Math 223B - Algebraic Number Theory

Home , Class field theory, Class formation, Global field, Local class field theory, Ray class field

Math 223b - Algebraic Number Theory

Taught by Alison Miller Notes by Dongryul Kim Spring 2018

The course was taught by Alison Miller on Mondays, Wednesdays, Fridays from 12 to 1pm. The textbook was Cassels and Fr¨ohlich’s Algberaic Number Theory. There were weekly assignments and a ﬁnal paper. The course assistance was Zijian Yao.

Contents

1 January 22, 2018 5 1.1 Global ﬁelds and adeles ...... 5

2 January 24, 2018 7 2.1 Compactness of the adele ...... 7

3 January 26, 2018 9 3.1 Compactness of the multiplicative adele ...... 9 3.2 Applications of the compactness ...... 10

4 January 29, 2018 11 4.1 The class group ...... 12

5 January 31, 2018 13 5.1 Classical theorems from compactness ...... 13

6 February 2, 2018 15 6.1 Statement of global class ﬁeld theory ...... 15 6.2 Ray class ﬁeld ...... 16

7 February 5, 2018 17 7.1 Image of the reciprocity map ...... 17 7.2 Identifying the ray class group ...... 18

8 February 7, 2018 19 8.1 Properties of the ray class ﬁeld ...... 19

1 Last Update: August 27, 2018 9 February 9, 2018 21 9.1 Ideles with ﬁeld extensions ...... 21 9.2 Formations ...... 22

10 February 12, 2018 23 10.1 Class formation ...... 24

11 February 14, 2018 25 11.1 Consequences of the class formation axiom ...... 25 11.2 Strategy for verifying the axioms ...... 26

12 February 16, 2018 28 12.1 Cohomology of the ideles ...... 28

13 February 21, 2018 31 13.1 First inequality ...... 31

14 February 23, 2018 33 14.1 Algebraic weak Chebotarev ...... 33 14.2 Reduction of the second inequality to Kummer extensions . . . . 34

15 February 26, 2018 35 15.1 Computing the size of the quotient ...... 35 15.2 Choosing the completely split places ...... 36

16 February 28, 2018 38 16.1 Finishing the second inequality ...... 38 16.2 Immediate corollaries ...... 39

17 March 2, 2018 41 17.1 Reduction to cyclic cyclotomic extensions ...... 41 17.2 Invariant and reciprocity maps ...... 42

18 March 5, 2018 44 18.1 θ-reciprocity for cyclotomic ﬁelds ...... 44 18.2 Invariant map on the class group ...... 45

19 March 7, 2018 46 19.1 CK form a class formation ...... 46 19.2 Consequences ...... 47

20 March 9, 2018 49 20.1 Classiﬁcation of normic subgroups ...... 49

21 March 19, 2018 51 21.1 Properties of the Hilbert class ﬁeld ...... 51

2 22 March 21, 2018 53 22.1 Class ﬁeld tower ...... 53 22.2 L-functions ...... 53

23 March 23, 2018 55 23.1 L-functions ...... 55 23.2 Convergence of Dirichlet series ...... 55

24 March 26, 2018 57 24.1 Meromorphic continuation of L-functions ...... 57

25 March 28, 2018 59 25.1 Densities of sets of primes ...... 60

26 March 30, 2018 61 26.1 Proof of second inequality ...... 61

27 April 2, 2018 63 27.1 Chebotarev density theorem ...... 63

28 April 4, 2018 65 28.1 Complex multiplication ...... 65 28.2 Elliptic curves ...... 66

29 April 6, 2018 67 29.1 Properties of the Weierstrass function ...... 67 29.2 j-invariant ...... 68

30 April 9, 2018 70 30.1 j-invariants of CM elliptic curves ...... 70

31 April 11, 2018 71 31.1 Modular forms and functions ...... 71

32 April 13, 2018 73 32.1 Congruence subgroups ...... 73

33 April 16, 2018 75 33.1 Modular polynomial ...... 75

34 April 18, 2018 77 34.1 Coeﬃcients of the modular polynomial ...... 77

35 April 20, 2018 79 35.1 Main theorem of complex multiplication ...... 79

3 36 April 23, 2018 81 36.1 Cube root of j ...... 81 36.2 Class number one problem ...... 82

37 April 25, 2018 83 37.1 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 a non-archemedian local field, (i.e., finite extensions of either Qp or Fp((t))), and L/K is Galois. What we showed last semester is that there exists a canonical isomorphism × × ab θL/K : K /NL → Gal(L/K) . The proof was via Galois cohomology. The left hand side is Hˆ 0(L/K, L×) and the right hand side is Hˆ −2(L/K, Z). We needed two essential facts: Hilbert 90 1 × 2 × ∼ 1 H (L/K, L ) = 0, and H (L/K, L ) = [L:K] Z/Z. This semester, we’re going to talk about global fields. Now K is a global field, and consider a finite L/K. We want a global reciprocity map

ab θL/K : CK /NCL → Gal(L/K) .

This CK is going to be the adelic class group or the idele class group and is defined as × × CK = AK /K . The shape of the argument is going to be exactly the same. We’re going to need the key lemmas on H1 and H2, and then everything else is going to be formal. This formalism that lets us do both local and global class field theory is 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 density theorem • complex multiplication for elliptic curves, abelian extensions of imaginary quadratic fields

1.1 Global fields and adeles This material is at the end of Chapter 2 of Cassels and Frölich. Let K be a global field. This can be define in two ways. First, it can be defined as a finite extension of Q or Fp(t). An equivalent definition is that every completion of K is 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 discrete valuation, 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 the residue field kv = Ov/(π). Then we define the normalized absolute value on Kv by −1 |π|v = |kv| . ∼ Also, if Kv = R then we define |a|v = |a|, and if Kv ∈ C then we define 2 |a|v = |a| . One of the motivations for defining normalization in this way is the product formula. For every a ∈ K×, we will have

Y |a|v = 1. v

We proved this last semester, by first doing it for K = Q, Fp(t), and then showing that both sides are preserved by taking finite extensions. Another motivation is that a locally compact abelian group has a Haar measure. So any local field + Kv has a Haar measure such that µ(E) = µ(a + E) for all a ∈ Kv , and such a measure is unique up to scaling. First we normalize it so that µ(Ov) = 1, in the non-archemedian case. We then consider |x|v = µ(xOv). It will follow that µ(aE) = |a|vµ(E) for any measurable E. This is because E 7→ µ(aE) is another Haar measure, so it is constant times the original Haar measure. Using this, we will give another proof of the product formula. Now let me define the adeles in 5 minutes. For K a global field, we consider the set {Kv} for v the places of K. The field K embeds into each Kv, and we are Q going to consider the embedding K,→ v Kv. 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

Y0 Xi = {(xi)i∈I : xi ∈ Yi for almost all i}. i Deﬁnition 1.2. We deﬁne the adele as

0 0 + Y + × Y × AK = Kv , AK = Kv v v

+ × where the restricted products are with respect to Ov and Ov . Math 223b Notes 7

2 January 24, 2018

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

Y0 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 the Q subspace topology from the product topology for i Xi. If S is any ﬁnite subset Q0 of I, the restricted product i Xi has an open cover of the form

Y0 Y Y Xi = Xi × Yi. S i∈S i/∈S

The subspace topology here is equal to the product topology.

2.1 Compactness of the adele Q0 Theorem 2.1. If Xi are locally compact and Yi are compact, then Xi is locally compact. Q0 Proof. Each of ( Xi)S is locally compact by Tychonoﬀ’s theorem. As we have deﬁned, the adele

Y0 AK = Kv v restricted with respect to Ov is a topological ring. The units

0 × Y × AK = Kv v

× restricted to Ov is a topological group. For the next few lectures, we are going to focus on the structure and properties. × The topology on AK is not the subspace topology from AK , by the way. This is a natural thing to do because the inverse map is not continuous with the subspace topology. In general, for R a topological ring, the inverse map on R× is not necessarily continuous. The correct way to topologize R× is to use the 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 this × topology on AK agrees with the topology by the restricted product. For local ﬁelds, we don’t need to worry about this issue. × We know that AK and AK are locally compact. There is a natural embedding

K,→ AK ; x 7→ x = (x)

× × Likewise, we have K ,→ AK . + + + + Proposition 2.2. K is discrete in AK , and is closed. Thus AK /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 topological rings. (In this case, we can define the tensor product in the following ∼ n way. Because L = K as vector spaces, we topologize AK ⊗K L by the product topology.) Q Proof. The important fact is that if v is a place of K, then Kv ⊗K L = v0 Lv0 for v0 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 for + n + n K. (This is because L ⊆ AL is equivalent to K ⊆ (AK ) .) So we can check for K = Q and K = Fp(t). Let’s only do this for K = Q. We take Y U = Zv × (−1, 1) ⊆ AQ. v6=R

Then a ∈ U ∩ Q implies that a ∈ Z, so a = 0. So a point is open in Q, which shows 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 Y h 1 1i D = × − , , Zv 2 2 v6=R this is compact by Tychonoff. Now the claim is that D +Q = AQ. We first want + for a ∈ AK , an r ∈ Q such that a − r ∈ D. There are finitely many finite places such that ap ∈/ Zp. We can make them all into Zp by adding a rational number. 1 1 Then we can add an integer so that the infinite component is in [− 2 , 2 ]. So D +/ + is a surjection. Because D is compact, the image is compact as AQ Q well. Math 223b Notes 9

3 January 26, 2018

+ + + We should last time that AK /K is compact. First of all, AK has a Haar measure because its a locally compact abelian group. This can also described Q + explicitly as the product of the local Haar measures, so that if v Ev ⊆ AK is + + such that Ev = Kv is measurable for each v and Ev = OV for almost all v, then Y Y µ Ev = µv(Ev). v v + + This Haar measure descends to a quotient AK /K (by using a fundamental + + + + domain). Because AK /K is compact, the measure µ(AK /K ) is going to be + + finite. As an exercise, show that µ(AK /K ) is the discriminant |disc K|. × Definition 3.1. For a = (ai) ∈ AK , we define its content as Y c(a) = |av|v. v

× + Proposition 3.2. If a ∈ AK and E ⊆ AK , 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 AK by multiplication. Then µ(aE) = + + + + c(a)µ(E) for any measurable E ⊆ AK /K . Taking E = AK /K shows that µ(E) = c(a)µ(E), and so c(a) = 1.

3.1 Compactness of the multiplicative adele × × What about the K ⊆ AK ? This is what this course will be ultimately care about.

× × Proposition 3.4. K is discrete inside AK . Proof. We have an embedding

× + + K ,→ AK ,→ AK × AK . + + + + × But K × K is already discrete inside AK × AK , so K should be discrete as well.

× × × Likewise, K is a closed subset. So AK /K is a Hausdorﬀ topological group. Is this going to be compact? The answer is no, because we have a surjective continuous map × × c : AK /K → R>0. (It can be shown that it is continuous.) So deﬁne

1 × AK = ker(c : AK → R>0). Math 223b Notes 10

1 × Theorem 3.5. AK /K is compact. We will prove this a bit later, but as two important corollaries: ﬁniteness of the class group, and Dirichlet’s unit theorem. (Neukirch actually goes the other direction to prove compactness.)

3.2 Applications of the compactness × Deﬁnition 3.6. For a ∈ AK , deﬁne

× + Sa = {x ∈ AK : |xv|v ≤ |av|v for all v} ⊆ AK . Theorem 3.7 (Minkowski). There exists a constant C > 1 (depending on K) × × such that for all a ∈ AK with c(a) > C, there exists x ∈ K ∩ Sa. Proof. Consider the set

Y + Y 1 + B = OV × B(0, 2 ) ⊆ AK . v ﬁn v inf

Then you can explicitly compute µ(B), and it is going to be µ(B) < ∞. If we + + deﬁne C = µ(AK /K )/µ(B), then we have

+ + µ(aB) = c(a)µ(B) > cµ(B) = µ(AK /K ).

