Galois Theory: A Study of Cyclotomic Field ... - Scripps College
Galois Theory: A Study of Cyclotomic Field ... - Scripps College 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
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<strong>Galois</strong> <strong>Theory</strong>:<br />
A <strong>Study</strong> <strong>of</strong> <strong>Cyclotomic</strong> <strong>Field</strong> Extensions<br />
Lisa Lambeth<br />
Pr<strong>of</strong>essor Christopher Towse, Advisor<br />
Pr<strong>of</strong>essor Anie Chaderjian, Reader<br />
Submitted to <strong>Scripps</strong> <strong>College</strong> in Partial Fulfillment<br />
<strong>of</strong> the Degree <strong>of</strong> Bachelor <strong>of</strong> Arts<br />
March 11, 2005<br />
Department <strong>of</strong> Mathematics
Abstract<br />
This expository thesis explores the ideas <strong>of</strong> <strong>Galois</strong> theory. We begin with<br />
a basic overview <strong>of</strong> field theory before exploring the necessary aspects <strong>of</strong><br />
<strong>Galois</strong> theory needed to study cyclotomic field extensions. The cyclotomic<br />
field extensions studied in detail are Q(ζ 19 ) and Q(ζ 31 ).
Contents<br />
Abstract<br />
Acknowledgments<br />
iii<br />
xi<br />
1 Introduction 1<br />
2 The Basics <strong>of</strong> <strong>Field</strong> <strong>Theory</strong> 3<br />
2.1 <strong>Field</strong> Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . 3<br />
2.2 Splitting <strong>Field</strong>s . . . . . . . . . . . . . . . . . . . . . . . . . . . 5<br />
3 <strong>Galois</strong> <strong>Theory</strong> 7<br />
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7<br />
3.2 The Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . 11<br />
3.3 Pro<strong>of</strong> <strong>of</strong> the Main Theorem <strong>of</strong> <strong>Galois</strong> <strong>Theory</strong>. . . . . . . . . . 12<br />
3.4 Intermediate <strong>Galois</strong> Extensions . . . . . . . . . . . . . . . . . 14<br />
4 <strong>Cyclotomic</strong> <strong>Field</strong> Extensions 17<br />
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17<br />
4.2 Structure <strong>of</strong> the <strong>Galois</strong> Group . . . . . . . . . . . . . . . . . . 18<br />
4.3 Group Generators . . . . . . . . . . . . . . . . . . . . . . . . . 19<br />
4.4 Subgroups and Fixed <strong>Field</strong>s . . . . . . . . . . . . . . . . . . . 21<br />
4.5 Real Subfields and Quadratic Extensions . . . . . . . . . . . 24<br />
4.6 Statement <strong>of</strong> the Kronecker-Weber Theorem . . . . . . . . . . 31<br />
Bibliography 33
List <strong>of</strong> Figures<br />
4.1 Group Lattice for Q(ζ 19 ) . . . . . . . . . . . . . . . . . . . . . 21<br />
4.2 <strong>Field</strong> Lattice for Q(ζ 19 ) . . . . . . . . . . . . . . . . . . . . . . 24<br />
4.3 Simplified <strong>Field</strong> Lattice for Q(ζ 19 ) . . . . . . . . . . . . . . . . 29<br />
4.4 Group Lattice for Q(ζ 31 ) . . . . . . . . . . . . . . . . . . . . . 30<br />
4.5 <strong>Field</strong> Lattice for Q(ζ 31 ) . . . . . . . . . . . . . . . . . . . . . . 31
List <strong>of</strong> Tables<br />
4.1 <strong>Field</strong> Generator for Q(ζ 19 ) . . . . . . . . . . . . . . . . . . . . 20<br />
4.2 Subfield Generators . . . . . . . . . . . . . . . . . . . . . . . . 20
Acknowledgments<br />
I would first like to thank my advisor Pr<strong>of</strong>essor Christopher Towse for not<br />
only introducing me to to this beautiful field <strong>of</strong> mathematics but also for<br />
his enthusiasm and support throughout the entire project. I would also<br />
like to thank my second reader, Pr<strong>of</strong>essor Anie Chaderjian, for her support.<br />
My thesis would not have had such a great title page, awesome formatting,<br />
or beautiful figures were it not for Eric Malm. For all <strong>of</strong> your help with<br />
L A TEX issues, thank you. Finally, I want to thank my friends and family for<br />
supporting me as I wrote this thesis.
Chapter 1<br />
Introduction<br />
<strong>Galois</strong> theory is the study <strong>of</strong> the roots <strong>of</strong> the irreducible polynomial<br />
f(x) = x n + a n−1 x n−1 + · · · + a 1 x + a 0 .<br />
This is done by not limiting ourselves to finding roots that exist merely in<br />
the rational numbers, denoted Q. Rather, we instead allow ourselves to<br />
adjoin non-rational numbers to Q in order to find roots <strong>of</strong> polynomials. For<br />
example, we can adjoin √ −1 to Q to find the roots <strong>of</strong> x 2 + 1. Specifically,<br />
we will concentrate on what are known as cyclotomic field extensions <strong>of</strong> Q,<br />
which will be defined later in the text.<br />
Before we discuss <strong>Galois</strong> theory and look at cyclotomic field extensions,<br />
we will begin with some basic background information on fields.
Chapter 2<br />
The Basics <strong>of</strong> <strong>Field</strong> <strong>Theory</strong><br />
A field F is a commutative ring with identity in which every nonzero element<br />
has a multiplicative inverse. The set <strong>of</strong> nonzero elements <strong>of</strong> a field F ,<br />
denoted F × , is an abelian group under multiplication.<br />
An important property in field theory is the characteristic <strong>of</strong> a field. The<br />
characteristic <strong>of</strong> a field, if it exists, is the smallest positive integer p such<br />
that, if 1 F denotes the multiplicative identity element in our field F , then<br />
p · 1 F = 0. It is well known that such a p is a prime number. If no such p<br />
exists, we say the field has characteristic 0. The rationals, Q, is such a field.<br />
We shall concentrate on fields that have characteristic 0.<br />
2.1 <strong>Field</strong> Extensions<br />
Say that K is a field that contains F . Then K is an extension field <strong>of</strong> F . We<br />
denote this by K/F and read “K over F.” For example, R is an extension<br />
field <strong>of</strong> Q.
