Galois Theory: A Study of Cyclotomic Field ... - Scripps College

Galois Theory:
A Study of Cyclotomic Field Extensions
Lisa Lambeth
Professor Christopher Towse, Advisor
Professor Anie Chaderjian, Reader
Submitted to Scripps College in Partial Fulfillment
of the Degree of Bachelor of Arts
March 11, 2005
Department of Mathematics

Abstract
This expository thesis explores the ideas of Galois theory. We begin with
a basic overview of field theory before exploring the necessary aspects of
Galois theory needed to study cyclotomic field extensions. The cyclotomic
field extensions studied in detail are Q(ζ 19 ) and Q(ζ 31 ).

Contents
Abstract
Acknowledgments
iii
xi
1 Introduction 1
2 The Basics of Field Theory 3
2.1 Field Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.2 Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3 Galois Theory 7
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3.2 The Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . 11
3.3 Proof of the Main Theorem of Galois Theory. . . . . . . . . . 12
3.4 Intermediate Galois Extensions . . . . . . . . . . . . . . . . . 14
4 Cyclotomic Field Extensions 17
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
4.2 Structure of the Galois Group . . . . . . . . . . . . . . . . . . 18
4.3 Group Generators . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.4 Subgroups and Fixed Fields . . . . . . . . . . . . . . . . . . . 21
4.5 Real Subfields and Quadratic Extensions . . . . . . . . . . . 24
4.6 Statement of the Kronecker-Weber Theorem . . . . . . . . . . 31
Bibliography 33

List of Figures
4.1 Group Lattice for Q(ζ 19 ) . . . . . . . . . . . . . . . . . . . . . 21
4.2 Field Lattice for Q(ζ 19 ) . . . . . . . . . . . . . . . . . . . . . . 24
4.3 Simplified Field Lattice for Q(ζ 19 ) . . . . . . . . . . . . . . . . 29
4.4 Group Lattice for Q(ζ 31 ) . . . . . . . . . . . . . . . . . . . . . 30
4.5 Field Lattice for Q(ζ 31 ) . . . . . . . . . . . . . . . . . . . . . . 31

List of Tables
4.1 Field Generator for Q(ζ 19 ) . . . . . . . . . . . . . . . . . . . . 20
4.2 Subfield Generators . . . . . . . . . . . . . . . . . . . . . . . . 20

Acknowledgments
I would first like to thank my advisor Professor Christopher Towse for not
only introducing me to to this beautiful field of mathematics but also for
his enthusiasm and support throughout the entire project. I would also
like to thank my second reader, Professor Anie Chaderjian, for her support.
My thesis would not have had such a great title page, awesome formatting,
or beautiful figures were it not for Eric Malm. For all of your help with
L A TEX issues, thank you. Finally, I want to thank my friends and family for
supporting me as I wrote this thesis.

Chapter 1
Introduction
Galois theory is the study of the roots of the irreducible polynomial
f(x) = x n + a n−1 x n−1 + · · · + a 1 x + a 0 .
This is done by not limiting ourselves to finding roots that exist merely in
the rational numbers, denoted Q. Rather, we instead allow ourselves to
adjoin non-rational numbers to Q in order to find roots of polynomials. For
example, we can adjoin √ −1 to Q to find the roots of x 2 + 1. Specifically,
we will concentrate on what are known as cyclotomic field extensions of Q,
which will be defined later in the text.
Before we discuss Galois theory and look at cyclotomic field extensions,
we will begin with some basic background information on fields.

Chapter 2
The Basics of Field Theory
A field F is a commutative ring with identity in which every nonzero element
has a multiplicative inverse. The set of nonzero elements of a field F ,
denoted F × , is an abelian group under multiplication.
An important property in field theory is the characteristic of a field. The
characteristic of a field, if it exists, is the smallest positive integer p such
that, if 1 F denotes the multiplicative identity element in our field F , then
p · 1 F = 0. It is well known that such a p is a prime number. If no such p
exists, we say the field has characteristic 0. The rationals, Q, is such a field.
We shall concentrate on fields that have characteristic 0.
2.1 Field Extensions
Say that K is a field that contains F . Then K is an extension field of F . We
denote this by K/F and read “K over F.” For example, R is an extension
field of Q.

4 The Basics of Field Theory
Definition. An element α ∈ K is algebraic over F if α is a root of a nonzero
polynomial f(x) ∈ F [x]. For example, √ 2 ∈ R is algebraic over Q since it is
a root of x 2 − 2.
Proposition 1. Let α be an algebraic element over the field F and f(x) be its
irreducible polynomial of degree n. Then {1, α, . . . , α n−1 } is a basis for F [α] as a
vector space over F .
Proof. See [A, p.495]
The degree of a field extension K/F is the dimension of K as a vector
space over F and is denoted [K : F ].
A reason to study field extensions is to understand roots of polynomials.
The polynomial x 2 + 1, for example, would not have any roots in R.
However, the complex numbers C, would be a field extension of R (and
thus an extension of Q) that would allow us to find roots of x 2 + 1. It is
easy to show that C is a field under normal addition and multiplication, so
every nonzero element in C has an inverse.
We can consider a field K in which the polynomial f(x) has its roots.
We will assume that the polynomial f(x) is irreducible over F [x] since if
there is a root of a factor of f(x), then it would certainly be a root of f(x)
itself.
Let K be a field generated by a single element α over F . That is to say,
every element β in K can be expressed as β = a 0 +a 1 α+a 2 α 2 +· · · a n α n for
some n and some a 0 , a 1 , a 2 , . . . , a n ∈ F. Then we can write K = F (α). We
say K is a simple extension of F and α is a primitive element for the extension.

