Algebraic Number Theory Notes

Math 223b : Algebraic Number Theory notes

Alison Miller

1 January 22

1.1 Where we are and what’s next Last semester we covered local class ﬁeld theory. The central theorem we proved was

Theorem 1.1. If L/K is a ﬁnite Galois extension of local ﬁelds there exists a canonical map, the local Artin map × × ab θL/K : K /NL (Gal(L/K)) We proved this by methods of Galois cohomology, intepreting both sides as Tate → cohomology groups. To prove this, we needed the following two lemmas:

• H1(L/K, L×) =∼ 0

• H2(L/K, L×) is cyclic of order [L : K].

This semester: we’ll do the analogous thing for L and K global ﬁelds. We will need × × × to replace K with CK = AK /K . Then the analogues of the crucial lemmas are true, but harder to prove. Our agenda this semester:

• start with discussion of global ﬁelds and adeles.

• then: class formations, axiomatize the parts of the argument that are common to the global and local cases.

• algebraic proofs of global class ﬁeld theory

• then we’ll take the analytic approach e.g. L-functions, class number formula, Ceb- otarev density, may mention the analytic proofs

• ﬁnally talk about complex multiplication, elliptic curves.

1 1.2 Global fields, valuations, and adeles Recall: have a notion of a global field K. equivalent definitions: • every completion of K is a local field.

• K is a finite extension of Q (number field) or of Fp((t)) (function field). The set of global fields is closed under taking finite extensions. We’ll primarily focus on the number field case in this class. Definition. A place v of a global field K is an equivalence class of absolute values on K. We say that v is finite / nonarchimedean if it comes from a discrete absolute value, ∼ otherwise v is infinite /archimedean (and Kv = R or C). We can pick out a distinguished element of this equivalence class, the normalized absolute value | · |v by requiring −1 |πv|v = |kv| 2 if Kv is nonarchimedean, and |a|R = a, |a|C = |a| . Recall from last semester:

Theorem 1.2 (Product Formula). If a ∈ K, then v |a|v = 1.

Sketch of how this was done last semester. Check forQ K = Q, Fp[t], then show that if the product formula holds for K, it holds for any finite extension L/K.( v0 extends v |a|v0 = |a|v.) Q + Remark. Kv has a Haar measure, unique up to scaling (if v is finite, normalize by µ(Ov) = 1, if v is infinite use the standard normalization on R and C.) + × For any measurable set E ⊂ Kv and any K ∈ Kv we have

µ(aE) = |a|vµ(E).

Deﬁnition. Suppose that {Xi}i∈I are topological groups, and for each i we have a subgroup Yi ⊂ Xi. Then the restricted topological product of the Xi with respect to the Yi is given by

Xi = {(xi)i∈I | xi ∈ Yi for all but finitely many i}. i aY The group structure here is clear. We take as a basis of open sets those sets of the form Ui × Yi i∈S Y Yi∈/S where S ⊂ I is an arbitrary finite set, and Ui ⊂ Xi are arbitrary open sets. (Note this is not the same as the subspace topology coming from viewing i Xi as a subspace of X .) i i `Q Q 2 Then we define + + AK = AK = Kv v aY + with respect to the family of subgroups Ov and

× × AK = Kv v aY × with respect to the family of subgroups Ov . + × Exercise: AK = AK is a topological ring with group of units AK .

2 January 24

2.1 Adeles

Definition. If K is a number field, define the topological ring AK as the restricted topological product v Kv of the Kv with respect to the Ov. This ring has open cover given by the sets `Q AK,S = Kv × Ov v∈S Y Yv∈/S for any finite set S containing the infinite places, and the subspace topology on each AK,S ⊂ AK agrees with the product topology. × × Likewise define AK as v Kv , where now the restricted product is with respect to O×. Define A× likewise. v K,S `Q × Here AK is equal to the group of units of AK, so the notation is consistent. × Caution! The topology on AK is not the subspace topology inherited from AK: indeed, it shouldn’t be, because the map x x−1 is not continuous in the subspace topology. In general, if you have a topological ring R, the correct topology to put on R× × −1 comes from the embedding of R , R × R given→ by x 7 (x, x ). × × × (Note though that if R = Kv or Ov , this topology on R agrees with the subspace topology coming from R. It’s only→ because of the infinite→ restricted product that we have issues.)