4 The Basics <strong>of</strong> <strong>Field</strong> <strong>Theory</strong><br />
Definition. An element α ∈ K is algebraic over F if α is a root <strong>of</strong> a nonzero<br />
polynomial f(x) ∈ F [x]. For example, √ 2 ∈ R is algebraic over Q since it is<br />
a root <strong>of</strong> x 2 − 2.<br />
Proposition 1. Let α be an algebraic element over the field F and f(x) be its<br />
irreducible polynomial <strong>of</strong> degree n. Then {1, α, . . . , α n−1 } is a basis for F [α] as a<br />
vector space over F .<br />
Pro<strong>of</strong>. See [A, p.495]<br />
The degree <strong>of</strong> a field extension K/F is the dimension <strong>of</strong> K as a vector<br />
space over F and is denoted [K : F ].<br />
A reason to study field extensions is to understand roots <strong>of</strong> polynomials.<br />
The polynomial x 2 + 1, for example, would not have any roots in R.<br />
However, the complex numbers C, would be a field extension <strong>of</strong> R (and<br />
thus an extension <strong>of</strong> Q) that would allow us to find roots <strong>of</strong> x 2 + 1. It is<br />
easy to show that C is a field under normal addition and multiplication, so<br />
every nonzero element in C has an inverse.<br />
We can consider a field K in which the polynomial f(x) has its roots.<br />
We will assume that the polynomial f(x) is irreducible over F [x] since if<br />
there is a root <strong>of</strong> a factor <strong>of</strong> f(x), then it would certainly be a root <strong>of</strong> f(x)<br />
itself.<br />
Let K be a field generated by a single element α over F . That is to say,<br />
every element β in K can be expressed as β = a 0 +a 1 α+a 2 α 2 +· · · a n α n for<br />
some n and some a 0 , a 1 , a 2 , . . . , a n ∈ F. Then we can write K = F (α). We<br />
say K is a simple extension <strong>of</strong> F and α is a primitive element for the extension.
Splitting <strong>Field</strong>s 5<br />
2.2 Splitting <strong>Field</strong>s<br />
Definition. If K is the extension field <strong>of</strong> a field F such that a polynomial<br />
f(x) ∈ F [x] factors completely into linear factors in K[x] (and f(x) does<br />
not factor into linear factors over any proper subfield <strong>of</strong> K that contains<br />
F ), then K is called the splitting field for f(x).<br />
Proposition. In any abstract splitting field for the polynomial x n − 1, the set<br />
<strong>of</strong> the nth roots <strong>of</strong> unity form a group under multiplication.<br />
Pro<strong>of</strong>. If α, β are both roots, α n β n = (αβ) n = 1, so the product αβ is<br />
in the set. Thus we have closure under multiplication. We know that 1<br />
is in our set since 1 n = 1. Given any root α, we can find a multiplicative<br />
inverse α −1 since (α) −1n<br />
= (α n ) −1 = 1 −1 = 1. We also have associativity<br />
because these roots are complex numbers. Thus we have a group under<br />
multiplication.<br />
□<br />
Definition. A primitive nth root <strong>of</strong> unity is a generator <strong>of</strong> the cyclic<br />
group <strong>of</strong> all nth roots <strong>of</strong> unity.<br />
We will denote the chosen primitive nth root <strong>of</strong> unity by ζ n . The other<br />
primitive nth roots <strong>of</strong> unity are then the elements ζn a where 1 ≤ a < n is<br />
an integer relatively prime to n, since these are the other generators for a<br />
cyclic group <strong>of</strong> order n. In particular there are precisely ϕ(n) primitive nth<br />
roots <strong>of</strong> unity, where ϕ(n) denotes the Euler ϕ-function.<br />
The splitting field <strong>of</strong> x n − 1 over Q is the field Q(ζ n ). The field Q(ζ n ) is<br />
called the cyclotomic field <strong>of</strong> nth roots <strong>of</strong> unity.
Chapter 3<br />
<strong>Galois</strong> <strong>Theory</strong><br />
3.1 Introduction<br />
<strong>Galois</strong> theory is the study <strong>of</strong> the roots <strong>of</strong> an irreducible polynomial<br />
f(x) = x n + a n−1 x n−1 + · · · + a 1 x + a 0 .<br />
We will study the symmetries between the roots <strong>of</strong> irreducible polynomials.<br />
A great deal can be understood from expanding the equation<br />
f(x) = (x − α 1 )(x − α 2 ) · · · (x − α n ).<br />
Comparing these two equations, we can see that the sum <strong>of</strong> the roots is<br />
α 1 + · · · + α n = −a n−1 and the product α 1 · · · α n = ±a 0 .<br />
Lagrange and <strong>Galois</strong> were the first mathematicians to see that the relationships<br />
between roots <strong>of</strong> polynomials could be understood in terms <strong>of</strong>
8 <strong>Galois</strong> <strong>Theory</strong><br />
symmetry [A, p.538]. Let us go back to our example <strong>of</strong> the polynomial<br />
x 2 + 1 = 0. We already know that our polynomial has no roots in Q, but it<br />
does have roots in C. Complex conjugation, the first model for the symmetric<br />
relationships we will be studying, gives us the roots ±i and leaves the<br />
rationals fixed. Any quadratic (degree 2) field extension will have similar<br />
symmetries.<br />
As an example, let us look at a degree 2 extension K/F generated by<br />
any α ∈ K which is not in F . Let f(x) = x 2 + bx + c be a polynomial with<br />
coefficients in F such that α is a root <strong>of</strong> f(x). We see that, b = −(α + α ′ ), so<br />
α ′ = −(α + b) is the other root. Over K, the polynomial f(x) splits into the<br />
linear factors (x − α)(x − α ′ ).<br />
Our symmetry comes from the fact that both α and α ′ are roots <strong>of</strong> f(x).<br />
Proposition 2. Let L = F (α) and L ′ = F (α ′ ) be two extension fields <strong>of</strong> F<br />
generated by the algebraic elements α and α ′ . Then there is an isomorphism <strong>of</strong><br />
fields<br />
σ : F (α) ˜→F (α ′ ),<br />
which is the identity on F and sends α → α ′ if and only if the irreducible polynomials<br />
for α and α ′ over F are equal.<br />
Pro<strong>of</strong>. See [A, p.496]<br />
According to the above proposition,<br />
σ : F (α) −→ F (α ′ )<br />
is an isomorphism which sends α → α ′ and is the identity on the field F .<br />