Splitting Fields 5
2.2 Splitting Fields
Definition. If K is the extension field of a field F such that a polynomial
f(x) ∈ F [x] factors completely into linear factors in K[x] (and f(x) does
not factor into linear factors over any proper subfield of K that contains
F ), then K is called the splitting field for f(x).
Proposition. In any abstract splitting field for the polynomial x n − 1, the set
of the nth roots of unity form a group under multiplication.
Proof. If α, β are both roots, α n β n = (αβ) n = 1, so the product αβ is
in the set. Thus we have closure under multiplication. We know that 1
is in our set since 1 n = 1. Given any root α, we can find a multiplicative
inverse α −1 since (α) −1n
= (α n ) −1 = 1 −1 = 1. We also have associativity
because these roots are complex numbers. Thus we have a group under
multiplication.
□
Definition. A primitive nth root of unity is a generator of the cyclic
group of all nth roots of unity.
We will denote the chosen primitive nth root of unity by ζ n . The other
primitive nth roots of unity are then the elements ζn a where 1 ≤ a < n is
an integer relatively prime to n, since these are the other generators for a
cyclic group of order n. In particular there are precisely ϕ(n) primitive nth
roots of unity, where ϕ(n) denotes the Euler ϕ-function.
The splitting field of x n − 1 over Q is the field Q(ζ n ). The field Q(ζ n ) is
called the cyclotomic field of nth roots of unity.

Chapter 3
Galois Theory
3.1 Introduction
Galois theory is the study of the roots of an irreducible polynomial
f(x) = x n + a n−1 x n−1 + · · · + a 1 x + a 0 .
We will study the symmetries between the roots of irreducible polynomials.
A great deal can be understood from expanding the equation
f(x) = (x − α 1 )(x − α 2 ) · · · (x − α n ).
Comparing these two equations, we can see that the sum of the roots is
α 1 + · · · + α n = −a n−1 and the product α 1 · · · α n = ±a 0 .
Lagrange and Galois were the first mathematicians to see that the relationships
between roots of polynomials could be understood in terms of

8 Galois Theory
symmetry [A, p.538]. Let us go back to our example of the polynomial
x 2 + 1 = 0. We already know that our polynomial has no roots in Q, but it
does have roots in C. Complex conjugation, the first model for the symmetric
relationships we will be studying, gives us the roots ±i and leaves the
rationals fixed. Any quadratic (degree 2) field extension will have similar
symmetries.
As an example, let us look at a degree 2 extension K/F generated by
any α ∈ K which is not in F . Let f(x) = x 2 + bx + c be a polynomial with
coefficients in F such that α is a root of f(x). We see that, b = −(α + α ′ ), so
α ′ = −(α + b) is the other root. Over K, the polynomial f(x) splits into the
linear factors (x − α)(x − α ′ ).
Our symmetry comes from the fact that both α and α ′ are roots of f(x).
Proposition 2. Let L = F (α) and L ′ = F (α ′ ) be two extension fields of F
generated by the algebraic elements α and α ′ . Then there is an isomorphism of
fields
σ : F (α) ˜→F (α ′ ),
which is the identity on F and sends α → α ′ if and only if the irreducible polynomials
for α and α ′ over F are equal.
Proof. See [A, p.496]
According to the above proposition,
σ : F (α) −→ F (α ′ )
is an isomorphism which sends α → α ′ and is the identity on the field F .
In our case, however, K can be generated by either of the two roots. That

Introduction 9
is, L = L ′ = K. Thus, σ is an automorphism of K. Since σ is the identity
map on F , b stays fixed (it is equal to −(α + α ′ )). Thus σ must send α to
α ′ . We can conclude that σ 2 is the identity map on K since σ(α) = α ′ and
σ(α ′ ) = α.
Definition. Let K be an extension field. An automorphism of K which
is the identity on F is an F-automorphism. It has the property that σ(a) = a
for all a ∈ F . The F -automorphisms of K form a group under composition.
The group of all F -automorphisms of the field extension K is known as the
Galois group of K over F , denoted G(K/F ).
Theorem 3. For any algebraic field extension K/F, the order of the Galois
group, |G(K/F )|, divides the degree of the extension [K : F ].
Proof. See [A, p.540]
We know for any σ ∈ G(K/F ), any a in F will be fixed by σ, that is,
σ(a) = a for all a in F . But a priori, there may be other elements of K fixed
by σ.
Definition. If the order of the Galois group is equal to the degree, then
the field extension K/F is a Galois extension.
Definition. The set of elements in a field K that are fixed by all the
elements of G, a group of automorphisms of K is known as a fixed field of
G and is denoted K G .
Corollary 4. Let K/F be a Galois extension. If its Galois group is G =
G(K/F ), then F is the fixed field of G.
Proof. Let L be the fixed field, K G , of G. Clearly, F ⊂ L. Since F ⊂ L,
every L-automorphism will also be an F -automorphism. This implies that
G(K/L) is contained in G. However, every element in G is by definition of

