Math 223b - Algebraic Number Theory · 2018-12-12 · Math 223b Notes 5 1 January 22, 2018 Last...

Math 223b - Algebraic Number Theory

Taught by Alison MillerNotes by Dongryul Kim

Spring 2018

The course was taught by Alison Miller on Mondays, Wednesdays, Fridaysfrom 12 to 1pm. The textbook was Cassels and Frohlich’s Algberaic NumberTheory. There were weekly assignments and a final paper. The course assistancewas Zijian Yao.

Contents

1 January 22, 2018 51.1 Global fields and adeles . . . . . . . . . . . . . . . . . . . . . . . 5

2 January 24, 2018 72.1 Compactness of the adele . . . . . . . . . . . . . . . . . . . . . . 7

3 January 26, 2018 93.1 Compactness of the multiplicative adele . . . . . . . . . . . . . . 93.2 Applications of the compactness . . . . . . . . . . . . . . . . . . 10

4 January 29, 2018 114.1 The class group . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

5 January 31, 2018 135.1 Classical theorems from compactness . . . . . . . . . . . . . . . . 13

6 February 2, 2018 156.1 Statement of global class field theory . . . . . . . . . . . . . . . . 156.2 Ray class field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

7 February 5, 2018 177.1 Image of the reciprocity map . . . . . . . . . . . . . . . . . . . . 177.2 Identifying the ray class group . . . . . . . . . . . . . . . . . . . 18

8 February 7, 2018 198.1 Properties of the ray class field . . . . . . . . . . . . . . . . . . . 19

1 Last Update: August 27, 2018

9 February 9, 2018 219.1 Ideles with field extensions . . . . . . . . . . . . . . . . . . . . . . 219.2 Formations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

10 February 12, 2018 2310.1 Class formation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

11 February 14, 2018 2511.1 Consequences of the class formation axiom . . . . . . . . . . . . . 2511.2 Strategy for verifying the axioms . . . . . . . . . . . . . . . . . . 26

12 February 16, 2018 2812.1 Cohomology of the ideles . . . . . . . . . . . . . . . . . . . . . . 28

13 February 21, 2018 3113.1 First inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

14 February 23, 2018 3314.1 Algebraic weak Chebotarev . . . . . . . . . . . . . . . . . . . . . 3314.2 Reduction of the second inequality to Kummer extensions . . . . 34

15 February 26, 2018 3515.1 Computing the size of the quotient . . . . . . . . . . . . . . . . . 3515.2 Choosing the completely split places . . . . . . . . . . . . . . . . 36

16 February 28, 2018 3816.1 Finishing the second inequality . . . . . . . . . . . . . . . . . . . 3816.2 Immediate corollaries . . . . . . . . . . . . . . . . . . . . . . . . . 39

17 March 2, 2018 4117.1 Reduction to cyclic cyclotomic extensions . . . . . . . . . . . . . 4117.2 Invariant and reciprocity maps . . . . . . . . . . . . . . . . . . . 42

18 March 5, 2018 4418.1 θ-reciprocity for cyclotomic fields . . . . . . . . . . . . . . . . . . 4418.2 Invariant map on the class group . . . . . . . . . . . . . . . . . . 45

19 March 7, 2018 4619.1 CK form a class formation . . . . . . . . . . . . . . . . . . . . . . 4619.2 Consequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

20 March 9, 2018 4920.1 Classification of normic subgroups . . . . . . . . . . . . . . . . . 49

21 March 19, 2018 5121.1 Properties of the Hilbert class field . . . . . . . . . . . . . . . . . 51

2

22 March 21, 2018 5322.1 Class field tower . . . . . . . . . . . . . . . . . . . . . . . . . . . 5322.2 L-functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

23 March 23, 2018 5523.1 L-functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5523.2 Convergence of Dirichlet series . . . . . . . . . . . . . . . . . . . 55

24 March 26, 2018 5724.1 Meromorphic continuation of L-functions . . . . . . . . . . . . . 57

25 March 28, 2018 5925.1 Densities of sets of primes . . . . . . . . . . . . . . . . . . . . . . 60

26 March 30, 2018 6126.1 Proof of second inequality . . . . . . . . . . . . . . . . . . . . . . 61

27 April 2, 2018 6327.1 Chebotarev density theorem . . . . . . . . . . . . . . . . . . . . . 63

28 April 4, 2018 6528.1 Complex multiplication . . . . . . . . . . . . . . . . . . . . . . . 6528.2 Elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

29 April 6, 2018 6729.1 Properties of the Weierstrass function . . . . . . . . . . . . . . . 6729.2 j-invariant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

30 April 9, 2018 7030.1 j-invariants of CM elliptic curves . . . . . . . . . . . . . . . . . . 70

31 April 11, 2018 7131.1 Modular forms and functions . . . . . . . . . . . . . . . . . . . . 71

32 April 13, 2018 7332.1 Congruence subgroups . . . . . . . . . . . . . . . . . . . . . . . . 73

33 April 16, 2018 7533.1 Modular polynomial . . . . . . . . . . . . . . . . . . . . . . . . . 75

34 April 18, 2018 7734.1 Coefficients of the modular polynomial . . . . . . . . . . . . . . . 77

35 April 20, 2018 7935.1 Main theorem of complex multiplication . . . . . . . . . . . . . . 79

3

36 April 23, 2018 8136.1 Cube root of j . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8136.2 Class number one problem . . . . . . . . . . . . . . . . . . . . . . 82

37 April 25, 2018 8337.1 Shimura reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . . 83

4

Math 223b Notes 5

1 January 22, 2018

Last semester, the big thing we did was local class field theory. Let K be anon-archemedian local field, (i.e., finite extensions of either Qp or Fp((t))), andL/K is Galois. What we showed last semester is that there exists a canonicalisomorphism

θL/K : K×/NL× → Gal(L/K)ab.

The proof was via Galois cohomology. The left hand side is H0(L/K,L×) andthe right hand side is H−2(L/K,Z). We needed two essential facts: Hilbert 90H1(L/K,L×) = 0, and H2(L/K,L×) ∼= 1

[L:K]Z/Z.

This semester, we’re going to talk about global fields. Now K is a globalfield, and consider a finite L/K. We want a global reciprocity map

θL/K : CK/NCL → Gal(L/K)ab.

This CK is going to be the adelic class group or the idele class group andis defined as

CK = A×K/K×.

The shape of the argument is going to be exactly the same. We’re going toneed the key lemmas on H1 and H2, and then everything else is going to beformal. This formalism that lets us do both local and global class field theoryis called class formations. So our agenda is:

• global fields, adeles, statement of global class field theory

• class formations

• algebraic proof of global class field theory

• L-functions, analytic proofs of global class field theory, Chebotarev densitytheorem

• complex multiplication for elliptic curves, abelian extensions of imaginaryquadratic fields

1.1 Global fields and adeles

This material is at the end of Chapter 2 of Cassels and Frolich. Let K be aglobal field. This can be define in two ways. First, it can be defined as a finiteextension of Q or Fp(t). An equivalent definition is that every completion of Kis a local field. A local field is valued field that is complete and locally compact.Local fields are classified by Fp((t)), Qp, their finite extensions, R, and C.

Definition 1.1. A place of K is an equivalence class of absolute values on K.It is usually denoted by v. We say that v is finite if v comes from a discretevaluation, and infinite otherwise, in which case Kv is either R or C.

Math 223b Notes 6

We can consider normalized absolute values. If v is finite, Kv is a non-archedemian local field, and so we can consider the uniformizer π ∈ Kv and theresidue field kv = Ov/(π). Then we define the normalized absolute value onKv by

|π|v = |kv|−1.

Also, if Kv∼= R then we define |a|v = |a|, and if Kv ∈ C then we define

|a|v = |a|2.One of the motivations for defining normalization in this way is the product

formula. For every a ∈ K×, we will have∏v

|a|v = 1.

We proved this last semester, by first doing it forK = Q,Fp(t), and then showingthat both sides are preserved by taking finite extensions. Another motivationis that a locally compact abelian group has a Haar measure. So any local fieldKv has a Haar measure such that µ(E) = µ(a + E) for all a ∈ K+

v , and sucha measure is unique up to scaling. First we normalize it so that µ(Ov) = 1, inthe non-archemedian case. We then consider |x|v = µ(xOv). It will follow thatµ(aE) = |a|vµ(E) for any measurable E. This is because E 7→ µ(aE) is anotherHaar measure, so it is constant times the original Haar measure. Using this, wewill give another proof of the product formula.

Now let me define the adeles in 5 minutes. For K a global field, we considerthe set {Kv} for v the places of K. The field K embeds into each Kv, and we aregoing to consider the embedding K ↪→

∏vKv. But this is too big. In general,

let {Xi} be a collection of topological groups and let Yi ⊆ Xi be a subgroup.The restricted topological product is defined as∏′

i

Xi = {(xi)i∈I : xi ∈ Yi for almost all i}.

Definition 1.2. We define the adele as

A+K =

∏′

v

K+v ,A

×K =

∏′

v

K×v

where the restricted products are with respect to O+v and O×v .

Math 223b Notes 7

2 January 24, 2018

Last time we defined the adeles, but let me go through this more carefully. Wefirst defined the restricted topological product. Let Yi ⊆ Xi be a collection oftopological spaces/groups/rings. We define the restricted product as

∏′Xi = {(ai)i∈I : ai ∈ Xi for all i and ai ∈ Yi for almost all i}.

The basis for the topology is given by

{(ai) : ai ∈ Ui}

where Ui ⊆ Xi are open subsets with Ui = Yi for almost all i. This is not thesubspace topology from the product topology for

∏iXi. If S is any finite subset

of I, the restricted product∏′iXi has an open cover of the form(∏′Xi

)S

=∏i∈S

Xi ×∏i/∈S

Yi.

The subspace topology here is equal to the product topology.

2.1 Compactness of the adele

Theorem 2.1. If Xi are locally compact and Yi are compact, then∏′

Xi islocally compact.

Proof. Each of (∏′

Xi)S is locally compact by Tychonoff’s theorem.

As we have defined, the adele

AK =∏′

v

Kv

restricted with respect to Ov is a topological ring. The units

A×K =∏′

v

K×v

restricted to O×v is a topological group. For the next few lectures, we are goingto focus on the structure and properties.

The topology on A×K is not the subspace topology from AK , by the way.This is a natural thing to do because the inverse map is not continuous withthe subspace topology. In general, for R a topological ring, the inverse map onR× is not necessarily continuous. The correct way to topologize R× is to usethe injection

R× ↪→ R×R; a 7→ (a, a−1)

Math 223b Notes 8

and use the induced subspace topology here. As an exercise, check that thistopology on A×K agrees with the topology by the restricted product. For localfields, we don’t need to worry about this issue.

We know that AK and A×K are locally compact. There is a natural embedding

K ↪→ AK ; x 7→ x = (x)

Likewise, we have K× ↪→ A×K .

Proposition 2.2. K+ is discrete in A+K , and is closed. Thus A+

K/K+ makes

sense, and it is compact and Hausdorff.

Let first prove the following.

Lemma 2.3. Let L/K be a finite extension. Then AL ∼= AK ⊗K L as topo-logical rings. (In this case, we can define the tensor product in the followingway. Because L ∼= Kn as vector spaces, we topologize AK ⊗K L by the producttopology.)

Proof. The important fact is that if v is a place of K, then Kv ⊗K L =∏v′ Lv′

for v′ extending v. We can apply this and check the topology.

Proof of Proposition 2.2. By the lemma, to show for L, it suffices to show forK. (This is because L ⊆ A+

L is equivalent to Kn ⊆ (A+K)n.) So we can check

for K = Q and K = Fp(t).Let’s only do this for K = Q. We take

U =∏v 6=R

Zv × (−1, 1) ⊆ AQ.

Then a ∈ U ∩ Q implies that a ∈ Z, so a = 0. So a point is open in Q, whichshows that Q is discrete. You can do Fp(t) in a similar way.

Now let us show that AQ/Q+ is compact. If we consider the space

D =∏v 6=R

Zv ×[−1

2,

1

2

],

this is compact by Tychonoff. Now the claim is that D+Q = AQ. We first wantfor a ∈ A+

K , an r ∈ Q such that a− r ∈ D. There are finitely many finite placessuch that ap /∈ Zp. We can make them all into Zp by adding a rational number.Then we can add an integer so that the infinite component is in [− 1

2 ,12 ]. So

D � A+Q/Q+ is a surjection. Because D is compact, the image is compact as

well.

Math 223b Notes 9

3 January 26, 2018

We should last time that A+K/K

+ is compact. First of all, A+K has a Haar

measure because its a locally compact abelian group. This can also describedexplicitly as the product of the local Haar measures, so that if

∏v Ev ⊆ A+

K issuch that Ev = K+

v is measurable for each v and Ev = O+V for almost all v,

then

µ

(∏v

Ev

)=∏v

µv(Ev).

This Haar measure descends to a quotient A+K/K

+ (by using a fundamentaldomain). Because A+

K/K+ is compact, the measure µ(A+

K/K+) is going to be

finite. As an exercise, show that µ(A+K/K

+) is the discriminant |discK|.

Definition 3.1. For a = (ai) ∈ A×K , we define its content as

c(a) =∏v

|av|v.

Proposition 3.2. If a ∈ A×K and E ⊆ A+K , then µ(aE) = c(a)µ(E).

Proof. This follows from the local case. It is clear that µ(avEv) = |av|vµ(Ev)in the local case.

Corollary 3.3. If a ∈ K× then c(a) = 1.

Proof. Consider the action on a on A+K by multiplication. Then µ(aE) =

c(a)µ(E) for any measurable E ⊆ A+K/K

+. Taking E = A+K/K

+ shows thatµ(E) = c(a)µ(E), and so c(a) = 1.

3.1 Compactness of the multiplicative adele

What about the K× ⊆ A×K? This is what this course will be ultimately careabout.

Proposition 3.4. K× is discrete inside A×K .

Proof. We have an embedding

K× ↪→ AK ↪→ A+K × A+

K .

But K+ ×K+ is already discrete inside A+K ×A+

K , so K× should be discrete aswell.

Likewise, K× is a closed subset. So A×K/K× is a Hausdorff topological group.Is this going to be compact? The answer is no, because we have a surjectivecontinuous map

c : A×K/K× → R>0.

(It can be shown that it is continuous.) So define

A1K = ker(c : A×K → R>0).

Math 223b Notes 10

Theorem 3.5. A1K/K

× is compact.

We will prove this a bit later, but as two important corollaries: finiteness ofthe class group, and Dirichlet’s unit theorem. (Neukirch actually goes the otherdirection to prove compactness.)

3.2 Applications of the compactness

Definition 3.6. For a ∈ A×K , define

Sa = {x ∈ A×K : |xv|v ≤ |av|v for all v} ⊆ A+K .

Theorem 3.7 (Minkowski). There exists a constant C > 1 (depending on K)such that for all a ∈ A×K with c(a) > C, there exists x ∈ K× ∩ Sa.

Proof. Consider the set

B =∏v fin

O+V ×

∏v inf

B(0, 12 ) ⊆ A+

K .

Then you can explicitly compute µ(B), and it is going to be µ(B) < ∞. If wedefine C = µ(A+

K/K+)/µ(B), then we have

µ(aB) = c(a)µ(B) > cµ(B) = µ(A+K/K

+).

So the quotient map aB → A+K/K

+ is non-injective. This shows that there areb1, b2 ∈ aB such that x = b1 − b2 ∈ aB ∩K×. By the non-archemedian triangleinequality, we have

|xv|v ≤ max{|(b1)v|v, |(b2)v|v} ≤ |av|v.

For the infinite components, we do the same thing and we get x ∈ aB ⊆ Sa.

Math 223b Notes 11

4 January 29, 2018

At the end of last time, we proved the adelic version of Minkowski’s lemma.

Theorem 4.1 (Minkowski). Let K be a global field. There exists a constant c(depending on K) such that for all a ∈ AK and c(a) > c such that there existsan x ∈ Sa ∩K× where

Sa = {x ∈ AK : |xv|v ≤ |av|v}.

There is another nice theorem that isn’t necessary but nice.

Theorem 4.2 (strong approximation). Let v0 be any place of K. The map

K ↪→ A+K/K

+v0 =

∏′

v 6=v0

Kv.

has dense image.

Proof. Let S be a finite set of places, and consider

B =∏v∈S

B(xv, ε)×∏v/∈S

O+V ,

where x ∈ A+K and ε > 0. This describes a neighborhood basis for x, and we

need to show that any such B contains an element of K+. We use compactnessof A+

K/K+ to get that there exists some a ∈ A×K such that (the image of) Sa

covers A+K/K

+. Now we can choose b ∈ A×K such that |bv|v < ε|av|v for v ∈ S

and c(b) > c. (c is from Minkowski.) Then there exists a z ∈ Sk ∩K×. Write

x

z= r + s ∈ A+

K = K+ + Sa,

where r ∈ K and s ∈ Sa. Now rz = x−sz with rz ∈ K. But x−sz ∈ B because|(sz)v|v ≤ |avbv|v < ε for v ∈ S. We also need to show that (x− sz)v ∈ Ov forv /∈ S, but we can enlarge S to make this true.