In our case, however, K can be generated by either <strong>of</strong> the two roots. That
Introduction 9<br />
is, L = L ′ = K. Thus, σ is an automorphism <strong>of</strong> K. Since σ is the identity<br />
map on F , b stays fixed (it is equal to −(α + α ′ )). Thus σ must send α to<br />
α ′ . We can conclude that σ 2 is the identity map on K since σ(α) = α ′ and<br />
σ(α ′ ) = α.<br />
Definition. Let K be an extension field. An automorphism <strong>of</strong> K which<br />
is the identity on F is an F-automorphism. It has the property that σ(a) = a<br />
for all a ∈ F . The F -automorphisms <strong>of</strong> K form a group under composition.<br />
The group <strong>of</strong> all F -automorphisms <strong>of</strong> the field extension K is known as the<br />
<strong>Galois</strong> group <strong>of</strong> K over F , denoted G(K/F ).<br />
Theorem 3. For any algebraic field extension K/F, the order <strong>of</strong> the <strong>Galois</strong><br />
group, |G(K/F )|, divides the degree <strong>of</strong> the extension [K : F ].<br />
Pro<strong>of</strong>. See [A, p.540]<br />
We know for any σ ∈ G(K/F ), any a in F will be fixed by σ, that is,<br />
σ(a) = a for all a in F . But a priori, there may be other elements <strong>of</strong> K fixed<br />
by σ.<br />
Definition. If the order <strong>of</strong> the <strong>Galois</strong> group is equal to the degree, then<br />
the field extension K/F is a <strong>Galois</strong> extension.<br />
Definition. The set <strong>of</strong> elements in a field K that are fixed by all the<br />
elements <strong>of</strong> G, a group <strong>of</strong> automorphisms <strong>of</strong> K is known as a fixed field <strong>of</strong><br />
G and is denoted K G .<br />
Corollary 4. Let K/F be a <strong>Galois</strong> extension. If its <strong>Galois</strong> group is G =<br />
G(K/F ), then F is the fixed field <strong>of</strong> G.<br />
Pro<strong>of</strong>. Let L be the fixed field, K G , <strong>of</strong> G. Clearly, F ⊂ L. Since F ⊂ L,<br />
every L-automorphism will also be an F -automorphism. This implies that<br />
G(K/L) is contained in G. However, every element in G is by definition <strong>of</strong>
10 <strong>Galois</strong> <strong>Theory</strong><br />
L an L-automorphism, which implies G(K/L) ⊃ G. We have containment<br />
in both directions, which means G = G(K/L). Since K/F was chosen to be<br />
a <strong>Galois</strong> extension, the order <strong>of</strong> G is equal to the degree <strong>of</strong> the extension K<br />
over F , or |G| = [K : F ]. If we now apply theorem 3, we know that the<br />
degree <strong>of</strong> the extension <strong>of</strong> K over L is divisible by the order <strong>of</strong> G. Since<br />
our original hypothesis was that F is contained in L, we are led to the<br />
conclusion that [K : F ] = [K : L], which implies F = L. Thus, F must also<br />
be a fixed field.<br />
□<br />
This corollary leads us into the following theorem relating fixed fields<br />
and splitting fields.<br />
Theorem 5. Let K be a splitting field <strong>of</strong> a polynomial f(x) whose coefficients<br />
are in the field F . Then K is a <strong>Galois</strong> extension <strong>of</strong> F . The reverse direction is also<br />
true; every <strong>Galois</strong> extension is a splitting field <strong>of</strong> some polynomial f(x) over F .<br />
Pro<strong>of</strong>. See [A, p.541]<br />
Corollary 6. Every finite field extension is contained in a <strong>Galois</strong> extension.<br />
Pro<strong>of</strong>. Let K/F be a finite extension. Suppose the generators <strong>of</strong> K over<br />
F are α 1 , . . . , α n . We need an irreducible polynomial for each α i over F ,<br />
call it f i (x). Let the product f 1 · · · f n be f. Then, let M be a field extension<br />
<strong>of</strong> K that is also a splitting field for f over K. This is done by adjoining any<br />
roots <strong>of</strong> some polynomial f i (x) to K if it was not already in K. If M is a<br />
splitting field for K, then it must also be a splitting field for F . By theorem<br />
5, M is a <strong>Galois</strong> extension. □<br />
Corollary 7. Let K/F be a <strong>Galois</strong> extension. Suppose L is an intermediate<br />
field F ⊂ L ⊂ K. Then K/L is also a <strong>Galois</strong> extension.<br />
Pro<strong>of</strong>. Let K be the splitting field <strong>of</strong> the polynomial f(x) over F . Since
The Main Theorem 11<br />
L ⊂ K, we know K is also a splitting field <strong>of</strong> f(x) over L. Then by theorem<br />
5, K is a <strong>Galois</strong> extension <strong>of</strong> L. □<br />
A critical part in the study <strong>of</strong> <strong>Galois</strong> theory is finding the intermediate<br />
fields L <strong>of</strong> an extension K/F . That is, such that F ⊂ L ⊂ K. We will be<br />
able to conclude from the Main Theorem <strong>of</strong> <strong>Galois</strong> theory that the intermediate<br />
fields <strong>of</strong> a <strong>Galois</strong> extension are in a bijective correspondence with the<br />
subgroups <strong>of</strong> the <strong>Galois</strong> group.<br />
3.2 The Main Theorem<br />
We will now return our attention to the Main Theorem <strong>of</strong> <strong>Galois</strong> <strong>Theory</strong>.<br />
Definition. Given an extension K/F , we will call any field L with F ⊂<br />
L ⊂ K an intermediate field.<br />
Theorem 8 (The Main Theorem <strong>of</strong> <strong>Galois</strong> <strong>Theory</strong>). If K is a <strong>Galois</strong> extension<br />
<strong>of</strong> a field F , and if G = G(K/F ) is its <strong>Galois</strong> group, then the function given<br />
by<br />
H ↦→ K H<br />
is a bijective map from the set <strong>of</strong> subgroups <strong>of</strong> G to the set <strong>of</strong> intermediate fields L.<br />
Its inverse function is defined by<br />
L ↦→ G(K/L).<br />
This corresponds to the property that if H = G(K/L), then [K : L] = |H|, hence<br />
the degree <strong>of</strong> the extension [L : F ] is equal to the index <strong>of</strong> the subgroup [G : H].