+ + So the quotient map aB → AK /K is non-injective. This shows that there are × b1, b2 ∈ aB such that x = b1 − b2 ∈ aB ∩ K . By the non-archemedian triangle inequality, 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 exists × an 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

Y0 K,→ + /K+ = K . AK v0 v v6=v0 has dense image.

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

Y Y + B = B(xv, ) × OV , v∈S v∈ /S

+ where x ∈ AK and > 0. This describes a neighborhood basis for x, and we need to show that any such B contains an element of K+. We use compactness + + × of AK /K to get that there exists some a ∈ AK such that (the image of) Sa + + × covers /K . Now we can choose b ∈ such that |bv|v < for v ∈ S AK AK |av |v × and c(b) > c.(c is from Minkowski.) Then there exists a z ∈ Sk ∩ K . Write x = r + s ∈ + = K+ + S , z AK a where r ∈ K and s ∈ Sa. Now rz = x−sz with rz ∈ K. But x−sz ∈ B because |(sz)v|v ≤ |avbv|v < for v ∈ S. We also need to show that (x − sz)v ∈ Ov for v∈ / S, but we can enlarge S to make this true.

1 × Proposition 4.3. AK /K is compact. 1 (Here, you’ll prove in the homework that AK has the same topology as a × 1 subspace of AK and as a subspace of AK . Also, AK is closed in AK .) 1 Proof. Let D = Sa ∩ AK , where c(a) > c. Note that Sa is closed in AK , and so 1 Sa is compact by Tychonoﬀ. So D = Sa ∩ AK is also compact. Now it suﬃces 1 × 1 to show that D AK /K is a surjection. This means for each x ∈ AK , there x × −1 exists a d ∈ D such that d ∈ K . For this, consider x Sa = Sx−1a. Here, c(x−1a) = c(x)−1c(a) = c(a) > c, and so there exists an element r ∈ K× such 1 that r ∈ Sx−1a. Then rx ∈ AK ∩ 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 K is deﬁned as

group of fractional ideals of OK Cl(K) = Cl(OK ) = . group of principal fractional ideals of OK More generally, let K be a global field and let S be the finite nonempty set of 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 of OK,S correspond to places not in S. If S is the set of archemedian places, then we 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 kernel × × is OK,S.) Next time we are going to show that ClS(K) is ﬁnite and OK,S is ﬁnitely generated of rank |S| − 1. Math 223b Notes 13

5 January 31, 2018

1 × Last time we showed that AK /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 all the infinite places. Then ClS(K) = Cl(OK,S) × is finite and the group of units OK,S is finitely generated of rank |S| − 1. Let us define

× Y × Y × AK,S = {v : |xv|v = 1 for v∈ / S} = Kv × Ov . v∈S v∈ /S

× This is the adelic version of OK,S. It is also one of the open sets that deﬁne × 1 1 1 AK . Also let us deﬁne AK,S = AK,S ∩ AK . This is also open in AK . We have a short exact sequence

1 × × 1 × 1 1 × 1 → AK,SK /K → AK /K → AK /AK,SK → 1.

1 × 1 The middle space is compact, and AK,SK is an open subgroup of AK . 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 to third map is continuous and the third space is compact as well. But when you quotient by an open subgroup, you get a discrete subgroup. This all shows that

1 1 × AK /(AK,S · K ) is a finite discrete group. 1 1 × ∼ Proposition 5.2. AK /(AK,SK ) = ClS(K). Proof. Consider the map 1 Y vp(av ) ϕ : AK → ClS(K); a 7→ p . v∈ /S where p is the prime ideal corresponding to the finite place v. The kernel of ϕ is precisely those that can be expressed as things coming from K× and those that 1 × give the fractional ideal (1). So it is ker ϕ = AK,S · K . On the other hand, ϕ surjective because we can just set av as we want. So we get the isomorphism. × × ∼ 1 × Proposition 5.3. (AK,SK )/K = AK,S/OK,S. × ∼ 1 × Proof. It suffices to show that OK,S = AK,S ∩ K . This is clear. Math 223b Notes 14

1 × We had that AK,S/OK,S is compact. So we are going to focus on what is happening with the S part. Lemma 5.4. The set

× 1 {x ∈ OK,S : |x|v ∈ [ 2 , 2] for all v ∈ S} is ﬁnite.

× Q × Q Proof. This is the same as the intersection of K and v∈ /S OV × 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 for all 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

1 Y + L : AK,S → R ;(L (a))v = log(|av|v). v∈S

Q + Proposition 5.6. L (OK,S) is a discrete subgroup of v∈S R , and the kernel × Q + of L : OK,S → v∈S R is ﬁnite.

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

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

1 × 1 × AK,S/OK,S → L (AK,S)/L (OK,S).

1 But the condition on L (AK,S) is that all components should add up to 1. It is also true that this is the only condition. This shows that

1 + s P L (AK,S) = H = {x ∈ (R ) : v xv = 0}.

1 × 1 × On the other hand, AK,S/OK,S is compact, and so L (AK,S)/L (OK,S) is × compact. This shows that L (OK,S) is a free group of rank |S| − 1. Therefore × OK,S is a ﬁnitely generated group of rank |S| − 1. Math 223b Notes 15

6 February 2, 2018

1 × We have shown that the group AK /K is compact. In class field theory, we × × will use CK = AK /K . How will we compare it to CK ? We have a short exact sequence relating the two. Suppose first that K is a number field. Then we have a short exact sequence

1 × × × c 1 → AK /K → AK /K −→ R>0 → 1.

× For each archemedian place v, we have a closed embedding R ,→ Kv , and then a × × × closed embedding Kv ,→ AK /K . This gives a splitting as topological groups. If K is a function ﬁeld over Fq, then we have

c 1 × × × Z 1 → AK /K → AK /K −→ q → 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

× × ab θ/K : CK = AK /K → Gal(K /K)

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

0 Y × 0 (AK ) = (Kv ) v inf which is easy to describe. There is also a local-to-global compatibility

θ × /Kv ab Kv Gal(Kv /Kv)

θ × × /K ab AK /K Gal(K /K).

This local-to-global compatibility speciﬁes θ/K uniquely because Kv generate it topologically. So θ/K induces a bijection

open subgroups of open subgroups of ←→ ab . Ck of ﬁnite index Gal(K /K) of ﬁnite index

Then the right hand side is the same as a ﬁnite abelian extension L/K. The map 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

× × Y NL/K : AL → AK ; a 7→ (Na)v = NLw/Kv (aw). w/v

It turns out that L× N K

× N × AL AK

× × commutes, so we get a map CL → CK . It turns out that NL/K (AL ) ⊆ AK is open.

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.

Q m(v) We write m = v v (so m = p1p2∞1 for instance.) For any m we have a congruence subgroup Y Y Y >0 × × Up,mp × Kv × R K /K ⊆ CK , p ﬁn v inf m(v)=0 v real m(v)=1 where mp Up,mp = {x ∈ OKp : x ≡ 1 (mod p )}.

Definition 6.3. The ray class field Lm is the fixed field of θ(Um), so that ∼ ∼ × × Y Y Gal(Lm/K) = CK /Um = AK /K Up,mp × (−) . p fin v fin

The punchline is that ∼ CK /Um = Clm(OK ) is the ray class group, which is the fractional ideals of OK relatively prime to m modulo the principal fractional ideals (a) with a ∈ Up,mp for all ﬁnite p and av > 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 global ab field. There exists a canonical θ/K : CK → Gal(K /K) such that

0 • ker θ/K = (CK ) is the connected component at 1, • im θ is dense in Gal(Kab/K) = lim Gal(L/K) (this means /K −→L/K ﬁn. ab. 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.

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

1 × Proof. We know that AK /K is compact. So it suﬃces to check that this sur- 0 jects onto CK /(CK ) . This is because we can change only the archemedian component to make the content 1, and this doesn’t change the connected component.

0 ab Now CK /(CK ) is compact, so the image of θ/K in Gal(K /K) is compact. This means that the image is closed, and also is dense. So it is the whole space Gal(Kab/K) for K a number field. Let us see what goes wrong when K is a function field. Then all of the × localizations are totally disconnected, so AK is totally disconnected and CK is totally disconnected. Then the map θ/K is injective. If θ/K were an isomorphism, we would have that CK is profinite and thus compact. Let k = Fq be the field of constants, and consider the field extension

k · K ⊆ Kab.

Then we get a homomorphism

ab Gal(K /K) → Zb =∼ Gal(k/k). It turns out that the square

ab ∼ 1 IK Gal(K /K) Zb = Gal(k/k) 1

∼= θ/K q7→1

1 × × × c Z 1 AK /K CK = AK /K q 1 commutes, and also both exact sequences split. Math 223b Notes 18

7.2 Identifying the ray class group Q m(v) Recall that for K a number ﬁeld, a modulus is a formal product m = v v . × Then we can deﬁne an open subset Um ⊆ AK

Y Y >0 Y × Um = Up,m(p) × R × Kv . p ﬁn v real,m(v)=1 v other

m Then the congruence subgroup CK ⊆ CK is deﬁned as

m × × × × CK = Um · K /K ⊆ AK /K = CK .

m We also defined the ray class field of modulus m as the fixed field of θ/K (CK ). In this case, we have

∼ ab m ∼ m Gal(Lm/K) = Gal(K /K)/θ/K (CK ) = CK /CK

m Proposition 7.3. CK /CK is isomorphic to the ray class group Clm(K), where frac. ideals rel. prime to m Clm(K) = . frac. ideals (a) with a ∈ Um Proof. Exercise. We can give the inverse map explicitly. If p is relatively prime to m then we × can send p to [(1,..., 1, π, 1,...)] where π ∈ Kp 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

∼ m θ/K Clm(K) = CK /CK −−→ Gal(Lm/K).

This sends p to Frobp lying in the decomposition group. This means that Lm is a 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 modulus m by the fixed field of

m × × θL/K (CK ) = 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. (For infinite places, we say that v0/v ramifies if v is is real and v0 is complex. Neukirch uses a different convention, that infinite places never ramify.) 0 (b) If m | m then Lm ⊆ Lm0 .

× m Op ,→ Um → CK .

