Class Field Theory

Anna Haensch

Spring 2012

These are my own notes put together from a reading of “Class Field Theory” by N. Childress [1],along with other references, [2], [4], and [6].

1 Goals and Origins of Class Field Theory

First we introduce some notation:

• K/Q a finite extension.

• OK the ring of algebraic integers over K.

• a a fractional ideal, i.e. a finitely generated OK-module with generators in K.

• A = AK is the multiplicative group of fractional ideals, OK = (1) is the identity element.

If a and b are fractional ideals, with

a = (α1, α2, ..., αt) and b = (β1, β2, ..., βs)

with αi, βj ∈ K, thenab = (..., αiβj , ...)

is the OK-module generated by the products of the generators of a and b. By the arithmetic of K,we mean the study of K, its subgroups, its factor groups, groups isomorphic to subgroups of K,and certain ideals of K. Motivated by this notion have the following goals of class field theory:

1. Describe all abelian extensions of K in terms of the arithmetic of K.

2. Find a canonical way to describe Gal(L/K) in terms of the arithmetic of K, whenever L/Kis abelian.

3. Describe the decomposition of a prime ideal from K to L in terms of the arithmetic of K,whenever L/K is abelian.

1.1 Some Terminology

We will briefly review some concepts from algebraic number theory:

• L/K is a degree n Galois extension.

• G := Gal(L/K).

1

• OK (resp. OL) is the ring of algebraic integers of K (resp. L).

• p is a prime ideal in OK

• Pi is a prime ideal in OL.

L

K

OL

OK p

P1e1 · · ·Pg

eg

In this case we use the following equivalent statements:

Pi divides p⇔ Pi | p⇔ P occurs in the factorization of pOL

Definition 1. ei is called the ramification index of Pi over p, and the degree of the field extension

fi = [OL/Pi : OK/p]

is called the inertia degree (or residue degree) of Pi over p.

The following can be found in [5, Chapter 1,§9].

Proposition 1.1. G acts transitively on the set of Pi lying a bove p.

Proof. Let P := P1. Suppose for some i, that Pi 6= σP for any σ ∈ G. By the Chinese Remaindertheorem, there exist x ∈ OL such that

x ≡ 0 mod Pi and x ≡ 1 mod σP

for all σ ∈ G. Then,

NL/K(x) =∏σ∈G

σ(x) ∈ Pi ∩ OK = p.

On the other hand, x 6∈ σP for any σ ∈ G, hence σ(x) 6∈ P for any σ ∈ G, consequently∏σ∈G6∈ P ∩Ok = p,

a contradiction.

So, Pi = σiP for some suitable σi ∈ G. Induced by σi, we have the following isomorphism

σi : OL/P −→ OL/σiPa mod P 7−→ σi(a) mod σi(P),

so thatfi = [OL/σi(P) : OK/p] = [OL/P : OK/p] = f1

2

for i = 1, ..., g. Furthermore, σi(pOL) = pOL,

Pν | pOL ⇔ σi(Pν) | σi(pOL)⇔ (σiP)ν | pOL,

so ei = e1 for all i = 1, ..., g. So in the Galois case,

pOL = (P1 · · ·Pg)e,

and the fundamental identity states

efg = n = [L : K].

Definition 2. Suppose L/K is Galois, and p splits from K to L into (P1 · · ·Pg)e, we say

“p is ramified in L”⇐⇒ e 6= 1

and“p is unramified in L”⇐⇒ e = 1

Definition 3. p splits completely from K to L if g = n.

This means, e = f = 1, so in particular p is unramified in L. Why do we care about thisparticular property?

Theorem 1.2. (Inclusion Theorem) L1, L2 Galois extenions on K. S1, S2 the set of primes whichsplit completely from K to L1, L2. Then

S1 ⊆∗ s2 ⇐⇒ L1 ⊇ L2

andS1 =∗ s2 ⇐⇒ L1 = L2

(⊆∗ means containment with finitely many exceptions).

So primes that split completely ”capture” (or describe) the Galois extension.

1.2 The Historical Setting for Class Field Theory

Gauss (1777-1855):

When does x2 − a ≡ 0 mod p have a solution x ∈ Z?

here p is a prime with p - a.

Theorem 1.3. (Gauss’ Theorem of Quadratic Reciprocity) If p.q are odd prime p, q - a, and p ≡ qmod 4a, then x2 − a ≡ 0 mod p has a solution if and only if x2 − a ≡ 0 mod q has a solution.

This is more concisely expressed using the Legendre symbol which is defined:(ap

)=

{1 if x2 − a ≡ 0 mod p is solvable in Z−1 otherwise

so Theorem 1.3 says: (ap

)= 1⇐⇒

(aq

)= 1.

3

Theorem 1.4. Let L = Q(√d) where d is a square free integer. p is an odd prime, p - d.(d

p

)= 1⇐⇒ p splits completely from Q to L.

Gauss relates the decomposition of primes, namely primes that split completely, to congruenceconditions in Z.

Example 1. Find all primes which split completely from Q to Q(√

2).

Q(√

2)

Q

OQ(√2)

ZFirst, which primes ramify? From a theorem of Minkowski,

p ramifies in OL ⇐⇒ p | dL

(See [3, Chapter 2]). Recall, if OL has the basis {ω1, ..., ωn}, then

dL =dOL=d({ω1, ..., ωn})

=∣∣σi(ωj)∣∣2 .

So, OQ(√2) = {a+ b

√2 : a, b ∈ Z} = Z +

√Z, and

dQ(√2) =

∣∣∣∣1 √2

1 −√

2

∣∣∣∣2 = (−√

2−√

2)2 = 8.

So only (2) ramifies in Q(√

2). Now, of the unramified primes, which split completely? For all oddprimes, p,

p ≡ 1, 3, 5, or 7 mod 8

(and notice that 1 ≡ 17 mod 8).

•(

217

)= 1 since 62 − 2 = 36− 2 = 34 ≡ 0 mod 17.

•(27

)= 1 since 32 − 2 = 9− 2 = 7 ≡ 7 mod 8.

•(23

)= −1 and

(25

)= −1.

So, p ≡ 1, 7 mod 8, then by Theorem 1.3(2p

)= 1, and by Theorem 1.4 p splits completely from

Q to Q(√

2). So the following split completely:

1, 1 + 8, 1 + 2 · 8, ...

4

and7, 7 + 8, 7 + 2 · 8, ...

and these progressions will contain infinitely many primes. So infinitely many ideals split com-pletely, and there is a clear relationship between the decomposition of ideals, and congruence con-ditions in Z.

Kronecker (1821-1891): Observed the following correspondence

Elliptic Curves

m

Automorphic Forms

⇓

Abelian extensions of imaginary quadratic number fields given by adjoining certain values of theautomorphic form.

Does this method give us all abelian extensions of a given number field (Kronecker’s Jugend-traum)? The followingt was conjectured by Kronecker and proved by Weber:

Theorem 1.5. (Kronecker-Weber) Every abelian extension of Q is contained in a cyclotomic ex-tension of Q.

Hilbert (1900): At the Paris, ICM, Hilbert posed the following two problems, which are the twomain questions of class field theory:

• Hilbert’s 9th: To develop the most general reciprocity law in an arbitrary number field,generalizing Gauss’ law of quadratic reciprocity.

• Hilbert’s 12th: To Generalize Kronecker’s Jugendtraum.

2 Dirichlet’s Theorem on Primes in Arithmetic Progression

2.1 Characters of Finite Abelian Groups

We will now begin a discussion of characters of finite abelian groups, which will lead us to Dirichletcharacters, and eventually a proof of Dirichlet’s Theorem on Primes in Arithmetic Progression.

Definition 4. Suppose G is a finite abelian group. A character, χ, of G is a multiplicative grouphomomorphism

χ : G −→ C×.