12 <strong>Galois</strong> <strong>Theory</strong><br />
Definition. Any two splitting fields K <strong>of</strong> a polynomial f(x) ∈ F [x] are<br />
isomorphic, so the <strong>Galois</strong> group G(K/F ) depends only on f up to isomorphism.<br />
It is called the <strong>Galois</strong> group <strong>of</strong> the polynomial over F .<br />
We are now almost ready to prove the Main Theorem <strong>of</strong> <strong>Galois</strong> theory.<br />
For convenience, the following corollary has been stated:<br />
Corollary 13. Let K/F be a algebraic field extension. The following properties<br />
are equivalent:<br />
(i) K is a <strong>Galois</strong> extension <strong>of</strong> F;<br />
(ii) K is the splitting field <strong>of</strong> an (irreducible) polynomial f(x) ∈ F [x];<br />
(iii) F is the fixed field for the action <strong>of</strong> the <strong>Galois</strong> group G(K/F ) on K;<br />
3.3 Pro<strong>of</strong> <strong>of</strong> the Main Theorem <strong>of</strong> <strong>Galois</strong> <strong>Theory</strong>.<br />
Pro<strong>of</strong> (<strong>of</strong> the Main Theorem). Let K/F be a <strong>Galois</strong> extension and L be an<br />
intermediate field. Let the corresponding subgroup <strong>of</strong> G be H = G(K/L).<br />
We know that H will act trivially on L, by definition. The intermediate field<br />
corresponding to a subgroup H <strong>of</strong> G(K/F ) is the fixed field K H <strong>of</strong> H. So,<br />
L is contained in K H .<br />
The following lemma assists us as the main counting argument in the<br />
pro<strong>of</strong> <strong>of</strong> the Main Theorem.<br />
Lemma 14. Let G be a group <strong>of</strong> automorphisms <strong>of</strong> a field K, and let the order<br />
<strong>of</strong> G be n. Then [K : K G ] = n.<br />
Pro<strong>of</strong>. [A. p. 554]
Pro<strong>of</strong> <strong>of</strong> the Main Theorem <strong>of</strong> <strong>Galois</strong> <strong>Theory</strong>. 13<br />
Since K/F is a <strong>Galois</strong> extension and F ⊂ L ⊂ K, the extension K/L is<br />
a <strong>Galois</strong> extension by Corollary 7. So, the degree <strong>of</strong> the extension [K : L] is<br />
equal to the order <strong>of</strong> H, denoted [K : L] = |H|. Using Lemma 14, we know<br />
that the order <strong>of</strong> H equals the degree <strong>of</strong> [K : K H ], or |H| = [K : K H ]. Since<br />
[K : L] = [K : K H ], and L ⊂ K H , we have L = K H .<br />
Now, suppose we start with H being a subgroup <strong>of</strong> G and we let L be<br />
the fixed field K H . Then H is a subgroup <strong>of</strong> G(K/L) since any element in<br />
H fixes L. That is, H ⊂ G(K/L). Now, the order <strong>of</strong> H is equal to the degree<br />
<strong>of</strong> the field extension from K H to K. But L = K H , so |H| = [K : K H ]. We<br />
know [K : L] is equal to |G(K/L)| since K/L is <strong>Galois</strong>. So we can conclude<br />
that [K : K H ] = [K : L] = |G(K/L)|. This gives us H = G(K/L).<br />
Now, we know that the two maps given by<br />
L ↦→ G(K/L) and H ↦→ K H<br />
are inverses. The Main Theorem <strong>of</strong> <strong>Galois</strong> <strong>Theory</strong>, as you may recall, stated<br />
that if K was a <strong>Galois</strong> extension <strong>of</strong> a field F , and G = G(K/F ) was its <strong>Galois</strong><br />
group, then there is a bijective map H ↦→ K H from the set <strong>of</strong> subgroups<br />
<strong>of</strong> G to the set <strong>of</strong> intermediate fields F ⊂ L ⊂ K. Further, its inverse function<br />
given by L ↦→ G(K/L). Further still, if H = G(K/L), then the degree<br />
<strong>of</strong> H is equal to [K : L], and [L : F ] = [G : H]. Since K/F was chosen to be<br />
a <strong>Galois</strong> extension <strong>of</strong> L, the degree <strong>of</strong> G is [K : F ]. Since |H| = [K : L] and<br />
[K : F ]/[K : L] = [L : K], with |G| being equal to [K : F ] and |H| equal<br />
to [K : L], we know [L : F ] = [G : H], and thus we have proved the Main<br />
Theorem <strong>of</strong> <strong>Galois</strong> <strong>Theory</strong>.<br />
□
14 <strong>Galois</strong> <strong>Theory</strong><br />
3.4 Intermediate <strong>Galois</strong> Extensions<br />
There are many important consequences <strong>of</strong> the relationship given by the<br />
Main Theorem <strong>of</strong> <strong>Galois</strong> theory. One concept is the idea <strong>of</strong> order reversing.<br />
That is to say, if L and L ′ are both intermediate fields and if the corresponding<br />
subgroups are H = (G/L) and H ′ = (K/L ′ ), then L ⊂ L ′ if and only if<br />
H ′ ⊂ H.<br />
Definition. Let L be an intermediate field. An F -automorphism σ <strong>of</strong> K<br />
will map L to some other intermediate field σL, which may or may not be<br />
the same as L. We define σL to be a conjugate subfield.<br />
Theorem 15. Let K/F be a <strong>Galois</strong> extension and L an intermediate field. Let<br />
H = G(K/L) be the corresponding subgroup <strong>of</strong> G = G(K/F ). Then we know<br />
the following are true:<br />
(a) Let σ ∈ G. The subgroup <strong>of</strong> G corresponding to the conjugate subfield<br />
σL is the conjugate subgroup σHσ −1 . That is to say, G(K/σL) =<br />
σHσ −1 .<br />
(b) L is a <strong>Galois</strong> extension <strong>of</strong> F if and only if H is a normal subgroup <strong>of</strong><br />
G. When H is a normal subgroup <strong>of</strong> G, then G(L/F ) is isomorphic to the<br />
quotient group G/H.<br />
Pro<strong>of</strong>. (a) Say that σL = L ′ and that τ ∈ H = G(K/L). Then στσ −1 ∈<br />
H ′ = G(K/L ′ ). To show this, we will need to prove that στσ −1 fixes any<br />
element in L ′ . Let α ′ ∈ L ′ . Then by definition, α ′ = σ(α) for some corresponding<br />
α ∈ L. Then στσ −1 (α ′ ) = στσ −1 (σ(α)) = στ(σ −1 )(σ)(α) =<br />
στ(α) = σ(α) = α ′ . Thus, στσ −1 fixes any element in L ′ . It follows that<br />
σHσ −1 ⊂ H ′ and, by symmetry with σ −1 and H −1 , we have σ −1 H ′ σ ⊂ H
Intermediate <strong>Galois</strong> Extensions 15<br />
if and only if H ′ ⊂ σHσ −1 , which gives us our equality H ′ = σHσ −1 .<br />
(b) ⇒ Suppose L/F is a <strong>Galois</strong> extension. Then L is a splitting field for<br />
some polynomial f(x) with coefficients in F . That is to say, L = F (β 1 , . . . , β n ),<br />
where the β i are roots <strong>of</strong> the polynomial f(x) in K. An F -automorphism σ<br />
<strong>of</strong> K permutes these roots and thus maps L to itself: σL = L. By (a) above,<br />
H = σHσ −1 . Thus, H is a normal subgroup.<br />
⇐ Let H be normal. Then for all σ ∈ G, we know H = σHσ −1 .<br />
So, G(K/L) = G(K/σL).<br />
This implies that for all σ, the composition<br />
σL = L. Every F -automorphism <strong>of</strong> K maps L to itself; it defines an F -<br />
automorphism <strong>of</strong> L by restriction. The restriction defines a homomorphism<br />
mapping G to G(L/F ), whose kernel is the set <strong>of</strong> all σ ∈ G which induces<br />
the identity on L, which we know to be H. We now can conclude that G/H<br />
is isomorphic to a subgroup <strong>of</strong> G(L/F ). We find that<br />
[L : F ] = |G/H| ≤ |G(L/F )|.<br />
Thus we can conclude that L is a <strong>Galois</strong> extension and G/H ≈ G(L/F ). □<br />
Definition. An extension K/F is an abelian extension if K/F is <strong>Galois</strong><br />
and the <strong>Galois</strong> group G(K/F ) is an abelian group.<br />
It is known that all subgroups <strong>of</strong> abelian groups are in fact normal subgroups.<br />
Further, we know all cyclic groups are abelian, but it is not necessarily<br />
true that all abelian groups are cyclic. We will now concentrate on<br />
extensions with cyclic <strong>Galois</strong> groups.