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

θ × /Kp Kp Gal((Lm)p0 /Kp)

θLm/K CK Gal(Lm/K).

m × Now we have CK = ker θLm/K and so Op ⊆ ker θ(Lm)p/Kp . This shows that θ (O× ) is trivial. By local class field theory, this just means that (Lm)p/Kp Kp × × (Lm)p/Kp is unramified. For infinite places, we use Kp instead of Op . m (b) and (c) just follow from the definition of Um and CK . For (d), observe × that any open subgroup of AK contains some Um. Then any open subgroup of m CK contains some CK . 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 a positive integer. First let us ﬁgure out Clm(K) and then ﬁgure out Lm. First, if m = m∞, frac. ideals rel. prime to m Cl ( ) = =∼ ( /m )×. m Q (a) where a > 0 and a ≡ 1 mod m Z Z

If m = m, then we would get (Z/mZ)×/{±1}. It now follows from the explicit description of θ/K for cyclotomic ﬁelds that −1 Lm∞ = Q(ζm). If we don’t allow ∞, we will get Lm = Q(ζm + ζm ). Math 223b Notes 20

Deﬁnition 8.3. Let L be a ﬁnite abelian extension of K. The minimal m such that 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

θL/K CK Gal(L/K)

f ∼ CK /CK = Clf(K) factors. We know that L is unramified outside the primes p - f. Then for each p - f, we have p = Frob ∈ Gal(L/K), L/K p and ( p ) is the restriction of ( p ) to Gal(L/K). Since ( p ) depends only L/K L/K Lp/K p on [p] ∈ Clf(K), same is true of ( L/K ). m Also, for ker θL/K = NL/K CL. So L ⊆ Lm is equivalent to NL/K CL ⊇ CK . 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, we have that √ q q Frobq(Q( p)/Q) = √ = Q( p)/Q p 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 places Q over 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 class field theory. Here, K is a local field. The key fact was homological: if L/K is finite Galois, then × × 0 × −2 ab K /NL =∼ Hˆ (L/K, L ) =∼ Hˆ (L/K, Z) =∼ Gal(L/K) . Then we took the direct limit over all finite L/K to get lim K×/NL× ∼ Gal(Kab/K). ←− = In characteristic zero, this gives Kb × =∼ 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 maps Gal(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, sep with L× = ((Ksep)×)Gal(K /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). We have closed embeddings × × AK ,→ AL, AK ,→ AL . × × × Gal(L/K) Proposition 9.1. AK is closed in AL . If L/K is Galois, then (AL ) = × × × AK and AK ∩ 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 AK /K → AL /L . That is, CK ,→ CL. Gal(L/K) Proposition 9.2. We have CL = CK . Proof. We use the short exact sequence × × 1 → L → AL → CL → 1. If we take Galois cohomology, we get × × Gal(L/K) 1 × 1 → K → AK → CL → H (L/K, L ) = 1. Gal(L/K) ∼ This shows that CL = CK . Gal(L/K) In general, Cl(OK ) → (Cl(OL)) 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 this case, open subgroups as finite extensions of the field. Going back to the general case, consider the set of open subgroup {GK : K ∈ X} of G. We are going to refer to these indices K as “fields”. In particular, we have a partial ordering of field, with K ⊆ L if and only if

GL ⊆ GK . There is a unique minimal field K0 corresponding to GK0 = G, and we 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/K is “normal” if GL C GK is a normal subgroup. We can define

GKL = GK ∩ GL,GK∩L = GK GL.

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

−1 GgK = gGK g .

Deﬁnition 9.3. A formation A is a G-module satisfying the following equivalent conditions: • G acts continuously on A (using the discrete topology on A)

S GK • A = K A sep sep × Example 9.4. Suppose that G = Gal(K0 /K0). Then A = (K0 ) is a formation. For global ﬁelds, we can instead look at A = lim CK . −→K/K0 ﬁn 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 called ﬁelds. A G-module was an abelian group A with a G-action such that the action is continuous. This can be explicitly described as [ [ A = AGK = AK K K

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

q q Definition 10.1. We define H (L/K) = H (GL/K ,AL). Likewise, se define ˆ q ˆ q H (L/K) = H (GL/K ,AL). We can also define profinite cohomology

Hq(/K) = Hq(G ,A) = lim Hq(G ,A ) = lim Hq(L/K). K −→ L/K L −→ L L

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

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

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

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

There is also conjugation action. If L/K is a normal layer and g ∈ G, we −1 can deﬁne GgK = gGK g . Then we have isomorphisms

∼ −1 GL/K = GgL/gK ;[x] 7→ [gxg ].

We also have isomorphisms

AL → AgL; a 7→ ga, and this induces g∗ : Hq(L/K) → Hq(gL/gK).

If g ∈ GK , then gK = K and gL = L because L/K is normal. In this case, you can 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 input something. Deﬁnition 10.2. (G, A) is a ﬁeld formation if for every layer L/K, H1(L/K) = 0.

sep Of course, the motivating example here is G = Gal(K0 /K0) and A = sep × (K0 ) . 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) = H2(L/K). −→ L/K

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

Deﬁnition 10.3. A is a class formation if for every L/K we provide an isomorphism 2 1 invL/K : H (L/K) → [L:K] Z/Z, with compatibility conditions

inv 2 L/K 1 H (L/K) [L:K] Z/Z

inf inv 2 E/K 1 H (E/K) [E:K] Z/Z for all normal E/L/K, and

inf 2 E/K 1 H (E/K) [E:K] Z/Z

res ×[L:K] inf 2 E/L 1 H (E/L) [E:L] Z/Z for E/L/K with E/K normal. Try to deduce the diagram for corestriction. Math 223b Notes 25

11 February 14, 2018

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, 2 ∼ 1 (ii) there are invariant maps inv : H (GL/K ,AL) = [L:K] Z/Z compatible with inﬂation and restriction (and thus correstriction)

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

1 1 H (Kn/Km) = H (nZ/mZ, Z) = 0 and 2 2 n H (Kn/Km) = H (nZ/mZ, Z) = Z/( m Z). 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,

sep [ G = Gal(K0 /K),A = CL L/K 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.

inv 2 E/L 1 H (E/L) [E:L] Z/Z

cor inv 2 E/K 1 H (E/K) [E:K] Z/Z

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

inv 2 L/K 1 H (L/K) [L:K] Z/Z

g∗ inv 2 L/K 1 H (gL/gK) [L:K] Z/Z 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 because g∗ : H2(L/K) → H2(L/K) is the identity map as we have shown. In general, 0 0 we have g ∈ GK0 = G, so we can ﬁnd L ⊇ L such that L /K is normal. Then 0 you compute the diagrams for L/K and L /K0. If (G, A) is a class formation, we deﬁne the fundamental class 1 u ∈ H2(L/K) by inv(u ) = . L/K L/K [L : K]

2 Clearly uL/K generates H (L/K). Proposition 11.4. Let E/L/K be ﬁelds with E/K normal (and L/K normal if necessary).

(a) res uE/K = uE/L.

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

Theorem 11.5. For any normal layer L/K, cup product with uL/K gives an isomorphism ˆ q ˆ q+2 H (GL/K , Z) → H (L/K). Proof. This is literally Tate’s theorem.

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

ab ˆ −2 ˆ 0 ∼ − ` uL/K : GL/K = H (GL/K , Z) → H (L/K) = AK /NL/K AL. sep S So if we can show that (Gal(K0 /K), L/K CL) is a class formation for any global base ﬁeld K0, then we get

ab ∼ Gal(L/K) = 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 invariant map. 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)| = [L : K]. |H1(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 to prove 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

2 1 invL/K : H (L/K) → [L:K] Z/Z by patching the local invariant maps together. Math 223b Notes 28

12 February 16, 2018

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

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. Then × AL is covered by Y Y × = × = L× × O×. AL,S AL,S w w w∈S w∈ /S We can then compute q × q YY × YY × H (L/K, AL,S) = H G, Lv × Ov v∈S v0|v v∈ /S v0|v Y q Y × Y q Y × = H G, Lv × H G, Ov . v∈S v0|v v∈ /S v0|v

For each v, choose w | v and let Gw be the decomposition group of w. We know that G acts on the set v0 | v transitively. So we can say that

Y Y Lgw 0 v |v g∈G/Gw is the coinduced module. So

q Q × ∼ q × H (G, v0|v Lv0 ) = H (Gw,Lw ). This isomorphism can be realized explicitly as

q Q × res q Q × π∗ q × ∼ q × H (G, v0|v Lv0 ) −−→ H (Gw, v0|v Lv0 ) −→ H (Gw,Lw ) = H (Lw/Kv,Lw ). Likewise, we have

q Q × ∼ q × ∼ q × H (G, v0|v Ov0 ) = H (Gw, Ow ) = H (Lw/Kv, Ow ).

The statements are also true for also Hˆ q. ˆ q × Suppose that v is unramiﬁed in L. Then H (Kw/Kv, Ow ) vanishes. In particular, if we assume that S contains all ramiﬁed places of K, then

q × ∼ Y q × H (L/K, AL,S) = H (Lw/Kv,Lw ). v∈S Now we can write × = lim × AL −→ AL,S S Math 223b Notes 29 where S runs over all ﬁnite sets containing all ramiﬁed places. Then

Hˆ q(L/K, ×) = lim Hq(L/K, × ) AL −→ AL,S Y M = lim Hq(L /K ,K×) = Hq(L /K ,L×). −→ w v w w v w S v∈S v

An immediate consequence is that

1 × ∼ H (L/K, AL ) = 1. That is, the ideles give a ﬁeld formation. That is, if we deﬁne

× [ × AKsep = AL , L/K

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

2 × ∼ M 2 × ∼ M 1 H (L/K, ) = H (Lw/Kv,L ) = / AL w [Lw:Kv ] Z Z v v and it is inﬁnite. So this is far oﬀ from what we need for a class formation. Later, we are going to see that

M 1 ∼ 2 × 2 ∼ 1 / = H (L/K, ) → H (L/K, Ck) = / [Lw:Kv ] Z Z AL [L:K] Z Z v is given by adding up the components. The other interesting thing we can compute is

× × ˆ 0 × ∼ M ˆ 0 × M × × AK /NAL = H (L/K, AL ) = H (Lw/Kv,Lw ) = Kv /NLw . v v This is equivalent to the norm theorem for ideles.

× × × Theorem 12.1. For (av) ∈ AK , we have a ∈ NAL if and only if av ∈ NLw for all v. There is an analogue for principal ideles, which requires the second inequality.

× × × Theorem 12.2 (Hasse). For a ∈ K with av ∈ Kv the v-component, a ∈ NL × if and only if av ∈ NLw for all v. × × Next, we want to get the Herbrand quotient of CL = AL /L when L/K is × cyclic. The problem is that the Herbrand quotient of AL is infinite, so we can’t × use this directly. So we are going to choose S large enough so that AL,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). Then we get a short exact sequence

× × 1 → OK,S → AL,S → CL → 1. So × Q h(AL,S) v∈S[Lw : Kv] h(CL) = × = × . h(OL,S) h(OL,S) × Next time we will calculate the Herbrand quotient of OL,S. Math 223b Notes 31

13 February 21, 2018

Our goal today is to get something for L/K cyclic extension of global ﬁelds. If G = Gal(L/K) and |G| = n = [L : K], we want to show that Hˆ 0(L/K) ≥ n.

13.1 First inequality We want to compute the Herbrand quotient

0 |Hˆ (G, CL)| h(CL) = . 1 |Hˆ (G, CL)| If S is a suﬃciently large ﬁnite set of places of K, we have a short exact sequence

× × 1 → OL,S → AL,S → CL → 1. Last time we showed that Q v∈S[Lw : Kv] h(AL,S) = × , h(OL,S)

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

× Y + L : OL,S → R ;(a 7→ (log|a|v0 )v0∈S v0∈S P whose image lies in the hyperplane v0 rv0 = 0. The image is a lattice, and so induces an isomorphism × ∼ OL,S ⊗Z R = H. Lemma 13.1. Let G be a finite group. Let L/K be an extension, with K is infinite. Two finite-dimensional K[G]-modules V1 and V2 are isomorphic if and only if V1 ⊗K L and V2 ⊗K L are isomorphic as L[G]-modules.

