 7.28. In other
words,
NEq3 = (#oE/3)z,
for any ideal 3 of oE. The cardinality #oE/^l is known as the absolute norm
(for ideals). We write for the absolute norm
3 = #oE/X
The absolute norm for ideals generalizes the absolute norm for elements in the
sense that Â£ =  (Â£)  for (6 E. See [10, p. 66], [15, p. 141] and [8, p. 26].
38
7.5 Cyclotomic Extensions of Q.
The following two results concerning cyclotomic extensions of Q will prove
useful. They are both proven in [6, p. 197].
Proposition 7.29 Assume that ( is a primitive nth root of unity, with p a
rational prime. Then an odd rational prime p is ramified m Q(C) iff pn; 2 is
ramified iff 4\n.
Proposition 7.30 A rational prime q is totally ramified in Q(fq).
39
8. The Different
Assume D is a Dedekind domain, k its quotient field, E a finite separable
extension of k, and D the integral closure of D in E.
If A is an additive subgroup of E, then the set of all Â£ E such that
TrEk(eyi) C 0
is a Dmodule, which we call the complementary set of A (relative to the
trace), and denote by A*.
Proposition 8.1 If A is a fractional ideal of D, then A* is also a fractional
ideal ofD.
Proof: See [8, pp. 57, 8].
Thus, in Dedekind domains, the complementary set of a fractional ideal is
invertible. If A, is a fractional ideal of Â£>, then we define the different of A to
be
DiffEk^=W1.
When the fields are understood, we drop the subscript. The operation A
A* is easily seen to be contravariant, meaning that if A C B then A* D B*. Since
the operation of inverting fractional ideals is also contravariant, the different is
monotone, i.e., if A C Â£, then Diffifi C DiffC.
Proposition 8.2 For each fractional ideal A of ID, we have Difhfi C A.
40
Proof: This follows from the containment R 1 C R*.
Thus, in particular, DiffEk(2)) is an integral ideal in D depending on k.
8.1 Dual Basis
This section is similar to the development in [15, p. 240].
The trace Tr^k induces a mapping ExE^kby (f,v)M TrEk(Â£u). This
is a symmetric kbilinear form. For Â£ Â£ E, let ip^ : E > k be the linear form
= TrEk(Â£f) Thus, ip^ 6 E', the dual kvector space of E. For a G k and
E, we have
from E > Eh Since the trace is nondegenerate for separable extensions, = 0
iff x = 0, and this mapping is not the zero map. Thus, ip is an isomorphism
between the kspaces E and E'.
If {Pi,..., Pm} is a kbasis of E then let {PI,..., /3*} C E be elements such
that
which certainly exists by surjectivity of
Moreover, {P{,..., /3*} is another kbasis for E, which we call the dual basis
(relative to the trace) of {Pi,..., /?}.
Theorem 8.3 Suppose a Â£ E and consider MQik(df) over k(a), where we may
write
1 if i = j, and
PpWj) TrElk(AA)
(8.1)
0 otherwise,
\
Maikpf) (X a)(70 + ~fiX + + 'YniXn *).
Then
41
is the dual basis to {1, a,..., an l} m k(o).
Proof: See [8, p. 58] or [10, pp. 94,5].
Proposition 8.4 The different is multiplicative in towers, i.e., i/kC EC F m
a tower of finite separable extensions, t) is a Dedekind domain whose quotient
field is k, D is the integral closure of D in E, and 94 the integral closure of c) in
F, then
DiffF,k(94) = DiffFE(94)(DiffEk(D))*.
Proof: See [15, p. 244].
We have an important application of the different for number fields:
Proposition 8.5 An ideal p of oE is ramified m Ek iff p divides DiffEk(oE).
Proof: See [15, p. 247]
Corollary 8.6 Only finitely many primes are ramified in a number field.
Lemma 8.7 Suppose thatty is totally ramified over p, and fix it E fp\fp2. Then
the exact power ofip dividing DiffEkS) is the exact power ofiP dividing the ideal
(Kr.kM)
Proof: See [15, p. 250].
There are also many interesting exercises in [10] related to complementary
sets and the different. See Exercises 3.33 through 3.40.
42
9. Galois Extensions
Prom now on assume that in our usual setting, the extension Ek is Galois
of degree n. Denote the Galois group of Ek by 9 and let fp be one of the primes
above p. If a G 9, then crip is maximal and hence a nonzero prime. Since 9
fixes t>, crip contains p, and aiP is a prime of D lying above p. Hence, the Galois
group acts on the primes above p.
Theorem 9.1 The action of the Galois group on the primes above p is transi
tive.
Proof: Assume that the lifting of p to T) has the factorization
Fix i G {1.. .g} and denote fPj by fp. Choose any a G ip \ We can
do this by the Chinese Remainder Theorem, as we now argue. There exists a
common solution to the set of congruences
X = 0 (mod
X = 1 (mod fpj) for j i.
Now, if X 1 G fPj and X G fPj, then 1 G fPj, a contradiction. Thus, if a is
such a solution, then a G (PKG.%
Now, Nei
NEkCt! G k fl fp = p. If j G {1.. .g}, then, since p C fp^, we have NEkQ; G iPj.
43
Assume j is fixed. Since fp., is prime, one of the factors, say yen, is in fpj. This
implies that a G y1^. Since y_1fpj is a prime above p, it follows by our choice
of a that y1fPj = ip, and yi]3 =
desired result.