Chapter 4<br />
<strong>Cyclotomic</strong> <strong>Field</strong> Extensions<br />
4.1 Introduction<br />
Definition. Recall from Chapter 2 that for any n ∈ N, the nth cyclotomic field<br />
is a subfield K <strong>of</strong> the complex numbers generated over Q by ζ n = e 2πi/n .<br />
We will denote ζ n simply as ζ when no confusion will arise. A cyclotomic<br />
extension <strong>of</strong> Q is Q(ζ n ) and it is the splitting field <strong>of</strong> x n − 1. The roots <strong>of</strong><br />
the polynomial are the powers <strong>of</strong> ζ, the nth roots <strong>of</strong> unity: 1, ζ, ζ 2 , . . . , ζ n−1 .<br />
Here we will concentrate on the cases when n is a prime number p.<br />
We know that we can factor the polynomial<br />
x p − 1 = (x − p)(x p−1 + x p−2 + · · · + x + 1),<br />
and the polynomial (x p−1 + x p−2 + · · · + x + 1) is irreducible over Q, and<br />
ζ = ζ p is one <strong>of</strong> its roots; it is the irreducible polynomial for ζ over Q and<br />
the roots are powers <strong>of</strong> ζ, ζ 2 , . . . , and ζ p−1 . Thus the order <strong>of</strong> the <strong>Galois</strong>
18 <strong>Cyclotomic</strong> <strong>Field</strong> Extensions<br />
group G(Q(ζ)/Q) is p − 1.<br />
4.2 Structure <strong>of</strong> the <strong>Galois</strong> Group<br />
Theorem 16. Let p ∈ Z be prime and ζ = ζ p . Then the <strong>Galois</strong> group <strong>of</strong> Q(ζ) over<br />
Q is isomorphic to the multiplicative group (Z/pZ) × <strong>of</strong> nonzero elements <strong>of</strong> the<br />
ring Z/pZ and is a cyclic group <strong>of</strong> order p − 1.<br />
Pro<strong>of</strong>. Let G be the <strong>Galois</strong> group <strong>of</strong> Q(ζ) over Q. We will need a map<br />
v : G → (Z/pZ) × as follows: Let σ be an automorphism in G. Then σ will<br />
carry ζ to another root <strong>of</strong> the polynomial x p + · · · + x + 1, call it ζ i . Since<br />
ζ has multiplicative order p, our exponent i will be an integer modulo p.<br />
We set v(σ) = i. We need to verify that v is multiplicative. If τ is another<br />
element <strong>of</strong> G such that v(τ) = j, that is, τ(ζ) = ζ j , then<br />
στ(ζ) = σ(ζ j ) = σ(ζ) j = ζ ij = ζ ij .<br />
The identity automorphism will send ζ to ζ, so v(id) = 1. We know that<br />
v is a homomorphism to (Z/pZ) × since v is multiplicative and v(σ) does<br />
not equal 0 for any σ. The homomorphism is injective since ζ generates<br />
Q(ζ) and the action <strong>of</strong> an automorphism is determined when we know<br />
its action on ζ. Then we know G is isomorphic to its image in (Z/pZ) × .<br />
This means that G is cyclic. The order <strong>of</strong> G is equal to that <strong>of</strong> (Z/pZ) × , or<br />
|G| = |(Z/pZ) × |, which equals p − 1. These two groups are isomorphic. □<br />
Since (Z/pZ) × is a cyclic group, so is every subgroup. The <strong>Galois</strong> group<br />
G <strong>of</strong> K = Q(ζ p ) over Q is a cyclic group <strong>of</strong> order p − 1, we know G has one
Group Generators 19<br />
subgroup <strong>of</strong> order k for each k ∈ N which divides p − 1. If (p − 1)/k = r<br />
and if σ is the generator <strong>of</strong> G, then the subgroup <strong>of</strong> order k is generated<br />
by σ r . Applying the Main Theorem <strong>of</strong> <strong>Galois</strong> theory, we know there will be<br />
exactly one intermediate field L with [L : Q] = r. These fields are generated<br />
by certain sums <strong>of</strong> powers <strong>of</strong> ζ = ζ p .<br />
Example. Consider the case when p = 19 and ζ = ζ 19 . Then by the<br />
above theorem, we know that the <strong>Galois</strong> group <strong>of</strong> Q(ζ 19 ) is isomorphic to<br />
(Z/19Z) × and that it is a cyclic group <strong>of</strong> order 18, which we shall denote<br />
C 18 . Then, we know that the intermediate fields will correspond to the nontrivial<br />
divisors <strong>of</strong> 18, i.e. 2, 3, 6, and 9, and Q and Q(ζ 19 ) will correspond to<br />
the trivial divisors <strong>of</strong> 18, namely 1 and 18, respectively. The Main Theorem<br />
<strong>of</strong> <strong>Galois</strong> theory then tells us there will be exactly one intermediate subfield<br />
generated by each <strong>of</strong> these divisors.<br />
4.3 Group Generators<br />
A group is cyclic if there is an element a in G which generates G. That is,<br />
if there exists some a ∈ G such that G = {a n |n ∈ Z}. Let us concentrate<br />
on the G = (Z/19Z) × . Through trial and error, we can find the generators<br />
<strong>of</strong> this group and its subgroups. First, we must find the generator <strong>of</strong> the<br />
entire group. Table 4.1 illustrates one such choice <strong>of</strong> generator.<br />
We can see that every element is generated by a = 2 and the order is:<br />
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 .<br />
Now that we know 2 is the generator <strong>of</strong> the entire group, we can find the<br />
generators <strong>of</strong> the intermediate subfields.