Proof. Suppose V1 ⊗K L and V2 ⊗K L are isomorphic as L[G]-modules. First, we can reduce to the case when L/K is finitely generated, because the isomorphism is going to be given by some finite number of coefficients. Then by Noether normalization, L is a finite extension of a transcendental extension of K, and the transcendental extension can be taken care easily by specialization. So we only 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 ⊗K L, V2 ⊗K L) → L, which is nonzero. But this is a polynomial function on HomK[G](V1,V2) ⊗K L. So there must be some ϕ ∈ HomK[G](V1,V2) such that det ϕ 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 the 1 ∼ product of factors GLn(L), and H (G, GLn(L)) = 1. This shows that there is G ∼ some ϕ ∈ (V1 ⊗ L, V2 ⊗ L) that gives an isomorphism V1 = V2. understand 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 index subgroup A0 ⊆ A/T (A) and B0 ⊆ B/T (B) such that A0 =∼ B0. Then h(A) = h(A0) = h(B0) = h(B) because finite groups have Herbrand quotient 1. We have an isomorphism

× ∼ Y + OL,S ⊗Z R = H ⊆ R v0∈S which extends to an isomorphism

× ∼ Y + (OL,S ⊕ Z) ⊗Z R = R v0∈S with (id ⊕1) ⊗ 1 7→ (1,..., 1). This shows that

× × h(OL,S ⊕ Z) = h(OL,S) · n is equal to Y h(Q ) = h(Q Q ) = h(coindG ) v0∈S Z v∈S v0|v Z Gw Z v∈S Y Y Y = h(Gw, Z) = |Gw| = [Lw : Kv]. v∈S v∈S v∈S Therefore Q [Lw : Kv] h(O× ) = v∈S L,S n and so × h(AL,S) h(CL) = = n. h(OL,S) The conclusion is that

0 1 |Hˆ (L/K, CL)| = [L : K]|H (L/K, CL)| ≥ [L : K]. Here, we have × × × CK /NCL = AK /K · NAL . Math 223b Notes 33

14 February 23, 2018

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

∼ × × × CK /NCL = AK /NAL · K .

14.1 Algebraic weak Chebotarev × × Corollary 14.1. Suppose L/K is abelian, and let D ⊆ NL/K AL ⊆ AK be a × × subgroup. If K D is dense in AK , then L = K. Proof. First we reduce to when L/K is cyclic. If L/K is abelian, there exists a L0 ⊆ L with L0/K cyclic. Then we can replace L by L0. × × × × × Observe that NL/K AL is open in AK . So K · NL/K A is open in AK , and × × × × hence closed. But K D is dense. This shows that K · NL/K AL = AK . So × × 1 = |AK /NL/K AL | ≥ [L : K]. So L = K. Corollary 14.2. Let L/K be a ﬁnite abelian extension with L 6= K. Then there are inﬁnitely many primes of K that do not split completely in L.

Proof. Suppose there are finitely many prime that are not split completely. Let S be this set of primes that don’t split completely, and also the infinite primes, and set × D = {x ∈ AK : xv = 1 for all v ∈ S}. × × × I need to check that D ⊆ NL/K AL . If v∈ / S then Lw = Kv so NLw = Kv . We × × × also need to check that K D is dense in AK . This is equivalent to K being Q × dense in v∈S Kv . 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 = Frob ∈ Gal(L/K) v v for v∈ / S.

Proof. Without loss of generality, assume that S contains all ramified and infinite primes. Let H ⊆ Gal(L/K) be generated by {Frobv : v∈ / S}. Let M ⊆ L be the subfield fixed by H. For any v∈ / S, we then have that Frobv fixes M. But Frobv is the generator of the decomposition group, and so v splits completely in M/K. This shows that H = Gal(L/K). Math 223b Notes 34

14.2 Reduction of the second inequality to Kummer extensions We now want to prove the second inequality. We want to prove that Proposition 14.4. If L/K and [L : K] = p, then

ˆ 0 |H (L/K, CL)| = |CK /NL/K CL| ≤ p.

First we reduce to the case where K contains µp. If [L : K] = p as above, let 0 0 0 L = L(ζp) and K = K(ζp). Because d = [K : K] divides p − 1, it is relatively prime to p and so [L0 : K0] = p and [L0 : L] = d. Now we have

NL0/L CL CL0 CL

NL/K NL0/K0 NL/K N 0 0 K /K CK CK CK

CK /NL/K CL CK0 /NL0/K0 CL0 CK /NL/K CL

1 1 1.

p But CK ⊆ NL/K CL and so CK /NL/K CL has exponent p. Also, the composite of two horizontal arrows is multiplication by d. So ×d : CK /NCL → CK /NCL is an isomorphism, and this shows that

|CK /NL/K CL| ≤ |CK0 /NL0/K0 CL0 |.

Therefore it is enough to show the second inequality for L0/K0. So we may assume that L/K is a cyclic extension with µp ⊆ K. By Kummer theory, we have √ L = K( p a) for some a ∈ K×. × × × × We need to bound [AK : K · AL ]. We will replace NAL with a smaller × × × subgroup F ⊆ NAL and then bound [AK : K F ]. Let S be a ﬁnite set of places of K containing all ramiﬁed places, and let v1, . . . , vk be places of K that split ∗ completely in L. Let S = S ∪ {v1, . . . , vk}. Then we take

Y Y Y F = (K )p · K× × O×. v vi v v∈S 1≤i≤k v∈ /S∗

Then F ⊆ N × because (K )p ⊆ NL× for v ∈ S, K = NL× , and O× ⊆ NL× AL v w vi wi v w because Lw/Kv is unramiﬁed. Math 223b Notes 35

15 February 26, 2018

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

× × |AK /NAL · K| ≤ p.

× × × We will choose F ⊆ NAL and show that |AK /K F | ≤ p. We will take

Y × p Y Y × F = (Kv ) × Kvi × OV . v∈S 1≤i≤k v∈S∗

Last time we checked that F ⊆ NL× provided that S contains all ramified primes in L/K. Proposition 15.1. Let K be a global field, and v a finite place of K. Assume × that K√v does not have residue characteristic p, with b ∈ K . Then v is ramified p × × p in K( b)/K if and only if b ∈ Ov · (Kv ) . Also, v is split if and only if × p b ∈ (Kv ) . 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 AK,S CK . Then

× × × AK,S∗ AK /K · F. The kernel is going to be

× × × × × AK,S∗ ∩ K F = F · (K ∩ AK,S∗ ) = F ·OK,S∗ . So we get × × × × 1 → F ·OK,S∗ → AK,S∗ → AK /K F → 1. In particular, × × × [AK : K F ] = |AK,S∗ /F ·OK,S∗ |. We also have

Let us ﬁrst look at the local component. We have

Y × × p |AK,S∗ /F | = Kv /(Kv ) .

v∈S But by the homework, we have

× × p 2 −1 |Kv /(Kv ) | = p · |p|v if µp ⊆ K. This shows that

× Y 2 −1 2|S| |AK,S∗ /F | = p |p|v = p , v∈S because S contained all v such that |p|v 6= 1. Now let us do the global computation. Observe that

× ∼ |S∗|−1 OK,S∗ = Z ⊕ torsion, where the torsion group contains µp. So we get

× × p |S∗| |S|+k |OK,S∗ /(OK,S∗ ) | = p = p . Combining everything we have, we get

× × × |AK,S∗ /F | |S|−k × × p |AK /K F | = × = p [F ∩ OK,S∗ :(OK,S∗ ) ]. |OK,S∗ /F ∩ OK,S∗ |

× The claim is that we can pick v1, . . . , vk such that k = |S| − 1 and F ∩ OK,S∗ = × p (OK,S∗ ) . Also, recall that v1, . . . , vk had to be split completely in L.

15.2 Choosing the completely split places √ p Proposition 15.3. Suppose v1, . . . , vk are places of K such that if K( b) is × p unramiﬁed outside v1, . . . , vk and split at all places of S, then b ∈ (K ) . Then × × p F ∩ OK,S∗ = (OK,S∗ ) . √ × p Proof. Suppose that b ∈ F ∩ O ∗ . By the exercise, we know that K( b)/K K,S √ is split at all v ∈ S, and also that K( p b) is split at all v ∈ S. Then b ∈ (K×)p, × × p and because b ∈ OK,S∗ , we have b ∈ (OK,S∗ ) . Math 223b Notes 37

Now we would like to choose v1, . . . , vk given only S. Consider the auxiliary extension q p × T = K( OK,S), which has exponent p. Then

∼ × × p ∼ |S| Gal(T/K) = Hom(OK,S/(OK,S) , µp) = (Z/pZ) . √ p Note that L = K( a) ⊆ T . Now we choose places w1, . . . , wk of L such that

T/L Frob(wi) = ∈ Gal(T/L) wi form a basis for the Fp-vector space Gal(T/L). Math 223b Notes 38

16 February 28, 2018

Last time we reduced the second inequality to, given S, ﬁnding additional places v1, . . . , vk with k = |S| − 1 such that

• v1, . . . , vk all split totally in L/K, and √ p • if K( 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 q p × T = K OK,S and tried to ﬁnd L inside T . Choose w1, . . . , wk be places of L such that ∼ k Frobw1 ,..., Frobwk form a basis for Gal(T/L) = (Z/pZ) . Then take vi to be places of K below wi. First of all, note that Frob = Frobei wi vi where ei is the inertia degree. But ei = 1 or ei = p, but it can’t be that ei = p because then Frob = Frobp = id. So we get e = 1 so that v is split. wi vi i i

So we have Frobvi = Frobwi inside Gal(T/K). So they generate the subgroup Gal(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). k+1 Y Lemma 16.1. O× /(O× )p → O× /(O× )p is bijective. K,S K,S vi vi i=1 Proof. For injectivity, we use Kummer theory. If [c] ∈ O× /(O× )p is in the √ K,S K,S kernel, then K( p c) is a Kummer extension. For 1 ≤ i ≤ k+1 and c ∈ (O× )p, so √ vi √ v splits completely in K( p c. This means that Frob is the identity on K( p c). i √ vi √ 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 × × p |S| k+1 |OK,S/(OK,S) | = p = p . By the homework, we have |O× /(O× )p| = p for all i. So the right hand side vi vi also has order pk+1.

k Y Corollary 16.2. O× /(O× )p O× /(O× )p is surjective. K,S K,S vi vi i=1 √ p Now let us check the second condition on vi. Suppose M = K( b) is a Kum- mer extension such that M/K is unramiﬁed outside v1, . . . , vk and completely split at S. We need to show that b ∈ (K×)p. Let Y Y Y D = K× × (K×)p × O×. v vi v v∈S vi v∈ /S∗ Math 223b Notes 39

× × Then D ⊆ NAM because we can check this locally. Note that D · K contains × × Q × Q × D ·OK,S. By the corollary, it includes AK,S = v∈S Kv × v∈ /S Ov . That is, × × D · K ⊇ AS . Recall that × × × AK,S AK /K × × by deﬁnition of S. Therefore D·K = AK . 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

0 |Hˆ (L/K, CL)| 1 = [L : K]. |H (L/K, CL)|

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

0 |Hˆ (L/K, CL)| ≤ [L : K] = p.

Then we immediately get

1 0 2 H (L/K, CL) = 1, |Hˆ (L/K, CL)| = |H (L/K, CL)| = p.

By the homework, this implies that

1 2 H (L/K, CL) = 1, |H (L/K, CL)| | [L : K] for all L/K. (This was done by ﬁrst showing for p-groups, and then showing it for all groups using Sylow subgroups.)

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

2 × 2 × ∼ M 2 × H (L/K, L ) → H (L/K, AL ) = H (Lw/Kv,Lw ) v is injective.

× × Proof. Use the sequence 1 → L → AL → 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 ∈ NLw for all primes w of L.

Proof. This is equivalent to the injectivity of

× × ∼ ˆ 0 × M ˆ 0 × ∼ M × × K /NL = H (L/K, L ) → H (Lw/Kv,Lw ) = Kv /NLw .