Other proofs are given in [10, p. 70] and [6, p. 182],
Now, notice that for er G S, crD = 2). Thus, for any ideal, 3 of 2), we have
D/crfJ = a'S)/a3 = '53/3, since a is an isomorphism. This is particularly useful
when 3 is prime. If *p and Q are primes above p, then 2)/H = 2D/fp.
Proposition 9.2 If ip and Q are two primes lying above the same prime p,
then e
Proof: The fact that e
rem 9.1. The fact that /
In other words, all the primes in the factorization of the lifting of p to 2)
occur with the same exponent, and have the same inertial degrees. We denote
the common ramification index by e, and the common inertial degree by /.
Corollary 9.3 If g denotes the number of distinct primes above p, then efg =
[E : k].
Proof: This is a direct result of Proposition 9.2 and Proposition 7.25.
9.1 Subgroups of the Galois Group
We now investigate some of the structure of the Galois group Sei< Some
very good treatments can be found in [8, I, Â§5], [9, VII, Â§2], and [10, Chapter 4].
44
9.1.1 The Decomposition Group
For each prime ideal ip of E we denote the stabilizer of ip in S by Dtpp or
D(ipp). That is,
D
and we call this subgroup of Sei< the decomposition group of ip over 0, or
over p, whichever is convenient. When there is no ambiguity, we write Dtp or
D(fp). If the prime ip is understood, we simply write D.
If fp and 0 are primes in ID lying above p, then their decomposition groups
are conjugate. Particularly, if crip = 0, then Dq = crDtpcr1.
The decomposition field is the fixed field of the decomposition group,
which we denote by Ed(
E( (
lying below 0 by 0D(
Proposition 9.4
[S : D] = [Ed :k]=g,
where g denotes the number of distinct primes above p.
Proof: In general, D need not be normal in S Nontheless, we may consider
the cosets modulo D, of which there are [S : D], We notice that erfP = rip iff 0
and r represent the same coset. Also, a = f iff crip rip. Using Theorem 9.1,
this provides us with a onetoone correspondence between the cosets modulo
D, and the primes lying above p, establishing [S : D] = g. The fact that [S :
] = [Ed : k] follows from Galois theory. h
45
Proposition 9.5 The field E is the smallest subfield T of E containing k such
that is totally ramified over *p D T.
Proof: This is proven in [9, p. 341].
Proposition 9.6 D/p = 2D/fp.
Proof: This is proven in [9, p. 341],
Corollary 9.7
ePPulp) = /CPdIp) = 1;
/(
e(WPl>) = eOPIf).
Since T)y fixes *p, T
fixes 5/p. Associating each a G with an automorphism a of 2D/
induces a homomorphism of 2D>p into the Galois group of 2D/
proof that the extension is Galois, see [8, p. 15].
9.1.2 The Inertia Group
The action of the Galois group on D induces an action on the cosets modulo
fp. The subgroup that induces the identity on the cosets, that is, stabilizes each
coset, is denoted Â£
Thus,
Â£
As with the decomposition group, we write Â£
there is no ambiguity. We notice that we have the tower
Â£ < D < S
46
of groups. In fact, Â£ < D, since Â£ is the kernel of the homomorphism of D into
the Galois group of the residue class field extension. The inertia field is the
fixed field of the inertia group, and following the rest of our notation, we denote
the inertia field by E^fp), and similarly we use 2)g and 0Â£(
In the case when t)/p is a finite field, for example, in the classical case, the
homomorphism into the Galois group of the residue class field extension is
onto, as proven in [15, p. 261]. Thus, we have
Proposition 9.8 /Â£ = S(S/fPh/p), so [:Â£] = /.
We summarize our development.
Theorem 9.9 In the classical case,
1. Ex> is the largest subfield T of E containing k such that e(fp fi T) = /(fP fl
T) = l;
2. ED is the smallest subfield T of E containing k such that fp is the only
prime lying over fp fl T;
3. Eg is the largest subfield T of E containing k such that e(ip fl T) = 1;
4 Eg is the smallest subfield T of E containing k such that is totally ram
ified in ET, i.e., e(
Proof: For a complete proof, see [10, p. 104]. See also [15, p. 262].
In other words, p does all of its splitting in Ex>k, all of the inertia happens
in the extension EgEx), and all of the ramification of tp over p happens in the
extension E]Eg. We point out that
#Â£(
47
Groups Fields Ideals Degrees Ramification Inertial
Indices Degrees
1 E V
e e 1
Â£ eÂ£ Vs
/ 1 /
T> Ed Vv
9 1 1
S k p
Table 9.1: Decomposition and Inertia Fields: Ramification and Inertial De
grees
Notice that in the classical case, f(Vsp) = / implies that Qz/Vs = 2)/fp.
It follows that 2>g contains a complete set of coset representatives of 2) modulo
f3, i.e., that 2) 2)g +
Lemma 9.10 Assume E and F are Galois extensions of k such that E D F = k.
Then if p is unramified in E and in F, then p is also unramified in EF.
Proof: See [15, p. 278].
9.1.3 The Ramification Groups
We define the ramification groups of tp over p by
Vm(spp) = {7 G SEk : Tu G 2), 7M = w (mod
for each m > 0. That is, Vm the stabilizer of 2) mod ^3m+1. Clearly, Vo = Â£
and (Vm) forms the descending tower
< vfc+1 < Vfc < < V, < Vo = Â£
48
of subgroups.
Proposition 9.11 The chain stops, i.e., there is a k such that V*, 1.
Proof: Since f]iexVl = {0}, as discussed in Chapter 7, we have f]iex^i = 1
The result follows.