Proposition 2.1. AK is a locally compact topological ring.

Proof. all AK,S are locally compact by Tychonoff. + Hence AK has a Haar measure also, which can be described as the product of the local Haar measures. + + × × Note that K embeds into AK and K embeds into AK , diagonally.

3 ∼ Proposition 2.2. If L/K is a ﬁnite extension, then AL = AK ⊗K L as topological rings. (Here ∼ n ∼ n we topologize AK ⊗K L as follows: by choosing a basis, identify L = K , so AK ⊗K L = AK, and n use the product topology on AK. One can check that this doesn’t depend on choice of basis.) ∼ L Proof. (Sketch): Use fact that Kv ⊗K L = v0 extends v Lv0 . See Chapter 2, Section 14 of Cassels-Fröhlich for details. (In fact, this is also an isomorphism of Gal(L/K)-modules set up correctly.) + + + + Proposition 2.3. For K a global ﬁeld, K is discrete (hence closed) in AK and AK /K is compact.

Proof. By the previous proposition, it’s enough to check this for K = Q and K = Fp(t). We’ll do Q; the proof for Fp(t) is similar. Discrete: the set U = v6=R Zv × (−1, 1) is open and Q ∩ U = Z ∩ (−1, 1) = {0}. Cocompact: Let D = Zv × [−1/2, 1/2]. We wish to show that D + Q = A . Let Qv6=R K a ∈ A+ be arbitrary: we want to show that a is congruent mod Q+ to some element of K Q D. Then there are ﬁnitely many primes p such that (a)p ∈/ Zp. For each such p, there exists rp ∈ Q such that rp ≡ ap (mod Z)p. Then let r = rp ∈ Q. Then a − r ∈ Zv × R. By subtracting off an appropriate s ∈ Z, get a − r − s ∈ Zv × v6=R P v6=R [−1/2, 1/2], as desired. Q Q

3 January 26

+ + Last time, we showed that AK /K is compact. + Note that AK has a Haar measure. It can be described by

µ( Ev) = µv(Ev) v v Y Y where Ev ⊂ Kv are measurable sets with Ev = Ov for almost all v. + + + Since K is discrete inside AK , we can push forward the measure on AK to get a + + measure on AK /K . To be more precise, let D be a measurable fundamental domain + + for the action of K on AK . (For example with K = Q we can take D = v6=R Zv × [−1/2, 1/2).) Then for E ⊂ A+ Borel define K Q −1 µAK/K(E) = µAK (π (E) ∩ D) where π : AK AK/K is the projection map. × Definition. If a = (av) ∈ A , then we define the content c(a) = |av| (this is defined → K v because |av| = 1 for all but finitely many v. Q 4 × >0 Exercise: The map c : AK R is a continuous homomorphism. Proposition 3.1. For any measurable set E ⊂ A+, and any a ∈ A×, then µ(aE) = aµ(E). → K K Proof. This follows from the description of µ as a product of local measures, and the local fact that µv(aEv) = |a|vµv(Ev). As a corollary, we get another proof of the product formula: if a ∈ K×, then mul- + tiplication by a gives an automorphism of AK which scales the Haar measure by c(a), and also sends K+ to itself. Hence multiplication by a scales the Haar measure on the + + quotient AK/K by c(a). On the other hand, µ(AK/K ) has finite measure, so c(a) must be 1. × × Now we move to the multiplicative situation: To show K is discrete in AK , use the × injection AK AK × AK. × × × × >0 Next question: is AK /K compact? No: c is a continuous map from AK /K R , and the latter→ is not compact. However, that’s the only obstruction: → Definition. Let 1 × × >0 AK = ker(c : AK /K R ) be the subgroup of multiplicative adeles of content 1. → 1 × × 1 Then AK is a closed subgroup of AK , and K is discrete in AK. 1 × We plan to show that AK/K is compact, and use this to deduce the finiteness of class group and Dirichlet’s units theorem. But first we’re going to develop the adelic version of Minkowski’s theorem. × + Definition. For a ∈ AK , let Sa ⊂ AK be the subset defined by + Sa = {x ∈ AK | |xv|v ≤ |av|v for all v}