−1 ∼ 1 This follows from Hˆ (L/K, CL) = H (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

2 1 inv : H (L/K, CL) → [L:K] Z/Z. we will construct this as the direct sum of local invariant maps, but it will take some work to show that this gives a well-deﬁned map. Here we are going to do for “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. For Kummer theory, we already needed characteristic zero. We have two recprocities we want to prove: 2 × 2 × • inv-reciprocity: if a ∈ H (L/K, L ) → H (Lw/Kv,Lw ) then X invv α = 0. v

• θ-reciprocity: if L/K is abelian, for a ∈ K× Y θLw/Kv (a) = 1. v Note that last time we showed for L/K ﬁnite Galois, we have

2 × M 2 × H (L/K, L ) ,→ H (Lw/Kv,Lw ). v If we put all L together, we get 2 × M 2 × M ∼ M Br(K) = H (K, K ) ,→ H (Kv, Kv ) = Br(Kv) = Q/Z. v v v

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

res Br(Kv) Br(Lw)

∼= invv ∼= invw

×[Lw:Kv ] Q/Z Q/Z.

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) we need [Lw : Kv] to be a multiple of the order of invv(α) ∈ Q/Z. It follows from the 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 an integer. There exists a totally complex cyclic cyclotomic extension L/K such that m | [Lw : Kv] for all v ∈ S.

Proof. We reduce to the case K = Q by replacing m with m · [K : Q]. Then r−1 for each prime `, take Q(ζ`r )/Q which contains a cyclic subﬁeld of order ` or `r−2, and combine them for diﬀerent `.

17.2 Invariant and reciprocity maps Deﬁnition 17.3. We deﬁne

2 × 1 X invL/K : H (L/K, AL ) → [L:K] Z/Z; c 7→ invLw/Kv c. v

2 × 2 × inv-reciprocity then states that inv vanishes on H (L/K, L ) ,→ H (L/K, AL ). 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,

X X inv (res c) = inv (res c) = [L : K ] inv c N/L L/K Nw0 /Lw Lw/Kv w v Nw0 /Kv w w X X X = [L : K ] inv (c) = [L : K] inv c w v Nw0 /Kv Nw0 /Kv v w|v v

= [L : K] invN/K c.

S × So L/K AL 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 not 1 surjective, because the image of invL/K is Z/Z. Take K = Q and √ √ lcm([Lw:Kv ]) L = Q( 13, 17) for instance. Deﬁnition 17.4. For L/K abelian, we deﬁne

× Y θL/K : AK → Gal(L/K); a 7→ θLw/Kv (a). v

Proposition 17.5. For any character χ : Gal(L/K) → Q/Z, considered as an element 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 product with 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-reciprocity for L/Q cyclic cyclotomic, deduce inv-reciprocity for L/K Galois, and then θ-reciprocity for L/K abelian. Math 223b Notes 44

18 March 5, 2018

We had two reciprocities we wanted to check: × Q • θ-reciprocity: for all a ∈ L , v αv(a) = 1. 2 × P • inv-reciprocity: for all L/K and a ∈ H (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. This means that L is contained in Q(ζn), but it is enough to check for L = Q(ζn) by the compatibility conditions, and further more, check for L = Q(ζ`r ). We want to check that

× Y θ : Q → Gal(Q(ζ`r )/Q); θ(a) = θv(a) v is zero. We can check this for a primes and −1. For a = p 6= `, we have 0 p θp0 (p) = 1 for p 6= p, `, θp(p) = Frobp(ζ 7→ ζ ) because Qv(ζ`r )/Q is unramiﬁed. On the other hand, because Q`(ζ`r ) is a Lubin–Tate extension, we have (p−1) θ`(p): ζ 7→ ζ .

Also, θ∞(p) = 1, so we get cancellation. 0 For a = `, we have θp0 (`) = 1 for p 6= ` and θ`(`) = 1 and θ∞(`) = 1. For 0 −1 a = −1, we have θp0 (−1) = 1 for p 6= ` and θ`(−1) : ζ 7→ ζ and θ∞(−1) is complex 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 [ 2 × Br(Q) = H (L/Q,L ) L/Q for L/Q cyclic cyclotomic. So we get inv-reciprocity for L/Q arbitrary. For arbitrary L/K, we can enlarge L to L0 with L0/Q Galois. Then we get inf cor H2(L/K),L×) H2(L0/K, (L0)×) H2(L0/Q, (L0)×)

inv inv inv=0

Q/Z Q/Z Q/Z. Because the horizontal maps on the top row are injective, we get that the invariant map H2(L/K, L×) → Q/Z is zero. This shows inv-reciprocity, and thus θ-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 = AK /NAL · K → Gal(L/K). Also, by inv-reciprocity, we have

2 × 2 × inv 1 H (L/K, L ) ,→ H (L/K, AL ) −−→ [L:K] Z/Z.

This is not always a short exact sequence, because we don’t always have surjectivity on the right. Proposition 18.1. If L/K is cyclic, then

2 × i∗ 2 × inv 1 1 → H (L/K, L ) −→ H (L/K, AL ) −−→ [L:K] Z/Z → 0 is exact. Proof. We ﬁrst show that the invariant map is surjective. We can describe 2 × ∼ L 2 × explicitly the image, because H (L/K, AL ) = v H (Lw/Kv,Lw ). 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 an exact sequence

2 × i∗ 2 × j∗ 2 3 × 1 → H (L/K, L ) −→ H (L/K, AL ) −→ H (L/K, CL) → H (L/K, L ) = 1. But we know that by the second inequality,

2 |H (L/K, CL)| ≤ [L : K].

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

2 ∼= 1 inv : H (L/K, CL) −→ [L:K] Z/Z.

This works to define the correct invariant map for L/K cyclic. We are going to do 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 L0/K be cyclic with [L0 : K] = 2 0 0 [L : K]. Then H (L/K, CL) and H (L /K, CL0 ) have the same image inside 2 S H (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. Last time we deﬁned the invariant map

2 1 inv : H (L/K, CL) → [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 L0/K is cyclic, and [L0 : K] = [L : K] then H2(L/K) and H2(L0/K) have the same image in H2(K/K).

Proof. First we want to show that inf(H2(L0/K)) ⊆ inf(H2(L/K)). Take K = L · L0 the compositum. Then L0/K is cyclic, the extension N/L is also cyclic with degree [N : L] | [L0 : K]. If c ∈ H2(L0/K), I need to show that

infN/L0 (c) ∈ im infN/L = ker resL/K .

2 0 × (Here, we are using the inﬂation-restriction sequence.) Choosec ˜ ∈ H (L /K, AL0 ) such that j∗(˜c) = c. Then we need to check that

1 invN/L resL/K infN/L0 j∗(˜c) = 0 ∈ [N:L] Z/Z.

But this thing is

∗ invN/L j resL/K infN/L0 (˜c) = invN/L resL/K infN/L(˜c) = [L : K] invL0/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(L0/K)) is contained in inf(H2(L/K)). But note that the size of H2(L0/K) is equal to [L0 : K] and the second inequality shows that the size of H2(L/K) is at most [L : K]. This shows the other containment as well. Now we can deﬁne 2 inv : H (K/K) → Q/Z by considering

H2(K/K) ∼ lim H2(L/K) ∼ lim H2(L0/K). = −→ = −→ L/K L0/K cyclic

For any L/K, this restricts to an isomorphism 1 H2(L/K) −−→inv / . [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 1 inv (u ) = . L/K L/K [L : K]

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

ˆ i ˆ i+2 ` uL/K : H (L/K, Z) → H (L/K, CL) is an isomorphism for all i. The i that we really care about is i = −2, and this gives us

ab ˆ −2 ˆ 0 ` uL/K : Gal(L/K) = H (L/K, Z) → H (L/K, CL) = CK /NCL. The inverse map is what we denote by

ab θ : CK /NCL → Gal(L/K) .

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 and a ∈ CK , we have an equality

χ(θL/K (a)) = invL/K (a ` δχ).

0 × × × This characterizes θL/K , so if I consider θL/K : CK /NCL = AK /K NAL → Gal(L/K) by the product of the local reciprocity maps, we previously checked that 0 χ(θ (a)) = inv(a ` δχ) 2 × 1 using the invariant map H (L/K, AL ) → [L:K] Z/Z. ab Note that θL/K : CK /NCL → Gal(L/K) is continuous using the adelic topology on CK . This is because NL/K CL are closed in CK . To see this, just note that ∼ 0 >0 ∼ 0 >0 CL = CL × R ,CK = CK × R 0 0 with the image of NL/K CL being closed in CK by compactness. Taking them all together gives a continuous map

θ : C → lim Gal(L/K)ab = Gal(Kab/K). /K K ←− L

ab This has dense image, because each CK → Gal(L/K) is surjective. The kernel is the intersection \ ker θ/K = NCL. L/K

So we need to ask again which subgroups A ⊆ CK are normic, i.e., equal to 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, this is going to be the same argument from last semester, which is doing a lot of Kummer theory. Note that if A is normic, so is A0 ⊇ A, because we have our finite reciprocity map θ. If A and B are normic, so is A ∩ B. What we are going to do is to pick a 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

ab θ/K : CK → Gal(K /K) that is continuous with dense image. Moreover, we can only take the image 0 of CK the content zero stuﬀ, because we can adjust the inﬁnite part to make anything have content zero. So the image is compact and hence closed, and this shows that θ/K is surjective. We also showed that the kernel is the intersection of all normic subgroups NL/K CL.

20.1 Classiﬁcation of normic subgroups

Proposition 20.1. A ⊆ CL is normic if and only if A ⊆ CL is finite index open. I justified last time that normic subgroups are finite index and open. If n A ⊆ CL is finite index, it is going to contain some CL. Because A is open, A also contains some subgroup of the form

Y Y × US = 1 × Ov ⊆ CK . v∈S v∈ /S

(This is not open, but at least A contains some US.) Combining these two facts, n we see that it is enough to show that CLUS is normic for large enough S. Lemma 20.2. Suppose S contains all inﬁnite primes and all primes dividing × n, and suppose AK,S CK . If K ⊇ µn then

n (CK ) US = NT/K T

pn n where T = K( OK,S). In general, (CK ) US is still normic. Proof. This is going to be a lot like what we did for the second inequality. We have ∼ × × n Gal(T/K) = Hom(OK,S/(OK,S) , µn). × × n ∼ S Here, we can non-canonically compute OK,S(OK,S) = (Z/nZ) . So |Gal(T/K)| = n|S|. We have a map θT/K : CK → Gal(T/K), n and (CK ) ⊆ ker θT/K . We also check that US ⊆ NT/K CT locally. Clearly × pn 1 ∈ Kv a norm for v ∈ S, and for v∈ / S, T = K( OK,S) implies that T/K is × × unramiﬁed, and hence Ov ⊆ NTw . This shows

n (CK ) US ⊆ NT/K CT . Math 223b Notes 50

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

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

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

× × n Y × × n n OK,S/(OK,S) → (Kv )/(Kv ) → CK /(CK ) · US → 1. v∈S

× (The second map is surjective because AK,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 suffices × × n × × n to check that (OK,S)/(OK,S) injects into Kv /(Kv ) . If a ∈ O× and a ∈ (K×)n for all v ∈ S, consider the extension L = √ K,S v K( n a)/K. This is unramified outside S and split completely . So todo Let us now look at the general case. If K does not contain µ, we let K0 = 0 0 0 0 K(µn) and S be the set of primes of K above S. Then there exists a L /K n 0 0 such that (CK0 ) US0 = NL0/K0 · CL0 by taking L ⊇ L with L /K Galois. WE then have

0 0 n n NL/K CL ⊆ NL0/K CL = NN 0/K CL = NK0/K (NL0/K0 CL0 ) = NK0/K (CK0 US0 ) ⊆ CK US.

n This shows that CK US is normic. Now we have shown that \ θ/K = U.