As we shall see, the tower of ramification groups is normal in each stage,
i.e.,
l=Vfc
Proposition 9.12 Assume that we are in the classical case. If it Â£ and
a Â£ Vmi, then a Â£ Vm iff a {it) = it (mod fpm+1).
1 Proof: Necessity being trivial, assume that o(it) = it (mod fpm+1). First,
we prove that for a Â£ (it), a (a) = a (mod fpm+1). Write a = Xit. Notice that
cr(A) A Â£ and it Â£ fp, so
(cr(A) A)7t G fpm+1.
Notice also that cr(A)(7r cr(7r)) Â£ ipm+1, so
a(X)a(it) Xit = a(X)it Xit (mod fpm+1).
In other words, cr(a) a = (
establishing the claim.
Next, we prove that for a Â£iP, a (a) = a (mod Write (it) = fpj;
thus + 3 = 2). Suppose that j3 Â£ X>Â£n(J \ (it)). Then [3 Â£ 3 implies /3a Â£ (it),
1In fact, this result holds for a G Â£, but we only need the weaker result. This stronger
result can be obtained by induction. See [10, Exercise 4.20],
49
and p Â£ 0Â£ implies a (ft) = p. Thus,
I
i
I
Pa = a(Pa) = Pa(a) (mod fpm+1).
Since P(a(a) a) Â£ fpm+1 and P P (because (7r) =
factorization of ideals that a (a) = a (mod *pm+1).
We need only argue that 0Â£ D (3 \ (ir)) ^ 0. Assume that 3 = 0i,..., 0t
are the primes dividing 3, and for each i denote by 0^; the prime of 0Â£ lying
below 0j. We argue that none of these primes lies below *p, whence we are done.
If 0Â£_i = ^Â£ for some i, then fp and 0; lie over the same prime p, but if that is
so, then fp 7^ 0^, by Theorem 9.9 We have established the second claim.
Now, for a Â£ D, recall that 2) = 2>Â£ + fp. Thus, we write a = S + p with
S Â£ 0Â£ and p Â£ ^3. Then, using the previous argument,
<7 (a) a = a (5 + p) 5 p
= a(6) + a(p) 6 p
= 5 + a(p) 5 p
=
which is in *pm+1 by the preeceding argument.
9.2 The Frobenius Automorphism in the Classical Case
Assume that k is a number field and E is a finite Galois extension with group
S Let p be a prime of Ok and fp a prime of Oe lying above p. Let p be the rational
prime lying below p. The residue class field Ok is an extension of degree /(pp)
over Z/pZ, and thus is isomorphic to the finite field Fp/(PP). The Galois group
50
of the extension Oe/^P over o^/p is cyclic and generated by the automorphism
In terms of congruences, $(Â£) = Â£HpH (mod fp). Any member of the coset
4>E
nius automorphism of ip over p. See also [8, p. 17] and [10, p. 64].
9.3 Embedding E/Vy (h/p)*
This section follows Exercise 4.21 in [10]. We construct a homomorphism on
Â£ with kernel Vi, showing that Vi < Â£. This homomorphism induces an embed
ding of the factor group E/Vy into the multiplicative group (!D/ip)x. We then
show that with an additional assumption, this same embedding actually embeds
E/Vy (D/p)x. Particularly, this works for abelian extensions. Throughout this
section, fix tt G fp \ ^P2.
Lemma 9.13 Suppose o G Vm_i, m > 1, and cr(7r) = pir (mod ipm+1). Then
cr(Â£) = /zÂ£ (mod fpm+1) for every Â£ G
Proof: We follow a similar procedure as in the first two steps of the proof of
Proposition 9.12. If If Â£ G (tt) say Â£ = Xtt, then notice that since a G Vm_i we
have cr(A) A G ipm. Moreover,
cr(Â£) pÂ£ = cr(A)cr(7r) p\i:
= o(X)fJLTT piXTT
= pn(a(X) A) (mod ipm+1).
Since cr(X) X G fpm and tt G ip, we have p.Tr(o(X) A) G i.e.,
<7(0/^e(pm+1.
51
Now, suppose Â£ G p. As before, write (tv) = pj, and find (3 G Sgfl (J\(7r)).
This implies that [3a G (ir), and a((3) = (3. Thus,
fi(3a = a (Pa) = Pa (a) (mod pm+1).
In other words, P(fia a (a)) G pm+1. Since (3 p (because (tv) = P fl J) we
must have a(a) fia E pm+1.
Lemma 9.14 For all a G D there exists G 2) such that
a(7r) = /r
Proof: If 7r G P \ P2, then we may write (n) = P3 with ip and 3 coprime.
Then ip2 and 3 are also coprime, so we may apply the Chinese Remainder
Theorem to find X such that
X = a(7r) (mod p2), and (9.1)
X = 0 (mod 3). (9.2)
Now, because a G V and tt G P we have a(ir) G p. Also X air G p2 C p so
X G p. In fact, X G P fl 3, which, since P + 3 = D, is the ideal (tt). Thus,
X = fjLa7r for some ^G, establishing our result.
Now, if /i7r = vtt (mod p2), then tt(/j, v) G p2. This means that
P21(7r)a)(/i I/), and by unique factorization, we get (i^Gp. Thus,
Po is uniquely determined modulo p.
We therefore have a welldefined map X> * 2)/P defined by a We notice
that 7r ^ p2 implies that /v ^ P, so the codomain is actually contained in
(/P)x
52
Now, assume that a G Â£. If (3 G ip, then, by Lemma 9.13, with m = 1, we
have cr(/3) = (mod ip2). We apply this to r(7r) for r G Â£. Indeed, r(7r) = 7r
(mod ip), so t(7t) G ip. Applying the Lemma 9.13, and the fact that t(7r) = /xT7r,
we have
or (it) = o(t(k)) = pcrr(7r) = (mod ip2).