The character group, denoted G, is the set of all characters on G. For any χ, ψ ∈ G,

χψ : G −→ C×

g 7−→ χ(g)ψ(g)

defines multiplication in the group G, and

χ0 : G −→ C×

g 7−→ 1

is the trivial character.

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Proposition 2.1. If G is a finite abelian group, then G ∼= G.

Proof. Since G is a finite abelian group, it is a direct sum of Z/mZ, where m ∈ Z. So, G is a

product of Z/mZ, and for any χ ∈ Z/mZ,

χ : Z/mZ −→ C×

1 7−→ χ(1).

Since Z/mZ is an additive group generated by 1, χ is completely determined by χ(1). Further,only χ0 sends 1 to the multiplicative identity in C×, so we have the following homomorphism

Z/mZ −→ C×

χ 7−→ χ(1)

which has trivial kernel, and hence is injective. Since Z/mZ is finite and additive with idenitity 0,

1 + 1 + ...+ 1︸ ︷︷ ︸m−times

= 0.

and for any χ ∈ Z/mZ, χ(0) = 1. Consequently,

1 = χ(0) = χ(1 + ...+ 1) = χ(1) · · ·χ(1)

so the image of this homomorphism is precisely the mth roots of unity in C×, which is isomorphicto a cyclic group of order m. Therefore,

Z/m/Z ∼= Z/mZ,

and hence G ∼= G.

Immediate from this, we haveG ∼= G, with the following canonical homomorphism sending any

g ∈ G to g, defined by:

g : G −→ C×

χ 7−→ χ(g)

Exercise 2.2. Show g 7→ g is a homomorphism.

Proposition 2.3. The map g 7→ g is an isomorphism G→ G.

Proof. Suppose g ∈ G, and χ(g) = 1 for all χ ∈ G (i.e. g is the trivial character). Let H = 〈g〉 < G.Then, every χ ∈ G can be expressed as a character of G/H, so

| G/H |≥| G |,

and by Proposition 2.1| G/H |≥| G |,

so | H |= 1. Therefore g = 1, and G ↪→ G is injective. But from Proposition 2.1 we also know that

| G |=| G |=| G |so the map surjective, and G ∼=

G.

6

Proposition 2.4. Let G be a finite abelian group. For H < G, let

H⊥ = {χ ∈ G : χ(h) = 1 for all h ∈ H}.

Then,

1. if H < G, then H⊥ ∼= G/H.

2. if H < G, then H ∼= G/H⊥.

3. (H⊥)⊥

= H (if we identifyG = G).

Proof. (Proof is repeated applications of Propositions 2.1 and 2.3. For a detailed proof, see [1,Proposition 1.3].)

The following proposition describes what are called the Orthogonality Relations of characters offinite abelian groups.

Proposition 2.5. 1. Fix a character χ of the finite abelian group G. Then

∑g∈G

χ(g) =

{0 if χ 6= χ0

| G | if χ = χ0

.

2. Fix an element g of the finite abelian group G. Then,

∑χ∈G

χ(g) =

{0 if g 6= 1

| G | if g = 1.

Proof. Let h ∈ G. Then, ∑g∈G

χ(g) =∑g∈G

χ(gh) = χ(h)∑g∈G

χ(g),

so(1− χ(h))

∑g∈G

χ(g) = 0

for every h ∈ G. If χ 6= χ0, then there exists h ∈ G such that χ(h) 6= 1, and then∑g∈G

χ(g) = 0.

If χ = χ0, then the expression becomes∑g∈G

χ0(g) =∑g∈G

1 =| G | .

For the second part, note that∑

χ∈G χ(g) =∑

χ∈G g(χ). Now use part (1).

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2.2 Dirichlet Characters

Definition 5. Let n be a positive integer. A Dirichlet Character modulo n is a character of theabelian group (Z/nZ)×. That is, a multiplicative homomorphism of the form

χ : (Z/nZ)× −→ C×.

We call n the modulus of χ.

Example 2. 1. Let p be an odd prime.

χ : (Z/mZ)× −→ C×

a 7−→(ap

)=

{1 if a− x2 ≡ 0 mod p for some x ∈ Z−1 otherwise

This particular character is called the Legendre Symbol.

2. Let i =√−1. Define

χ : (Z/5Z)× −→ C×

1 7−→ 1

2 7−→ i

3 7−→ −i4 7−→ −1

Suppse χ is a Dirichlet character of modulus n and n | m. Then, by the natural homomorphism,

ϕ : (Z/mZ)× −→ (Z/nZ)×

define χ′ = χ ◦ ϕ. Here χ′ is a Dirichlet character of modulus m. We say χ′ is induced by χ.

Definition 6. Let fχ be the minimal modulus for the character χ. That is, χ is not induced byany Dirichlet character of modulus less that fχ. Then we call fχ the conductor of χ. A Dirichletcharacter defined modulo its conductor is called primitive.

Note, for any Dirichlet character χ, the following is always true,

χ : (Z/nZ)× −→ C×

1 7−→ 1,

and since for any a ∈ (Z/nZ)×, we know aϕ(n) = 1, it follows that

1 = χ(aϕ(n)) = χ(a)ϕ(n)

so the image of χ is always a finite set of ϕ(n)th roots of unit in C×.

Example 3. 1. Define

χ : (Z/12Z)× −→ C×

1 7−→ 1

5 7−→ −1

7 7−→ 1

11 7−→ −1.

8

Notice, χ(a+ 3k) = χ(a), so in fact χ is induced by

ψ : (Z/3Z)× −→ C×

1 7−→ 1

2 7−→ −1.

ψ is irreducible, and fχ = 3.

2. Define

χ : (Z/12Z)× −→ C×

1 7−→ 1

5 7−→ −1

7 7−→ −1

11 7−→ 1.

Here χ is primitive so fχ = 12. To verify, just check possible maps on (Z/mZ)× where m | 12.

Definition 7. Let χ0 denote the trivial character, χ and ψ primitive Dirichlet characters of con-ductor fχ and fψ. Let n = lcm(fχ, fψ). Then χψ is the primitive Dirichlet character that inducesη defined

η : (Z/nZ)× −→ C×

a 7−→ χ(a)ψ(a).

Note, fχψ | fχ · fψ, and the set of Dirichlet characters is closed under this multiplication.Recall now, that for any root of unit in C, its complex conjugate is its inverse. Define

χ : (Z/nZ)× −→ C×

a 7−→ χ(a) = χ(a) = χ(a)−1.

Exercise 2.6. Show that the set of all Dirichlet characters form a commutative group under thisnotion of multiplication, with χχ = χ0 as the identity element.

Let’s see an example of how this η works.

Example 4. Define χ and ψ as the following,

χ : (Z/12Z)× −→ C×

1 7−→ 1

5 7−→ −1

7 7−→ −1

11 7−→ 1

and

ψ : (Z/4Z)× −→ C×

1 7−→ 1

3 7−→ −1.

9

Then n = 12 = lcm(12, 4), and we get the following map:

η : (Z/12Z)× −→ C×

1 7−→ χ(1)ψ(1) = 1

5 7−→ χ(3)ψ(3) = −1

7 7−→ χ(5)ψ(5) = 1

11 7−→ χ(11)ψ(11) = −1

We can see that η is imprimitive, since it is induced by the map

(Z/3Z)× −→ C×

1 7−→ 1

2 7−→ −1.

Since this map is primitive (clearly) and induces η, it must be the product χψ. Note that thisdefinition does not mean that χ(x)ψ(x) = χψ(x), since χψ(2) = −1, but χ and ψ are not definedfor 2.