U⊆CK ﬁnite index

We claim that this is equal to Y >0 Y × DK = connected comp. of 1 ∈ CK = closure of R C . v real v complex T 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

\ n DK = divisible elements of CK = (CK ) . n Math 223b Notes 51

21 March 19, 2018

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

∼= ab θL/K : CK /NCL −→ Gal(L/K) . The map is natural in L and K with inﬂation and restriction:

θ L/K ab CK /NL/K CL Gal(L/K)

transfer θ L/M ab CM /NL/M CL Gal(L/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 is the maximal abelian extension of K that is unramified at all places (including infinite 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 and only 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 and only if p splits completely in OH .

Proof. If (p) does not split completely in OK , then it is not a norm in OK and also 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 if and only if (p) splits completely in OH . √ 2 2 Example 21.3. Let K =√ Q( −5). We have√ that p = x + 5y 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 fractional ideal of OK , then aOK is principal.

This property does not uniquely determine OH . This also does not mean that OH have class number 1. The theorem follows from the following group theory fact. Math 223b Notes 52

Theorem 21.5. Let G be a group, G0 = [G, G], and G00 = [G0,G0]. Then the transfer map G/G0 → G0/G00 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 abelian subextension of H, because H1/K is unramiﬁed. Let G = Gal(H1/K) so that ab 0 Gal(H/K) = G and Gal(H1/H) = G . We have

θ H1/K ab ab CK /NH1/K CH1 Gal(H1/K) = G

transfer θ H1/H ab 0 ab CH /NH1/H CH1 Gal(H1/H) = (G ) .

ab ∼ ∼ But note that Gal(H1/K) = Gal(H/K) = Cl(OK ). So we get that the map Cl(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 such that 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 some Hn which has Cl(Hn) = 1, in which case the tower stabilizes. Actually this is a natural thing to do. Lemma 22.2. If L/K is such that Cl(L) = 1, then L contains H the Hilbert class 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 = L and 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 fixed field of the p-group of H0, and so on K ⊆ H0,p ⊆ H1,p ⊆ · · · . This is called the p-class field tower of K. Theorem 22.3 (Golod–Shafarevich). If the p-class field tower stabilizes, then p it 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 the 2-rank of Cl(K) to be at least 6. This holds for any imaginary quadratic fields with ≥ 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

X an . ns n≥1

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 functional equation. We can write Y 1 ζ(s) = , 1 − p−s p and more generally if K is a number ﬁeld, an Euler product is a product of the form Y 1 −s −s (1 − a1,p(Np) ) ··· (1 − akp,p(Np) ) p⊆OK where Np = |OK /p|. L-functions are Dirichlet series with Euler products. For instance, the Dedekind ζ-function of a number ﬁeld K is

X 1 Y 1 ζ (s) = = . K |Na|−s 1 − (Np)−s a⊆OK p⊂−OK

We can actually do this for function ﬁelds. Let us take K = Fq(t) for instance. Let us take OK,∞ = Fq[t]. Then we are looking at X 1 . (Na)−s a⊆Fq [t]

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

X qd 1 ζ (s) = = K q−ds 1 − q(1−s) d≥0

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

Y 1 1 1 −s = −s ζK (s) = −s −s . 1 − |kv| 1 − q (1 − q )(1 − qq ) v of Fq (t)

This is the zeta function of 1 . It is a rational function in t = q−s. For any PFq function ﬁeld K(C), the zeta function of C is going to be of the form

Q −s i(1 − αiq ) (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

X 1 Y 1 = . (Na)s 1 − (Np)s a⊆OK p⊆OK

23.1 L-functions

Then we can deﬁne the Dirichlet L-function over OK as the following. First choose 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-function as X χ(a) Y 1 L(s, χ) = = . (Na)s 1 − χ(p)(Np)s (a,m)=1 p⊆OK ,p-m ∼ × For instance, if K = Q and m = m∞ then Clm(Z) = (Z/mZ) and χ : (Z/mZ)× → S0. Then X χ(a) L(s, χ) = . as (a,m)=1

0 1 We can consider χ : Clm(OK ) → S as a character of CK → S , because 1 there is a map CK S . Hecke L-functions generalize Dirichlet L-functions 1 by allowing arbitrary characters χ : CK → S with non-discrete image. You can prove using this that primes in Z[i] have equidistributed arguments, for instance. 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 trace χV and we can deﬁne the Artin L-series

Y 1 . 1 − χ(Frob )(Np)−s p p

23.2 Convergence of Dirichlet series b P −s If an is O(n ), then the series ann converges locally uniformly and abso- lutely in the half-plane <(s) > b + 1.

P b P s Proposition 23.1. Let S(x) = n≤x an. If S(x) = O(x ) then ann can be extended 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 by parts and write X an X 1 1 = s − . ns n ns (n + 1)s n≥0 n≥0

b −s−1 Here sn = O(n ) and the diﬀerence is O(n ). Proposition 23.2. ζ(s) has a meromorphic continuation to <(s) > 0, with a pole at s = 1 an no other poles.

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

X (−1)n+1 ζ (s) = 2 ns n≥1 on <(s) > 1. By our proposition, this can be extended to <(s) > 0 holomorphi- 2πi cally, so ζ(s) is meromorphic on <(s) > 0 with poles possibly as s = 1 + k log 2 . If we deﬁne ζ3(s) similarly, we see that ζ(3) can have a pole only at s = 1. It actually 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

X 1 Z ∞ 1 1 X 1 ≥ dx = ≥ = ζ(s) − 1, ns xs s − 1 ns n≥1 x=1 n≥2 so the residue should be 1.

For the Dedekind zeta function, we will show that ζK (s) also have a simple pole at s = 1 and will give a formula for the residue involving the class number. Math 223b Notes 57

24 March 26, 2018

P b We showed the following last time. If S(x) = n≤x an has S(x) = O(x ), then P −s ann converges to a holomorphic function on <(s) > b. Using this, we showed that X 1 ζ(s) = ns has a meromorphic continuation to <(s) > 0 with a simple pole at s = 1 with residue 1.

24.1 Meromorphic continuation of L-functions b Proposition 24.1. If there exists some a0 such that S(n) = a0n + O(x ) with P −s 0 ≤ b < 1, then ann can be analytically continued to a meromorphic function on <(s) > b with a simple pole at s = 1 having residue a0.

P −s P −s Proof. We have ann = a0ζ(s) + (an − a0)n .

1 We will apply this to ζK , and more generally, L(s, χ) for χ : Clm(OK ) → S given by X χ(a) L(s, χ) = . Nas a⊆OK ,(a,m)=1

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

X 1 ζ(s, K) = . (Na)s a∈K

Then we can write X L(s, χ) = χ(K)ζ(s, K).

K∈Clm(OK ) The goal is to apply the proposition to ζ(s, K). Consider the partial sum function S(x, K) = #{a ∈ K : Na ≤ x}.

(1− 1 ) Proposition 24.3. We have S(x, K) = gmx + O(x d ). Thus, ζ(s, K) has a pole or residue gm at s = 1. Here, 2r(2π)s reg(m) gm = p , 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 OK,m = {a ∈ OK : a ≡ 1 mod m, a positive at reals of m}, Nm is the norm of m as an ideal (ignoring places at inﬁnity), disc(K) is the discriminant. Math 223b Notes 58

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

× Y + L : OK → H,→ R ; a 7→ (log|a|v1 ,..., log|a|vr+s ). v|∞

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

If m = 1, we also say reg(m) = reg(OK ). If K is real quadratic, we have reg(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. First choose c ∈ K. Then any a ∈ K is going to look like a = ac where a ≡ 1 mod m and a ∈ c−1. Then Na = Nc−1Na ≤ (Na)x for some x > 0. This means that a belongs to an additive coset of mc−1, which is going to be some lat- Q tice in v|∞ Kv. Then the number of these points in some ball is going to be approximately

−1 vol(ball) 1− 1 x · Nc · (··· ) 1− 1 + O(x d ) = + O(x d ). covolume of lattice pdisc(K)NmNc−1

If you work out the coeﬃcients out, you are going to get that. If you have a real quadratic ﬁeld, for instance, you actually have a hyperbola, and you have to 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.

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

2r(2π)s reg(m) gm = p wmNm |disc K|

rm and Nm = Nmﬁn2 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 then Q some lattice in v|∞ Kv interesected with a fundamental domain for the action × of OK,m. Theorem 25.2. Let χ be a Dirichlet character of modulus m. If χ 6= 1, then L(s, χ) is holomorphic at s = 1. If χ = 1, then L(s, χ) has a pole of order 1 at s = 1, with reseidue |Clm(OK )|gm. Proof. This immediately follows from

X L(s, χ) = χ(K)ζ(s, K).

K∈Clm

Here, each ζ(s, K) has a pole of order 1 at s = 1, with residue gm. So if χ is not the trivial character, we get cancellation, and if χ is trivial, we get the correct thing.

Corollary 25.3. For m = 1, we get that ζK,s(s) = L(s, 1) has pole at s = 1 with residue given by r s 2 (2π) reg(K)hK hK g1 = p . wK |disc(K)| For instance, if K is imaginary quadratic, then the residue is 2πh √ K . wK −∆ If K is real quadratic, then the residue is

2 log|u|h √ K . 2 ∆ There’s something called the Stark’s conjecture that generalize this to other L-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 primes of OK . • The Dirichlet density is a number δ satisfying X 1 1 = δ log + O(1) (Np)s 1 − s p∈T as s → 1+. • The natural density is #{p ∈ T : Np ≤ x} δ = lim . x→∞ {p : Np ≤ x}

• If the Euler product n Y 1 −1 1 − Nps p∈T has an analytic continuation with pole of order m at s = 1, we say that m its polar density is 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 and they are equal. If polar density exists, so does Dirichlet density and they are equal. There are sets like primes whose first digit of p is 1, whose Dirichlet density exists but but natural density does not exist. Proof. Proof of natural implies Dirichlet is basically partial summation. For the proof of polar implies Dirichlet, we can use −1 Y 1 X log = (Np)s) + O(1) 1 − Nps p∈T p∈T 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 three definitions. Proof. This is obvious for natural density. For polar density, you can use that ζK (s) has pole of order 1 at s = 1. You can do this for Dirichlet density as well. Math 223b Notes 61

26 March 30, 2018

Recall that we deﬁned Dirichlet density of T as X 1 1 = δ log + O(1), Nps 1 − s p∈T

m and we deﬁned polar density as n if n Y 1 −1 1 − Nps p∈T has a pole of order m at s = 1.