On the other hand, err G Â£, so
crr(7r) = Pottt (mod ip2).
By the uniqueness modulo ip proved earlier, we have
crr(7r) = (mod ip2), and [iar = ixa^ir (mod ip).
In other words, our map is a homomorphism. It is obvious that the kernel is W.
Thus, we have proven the following:
Theorem 9.15 There is an embedding Â£/Vi c> (3D/ip)x Consequently, in the
classical case, Â£/Vi is cyclic of order dividing ip 1.
9.3.1 A Better Result for Abelian Extensions in the Classical Case
We have seen that in the classical case D/Â£ is abelian. If the commutator
subgroup of T> lies strictly between Â£ and then V/Vy is not abelian.
In this section, we prove the following:
Theorem 9.16 In the classical case, if V/Vy is abelian, then the embedding
Â£/Vi > (3D/ip)x actually sends 8./Vy into the subgroup (D/p)x.
This will be useful to us because of the following corollary:
53
Corollary 9.17 In the classical case, Z/V\ is cyclic of order dividing p 1.
In particular, this is true when Ek is abelian. We first prove a lemma about the
Frobenius automorphisms. Recall that n G ip \ ip2.
Lemma 9.18 Let (f> denote a Frobenius automorphism off,P over p. Fix a G Â£,
and suppose that a(n) = git (mod fp2). Then
7r) = g^n (mod fp2).
Proof: Notice that (ft, 4>~l, and a G T, so g^, g^g and ga are defined as in
Lemma 9.14. Moreover, 4>~l(ir) G *P since 7T G fp. Thus, by Lemma 9.13, since
a G Â£ = Vq, we have
a
= gag4>i'K (modip2).
Thus,
~l(n) = i)g4lT: (mod tp2). (9.3)
Now, 4>(ga) = gtP^ (mod ip), by definition of
we have
= Fla11 + Pa
in addition, we have
#1(7T) = 7r)
= (,Kp
= (p{g'1T (mod *p2).
54
Since we also have 00 1(7r) = 7r (mod ip2), the uniqueness assertion made fol
lowing Lemma 9.14 implies
0(/^i)/^ = 1 (mod ip).
We therefore assume that
^(/r^i)^ = 1 + p^, with Â£ ip.
Now, equation 9.3 becomes
l( 7r) = (/ij[Pl1 + pa)( 1 + p^TT
= /4pII7t + pa 7T + /4p%7T + pap4,rr
= p.^n (mod ip2),
since all other terms are in ip2.
We now prove the theorem.
Proof: Let a Â£ Â£, and again let 0 be a Frobenius automorphism of ip over
p. By Lemma 9.18, 0<70_1(7r) = plP^ir (mod ip2). Since 0
4>a(f)~l(n) Â£ ip, so by Lemma 9.13,
<710CT01(7r) = //(Tl0O'0_1(7r)
= Pvip^n (mod ip2).
Now assume that V/Vi is abelian. Thus, the commutator subgroup of D
is contained in Vi, so o~l4>o(j)~l Â£ W, and cr_10cr0_1(7r) = 11 (mod ip2). By
the uniqueness assertion following Lemma 9.14, we must have p^ipl^ = 1
(mod ip). Now, also paipa = 1 (mod ip), so
Paip.v Pa^lPl1 e ip.
55
Since na1 ^ ip, we must have fia = (mod ip). Considered in the field
extension Â£>/ip()/p, we see that is fixed under the action of the Galois group.
Thus, ixa G D/p.
9.4 Embedding ^ Â£>/<Â£
In this section we follow Exercise 4.22 in [10]. We prove that Vm_i < Vm,
by constructing a homomorphism on VTO_i having kernel Vm. Moreover, this
induces an embedding of <* (D/ip)+.
Again fix n G ip \ ip2. Notice that nm G ipm \ ipm+1 by unique factorization
of ideals (particularly (7r)m).
We first prove the following:
Lemma 9.19 For each a G Vmi there exists aa G D such that a(n) = 7r +
ao7rm (mod ipm+1).
Proof: We have proven this result for m= 1 (a = fi 1), so we assume tacitly
that m > 2. The argument is similar to the proof of Lemma 9.14. Write (7rm) =
fpm9 (so ipm and 3 are coprime). Apply the Chinese Remainder Theorem to
find X G D such that
X = a(n) 7r (mod ipm+1), and (9.4)
A = 0 (mod 9). (9.5)
First, X = <7(7r) 7r (mod ipm+1) implies X = a(n) n (mod ipm), so X G ipm
and hence X G ipm fl 3 = (n) Writing X = aanm, we have aa7rm = a(n) n
(mod ipm+1), or
an = 7r + acr7rrn (mod ipm+1).
56
Now, if 7r + airm = 7r + /?7rm (mod fpm+1), then (a P)TTm G ipm+1. Since
7rm G fpm \ fpm+1, unique factorization forces a (3 G fp. Thus,
zs unique modulo fp.
We therefore have a welldefined map of Vm_i > fD/fp. We apply Lemma 9.13
to t(7r) for r G Vm_i. Since
cr(7r) = 7r + aCT7rm = (1 + ct(J7rm_1)7r (mod fpm+1),
by Lemma 9.13, for f3 G ip, we have that cr(/3) = (1 + aa7rm~i)(3 (mod fpm+1).