Definition 8. The order of a Dirichlet character is its order as an element in the group of allDirichlet characters. This order will always be finite, and if χ has conductor n, then the order ofχ must divide ϕ(n). A Dirichlet character of order 2 is called a quadratic Dirichlet character.

For any Dirichlet character, we must have χ(−1) = ±1. We say

χ is even⇐⇒ χ(−1) = 1

χ is odd⇐⇒ χ(−1) = −1.

Exercise 2.7. The set of even Dirichlet characters is a subgroup of the group of all Dirichletcharacters under multiplication.

2.3 Dirichlet Characters and Q(ζn)

Fix an integer n. The set of all Dirichlet characters such that the conductor divides n form a group.In fact,

G := Gal(Q(ζn)/Q) ∼= (Z/nZ)×,

so the group of Dirichlet characters mod n can be seen as the characters of that Galois group.

Definition 9. Letχ : G −→ C×,

and let K be the fixed field of the kernel of χ. Then K is the field associated to χ.

Example 5. Let G := Gal(Q(ζ12)/Q), and define χ as follows:

χ : G −→ C×

σ1 7−→ 1

σ5 7−→ −1

σ7 7−→ 1

σ11 7−→ −1

10

where σj : ζ12 7→ ζj12. Then ker(χ) = {σ1 = id, σ7}, and we have the following Galois correspon-dence:

Q(ζ12)

Q(ζ3)

Q G

〈σ7〉

{1}

Note that σ7(ζ3) = ζ73 = ζ33ζ33ζ3 = ζ3, so Q(ζ3) is the fixed field of 〈σ7〉. Since

| Gal(Q(ζ12)/Q(ζ3) |= [Q(ζ12) : Q(ζ3)] =4

2= 2

we knowker(χ) = Gal(Q(ζ12)/Q(ζ3)),

so Q(ζ3) is the field associate to χ. So we may view χ as the following:

χ : G/ ker(χ) −→ C×

sinceG/ ker(χ) ∼= Gal(Q(ζ3)/Q) ∼= (Z/3Z)×.

Note that fχ = 3.

More generally, letX = {finite group of Dirichlet characters},

let n = lcm(fxi). Now X ≤ G, G = Gal(Q(ζn)/Q). Let

H =⋂χ∈X

kerχ,

so H ≤ G. Let K be the fixed field of H, then we call K the fixed field associated to X.Notice,

χ ∈ X =⇒ H ≤ ker(χ).

So,{field associated to χ} ⊆ {field associated to X}.

If X = 〈χ〉, then {field associated to χ} = {field associated to X}.

Q(ζn)

{field associated to χ}

Q

{field associated to X}

G

ker(χ)

{1}

H

11

Example 6.

χ : (Z/15Z)× −→ C×

1 7−→ 1

2 7−→ −1

4 7−→ 1

7 7−→ −1

8 7−→ −1

11 7−→ 1

13 7−→ −1

14 7−→ 1

Here | ker(χ) |= 4, so if K is the fixed field of ker(χ), then [K : Q] = 2. Let G = Gal(Q(ζ15)/Q),and consider χ : G→ C×. By above, we know σ14 ∈ ker(χ), so σ14(k) = k for any k ∈ K. But σ14corresponds to complex conjugation, so its fixed field must be real. Therefore, K = Q(

√5), since

this is the only real subfield in Q(ζ5).Note: dK/Q = 5, and χ = χ′ ◦ ϕ, where

χ′ : (Z/5Z)× −→ C×

1 7−→ 1

2 7−→ −1

3 7−→ −1

4 7−→ 1

since χ′ is primitive, fχ = 5.

Example 7. Let

ψ : (Z/15Z)× −→ C×

1 7−→ 1

2 7−→ −1

4 7−→ 1

7 7−→ 1

8 7−→ −1

11 7−→ −1

13 7−→ 1

14 7−→ −1

12

and

θ : (Z/15Z)× −→ C×

1 7−→ 1

2 7−→ 1

4 7−→ 1

7 7−→ −1

8 7−→ 1

11 7−→ −1

13 7−→ −1

14 7−→ −1.

In each of these, | ker(ψ) |=| ker(θ) |= 4, so the field associated to each of these characters must bea degree 2 extension of Q. Immediately, we notice that ψ factors through ψ′, where

ψ′ : (Z/3Z)× −→ C×

1 7−→ 1

2 7−→ −1.

So, fψ = 3. It is not too difficult to see now that θ is primitive, so fθ = 15. So what are the quadraticfields associated to these character groups? Since σa+3k is in the kernel of ψ, a reasonable conclusionis that Kψ = Q(

√−3). Now there is only one remaining quadratic extension, so Kθ = Q(

√−15).

Notice, there seems to be some connection between the conductor of a character, and the dis-criminant of its associated field. Let’s explore that further. But first, one more example.

Example 8. Let G := Gal(Q(ζn)/Q) ∼= (Z/nZ)×. Let X be the set of all even characters in G.Then X < G, by Exercise 2.7, and in particular it is an index 2 subgroup, since the product of anytwo odd characters is an even character. Further, χ(−1) = 1 for every χ ∈ X, and so σ−1 is inthe kernel of χ (here we have identified (Z/nZ)× ∼= G). Since σ−1 is complex conjugation, the fieldassociated to it must be real. Notice, that for any character χ, the field associated to χ is real ifand only if χ is even. Here, X is the largest subgroup made entirely of even characters, so a goodguess would be that is associated field is the maximal real subfield Q(ζn + ζ−1n ) of Q(ζn). We know,by definition, that the field associated to X is the fixed field of

H =⋂χ∈X

ker(χ).

We know {σ1, σ−1} ∈ H. Now, suppose σr ∈ H, and σr is non-trivial. Then σr given by G ↪→ G

is nontrivial, by Proposition 2.3. Then, there exists some ψ ∈ G such that ψ(σr) 6= 1. But thenψ 6∈ X, or else σr ∈ ker(ψ). So ψ is odd, and ψ2 is even, and so ψ2(σr) = 1, and hence ψ(σr) = −1(since ψ(σr) 6= 1). Now, since [G : X] = 2, any odd character has the form ψχ for some χ ∈ X.So,

σr : G −→ C×

ψχ 7−→ ψ(σr)χ(σr) = −1

χ 7−→ χ(σ1) = 1,

13

So σr = σ−1. Therefore, {σ1, σ−1} = H. Now the field associated to X is precisely the fixed fieldof H, so it must be real, and it must be of index 2 in Q(ζn). So there is only one choice, namelyQ(ζn + ζ−1n ).

Theorem 2.8. (Conductor Discriminant Formula) Let X be a finite group of Dirichlet charactersof K its associated field. Then,

dK/Q = (−1)r2∏χ∈X

fχ,

where r2 is the number of pairs of imaginary embeddings of K.

Exercise 2.9. 1. Let p > 2 be prime. Use Theorem 2.8 to find the dicriminant of the maximalreal subfield of Q(ζn). (Hint: remember that only the prime p can ramify here, if at all.)

2. Let p > 2 be prime. Use Theorem 2.8 to find the discriminant of Q(ζpn) over Q, where n > 0.

Our next goal is to establish a full “Galois Type” correspondence between character groups andassociated fields. From the fundamental theorem of Galois Theory, we already have the following:

K

L

Q G = Gal(K/Q)

H = Gal(K/L)

{1}

Let XK be the group of characters of Gal(K/Q), so XK = G. Then, we may define the followingsubset of XK

{χ ∈ XK : χ(g) = 1,∀g ∈ Gal(K/L)} = Gal(K/L)⊥

= Gal(K/Q)/Gal(K/L)

= Gal(L/Q)(by the F.T. of G.T.)

= XL.