10 Galois Theory
L an L-automorphism, which implies G(K/L) ⊃ G. We have containment
in both directions, which means G = G(K/L). Since K/F was chosen to be
a Galois extension, the order of G is equal to the degree of the extension K
over F , or |G| = [K : F ]. If we now apply theorem 3, we know that the
degree of the extension of K over L is divisible by the order of G. Since
our original hypothesis was that F is contained in L, we are led to the
conclusion that [K : F ] = [K : L], which implies F = L. Thus, F must also
be a fixed field.
□
This corollary leads us into the following theorem relating fixed fields
and splitting fields.
Theorem 5. Let K be a splitting field of a polynomial f(x) whose coefficients
are in the field F . Then K is a Galois extension of F . The reverse direction is also
true; every Galois extension is a splitting field of some polynomial f(x) over F .
Proof. See [A, p.541]
Corollary 6. Every finite field extension is contained in a Galois extension.
Proof. Let K/F be a finite extension. Suppose the generators of K over
F are α 1 , . . . , α n . We need an irreducible polynomial for each α i over F ,
call it f i (x). Let the product f 1 · · · f n be f. Then, let M be a field extension
of K that is also a splitting field for f over K. This is done by adjoining any
roots of some polynomial f i (x) to K if it was not already in K. If M is a
splitting field for K, then it must also be a splitting field for F . By theorem
5, M is a Galois extension. □
Corollary 7. Let K/F be a Galois extension. Suppose L is an intermediate
field F ⊂ L ⊂ K. Then K/L is also a Galois extension.
Proof. Let K be the splitting field of the polynomial f(x) over F . Since

The Main Theorem 11
L ⊂ K, we know K is also a splitting field of f(x) over L. Then by theorem
5, K is a Galois extension of L. □
A critical part in the study of Galois theory is finding the intermediate
fields L of an extension K/F . That is, such that F ⊂ L ⊂ K. We will be
able to conclude from the Main Theorem of Galois theory that the intermediate
fields of a Galois extension are in a bijective correspondence with the
subgroups of the Galois group.
3.2 The Main Theorem
We will now return our attention to the Main Theorem of Galois Theory.
Definition. Given an extension K/F , we will call any field L with F ⊂
L ⊂ K an intermediate field.
Theorem 8 (The Main Theorem of Galois Theory). If K is a Galois extension
of a field F , and if G = G(K/F ) is its Galois group, then the function given
by
H ↦→ K H
is a bijective map from the set of subgroups of G to the set of intermediate fields L.
Its inverse function is defined by
L ↦→ G(K/L).
This corresponds to the property that if H = G(K/L), then [K : L] = |H|, hence
the degree of the extension [L : F ] is equal to the index of the subgroup [G : H].

12 Galois Theory
Definition. Any two splitting fields K of a polynomial f(x) ∈ F [x] are
isomorphic, so the Galois group G(K/F ) depends only on f up to isomorphism.
It is called the Galois group of the polynomial over F .
We are now almost ready to prove the Main Theorem of Galois theory.
For convenience, the following corollary has been stated:
Corollary 13. Let K/F be a algebraic field extension. The following properties
are equivalent:
(i) K is a Galois extension of F;
(ii) K is the splitting field of an (irreducible) polynomial f(x) ∈ F [x];
(iii) F is the fixed field for the action of the Galois group G(K/F ) on K;
3.3 Proof of the Main Theorem of Galois Theory.
Proof (of the Main Theorem). Let K/F be a Galois extension and L be an
intermediate field. Let the corresponding subgroup of G be H = G(K/L).
We know that H will act trivially on L, by definition. The intermediate field
corresponding to a subgroup H of G(K/F ) is the fixed field K H of H. So,
L is contained in K H .
The following lemma assists us as the main counting argument in the
proof of the Main Theorem.
Lemma 14. Let G be a group of automorphisms of a field K, and let the order
of G be n. Then [K : K G ] = n.
Proof. [A. p. 554]

Proof of the Main Theorem of Galois Theory. 13
Since K/F is a Galois extension and F ⊂ L ⊂ K, the extension K/L is
a Galois extension by Corollary 7. So, the degree of the extension [K : L] is
equal to the order of H, denoted [K : L] = |H|. Using Lemma 14, we know
that the order of H equals the degree of [K : K H ], or |H| = [K : K H ]. Since
[K : L] = [K : K H ], and L ⊂ K H , we have L = K H .
Now, suppose we start with H being a subgroup of G and we let L be
the fixed field K H . Then H is a subgroup of G(K/L) since any element in
H fixes L. That is, H ⊂ G(K/L). Now, the order of H is equal to the degree
of the field extension from K H to K. But L = K H , so |H| = [K : K H ]. We
know [K : L] is equal to |G(K/L)| since K/L is Galois. So we can conclude
that [K : K H ] = [K : L] = |G(K/L)|. This gives us H = G(K/L).
Now, we know that the two maps given by
L ↦→ G(K/L) and H ↦→ K H
are inverses. The Main Theorem of Galois Theory, as you may recall, stated
that if K was a Galois extension of a field F , and G = G(K/F ) was its Galois
group, then there is a bijective map H ↦→ K H from the set of subgroups
of G to the set of intermediate fields F ⊂ L ⊂ K. Further, its inverse function
given by L ↦→ G(K/L). Further still, if H = G(K/L), then the degree
of H is equal to [K : L], and [L : F ] = [G : H]. Since K/F was chosen to be
a Galois extension of L, the degree of G is [K : F ]. Since |H| = [K : L] and
[K : F ]/[K : L] = [L : K], with |G| being equal to [K : F ] and |H| equal
to [K : L], we know [L : F ] = [G : H], and thus we have proved the Main
Theorem of Galois Theory.
□