20 <strong>Cyclotomic</strong> <strong>Field</strong> Extensions<br />
a = 2 is a generator for G:<br />
a = 2 a 6 = 7 a 11 = 15 a 16 = 5<br />
a 2 = 4 a 7 = 14 a 12 = 11 a 17 = 10<br />
a 3 = 8 a 8 = 9 a 13 = 3 a 18 = 1<br />
a 4 = 16 a 9 = 18 a 14 = 6 a 19 = 2<br />
a 5 = 13 a 10 = 17 a 15 = 12<br />
Table 4.1: <strong>Field</strong> Generator for Q(ζ 19 )<br />
a 9 is a generator: a 6 is a generator: a 3 is a generator: a 4 is a generator:<br />
a 9 = 18 a 6 = 7 a 3 = 8 a 4 = 16<br />
a 18 = 1 a 12 = 11 a 6 = 7 a 8 = 9<br />
a 18 = 1 a 9 = 18 a 12 = 11<br />
a 12 = 11 a 16 = 5<br />
a 15 = 12 a 2 = 4<br />
a 18 = 1 a 6 = 7<br />
a 10 = 17<br />
a 14 = 6<br />
a 18 = 1<br />
Table 4.2: Subfield Generators<br />
Let us name the above subgroups as follows. Since a 9 a subgroup <strong>of</strong><br />
order 2, we will denote this subgroup <strong>of</strong> C 18 as C 2 . Similarly, a 6 generates<br />
a subgroup <strong>of</strong> order 3, denoted C 3 . Next, a 3 generates a subgroup <strong>of</strong> order<br />
6, so we will denote this subgroup C 6 . Finally, a 4 generates a subgroup <strong>of</strong><br />
order 9, so this subgroup will be denoted C 9 .<br />
The subgroups <strong>of</strong> the cyclic group C 18 correspond to H 2 , H 3 , H 6 , and<br />
H 9 , which are subgroups <strong>of</strong> the group G with powers <strong>of</strong> σ as generators.<br />
For example, C 2 = {e, a 9 } corresponds to H 2 = {id, σ 9 }.<br />
Figure 4.1 gives a lattice representation <strong>of</strong> the subgroups. The lines in-
Subgroups and Fixed <strong>Field</strong>s 21<br />
dicate containment. For example, the line connecting H 2 and H 6 represents<br />
[H 6 : H 3 ] = 3<br />
3<br />
{e}<br />
2<br />
H 3<br />
H 2<br />
3 2 3<br />
H 9<br />
2 3<br />
H 6<br />
G<br />
Figure 4.1: Group Lattice for Q(ζ 19 )<br />
4.4 Subgroups and Fixed <strong>Field</strong>s<br />
The intermediate fields are generated by sums <strong>of</strong> the powers <strong>of</strong> ζ = ζ 19 .<br />
We have found that a 3 yields a subgroup <strong>of</strong> order 6, a 4 one <strong>of</strong> order 9, a 6<br />
<strong>of</strong> order 3, and finally a 9 generates a subgroup <strong>of</strong> order 2. Thus our corresponding<br />
fixed fields will have degrees 6, 9, 3, and 2 over Q, respectively.<br />
We have found a = 2 to be a generator <strong>of</strong> (Z/19Z) × . This corresponds to<br />
σ ∈ G(Q(ζ 19 )/Q), where σ maps ζ to ζ 2 . We can now find generators in the<br />
field in terms <strong>of</strong> ζ 19 .<br />
We know that C 2 = {e, a 9 } corresponds to H 2 = {id, σ 9 }. Consider<br />
ζ + σ 9 ζ.
22 <strong>Cyclotomic</strong> <strong>Field</strong> Extensions<br />
Then<br />
σ 9 (ζ + σ 9 ζ) = σ 9 ζ + σ 18 ζ = σ 9 ζ + ζ.<br />
So<br />
ζ + σ 9 ζ = ζ + ζ 29 = ζ + ζ −1<br />
is fixed by all <strong>of</strong> H 2 . Thus, H 2 fixes Q(ζ + ζ −1 ). Since no other σ in G fixes<br />
ζ + ζ −1 , this is the fixed field <strong>of</strong> H 2 . That is, Q(ζ + ζ −1 ) is the fixed field <strong>of</strong><br />
H 2 .<br />
Next, consider the field H 3 which corresponds to C 3 = {e, a 6 , a 12 }. Since<br />
H 3 corresponds to C 3 , we know H 3 = {id, σ 6 , σ 12 }. Now we can consider<br />
ζ + σ 6 ζ + σ 12 ζ. We can check that every element in H 3 fixes ζ + σ 6 ζ + σ 12 ζ.<br />
First, we see σ 6 fixes every element.<br />
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.<br />
Then, we find that σ 12 also fixes every element.<br />
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.<br />
So this means ζ + σ 6 ζ + σ 12 ζ = ζ + ζ 26 + ζ 212 = ζ + ζ 7 + ζ 11 is fixed by all <strong>of</strong><br />
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 )<br />
is the fixed field <strong>of</strong> H 3 .<br />
Then we can look at H 6 , which corresponds to C 6 . The elements <strong>of</strong> H 6 =<br />
{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 <strong>Field</strong>s 23<br />
Then this means<br />
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<br />
= ζ + ζ 8 + ζ 7 + ζ 18 + ζ 11 + ζ 12<br />
is fixed by all <strong>of</strong> H 6 and Q(ζ + ζ 8 + ζ 7 + ζ 18 + ζ 11 + ζ 12 ) is the fixed field<br />
<strong>of</strong> H 6 .<br />
Finally, we consider the corresponding field for C 9 , namely H 9 . The<br />
elements <strong>of</strong> H 9 = {id, σ 2 , σ 4 , σ 6 , σ 8 , σ 10 , σ 12 , σ 14 , σ 16 } fix all elements in<br />
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.<br />
This means<br />
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<br />
= ζ + ζ 22 + ζ 24 + ζ 26 + ζ 28 + ζ 210 + ζ 212 + ζ 214 + ζ 216<br />
= ζ + ζ 4 + ζ 16 + ζ 7 + ζ 9 + ζ 17 + ζ 11 + ζ 6 + ζ 5<br />
is fixed by all <strong>of</strong> H 9 . Again, no other σ in G fixes<br />
ζ + ζ 4 + ζ 16 + ζ 7 + ζ 9 + ζ 17 + ζ 11 + ζ 6 + ζ 5 ,<br />
so Q(ζ + ζ 4 + ζ 16 + ζ 7 + ζ 9 + ζ 17 + ζ 11 + ζ 6 + ζ 5 ) is the fixed field <strong>of</strong> H 9 .<br />
This can be more easily understood in the form <strong>of</strong> a field extension lattice.<br />
See Figure 4.2<br />
In the lattice representation <strong>of</strong> the field extension, we can clearly see the