26.1 Proof of second inequality i Proposition 26.1. The set T = {p ⊆ OK : Np = p for i ≥ 2} has polar density 0.

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

d Y 1 −1 Y Y js 1 − = (1 − p−js)#{p:Np=p }. Nps p∈T j=2 p

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

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

1 which still is holomorphic on 2 . 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 f some pL lies above some pK , then NpL = pK . So T is the set of primes p such that Np is prime, up to a finite number of ramified primes. This implies that the polar density δ(T ) = 1, because the set it misses can be ignored. Now we observe that Y 1 Y 1 [L:K] 1 − s = 1 − s . NpL NpK pL∈T pk∈Spl(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 in Clm = Clm(OK ) fix a subgroup H of Clm. Consider L(1, χ) for χ a nontrivial Dirichlet 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 X 1 X X log(L(s, χ)) = − log(1 − χ(p)Np−s) |G| |G| χ χ p 1 X X = (χ(p)Np−s) + O(1) = Np−s + O(1). |G| χ,p [p]∈H~~

~~But the left hand side is just~~

~~1 1 X 1 log − ord L(s, χ) log + O(1) |G| 1 − s s=1 1 − s χ shows that the Dirichlet density of {p :[p] ∈ H} is just~~

~~1 X 1 − ord (L(s, χ)) . |G| s=1 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~~

~~× × |CK /(NL/K CL )Um| ≤ [L : K].~~

~~This implies the second inequality because we can take m so that Um ⊆ × × × NL/K CL . Then |CK /NL/K CL | ≤ [L : K].~~

Proof. Consider G = Clm(OK )/H where H is the subgroup of Clm(OK ) generated by ideals that are norms from OL. We know that δ(p :[p] ∈ H) is either 0 or 1/|G|. But if p splits completely in L, then p is a norm form OL so p ∈ H. This shows that 1 δ(p :[p] ∈ H) ≥ δ(Spl(L/K)) = . [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 only shows that L(1, χ) 6= 0 only for χ that vanish on H. So we use an existence × theorem. There exists a L/K Galois such that NCL ⊆ Um, by class ﬁeld theory. 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 extension of 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 decomposition group, and then the prime splits completely. The strategy is to ﬁrst do L/K abelian 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~~

X 1 1 X s = χ(K) log L(s, χ) + O(1) Np |Clm| p∈K χ as s → 1+. But we know that all log L(1, χ) is a number for χ nontrivial, and 1 L(s, χ) behaves like log 1−s as s → 1. Note that for K = Q, this is just Dirichlet’s theorem for primes in arithmetic progressions. 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 subgroup H. 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 = Lhgi. Then we have a tower L/M/K where L/M is cyclic of degree m. What does it mean for p ∈ Tg. This means that there exists a pL above p such that FrobpL = g. Then hgi = 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. So we have 1 δ(T ) = δ(p of M : Frob = g ∈ Gal(L/M)) = . M,g pM m We can also understand the ﬁber of this surjection as well. Given p, how may primes pM ∈ TM,g lie over p? In other words, how may pL are there above p such that FrobpL = g ∈ Gal(L/K)? Any such pL above p are going to look like pL = hpL,0, for h ∈ Gal(L/K). Then we have

~~−1 −1 FrobpL = h FrobpL,0 h = hgh .~~

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

~~X 1 m|[g]| X 1 |[g]| δ(Tg) = = = log(1 − s) + O(1). Np−s n Np−s |Gal(L/K)| p∈Tg pM ∈TM,g M~~

This gives the correct density. Here is one application. Let L/K be an arbitrary ﬁnite extension. Then we deﬁned Spl(L/K) as the prime of K that split completely in L. You can show that if L0 is the Galois closure of L, then Spl(L/K) = Spl(L0/K). So 1 δ(Spl(L/K)) = δ(Spl(L0/K)) = . [L0 : K]

~~Proposition 27.4. If L and L0 Galois have the same splitting set, then L = L0.~~

~~Proof. Consider the compositum LL0/K. This has splitting set density equal to 1 over [L : K] and of [LL0 : K]. Math 223b Notes 65~~

~~28 April 4, 2018~~

Now we would like to do explicit class field theory for Q and imaginary quadratic ab S fields. By Kronecker–Weber, we know that Q = n Q(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 the maximal 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 instance, we take E/K an elliptic curve for every n, consider the n-torsion points E[n](Q) =∼ (Z/nZ)2, and take K(E[n])/K. The problem is that this is not necessarily 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 √ ∼ n 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 multiplication by O. In this case, for any a ∈ O we can look at the kernel E[a] of a : E → E and 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 for K an imaginary quadratic field? Unless h(K) = 1, it turns out that there are no CM elliptic curves defined over K. So we are going to enlarge K to H so that there are CM elliptic curves defined over H with 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 look at the j-invariants j(E) ∈ H. This is going to have minimal polynomial Y (X − j(E)) E CM by K 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 should correspond to a curve over C. We want to make this isomorphism C/L → E explicit.~~

~~Deﬁnition 28.1. An elliptic function is a meromorphic function fn on C/L. There is the Weierstrass ℘-function deﬁned by~~

~~1 X 1 1 ℘(z, L) = + − . z2 (z − w)2 w2 w∈L\{0}~~

~~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 meromorphic function f : C → C that is L-periodic. We know that C/L is an elliptic curve. We are interested in what elliptic function this is. Last time we deﬁned

~~1 X 1 1 ℘(z, L) = + − . z2 (z − w)2 w2 w∈L\{0}~~

~~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 we observe 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 all z. Now note that we can just differentiate and get

X 1 ℘0(z) = −2 . (z − w)3 w∈L This is again an elliptic function, and it now has triple poles at points of L. But because I am on a curve, there should be an algebraic relation between these two functions. It is useful to have a power series for ℘. 1 X Lemma 29.1. ℘(z) = + (2n+1)G z2n where G (L) = P 1 . z2 2n+2 2n+2 w∈L\{0} w2n+2 n≥1 Proof. I just expand

~~1 1 X − = (m + 1)w−m−2zm. (z − w)2 w2 m≥1~~

Here, note that if m is odd, then Gm+2 = 0. We can now use this to ﬁnd a polynomial P in ℘ and ℘0 such that P (℘, ℘0) = P n 0 z≥1 cnz . This would imply that P (℘, ℘ ) = 0. If you work this out, you get

~~0 2 3 (℘ (z)) = 4℘(z) − g2℘(z) − g3, g2 = 60G4, g3 = 120G6.~~

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

~~2 3 0 C/L → E : 4y = x − 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 to show 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. We know that the function has double poles at z = 0, and zeros at ±w (unless w is 2-torsion). So there can’t be any other zeros. If w is 2-torsion, it is going to have a double zero because it is going to be even. Using this, we see that (℘(z), ℘0(z)) = (℘(w), ℘0(w)) can only happen when w = −z or w = z. In the ﬁrst case, we see that ℘0(z) = ℘0(w). But ℘0(z) has 1 zeros as 3 nonzero point of 2 L/L. So we have an isomorphism

C/L =∼ E of Riemann surfaces. If you know enough theory, it follows that this is also an isomorphism of complex Lie groups. But we can do this explicitly. 1 ℘0(z) − ℘0(w)2 Theorem 29.3 (Addition theorem). ℘(z+w) = −℘(z)−℘(w)+ . 4 ℘(z) − ℘(w) 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~~

~~3 2 ∆(L) = g2 − 27g3~~

3 3 which is the discriminant of the polynomial 4x − g2x − g3. Then deﬁne the j-invariant as 1728g3 j(L) = 2 . ∆ Theorem 29.4. For two lattices L, L0 ⊆ C, j(L) = j(L0) if and only if L0 is homothetic to L.

0 0 −4 0 −6 Proof. If L = λL then g2(L ) = λ g2(L) and g3(L ) = λ g3(L). If j(L) = 0 −2 0 −3 0 j(L ), then there exist λ such that g2(L) = λ g2(L ) and g3(L) = λ g3(L ). −2n 0 Because G2n are polynomials in g2 and g3, we get G2n(L) = λ G2n(L ). So

~~℘(z, L) = ℘(λ−1z, L0).~~

Comparing poles, we recover L and L0. 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 K such 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 a be a proper fractional O-ideal. Then j(a) = j(C/a) is an algebraic number of degree 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 Galois orbits, 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, in general we can describe K(j[a]) as follows. We deﬁne~~

~~∼ Y × Gal(LO/K) = CK / Uv v ﬁn 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. Then End(E) = O )Z, and we say that L has complex multiplication by O. In that case, L ∼ a for some invertible fraction ideal O. We saw that j(L) is an algebraic number of degree at most h(L). Theorem 31.1 (main theorem of complex multiplication). The j-invariant j(L) is an algebraic integer of degree h(L) and K(j(L)) = LO (which is the ring class ﬁeld of O). Deﬁnition 31.2. A cyclic sublattice of index m is a sublattice L0 ⊆ L such that L/L0 =∼ Z/mZ. Proposition 31.3. If L has complex multiplication, then there exists a cyclic sublattice L0 ⊆ L such that L0 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 in 0 OK 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 a 1 homothety on the lattice by taking L to cτ+d L. So we get 4 6 g2(γτ) = (cτ + d) g2(τ), g3(γτ) = (cτ + d) g3(τ).~~

This is saying that g2 is a modular form of weight 4 and g3 is a modular form of 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 1 1 {τ ∈ H : − 2 < <(τ) < 2 , |τ| > 1} and then you can see that this is a sphere minus one point, but with one order 2 orbifold point and one 3 orbifold point. This is not compact, so we compactify it to get a Riemann surface X(1) = Y (1) ∪ {∞}. Around inﬁnity, the local coordinates are going to be given by q = e2πiz. Math 223b Notes 72

Deﬁnition 31.4. A modular function for SL2(Z (a meromorphic modular form of weight 0) is a meromorphic function f on H such that f(γτ) = f(τ) for P n all τ ∈ H and γ ∈ SL2(Z), and f can be written as n≥−k cnq converging for =τ 0. For example j(τ) is a modular function. Next time we are going to try and get 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 meromorphic function on X(1) = SL2(Z) \ H ∪ {∞}. This is a compact Riemann surface of genus 0. An alternative description of a modular function is a meromorphic function f on H such that f(γτ) = f(τ) P n for all γ ∈ SL2(Z) and f = n≥−k cnq . We now check that j(τ) actually has a q-expansion. We have

k (2πi)2 X X G (τ) = 2ζ(2k) + 2 σ (n)qn, σ (n) = d2k−1. 2k (2k − 1)! 2k−1 2k−1 n≥1 d|n So 4 X g (τ) = 60G (π) = π4 1 + 240 σ (n)qn , 2 4 3 3 n≥1 8 X g (τ) = 120G (π) = π6 1 + 504 σ (n)qn , 3 6 27 5 n≥1 X ∆(τ) = (2π)12 τ(n)qn, n≥1 1 X 1 j(τ) = + c(n)qn = + 744 + 196884q + ··· . q q n≥0 The important thing here is that the coeﬃcients c(n) are all integers. This shows that j(τ) is indeed a modular function. It has a simple pole at ∞ and no other poles. It has the only 0 at τ = ω where ω3 = 1. It follows that the ﬁeld of modular functions for SL2(Z) is C(j(z)) and the subring of modular functions that 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. We 0 want to look at a finer group, so that Γ0(m) \ H parametrizes pairs (L, L ) up to homothety, where L ⊆ C is a lattice and L0 ⊆ L is cyclic of index m. Definition 32.1. For m ≥ 1, we define a b Γ (m) = ∈ SL ( ): c ≡ 0 (mod m) . 0 c d 2 Z

So if [τ] ∈ Γ0(m) \ H, we look at the corresponding [1, τ] and [1, mτ]. Up to a b homothety, the lattice acted by ( c 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 by adding 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 a Laurent series in q1/m.~~

~~Examples include j(τ) or even j(mτ). Note that the projection map X0(m) → X(1) has degree~~

~~Y 1 [SL ( ):Γ (m)] = m 1 + = d , 2 Z 0 p m p|m~~

~~2 where dm counts the number of cyclic index m sublattices of Z . Then we are going 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 of SL2.~~

~~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. For every 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) and j(mτ). We are going to leverage this to prove stuﬀ about the j-invariant. In particular, 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 rational functions on X0(m) is C(j(τ), j(mτ)) and the rational functions on X(1) is C(j(τ)). So we have a ﬁeld extension, and there is going to be an algebraic relation 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~~

~~Y Y 0 fm(X, τ) = (X − j(mγτ)) = (X − j(L )). 0 γ∈Γ0(m)\SL2(Z) L ⊆[1,τ] cyclic~~