14 Galois Theory
3.4 Intermediate Galois Extensions
There are many important consequences of the relationship given by the
Main Theorem of Galois theory. One concept is the idea of order reversing.
That is to say, if L and L ′ are both intermediate fields and if the corresponding
subgroups are H = (G/L) and H ′ = (K/L ′ ), then L ⊂ L ′ if and only if
H ′ ⊂ H.
Definition. Let L be an intermediate field. An F -automorphism σ of K
will map L to some other intermediate field σL, which may or may not be
the same as L. We define σL to be a conjugate subfield.
Theorem 15. Let K/F be a Galois extension and L an intermediate field. Let
H = G(K/L) be the corresponding subgroup of G = G(K/F ). Then we know
the following are true:
(a) Let σ ∈ G. The subgroup of G corresponding to the conjugate subfield
σL is the conjugate subgroup σHσ −1 . That is to say, G(K/σL) =
σHσ −1 .
(b) L is a Galois extension of F if and only if H is a normal subgroup of
G. When H is a normal subgroup of G, then G(L/F ) is isomorphic to the
quotient group G/H.
Proof. (a) Say that σL = L ′ and that τ ∈ H = G(K/L). Then στσ −1 ∈
H ′ = G(K/L ′ ). To show this, we will need to prove that στσ −1 fixes any
element in L ′ . Let α ′ ∈ L ′ . Then by definition, α ′ = σ(α) for some corresponding
α ∈ L. Then στσ −1 (α ′ ) = στσ −1 (σ(α)) = στ(σ −1 )(σ)(α) =
στ(α) = σ(α) = α ′ . Thus, στσ −1 fixes any element in L ′ . It follows that
σHσ −1 ⊂ H ′ and, by symmetry with σ −1 and H −1 , we have σ −1 H ′ σ ⊂ H

Intermediate Galois Extensions 15
if and only if H ′ ⊂ σHσ −1 , which gives us our equality H ′ = σHσ −1 .
(b) ⇒ Suppose L/F is a Galois extension. Then L is a splitting field for
some polynomial f(x) with coefficients in F . That is to say, L = F (β 1 , . . . , β n ),
where the β i are roots of the polynomial f(x) in K. An F -automorphism σ
of K permutes these roots and thus maps L to itself: σL = L. By (a) above,
H = σHσ −1 . Thus, H is a normal subgroup.
⇐ Let H be normal. Then for all σ ∈ G, we know H = σHσ −1 .
So, G(K/L) = G(K/σL).
This implies that for all σ, the composition
σL = L. Every F -automorphism of K maps L to itself; it defines an F -
automorphism of L by restriction. The restriction defines a homomorphism
mapping G to G(L/F ), whose kernel is the set of all σ ∈ G which induces
the identity on L, which we know to be H. We now can conclude that G/H
is isomorphic to a subgroup of G(L/F ). We find that
[L : F ] = |G/H| ≤ |G(L/F )|.
Thus we can conclude that L is a Galois extension and G/H ≈ G(L/F ). □
Definition. An extension K/F is an abelian extension if K/F is Galois
and the Galois group G(K/F ) is an abelian group.
It is known that all subgroups of abelian groups are in fact normal subgroups.
Further, we know all cyclic groups are abelian, but it is not necessarily
true that all abelian groups are cyclic. We will now concentrate on
extensions with cyclic Galois groups.

Chapter 4
Cyclotomic Field Extensions
4.1 Introduction
Definition. Recall from Chapter 2 that for any n ∈ N, the nth cyclotomic field
is a subfield K of the complex numbers generated over Q by ζ n = e 2πi/n .
We will denote ζ n simply as ζ when no confusion will arise. A cyclotomic
extension of Q is Q(ζ n ) and it is the splitting field of x n − 1. The roots of
the polynomial are the powers of ζ, the nth roots of unity: 1, ζ, ζ 2 , . . . , ζ n−1 .
Here we will concentrate on the cases when n is a prime number p.
We know that we can factor the polynomial
x p − 1 = (x − p)(x p−1 + x p−2 + · · · + x + 1),
and the polynomial (x p−1 + x p−2 + · · · + x + 1) is irreducible over Q, and
ζ = ζ p is one of its roots; it is the irreducible polynomial for ζ over Q and
the roots are powers of ζ, ζ 2 , . . . , and ζ p−1 . Thus the order of the Galois