Proposition 4.3. A1K/K

× is compact.

(Here, you’ll prove in the homework that A1K has the same topology as a

subspace of A×K and as a subspace of AK . Also, A1K is closed in AK .)

Proof. Let D = Sa ∩A1K , where c(a) > c. Note that Sa is closed in AK , and so

Sa is compact by Tychonoff. So D = Sa ∩ A1K is also compact. Now it suffices

to show that D � A1K/K

× is a surjection. This means for each x ∈ A1K , there

exists a d ∈ D such that xd ∈ K×. For this, consider x−1Sa = Sx−1a. Here,

c(x−1a) = c(x)−1c(a) = c(a) > c, and so there exists an element r ∈ K× suchthat r ∈ Sx−1a. Then rx ∈ A1

K ∩ Sa = D.

Math 223b Notes 12

4.1 The class group

If K is a number field, it has a ring of integers OK , and the class group of Kis defined as

Cl(K) = Cl(OK) =group of fractional ideals of OK

group of principal fractional ideals of OK.

More generally, let K be a global field and let S be the finite nonempty setof places of K such that S contains all the archemedian places. If we define

OK,S = {x ∈ K : |xv|v ≤ 1 for v /∈ S}.

It can be shown that OK,S is a Dedekind domain and the nonzero primes ofOK,S correspond to places not in S. If S is the set of archemedian places, thenwe can define OK = OK,S . As before, we can define the class group

ClS(K) = coker(K× → I(K,S))

where I(K,S) is the group of fractional ideals of OK,S . (By the way, the kernelis O×K,S .) Next time we are going to show that ClS(K) is finite and O×K,S isfinitely generated of rank |S| − 1.

Math 223b Notes 13

5 January 31, 2018

Last time we showed that A1K/K

× is compact, if K is a global field.

5.1 Classical theorems from compactness

We will prove the following theorem:

Theorem 5.1. Let S be a nonempty finite set of places of K, containing allthe infinite places. Then

ClS(K) = Cl(OK,S)

is finite and the group of units O×K,S is finitely generated of rank |S| − 1.

Let us define

A×K,S = {v : |xv|v = 1 for v /∈ S} =∏v∈S

K×v ×∏v/∈S

O×v .

This is the adelic version of O×K,S . It is also one of the open sets that define

A×K . Also let us define A1K,S = AK,S ∩A1

K . This is also open in A1K . We have a

short exact sequence

1→ A1K,SK

×/K× → A1K/K

× → A1K/A1

K,SK× → 1.

The middle space is compact, and A1K,SK

× is an open subgroup of A1K .

So the first thing is an open subgroup of the second thing, which is compact.Therefore it is also a closed subgroup, and hence compact. Then the second tothird map is continuous and the third space is compact as well. But when youquotient by an open subgroup, you get a discrete subgroup. This all shows that

A1K/(A1

K,S ·K×)

is a finite discrete group.

Proposition 5.2. A1K/(A1

K,SK×) ∼= ClS(K).

Proof. Consider the map

ϕ : A1K → ClS(K); a 7→

[∏v/∈S

pvp(av)

].

where p is the prime ideal corresponding to the finite place v. The kernel of ϕ isprecisely those that can be expressed as things coming from K× and those thatgive the fractional ideal (1). So it is kerϕ = A1

K,S ·K×. On the other hand, ϕsurjective because we can just set av as we want. So we get the isomorphism.

Proposition 5.3. (AK,SK×)/K× ∼= A1K,S/O

×K,S.

Proof. It suffices to show that O×K,S ∼= A1K,S ∩K×. This is clear.

Math 223b Notes 14

We had that A1K,S/O

×K,S is compact. So we are going to focus on what is

happening with the S part.

Lemma 5.4. The set

{x ∈ O×K,S : |x|v ∈ [ 12 , 2] for all v ∈ S}

is finite.

Proof. This is the same as the intersection of K× and∏v/∈S O

×V ×

∏v∈S Bv

which is discrete and compact. So it is finite.

Corollary 5.5. A number a ∈ K× is a root of unity if and only if |a|v = 1 forall v.

Proof. One direction is obvious. For the other direction, we note that the group{a ∈ K : |a|v = 1 for all v} is finite, and hence torsion.

Now define a homomorphism

L : A1K,S →

∏v∈S

R+; (L (a))v = log(|av|v).

Proposition 5.6. L (OK,S) is a discrete subgroup of∏v∈S R+, and the kernel

of L : O×K,S →∏v∈S R+ is finite.

Proof. This follows from the lemma, because it is saying that L −1 of somecompact region is finite. Then the image has to be discrete, and the kernelshould be finite.

Now we have L : A1K,S → (R+)|S|, and this induces a map

A1K,S/O×K,S → L (A1

K,S)/L (O×K,S).

But the condition on L (A1K,S) is that all components should add up to 1. It is

also true that this is the only condition. This shows that

L (A1K,S) = H = {x ∈ (R+)s :

∑v xv = 0}.

On the other hand, A1K,S/O

×K,S is compact, and so L (A1

K,S)/L (O×K,S) is

compact. This shows that L (O×K,S) is a free group of rank |S| − 1. Therefore

O×K,S is a finitely generated group of rank |S| − 1.

Math 223b Notes 15

6 February 2, 2018

We have shown that the group A1K/K

× is compact. In class field theory, wewill use CK = A×K/K×. How will we compare it to CK?

We have a short exact sequence relating the two. Suppose first that K is anumber field. Then we have a short exact sequence

1→ A1K/K

× → A×K/K× c−→ R>0 → 1.

For each archemedian place v, we have a closed embedding R ↪→ K×v , and then aclosed embedding K×v ↪→ A×K/K×. This gives a splitting as topological groups.

If K is a function field over Fq, then we have

1→ A1K/K

× → A×K/K× c−→ qZ → 1.

Again, there is a splitting, not in a particularly natural way.

6.1 Statement of global class field theory

Theorem 6.1 (Main theorem of global class field theory). If K is a global field,we have a reciprocity map

θ/K : CK = A×K/K× → Gal(Kab/K)

with dense image, and ker(θ/K) = (CK)0 the connected component of CK . Thiscan be described as the closure of the image of

(AK)0 =∏v inf

(K×v )0

which is easy to describe. There is also a local-to-global compatibility

K×v Gal(Kabv /Kv)

A×K/K× Gal(Kab/K).

θ/Kv

θ/K

This local-to-global compatibility specifies θ/K uniquely becauseKv generateit topologically. So θ/K induces a bijection{

open subgroups ofCk of finite index

}←→

{open subgroups of

Gal(Kab/K) of finite index

}.

Then the right hand side is the same as a finite abelian extension L/K. Themap from extensions L/K to subgroups of CK is going to be given by

L 7→ NL/K(CL) ⊆ CK .

Math 223b Notes 16

Here, we define

NL/K : A×L → A×K ; a 7→ (Na)v =∏w/v

NLw/Kv(aw).

It turns out that

L× K

A×L A×K

N

N

commutes, so we get a map CL → CK . It turns out that NL/K(A×L ) ⊆ A×K isopen.

6.2 Ray class field

Definition 6.2. A modulus of a field K is a function m : {places of K} → Z≥0

with the property

• m(v) = 0 for all but finitely many v,

• m(v) = 0 or 1 if v is real,

• m(v) = 0 if v is complex.

We write m =∏v v

m(v) (so m = p1p2∞1 for instance.)

For any m we have a congruence subgroup(∏p fin

Up,mp×

∏v inf m(v)=0

Kv ×∏

v real m(v)=1

R>0

)K×/K× ⊆ CK ,

whereUp,mp

= {x ∈ OKp: x ≡ 1 (mod pmp)}.

Definition 6.3. The ray class field Lm is the fixed field of θ(Um), so that

Gal(Lm/K) ∼= CK/Um∼= A×K/K

×(∏

p fin

Up,mp×∏v fin

(−)

).

The punchline is that

CK/Um∼= Clm(OK)

is the ray class group, which is the fractional ideals of OK relatively prime tom modulo the principal fractional ideals (a) with a ∈ Up,mp

for all finite p andav > 0 for v real with m(v) = 1.

Math 223b Notes 17

7 February 5, 2018

Last Friday I started introducing the adelic statement of global class field theory.

Theorem 7.1 (Main theorem of global class field theory). Let K be a globalfield. There exists a canonical θ/K : CK → Gal(Kab/K) such that

• ker θ/K = (CK)0 is the connected component at 1,

• im θ/K is dense in Gal(Kab/K) = lim−→L/K fin. ab.Gal(L/K) (this means

that θL/K : CK → Gal(L/K) is surjective for any L/K).

7.1 Image of the reciprocity map

This map θ/K that has dense image is actually surjective for K a number field,but not surjective for K a function field.

Proposition 7.2. If K is a number field, then CK/(CK)0 is compact.

Proof. We know that A1K/K

× is compact. So it suffices to check that this sur-jects onto CK/(CK)0. This is because we can change only the archemediancomponent to make the content 1, and this doesn’t change the connected com-ponent.

Now CK/(CK)0 is compact, so the image of θ/K in Gal(Kab/K) is compact.This means that the image is closed, and also is dense. So it is the whole spaceGal(Kab/K) for K a number field.

Let us see what goes wrong when K is a function field. Then all of thelocalizations are totally disconnected, so A×K is totally disconnected and CK istotally disconnected. Then the map θ/K is injective. If θ/K were an isomor-phism, we would have that CK is profinite and thus compact. Let k = Fq bethe field of constants, and consider the field extension

k ·K ⊆ Kab.

Then we get a homomorphism

Gal(Kab/K)→ Z ∼= Gal(k/k).

It turns out that the square

1 IK Gal(Kab/K) Z ∼= Gal(k/k) 1

1 A1K/K

× CK = A×K/K× qZ 1

∼=

c

θ/K q 7→1

commutes, and also both exact sequences split.

Math 223b Notes 18

7.2 Identifying the ray class group

Recall that for K a number field, a modulus is a formal product m =∏v v

m(v).Then we can define an open subset Um ⊆ A×K

Um =∏p fin

Up,m(p) ×∏

v real,m(v)=1

R>0 ×∏

v other

K×v .

Then the congruence subgroup CmK ⊆ CK is defined as

CmK = Um ·K×/K× ⊆ A×K/K

× = CK .

We also defined the ray class field of modulus m as the fixed field of θ/K(CmK).

In this case, we have

Gal(Lm/K) ∼= Gal(Kab/K)/θ/K(CmK) ∼= CK/C

mK

Proposition 7.3. CK/CmK is isomorphic to the ray class group Clm(K), where

Clm(K) =frac. ideals rel. prime to m

frac. ideals (a) with a ∈ Um.

Proof. Exercise.

We can give the inverse map explicitly. If p is relatively prime to m then wecan send p to [(1, . . . , 1, π, 1, . . .)] where π ∈ K×p is an arbitrary uniformizer.

Proposition 7.4. The field Lm/K is unramified at all p relatively prime to m.

Proof. We have an isomorphism

Clm(K) ∼= CK/CmK

θ/K−−→ Gal(Lm/K).

This sends p to Frobp lying in the decomposition group. This means that Lm isa class field in the classical sense.

Math 223b Notes 19

8 February 7, 2018

Recall that for K a number field, we defined the ray class field Lm of modulusm by the fixed field of

θL/K(CmK) = θL/K(UmK

×/K×).

8.1 Properties of the ray class field

Proposition 8.1. (a) Lm is unramified at all places not dividing m. (Forinfinite places, we say that v′/v ramifies if v is is real and v′ is complex.Neukirch uses a different convention, that infinite places never ramify.)

(b) If m | m′ then Lm ⊆ Lm′ .

(c) Lm1∩ Lm2

= Lgcd(m1,m2).

(d)⋃

m Lm = Kab.

Proof. (a) First use the fact that if p - m then

O×p ↪→ Um → CmK .

For p′ any prime of Lm over p, we have

K×p Gal((Lm)p′/Kp)

CK Gal(Lm/K).

θ/Kp

θLm/K

Now we have CmK = ker θLm/K and so O×p ⊆ ker θ(Lm)p/Kp

. This shows that

θ(Lm)p/Kp(O×Kp

) is trivial. By local class field theory, this just means that

(Lm)p/Kp is unramified. For infinite places, we use K×p instead of O×p .(b) and (c) just follow from the definition of Um and Cm

K . For (d), observethat any open subgroup of A×K contains some Um. Then any open subgroup ofCK contains some Cm

K . Using global class field theory and Galois correspon-dence, we see that any finite abelian extension lies in some Lm.

Example 8.2. Let K = Q. We can take m = m or m = m∞ where m is apositive integer. First let us figure out Clm(K) and then figure out Lm. First,if m = m∞,

Clm(Q) =frac. ideals rel. prime to m

(a) where a > 0 and a ≡ 1 mod m∼= (Z/mZ)×.

If m = m, then we would get (Z/mZ)×/{±1}.It now follows from the explicit description of θ/K for cyclotomic fields that

Lm∞ = Q(ζm). If we don’t allow ∞, we will get Lm = Q(ζm + ζ−1m ).

Math 223b Notes 20

Definition 8.3. Let L be a finite abelian extension of K. The minimal m suchthat L ⊆ Lm is called the conductor of L, and is written f = fL/K .

Another way to describe f is that f is minimal such that

CK Gal(L/K)

CK/CfK∼= Clf(K)

θL/K

factors.We know that L is unramified outside the primes p - f. Then for each p - f,

we have ( p

L/K

)= Frobp ∈ Gal(L/K),

and ( pL/K ) is the restriction of ( p

L/K ) to Gal(L/K). Since ( pLp/K

) depends only

on [p] ∈ Clf(K), same is true of ( pL/K ).

Also, for ker θL/K = NL/KCL. So L ⊆ Lm is equivalent to NL/KCL ⊇ CmK .

This means that f is the minimal m that makes this true.You can show as an exercise that p ramifies if and only if p | fL/K .

Example 8.4. Take K = Q, and L = Q(√p) where p ≡ 1 (mod 4) and p is

prime. Then L is ramified only at p and fL/Q = p. For q any other prime, wehave that

Frobq(Q(√p)/Q) =

( q

Q(√p)/Q

)=(qp

)only depends on the image of q in (Z/pZ)×/(±1).

Let K be an arbitrary field, and let m = 1. Then

Clm(K) = Cl(OK).

Here, L1 = H is called the Hilbert class field that is unramified at all placesover K, including the infinite places. For m =

∏v real v, we can similarly define

Lm = H+ the narrow Hilbert class field.

Math 223b Notes 21

9 February 9, 2018

The goal is to set up the abstract theory of class formations. Recall local classfield theory. Here, K is a local field. The key fact was homological: if L/K isfinite Galois, then

K×/NL× ∼= H0(L/K,L×) ∼= H−2(L/K,Z) ∼= Gal(L/K)ab.

Then we took the direct limit over all finite L/K to get

lim←−K×/NL× ∼= Gal(Kab/K).

In characteristic zero, this gives K× ∼= Gal(Kab/K).Here are the inputs we put into the Galois cohomology machinery:

• for every finite L/K, a Gal(L/K)-structure on L×

• the collection of groups {Gal(L/K)} for L finite Galois, and quotient mapsGal(E/K)� Gal(L/K) when E/L/K

• compatibility conditions on the Gal-module structures with E× = (L×)Gal(E/L).

We can also package this whole thing into G = Gal(Ksep/K) and A = (Ksep)×.Then our collection is exactly quotients of G by finite index open subgroups,with L× = ((Ksep)×)Gal(Ksep/L).

The goal is to be able to replace K× with CK everywhere.

9.1 Ideles with field extensions

For L/K Galois, we know already that K× ↪→ L× and K× = (L×)Gal(L/K).We will need an analogous statement for CK and CL.

Let L/K be a finite extension of global fields (not necessarily Galois). Wehave closed embeddings

AK ↪→ AL, A×K ↪→ A×L .

Proposition 9.1. A×K is closed in A×L . If L/K is Galois, then (A×L )Gal(L/K) =AK and A×K ∩ L× = K×.

Proof. All follows immediately form AL ∼= AK ⊗K L as topological rings. Be-cause L/K is Galois, they are isomorphic as Galois modules.

So we get an inclusion A×K/K× → A×L/L×. That is, CK ↪→ CL.

Proposition 9.2. We have CGal(L/K)L = CK .

Proof. We use the short exact sequence

1→ L× → A×L → CL → 1.

If we take Galois cohomology, we get

1→ K× → A×K → CGal(L/K)L → H1(L/K,L×) = 1.