So XL ≤ XK . But from this definition, we see that given any χ ∈ XL, χ(g) = 1 for everyg ∈ Gal(K/L), so in particular, ⋂

χ∈XL

ker(χ) = Gal(K/L)

and by the fundamental theorem of Galois Theory, L is the fixed field associated to Gal(K/L). So,for every subgroup of XK , we have an associated fixed field in K.

Suppose Y ≤ XK , and L is the fixed field of Y ⊥. Then

Y ⊥ = Gal(K/L)

by the fundamental theorem of Galois Theory. So,

(Y ⊥)⊥ = (Gal(K/L))⊥ = XL

14

where the last equality comes from above. So to enhance our existing picture,

K

L

Q G = Gal(K/Q)

H = Gal(K/L)

{1} XK = G

XL = H⊥ = G/H

{χ0} = G⊥

So this is a containment preserving correspondence of

{Subgroups of XK} ←→ {Subfields of K}

given byY ←→ {Fixed field of Y ⊥}

or byXL = Gal(K/L)⊥ ←→ L

So let’s revisit Example 8. Here X was the set of all even Dirichlet characters whose conductorsdivide n. We know that the field associated to X is real (since it is fixed by complex conjugation).Suppose L is any real subfield of Q(ζn), then L is fixed by σ−1. By the correspondence given above,L is precisely the fixed field of X⊥L , so σ−1 ∈ X⊥L . But

X⊥L = {g ∈ G : g(χ) = 1∀χ ∈ XL}

or equivalently,X⊥L = {g ∈ G : χ(g) = 1∀χ ∈ XL}.

So,σ−1 ∈ X⊥L =⇒ χ(−1) = 1∀χ ∈ XL,

in other words, every element of XL is even. But then, XL ≤ X, so the field associated to X mustbe the maximal real subfield, since any other real subfield corresponds to a subgroup of X.

2.4 Ramification

Just as we describe the Galois correspondence for finite extensions of Q in terms of Dirichletcharacters, we can describe the ramification behavior of finite extensions.

Definition 10. Let n be a positive integer, and suppose

n =m∏j=1

pajj ,

where pj are distinct primes, and aj > 0. Then

(Z/nZ)× ∼=m∏j=1

(Z/pajj Z)×,

15

so any Dirichlet character modulo n, can be written as a

χ =m∏j=1

χpj ,

where χpj is a Dirichlet character modulo pajj . For any group of characters X mod n, we will let

Xpj = {χpj : χ ∈ X}.

Notice, that for any χpj ∈ Xpj , the conductor is some power of pj, so in particular, p always dividesthe conductor.

Example 9. Consider the character

θ : (Z/15Z)× −→ C×

1 7−→ 1

2 7−→ 1

4 7−→ 1

7 7−→ −1

8 7−→ 1

11 7−→ −1

13 7−→ −1

14 7−→ −1.

This may be rewritten as θ = θ3θ5, where

θ3 : (Z/3Z)× −→ C×

1 7−→ 1

2 7−→ −1

and

θ5 : (Z/5Z)× −→ C×

1 7−→ 1

2 7−→ −1

3 7−→ −1

4 7−→ 1.

In our example above, if X = {θ}, then X5 = {θ5} and X3 = {θ3}.

Theorem 2.10. Suppose X is a group of Dirichlet characters with associated field K. If p ∈ Z isa prime, then the ramification index of p in K/Q is e =| Xp |.

Proof. (See [1, Theorem 2.2])

Sometimes it will help us to view a Dirichlet character as a function χ : Z→ C given by

χ(a) =

{χ(a mod fχ) if (a, fχ) = 1

0 if (a, fχ) 6= 1.

But note that this is equivalent to our previous construction, except now χ(fχ) = 0.

16

Corollary 2.11. Let X be a group of Dirichlet characters and let K be its associated field. Aprime p is unramified in K/Q if and only if χ(p) 6= 0 for all χ ∈ X.

Proof. Suppose p ramifies in K/Q. We then have e =| Xp |6= 1. So Xp contains some non-trivialelement χ. Since χ ∈ Xp, we know p | fχ, so in particular, χ(p) = 0.

Conversely, if χ(p) = 0 for some element of X, then the conductor of χ must be divisible by p.Thus, Xp must be non-trivial. But this implies that e =| Xp |> 1, so p is ramified.

Example 10. Consider K = Q(ζ12) and L = Q(i), so L ⊆ K. By our previous discussion, L hasthe associated field

XL = Gal(K/L)⊥ = {χ ∈ (Z/12Z)× : χ(σ) = 1∀σ ∈ Gal(K/L)}.

But, Gal(K/L) must have size 2, since [K : Q] = 4. So Gal(K/L) = {1, σ}, and σ fixes L.Consider the element i = ζ312 ∈ L. Suppose σ is an element in Gal(K/Q)) fixing this element, then

σ : (Z/12Z)× −→ (Z/12Z)×

ζ12 7−→ ζ512

ζ312 7−→ (ζ312)5 = ζ1512 = ζ312 = i.

So, χ(σ) = 1 if and only if χ(5) = 1 (recall σ = σ5 and 5 correspond under the canonical isomor-phism Gal(K/Q) ∼= (Z/12Z)×.) So, XL = {1, θ}, where

θ : (Z/12Z)× −→ C×

1 7−→ 1

5 7−→ 1

7 7−→ −1

11 7−→ −1.

However, this factors through θ′ in the following way

θ : (Z/12Z)× −→ (Z/4Z)× −→ C×

1 7−→ 1 7−→ 1

5 7−→ 1

7 7−→ 3 7−→ −1

11 7−→ 3.

so fθ = 4. We know θ(4) = 0, so by Corollary 2.11, we know that 2 is the only prime that ramifiesin Q(i)/Q (thus, confirming something we already know...but in a much cooler way.)

Before proving the following theorem, let’s recall what we know about inertia and decompositiongroups from algebraic number theory. Suppose we have an abelian extension, K/L and G :=Gal(K/L). Suppose p is a prime in L, which splits in K as

pOK = (P1 · · ·Pn)e.

Recall from the first lecture, we know that the ramification index is identical for each prime sittingabove p. We define the decomposition group of P over L as

GP = {σ ∈ G : σ(P) = P}.

17

Recall, that since the primes over p are Galois conjugates of one another, and it is easy to showthat

τ ∈ GσP ⇔ τ ∈ σGPσ−1.

Also, since GP is the stabilizer of P under the action of the Galois group, the index of GP in G isprecisely the size of the orbit of containing P. So we know that

[G : GP] = g,

so from the fundamental identity, we conclude

[K : L] = efg ⇒| GP |=n

g= ef.

The fixed field of GP, denoted ZP is the decomposition field. Define, κ(P) = OK/P and κ(p) =OL/p, then

[κ(P) : κ(p)] = f,

the residue degree. But κ(P)/κ(p) is a finite extension of a finite field, and as such, we can easilyidentify the Galois group of this extension. The following two lemmas will be stated without proof,but the reader may refer to [5, §9].

Lemma 2.12. Fixing a prime P above p, let PZ = P ∩ ZP. Then,

1. P is the only prime above PZ .

2. The ramification index and the inertia degree of PZ over p are both equal to 1.

Lemma 2.13. The map GP → Gal(κ(P)/κ(p)) is a surjective homomorphism.

The kernel of this homomorphism is called the inertia group, denoted I. It is known that thefixed field of I, which we call the inertia field, and denote TP, is the largest intermediate field suchthat e = 1. Now we can prove the theorem.

Theorem 2.14. Let X be a group of Dirichlet characters with associated field K. Let p be prime,and define the subgroups

Y = {χ ∈ X : χ(p) 6= 0}

Y1 = {χ ∈ X : χ(p) = 1}.