18 Cyclotomic Field Extensions
group G(Q(ζ)/Q) is p − 1.
4.2 Structure of the Galois Group
Theorem 16. Let p ∈ Z be prime and ζ = ζ p . Then the Galois group of Q(ζ) over
Q is isomorphic to the multiplicative group (Z/pZ) × of nonzero elements of the
ring Z/pZ and is a cyclic group of order p − 1.
Proof. Let G be the Galois group of Q(ζ) over Q. We will need a map
v : G → (Z/pZ) × as follows: Let σ be an automorphism in G. Then σ will
carry ζ to another root of the polynomial x p + · · · + x + 1, call it ζ i . Since
ζ has multiplicative order p, our exponent i will be an integer modulo p.
We set v(σ) = i. We need to verify that v is multiplicative. If τ is another
element of G such that v(τ) = j, that is, τ(ζ) = ζ j , then
στ(ζ) = σ(ζ j ) = σ(ζ) j = ζ ij = ζ ij .
The identity automorphism will send ζ to ζ, so v(id) = 1. We know that
v is a homomorphism to (Z/pZ) × since v is multiplicative and v(σ) does
not equal 0 for any σ. The homomorphism is injective since ζ generates
Q(ζ) and the action of an automorphism is determined when we know
its action on ζ. Then we know G is isomorphic to its image in (Z/pZ) × .
This means that G is cyclic. The order of G is equal to that of (Z/pZ) × , or
|G| = |(Z/pZ) × |, which equals p − 1. These two groups are isomorphic. □
Since (Z/pZ) × is a cyclic group, so is every subgroup. The Galois group
G of K = Q(ζ p ) over Q is a cyclic group of order p − 1, we know G has one

Group Generators 19
subgroup of order k for each k ∈ N which divides p − 1. If (p − 1)/k = r
and if σ is the generator of G, then the subgroup of order k is generated
by σ r . Applying the Main Theorem of Galois theory, we know there will be
exactly one intermediate field L with [L : Q] = r. These fields are generated
by certain sums of powers of ζ = ζ p .
Example. Consider the case when p = 19 and ζ = ζ 19 . Then by the
above theorem, we know that the Galois group of Q(ζ 19 ) is isomorphic to
(Z/19Z) × and that it is a cyclic group of order 18, which we shall denote
C 18 . Then, we know that the intermediate fields will correspond to the nontrivial
divisors of 18, i.e. 2, 3, 6, and 9, and Q and Q(ζ 19 ) will correspond to
the trivial divisors of 18, namely 1 and 18, respectively. The Main Theorem
of Galois theory then tells us there will be exactly one intermediate subfield
generated by each of these divisors.
4.3 Group Generators
A group is cyclic if there is an element a in G which generates G. That is,
if there exists some a ∈ G such that G = {a n |n ∈ Z}. Let us concentrate
on the G = (Z/19Z) × . Through trial and error, we can find the generators
of this group and its subgroups. First, we must find the generator of the
entire group. Table 4.1 illustrates one such choice of generator.
We can see that every element is generated by a = 2 and the order is:
a 18 , a 19 , a 13 , a 2 , a 16 , a 14 , a 6 , a 3 , a 8 , a 17 , a 12 , a 15 , a 5 , a 7 , a 11 , a 4 , a 10 , a 9 .
Now that we know 2 is the generator of the entire group, we can find the
generators of the intermediate subfields.

20 Cyclotomic Field Extensions
a = 2 is a generator for G:
a = 2 a 6 = 7 a 11 = 15 a 16 = 5
a 2 = 4 a 7 = 14 a 12 = 11 a 17 = 10
a 3 = 8 a 8 = 9 a 13 = 3 a 18 = 1
a 4 = 16 a 9 = 18 a 14 = 6 a 19 = 2
a 5 = 13 a 10 = 17 a 15 = 12
Table 4.1: Field Generator for Q(ζ 19 )
a 9 is a generator: a 6 is a generator: a 3 is a generator: a 4 is a generator:
a 9 = 18 a 6 = 7 a 3 = 8 a 4 = 16
a 18 = 1 a 12 = 11 a 6 = 7 a 8 = 9
a 18 = 1 a 9 = 18 a 12 = 11
a 12 = 11 a 16 = 5
a 15 = 12 a 2 = 4
a 18 = 1 a 6 = 7
a 10 = 17
a 14 = 6
a 18 = 1
Table 4.2: Subfield Generators
Let us name the above subgroups as follows. Since a 9 a subgroup of
order 2, we will denote this subgroup of C 18 as C 2 . Similarly, a 6 generates
a subgroup of order 3, denoted C 3 . Next, a 3 generates a subgroup of order
6, so we will denote this subgroup C 6 . Finally, a 4 generates a subgroup of
order 9, so this subgroup will be denoted C 9 .
The subgroups of the cyclic group C 18 correspond to H 2 , H 3 , H 6 , and
H 9 , which are subgroups of the group G with powers of σ as generators.
For example, C 2 = {e, a 9 } corresponds to H 2 = {id, σ 9 }.
Figure 4.1 gives a lattice representation of the subgroups. The lines in-

Subgroups and Fixed Fields 21
dicate containment. For example, the line connecting H 2 and H 6 represents
[H 6 : H 3 ] = 3
3
{e}
2
H 3
H 2
3 2 3
H 9
2 3
H 6
G
Figure 4.1: Group Lattice for Q(ζ 19 )
4.4 Subgroups and Fixed Fields
The intermediate fields are generated by sums of the powers of ζ = ζ 19 .
We have found that a 3 yields a subgroup of order 6, a 4 one of order 9, a 6
of order 3, and finally a 9 generates a subgroup of order 2. Thus our corresponding
fixed fields will have degrees 6, 9, 3, and 2 over Q, respectively.
We have found a = 2 to be a generator of (Z/19Z) × . This corresponds to
σ ∈ G(Q(ζ 19 )/Q), where σ maps ζ to ζ 2 . We can now find generators in the
field in terms of ζ 19 .
We know that C 2 = {e, a 9 } corresponds to H 2 = {id, σ 9 }. Consider
ζ + σ 9 ζ.