This shows that CGal(L/K)L

∼= CK .

In general, Cl(OK)→ (Cl(OL))Gal(L/K) is neither injective nor surjective.

Math 223b Notes 22

9.2 Formations

Let G be a profinite group. Think of G as some absolute Galois group. In thiscase, open subgroups as finite extensions of the field. Going back to the generalcase, consider the set of open subgroup {GK : K ∈ X} of G. We are going torefer to these indices K as “fields”.

In particular, we have a partial ordering of field, with K ⊆ L if and only ifGL ⊆ GK . There is a unique minimal field K0 corresponding to GK0

= G, andwe call K0 the “base field”. A pair (K,L) with K ⊆ L is called a “layer” L/K.The “degree” of the layer is defined as [L : K] = [GK : GL]. We say that L/Kis “normal” if GL C GK is a normal subgroup. We can define

GKL = GK ∩GL, GK∩L = GKGL.

The group G acts on the set X of all fields by

GgK = gGKg−1.

Definition 9.3. A formation A is a G-module satisfying the following equiv-alent conditions:

• G acts continuously on A (using the discrete topology on A)

• A =⋃K A

GK

Example 9.4. Suppose that G = Gal(Ksep0 /K0). Then A = (Ksep

0 )× is aformation. For global fields, we can instead look at A = lim−→K/K0 fin

CK .

Math 223b Notes 23

10 February 12, 2018

We were talking about class formations. This was a profinite group G with{GK : K ∈ X} indexing the open subgroups of G. These K ∈ X were calledfields. A G-module was an abelian group A with a G-action such that the actionis continuous. This can be explicitly described as

A =⋃K

AGK =⋃K

AK

where we define AK = AGK . We also say that L/K is a normal layer if GL C Gkis normal. Then we write GL/K = GK/GL. Then GL/K acts on AL and AL isa GL/K-module.

Definition 10.1. We define Hq(L/K) = Hq(GL/K , AL). Likewise, se define

Hq(L/K) = Hq(GL/K , AL). We can also define profinite cohomology

Hq(/K) = Hq(GK , A) = lim−→L

Hq(GL/K , AL) = lim−→L

Hq(L/K).

If E/L/K, with E/K and L/K both normal layers, we have inflation maps

inf : Hq(L/K) = Hq(GL/K , A) = Hq(GL/K , (AE)GL/GE )→ Hq(GE/K , AE) = Hq(E/K).

The are also restriction and corestriction. If E/L/K and E/K is normal thenE/L is normal. Then we have GE/K ⊇ GE/L. So we get restriction andcorestriction maps

res : Hq(E/K)→ Hq(E/L), cores : Hq(E/L)→ Hq(E/K).

There is also conjugation action. If L/K is a normal layer and g ∈ G, wecan define GgK = gGKg

−1. Then we have isomorphisms

GL/K ∼= GgL/gK ; [x] 7→ [gxg−1].

We also have isomorphisms

AL → AgL; a 7→ ga,

and this inducesg∗ : Hq(L/K)→ Hq(gL/gK).

If g ∈ GK , then gK = K and gL = L because L/K is normal. In this case, youcan check that

g∗ : Hq(L/K)→ Hq(L/K)

is the identity by checking at q = 0 and then dimension shifting.

Math 223b Notes 24

10.1 Class formation

To say something other than formalities with cohomology, you need to inputsomething.

Definition 10.2. (G,A) is a field formation if for every layer L/K, H1(L/K) =0.

Of course, the motivating example here is G = Gal(Ksep0 /K0) and A =

(Ksep0 )×. From this, we get an inflation-restriction sequence on the second

cohomology:0→ H2(L/K)→ H2(E/K)→ H2(E/L)

whenever E/L/K with E/K and L/K normal. Then we have

H2(/K) = lim−→H2(L/K) =⋃L/K

H2(L/K).

For local fields, we also had the fact that H2 is cyclic. But we want to saythis carefully.

Definition 10.3. A is a class formation if for every L/K we provide anisomorphism

invL/K : H2(L/K)→ 1[L:K]Z/Z,

with compatibility conditions

H2(L/K) 1[L:K]Z/Z

H2(E/K) 1[E:K]Z/Z

invL/K

inf

invE/K

for all normal E/L/K, and

H2(E/K) 1[E:K]Z/Z

H2(E/L) 1[E:L]Z/Z

res

infE/K

×[L:K]

infE/L

for E/L/K with E/K normal.

Try to deduce the diagram for corestriction.

Math 223b Notes 25


A class formation is a formation (G, (GK)K∈X , A) satisfying the axioms

(i) (Hilbert 90) H1(L/K) = 0 for all normal L/K,

(ii) there are invariant maps inv : H2(GL/K , AL) ∼= 1[L:K]Z/Z compatible with

inflation and restriction (and thus correstriction)

Example 11.1. Suppose G = Z and suppose A = Z with the trivial action.The fields are Kn = nZ with AKn = Z. Then for m | n,

H1(Kn/Km) = H1(nZ/mZ,Z) = 0

andH2(Kn/Km) = H2(nZ/mZ,Z) = Z/( nmZ).

You can of this as something over finite fields.

Example 11.2. By last semester, we know that for K a local field, G =Gal(Ksep/K) with A = (Ksep)× is a class formation.

Our goal is to show that for K0 global,

G = Gal(Ksep0 /K), A =

⋃L/K

CL

is a class formation. In this case, AL = CL.

11.1 Consequences of the class formation axiom

Proposition 11.3. Let A be a class formation.

(a) For E/L/K with E/K normal, the following commutes.

H2(E/L) 1[E:L]Z/Z

H2(E/K) 1[E:K]Z/Z

cor

invE/L

invE/K

(b) For normal L/K and g ∈ G, the following commutes.

H2(L/K) 1[L:K]Z/Z

H2(gL/gK) 1[L:K]Z/Z

g∗

invL/K

invL/K

Math 223b Notes 26

Proof. For (a), we simply note that

H2(E/K)res−−→ H2(E/L)

cor−−→ H2(E/K)

is multiplication by [L : K]. For (b), if g ∈ GK then we are done becauseg∗ : H2(L/K) → H2(L/K) is the identity map as we have shown. In general,we have g ∈ GK0 = G, so we can find L′ ⊇ L such that L′/K is normal. Thenyou compute the diagrams for L/K and L′/K0.

If (G,A) is a class formation, we define the fundamental class

uL/K ∈ H2(L/K) by inv(uL/K) =1

[L : K].

Clearly uL/K generates H2(L/K).

Proposition 11.4. Let E/L/K be fields with E/K normal (and L/K normalif necessary).

(a) resuE/K = uE/L.

(b) inf uL/K = [E : L]uE/K .

(c) coruE/L = [L : K]uE/K .

(d) g∗(uE/K) = ugE/gL.

Theorem 11.5. For any normal layer L/K, cup product with uL/K gives anisomorphism

Hq(GL/K ,Z)→ Hq+2(L/K).

Proof. This is literally Tate’s theorem.

In particular, for q = −2 we get an isomorphism

− ` uL/K : GabL/K = H−2(GL/K ,Z)→ H0(L/K) ∼= AK/NL/KAL.

So if we can show that (Gal(Ksep0 /K),

⋃L/K CL) is a class formation for any

global base field K0, then we get

Gal(L/K)ab ∼= CK/NCL

for all L/K normal.

11.2 Strategy for verifying the axioms

We will need to check Hilbert 90, and we will also need to construct this invariantmap. Let me outline the strategy.

Lemma 11.6 (first inequality). For L/K cyclic, |H2(L/K)| ≥ [L : K].

Math 223b Notes 27

We will prove this by looking at the Herbrand quotient. In fact, we will show

|H2(L/K)||H1(L/K)|

= [L : K].

Lemma 11.7 (second inequality). For L/K cyclic and [L : K] prime, |H2(L/K)| ≤[L : K].

This is actually easier to prove analytically. But Chevalley managed toprove this algebraically in the 1940s. This formally implies |H1(L/K)| = 1 and|H2(L/K)| ≤ [L : K] for all L/K. We will then construct an invariant map

invL/K : H2(L/K)→ 1[L:K]Z/Z

by patching the local invariant maps together.

Math 223b Notes 28


Our goal is now to find Hq(L/K,CL) for q = 1, 2. The first step is to look atHq(L/K,A×L ).

12.1 Cohomology of the ideles

Let S be a finite set of places of K, and S be the places of L lying over S. ThenA×L is covered by

A×L,S = A×L,S

=∏w∈S

L×w ×∏w/∈S

O×w .

We can then compute

Hq(L/K,A×L,S) = Hq

(G,∏v∈S

(∏v′|v

L×v

)×∏v/∈S

(∏v′|v

O×v))

=∏v∈S

Hq

(G,∏v′|v

L×v

)×∏v/∈S

Hq

(G,∏v′|v

O×v).

For each v, choose w | v and let Gw be the decomposition group of w. Weknow that G acts on the set v′ | v transitively. So we can say that∏

v′|v

∏g∈G/Gw

Lgw

is the coinduced module. So

Hq(G,∏v′|v L

×v′)∼= Hq(Gw, L

×w).

This isomorphism can be realized explicitly as

Hq(G,∏v′|v L

×v′)

res−−→ Hq(Gw,∏v′|v L

×v′)

π∗−→ Hq(Gw, L×w) ∼= Hq(Lw/Kv, L

×w).

Likewise, we have

Hq(G,∏v′|v O

×v′)∼= Hq(Gw,O×w ) ∼= Hq(Lw/Kv,O×w ).

The statements are also true for also Hq.Suppose that v is unramified in L. Then Hq(Kw/Kv,O×w ) vanishes. In

particular, if we assume that S contains all ramified places of K, then

Hq(L/K,A×L,S) ∼=∏v∈S

Hq(Lw/Kv, L×w).

Now we can writeA×L = lim−→

S

A×L,S

Math 223b Notes 29

where S runs over all finite sets containing all ramified places. Then

Hq(L/K,A×L ) = lim−→Hq(L/K,A×L,S)

= lim−→S

∏v∈S

Hq(Lw/Kv,K×w ) =

⊕v

Hq(Lw/Kv, L×w).

An immediate consequence is that

H1(L/K,A×L ) ∼= 1.

That is, the ideles give a field formation. That is, if we define

A×Ksep =⋃L/K

A×L ,

then (Gal(Ksep,K),A×Ksep) form a field formation.But this is not a class formation, because we can also compute

H2(L/K,A×L ) ∼=⊕v

H2(Lw/Kv, L×w) ∼=

⊕v

1[Lw:Kv]Z/Z

and it is infinite. So this is far off from what we need for a class formation.Later, we are going to see that⊕

v

1[Lw:Kv ]Z/Z ∼= H2(L/K,A×L )→ H2(L/K,Ck) ∼= 1

[L:K]Z/Z

is given by adding up the components.The other interesting thing we can compute is

A×K/NA×L = H0(L/K,A×L ) ∼=⊕v

H0(Lw/Kv, L×w) =

⊕v

K×v /NL×w .

This is equivalent to the norm theorem for ideles.

Theorem 12.1. For (av) ∈ A×K , we have a ∈ NA×L if and only if av ∈ NL×wfor all v.

There is an analogue for principal ideles, which requires the second inequal-ity.

Theorem 12.2 (Hasse). For a ∈ K× with av ∈ K×v the v-component, a ∈ NL×if and only if av ∈ NL×w for all v.

Next, we want to get the Herbrand quotient of CL = A×L/L× when L/K iscyclic. The problem is that the Herbrand quotient of A×L is infinite, so we can’tuse this directly. So we are going to choose S large enough so that A×L,S → CL is

surjective. This can be done by making S generate Cl(OL). Also, let S contain

Math 223b Notes 30

all the ramified primes, all infinite primes, and make S generate Cl(K). Thenwe get a short exact sequence

1→ O×K,S → A×L,S → CL → 1.

So

h(CL) =h(A×L,S)

h(O×L,S)=

∏v∈S [Lw : Kv]

h(O×L,S).

Next time we will calculate the Herbrand quotient of O×L,S .

Math 223b Notes 31


Our goal today is to get something for L/K cyclic extension of global fields. IfG = Gal(L/K) and |G| = n = [L : K], we want to show that H0(L/K) ≥ n.

13.1 First inequality

We want to compute the Herbrand quotient

h(CL) =|H0(G,CL)||H1(G,CL)|

.

If S is a sufficiently large finite set of places of K, we have a short exact sequence

1→ O×L,S → A×L,S → CL → 1.

Last time we showed that

h(AL,S) =

∏v∈S [Lw : Kv]

h(O×L,S),

so we need to find h(O×L,S).Recall that we have the log map of G-modules

L : O×L,S →∏v′∈S

R+; (a 7→ (log|a|v′)v′∈S

whose image lies in the hyperplane∑v′ rv′ = 0. The image is a lattice, and so

induces an isomorphismO×L,S ⊗Z R ∼= H.

Lemma 13.1. Let G be a finite group. Let L/K be an extension, with K isinfinite. Two finite-dimensional K[G]-modules V1 and V2 are isomorphic if andonly if V1 ⊗K L and V2 ⊗K L are isomorphic as L[G]-modules.

Proof. Suppose V1⊗KL and V2⊗KL are isomorphic as L[G]-modules. First, wecan reduce to the case when L/K is finitely generated, because the isomorphismis going to be given by some finite number of coefficients. Then by Noethernormalization, L is a finite extension of a transcendental extension of K, andthe transcendental extension can be taken care easily by specialization. So weonly need to consider when L/K is finite.

Now use that there is an isomorphism

HomL[G](V1 ⊗K L, V2 ⊗K L) ∼= HomK[G](V1, V2)⊗K L.

Because V1 ⊗K L ∼= V2 ⊗K L, there is a determinant map HomL[G](V1 ⊗KL, V2 ⊗K L) → L, which is nonzero. But this is a polynomial function onHomK[G](V1, V2)⊗K L. So there must be some ϕ ∈ HomK[G](V1, V2) such thatdetϕ 6= 0.

Math 223b Notes 32

There is another proof using Galois descent. Consider

IsomL[G](V1 ⊗ L, V2 ⊗ L)

which is a torsor for the group AutL[G](V1 ⊗ L). But this is isomorphic to theproduct of factors GLn(L), and H1(G,GLn(L)) ∼= 1. This shows that there issome ϕ ∈ (V1 ⊗ L, V2 ⊗ L)G that gives an isomorphism V1

∼= V2. understandunderstand

Proposition 13.2. If A, B are G-modules, finitely generated as abelian groups,and A⊗Z R ∼= B ⊗Z R, then h(A) = h(B).

Proof. The lemma gives A ⊗Z Q ∼= B ⊗Z Q. Then there exists a finite indexsubgroup A′ ⊆ A/T (A) and B′ ⊆ B/T (B) such that A′ ∼= B′. Then

h(A) = h(A′) = h(B′) = h(B)

because finite groups have Herbrand quotient 1.

We have an isomorphism

O×L,S ⊗Z R ∼= H ⊆∏v′∈S

R+

which extends to an isomorphism

(O×L,S ⊕ Z)⊗Z R ∼=∏v′∈S

R+

with (id⊕1)⊗ 1 7→ (1, . . . , 1). This shows that

h(O×L,S ⊕ Z) = h(O×L,S) · n

is equal to

h(∏v′∈S Z) = h(

∏v∈S

∏v′|v Z) =

∏v∈S

h(coindGGwZ)

=∏v∈S

h(Gw,Z) =∏v∈S|Gw| =

∏v∈S

[Lw : Kv].

Therefore

h(O×L,S) =

∏v∈S [Lw : Kv]

nand so

h(CL) =h(A×L,S)

h(OL,S)= n.

The conclusion is that

|H0(L/K,CL)| = [L : K]|H1(L/K,CL)| ≥ [L : K].

Here, we haveCK/NCL = A×K/K

× ·NA×L .

Math 223b Notes 33


We showed that |H0(L/K,CL)| ≥ [L : K] for L/K a cyclic extension of globalfields. By the way we have

CK/NCL ∼= A×K/NA×L ·K×.

14.1 Algebraic weak Chebotarev

Corollary 14.1. Suppose L/K is abelian, and let D ⊆ NL/KA×L ⊆ A×K be a

subgroup. If K×D is dense in A×K , then L = K.

Proof. First we reduce to when L/K is cyclic. If L/K is abelian, there exists aL′ ⊆ L with L′/K cyclic. Then we can replace L by L′.

Observe that NL/KA×L is open in A×K . So K× ·NL/KA× is open in A×K , and

hence closed. But K×D is dense. This shows that K× · NL/KA×L = A×K . So

1 = |A×K/NL/KA×L | ≥ [L : K]. So L = K.

Corollary 14.2. Let L/K be a finite abelian extension with L 6= K. Then thereare infinitely many primes of K that do not split completely in L.