In particular, if r G Vm_i, then t(tt) G fp, so
crr(7r) = cr(r(7r))
= (1 + aff7rm_1)r(7r)
ee {l + aairml){Tr + aT7rm)
= 7r + ao7rm + aTnm + ctoaTvr27711 (mod fpm+1).
Now, m > 2 => 2m 1 > m + 1, so a(TaTTT2m~l G fpm+1, and we have
crr(7r) = 7T + Qo7rm + ar7Tm
= 7T + (aff + ar)7rm (mod fpm+l).
Thus,
ar(7r) = 7r + (a + ar)nm (mod fpm+1).
Uniqueness modulo fp implies
aa + aT = aaT (mod ip).
57
In other words, our map is a homomorphism. Clearly the kernel is Vm, and we
have the following:
Theorem 9.20 There is an embedding Vm^i/Vm c (3D/
the classical case, #'Vrn_1/'Vm divides p.
Notice that we have now established that the tower
1 = V* < Vfc_! < < Vi < V0 = Â£
is indeed normal in each stage.
Proposition 9.21 In the classical case, Vi is a pgroup. Hence, Vi is a p
group for each i > 1.
Proof: 8,/Vi is cyclic and Vi/Vi+i is an elementary abelian pgroup, so Â£ is
solvable and Vi is a pgroup. [15, p. 269].
Proposition 9.22 In the classical case, V is solvable.
Proof: We have D > Â£ > 1, with Â£ is solvable, and D/Â£ cyclic.[15, p. 269]
9.5 Hilberts Formula for Totally Ramified Primes in the Classical
Case
Assume that is totally ramified over p and that we are in the classical
case. Then S = T> = Â£ by Theorem 9.9. Let t denote the exact power of ip
dividing DiscEk(oE) In this section we find an explicit formula for t.
Lemma 9.23 If it Â£ ($\ fp2, then fps is an exact divisor of (M^k(7r))j).
Proof: A direct corollary to Lemma 8.7.
58
Lemma 9.24 If a & VTO_i \ Vm, then the ideal {7r a(7r))xi is exactly divisible
by
Proof: If a G Vm_i, then n <7(77) G by definition of the group Vm_i. If
cr ^ Vm, then by Proposition 9.12, 7r a(tt) ^
Theorem 9.25
t = 5^(#v 1).
m>0
Proof: Write
M'1r,k(7r) =
7S
Since ^ is totally ramified, S = V0, and 5 U Vmi \ Vm, it follows that
t = Y,m 1 \v) = x)
m>l m> 0
The formula in Theorem 9.25 is known as Hilberts formula.
59
10. Abelian Extensions and the KroneckerWeber Theorem
We wish now to discuss abelian extensions. With our usual situation, assume
that Ek is an abelian extension, that is, Galois with abelian group, which we
shall denote by S. Let p be a prime of h and fp a prime of D lying over p. Let D
and Â£ be the decomposition and inertia groups of
prime 0 over p can be written as crip for some a S, and the decomposition
and inertia groups of 0p are the conjugates aT>a~1 and crÂ£cr_1, respectively;
see [8, p. 18]. When Ek is abelian, then these conjugations have no effect, and
it makes sense to speak of the decomposition and inertia groups of p itself. See
[7, P 1].
We devote the remainder of this chapter to proving our main theorem:
Theorem 10.1 (KroneckerWeber) Every finite abelian extension of Q is
contained in a cyclotomic extension of Q.
10.1 Reduction to Prime Power Degree
Recall that in Chapter 4 we proved that any abelian extension is the com
positum of abelian extensions of prime power degree; see Theorem 4.10. We
wish to apply this to our present situation.
Theorem 4.10 reduces our problem to the case in which E is an abelian
extension of prime power degree over Q. If E is the compositum Ki Ks, and
for each i we have K* C Q((mJ, then E C Q(Â£mi) Q(CmJ ^ Q(Cicm[m1,...,ms])
60
10.2 Reduction to One (Particular) Ramified Prime
At this point we are forced to forfeit some generality. The theorem that we
wish to prove is not true in more general a context than when the base field is
Q. Prom here on, all of our extensions will be finite and abelian over Q.
Assume that KQ is a finite abelian extension. Considerring Theorem 4.10
(and Section 10.1), we may assume that [K : Q] = pm for some rational prime
p, which we do. Assume that ip is a prime of o lying over p. In this section, we
shall reduce our situation further to one in which only p is ramified.
Suppose that q is a prime ^ p that is ramified in K, and assume that Q is
a prime of o lying over q. Now, 'Vi(H) is a ggroup, by Proposition 9.21. As a
subgroup of S, which has order pm, Vi(H) is also a pgroup. Since p and q are
distinct, we must have
VrCQHi.
It follows from this and Theorem 9.16 that
eQ\q 1.
Now, Q((q) has degree q 1 over Q, and the Galois group of Q(Cq)Q is cyclic.
By the Fundamental Theorem of Galois Theory and Theorem 4.4, Q(Cq) has a
unique subfield L of degree over Q. We now consider how q splits in L.
Assume O is a prime of KL lying over Q, and let K' denote the inertia field
(Kh)g(Q)
The prime q is unramified in K'Q and every rational prime that is
unramified in KQ is also unramified in K'Q.
61
!
j
!
I
i
Figure 10.1: The behavior of q ^ p
Proof: First, q is unramified in K', by Theorem 9.9. Let r be a rational
prime ^ q. Then r is unramified in Q(C9) because DiscQ(Â£g) = qq~2. Since
L C Q(Cq), t is unramified in L, by multiplicativity of the ramification indecies,
Proposition 7.20. Now, if r is unramified in K, then r is unramified in KL, by
Lemma 9.10. Since K' C KL, r is unramified in K'.