22 Cyclotomic Field Extensions
Then
σ 9 (ζ + σ 9 ζ) = σ 9 ζ + σ 18 ζ = σ 9 ζ + ζ.
So
ζ + σ 9 ζ = ζ + ζ 29 = ζ + ζ −1
is fixed by all of H 2 . Thus, H 2 fixes Q(ζ + ζ −1 ). Since no other σ in G fixes
ζ + ζ −1 , this is the fixed field of H 2 . That is, Q(ζ + ζ −1 ) is the fixed field of
H 2 .
Next, consider the field H 3 which corresponds to C 3 = {e, a 6 , a 12 }. Since
H 3 corresponds to C 3 , we know H 3 = {id, σ 6 , σ 12 }. Now we can consider
ζ + σ 6 ζ + σ 12 ζ. We can check that every element in H 3 fixes ζ + σ 6 ζ + σ 12 ζ.
First, we see σ 6 fixes every element.
σ 6 (ζ + σ 6 ζ + σ 12 ζ) = σ 6 ζ + σ 12 ζ + σ 18 ζ = σ 6 ζ + σ 12 ζ + ζ.
Then, we find that σ 12 also fixes every element.
σ 12 (ζ + σ 6 ζ + σ 12 ζ) = σ 12 ζ + σ 18 ζ + σ 6 ζ = σ 12 ζ + ζ + σ 6 ζ.
So this means ζ + σ 6 ζ + σ 12 ζ = ζ + ζ 26 + ζ 212 = ζ + ζ 7 + ζ 11 is fixed by all of
H 3 . Since no other σ in G will fix ζ + σ 6 ζ + σ 12 ζ, we know Q(ζ + ζ 7 + ζ 11 )
is the fixed field of H 3 .
Then we can look at H 6 , which corresponds to C 6 . The elements of H 6 =
{id, σ 3 , σ 6 , σ 9 , σ 12 , σ 15 } fix all elements in ζ + σ 3 ζ + σ 6 ζ + σ 9 ζ + σ 12 ζ + σ 15 ζ.

Subgroups and Fixed Fields 23
Then this means
ζ + σ 3 ζ + σ 6 ζ + σ 9 ζ + σ 12 ζ + σ 15 ζ = ζ + ζ 23 + ζ 26 + ζ 29 + ζ 212 + ζ 215
= ζ + ζ 8 + ζ 7 + ζ 18 + ζ 11 + ζ 12
is fixed by all of H 6 and Q(ζ + ζ 8 + ζ 7 + ζ 18 + ζ 11 + ζ 12 ) is the fixed field
of H 6 .
Finally, we consider the corresponding field for C 9 , namely H 9 . The
elements of H 9 = {id, σ 2 , σ 4 , σ 6 , σ 8 , σ 10 , σ 12 , σ 14 , σ 16 } fix all elements in
ζ + σ 2 ζ + σ 4 ζ + σ 6 ζ + σ 8 ζ + σ 10 ζ + σ 12 ζ + σ 14 ζ + σ 16 ζ.
This means
ζ + σ 2 ζ + σ 4 ζ+σ 6 ζ + σ 8 ζ + σ 10 ζ + σ 12 ζ + σ 14 ζ + σ 16 ζ
= ζ + ζ 22 + ζ 24 + ζ 26 + ζ 28 + ζ 210 + ζ 212 + ζ 214 + ζ 216
= ζ + ζ 4 + ζ 16 + ζ 7 + ζ 9 + ζ 17 + ζ 11 + ζ 6 + ζ 5
is fixed by all of H 9 . Again, no other σ in G fixes
ζ + ζ 4 + ζ 16 + ζ 7 + ζ 9 + ζ 17 + ζ 11 + ζ 6 + ζ 5 ,
so Q(ζ + ζ 4 + ζ 16 + ζ 7 + ζ 9 + ζ 17 + ζ 11 + ζ 6 + ζ 5 ) is the fixed field of H 9 .
This can be more easily understood in the form of a field extension lattice.
See Figure 4.2
In the lattice representation of the field extension, we can clearly see the