Proof. Suppose there are finitely many prime that are not split completely. LetS be this set of primes that don’t split completely, and also the infinite primes,and set

D = {x ∈ A×K : xv = 1 for all v ∈ S}.

I need to check that D ⊆ NL/KA×L . If v /∈ S then Lw = Kv so NL×w = K×v . We

also need to check that K×D is dense in A×K . This is equivalent to K× beingdense in

∏v∈S K

×v . This is the weak approximation. So the technical lemma

implies that L = K.

Corollary 14.3. If S is a finite set of places of K, and L/K is finite abelian,then Gal(L/K) is generated by(L/K

v

)= Frobv ∈ Gal(L/K)

for v /∈ S.

Proof. Without loss of generality, assume that S contains all ramified and infi-nite primes. Let H ⊆ Gal(L/K) be generated by {Frobv : v /∈ S}. Let M ⊆ Lbe the subfield fixed by H. For any v /∈ S, we then have that Frobv fixes M . ButFrobv is the generator of the decomposition group, and so v splits completelyin M/K. This shows that H = Gal(L/K).

Math 223b Notes 34

14.2 Reduction of the second inequality to Kummer ex-tensions

We now want to prove the second inequality. We want to prove that

Proposition 14.4. If L/K and [L : K] = p, then

|H0(L/K,CL)| = |CK/NL/KCL| ≤ p.

First we reduce to the case where K contains µp. If [L : K] = p as above, letL′ = L(ζp) and K ′ = K(ζp). Because d = [K ′ : K] divides p− 1, it is relativelyprime to p and so [L′ : K ′] = p and [L′ : L] = d. Now we have

CL CL′ CL

CK C ′K CK

CK/NL/KCL CK′/NL′/K′CL′ CK/NL/KCL

1 1 1.

NL/K NL′/K′

NL′/L

NL/K

NK′/K

But CpK ⊆ NL/KCL and so CK/NL/KCL has exponent p. Also, the compositeof two horizontal arrows is multiplication by d. So ×d : CK/NCL → CK/NCLis an isomorphism, and this shows that

|CK/NL/KCL| ≤ |CK′/NL′/K′CL′ |.

Therefore it is enough to show the second inequality for L′/K ′. So we mayassume that L/K is a cyclic extension with µp ⊆ K. By Kummer theory, wehave

L = K( p√a)

for some a ∈ K×.We need to bound [A×K : K× · A×L ]. We will replace NA×L with a smaller

subgroup F ⊆ NA×L and then bound [A×K : K×F ]. Let S be a finite set of placesof K containing all ramified places, and let v1, . . . , vk be places of K that splitcompletely in L. Let S∗ = S ∪ {v1, . . . , vk}. Then we take

F =∏v∈S

(Kv)p ·

∏1≤i≤k

K×vi ×∏v/∈S∗

O×v .

Then F ⊆ NA×L because (Kv)p ⊆ NL×w for v ∈ S, Kvi = NL×wi

, andO×v ⊆ NL×wbecause Lw/Kv is unramified.

Math 223b Notes 35


Let L/K be cyclic with [L : K] = p. Assume K ⊇ µp. We need to show that

|A×K/NA×L ·K| ≤ p.

We will choose F ⊆ NA×L and show that |A×K/K×F | ≤ p. We will take

F =∏v∈S

(K×v )p ×∏

1≤i≤k

Kvi ×∏v∈S∗

O×V .

Last time we checked that F ⊆ NL× provided that S contains all ramifiedprimes in L/K.

Proposition 15.1. Let K be a global field, and v a finite place of K. Assumethat Kv does not have residue characteristic p, with b ∈ K×. Then v is ramifiedin K( p

√b)/K if and only if b ∈ O×v · (K×v )p. Also, v is split if and only if

b ∈ (K×v )p.

Proof. Exercise.

Corollary 15.2. If we choose S such that S contains primes of K dividing p,and also all v such that |a|v 6= 1, then L/K is unramified outside S.

15.1 Computing the size of the quotient

We also want S large enough that A×K,S � CK . Then

A×K,S∗ � A×K/K× · F.

The kernel is going to be

A×K,S∗ ∩K×F = F · (K× ∩ A×K,S∗) = F · O×K,S∗ .

So we get1→ F · O×K,S∗ → A×K,S∗ → A×K/K

×F → 1.

In particular,[AK : K×F ] = |A×K,S∗/F · O

×K,S∗ |.

We also have

1→ OK,S∗/(F ∩OK,S∗) ∼= F · OK,S∗/F → A×K,S∗/F → A×K,S∗/F ∩O×K,S∗ → 1.

That is,

|A×K/K×F | =

|A×K,S∗/F ||O×K,S∗/F ∩ O

×K,S∗ |

.

So we have broken up this into a local computation and a global computation.

Math 223b Notes 36

Let us first look at the local component. We have

|AK,S∗/F | =∣∣∣∣∏v∈S

K×v /(K×v )p

∣∣∣∣.But by the homework, we have

|K×v /(K×v )p| = p2 · |p|−1v

if µp ⊆ K. This shows that

|A×K,S∗/F | =∏v∈S

p2|p|−1v = p2|S|,

because S contained all v such that |p|v 6= 1.Now let us do the global computation. Observe that

F ∩ O×K,S∗ ⊇ OpK,S∗ .

Then

|O×K,S∗/F ∩ O×K,S∗ | =

|O×K,S∗/(OpK,S∗ |

|F ∩ OK,S∗/(O×K,S∗)p|.

By Dirichlet’s unit theorem, we have

O×K,S∗ ∼= Z|S∗|−1 ⊕ torsion,

where the torsion group contains µp. So we get

|O×K,S∗/(O×K,S∗)

p| = p|S∗| = p|S|+k.

Combining everything we have, we get

|A×K/K×F | =

|A×K,S∗/F ||O×K,S∗/F ∩ OK,S∗ |

= p|S|−k[F ∩ O×K,S∗ : (O×K,S∗)p].

The claim is that we can pick v1, . . . , vk such that k = |S| − 1 and F ∩O×K,S∗ =

(O×K,S∗)p. Also, recall that v1, . . . , vk had to be split completely in L.

15.2 Choosing the completely split places

Proposition 15.3. Suppose v1, . . . , vk are places of K such that if K( p√b) is

unramified outside v1, . . . , vk and split at all places of S, then b ∈ (K×)p. ThenF ∩ O×K,S∗ = (O×K,S∗)p.

Proof. Suppose that b ∈ F ∩ O×K,S∗ . By the exercise, we know that K( p√b)/K

is split at all v ∈ S, and also that K( p√b) is split at all v ∈ S. Then b ∈ (K×)p,

and because b ∈ O×K,S∗ , we have b ∈ (O×K,S∗)p.

Math 223b Notes 37

Now we would like to choose v1, . . . , vk given only S. Consider the auxiliaryextension

T = K( p

√O×K,S),

which has exponent p. Then

Gal(T/K) ∼= Hom(O×K,S/(O×K,S)p, µp) ∼= (Z/pZ)|S|.

Note that L = K( p√a) ⊆ T . Now we choose places w1, . . . , wk of L such that

Frob(wi) =(T/Lwi

)∈ Gal(T/L)

form a basis for the Fp-vector space Gal(T/L).

Math 223b Notes 38


Last time we reduced the second inequality to, given S, finding additional placesv1, . . . , vk with k = |S| − 1 such that

• v1, . . . , vk all split totally in L/K, and

• if K( p√bi) is unramified outside v1, . . . , vk and completely splits at all

v ∈ S then b ∈ (K×)p.

16.1 Finishing the second inequality

To do this, we looked at the auxiliary extension

T = K(

p

√O×K,S

)and tried to find L inside T . Choose w1, . . . , wk be places of L such thatFrobw1 , . . . ,Frobwk

form a basis for Gal(T/L) ∼= (Z/pZ)k. Then take vi tobe places of K below wi. First of all, note that

Frobwi= Frobeivi

where ei is the inertia degree. But ei = 1 or ei = p, but it can’t be that ei = pbecause then Frobwi

= Frobpvi = id. So we get ei = 1 so that vi is split.So we have Frobvi = Frobwi

inside Gal(T/K). So they generate the subgroupGal(T/L) which is of index p in Gal(T/K). Let vk+1 be a place such that(Frobvi)1≤i≤k+1 generate Gal(T/K).

Lemma 16.1. O×K,S/(O×K,S)p →

k+1∏i=1

O×vi/(O×vi)

p is bijective.

Proof. For injectivity, we use Kummer theory. If [c] ∈ O×K,S/(O×K,S)p is in the

kernel, then K( p√c) is a Kummer extension. For 1 ≤ i ≤ k+1 and c ∈ (O×vi)

p, sovi splits completely in K( p

√c. This means that Frobvi is the identity on K( p

√c).

So all of Gal(T/K) acts trivially on K( p√c), and this shows that K( p

√c) = K.

For surjectivity, we count the size. We have

|O×K,S/(O×K,S)p| = p|S| = pk+1.

By the homework, we have |O×vi/(O×vi)

p| = p for all i. So the right hand sidealso has order pk+1.

Corollary 16.2. O×K,S/(O×K,S)p �

k∏i=1

O×vi/(O×vi)

p is surjective.

Now let us check the second condition on vi. Suppose M = K( p√b) is a Kum-

mer extension such that M/K is unramified outside v1, . . . , vk and completelysplit at S. We need to show that b ∈ (K×)p. Let

D =∏v∈S

K×v ×∏vi

(K×vi)p ×

∏v/∈S∗

O×v .

Math 223b Notes 39

Then D ⊆ NA×M because we can check this locally. Note that D ·K× containsD · O×K,S . By the corollary, it includes A×K,S =

∏v∈S K

×v ×

∏v/∈S O×v . That is,

D ·K× ⊇ A×S . Recall thatA×K,S � A×K/K

×

by definition of S. Therefore D ·K× = A×K . By the corollary the first inequality,we get [M : K] = 1 as needed. This completes the proof of the second inequality.

16.2 Immediate corollaries

In the first inequality, for L/K cyclic, we showed that

|H0(L/K,CL)||H1(L/K,CL)|

= [L : K].

From the second inequality, for L/K cyclic and prime, we showed that

|H0(L/K,CL)| ≤ [L : K] = p.

Then we immediately get

H1(L/K,CL) = 1, |H0(L/K,CL)| = |H2(L/K,CL)| = p.

By the homework, this implies that

H1(L/K,CL) = 1, |H2(L/K,CL)| | [L : K]

for all L/K. (This was done by first showing for p-groups, and then showing itfor all groups using Sylow subgroups.)

Corollary 16.3. If L/K is abelian, then the map

H2(L/K,L×)→ H2(L/K,A×L ) ∼=⊕v

H2(Lw/Kv, L×w)

is injective.

Proof. Use the sequence 1→ L× → A×L → CL → 1.

Corollary 16.4 (Hasse norm theorem). Let L/K be a cyclic extension of num-ber fields, and a ∈ K×. Then a ∈ NL× if and only if a ∈ NL×w for all primesw of L.

Proof. This is equivalent to the injectivity of

K×/NL× ∼= H0(L/K,L×)→⊕

H0(Lw/Kv, L×w) ∼=

⊕K×v /NL

×w .

This follows from H−1(L/K,CL) ∼= H1(L/K,CL) = 0.

Math 223b Notes 40

Note that this only true for cyclic extensions. Here is a counter example.Take K = Q and L = Q(

√13,√

17), and a = −1.Next, we will build the invariant map

inv : H2(L/K,CL)→ 1[L:K]Z/Z.

we will construct this as the direct sum of local invariant maps, but it will takesome work to show that this gives a well-defined map. Here we are going to dofor “nice extensions”, i.e., cyclic cyclotomic extensions.

Math 223b Notes 41

17 March 2, 2018

From this point on, all fields are number fields unless stated otherwise. ForKummer theory, we already needed characteristic zero. We have two recprocitieswe want to prove:

• inv-reciprocity: if a ∈ H2(L/K,L×)→ H2(Lw/Kv, L×w) then∑

v

invv α = 0.

• θ-reciprocity: if L/K is abelian, for a ∈ K×∏v

θLw/Kv(a) = 1.

Note that last time we showed for L/K finite Galois, we have

H2(L/K,L×) ↪→⊕v

H2(Lw/Kv, L×w).

If we put all L together, we get

Br(K) = H2(K,K×

) ↪→⊕v

H2(Kv,K×v ) =

⊕v

Br(Kv) ∼=⊕v

Q/Z.

That is, α ∈ Br(K) is zero if and only if invv(α) = 0 for all v.

17.1 Reduction to cyclic cyclotomic extensions

Proposition 17.1. Let K be a number field, and α ∈ Br(K) = H2(K,K×

).Then there exists a cyclic cyclotomic extension L/K such that resL/K α = 0 ∈BrL.

We are going to say that “L splits α” in this case. By the inflation-restrictionsequence, this is equivalent to α being in the image of inf : H2(L/K,L×) →Br(K).

Proof. Consider

Br(K) Br(L)

Br(Kv) Br(Lw)

Q/Z Q/Z.

res

res

invv∼= invw

∼=×[Lw:Kv]

So for L to split α, we just need [Lw : Lv] invv(α) = 0 for all v. In other words,for any v with invv(α) 6= 0 in Q/Z (this happens for only finitely many v) weneed [Lw : Kv] to be a multiple of the order of invv(α) ∈ Q/Z. It follows fromthe following lemma.

Math 223b Notes 42

Lemma 17.2. Let S be a finite set of finite primes of K, and let m > 0 be aninteger. There exists a totally complex cyclic cyclotomic extension L/K suchthat m | [Lw : Kv] for all v ∈ S.

Proof. We reduce to the case K = Q by replacing m with m · [K : Q]. Thenfor each prime `, take Q(ζ`r )/Q which contains a cyclic subfield of order `r−1

or `r−2, and combine them for different `.

17.2 Invariant and reciprocity maps

Definition 17.3. We define

invL/K : H2(L/K,A×L )→ 1[L:K]Z/Z; c 7→

∑v

invLw/Kvc.

inv-reciprocity then states that inv vanishes onH2(L/K,L×) ↪→ H2(L/K,A×L ).There are compatibilities

• invN/K(infN/L c) = invL/K(c),

• invN/L(resL/K c) = [L : K] invN/K(c),

• invN/L(corL/K c) = invN/K(c),

following from the local compatibilities. For instance,

invN/L(resL/K c) =∑w

invNw′/Lw(resLw/Kv

c) =∑w

[Lw : Kv] invNw′/Kvc

=∑v

∑w|v

[Lw : Kv] invNw′/Kv(c) =

∑v

[L : K] invNw′/Kvc

= [L : K] invN/K c.

So⋃L/K A×L looks almost like a class formation, except that the map inv is not

an isomorphism. The inv is clearly not injective, and it even is sometimes notsurjective, because the image of invL/K is 1

lcm([Lw:Kv])Z/Z. Take K = Q and

L = Q(√

13,√

17) for instance.

Definition 17.4. For L/K abelian, we define

θL/K : A×K → Gal(L/K); a 7→∏v

θLw/Kv(a).

Proposition 17.5. For any character χ : Gal(L/K)→ Q/Z, considered as anelement of H1(L/K,Q/Z), we have

χ(θL/K(a)) = invL/K([a] ` δχ)

Proof. It follows from the local case.

Math 223b Notes 43

Corollary 17.6. If L/K is abelian, then inv-reciprocity implies θ-reciprocity.If L/K is cyclic, then θ-reciprocity implies inv-reciprocity.

Proof. Suppose L/K is abelian and we have inv-reciprocity. Then

χ(θL/K(a)) = 0

for all a ∈ K× and χ, and so θL/K(a) = 1.If L/K is cyclic and we have θ-reciprocity, take χ to be a generator of

H1(L/K,Q/Z). Then δχ generates H2(L/K,Z) and so taking cup productwith δχ is an isomorphism. θ-reciprocity gives that

invL/K([a] ` δχ) = 0

for all a ∈ K×, and so invL/K = 0.

We will first prove θ-reciprocity for L/Q cyclotomic, deduce inv-reciprocityfor L/Q cyclic cyclotomic, deduce inv-reciprocity for L/K Galois, and thenθ-reciprocity for L/K abelian.

Math 223b Notes 44

18 March 5, 2018

We had two reciprocities we wanted to check:

• θ-reciprocity: for all a ∈ L×,∏v αv(a) = 1.

• inv-reciprocity: for all L/K and a ∈ H2(L/K,L×),∑v invv a = 0.

We proved two things:

• if L/K is cyclic, then θ-reciprocity implies inv-reciprocity.

• if L/K is abelian, then inv-reciprocity implies θ-reciprocity.