24 <strong>Cyclotomic</strong> <strong>Field</strong> Extensions<br />
Q(ζ 19 )<br />
3<br />
2<br />
Q(ζ + ζ 7 + ζ 11 )<br />
Q(ζ + ζ −1 )<br />
3 2<br />
3<br />
Q(ζ + ζ 4 + ζ 5 + ζ 6 + ζ 7 + ζ 9 + ζ 11 + ζ 16 + ζ 17 )<br />
2<br />
Q(ζ + ζ 7 + ζ 8 + ζ 12 + ζ 18 )<br />
3<br />
Q<br />
Figure 4.2: <strong>Field</strong> Lattice for Q(ζ 19 )<br />
degree 2 and degree 3 extensions from Q to Q(ζ 19 ). Fortunately, this is not<br />
all that is known about the subfields. In the following section, we will get<br />
a better understanding <strong>of</strong> these two intermediate fields.<br />
4.5 Real Subfields and Quadratic Extensions<br />
Proposition 17. The subfield L max <strong>of</strong> K = Q(ζ p ) whose degree over Q is 1 2 (p−1)<br />
is generated over Q by the element η = ζ +ζ p−1 = 2 cos 2π/p. Moreover, L max =<br />
K ∩ R, so L max is called the maximal real subfield <strong>of</strong> K.<br />
Pro<strong>of</strong>. Consider the quadratic polynomial x 2 − ηx + 1, which has coefficients<br />
in Q(η) and η = ζ + ζ −1 . Note that ζ is a root. So we know that
Real Subfields and Quadratic Extensions 25<br />
[K : Q(ζ)] ≤ 2. Since η is a real number and ζ is not, Q(η) ⊂ K. So,<br />
[K : Q(η)] = 2, and Q(η) = K ∩ R. Thus, [Q(η) : Q] = 1 2<br />
(p − 1). □<br />
Thus in our example <strong>of</strong> Q(ζ 19 ), we can simplify our representation <strong>of</strong><br />
our real subfield <strong>of</strong> Q(ζ 19 ):<br />
( )<br />
However, if 2 cos 2π<br />
19<br />
( ( )) 2π<br />
Q(ζ + ζ −1 ) = Q 2 cos .<br />
19<br />
( )<br />
is in L max , then certainly cos 2π<br />
19<br />
will also be in the<br />
field. So we can simplify our representation <strong>of</strong> the real subfield further by<br />
( ( ))<br />
simply writing it as L max = Q cos 2π<br />
19<br />
. The maximal real subfield is just<br />
a degree 2 subfield <strong>of</strong> Q(ζ 19 ). In other words, [Q(ζ 19 ) : L max ] = 2.<br />
4.5.1 Discriminant <strong>of</strong> a Quadratic Extension<br />
Given any f(x) ∈ Q[x] and a splitting field K, we write f(x) = (x −<br />
α 1 ) · · · (x − α n ). Since K is a splitting field for f(x), all the roots <strong>of</strong> f(x)<br />
are going to be in K. Note that any σ ∈ G(K/Q) permutes these roots.<br />
Definition.<br />
The discriminant <strong>of</strong> a polynomial f(x) = (x − α 1 )(x −<br />
α 2 ) · · · (x − α n ) is defined to be<br />
D = (α 1 − α 2 ) 2 (α 1 − α 3 ) 2 · · · (α n−1 − α n ) 2<br />
= ∏ (α i − α j ) 2 = ± ∏ i − α j ).<br />
i
26 <strong>Cyclotomic</strong> <strong>Field</strong> Extensions<br />
Q contained in the cyclotomic field Q(ζ 19 ). Then<br />
L quad = Q( √ −19),<br />
where the sign was determined (−1) (p−1)/2 = (−1) 9 = −1.<br />
Pro<strong>of</strong>. We shall use the discriminant <strong>of</strong> the polynomial to find a generator<br />
for L quad . Let D be the discriminant <strong>of</strong> the polynomial<br />
x 19 − 1.<br />
This discriminant can be computed directly in terms <strong>of</strong> the roots {1, ζ, ζ 2 , . . . , ζ 18 },<br />
but the following lemma gives us a formula which is easier to use:<br />
Lemma 19. Let f(x) = (x − α 1 ) · · · (x − α n ). The discriminant <strong>of</strong> f is<br />
D = ±f ′ (α 1 ) · · · f ′ (α n ) = ± ∏ i<br />
f ′ (α i ),<br />
where f ′ is the derivative <strong>of</strong> f.<br />
Pro<strong>of</strong> <strong>of</strong> Lemma. By the product rule <strong>of</strong> differentiation,<br />
n∑<br />
f ′ (x) = (x − α 1 ) · · · (x − α i−1 )(x − α i+1 ) · · · (x − α n ).<br />
i=1<br />
Setting x = α i , we get<br />
f ′ (α i ) = (α i − α 1 ) · · · (α i − α i−1 )(α i − α i+1 ) · · · (α i − α n ). (4.1)<br />
i ≠ j.<br />
This is the product <strong>of</strong> the differences (α i − α j ), with the given i and
Real Subfields and Quadratic Extensions 27<br />
Thus,<br />
∏<br />
f ′ (α i ) = ∏ i − α j ) = ±D.<br />
i<br />
i≠j(α<br />
When we apply this lemma to the polynomial x 19 − 1, its derivative is<br />
19x 18 . The discriminant is given by<br />
□<br />
±D =<br />
18∏<br />
i=0<br />
19ζ i(18) = ζ N 19 19 ,<br />
where the exponent N is some integer. To determine ζ N , we note that ±D<br />
is a rational number, because x p − 1 ∈ Q[x] are rational. The reason D<br />
is rational is because it is in the splitting field K over Q. Because D was<br />
symmetrically defined, it is clearly in the fixed field <strong>of</strong> the whole <strong>Galois</strong><br />
group. That is, any Q-automorphism <strong>of</strong> K must permute the roots <strong>of</strong> f;<br />
any permutation <strong>of</strong> roots will fix D. However, the fixed field K G where G<br />
is the whole <strong>Galois</strong> group G(K/Q) is itself just Q. Thus, D is in Q. The only<br />
power <strong>of</strong> ζ which is rational is 1. So, ζ N = 1 and<br />
±D = 19 19 .<br />
The sign <strong>of</strong> the discriminant is determined to be negative by carefully<br />
comparing the definition <strong>of</strong> the derivitive and Equation 4.1. So the square<br />