Note that coeﬃcients of fm(X, τ) are holomorphic functions of τ ∈ H. It is also SL2(Z)-invariant. So once we check that they have q-expansions, we will get that 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 L0 ⊆ L, take m d ∈ Z>0 be minimal with d ∈ L. Then d | m and so we can take a = d . Then there is some b with 0 ≤ b < d and aτ + b ∈ L. In this case, L0 = [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 a b σ ∈ C = : a, b, d as before . m 0 d

One can equivalently classify cosets Γ0(m) \ SL2(Z) as −1 SL2(Z) ∩ σ SL2(Z)σ m 0 −1 for σ ∈ Cm. For instance, if we consider σ0 = ( 0 1 ) then σ0 SL2(Z)σ0 = ( a b/m ) and so the intersection is Γ (m). Then cm d 0 Y fm(X, τ) = (X − j(στ)),

~~σ∈Cm with the coeﬃcients of Xi being symmetric functions in j(στ). We know that each j(στ) is aτ + b j(στ) = j , d which is a Laurent series in~~

2πi(aτ+b)/d a2 ab e = Q ζm Math 223b Notes 76 where Q = q1/m and Q = 2πiτ. Then it is a Laurent series in Q = q1/m and invariant in τ 7→ τ + 1. So it is a power series in q = e2πiτ . This now shows that 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 coefficients of 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 −a2 −ab X a2k abk Q ζm + ckQ ζm : σ ∈ Cm k≥0 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 property 0 0 that Φm(j(L ), j(L)) = 0 if and only if L is a cyclic index m sublattice of L. So 0 0 Φm(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. p p (c) (Kronecker congruence) Φp(X,Y ) ≡ (X − Y )(X − Y ) mod p. Proof. For (a), we show ﬁrst that the zero locus is symmetric. So we show that L0 is similar to a cyclic index m sublattice of L if and only if L is similar to a cyclic index m sublattice of L0. But if L0 ⊆ L is cyclic of index m, then mL ⊆ L0 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

~~−m X −k Φm(j(τ), j(τ)) = cmq ckq . k>−m~~

~~We want to show that cm is ±1. But we see that Y Φm(j(τ), j(τ)) = (j(τ) − j(στ))~~

~~σ∈Cm~~

2 where j(τ) − j(στ) = (Q−m + holomorphic) + (ζ−abQ−a + holomorphic). But if m is not a square, whatever the leading term is, it is going to have coeﬃcient a root of unity. So cm should be a product of roots of unity, which is also a integer. For (c), we need the explicit description of Cp. It is going to be

~~1 i p 0 Cp = {σi = ( 0 p ) : 0 ≤ i < p} ∪ {σ∞ = ( 0 1 )}. Then we have~~

~~−p X pk p j(σ∞τ) = j(pτ) = q + ckq ≡ (j(τ)) (mod p). k≥0~~

~~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~~

~~τ + i X X j(σ τ) = j = ζ−iQ−1 + c ζikQk ≡ Q−1 + c Qk (mod 1 − ζ ). i p p k k p k≥0 k≥0 Math 223b Notes 78~~

~~Then we can compute~~

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

p p and so Φm(X,Y ) ≡ (X − Y )(X − 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 L0 ⊆ L cyclic of index m 0 such that Φm(j(L), j(L)) = 0. Here, we were able to choose L = πL for some principal prime ππ¯ = p. So Φm(j(L), j(L)) = 0 and so j(L) is a root of a monic polynomial.

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 then M = 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 of 0 M is contained in L. For the other half, we will show that Spl (M/Q) (which has at least one completely split factor in M) is contained in Spl(L/Q). That implies 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 + fOK where f is called the conductor.~~

~~35.1 Main theorem of complex multiplication Theorem 35.1 (main theorem of complex multiplication). If a is any invertible ideal 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 many exceptions. 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 p implies that p = Nπ = ππ¯ for some π ∈ O. Then we have πa ⊆ a a cyclic sublattice of index p. This implies that

~~Φp(j(a), j(a)) = 0~~

p p and so j(a) is a root of Φp(x, x). Observe that Φp(x, x) ≡ (x − x)(x − x ) modulo p. Choose an arbitrary prime pM of M = K(j(a)) above p. Module pM , we have p p (j(a) − j(a))(j(a) − j(a) ) ≡ 0 (mod pM ).

So j(a) is in pM . But the hypothesis that p - [OM : OK [j(aj)]] implies that OK [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. 0 The next step is to show Spl (M/Q) ⊆ Spl(L/Q) again up to ﬁnitely many primes p. Here, Spl0 is the set of primes with at least one prime with no in- 0 teria. Consider p ∈ Spl (M/Q) and exclude p | f and p sharing a factor with Q i

~~If we pick pL over pM and work in OL/pL, we get p p 0 = (j(a0) − j(a))(j(a) − j(a0))~~

~~0 inside OL/pL. But j(a) is actually in OM /pM = Fp. So we get j(a ) = j(a) Q modulo pL. Because we have set our p so that pL does not divide i~~

~~0 j(aj)), this implies that j(a) = j(a ). This shows that there is some π such that p ∩ O = πO. The conclusion is that K(j(a)) = LO. Then we have~~

~~h(O) ≥ [Q(j(a)) : Q] ≥ [K(j(a)) : K] = [LO : K] = h(O) and so LO = K(j(a)) h(O) 2~~

~~K Gal Q(j(a))~~

~~2 h(O) Q~~

~~We can talk about Shimura reciprocity which says that if p is a prime of OK and (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 for j(a) − j(a0). We can also talk about the class number 1 problem. If h(O) = 1 then j(O) ∈ Z and actually j(O) is always a cube. Then the problem comes down 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 √ 3 3 For j = g2/∆, we deﬁne γ2 = j in the q-expansion, so that~~

1/3 X k γ2(τ) = q + akq + ··· k where ak are rational. This is no longer going to be SL2-invariant, but we will have r2(γτ) = ζr2(τ) 3 for some ζ = 1. It turns out that r2(γψ) = r2(ψ) if and only if a ≡ d ≡ 0 mod 3 or 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 √ 3+ −m not divide Disc(O), deﬁne τ0 = −m if Disc(O) = −4m and τ0 = 2 otherwise (so that τ0 is relatively prime to 3). Then Q(r2(τ0)) = Q(j(τ0)) and K(r2(τ0)) = K(j(τ0)) = LO.

τ0 τ0 Proof. We have seen that r2(τ0) ∈ Q(j( 3 ), j(3τ0)). But both lattices [1, 3 ] and 0 [1, 3τ0] has complex multiplication by exactly O = Z[3τ0], by the coprimality conditions. This immediately shows that r2(τ0) ∈ LO0 . So we need to know how passing to a smaller order aﬀects. In general, let O be an order in an imaginary quadratic ﬁeld, and let p - Disc(O). Suppose O0 ⊆ O is the unique index p suborder, which is O0 = Z + pO. In this case, we have the Galois group of LO0 /LO is going to be (O/pO)×/(Z/pZ)× which is a cyclic group of order either p−1 or p+1, depending 2 on whether O/pO is Fp2 or Fp. × So we have LO0 /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 know 3 that r2(τ0) ∈ LO. This shows that r2(τ0) ∈ LO.

~~1/48 There are Weber functions f, f1, f2 ∈ Q((q )) and SL2(Z) ﬁxes the set 48 48 48 {f , f1 , f2 }. Here, we have that √ f(τ)f1(τ)f2(τ) = f1(2τ)f2(τ) = 2.~~

8 8 8 3 It is also true that f , −f1, −f2 are roots of X − γ2X − 16. Proposition 36.2. If m ≡ −3 mod 8 then √ K(f( −m)2) = L √ , Z[− m] √ √ where Z[ −m] is of index 2 in O −m. 6 Proof. Use the fact that f(84) 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 is some prime ideal p ⊆ OK with Np = 2. Because p has to be a prime, this can happen 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 number of prime√ factors of d minus 1. This shows that d = −2, and 2 being inert in Q(− d) shows that p ≡ 3 (mod 8). Setting aside the case 3 = p, we may assume that 3 p. Consider O = √ - O √ and O0 = [ −p]. Then [O : O0] = 2 and because 2 is inert, we have Q( −p) Z that LO0 /LO is of degree 2 + 1 = 3. Then we have the following.

~~√ √ 2 √ 2 LO0 = Q( −p)(f( −p) ) Q(f( −p) )~~

3 3 √ 2 LO = Q( −p) Q √ You can check that f( −p) is some real number. √ √ Let τ = 3+ −p and let α = ζ−2f (τ )2 = 2/f( −p)2. (This is from some 0 2 8 2 0 √ Weber function identity.) Because α generates Q(f( −p)2), α4 generates that 4 4 8 ﬁeld. What is the minimal polynomial of α over Q? Because α = −f2(τ0) , 3 its minimal polynomial must be X − γ2(τ)X − 16. On the other hand, this is very restrictive, because we have an algebraic integer that is a fourth power in a cubic ﬁeld. α is an algebraic integer of degree 3 and take X3 + aX2 + bX + c the minimal polynomial of α. You can show that α2 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 + (f 2 − 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 list of 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 . Assume 2 that p - Disc(O) = f Disc(OK ). The thing we want to show is that~~

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

~~37.1 Shimura reciprocity Here is the elliptic curve perspective. For O ⊆ K an order in an imaginary quadratic 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~~

~~−1 a ∗ (C/L) = C/(a L).~~

~~The claim is that a ∗ E = HomO(a,E) as O-modules. The reason is that if E = C/b then 0 → b → C → E → 0. Because a is a projective O-module, we get~~

~~−1 ∼ 0 → a b = 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 on HomO(M,E). If M is free, this is clear. If not, take a finite presentation Ob → Oa → M → 0 and define

a b HomO(M,E) = ker(E → E ) given by the matrix. If M = a, we are going to get another elliptic curve E0 = 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 another elliptic 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 that g(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 has good reduction at pL, where pL is the prime of L above p. Write E˜ the reduction of E over L. Let E˜p be the elliptic curve given by y2 = x3apx + bp. Then we have j(E˜p) = j(E˜)p. Now the claim is that

~~E˜Np =∼ p ∗ E˜ = p]∗ E.~~

~~(The second equality follows from the deﬁnition of ∗.) If we have this, then~~

Np Np j(p ∗ E) = j(p]∗ E) = j(E˜) = j(E) (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 p E˜ −→ p ∗ E˜ of degree p. We also have an isogeny E˜ → E˜p given by Frobinius. We can show that both maps are inseparable isogenies of degree p. Then there is a unique such isogeny, so we get that the target are isomorphic. Now suppose that Np = p2, which means that p = (p). Then the reduction E˜ has no p-torsion and they are called supersingular. Then there exists a unique isogeny from E˜ → E˜0 of degree p2. Then again I can there are both the Frobenius and restriction are degree p2 isogenies. Index adele, 6 j-invariant, 68 adelic class group, 5 Kronecker congruence, 77 Chebotarev density theorem, 63 class formation, 24 main theorem of complex class group, 12 multiplication, 79 complex multiplication, 65 Minkowski’s theorem, 10 conductor, 20 modular function, 72, 74 content, 9 modulus, 16

Dedekind ζ-function, 54 narrow Hilbert class field, 20 Dirichlet density, 60 natural density, 60 Dirichlet L-function, 55 normalized absolute value, 6 Dirichlet series, 53 discriminant of elliptic curve, 68 place, 5 polar density, 60 elliptic function, 66 principal ideal theorem, 51 Euler product, 54 ray class field, 16 field formation, 24 ray class group, 16 formation, 22 regulator, 58 global class field theory, 15, 17 restricted product, 6 global field, 5 strong approximation, 11 Hasse norm theorem, 39 Hilbert class field, 20 Weber functions, 81 Weierstrass form, 67 idele class group, 5 Weierstrass ℘-function, 66