24 Cyclotomic Field Extensions
Q(ζ 19 )
3
2
Q(ζ + ζ 7 + ζ 11 )
Q(ζ + ζ −1 )
3 2
3
Q(ζ + ζ 4 + ζ 5 + ζ 6 + ζ 7 + ζ 9 + ζ 11 + ζ 16 + ζ 17 )
2
Q(ζ + ζ 7 + ζ 8 + ζ 12 + ζ 18 )
3
Q
Figure 4.2: Field Lattice for Q(ζ 19 )
degree 2 and degree 3 extensions from Q to Q(ζ 19 ). Fortunately, this is not
all that is known about the subfields. In the following section, we will get
a better understanding of these two intermediate fields.
4.5 Real Subfields and Quadratic Extensions
Proposition 17. The subfield L max of K = Q(ζ p ) whose degree over Q is 1 2 (p−1)
is generated over Q by the element η = ζ +ζ p−1 = 2 cos 2π/p. Moreover, L max =
K ∩ R, so L max is called the maximal real subfield of K.
Proof. Consider the quadratic polynomial x 2 − ηx + 1, which has coefficients
in Q(η) and η = ζ + ζ −1 . Note that ζ is a root. So we know that

Real Subfields and Quadratic Extensions 25
[K : Q(ζ)] ≤ 2. Since η is a real number and ζ is not, Q(η) ⊂ K. So,
[K : Q(η)] = 2, and Q(η) = K ∩ R. Thus, [Q(η) : Q] = 1 2
(p − 1). □
Thus in our example of Q(ζ 19 ), we can simplify our representation of
our real subfield of Q(ζ 19 ):
( )
However, if 2 cos 2π
19
( ( )) 2π
Q(ζ + ζ −1 ) = Q 2 cos .
19
( )
is in L max , then certainly cos 2π
19
will also be in the
field. So we can simplify our representation of the real subfield further by
( ( ))
simply writing it as L max = Q cos 2π
19
. The maximal real subfield is just
a degree 2 subfield of Q(ζ 19 ). In other words, [Q(ζ 19 ) : L max ] = 2.
4.5.1 Discriminant of a Quadratic Extension
Given any f(x) ∈ Q[x] and a splitting field K, we write f(x) = (x −
α 1 ) · · · (x − α n ). Since K is a splitting field for f(x), all the roots of f(x)
are going to be in K. Note that any σ ∈ G(K/Q) permutes these roots.
Definition.
The discriminant of a polynomial f(x) = (x − α 1 )(x −
α 2 ) · · · (x − α n ) is defined to be
D = (α 1 − α 2 ) 2 (α 1 − α 3 ) 2 · · · (α n−1 − α n ) 2
= ∏ (α i − α j ) 2 = ± ∏ i − α j ).
i

26 Cyclotomic Field Extensions
Q contained in the cyclotomic field Q(ζ 19 ). Then
L quad = Q( √ −19),
where the sign was determined (−1) (p−1)/2 = (−1) 9 = −1.
Proof. We shall use the discriminant of the polynomial to find a generator
for L quad . Let D be the discriminant of the polynomial
x 19 − 1.
This discriminant can be computed directly in terms of the roots {1, ζ, ζ 2 , . . . , ζ 18 },
but the following lemma gives us a formula which is easier to use:
Lemma 19. Let f(x) = (x − α 1 ) · · · (x − α n ). The discriminant of f is
D = ±f ′ (α 1 ) · · · f ′ (α n ) = ± ∏ i
f ′ (α i ),
where f ′ is the derivative of f.
Proof of Lemma. By the product rule of differentiation,
n∑
f ′ (x) = (x − α 1 ) · · · (x − α i−1 )(x − α i+1 ) · · · (x − α n ).
i=1
Setting x = α i , we get
f ′ (α i ) = (α i − α 1 ) · · · (α i − α i−1 )(α i − α i+1 ) · · · (α i − α n ). (4.1)
i ≠ j.
This is the product of the differences (α i − α j ), with the given i and

Real Subfields and Quadratic Extensions 27
Thus,
∏
f ′ (α i ) = ∏ i − α j ) = ±D.
i
i≠j(α
When we apply this lemma to the polynomial x 19 − 1, its derivative is
19x 18 . The discriminant is given by
□
±D =
18∏
i=0
19ζ i(18) = ζ N 19 19 ,
where the exponent N is some integer. To determine ζ N , we note that ±D
is a rational number, because x p − 1 ∈ Q[x] are rational. The reason D
is rational is because it is in the splitting field K over Q. Because D was
symmetrically defined, it is clearly in the fixed field of the whole Galois
group. That is, any Q-automorphism of K must permute the roots of f;
any permutation of roots will fix D. However, the fixed field K G where G
is the whole Galois group G(K/Q) is itself just Q. Thus, D is in Q. The only
power of ζ which is rational is 1. So, ζ N = 1 and
±D = 19 19 .
The sign of the discriminant is determined to be negative by carefully
comparing the definition of the derivitive and Equation 4.1. So the square
root of this discriminant is δ = √ −19 19 . By definition of D, it is clear that δ
in the field Q(ζ). Since the square factors can be pulled out of a square root,