Now, we wish to apply Theorem 4.11 to our present context. We have an
embedding
SkL(Q SkQ x SlQ
Since Eq < SklQj we can consider the image of Â£g under this embedding.
Proposition 10.2 This embedding sends Â£^ into Â£q x Slq
Proof: Let a Eq. What needs to be shown is that a & Eq. First, for
a G Okli we have tr(o:) a Â£ Â£3, by definition of the group. Restricting a to K
leads us to consider a G Ok, since ol Fl K = Ok Now, cr(a) a G K because
KQ is assumed normal. Moreover, Â£2 fl K = Q. Thus, cr (a) aGQ.
62
Recall that [L : Q] = eQ, which divides pm. By this and the preceding
result, Â£(0g) is a pgroup. Thus, Vx(0g) is also a pgroup. Of course, by
Proposition 9.21, Vi(0g) is also a ggroup. Again, p 7^ g, so
Vi(Qg) = l.
The group Â£(0g) is therefore cyclic, by Lemma 9.16. Also, Â£(0g) may be
regarded as a subgroup of Â£q x SlQi as argued above. In fact, Â£(0g) may be
regarded either as a subgroup of Â£q or of SlQi by Lemma 4.9. In either case,
we conclude that eQeQ. On the other hand, by multiplicativity of the
ramification index. Hence,
eQ = eQ.
Now, q is totally ramified in LQ, by Proposition 7.30, since L C Q(Cg) Since
[L : Q] = eQ, the equality eq = eQ, implies that q does not pick up any more
ramification in KLL; more precisely,
0 is unramified over L.
By multiplicativity of the ramification indicies in the tower L C K'L C KL, we
conclude the more important fact that
0 is unramified over K'L.
Also, by Theorem 9.9, 0 is totally ramified over K'. By considering the tower
K' C K'L C KL and using Proposition 7.24, we conclude that
0 is totally ramified over KT.
63
Thus, H is both unramified and totally ramified over K'L. The extension KLK'L
must therefore be trivial, i.e.,
K'L = KL.
Since L C Q(Â£g), if K' C Q(6) then K C KL = K'L C Qfo)Q(&) C
Q(icm(q,t)) Thus, if K' is contained in a cyclotomic field, then so is K.
Notice that [K' : Q] is a power of p because [L : Q] and [K : Q] are both
powers of p, and K' C KL. Also, recall that all primes unramified in KQ are
unramified in K'Q. Thus, we have reduced the number of ramified primes p
by at least 1. Since there are only finitely many ramified rational primes in K,
we can repeat this process until no primes other than p are ramified. We have
therefore reduced the problem to the case when the extension is prime power
degree, and that prime is the only ramified rational prime. Indeed, since we
show in Appendix B that at least one prime is ramified unless the extension is
trivial, we may assume that this prime is ramified. We henceforth assume this
to be the case.
10.3 The Case for 2
1 Assume that K is an abelian extension of degree 2m over Q.
Lemma 10.3 If R is a real abelian extension of degree 2m over Q, and 2 is the
only ramified prime in R, then R contains the unique quadratic subfield Q(\/2).
Consequently, Srq is cyclic.
Proof: If m = 1, then by the formula for the discriminant, R = (\/2). The
Galois group is cyclic because its order is the prime 2.
^ur treatment of this case is very similar to that in [15, pp. 278, 279],
64
Assume m > 1. By Lemma 4.5, the Galois group Srq contains a subgroup of
order 2m_1, i.e., a subgroup of index 2. By the Fundamental Theorem of Galois
Theory, R contains a subfield F such that [F : Q] = 2. Now, 2 is the only ramified
rational prime in F. Thus, DiscpQ(oF) is a power of 2, which, by the formula for
discriminants of quadratic number fields, leaves F = Q(\/2), Q(i), or Q(n/=2)
The only real field among these is Q(\/2). Thus, the subgroup of Srq of order
27711 is unique, and we conclude that Srjq is cyclic.
Lemma 10.4 For each m, there is exactly one real abelian extension of degree
2m over Q such that 2 is the only ramified prime.
Proof: Assume that R and R' are two such extensions. Then the compositum
RR' as degree a power of 2, and only 2 is ramified by Lemma 10.3. Thus Srr'q
is cyclic. The product
SrR'Q = 0RRnR' X SR'IRnR'
must therefore have one of its factors trivial. In either case, R = R'.
Now, let C be a primitive root of unity of order 2m+2. Set L = Q(C), so
[L : Q] = 4>{2m+2) = 2m+1. Let R = L D R. If Â£ = Â£ + iv, with Â£, v G M, then
Â£ iv G L by normality, so the sum 2Â£ and difference 2iv are in L. Thus, Â£ Â£ L,
and since i G L, also v 6 L. Since also Â£, v E E, we have Â£, v G R. Thus, f G R(i);
hence L C R. On the other hand, i G L and RCL together imply that R(i) C L.
Thus,
R(<) = L.
65
Since z ^ R, we have that [L : R] ^ 1. Since i satisfies the quadratic equation
X2 + 1 = 0 over L, we have [L : R] = 2, and
[R : Q] = 2m
Thus, R is the unique real abelian extension of Q of degree 2m with only 2 ram
ified. We notice that R is contained in a cyclotomic extension. The important
fact, which we have proven, is this:
Lemma 10.5 The unique real abelian extension of degree 2m with only 2 ram
ified is contained in a cyclotomic field, namely Q(C2m+2)
Now we return our attention to K. Since Q(z)Q is abelian, the compositum
K(z) is abelian over Q. Also, since 2 is the only ramified prime in Q(z) as well as
in K, by Lemma 9.10, only 2 is ramified in K(z). Moreover, [K(z) : Q] is a power
of 2.