Then,

1. X/Y is isomorphic to the inertia subgroup for p.

2. X/Y1 is isomorphic to the decomposition group for p.

3. Y/Y1 is cyclic of order f .

Proof. [1, Theorem 2.4]

18

2.5 Dirichlet Series

Definition 11. A Dirichlet series is a series of the form

f(s) =∞∑n=1

anns

where an ∈ C for all n and s is a complex variable.

We will first introduce some elementary facts about Dirichlet series, beginning with Abel’sLemma.

Exercise 2.15. Prove Abel’s Lemma: Let (an) and (bn) be sequences, and for r ≥ m put Am,r =∑rn=m = an and Sm,r =

∑rn=m anbn. Then,

Sm,r =r−1∑n=m

Am,n(bn − bn+1) +Am,rbr.

Exercise 2.16. Let A be am open subset of C and let (fn) be a sequence of holomorphic functionson A that converge uniformly on every compact subset to a function f . Show that f is holomorphicon A and the sequence of derivatives (f ′n) converges uniformly on all compact subsets to f ′.

Lemma 2.17. If f(s) =∑∞

n=1anns converges for s = s0, then it converges uniformly in every

domain of the form {s : Re(s− s0) ≥ 0, | Arg(s− s0) |≤ θ} with θ < π2 .

Proof. Translating if necessary, we may assume s0 = 0. Then we have that

∞∑n=1

an

converges, and we must show that f(s) converges uniformly in every domain of the form {s :Re(s) ≥ 0, | Arg(s) |≤ θ} for θ < π

2 . Equivalently, we must show that f(s) converges uniformly inevery domain of the form

{s : Re(s) ≥ 0,| s |

Re(s)≤M}.

Let ε > 0, and let

Am,r =

r∑n=m

an

as in Abel’s Lemma. Since∑∞

n=1 an converges, there is some sufficiently large number N , so thatif r > m ≥ N , then we have | Am,r |< ε. Now, let bn = n−s and apply Abel’s Lemma to get

Sm,r =

r−1∑n=m

Am,n(n−s − (n+ 1)−s) +Am,rr−s.

First, we notice the following equivalence obtained by taking the real integral:

s

∫ d

ce−tsdt = e−cs − e−ds.

19

So, taking the absolute value, we get

| e−cs − e−ds |=| s |∫ d

ce−tRe(s)dt =

| s |Re(s)

(e−cRe(s) − e−dRe(s)).

So, taking c = ln(n), that is, n = ec, and similarly, d = ln(n+ 1).we obtain

| Sm,r | ≤ ε | r−s | +r−1∑n=m

ε | n−s − (n+ 1)−s |

≤ ε(| r−s | +

r−1∑n=m

| n−s − (n+ 1)−s |)

≤ ε(

1 +| s |

Re(s)

r−1∑n=m

(n−Re(s) − (n+ 1)−Re(s))).

But, this sum is telescoping, so simplifying, we get

| Sm,r |≤ ε(1 +M(m−Re(s) − r−Re(s))),

and m, r ∈ N and Re(s) > 0, so in fact

| Sm,r |≤ ε(1 +M).

Of course, Sm,r is just a difference of partial sums of our Dirichlet series, so the uniform convergence

of f(s) on {s : Re(s) ≥ 0, |s|Re(s) ≤M} follows.

Theorem 2.18. If the Dirichlet series f(s) =∑∞

n=1anns converges for s = s0, then it converges

(not necessarily absolutely) for Re(s) > Re(s0) to a function that is holomorphic there.

Proof. Clear from Lemma 2.17 and Exercise 2.16.

Corollary 2.19. Let f(s) =∑∞

n=1anns be a Dirichlet series.

i. If the an are bounded, then f(s) converges absolutely for Re(s) > 1.

ii. If An = a1 + ... + an is a bounded sequence, then f(s) converges (though not necessarilyabsolutely) for Re(s) > 0.

iii. If f(s) =∑∞

n=1anns converges at s = s0, then it converges absolutely for Re(s) > Re(s0) + 1.

Proof. i. Suppose the an are bounded, so say that | an |≤ M . Also suppose that Re(s) < 1.Then, ∣∣∣∣∣

r∑n=1

anns

∣∣∣∣∣ ≤r∑

n=1

| an || ns |

≤Mr∑

n=1

| n−s |

= Mr∑

n=1

| n−Re(s) | · | (ni)−Im(s) |

= M

r∑n=1

| n−Re(s) | .

But since Re(s) > 1, this∑∞

n=1 n−Re(s) converges, and the result follows.

20

ii. Suppose that the Am are bounded. For r > m let Am,r =∑r

n=m an as in Abel’s Lemma.We have that the Am,r are bounded, let’s say | Am,r |≤ M . Applying Abel’s Lemma, withbn = n−s, as in the proof of Lemma 2.17, we get

| Sm,r | =

∣∣∣∣∣r∑

n=m

Am,n

( 1

ns− 1

(n+ 1)s

)+Am,r

( 1

rs

)∣∣∣∣∣≤

r∑n=m

∣∣∣∣Am,n( 1

ns− 1

(n+ 1)s

)∣∣∣∣+

∣∣∣∣Am,r( 1

rs

)∣∣∣∣≤M

( r∑n=m

∣∣∣∣ 1

ns− 1

(n+ 1)s

∣∣∣∣+

∣∣∣∣ 1

rs

∣∣∣∣ ).By Theorem 2.18, it suffices to show that f(σ) converges for a real value of σ, since then wemay conclude that f(s) converges whenever Re(s) > σ. So, suppose s = σ is real, then we have

| Sm,r |≤M

mσ,

which is a Cauchy sequence whenever σ > 0. So f(s) converges whenever Re(s) > σ > 0, andwe are done.

iii. Suppose f(s) =∑∞

n=1anns converges for s = s0. Let

g(s) = f(s+ s0) =∑

(anns0

)(1

ns).

Since f(s0) converges, bn = anns0 approaches 0 as n → ∞. Therefore, {bn} and hence {bn( 1

ns )}is bounded if Re(s) > 1. By part (i), g(s) converges absolutely for Re(s) > 1. Thus, f(s) =g(s− s0) converges absolutely for Re(s− s0) > 1, that is, Re(s) > 1 + Re(s0).

Definition 12. Let χ :(Z/mZ

)×→ C×, be a Dirichlet character. The Dirichlet L-function

associated to χ is

L(s, χ) =∞∑n=1

χ(n)

ns.

The Riemann zeta function is

ζ(s) =∞∑n=1

1

ns.

Suppose χ 6= χ0. Let An = χ(1) + ...+ χ(n) and write n = mk + r where 0 ≤ r ≤ m− 1. Then,

An = [χ(1)+...+χ(m)]+[χ(m+1)+...+χ(2m)]+...+[χ(km+1)+...+χ(km+r)] = [χ(km+1)+...+χ(km+r)]

by our orthogonality relations. Therefore,

| An |≤| χ(km+ 1) | +...+ | χ(km+ r) |= r < m.

Now, use Corollary 2.19. If χ 6= χ0, then we’ve shown An is a bounded sequence, so L(s, χ)converges for Re(s) > 0, by part (ii). Then by part (iii), L(s, χ) converges absolutely for anyRe(s) > 1.

21

Lemma 2.20. L(s, χ) has a so-called Euler product:

L(s, χ) =∏

p prime

(1− χ(p)p−s)−1

for Re(s) > 1.

Proof. Fix s, with Re(s) > 1. We want to show:

limN→∞

∏p≤N

(1− χ(p)

ps

)−1= L(s, χ).

Say p1, ..., pN are the primes less than N . Then, we have the following infinite geometric series

k∏i=1

(1− χ(pi)

psi

)−1=

k∏i=1

(1 +

χ(pi)

pi+ ...+

χ(pi)m

(psi )m

+ ...

)

=k∏i=1

(1 +

χ(pi)

pi+ ...+

χ(pmi )

psmi+ ...