18.1 θ-reciprocity for cyclotomic fields

We are going to first check θ-reciprocity for K = Q and L cyclotomic. Thismeans that L is contained in Q(ζn), but it is enough to check for L = Q(ζn) bythe compatibility conditions, and further more, check for L = Q(ζ`r ). We wantto check that

θ : Q× → Gal(Q(ζ`r )/Q); θ(a) =∏v

θv(a)

is zero. We can check this for a primes and −1.For a = p 6= `, we have

θp′(p) = 1 for p′ 6= p, `, θp(p) = Frobp(ζ 7→ ζp)

because Qv(ζ`r )/Q is unramified. On the other hand, because Q`(ζ`r ) is aLubin–Tate extension, we have

θ`(p) : ζ 7→ ζ(p−1).

Also, θ∞(p) = 1, so we get cancellation.For a = `, we have θp′(`) = 1 for p′ 6= ` and θ`(`) = 1 and θ∞(`) = 1. For

a = −1, we have θp′(−1) = 1 for p′ 6= ` and θ`(−1) : ζ 7→ ζ−1 and θ∞(−1) iscomplex conjugation.

So we have checked θ-reciprocity for L/Q cyclic cyclotomic, and so inv-recipcority for L/Q cyclic cyclotomic. But from last time we have that

Br(Q) =⋃L/Q

H2(L/Q, L×)

for L/Q cyclic cyclotomic. So we get inv-reciprocity for L/Q arbitrary.For arbitrary L/K, we can enlarge L to L′ with L′/Q Galois. Then we get

H2(L/K), L×) H2(L′/K, (L′)×) H2(L′/Q, (L′)×)

Q/Z Q/Z Q/Z.

inf

inv

cor

inv inv=0

Because the horizontal maps on the top row are injective, we get that the in-variant map H2(L/K,L×)→ Q/Z is zero. This shows inv-reciprocity, and thusθ-reciprocity for any L/K abelian.

Math 223b Notes 45

18.2 Invariant map on the class group

Now we have (for L/K abelian) a map

θL/K : CK/NCL = A×K/NA×L ·K× → Gal(L/K).

Also, by inv-reciprocity, we have

H2(L/K,L×) ↪→ H2(L/K,A×L )inv−−→ 1

[L:K]Z/Z.

This is not always a short exact sequence, because we don’t always have surjec-tivity on the right.

Proposition 18.1. If L/K is cyclic, then

1→ H2(L/K,L×)i∗−→ H2(L/K,A×L )

inv−−→ 1[L:K]Z/Z→ 0

is exact.

Proof. We first show that the invariant map is surjective. We can describeexplicitly the image, because H2(L/K,A×L ) ∼=

⊕vH

2(Lw/Kv, L×w). Then it is

enough to show that lcm([Lw : Kv]) = [L : K]. Recall that Frobv ∈ Gal(L/K)generate, because the order of Frobv is equal to [Lw : Kv], we get lcm = [L : K].

Now we show that the kernel of inv is equal to the image of i∗. We have anexact sequence

1→ H2(L/K,L×)i∗−→ H2(L/K,A×L )

j∗−→ H2(L/K,CL)→ H3(L/K,L×) = 1.

But we know that by the second inequality,

|H2(L/K,CL)| ≤ [L : K].

By comparing the two sequences, we get that both sequences are exact.

Moreover, we get an isomorphism

inv : H2(L/K,CL)∼=−→ 1

[L:K]Z/Z.

This works to define the correct invariant map for L/K cyclic. We are going todo the same thing we did last time, which is to first define the map for special(last semester, for unramified extensions) and then extend it to the general case.

Lemma 18.2. Let L/K be Galois, and let L′/K be cyclic with [L′ : K] =[L : K]. Then H2(L/K,CL) and H ′(L′/K,CL′) have the same image insideH2(K/K,CK =

⋃M CM ).

Math 223b Notes 46

19 March 7, 2018

Today we are going to finish proving that everything is a class formation. Lasttime we defined the invariant map

inv : H2(L/K,CL)→ 1[L:K]Z→ Z

for L/K cyclic. We will extend this to all L/K finite Galois.

19.1 CK form a class formation

Lemma 19.1. If L/K is Galois and L′/K is cyclic, and [L′ : K] = [L : K]then H2(L/K) and H2(L′/K) have the same image in H2(K/K).

Proof. First we want to show that inf(H2(L′/K)) ⊆ inf(H2(L/K)). Take K =L · L′ the compositum. Then L′/K is cyclic, the extension N/L is also cyclicwith degree [N : L] | [L′ : K]. If c ∈ H2(L′/K), I need to show that

infN/L′(c) ∈ im infN/L = ker resL/K .

(Here, we are using the inflation-restriction sequence.) Choose c ∈ H2(L′/K,A×L′)such that j∗(c) = c. Then we need to check that

invN/L resL/K infN/L′ j∗(c) = 0 ∈ 1[N :L]Z/Z.

But this thing is

invN/L j∗ resL/K infN/L′(c) = invN/L resL/K infN/L(c) = [L : K] invL′/K(c) = 0

because the invariant map invN/L from the second term is on A× and so is local.This shows that inf(H2(L′/K)) is contained in inf(H2(L/K)). But note thatthe size of H2(L′/K) is equal to [L′ : K] and the second inequality shows thatthe size of H2(L/K) is at most [L : K]. This shows the other containment aswell.

Now we can defineinv : H2(K/K)→ Q/Z

by considering

H2(K/K) ∼= lim−→L/K

H2(L/K) ∼= lim−→L′/K cyclic

H2(L′/K).

For any L/K, this restricts to an isomorphism

H2(L/K)inv−−→ 1

[L : K]Z/Z.

Thus we have verified all the axioms of class formations.

Math 223b Notes 47

19.2 Consequences

We can identify a fundamental class uL/K with

invL/K(uL/K) =1

[L : K].

We then have, by Tate’s theorem, an isomorphism

` uL/K : Hi(L/K,Z)→ Hi+2(L/K,CL)

is an isomorphism for all i. The i that we really care about is i = −2, and thisgives us

` uL/K : Gal(L/K)ab = H−2(L/K,Z)→ H0(L/K,CL) = CK/NCL.

The inverse map is what we denote by

θ : CK/NCL → Gal(L/K)ab.

This map is actually going to be the same as the product of the local θs.By the argument as last semester, for all characters χ : Gal(L/K) → Q/Z anda ∈ CK , we have an equality

χ(θL/K(a)) = invL/K(a ` δχ).

This characterizes θL/K , so if I consider θ′L/K : CK/NCL = A×K/K×NA×L →Gal(L/K) by the product of the local reciprocity maps, we previously checkedthat

χ(θ′(a)) = inv(a ` δχ)

using the invariant map H2(L/K,A×L )→ 1[L:K]Z/Z.

Note that θL/K : CK/NCL → Gal(L/K)ab is continuous using the adelictopology on CK . This is because NL/KCL are closed in CK . To see this, justnote that

CL ∼= C0L × R>0, CK ∼= C0

K × R>0

with the image of NL/KC0L being closed in C0

K by compactness. Taking themall together gives a continuous map

θ/K : CK → lim←−L

Gal(L/K)ab = Gal(Kab/K).

This has dense image, because each CK → Gal(L/K)ab is surjective. The kernelis the intersection

ker θ/K =⋂L/K

NCL.

So we need to ask again which subgroups A ⊆ CK are normic, i.e., equalto NCL for some L/K finite Galois. Note that it must be a finite index open

Math 223b Notes 48

subgroup. We are going to show that this is actually enough. Roughly, thisis going to be the same argument from last semester, which is doing a lot ofKummer theory.

Note that if A is normic, so is A′ ⊇ A, because we have our finite reciprocitymap θ. If A and B are normic, so is A∩B. What we are going to do is to picka sequence of open subgroups and show that each of them are normic.

Math 223b Notes 49

20 March 9, 2018

Recall that we now have the reciprocity map

θ/K : CK → Gal(Kab/K)

that is continuous with dense image. Moreover, we can only take the imageof C0

K the content zero stuff, because we can adjust the infinite part to makeanything have content zero. So the image is compact and hence closed, and thisshows that θ/K is surjective. We also showed that the kernel is the intersectionof all normic subgroups NL/KCL.

20.1 Classification of normic subgroups

Proposition 20.1. A ⊆ CL is normic if and only if A ⊆ CL is finite indexopen.

I justified last time that normic subgroups are finite index and open. IfA ⊆ CL is finite index, it is going to contain some CnL. Because A is open, Aalso contains some subgroup of the form

US =∏v∈S

1×∏v/∈S

O×v ⊆ CK .

(This is not open, but at least A contains some US .) Combining these two facts,we see that it is enough to show that CnLUS is normic for large enough S.

Lemma 20.2. Suppose S contains all infinite primes and all primes dividingn, and suppose A×K,S � CK . If K ⊇ µn then

(CK)nUS = NT/KT

where T = K( n√OK,S). In general, (CK)nUS is still normic.

Proof. This is going to be a lot like what we did for the second inequality. Wehave

Gal(T/K) ∼= Hom(O×K,S/(O×K,S)n, µn).

Here, we can non-canonically computeO×K,S(O×K,S)n ∼= (Z/nZ)S . So |Gal(T/K)| =n|S|. We have a map

θT/K : CK → Gal(T/K),

and (CK)n ⊆ ker θT/K . We also check that US ⊆ NT/KCT locally. Clearly

1 ∈ K×v a norm for v ∈ S, and for v /∈ S, T = K( n√OK,S) implies that T/K is

unramified, and hence O×v ⊆ NT×w . This shows

(CK)nUS ⊆ NT/KCT .

Math 223b Notes 50

Now we compute the index of both groups and show that they are equal.We first immediately see

[CK : NT/KCT ] = |Gal(T/K)| = n|S|.

For the other inclusion, we note that there is an exact

O×K,S/(O×K,S)n →

∏v∈S

(K×v )/(K×v )n → CK/(CK)n · US → 1.

(The second map is surjective because A×K,S � CK is assumed to be surjective.)

But the first term has size n|S| and the second term has size n2|S|. So it sufficesto check that (O×K,S)/(O×K,S)n injects into K×v /(K

×v )n.

If a ∈ O×K,S and a ∈ (K×v )n for all v ∈ S, consider the extension L =

K( n√a)/K. This is unramified outside S and split completely . So todotodo

Let us now look at the general case. If K does not contain µ, we let K ′ =K(µn) and S′ be the set of primes of K ′ above S. Then there exists a L′/K ′

such that (CK′)nUS′ = NL′/K′ · CL′ by taking L ⊇ L′ with L′/K Galois. WE

then have

NL/KCL ⊆ NL′/KC ′L = NN ′/KC′L = NK′/K(NL′/K′CL′) = NK′/K(CnK′US′) ⊆ CnKUS .

This shows that CnKUS is normic.

Now we have shown that

θ/K =⋂

U⊆CK finite index

U.

We claim that this is equal to

DK = connected comp. of 1 ∈ CK = closure of

( ∏v real

R>0

)( ∏v complex

C×).

It is clear that DK ⊆⋂U , and on the other hand CK/DK is profinite because

it is compact and totally disconnected. We also can characterize this as

DK = divisible elements of CK =⋂n

(CK)n.

Math 223b Notes 51

21 March 19, 2018

For L/K a finite extension of number fields, we now have

θL/K : CK/NCL∼=−→ Gal(L/K)ab.

The map is natural in L and K with inflation and restriction:

CK/NL/KCL Gal(L/K)ab

CM/NL/MCL Gal(L/M)ab

θL/K

transfer

θL/M

21.1 Properties of the Hilbert class field

The Hilbert class field H of a number field K is the ray class field of (1). This isthe maximal abelian extension of K that is unramified at all places (includinginfinite fields). Then there is a canonical isomorphism

Cl(OK)∼=−→ Gal(H/K)

given by the Artin map.

Example 21.1. For K = Q, we have H = Q. If K = Q(√−5), then H =

Q(√−5, i). For K = Q(

√−23), we get H = K(α) where α3 − α+ 1 = 0. (This

has discriminant 23.)

Because this extension H/K is unramified at every prime, a prime p ∈Cl(OK) is mapped to the Frobenius Frobp ∈ Gal(H/K). So p is principal if andonly if Frobp ∈ Gal(H/K) if and only if p splits completely in H/K.

Corollary 21.2. Given p a prime of Q, the prime p is a norm from OK if andonly if p splits completely in OH .

Proof. If (p) does not split completely in OK , then it is not a norm in OK andalso it does not split completely in OH . If (p) does split completely in OK , write(p) = Np. Then p is a principal ideal if and only if p splits completely in OH ifand only if (p) splits completely in OH .

Example 21.3. Let K = Q(√−5). We have that p = x2 + 5y2 if and only if

p splits completely in Q(√−5, i). Because Q(

√−5, i) ⊆ Q(ζ20), this is going to

be a condition on p modulo 20.

Theorem 21.4 (principal ideal theorem of class field theory). If a is any frac-tional ideal of OK , then aOK is principal.

This property does not uniquely determine OH . This also does not meanthat OH have class number 1. The theorem follows from the following grouptheory fact.

Math 223b Notes 52

Theorem 21.5. Let G be a group, G′ = [G,G], and G′′ = [G′, G′]. Then thetransfer map G/G′ → G′/G′′ is trivial.

Proof. Let us start out with H/K and let H1 be the Hilbert class field of H,so that Gal(H1/H) = Cl(H). Now H is going to be the maximal abeliansubextension of H, because H1/K is unramified. Let G = Gal(H1/K) so thatGal(H/K) = Gab and Gal(H1/H) = G′. We have

CK/NH1/KCH1 Gal(H1/K)ab = Gab

CH/NH1/HCH1Gal(H1/H)ab = (G′)ab.

θH1/K

transfer

θH1/H

But note that Gal(H1/K)ab ∼= Gal(H/K) ∼= Cl(OK). So we get that the mapCl(OK)→ Cl(OH) is trivial.

Math 223b Notes 53

22 March 21, 2018

Last time we showed the following:

Theorem 22.1 (principal ideal theorem). If H is the Hilbert class field of K,then Cl(K)→ Cl(H) is trivial.

22.1 Class field tower

Then people started wondering about whether one can always find L/K suchthat Cl(L) is trivial. Examples last time had Cl(H) trivial, so L = H works.Historically, this was to enlarge the field so that unique factorization works.One thing you can do is to take the class field tower K ⊆ H ⊆ H1 ⊆ H2 ⊆ · · · .For small cases, this appears to always terminate and you eventually hit someHn which has Cl(Hn) = 1, in which case the tower stabilizes. Actually this is anatural thing to do.

Lemma 22.2. If L/K is such that Cl(L) = 1, then L contains H the Hilbertclass field of K. (So L ⊇ Hn for all n.)

So such L exists if and only if the class field tower stabilizes.

Proof. Note that HL/L is an unramified extension but Cl(L) = 1. So H0L = Land H0 ⊆ L.

This is related to Gal(Kunr/K) = G, where Kunr is the maximal (not-abelian) everywhere unramified extension. The entire tower then lies in Kunr.Also, H0 is the fixed field of G1 = [G,G], then H1 is the fixed field of [G1, G1],and so on. It is easier to study p-groups for a fixed p. Let H0,p be the fixedfield of the p-group of H0, and so on K ⊆ H0,p ⊆ H1,p ⊆ · · · . This is called thep-class field tower of K.

Theorem 22.3 (Golod–Shafarevich). If the p-class field tower stabilizes, thenit must be that the p-rank of Cl(K) is at most 2 + 2

√[K : Q] + 1.

If the latter condition fails, then it should have infinite p-class field tower,and hence infinite class field tower. For instance we just need it to have the2-rank of Cl(K) to be at least 6. This holds for any imaginary quadratic fieldswith ≥ 6 ramified primes or real quadratic fields with ≥ 8 ramified primes.(This is genus theory.)

22.2 L-functions

A Dirichlet series is a series of the form∑n≥1

anns.

Then ζ(s) is the case an = 1 for all n. This series generally converges for<(s) � 0. For ζ(s), the series converges for <(s) > 1. But one important

Math 223b Notes 54

thing is that there is a meromorphic continuation to all of C, using a functionalequation.

We can write

ζ(s) =∏p

1

1− p−s,

and more generally if K is a number field, an Euler product is a product ofthe form ∏

p⊆OK

1

(1− a1,p(Np)−s) · · · (1− akp,p(Np)−s)

where Np = |OK/p|. L-functions are Dirichlet series with Euler products. Forinstance, the Dedekind ζ-function of a number field K is

ζK(s) =∑

a⊆OK

1

|Na|−s=

∏p⊂−OK

1

1− (Np)−s.

We can actually do this for function fields. Let us take K = Fq(t) forinstance. Let us take OK,∞ = Fq[t]. Then we are looking at∑

a⊆Fq [t]

1

(Na)−s.

But every a is principal and if we write a = (f) then Na = q− deg f . So

ζK(s) =∑d≥0

qd

q−ds=

1

1− q(1−s)

This kind of thing always happens for function fields. If we take the Eulerproduct formula, we get∏

v of Fq(t)

1

1− |kv|−s=

1

1− q−sζK(s) =

1

(1− q−s)(1− qq−s).