Let T = K(z) flM. Thus, T is a real abelian extension of , and as a subfield
of K(z), T has degree a power of 2 over Q and only 2 is ramified. Assume by
Lemma 10.5 that T C Q(cu) with u> a primitive root of unity.
Now, by the Primitive Element Theorem, Proposition 3.21, we may assume
that K(z) = T(a + ifi) with a, ft Â£ M. Then a ifd Â£ K(z) by normality, so a, ifi,
and thus 2 are in K(z). We now have a and fif2 Â£ K(z) fl R = T. Consequently,
a + i(3 satisfies the polynomial X2 2aX + (a2 + (32) with coefficients in T, and
hence [K(z) : T] = 2. Thus, K(z) = T(z), and K C K(i) = T(z) C Q(u;,z), which is
contained in a cyclotomic extension of Q, since both u and i are roots of unity.
66
10.4 The Case When p is Odd and m 1
By assumption, ip is ramified over p, that is, e 1. If g denotes the number
of primes above p, then e
Let 7r G fp \ fp2. Notice that ip D Q = pZ C ipp C ip2 because p > 2; hence,
7r ^ Q. Since the degree of 7r divides [K : Q] = p, 7r must have degree p over Q.
Denote
MWjq(X) = Xp + apiXp 1 + + Oo
Thus, a; G Z for each i 1,... ,p, and
7TP + Qp_i7TP ^ T + Qi7T + Go = 0.
Hence,
Go = 7TP f ap_i7TP T" + Gi7T.
Since 7r G ip, it follows that Go G ip. Thus, we have ao G Z fl ip = pZ, which is
contained in fpp, so, in fact, ao G ipp.
Now consider congruence modulo ip2. Since
Gi7r Go = 7TP + dp17TP + + G27T2,
it follows that + a0 G ip2. Since p > 2, we have a0 G ip2. Thus, ai7r G ip2.
Now, 7r ^ ip2; by unique factorization of the ideal (ai7r) we must have d\ G ip.
As with g0, we see that d\ G ipp. A simple induction argument shows that g^ G P
for each z = l,...,p 1. 2 Consequently,
2Notice that, in fact, we showed that l,7r,... 7rp_1 are independent modulo p. This result
can be obtained more generally. See [10, Ex. 3.20].
67
a,i is divisible by p for each i 0,.. .p 1.
Let fpfc be the exact power of fp dividing the principal ideal (M'(7r)). Recall
Hilberts formula from Theorem 9.25:
k = Â£(#Vm 1).
m> 0
Since [K : Q] = p, for each m > 0, Vm has order 1 or p. Thus, each nonzero term
is p 1, and
k is a multiple of p 1.
Consider the exact power of fp dividing each ideal
(p7rp_1>, ((p l)apiTrp~2), , (2a27r), (ai)
generated by a term of M'(7r). Now, if pbi\\ai, then fpfciP (aj) because (p) = *pp.
Now p \ 1,..., p 1 implies fp \ (1),..., {p 1), and the only other power of fp
that can occur in the zth term (from the right) is that in (tt1"1} = (tt)*1 Now,
7r Â£ fp \ fp2 implies fp*_1 (7r)I_1, and the ith term has exact divisor
for i = 1,... ,p 1. The pth term clearly has exact divisor fp2p_1. Notice that
these exponents are all distinct modulo p; hence they are all distinct.
We now apply Lemma 7.14. Because the exponents considered above are all
distinct, it follows by Lemma 7.14 that k is the minimum of these exponents,
and since each exponent is at least p, we have k > p. On the other hand, the
exponent in the term p7rp_1 was explicitly calculated to be 2p 1. Hence,
p < k < 2p 1.
68
Now since p > 2,
1 <
P
P~ 1
<
k
P 1
<
2 pl n 1
= 2 +
p 1 p 1
< 3.
Thus, k 2{jp 1). Since only ip is ramified, by Proposition 8.5,
DiffKQ(oK) = qJ2(p"1).
10.4.1 The Case p is odd and m = 2
Since p is unramified in Kg and no other rational prime is ramified in K, we
conclude that no rational prime is ramified in Kg. By Theorem B.2, Kg = Q,
and by Theorem 9.9,
ip is totally ramified over p.
In other words, e(fpp) p2. Thus, Â£(ipp) has order p2. Now, by Theorem 9.16,
Â£(iPp)/'V1(ipp) has order dividing p 1, but this group is also a pgroup since
both Â£ and Vi are pgroups. The only power of p that divides p 1 is 1; we
conclude that Vi(fpp) has order p2.
Let V7.(ipp) be the first ramification group having order < p2. By Theo
rem 9.20, Vri/VT has order dividing #0e/^P, which is p because f
Vr has order p.
Now, let K be any subgroup of Skq having order p, and let K;h be the fixed
field of TC. Then ip is totally ramified in the extension KK^. By our calculation
in the case when m 1, we have
DiffKMQ(oKM) = (?pnM2^1>.
69
Since ^ is totally ramified in KK^, which is of degree p, we have
K =
By multiplicativity of the different in the tower Q C C K, we have
DiffKQ(0K) = DiffKIK^lOK)^1^.
Thus, DiffxiKj^ (ok) is independent of ff{. On the other hand, the exponent of ^
in Diff^iK^ (o) is strictly maximized when ffC = Vr, which we now prove.