28 Cyclotomic Field Extensions
that is, √ −19 19 = 19 9√ −19, so we see
Q(δ) = Q( √ −19).
This gives us a quadratic subfield of Q(ζ). However, L is the only quadratic
subfield, which means Q(δ) = L.
□
Note that the above theorem can be applied to any odd prime.
Applying this theorem to Q(ζ 19 ), we can represent our unique quadratic
extension of Q as:
Q(ζ + ζ 4 + ζ 5 + ζ 6 + ζ 7 + ζ 9 + ζ 11 + ζ 16 + ζ 17 ) = Q( √ −19)
Then if we re-visit our field lattice extension, we see
Notice that L quad is not contained in L max . This corresponds to the fact
that (p − 1)/2 is odd. Had it been even, then our discriminant D would
have been a positive value, making δ real. Then, L quad would have to be a
subfield of L max .
The reader might have noticed that the prime 19 was of the form 19 =
2·3 2 +1, or rather twice the square of a prime plus one. Let us now see how
such a prime differs from a prime of the form 2 · p 1 · p 2 + 1, or twice two
unique primes plus one. We will briefly explore the field extension Q(ζ 31 )
over Q.
Example. The subfields will correspond to the subgroups of (Z/31Z) × .
The generator of this group is a = 3 and the generator for this cyclic group
then the automorphism σ = σ 3 which maps ζ 31 to ζ31 3 . The subgroups of G

Real Subfields and Quadratic Extensions 29
Q(ζ 19 )
3
2
Q(ζ + ζ 7 + ζ 11 )
( ( ))
Q cos 2π
19
3 2
3
Q( √ −19)
Q(ζ + ζ 7 + ζ 8 + ζ 12 + ζ 18 )
2
3
Q
Figure 4.3: Simplified Field Lattice for Q(ζ 19 )
are
H 2 = 〈σ 15 〉 = {1, σ 15 };
H 3 = 〈σ 10 〉 = 〈σ 20 〉 = {1, σ 10 , σ 20 };
H 5 = 〈σ 6 〉 = 〈σ 12 〉 = 〈σ 18 〉 = 〈σ 24 〉 = {1, σ 6 , σ 12 , σ 18 , σ 24 };
H 6 = 〈σ 5 〉 = 〈σ 25 〉 = {1, σ 5 , σ 10 , σ 15 , σ 20 , σ 25 };
H 10 = 〈σ 3 〉 = 〈σ 9 〉 = 〈σ 21 〉 = 〈σ 27 〉 = {1, σ 3 , σ 6 , σ 9 , σ 12 , σ 15 , σ 18 , σ 21 , σ 24 , σ 27 };
H 15 = 〈σ 2 〉 = 〈σ 4 〉 = 〈σ 8 〉 = 〈σ 14 〉 = 〈σ 16 〉 = 〈σ 22 〉 = 〈σ 26 〉 = 〈σ 28 〉 =
{1, σ 2 , σ 4 , σ 6 , σ 8 , σ 10 , σ 12 , σ 14 , σ 16 , σ 18 , σ 20 , σ 22 , σ 24 , σ 26 , σ 28 , }
having orders 2, 3, 5, 10, and 15, respectively. The group lattice for Q(ζ 31 )
would then be as shown in Figure 4.4. A degree 2 extension is represented
with a red line, a degree 3 extension with a blue line, and a degree 5 exten-

30 Cyclotomic Field Extensions
sion with a black line.
{e}
H 3 H 5 H 2
H 15 H 6 H 10
C 30
Figure 4.4: Group Lattice for Q(ζ 31 )
We then will go back and apply what we know about the real subfield
and the unique quadratic extension to our lattice. The real subfield, ζ + ζ −1
( )
can be replaced by cos 2π
31
and the quadratic extension by Q( √ −31). Our
field lattice is then shown in Figure 4.5. Once again, a degree 2 extension
is represented with a red line, a degree 3 extension with a blue line, and a
degree 5 extension with a black line.

Statement of the Kronecker-Weber Theorem 31
Q(ζ 31 )
( ( ))
Q(ζ + ζ 5 + ζ 25 ) Q(ζ + ζ 2 + ζ 4 + ζ 8 + ζ 16 ) Q cos 2π
31
Q( √ −31)
Q(ζ + ζ −1 + ζ 5
+ζ 6 +ζ 25 +ζ 26 )
Q(ζ + ζ −1 + ζ 2 + ζ 4 + ζ 8
+ ζ 15 + ζ 16 + ζ 23 + ζ 27 + ζ 29 )
Q
Figure 4.5: Field Lattice for Q(ζ 31 )
4.6 Statement of the Kronecker-Weber Theorem
Recall at the end of Chapter 3, we defined an abelian extension K/F to be
an extension whose Galois group G(K/F ) was an abelian group. Having
studied the cyclotomic fields, we conclude with this surprising result.
Theorem 20. Every Galois extension K of Q whose Galois group is
abelian is contained in one of the cyclotomic fields Q(ζ n ).
The proof of this theorem is beyond the scope of this thesis, however, L.
Washington presents a proof in his book Introduction to Cyclotomic Fields.

Bibliography
Artin, M. (1991). Algebra. Prentice-Hall, Inc., Upper Saddle River, New
Jersey.
Dummit, D. S. and Foote, R. M. (1991). Abstract Algebra. Prentice-Hall, Inc,
Englewood Cliffs, New Jersey.
Gallian, J. A. (2002). Contemporary Abstract Algebra. Houghton Mifflin Company,
Boston, Massachusetts.
Washington, L. C. (1982). Introduction to Cyclotomic Fields. Number 83 in
Graduate Texts in Mathematics. Springer-Verlag, New York.