root <strong>of</strong> this discriminant is δ = √ −19 19 . By definition <strong>of</strong> D, it is clear that δ<br />
in the field Q(ζ). Since the square factors can be pulled out <strong>of</strong> a square root,
28 <strong>Cyclotomic</strong> <strong>Field</strong> Extensions<br />
that is, √ −19 19 = 19 9√ −19, so we see<br />
Q(δ) = Q( √ −19).<br />
This gives us a quadratic subfield <strong>of</strong> Q(ζ). However, L is the only quadratic<br />
subfield, which means Q(δ) = L.<br />
□<br />
Note that the above theorem can be applied to any odd prime.<br />
Applying this theorem to Q(ζ 19 ), we can represent our unique quadratic<br />
extension <strong>of</strong> Q as:<br />
Q(ζ + ζ 4 + ζ 5 + ζ 6 + ζ 7 + ζ 9 + ζ 11 + ζ 16 + ζ 17 ) = Q( √ −19)<br />
Then if we re-visit our field lattice extension, we see<br />
Notice that L quad is not contained in L max . This corresponds to the fact<br />
that (p − 1)/2 is odd. Had it been even, then our discriminant D would<br />
have been a positive value, making δ real. Then, L quad would have to be a<br />
subfield <strong>of</strong> L max .<br />
The reader might have noticed that the prime 19 was <strong>of</strong> the form 19 =<br />
2·3 2 +1, or rather twice the square <strong>of</strong> a prime plus one. Let us now see how<br />
such a prime differs from a prime <strong>of</strong> the form 2 · p 1 · p 2 + 1, or twice two<br />
unique primes plus one. We will briefly explore the field extension Q(ζ 31 )<br />
over Q.<br />
Example. The subfields will correspond to the subgroups <strong>of</strong> (Z/31Z) × .<br />
The generator <strong>of</strong> this group is a = 3 and the generator for this cyclic group<br />
then the automorphism σ = σ 3 which maps ζ 31 to ζ31 3 . The subgroups <strong>of</strong> G
Real Subfields and Quadratic Extensions 29<br />
Q(ζ 19 )<br />
3<br />
2<br />
Q(ζ + ζ 7 + ζ 11 )<br />
( ( ))<br />
Q cos 2π<br />
19<br />
3 2<br />
3<br />
Q( √ −19)<br />
Q(ζ + ζ 7 + ζ 8 + ζ 12 + ζ 18 )<br />
2<br />
3<br />
Q<br />
Figure 4.3: Simplified <strong>Field</strong> Lattice for Q(ζ 19 )<br />
are<br />
H 2 = 〈σ 15 〉 = {1, σ 15 };<br />
H 3 = 〈σ 10 〉 = 〈σ 20 〉 = {1, σ 10 , σ 20 };<br />
H 5 = 〈σ 6 〉 = 〈σ 12 〉 = 〈σ 18 〉 = 〈σ 24 〉 = {1, σ 6 , σ 12 , σ 18 , σ 24 };<br />
H 6 = 〈σ 5 〉 = 〈σ 25 〉 = {1, σ 5 , σ 10 , σ 15 , σ 20 , σ 25 };<br />
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 };<br />
H 15 = 〈σ 2 〉 = 〈σ 4 〉 = 〈σ 8 〉 = 〈σ 14 〉 = 〈σ 16 〉 = 〈σ 22 〉 = 〈σ 26 〉 = 〈σ 28 〉 =<br />
{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 , }<br />
having orders 2, 3, 5, 10, and 15, respectively. The group lattice for Q(ζ 31 )<br />
would then be as shown in Figure 4.4. A degree 2 extension is represented<br />
with a red line, a degree 3 extension with a blue line, and a degree 5 exten-
30 <strong>Cyclotomic</strong> <strong>Field</strong> Extensions<br />
sion with a black line.<br />
{e}<br />
H 3 H 5 H 2<br />
H 15 H 6 H 10<br />
C 30<br />
Figure 4.4: Group Lattice for Q(ζ 31 )<br />
We then will go back and apply what we know about the real subfield<br />
and the unique quadratic extension to our lattice. The real subfield, ζ + ζ −1<br />
( )<br />
can be replaced by cos 2π<br />
31<br />
and the quadratic extension by Q( √ −31). Our<br />
field lattice is then shown in Figure 4.5. Once again, a degree 2 extension<br />
is represented with a red line, a degree 3 extension with a blue line, and a<br />
degree 5 extension with a black line.
Statement <strong>of</strong> the Kronecker-Weber Theorem 31<br />
Q(ζ 31 )<br />
( ( ))<br />
Q(ζ + ζ 5 + ζ 25 ) Q(ζ + ζ 2 + ζ 4 + ζ 8 + ζ 16 ) Q cos 2π<br />
31<br />
Q( √ −31)<br />
Q(ζ + ζ −1 + ζ 5<br />
+ζ 6 +ζ 25 +ζ 26 )<br />
Q(ζ + ζ −1 + ζ 2 + ζ 4 + ζ 8<br />
+ ζ 15 + ζ 16 + ζ 23 + ζ 27 + ζ 29 )<br />
Q<br />
Figure 4.5: <strong>Field</strong> Lattice for Q(ζ 31 )<br />
4.6 Statement <strong>of</strong> the Kronecker-Weber Theorem<br />
Recall at the end <strong>of</strong> Chapter 3, we defined an abelian extension K/F to be<br />
an extension whose <strong>Galois</strong> group G(K/F ) was an abelian group. Having<br />
studied the cyclotomic fields, we conclude with this surprising result.<br />
Theorem 20. Every <strong>Galois</strong> extension K <strong>of</strong> Q whose <strong>Galois</strong> group is<br />
abelian is contained in one <strong>of</strong> the cyclotomic fields Q(ζ n ).<br />
The pro<strong>of</strong> <strong>of</strong> this theorem is beyond the scope <strong>of</strong> this thesis, however, L.<br />
Washington presents a pro<strong>of</strong> in his book Introduction to <strong>Cyclotomic</strong> <strong>Field</strong>s.
Bibliography<br />
Artin, M. (1991). Algebra. Prentice-Hall, Inc., Upper Saddle River, New<br />
Jersey.<br />
Dummit, D. S. and Foote, R. M. (1991). Abstract Algebra. Prentice-Hall, Inc,<br />
Englewood Cliffs, New Jersey.<br />
Gallian, J. A. (2002). Contemporary Abstract Algebra. Houghton Mifflin Company,<br />
Boston, Massachusetts.<br />
Washington, L. C. (1982). Introduction to <strong>Cyclotomic</strong> <strong>Field</strong>s. Number 83 in<br />
Graduate Texts in Mathematics. Springer-Verlag, New York.