This is the zeta function of P1Fq

. It is a rational function in t = q−s. For any

function field K(C), the zeta function of C is going to be of the form∏i(1− αiq−s)

(1− qq−s)(1− q−s)

where all αi has |αi| =√q. This is the Riemann hypothesis for curves over

finite fields.

Math 223b Notes 55

23 March 23, 2018

We defined the Dedekind ζ function of Ok as∑a⊆OK

1

(Na)s=∏

p⊆OK

1

1− (Np)s.

23.1 L-functions

Then we can define the Dirichlet L-function over OK as the following. Firstchoose a module m and look at

Clm(OK) = Fracm(OK)/Prinm(OK)

and choose a character χ of Clm(OK). Now we define the Dirichlet L-functionas

L(s, χ) =∑

(a,m)=1

χ(a)

(Na)s=

∏p⊆OK ,p-m

1

1− χ(p)(Np)s.

For instance, if K = Q and m = m∞ then Clm(Z) ∼= (Z/mZ)× and χ :(Z/mZ)× → S′. Then

L(s, χ) =∑

(a,m)=1

χ(a)

as.

We can consider χ : Clm(OK) → S′ as a character of CK → S1, becausethere is a map CK � S1. Hecke L-functions generalize Dirichlet L-functionsby allowing arbitrary characters χ : CK → S1 with non-discrete image. Youcan prove using this that primes in Z[i] have equidistributed arguments, forinstance.

There is another point of view. Note that there is a map

Clm(OK)→ Gal(Lm/K); p 7→ Frobp .

So if V is a finite-dimensional representation of Gal(L/K), we can take the traceχV and we can define the Artin L-series∏

p

1

1− χ(Frobp)(Np)−s.

23.2 Convergence of Dirichlet series

If an is O(nb), then the series∑ann

−s converges locally uniformly and abso-lutely in the half-plane <(s) > b+ 1.

Proposition 23.1. Let S(x) =∑n≤x an. If S(x) = O(xb) then

∑ann

s can beextended to a holomorphic function on <(s) > b.

Math 223b Notes 56

Proof. This is just analysis. The basic idea is that we can do summation byparts and write ∑

n≥0

anns

=∑n≥0

sn

( 1

ns− 1

(n+ 1)s

).

Here sn = O(nb) and the difference is O(n−s−1).

Proposition 23.2. ζ(s) has a meromorphic continuation to <(s) > 0, with apole at s = 1 an no other poles.

Proof. Consider ζ2(s) = ζs(1− 2−s). Then

ζ2(s) =∑n≥1

(−1)n+1

ns

on <(s) > 1. By our proposition, this can be extended to <(s) > 0 holomorphi-cally, so ζ(s) is meromorphic on <(s) > 0 with poles possibly as s = 1 + k 2πi

log 2 .

If we define ζ3(s) similarly, we see that ζ(3) can have a pole only at s = 1. Itactually does have a pole.

Proposition 23.3. ζ(s) has a pole at s = 1 of residue 1.

Proof. We look at ζ(s) for s > 1. We have∑n≥1

1

ns≥∫ ∞x=1

1

xsdx =

1

s− 1≥∑n≥2

1

ns= ζ(s)− 1,

so the residue should be 1.

For the Dedekind zeta function, we will show that ζK(s) also have a simplepole at s = 1 and will give a formula for the residue involving the class number.

Math 223b Notes 57

24 March 26, 2018

We showed the following last time. If S(x) =∑n≤x an has S(x) = O(xb), then∑

ann−s converges to a holomorphic function on <(s) > b. Using this, we

showed that

ζ(s) =∑ 1

ns

has a meromorphic continuation to <(s) > 0 with a simple pole at s = 1 withresidue 1.

24.1 Meromorphic continuation of L-functions

Proposition 24.1. If there exists some a0 such that S(n) = a0n+O(xb) with0 ≤ b < 1, then

∑ann

−s can be analytically continued to a meromorphicfunction on <(s) > b with a simple pole at s = 1 having residue a0.

Proof. We have∑ann

−s = a0ζ(s) +∑

(an − a0)n−s.

We will apply this to ζK , and more generally, L(s, χ) for χ : Clm(OK)→ S1

given by

L(s, χ) =∑

a⊆OK ,(a,m)=1

χ(a)

Nas.

Definition 24.2. We define, for K ∈ Clm(OK),

ζ(s,K) =∑a∈K

1

(Na)s.

Then we can write

L(s, χ) =∑

K∈Clm(OK)

χ(K)ζ(s,K).

The goal is to apply the proposition to ζ(s,K). Consider the partial sumfunction

S(x,K) = #{a ∈ K : Na ≤ x}.

Proposition 24.3. We have S(x,K) = gmx + O(x(1− 1d )). Thus, ζ(s,K) has a

pole or residue gm at s = 1. Here,

gm =2r(2π)s reg(m)

wm(Nm)√|disc(K)|

,

where r is the number of real places, s is the number of complex places, reg(m)is the regulator, wm is the number of roots or unity in O×K,m = {a ∈ O×K : a ≡1 mod m, a positive at reals of m}, Nm is the norm of m as an ideal (ignoringplaces at infinity), disc(K) is the discriminant.

Math 223b Notes 58

Definition 24.4. The regulator is the “size of O×K,m”. Consider the log map

L : O×K → H ↪→∏v|∞

R+; a 7→ (log|a|v1 , . . . , log|a|vr+s).

Then L(O×K,m) is a lattice in H, and now we take reg(m) as the covolume of

L(O×K,m) inside H.

If m = 1, we also say reg(m) = reg(OK). If K is real quadratic, we havereg(m) = log|u| where u is the generator of the unit with |u| > 1.

Proof. For a completely proof, read Lang’s Algebraic number theory. Firstchoose c ∈ K. Then any a ∈ K is going to look like a = ac where a ≡ 1 mod mand a ∈ c−1. Then Na = Nc−1Na ≤ (Na)x for some x > 0. This meansthat a belongs to an additive coset of mc−1, which is going to be some lat-tice in

∏v|∞Kv. Then the number of these points in some ball is going to be

approximately

vol(ball)

covolume of lattice+O(x1− 1

d ) =x ·Nc−1 · (· · · )√disc(K)NmNc−1

+O(x1− 1d ).

If you work out the coefficients out, you are going to get that. If you have areal quadratic field, for instance, you actually have a hyperbola, and you haveto quotient by the unit group. Here is where you get the regulator.

Math 223b Notes 59

25 March 28, 2018

We were counting

S(x,K) = #{ideals a ∈ K : Na ≤ x},

where K ∈ Clm(OK). We had the following proposition.

Proposition 25.1. S(x,K) = gmx+O(x1− 1d ), where

gm =2r(2π)s reg(m)

wmNm√|discK|

and Nm = Nmfin2rm for rm the number of real places in m.

Proof. We first chose c ∈ K so that a = ac. Here, the set of possible a is thensome lattice in

∏v|∞Kv interesected with a fundamental domain for the action

of O×K,m.

Theorem 25.2. Let χ be a Dirichlet character of modulus m. If χ 6= 1, thenL(s, χ) is holomorphic at s = 1. If χ = 1, then L(s, χ) has a pole of order 1 ats = 1, with reseidue |Clm(OK)|gm.

Proof. This immediately follows from

L(s, χ) =∑

K∈Clm

χ(K)ζ(s,K).

Here, each ζ(s,K) has a pole of order 1 at s = 1, with residue gm. So if χ is notthe trivial character, we get cancellation, and if χ is trivial, we get the correctthing.

Corollary 25.3. For m = 1, we get that ζK,s(s) = L(s, 1) has pole at s = 1with residue given by

hKg1 =2r(2π)s reg(K)hK

wK√|disc(K)|

.

For instance, if K is imaginary quadratic, then the residue is

2πhK

wK√−∆

.

If K is real quadratic, then the residue is

2 log|u|hK2√

∆.

There’s something called the Stark’s conjecture that generalize this to otherL-functions. This is also very similar to the Birch–Swinnerton-Dyer conjecture.

Math 223b Notes 60

25.1 Densities of sets of primes

I’m going to give you three different notions of density. Let T be a set of primesof OK .

• The Dirichlet density is a number δ satisfying∑p∈T

1

(Np)s= δ log

1

1− s+O(1)

as s→ 1+.

• The natural density is

δ = limx→∞

#{p ∈ T : Np ≤ x}{p : Np ≤ x}

.

• If the Euler product (∏p∈T

(1− 1

Nps

)−1)n

has an analytic continuation with pole of order m at s = 1, we say thatits polar density is m

n .

Clearly, all of them are

• monotone where defined,

• finitely additive,

• subsets of density 0 have density 0,

• finite sets have density 0.

Proposition 25.4. If natural density exists, then so does Dirichlet density andthey are equal. If polar density exists, so does Dirichlet density and they areequal.

There are sets like primes whose first digit of p is 1, whose Dirichlet densityexists but but natural density does not exist.

Proof. Proof of natural implies Dirichlet is basically partial summation. For theproof of polar implies Dirichlet, we can use

log

(∏p∈T

1

1−Nps

)−1

=∑p∈T

(Np)s) +O(1)

as s→ 1, which is obtained by looking at the Taylor expansion.

Proposition 25.5. The set of all primes has density 1, with respect to all threedefinitions.

Proof. This is obvious for natural density. For polar density, you can use thatζK(s) has pole of order 1 at s = 1. You can do this for Dirichlet density aswell.

Math 223b Notes 61

26 March 30, 2018

Recall that we defined Dirichlet density of T as∑p∈T

1

Nps= δ log

( 1

1− s

)+O(1),

and we defined polar density as mn if(∏

p∈T

(1− 1

Nps

)−1)n

has a pole of order m at s = 1.

26.1 Proof of second inequality

Proposition 26.1. The set T = {p ⊆ OK : Np = pi for i ≥ 2} has polardensity 0.

Proof. Let d = [K : Q]. Write

∏p∈T

(1− 1

Nps

)−1

=

d∏j=2

∏p

(1− p−js)#{p:Np=pjs}.

Then it is clear that #{p : Np = pj} ≤ d. So in the worst case, we get

d∏j=2

∏p

(1− p−js)d = ζ(2s)dζ(3s)d · · · ζ(ds)d

which still is holomorphic on <s > 12 .

Proposition 26.2. Let L/K be finite Galois. Then Spl(L/K) = {p : p splits completely in L}has polar density 1/[L : K].

Proof. Let T be the set of primes of L lying above primes of Spl(L/K). Then if

some pL lies above some pK , then NpL = pfK . So T is the set of primes p suchthat Np is prime, up to a finite number of ramified primes. This implies thatthe polar density δ(T ) = 1, because the set it misses can be ignored.

Now we observe that∏pL∈T

(1− 1

NpsL

)=

∏pk∈Spl(L/K)

(1− 1

NpsK

)[L:K]

.

This implies that Spl(L/K) has polar density 1/[L : K].

We are working towards the second inequality. Fix m a modulus. Then inClm = Clm(OK) fix a subgroup H of Clm. Consider L(1, χ) for χ a nontrivialDirichlet character of modulus m that vanishes on H.

Math 223b Notes 62

Theorem 26.3. At most one character χ of H has L(1, χ) = 0.

This will imply that the Dirichlet density of {p : [p] ∈ H} is equal to 1/|G|if L(1, χ) 6= 0 for all such χ, and L(1, χ) = 0 for one χ.

Proof. We consider the sum

1

|G|∑χ

log(L(s, χ)) =1

|G|∑χ

∑p

− log(1− χ(p)Np−s)

=1

|G|∑χ,p

(χ(p)Np−s) +O(1) =∑

[p]∈H

Np−s +O(1).

But the left hand side is just

1

|G|

(log

(1

1− s

)−∑χ

ords=1 L(s, χ) log( 1

1− s

))+O(1)

shows that the Dirichlet density of {p : [p] ∈ H} is just

1

|G|

(1−

∑χ

ords=1(L(s, χ))

).

This shows that the sum of the orders is either 0 or 1.

Theorem 26.4. Let L/K be Galois and m be a modulus of K. Then

|C×K/(NL/KC×L )Um| ≤ [L : K].

This implies the second inequality because we can take m so that Um ⊆NL/KC

×L . Then |C×K/NL/KC

×L | ≤ [L : K].

Proof. Consider G = Clm(OK)/H where H is the subgroup of Clm(OK) gener-ated by ideals that are norms from OL. We know that δ(p : [p] ∈ H) is either 0or 1/|G|. But if p splits completely in L, then p is a norm form OL so p ∈ H.This shows that

δ(p : [p] ∈ H) ≥ δ(Spl(L/K)) =1

[L : K].

So δ(p : [p] ∈ H) = 1/|G|. This shows that |G| ≤ [L : K].

Corollary 26.5. If χ is a nontrivial character of Clm(OK), then L(1, χ) 6= 0.

Proof. Here, we need to be a bit careful because the previous argument onlyshows that L(1, χ) 6= 0 only for χ that vanish on H. So we use an existencetheorem. There exists a L/K Galois such that NC×L ⊆ Um, by class fieldtheory.

Math 223b Notes 63

27 April 2, 2018

We want to prove the Chebotarev density theorem.

Theorem 27.1 (Chebotarev density theorem). Let L/K be a Galois extensionof fields. There is a map

{unramified primes of K} → {conjugacy classes of Gal(L/K)}, p 7→ [Frobp].

(Note that Frobp is only defined up to conjugacy.) Then for any conjugacy class[g],

Tg = {p : [Frobp] = [g]}

has Dirichlet density |[g]|/|Gal(L/K)|.

27.1 Chebotarev density theorem

Note that we have already proved this for g = 1. Note that we have

T1 = Spl(L/K)

because having trivial Frobenius means that you have trivial decompositiongroup, and then the prime splits completely. The strategy is to first do L/Kabelian and then use this to do it in the general case.

Proposition 27.2. Let C be any subset of Clm. Then the density of {p : [p] ∈C} is |C|/|Clm|.

Proof. By additivity, assume that C = {K}. Then∑p∈K

1

Nps=

1

|Clm|∑χ

χ(K) logL(s, χ) +O(1)

as s → 1+. But we know that all logL(1, χ) is a number for χ nontrivial, andL(s, χ) behaves like log 1

1−s as s→ 1.

Note that for K = Q, this is just Dirichlet’s theorem for primes in arithmeticprogressions.

Corollary 27.3. L/K satisfies Chebotarev for L/K abelian.

Proof. Choose m with L ⊆ Lm. Then Gal(L/K) ∼= Clm /H for some subgroupH. Then Frobp = g if and only if [p] ∈ gH.

Let’s now do the non-abelian case.

Proof of Chebotarev density. Ignore all ramified primes, and let [L : K] = n.Let g ∈ Gal(L/K) have order m and consider M = L〈g〉. Then we have a towerL/M/K where L/M is cyclic of degree m. What does it mean for p ∈ Tg. Thismeans that there exists a pL above p such that FrobpL

= g. Then 〈g〉 = DpL.

Math 223b Notes 64

Here, we let pM lies below pL. Then we have a surjection

TM,g = {pM : fM/K = 1,FrobpM= g ∈ Gal(L/M)}� Tg.

Now we can use abelian Chebotarev to compute Dirichlet density of TM,g. Here,we can ignore fM/K = 1 because other primes will have density 0 anyways. Sowe have

δ(TM,g) = δ(p of M : FrobpM= g ∈ Gal(L/M)) =

1

m.

We can also understand the fiber of this surjection as well. Given p, howmay primes pM ∈ TM,g lie over p? In other words, how may pL are there abovep such that FrobpL

= g ∈ Gal(L/K)? Any such pL above p are going to looklike

pL = hpL,0,

for h ∈ Gal(L/K). Then we have

FrobpL= hFrobpL,0

h−1 = hgh−1.

So hgh−1 = g if and only if h is in the centralizer. If you compute the densityusing orbit-stablizer and stuff, we find

δ(Tg) =∑p∈Tg

1

Np−s=m|[g]|n

∑pM∈TM,g

1

Np−sM=

|[g]||Gal(L/K)|

log(1− s) +O(1).

This gives the correct density.

Here is one application. Let L/K be an arbitrary finite extension. Then wedefined Spl(L/K) as the prime of K that split completely in L. You can showthat if L′ is the Galois closure of L, then Spl(L/K) = Spl(L′/K). So

δ(Spl(L/K)) = δ(Spl(L′/K)) =1

[L′ : K].

Proposition 27.4. If L and L′ Galois have the same splitting set, then L = L′.

Proof. Consider the compositum LL′/K. This has splitting set density equalto 1 over [L : K] and of [LL′ : K].