)=

∑m1,...,mk≥0

χ(pm11 · · · p

mkk )

(pm11 · · · p

mkk )s

=∑n∈Jn

χ(n)

ns,

where Jn = {n ∈ Z : n > 0 and n is not divisible by any prime p > N}.Now we have

L(s, χ)−∏p≤N

(1− χ(p)

ps

)−1=

∑n∈(Z+\Jn)

χ(n)

ns.

Taking absolute values and applying the triangle inequality, we get∣∣∣∣∣∣∑

n∈(Z+\Jn)

χ(n)

ns

∣∣∣∣∣∣ ≤∑

n∈(Z+\Jn)

∣∣∣∣χ(n)

ns

∣∣∣∣=

∑n∈(Z+\Jn)

1

| ns |

=∑

n∈(Z+\Jn)

1

| nσ |where σ = Re(s)

≤∑n≥N

1

nσ,

since clearly (Z+ \ Jn) ( {n ≥ N}. However, the series∑

n>N1nσ converges for σ > 1,∑

n≥N

1

nσ→ 0,

22

as N →∞. So,

limN→∞

L(s, χ)−∏p≤N

(1− χ(p)

ps

)−1 = 0,

and so we are done.

Note that when Re(s) > 1, then we have

L(s, χ) =∏

p prime

(1− χ(p)p−s)−1 =∏

p prime

ps

ps − χ(p)6= 0

since ps 6= 0.Taking the log of L(s, χ) for Re(s) > 1, we get

log(L(s, χ)) = log

∏p prime

(1− χ(p)

ps)−1

= −∑p

log(1− χ(p)p−s).

Let this be the branch of log defined on the upper half plane, such that log(L(s, χ))→ 0 as s→∞.Recall, for a complex valued T with Re(T ) < 1, we have the following Taylor series expansion

log(1− T ) = −∞∑n=1

Tn

n.

Since | χ(p)ps |=1|ps| ≤ 1 when Re(s) > 1, we may use this expansion. Letting T = χ(p)

ps , we get

log(L(s, χ)) =∑p

− log

(1− χ(p)

ps

)=∑p

∑n≥1

χ(p)np−ns

n

=∑p

∑n≥1

χ(p)n

npns.

Now, ∣∣∣∣χ(p)n

npns

∣∣∣∣ ≤ ∣∣∣∣ 1

pns

∣∣∣∣ =1

pnσ,

where σ = Re(s). So, ∑p

∑n≥1

∣∣∣∣χ(p)n

npns

∣∣∣∣ ≤∑p

∑n≥1

1

pnσ≤∑m≥1

1

mσ, (1)

which converges for σ > 1. Therefore, the above expression for log(L(s, χ)) is absolutely convergentfor Re(s) > 1 (since the sum of the absolute values is convergent). So we may rearrange the termsto get

log(L(s, χ)) =∑p

∑n≥1

χ(p)n

npns=∑p

χ(p)

ps+∑p

∑n≥2

χ(p)n

npns.

23

Let

β(s, χ) =∑p

∑n≥2

χ(p)n

npns.

From equation (1) we know that β(s, χ) is absolutely convergent Re(s) > 12 , so in particular, it

takes a finite value whenever Re(s) > 1.

2.6 Dirichlet’s Theorem on Primes in Arithmetic Progressions

Consider Euler’s classical proof of the existence of infinitely many primes, which uses the Riemannzeta function. We have the Euler product

ζ(s) =∞∑n=1

1

ns=

∏p prime

(1− p−s)−1.

Suppose that there are only finitely many primes, say p1, ..., pn ∈ Z. Then,

ζ(s) =n∏j=1

(1− p−sj )−1 =n∏j=1

1

1− 1psj

.

Taking the limit as s→ 1, we get

lims→1

ζ(s) =

n∏j=1

(1

1− 1pj

),

which is rational. However, it is clear that

lims→1

∞∑n=1

1

ns=∞,

which is a contradiction.The proof of Dirichlet’s theorem of primes in arithmetic progression will be a generalization of

this idea, except here the Riemann zeta function is replaced by the Dirichlet L-function.

Theorem 2.21. (Dirichlet’s Theorem on Primes in Arithmetic Progressions) If m is apositive integer and s is an integer for which (a,m) = 1, then there are infinitely many primes psatisfying p ≡ a mod m.

Proof. Let (a,m) = 1, and consider the set of all Dirichlet characters modulo m. Then,∑χ∈ (Z/mZ)×

χ(a)−1 log(L(s, χ)) =∑χ

χ(a)−1(∑

p

χ(p)

ps+ β(s, χ)

)=∑p

1

ps

∑χ

χ(a)−1χ(p) +∑χ

χ(a)−1β(s, χ)

=∑p

1

ps

∑χ

χ(pa−1) +∑χ

χ(a)−1β(s, χ).

24

Recall from our orthogonality relations,

∑χ

χ(pa−1) =

{ϕ(m) if pa−1 ≡ 1 mod m

0 if pa−1 6≡ 1 mod m,

in other words, ∑χ

χ(pa−1) =

{ϕ(m) if p ≡ a mod m

0 otherwise.

So, ∑χ

χ(a)−1 log(L(s, χ)) = ϕ(m)∑

p≡a mod m

p−s + (∗). (2)

Here (∗) =∑

χ χ(a)−1β(s, χ), and as shown previously, for Re(s) > 12 , β(s, χ) is finite, say M , so∣∣∣∣∣∑

χ

χ(a)−1β(s, χ)

∣∣∣∣∣ ≤∑χ

∣∣∣∣β(s, χ)

χ(a)

∣∣∣∣ =∑χ

M.

So, (∗) is something that converges absolutely for Re(s) > 12 .

Now let s→ 1, then on the right hand side of (2), we have

lims→1

ϕ(m)∑

p≡a mod m

p−s + ( a finite constant ),

which would be finite if there were only finitely many primes p with p ≡ a mod m. Now, the proofwill be complete if we can show that the left side of (2) is infinite as s→ 1.

If χ = χ0 (with modulus m), then

L(s, χ0) =∏p

(1− χ0(p)

ps)−1 = ζ(s)

∏p|m

(1− p−s)→∞ as s→ 1.

So, L(s, χ0)→∞ as s→ 1.Now, suppose χ 6= χ0. We have

L(s, χ) =∞∑n=1

χ(n)

ns.

and any partial sum,

|χ(1) + χ(2) + ...+ χ(n)| ≤ |χ(1)|+ |χ(2)|+ ...+ |χ(n)| ≤ n

is bounded, so from (ii) of Corollary 2.19 we know that L(s, χ) converges for any Re(s) > 0.Further, L(s, χ) is analytic, from our ”complex analysis fact.” We know that L(1, χ) is defined forχ 6= χ0, so log(L(s, χ)) will be finite if we can show that L(1, χ) 6= 0 for χ 6= χ0. Given this, we’llhave ∑

χ

χ(a)−1 log(L(s, χ))→∞

as s→ 1+, and the proof will be complete.Now we need to show that L(s, χ) 6= 0, when χ 6= χ0.

25

Definition 13. Let K be an algebraic number field, and let a very through the nonzero integralideals OK , (so that we may view Na as a positive integer). Define the Dedekind zeta function ofK as

ζK(s) =∑a

1

NK/Q(as)

By an argument similar to the one for the L-function, we have

ζk(s) =∏p

(1−NK/Q(ps)−1,

where p runs over the prime ideals in OK . (The proof uses uniques factorization of prime ideals.)It is easy to see that

ζk(s) =

∞∑n=1

γnns

where γn = #{a : NK/Q(a) = n}.

Exercise 2.22. Show that ζK(s) is absolutely convergent for Re(s) > 1. (compare this to (i) ofCorollary 2.19...are the γn bounded?)