Proof: By Hilberts formula, Theorem 9.25, if t! is this exponent, then t' =
1) We consider each term. Notice that
by applying the definition. Moreover, #Vm D fff < p, and is maximized (= p)
iff fff < Vm For m such that = p2, this maximum is certainly achieved
when CK = Vr. On the other hand, for m such that = p, this maximum is
achieved if and only if Vr = ffC. Thus we have the desired result.
We conclude that Skq has only one subgroup of index p, namely Vr; hence
Skiq is cyclic.
10.4.2 Back to m 1
Now, if K and K7 are two extensions of Q of degree p, and p is the only prime
ramified in either K or K', then the compositum KK7 is an extension of degree
p2 over Q, and p is the only prime ramified in KK7, as argued earlier.
By Subsection 10.4.1, KK7 has only one subfield of degree p over Q. This
implies that K7 = K, i.e., that K is unique.
Now, the p2th cyclotomic field has cyclic Galois group of order p(p 1),
which has a subgroup of index p. The fixed field of this subgroup has degree p
70
over Q. We conclude that K is this unique subfield of the p2th cyclotomic field
having degree p over Q.
10.5 The Case p is odd and m > 1
Now, the extension Q(CP+i)Q has Galois group (Zpm+i)x, which is cyclic of
order 4>{prn+l) = prn(p~ 1). As a cyclic group, there is a unique subgroup of order
p 1. Let L be the fixed field of this subgroup. Thus, L is the unique subfield of
Q((pm+i) of degree pm over Q. Then Slq has order pm and is cyclic as a subgroup
of a cyclic group (another way of seeing this is by using Subsection 10.4.2).
Now, [KL : Q] is a power of p, and p is the only ramified rational prime.
Since the extension is abelian, KL contains a subfield of degree p over (Q>, and
as discussed in Section 10.4.2, this field is unique. We conclude that the Galois
group of KL is cyclic.
Thus, the product
SLKnL SKKnL X Sl_KnL
must have one of its factors trivial. In either case, we conclude that K = L.
This completes the proof of the theorem of Kronecker and Weber.
71
Appendix A. Notation
Notation Meaning and Page Where Introduced
MQjk The minimal polynomial of a over k, p. 13
SEk or S(Ek) The Galois group of the extension Ek, p. 15
Ear The fixed field of IK, where !H < SEk
kal The algebraic closure of the field k, p. 15
Ok The algebraic integers in k, p. 16
eOPp), e
fmp), h The inertial degree of ip over p, p. 36
m\p) The inertia group of fp over p, p. 46
opip) The decomposition group of fp over p, p. 45
Vm(Wp) The ramification groups of fp over p, p. 48
$ 771 The mth cyclotomic polynomial, p. 23
4> The Euler totient function, p. 22
{^l, , The ideal generated by an,..., an, p. 4
(5)c The ideal generated by S in t, p. 4
DiscEk The discriminant relative to the extension Ek, p. 26
TrEk The trace relative to the extension Ek, p. 25
DiffEk The different relative to the extension Ek, p. 40
Cn Any primitive nth root of unity, p. 22
72
Re(0 The real part of Â£, p. 21
MO The imaginary part of Â£, p. 21
Arg(C) The argument of Â£, p. 21
ICI The modulus of Â£, p. 21
pa\\n pa is an exact divisor of n
(ai,j) The matrix whose (i,j) entry is a,
Det The determinant
x< s 3i is a subgroup of S
TC is a proper subgroup of S
IN is a normal subgroup of S
1 The group with one element
SCT 5 is a subset of T
#s The cardinality of S
SuT The disjoint union of S and T
73
Appendix B. The Minkowski Bound
In a Dedekind domain D it is common to define a relation ~ on the class of
ideals by
(3a,0e)a3 = P3.
It is not difficult to show that this is an equivalence relation, and hence partitions
the class of ideals of V into equivalence classes. These equivalence classes are
known as ideal classes. Under multiplication of ideals, the ideal classes can be
shown to form a group known as the ideal class group or just class group.
An interesting result is that for any abelian group there is a Dedekind domain
with an isomorphic class group [3, p. 735]. Our concern here is only with one
result concerning the class groups of number rings.
Assume that K is an extension of degree n > 1 over Q, and that a\,... ,aT
are the monomorphisms K > C that in fact embed K > E. Then the remaining
n r embeddings K C occur in conjugate pairs. The number of these pairs
we denote by s (thus n = r + 2s).
Theorem B.l Every ideal class contains an ideal 3 having norm satisfying
II3II < ^ () VDiscoK.
nn \7rJ
The number on the right of the inequality is known as the Minkowski bound
since it is due to his lemma. For a proof of this result, we refer the reader to
[10]. Our concern is only with one application of this result.
74
Theorem B.2 Every number field other than Q has a ramified rational prime.
Proof: It follows that the Minkowski bound must be at least 1, so
/TTr nn /"n\s nn /7r\n/2
^^^(4)^(4)
Denote this last quantity by bn. First,
, 22 /7T\2/2
2 2\ (4/
7r
= > 1.
Next,
bn+1
Noting that for n > 1,
we see that
(n+l)n+1 i'Â£\(n+1)/2
(n+1)! UI
nn / 74 \ n/2
n! V 4 /
1 1 \ n
n + 1 \
n
n + 1
n
>2,
^n+l
bn
> 2
> 1,
so 6n+i > bn for each n. Hence, for n > 2 we have bn > 1 and Disco > 1,
that is, Disco > 1, which means that there is a prime p dividing Disco, and
hence ramified in KQ.
75
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