Math 223b Notes 65

28 April 4, 2018

Now we would like to do explicit class field theory for Q and imaginary quadraticfields. By Kronecker–Weber, we know that Qab =

⋃nQ(ζn). These Q(ζn) are

the ray class fields of modulus n∞.Consider the algebraic group Gm over Q. Its endomorphism ring is End(Qm) ∼=

Z, with n corresponding to a 7→ an, and for each n, the points of ker(a → an)generate Q(ζn). This is great for Q, but if K 6= Q, then K(ζ∞) is not themaximal abelian extension. So we need new sources of abelian extensions.

28.1 Complex multiplication

The idea is to take different 1-dimensional algebraic groups over K. For in-stance, we take E/K an elliptic curve for every n, consider the n-torsion pointsE[n](Q) ∼= (Z/nZ)2, and take K(E[n])/K. The problem is that this is notnecessarily abelian. Still there is an embedding

Gal(K(E[n])/K) ↪→ Aut(E[n]) = GL2(Z/nZ).

So we only want to look at special elliptic curves.If E is an elliptic curve over K, then

EndK(E) ∼={Z O ⊆ Q(

√−D).

We say that E is of complex multiplication if it falls in the second category.Moreover, for any O, there are only finitely many E with complex multiplicationby O. In this case, for any a ∈ O we can look at the kernel E[a] of a : E → Eand see that E[a] ∼= O/(a) as O-modules. Then we have

Gal(K(E[a])) ↪→ AutO(E[a]) ∼= (O/(a))×.

What happens if I do this forK an imaginary quadratic field? Unless h(K) =1, it turns out that there are no CM elliptic curves defined over K. So we aregoing to enlarge K to H so that there are CM elliptic curves defined over Hwith CM by OK . Here, we can take H to be the Hilbert class field of K.

Note that there is a bijection

{E elliptic curve over C with CM over K} ←→ Cl(K),

because for any ideal a we can take C/a as an elliptic curve. Then we can lookat the j-invariants j(E) ∈ H. This is going to have minimal polynomial∏

E CM by K

(X − j(E))

over K and then j(E) generates the Hilbert class field.Let m be a modulus of K. (There are no infinity parts, because K has only

a complex place.) Then we can define

E[m] = {P ∈ E : aP = 0 for all a ∈ m} ∼= OK/m.

Then Lm = H(E[m]) is the ray class field corresponding to m.

Math 223b Notes 66

28.2 Elliptic curves

Let L ⊆ C, and consider C/L. This is a Riemann surface of genus 1, so it shouldcorrespond to a curve over C. We want to make this isomorphism C/L → Eexplicit.

Definition 28.1. An elliptic function is a meromorphic function fn on C/L.

There is the Weierstrass ℘-function defined by

℘(z, L) =1

z2+

∑w∈L\{0}

( 1

(z − w)2− 1

w2

).

Then ℘ is an elliptic function, with double poles at points of L.

Math 223b Notes 67

29 April 6, 2018

Let L ⊆ C be a lattice. Recall that we defined an elliptic function as a mero-morphic function f : C→ C that is L-periodic. We know that C/L is an ellipticcurve. We are interested in what elliptic function this is. Last time we defined

℘(z, L) =1

z2+

∑w∈L\{0}

( 1

(z − w)2− 1

w2

).

We showed that this as double poles at L and elsewhere holomorphic.

29.1 Properties of the Weierstrass function

Let us fix L and write ℘(z) = ℘(z, L). We first note that ℘(−z) = ℘(z). Now weobserve that ℘(z)− ℘(z +w) is a bounded entire function, and hence constant.But for z = −w/2, we have ℘(z) = ℘(z +w). This shows that this holds for allz.

Now note that we can just differentiate and get

℘′(z) = −2∑w∈L

1

(z − w)3.

This is again an elliptic function, and it now has triple poles at points of L. Butbecause I am on a curve, there should be an algebraic relation between thesetwo functions. It is useful to have a power series for ℘.

Lemma 29.1. ℘(z) =1

z2+∑n≥1

(2n+1)G2n+2z2n where G2n+2(L) =

∑w∈L\{0}

1w2n+2 .

Proof. I just expand

1

(z − w)2− 1

w2=∑m≥1

(m+ 1)w−m−2zm.

Here, note that if m is odd, then Gm+2 = 0.

We can now use this to find a polynomial P in ℘ and ℘′ such that P (℘, ℘′) =∑z≥1 cnz

n. This would imply that P (℘, ℘′) = 0. If you work this out, you get

(℘′(z))2 = 4℘(z)3 − g2℘(z)− g3, g2 = 60G4, g3 = 120G6.

This is called the Weierstrass form of the elliptic curve. Then we get a map

C/L→ E : 4y2 = x3 − g2x− g3; z 7→ (℘(z), ℘′(z)).

In fact, this implies that for all n, G2n is a polynomial in g2 and g3 of degree n,where we consider g2 as having degree 2 and g3 as having degree 3. We want toshow that this map is an isomorphism.

Math 223b Notes 68

Proposition 29.2. ℘(z) = ℘(w) if and only if z ∼= ±w mod L.

Proof. Let’s take the difference ℘(z) − ℘(w) as a function of z on C/L. Weknow that the function has double poles at z = 0, and zeros at ±w (unless wis 2-torsion). So there can’t be any other zeros. If w is 2-torsion, it is going tohave a double zero because it is going to be even.

Using this, we see that (℘(z), ℘′(z)) = (℘(w), ℘′(w)) can only happen whenw = −z or w = z. In the first case, we see that ℘′(z) = ℘′(w). But ℘′(z) haszeros as 3 nonzero point of 1

2L/L. So we have an isomorphism

C/L ∼= E

of Riemann surfaces. If you know enough theory, it follows that this is also anisomorphism of complex Lie groups. But we can do this explicitly.

Theorem 29.3 (Addition theorem). ℘(z+w) = −℘(z)−℘(w)+(1

4

℘′(z)− ℘′(w)

℘(z)− ℘(w)

)2

.

Proof. Differentiate both sides in z. It is elliptic, entire and 0 at z = 0.

This shows that C/L ∼= E is indeed an isomorphism of complex Lie groups.

29.2 j-invariant

Define the discriminant of an elliptic curve by

∆(L) = g32 − 27g2

3

which is the discriminant of the polynomial 4x3 − g32x − g3. Then define the

j-invariant as

j(L) =1728g3

2

∆.

Theorem 29.4. For two lattices L,L′ ⊆ C, j(L) = j(L′) if and only if L′ ishomothetic to L.

Proof. If L′ = λL then g2(L′) = λ−4g2(L) and g3(L′) = λ−6g3(L). If j(L) =j(L′), then there exist λ such that g2(L) = λ−2g2(L′) and g3(L) = λ−3g3(L′).Because G2n are polynomials in g2 and g3, we get G2n(L) = λ−2nG2n(L′). So

℘(z, L) = ℘(λ−1z, L′).

Comparing poles, we recover L and L′.

Now which L give E that have complex multiplication? If we have an isogeny(a nonzero homomorphism of complex Lie groups) ϕ : E → E, on C/L→ C/L,it is going to look like multiplication by some complex number.

Theorem 29.5. Let L be a lattice and consider ℘(z) = ℘(z, L). Take α ∈ C\Z.The following are equivalent:

Math 223b Notes 69

(1) multiplication by α : C→ C induces an isogeny of C/L(2) αL ⊆ L(3) ℘(αz) is a rational function in ℘(z)

(4) there exists a O the ring of integers of some imaginary quadratic field Ksuch that α ∈ O and a ⊆ O is homothetic to a.

Math 223b Notes 70

30 April 9, 2018

I missed 40 minutes of class.

30.1 j-invariants of CM elliptic curves

Theorem 30.1. Let O be the order in an imaginary quadratic field, and let abe a proper fractional O-ideal. Then j(a) = j(C/a) is an algebraic number ofdegree at most h(O).

Proof. We consider {j(E) : End(E) ∼= O} ⊆ C. This is the same as {j(a)}where a runs over invertible fractional ideals of O. This is a union of Galoisorbits, which is set of size h(O).

We will be showing that actually

[Q(j(a)) : Q] = h(O)

and also that j([a]) is an algebraic integer. When O = OK , an K(j[a]) = H, ingeneral we can describe K(j[a]) as follows. We define

Gal(LO/K) ∼= CK/∏v fin

U×v

where Uv is the closure of O in Ov. (Then if O = Z + fOK then Uv = (Zp +fOv)×. This means that LO ⊆ Lf .)

Math 223b Notes 71

31 April 11, 2018

Let L ⊆ C be a lattice and let E = C/L have complex multiplication. ThenEnd(E) = O ) Z, and we say that L has complex multiplication by O. Inthat case, L ∼ a for some invertible fraction ideal O. We saw that j(L) is analgebraic number of degree at most h(L).

Theorem 31.1 (main theorem of complex multiplication). The j-invariantj(L) is an algebraic integer of degree h(L) and K(j(L)) = LO (which is the ringclass field of O).

Definition 31.2. A cyclic sublattice of index m is a sublattice L′ ⊆ L suchthat L/L′ ∼= Z/mZ.

Proposition 31.3. If L has complex multiplication, then there exists a cyclicsublattice L′ ⊆ L such that L′ is homothetic to L.

Proof. Without loss of generality, let a be an invertible ideal of O = Z + fOK .Choose m = p such that p is relatively prime to a, f , and p completely splits inOK as p = ππ. Take L′ = πL so that

L/πL = a/pa ∼= OK/πOK ∼= Z/pZ.

31.1 Modular forms and functions

We defined the j-invariant as a function for a lattice invariant under homothety,but the classical way of looking at it is as a function on the upper half plane H.For τ ∈ H, the free abelian group [1, τ ] ⊆ C is a lattice. So we can define

g2(τ) = g2[1, τ ], g3(τ) = g3[1, τ ], ∆(τ) = ∆[1, τ ], j(τ) = j[1, τ ].

Here, there is an action of SL2(Z) on H, and the action on τ only does ahomothety on the lattice by taking L to 1

cτ+dL. So we get

g2(γτ) = (cτ + d)4g2(τ), g3(γτ) = (cτ + d)6g3(τ).

This is saying that g2 is a modular form of weight 4 and g3 is a modular formof weight 4. For the j-invariant, we will have

j(γτ) = j(τ)

for all j ∈ SL2(Z). This means that j descends to a function on Y (1) =SL2(Z) \H. There is a fundamental domain

{τ ∈ H : − 12 < <(τ) < 1

2 , |τ | > 1}

and then you can see that this is a sphere minus one point, but with one order 2orbifold point and one 3 orbifold point. This is not compact, so we compactifyit to get a Riemann surface X(1) = Y (1) ∪ {∞}. Around infinity, the localcoordinates are going to be given by q = e2πiz.

Math 223b Notes 72

Definition 31.4. A modular function for SL2(Z (a meromorphic modularform of weight 0) is a meromorphic function f on H such that f(γτ) = f(τ) forall τ ∈ H and γ ∈ SL2(Z), and f can be written as

∑n≥−k cnq

n converging for=τ � 0.

For example j(τ) is a modular function. Next time we are going to try andget a q-series for the j function.

Math 223b Notes 73

32 April 13, 2018

Last time we saw that a modular function for SL2(Z) is just a meromorphicfunction on X(1) = SL2(Z) \ H ∪ {∞}. This is a compact Riemann surfaceof genus 0. An alternative description of a modular function is a meromorphicfunction f on H such that

f(γτ) = f(τ)

for all γ ∈ SL2(Z) and f =∑n≥−k cnq

n.We now check that j(τ) actually has a q-expansion. We have

G2k(τ) = 2ζ(2k) + 2(2πi)2k

(2k − 1)!

∑n≥1

σ2k−1(n)qn, σ2k−1(n) =∑d|n

d2k−1.

So

g2(τ) = 60G4(π) =4

3π4

(1 + 240

∑n≥1

σ3(n)qn),

g3(τ) = 120G6(π) =8

27π6

(1 + 504

∑n≥1

σ5(n)qn),

∆(τ) = (2π)12∑n≥1

τ(n)qn,

j(τ) =1

q+∑n≥0

c(n)qn =1

q+ 744 + 196884q + · · · .

The important thing here is that the coefficients c(n) are all integers. Thisshows that j(τ) is indeed a modular function. It has a simple pole at ∞ and noother poles. It has the only 0 at τ = ω where ω3 = 1. It follows that the field ofmodular functions for SL2(Z) is C(j(z)) and the subring of modular functionsthat are holomorphic away from ∞ is C[j(z)].

32.1 Congruence subgroups

Recall that SL2(Z) \ H corresponds to lattices L ⊆ C up to homothety. Wewant to look at a finer group, so that Γ0(m) \H parametrizes pairs (L,L′) upto homothety, where L ⊆ C is a lattice and L′ ⊆ L is cyclic of index m.

Definition 32.1. For m ≥ 1, we define

Γ0(m) =

{(a bc d

)∈ SL2(Z) : c ≡ 0 (mod m)

}.

So if [τ ] ∈ Γ0(m) \H, we look at the corresponding [1, τ ] and [1,mτ ]. Up tohomothety, the lattice acted by ( a bc d ) is [cτ + d,m(aτ + b)], and this is equal to[1,mτ ] because c is a multiple of m. Then we can look at the modular curve

Y0 = Γ(m) \H.

Math 223b Notes 74

This is going to be a non-compact Riemann surface, and we compactify it byadding cusps. This can be compactified to a modular curve X0(m).

Definition 32.2. A modular function for Γ0(m) is a

• meromorphic function on X0(m),

• a meromorphic function f on H such that f(γτ) = f(τ) for all γ ∈ Γ0(m),and the q-series such that for all γ ∈ SL2(Z), the function f(γ(τ)) is aLaurent series in q1/m.

Examples include j(τ) or even j(mτ). Note that the projection mapX0(m)→X(1) has degree

[SL2(Z) : Γ0(m)] = m∏p|m

(1 +

1

p

)= dm,

where dm counts the number of cyclic index m sublattices of Z2. Then we aregoing to have a field extension

(modular functions for Γ0(m))/(modular functions for SL2(Z)) = C(j(τ))

of degree dm. But this is not Galois because Γ0(m) is not a normal subgroup ofSL2.

Theorem 32.3. The modular functions for Γ0(m) are C(j(z), j(mz)).

Proof. We will show that [C(j(z), j(mz)) : C(j(z))] is at least dm. To show that[L : K] ≥ k, we exhibit a field K ↪→ E with k distinct extensions L ↪→ E. Forevery γ ∈ SL2(Z)/Γ0(m), we embed

C(j(z), j(mz)) ↪→ E; j(z) 7→ j(γz) = j(z), j(mz) 7→ j(mγz).

This shows that j(z) and j(mz) exhaust the field of modular functions.

This tells us that there exists some algebraic relation between j(τ) andj(mτ). We are going to leverage this to prove stuff about the j-invariant. Inparticular, we are going to explicitly construct Φm such that Φm(j(τ), j(mτ)) =0.

Math 223b Notes 75

33 April 16, 2018

Last time we looked at X0(m) over X(1), which has degree dm. The rationalfunctions on X0(m) is C(j(τ), j(mτ)) and the rational functions on X(1) isC(j(τ)). So we have a field extension, and there is going to be an algebraicrelation between the two.

33.1 Modular polynomial

The goal today is to find a polynomial Φm(x, y) such that Φ(j(mτ), j(τ)) = 0.Then we will have Φm(j(mγτ), j(γτ)) = 0. The strategy is to look at

fm(X, τ) =∏

γ∈Γ0(m)\SL2(Z)

(X − j(mγτ)) =∏

L′⊆[1,τ ] cyclic

(X − j(L′)).

Note that coefficients of fm(X, τ) are holomorphic functions of τ ∈ H. It is alsoSL2(Z)-invariant. So once we check that they have q-expansions, we will getthat they are modular functions for SL2(Z). Then the coefficients lie in C[j(τ)]and so we can write

fm(X, τ) = Φm(X, j(τ)).

This will be our Φm.We first study cyclic index m sublattice of L = [1, τ ]. For L′ ⊆ L, take

d ∈ Z>0 be minimal with d ∈ L. Then d | m and so we can take a = md . Then

there is some b with 0 ≤ b < d and aτ + b ∈ L. In this case, L′ = [d, aτ + b]with ad = m, 0 ≤ b < d, gcd(a, b, d) = 1. So the cyclic index m sublattices of[1, τ ] are (up to homothety) the lattices [1, στ ] for

σ ∈ Cm =

{(a b0 d

): a, b, d as before

}.

One can equivalently classify cosets Γ0(m) \ SL2(Z) as

SL2(Z) ∩ σ−1 SL2(Z)σ

for σ ∈ Cm. For instance, if we consider σ0 = (m 00 1 ) then σ−1

0 SL2(Z)σ0 =( a b/mcm d

) and so the intersection is Γ0(m). Then

fm(X, τ) =∏σ∈Cm

(X − j(στ)),

with the coefficients of Xi being symmetric functions in j(στ).We know that each j(στ) is

j(στ) = j(aτ + b

d

),

which is a Laurent series in

e2πi(aτ+b)/d = Qa2

ζabm

Math 223b Notes 76

where Q = q1/m and Q = 2πiτ . Then it is a Laurent series in Q = q1/m andinvariant in τ 7→ τ + 1. So it is a power series in q = e2πiτ . This now showsthat the coefficients of fm(τ) actually line in C[j(τ)]. So fm(τ) = Φm(X, j(τ))for Φm ∈ C[X, τ ].