Now we will state without proof a theorem of Dedekind:

Theorem 2.23. ζK(s) can be analytically continued to C− {1} with a simple pole at s = 1, i.e.

ζK(s) =ρ(K)

s− 1+ { something entire }

Moreover,

ρ(K) =2r1(2π)r2hKRK

ωK√| dK/Q |

where

r1 = # real embeddings of K

r2 = # pairs of complex embeddings of K

hK = #CK = Class number of K

RK = regulator of K

ωk = # roots of unity in K

dK/Q = the discriminant

Now we can use the Dedekind zeta funtion to show that L(1, χ) 6= 0 when χ 6= χ0, completingthe proof of Primes in Arithmetic Progressions. Take K = Q(ζm), where m is the modulus inDirichlet’s theorem, so also the modulus of the characters χ. We have

ζK(s) =∏p

(1−NK/Q(ps)−1

=∏p

∏p|p

(1−NK/Q(ps)−1

=

∏p|m

∏p|p

(1−NK/Q(ps))−1

∏p-m

∏p|p

(1−NK/Q(ps))−1

.

26

Now, recall that NK/Q(p) = pf . Since K/Q is Galois, ϕ(m) = fg, and f = #Z(p) is the smallest

integer so that pf ≡ 1 mod m, and g = ϕ(m)f .

Lemma 2.24. If p - m then

(1− T f )ϕ(m)f =

∏χ mod m

(1− χ(p)T ).

Proof. Let G = Gal(K/Q), where K = Q(ζm) as before. Let Z = Z(p) be the decomposition groupfor p in K/Q and define a map

G −→ Z

χ 7−→ χ |Z .

First observe that{χ mod m} = {χ ∈ (Z/mZ)×} = {χ ∈ G}.

Then, ∏χ∈G

(1− χ(p)T ) =∏ψ∈Z

∏χ∈G,χ|Z=ψ

(1− χ(p)T ) =∏ψ∈Z

(1− ψ(p)T )`(ψ)

where

`(ψ) =#{χ ∈ G : χ |Z ψ}

=# ker(G→ Z)

=#G

#Z

=ϕ(m)

f

=g.

Thus, ∏χ mod m

(1− χ(p)T ) =∏χ∈G

(1− χ(p)T ) =∏ψ∈Z

(1− ψ(p)T )g,

and we know g = ϕ(m)f . It only remains to be shown that 1 − T f =

∏ψ∈Z(1 − ψ(p)T ). As a

subgroup of G, Z is generated by the unique Frobenius element, F (p | p), where

F (p | p)(a) ≡ ap mod p.

Alternatively, we know that in the unramified case, Z ∼= Gal(Fpn/Fp), a cyclic group generated by

σp : Fpn → Fpna 7→ ap.

So this generator must pull back to a generator of Z, giving us our unique Frobenius element, σp.Viewing Z as a subgroup of (Z/mZ)×, Z is generated by p mod m. Consider the following map:

Z −→ µf = {f th roots of unity }ψ 7−→ ψ(p).

27

Since p is a generator for Z, this map is an isomorphism. Thus,∏ψ∈Z

(1− ψ(p)T ) =∏η∈µf

(1− ηT ) = 1− T f .

So we are done.

Now, in Lemma 2.24, put T = p−s, giving

(1− p−sf )ϕ(m)/f =∏

χ mod m

(1− χ(p)p−s).

taking the product over all primes p not dividing m, we get∏p-m

(1− p−sf )−ϕ(m)/f =∏χ

∏p-m

(1− χ(p)p−s)−1.

Now, if p | m, then χ(p) = 0, so ∏p|m

(1− χ(p)p−s)−1 = 1.

Putting this all together,

L(s, χ) =∏p

(1− χ(p)p−s)−1∏χ

L(s, χ) =∏χ

∏p

(1− χ(p)p−s)−1

=∏χ

∏p|m

(1− χ(p)p−s)−1 ·∏p-m

(1− χ(p)p−s)−1

=∏χ

∏p-m

(1− χ(p)p−s)−1

=∏p-m

(1− p−sf )−ϕ(m)/f .

On the other hand, ∏p|p

(1−N(p)−s)−1 =∏p|p

(1− p−sf )−1

=(1− p−sf )−ϕ(m)/f ,

since there are g = ϕ(m)/f primes p contained in p. This gives us∏p-m

(1− p−sf )ϕ(m)/f =∏p-m

∏p|p

(1−N(p)−s)−1

=ζk(s)∏p|m

∏p|p

(1−N(p−s),

28

thus

ζk(s)∏p|m

∏p|p

(1−N(p−s)) =∏χ

L(s, χ)

=L(s, χ0)∏χ 6=χ0

L(s, χ)

=ζ(s)∏p|m

(1− p−s)∏χ 6=χ0

L(s, χ).

We get ∏χ 6=χ0

L(s, χ) =ζk(s)

∏p|m∏p|p(1−N(p−s))−1

ζ(s)∏p|m(1− p−s)

.

Now, ∏p|m

∏p|p

(1−N(p−s))−1

and ∏p|m

(1− p−s)

are non-zero constants, while each of the ζK(s) and ζ(s) has a simple pole at s = 1. Let s→ 1, theexpression on the left side approaches a constant, hence∏

χ 6=χ0

L(s, χ)

does too. This shows that L(1, χ) 6= 0 for all χ 6= χ0, completing the proof of Dirichlet’s Theorem.

2.7 Dirichlet Density

Suppose f(s) and g(s) are defined for s ∈ R, s > 1. We will write f(s) ∼ g(s) if f(s) − g(s) isbounded as s→ 1+. Using this notation, we will reformulate our proof of Dirichlet’s Theorem. Forany χ,

log(L(s, χ)) =∑p

χ(p)

ps+ {something converging for Re(s) >

1

2}

(see the equality on the bottom of page 23). So,

log(L(s, χ))−∑p

χ(p)

ps= {something converging for Re(s) >

1

2},

which is bounded as s→ 1+. So,

log(L(s, χ)) ∼∑p

χ(p)

ps.

Assuming we feel comfortable with L(1, χ) 6= 0 when χ 6= χ0, we have that log(L(1, χ)) is finitewhen χ 6= χ0, so∑

χ

χ(a−1) log(L(s, χ)) = log(L(s, χ0)) +∑χ 6=χ0

χ(a−1) log(L(s, χ)).

29

As s→ 1+, the term on the far right is finite, so we conclude that∑χ

χ(a−1) log(L(s, χ)) ∼ log(L(s, χ0)).

Furthermore, ∑χ

χ(a−1) log(L(s, χ)) ∼∑χ

χ(a−1)∑p

χ(p)

ps

=∑p

1

ps

∑χ

χ(a−1p)

=∑p

1

ps·

{ϕ(m) if p ≡ a mod m

0 otherwise

=∑

p≡a mod m

ϕ(m)

ps.

Combining these expressions, we get

log(L(s, χ0)) ∼∑

p≡a mod m

ϕ(m)

ps.

Using our well-known Euler product, we have

L(s, χ0) =∏p

(1− χ0(p)

ps

)−1= ζ(s)

∏p|m

(1− p−s)

.

Taking the log of both sides, we get

log(L(s, χ0)) = log(ζ(s)) + log

∏p|m

(1− p−s)

,

and since the term on the far right is clearly finite,

log(L(s, χ0)) ∼ log(ζ(s)).

So, we get ∑p≡a mod m

1

ps∼ 1

ϕ(m)log(ζ(s)).

Clearly,

lims→1+

log(ζ(s)) = log

( ∞∑n=1

1

n

)→∞.

So,∑

p≡a mod m1ps diverges, which proves Dirichlet’s theorem.

It is a nontrivial fact thatlims→1+

(s− 1)ζ(s) = 1.

30

This is a consequence of ζ(s) having a simple pole at s = 1, a fact which we discussed earlier. So,we can see that

log(ζ(s)) = log

((s− 1)ζ(s)

(s− 1)

)= log((s− 1)ζ(s)) + log

(1

s− 1

).

So,1

ϕ(m)log(ζ(s)) =

1

ϕ(m)

(log((s− 1)ζ(s)) + log

(1

s− 1

)),

and

lims→1+

1

ϕ(m)

(log(ζ(s))− log

(1

s− 1

))= 0.

Therefore, ∑p≡a mod m

p−s ∼ 1

ϕ(m)log

(1

s− 1

),

and

lims→1+

=

∑p≡a mod m p

−s

log(

1s−1

) = bounded.

Definition 14. Let S be any set of primes. If

lims→1+

=

∑p∈S p

−s

log(

1s−1

) = δ

exists, then we say that S has Dirichlet density δ = δ(S).

Theorem 2.25. Let K/Q be Galois, and let

SK = {p ∈ Z : p splits completely in K/Q}.

Then δ(SK) = 1[K:Q] .

Proof. Let ζK(s) =∏

p(1−Np−s)−1 for Re(s) > 1, be the Dedekind zeta function for K. Considers ∈ R, s > 1. We have

log(ζK(s)) = −∑p

log(1−Np−s) =∑p

∑n

1

nNp−ns,

and

log(ζK(s)) = log((s− 1)ζK(s)) + log

(1

s− 1

),

so

log ζK(s) ∼ log

(1

s− 1

)∼∑p

∑n

1

nNp−ns

∼∑p

Np−s +∑p

∞∑n=2

1

nNp−ns

∼∑p

Np−s.

31

since∑

p

∑∞n=2

1nNp−ns is bounded as s→ 1+. Then,

log ζK(s) ∼ log

(1

s− 1

)∼∑p

Np−s

=∑

f(p|p)=1=e(p|p)

p−s +∑

f(p|p)>1

p−fs +∑

f(p|p)=1,e(p|p)>1

p−s,

but the second term is bounded as s → 1+, and the third term is finite, since only finitely manyprimes ramify in any given extension. So,

log ζK(s) ∼ log

(1

s− 1

)=∑p∈SK

g(p)p−s,

where g(p) counts the number of primes p sitting over p. For p ∈ SK , we know that g(p) = [K : Q],so

log ζK(s) ∼ log

(1

s− 1

)∼∑p∈SK

[K : Q]p−s.

Therefore,

log

(1

s− 1

)= [K : Q]

∑p∈SK

p−s + b(s),

where b(s) is something bounded as s→ 1+. Now, computing SK ,

δ(SK) = lims→1+

∑p∈SK p

−s

log(

1s−1

)= lims→1+

log(

1s−1

)∑

p∈SK p−s

−1

= lims→1+

([K : Q]

∑p∈SK p

−s + b(s)∑p∈SK p

−s

)−1=

1

[K : Q].

More generally, we could let S be the set of prime ideals of OF , where F is a number field. Then,if

lims→1+

∑p∈S p

−s

log(

1s−1

) = δ

exists, then S has Dirichlet density δ = δF (S).

Corollary 2.26. Let K/F be Galois, and SK/F = {p ∈ OF : p splits completely in K/F}, then

δF (SK/F ) =1

[K : F ].

32

This proof follows from the previous theorem. Let S and T be sets of prime ideals in OF , thenwe say

S ≺ T ⇐⇒ δF (S \ T ) = 0

andS ≈ T ⇐⇒ S ≺ T ≺ S.

Theorem 2.27. E and K are number fields, both Galois over Q. Then,

SK ≺ SE ⇐⇒ E ⊆ K.

Proof. ⇐=) obvious.=⇒) For this direction, suppose we have two sets of primes S and T , so that S ∩ T = ∅. Then

δ(S ∪ T ) = lims→1+

∑p∈S∪T p

−s

log(

1s−1

) = lims→1+

∑p∈S p

−s

log(

1s−1

) +

∑p∈T p

−s

log(

1s−1

) = δ(S) + δ(T ).

But we haveSKE = SK ∩ SE ,

and(SK ∩ SE) ∩ (SK \ SE) = ∅,

and(SK ∩ SE) ∪ (SK \ SE) = SK .

So,δ(SKE) = δ (SK ∩ SE) = δ (SK ∩ SE) + δ (SK \ SE) = δ(SK).

So, from Theorem 2.25, we get

[KE : Q] = δ(SKE)−1 = δ(SK)−1 = [K : Q],

so E ⊆ K.

3 Ray Class Groups

3.1 Approximation Theorems and Infinite Primes

Theorem 3.1. Let | · |1 · · · | · |n be non-trivial pairwise inequivalent absolute values on a numberfield F , and let β1, ..., βn be non-zero elements of F . For any ε > 0, there is an element α ∈ F suchthat | α− βj |< ε, for each j = 1, ..., n.

Recall, given a number field F/Q of degree n, there are d real embeddings of F into R, given bythe real roots of the minimal polynomial for F/Q. For any such embedding, σi : F → R, we getthe absolute value

| · |σi : F ↪→ Rx 7→| σi(x) |,

where | · | is the usual absolute value on R. To each of these places, we will associate the followingformal object, ∞i, given by

α ≡ β mod ∞i ⇔ σi(α

β) > 0,

and we call this an infinite real prime.

33

Definition 15. A modulus (or divisor) m of F is a formal product

m = m0 ·m∞

where m0 is a product of finite primes, and m∞ is a product of infinite primes. In fact,

m0 =∏p

pordp(m0),

can just be thought of as a finite non-zero integral ideal of OF . Similarly, we define the formalproduct

m∞ =∏∞i

∞it

where t = 0, 1.

Example 11. If F = Q, then there are only two types of moduli,

m = (n),

where n ∈ Z, orm = (n) · ∞,

where ∞ is the unique embedding of Q into R.

3.2 Ray Class Groups and the Universal Norm-Index Inequality

Recall the followingIF = { nonzero fractional ideals of F},

andPF = { principal fractional ideals in IF }.

From this, we get the ideal class group

CF = IF /PF .

Now define

IF (m) = { free abelian group generated by finite places not dividing m}.

So, in particular, if m = (1), thenIF (m) = IF ,

and if m | m′, thenIm ⊆ I ′m.

Next, we define the ray

PF (m) = 〈{(α) : such that (i) and (ii) hold}〉

(i) for all finite p | m, ordp(α− 1) ≥ ordp(m),

(ii) for all ∞i | m∞, σi(α) > 0.

34

and the narrow ray

P+F (m) = 〈{(α) : such that (i) holds, and σi(α) > 0∀i}〉,

observing thatP+F (m) ≤ PF (m) ≤ IF (m).

Definition 16. We define the ray class group of F for m as

RF (m) = IF (m)/PF (m),

and the narrow ray class group as

R+F (m) = IF (m)/P+

F (m),

Notice, if we let

m∞ =∏∞i

∞i,

for some F , and let m = m0 ·m∞, then

P+F (m0) = PF (m).

References

[1] N. Childress, Class Field Theory, Universitext, Springer-Verlag, New York, 2009.

[2] D. Garbanati, Class Field Theory Summarized, Rocky Mountain Journal of Mathematics, 11,No. 2, (1981). 195-225.

[3] D. Marcus, Number Fields, Universitext, Springer-Verlag, New York, 1977.

[4] J. S. Milne, Class Field Theory, v. 4.01.

[5] J. Neukirch, Algebraic Number Theory, Springer-Verlag, 1999.

[6] T. Shemanski, An Overview of Class Field Theory, http://www.math.dartmouth.edu/~trs/expository-papers/tex/CFT.pdf

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