To show that the coefficients of Φm are integers, we show that the coefficientsof Φm(X, j(τ)) are in Z((q)).

Proposition 33.1. If p ∈ C[x] and p(j(τ)) ∈ Z((q)) then p ∈ Z[x].

Proof. Exercise.

Now it is obvious that the coefficients of Φm(X, j(τ)) are algebraic integers,because we are multiplying algebraic integer coefficient power series together.But we also note that the set{

Q−a2

ζ−abm +∑k≥0

ckQa2kζabkm : σ ∈ Cm

}

is invariant under Gal(Q(ζm)/Q). This shows that the coefficients are in Z.

Math 223b Notes 77

34 April 18, 2018

We constructed Φm(X,Y ) ∈ Z[X,Y ]. By construction, this has the propertythat Φm(j(L′), j(L)) = 0 if and only if L′ is a cyclic index m sublattice of L. SoΦm(j(E′), j(E)) = 0 if and only if there exists a cyclic m-isogeny φ : E′ → E(isogeny with ker(φ) cyclic of order m). This is true even in characteristic p.

34.1 Coefficients of the modular polynomial

Proposition 34.1. (a) Φm(X,Y ) = Φm(Y,X).

(b) If m is not a square, then Φm(X,X) has leading coefficient ±1.

(c) (Kronecker congruence) Φp(X,Y ) ≡ (Xp − Y )(X − Y p) mod p.

Proof. For (a), we show first that the zero locus is symmetric. So we show thatL′ is similar to a cyclic index m sublattice of L if and only if L is similar to acyclic index m sublattice of L′. But if L′ ⊆ L is cyclic of index m, then mL ⊆ L′is cyclic of index m. This shows that Φm(X,Y ) = cΦm(Y,X), where c = ±1,but c = 1 implies Φm(X,X) = 0, which is not true.

For (b), we look at

Φm(j(τ), j(τ)) = cmq−m

∑k>−m

ckq−k.

We want to show that cm is ±1. But we see that

Φm(j(τ), j(τ)) =∏σ∈Cm

(j(τ)− j(στ))

where j(τ)− j(στ) = (Q−m + holomorphic) + (ζ−abQ−a2

+ holomorphic). Butif m is not a square, whatever the leading term is, it is going to have coefficienta root of unity. So cm should be a product of roots of unity, which is also ainteger.

For (c), we need the explicit description of Cp. It is going to be

Cp = {σi = ( 1 i0 p ) : 0 ≤ i < p} ∪ {σ∞ = ( p 0

0 1 )}.

Then we have

j(σ∞τ) = j(pτ) = q−p +∑k≥0

ckqpk ≡ (j(τ))p (mod p).

For j(σiτ), the coefficients are going to be in Z[ζp]. So we work at the prime(1− ζp) that lies over p. Then

j(σiτ) = j(τ + i

p

)= ζ−ip Q−1 +

∑k≥0

ckζikQk ≡ Q−1 +

∑k≥0

ckQk (mod 1− ζp).

Math 223b Notes 78

Then we can compute

Φm(X, j(τ)) ≡ (X−j(τ)p)(X−j(τ/p))p ≡ (X−j(τ)p)(Xp−j(τ)) (mod 1−ζp)

and so Φm(X,Y ) ≡ (X − Y p)(Xp − Y ) modulo p.

Theorem 34.2. If L is a lattice with complex multiplication by O, then j(L)is an algebraic integer.

Proof. We have previously seen that there exists an L′ ⊆ L cyclic of index msuch that Φm(j(L), j(L)) = 0. Here, we were able to choose L′ = πL for someprincipal prime ππ = p. So Φm(j(L), j(L)) = 0 and so j(L) is a root of a monicpolynomial.

Recall that LO the ring class field of O was defined as LO/K = Frac(O)of degree Cl(O). We want to prove that if a is an invertible ideal of O thenM = K(j(a)) = LO.

Next time, we will show M ⊆ LO by looking at splitting behavior of primes.First we will show that Spl(L/Q) ⊆ Spl(M/Q) so that the Galois closure ofM is contained in L. For the other half, we will show that Spl′(M/Q) (whichhas at least one completely split factor in M) is contained in Spl(L/Q). Thatimplies that L ⊆M .

Math 223b Notes 79

35 April 20, 2018

Let O be an order in an imaginary quadratic field K, and write O = Z + fOKwhere f is called the conductor.

35.1 Main theorem of complex multiplication

Theorem 35.1 (main theorem of complex multiplication). If a is any invertibleideal of O, then M = K(j(a)) is the ring class field L = LO.

The first step is to show that Spl(L/Q) ⊆ Spl(M/Q) up to finitely manyexceptions. Take p ∈ Spl(L/Q) and assume p - f and p - a and p - [OM :OK [j(a)]]. You are going to show in the homework that the hypothesis for pimplies that p = Nπ = ππ for some π ∈ O. Then we have πa ⊆ a a cyclicsublattice of index p. This implies that

Φp(j(a), j(a)) = 0

and so j(a) is a root of Φp(x, x). Observe that Φp(x, x) ≡ (xp − x)(x − xp)modulo p.

Choose an arbitrary prime pM of M = K(j(a)) above p. Module pM , wehave

(j(a)p − j(a))(j(a)− j(a)p) ≡ 0 (mod pM ).

So j(a) is in pM . But the hypothesis that p - [OM : OK [j(aj)]] implies thatOK [j(a)] surjects to OM/pM . Then

OM/pM ∼= (OK/pK)[j(a)] ∼= Fp,

so this means that p splits completely in M/Q. This shows that M ⊆ L.The next step is to show Spl′(M/Q) ⊆ Spl(L/Q) again up to finitely many

primes p. Here, Spl′ is the set of primes with at least one prime with no in-teria. Consider p ∈ Spl′(M/Q) and exclude p | f and p sharing a factor with∏i<j(j(ai) − j(aj)) ∈ OL. (Here, a1, . . . , aH(O) are the representatives for all

ideal classes of O.By assumptions, there exists a pM above p such that p = NpM . Let p be

the prime in K below pM . Our goal is to show p ∩ O = πO for some π. Thiswill imply

p = [OK : p] = [O : p ∩ O] = [O : πO] = Nπ

and so p ∈ Spl(L/Q). Let a′ = p∩O. Then a′ is a cyclic sublattice of a of indexp. Then we have

Φp(j(a′), j(a)) = 0.

If we pick pL over pM and work in OL/pL, we get

0 = (j(a′)p− j(a))(j(a)

p− j(a′))

inside OL/pL. But j(a) is actually in OM/pM = Fp. So we get j(a′) = j(a)modulo pL. Because we have set our p so that pL does not divide

∏i<j(j(ai)−

Math 223b Notes 80

j(aj)), this implies that j(a) = j(a′). This shows that there is some π such thatp ∩ O = πO.

The conclusion is thatK(j(a)) = LO.

Then we have

h(O) ≥ [Q(j(a)) : Q] ≥ [K(j(a)) : K] = [LO : K] = h(O)

and soLO = K(j(a))

K Q(j(a))

Q

h(O) 2

2 h(O)

Gal

We can talk about Shimura reciprocity which says that if p is a prime of OKand (p, f) = 1 with no ramification, then

Frobp(j(a)) = j((p ∩ O)a).

Another thing we can talk about is Gross–Zagier, which gives a formula forj(a) − j(a′). We can also talk about the class number 1 problem. If h(O) = 1then j(O) ∈ Z and actually j(O) is always a cube. Then the problem comesdown to showing finitely many rational points on elliptic curves.

Math 223b Notes 81

36 April 23, 2018

Today we will look at Heegner’s proof of the class number 1 problem.

36.1 Cube root of j

For j = g32/∆, we define γ2 = 3

√j in the q-expansion, so that

γ2(τ) = q1/3 +∑k

akqk + · · ·

where ak are rational. This is no longer going to be SL2-invariant, but we willhave

r2(γτ) = ζr2(τ)

for some ζ3 = 1. It turns out that r2(γψ) = r2(ψ) if and only if a ≡ d ≡ 0 mod 3or b ≡ c ≡ 0 mod 3. So γ2(3τ) is a modular function for Γ0(9). So r2(3τ) ∈Q(j(τ), j(3τ)). )

Theorem 36.1. If O is an order in an imaginary quadratic field, and 3 does

not divide Disc(O), define τ0 =√−m if Disc(O) = −4m and τ0 = 3+

√−m

2otherwise (so that τ0 is relatively prime to 3). Then Q(r2(τ0)) = Q(j(τ0)) andK(r2(τ0)) = K(j(τ0)) = LO.

Proof. We have seen that r2(τ0) ∈ Q(j( τ03 ), j(3τ0)). But both lattices [1, τ03 ] and[1, 3τ0] has complex multiplication by exactly O′ = Z[3τ0], by the coprimalityconditions. This immediately shows that r2(τ0) ∈ LO′ . So we need to knowhow passing to a smaller order affects.

In general, let O be an order in an imaginary quadratic field, and let p -Disc(O). Suppose O′ ⊆ O is the unique index p suborder, which is O′ =Z + pO. In this case, we have the Galois group of LO′/LO is going to be(O/pO)×/(Z/pZ)× which is a cyclic group of order either p−1 or p+1, dependingon whether O/pO is Fp2 or F2

p.So we have LO′/LO of degree either 2 or 4 (assuming O× = ±1). So the

degree of the minimal polynomial of r2(τ0) over LO divides 4. But we knowthat r2(τ0)3 ∈ LO. This shows that r2(τ0) ∈ LO.

There are Weber functions f, f1, f2 ∈ Q((q1/48)) and SL2(Z) fixes the set{f48, f48

1 , f482 }. Here, we have that

f(τ)f1(τ)f2(τ) = f1(2τ)f2(τ) =√

2.

It is also true that f8,−f81,−f82 are roots of X3 − γ2X − 16.

Proposition 36.2. If m ≡ −3 mod 8 then

K(f(√−m)2) = LZ[−

√m],

where Z[√−m] is of index 2 in O√−m.

Proof. Use the fact that f(84)6 is a modular function for Γ0(64).

Math 223b Notes 82

36.2 Class number one problem

Theorem 36.3. Let K be an imaginary quadratic field, and let dK = disc(K).Then h(OK) = 1 if and only if dK ∈ {−3,−5,−7,−8,−11,−19,−43,−67,−163}.

Proof. First, consider the case when 2 is split or ramified in OK . Then there issome prime ideal p ⊆ OK with Np = 2. Because p has to be a prime, this canhappen only for d = −4,−7,−8. So assume that d is odd.

By one of the homework, we have Cl(K)[2] = (Z/2Z)k where k is the numberof prime factors of d minus 1. This shows that d = −2, and 2 being inert inQ(−√d) shows that p ≡ 3 (mod 8).

Setting aside the case 3 = p, we may assume that 3 - p. Consider O =OQ(

√−p) and O′ = Z[

√−p]. Then [O : O′] = 2 and because 2 is inert, we have

that LO′/LO is of degree 2 + 1 = 3. Then we have the following.

LO′ = Q(√−p)(f(

√−p)2) Q(f(

√−p)2)

LO = Q(√−p) Q

3 3

2

You can check that f(√−p) is some real number.

Let τ0 = 3+√−p

2 and let α = ζ−28 f2(τ0)2 = 2/f(

√−p)2. (This is from some

Weber function identity.) Because α generates Q(f(√−p)2), α4 generates that

field. What is the minimal polynomial of α4 over Q? Because α4 = −f2(τ0)8,its minimal polynomial must be X3 − γ2(τ)X − 16.

On the other hand, this is very restrictive, because we have an algebraicinteger that is a fourth power in a cubic field. α is an algebraic integer of degree3 and take X3 +aX2 + bX+ c the minimal polynomial of α. You can show thatα2 has minimal polynomial X3 + eX2 + fX + g for

e = 2b− a2, f = b2 − ac, g = c2,

and so α4 has minimal polynomial

X3 + (2f − e2)X2 + (f2 − eg)X − g2.

This shows that 2f = e2, g2 = 16. So g = −4 and (assuming c = 2) 2f = e2

becomes 2(b2 − 4a)2 = (2b− a2)2. Change of variables gives Y 2 = 2X(X3 + 1)and you end up getting a finite number of integral points. These gives the listof possible j-invariants.

Math 223b Notes 83

37 April 25, 2018

Let a be an invertible ideal in O ⊆ K, and let p be any prime of OK . Assumethat p - Disc(O) = f2 Disc(OK). The thing we want to show is that

Frobp(j(a)) = j((p ∩ O)a) = j((p ∩ O)−1a).

37.1 Shimura reciprocity

Here is the elliptic curve perspective. For O ⊆ K an order in an imaginaryquadratic field, we have

Ell(O) = {isomorphism classes of E/Q with End(E) ∼= O}.

Then we have a bijection Ell(O) with Cl(O), by sending a to C/a.Let Cl(O) act on Ell(O), define analytically by

a ∗ (C/L) = C/(a−1L).

The claim is that a ∗ E = HomO(a, E) as O-modules. The reason is that ifE = C/b then 0 → b → C → E → 0. Because a is a projective O-module, weget

0→ a−1b = HomO(a, b)→ HomO(a,C) ∼= C→ HomO(a, E)→ 0.

This shows that a ∗ E = HomO(a, E).But this is analytic. How do you make a ∗ E into a group variety? For

any finitely generated O-module M , we can put a group variety structure onHomO(M,E). If M is free, this is clear. If not, take a finite presentationOb → Oa →M → 0 and define

HomO(M,E) = ker(Ea → Eb)

given by the matrix. If M = a, we are going to get another elliptic curveE′ = a ∗ E.

For any elliptic curve E over a number field L, such that O ↪→ EndL(E),we define a ∗ E = HomO(a, E). For any invertible ideal a of O, we get anotherelliptic curve

O ↪→ EndL(a ∗ E).

Now Gal(LO/K) acts on Ell(O), and also the class group Cl(O) acts on Ell(O).

Proposition 37.1. We have g(a ∗ E) = a ∗ g(E) for all g ∈ Gal(LO/K).

So we are going to get a map φ : Gal(LO/K) → Cl(O), defined so thatg(E) = φ(g) ∗ E. We will show that φ is inverse to the map

θ : Cl(O)→ Gal(LO/K).

Math 223b Notes 84

Proof. We check this for [p] ∈ Cl(O). Take p ⊆ OK and look at pO = p ∩ O,which we are going to refer to as p as well. Take E ∈ Ell(O) such that E hasgood reduction at pL, where pL is the prime of L above p. Write E the reductionof E over L.

Let Ep be the elliptic curve given by y2 = x3apx + bp. Then we havej(Ep) = j(E)p. Now the claim is that

ENp ∼= p ∗ E = p ∗ E.

(The second equality follows from the definition of ∗.) If we have this, then

j(p ∗ E) = j(p ∗ E) = j(E)Np = j(E)Np (mod pL)

shows that Frobp(j(E)) = j(p ∗ E).First suppose that Np = p. Then restriction Hom(O, E)→ Hom(p, E) is an

isogeny

Ep−→ p ∗ E

of degree p. We also have an isogeny E → Ep given by Frobinius. We can showthat both maps are inseparable isogenies of degree p. Then there is a uniquesuch isogeny, so we get that the target are isomorphic.

Now suppose that Np = p2, which means that p = (p). Then the reductionE has no p-torsion and they are called supersingular. Then there exists a uniqueisogeny from E → E′ of degree p2. Then again I can there are both the Frobeniusand restriction are degree p2 isogenies.

Index

adele, 6adelic class group, 5

Chebotarev density theorem, 63class formation, 24class group, 12complex multiplication, 65conductor, 20content, 9

Dedekind ζ-function, 54Dirichlet density, 60Dirichlet L-function, 55Dirichlet series, 53discriminant of elliptic curve, 68

elliptic function, 66Euler product, 54

field formation, 24formation, 22

global class field theory, 15, 17global field, 5

Hasse norm theorem, 39Hilbert class field, 20

idele class group, 5

j-invariant, 68

Kronecker congruence, 77

main theorem of complexmultiplication, 79

Minkowski’s theorem, 10modular function, 72, 74modulus, 16

narrow Hilbert class field, 20natural density, 60normalized absolute value, 6

place, 5polar density, 60principal ideal theorem, 51

ray class field, 16ray class group, 16regulator, 58restricted product, 6

strong approximation, 11

Weber functions, 81Weierstrass form, 67Weierstrass ℘-function, 66

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