Theorem 3.2 (Adelic Minkowski). There exists a constant C > 1, depending on K, such that × × for every a = (av) ∈ AK with c(a) > C, there exists some x ∈ K ∩ Sa (in particular, x 6= 0). × 1 + µ(AK/K Proof. Let B = v ﬁnite Ov × v inﬁnite{x ∈ Kv | x ≤ 2 } ⊂ AK . Then let C = µ(B) . + + Now, µ(aB) = µ(B)c(a) > µ(AK /K ). Hence there exists nonzero x1, x2 ∈ ∩aB, Q Q + x1 6= x2 such that x = x1 − x2 ∈ K . By the nonarchimedean and archimedean triangle inequalities, x ∈ Sa.

4 January 29

4.1 Applications of Adelic Minkowski v K× A+ K = K Theorem 4.1 (Strong Approximation). For any place 0, is dense in K / v0 v6=v0 v `Q 5 Proof. Enough to show that for any ﬁnite set S containing all inﬁnite places, and any > 0 B(x ) × O K+ v∈S , v∈/S,v6=v0 v contains a point of . By compactness exists some a ∈ K× such that S surjects onto A+/K+. Q Q a K Expand S as necessary to include all places v such that |av|v 6= 1 or |xv|v 6= 1. × Choose b ∈ AK such that |bv|v < /|av|v for v ∈ S, |bv|v = 1 for all other v 6= v0, and

|bv0 |v0 is large enough to ensure c(b) > C, where C is the constant from Minkowski’s theorem. Have z ∈ Sb ∩ K. Can now write x/z = r + s where r ∈ K, s ∈ Sa. Then sz ∈ SaSb = Sab implies that rz = x − sz B(x, ) × Ov, v∈S Y v∈/SY,v6=v0 but also rz ∈ K+ as desired.

1 × Theorem 4.2. AK/K is compact. × Proof. Let C be as in the Adelic Minkowski, and let a ∈ AK be any idele of content > C. Then let 1 1 D = AK ∩ Sa = {x ∈ AK | |xv| ≤ |av| for all v}. 1 × 1 We claim that D surjects onto AK/K . For this, let x ∈ AK be arbitrary: we need to show that x−1D contains an element of K×. −1 −1 For this, note that x Sa = Sx−1a, and c(x a) = c(a) > c, so by adelic Minkowski, −1 × x Sa contains some x ∈ K , and this x also lies in D.

4.2 Class groups and S-class groups Recall that if K is a number field, then the class group of K is Cl(K) is the cokernel of the × map K I(K) where I(K) denotes the group of fractional ideals of OK. We’ll now generalize this slightly in a way that also works for local fields. → Definition. K is a global field, S a nonempty finite set of places including all archimedean ones. Let OK,S = {x ∈ K | |x|v ≤ 1 for all v ∈/ S}.

If K is a number ﬁeld, and S is the set of archimedean places of K, then OK,S = OK. The ring OK,S is a Dedekind domain, and the (nonzero) primes of OK,S are in bijection with places v ∈/ S. If K is a number ﬁeld, and S = {archimedean places} ∩ {p1, ... , pn}, then OK,S is the localization of K made by inverting those elements a ∈ OK such that all prime factors of the ideal (a) are contained in the set {p1, ... , pn}.

6 Definition. Let IK,S be. the group of fractional ideals of OK,S. Then we define the class × group Cl(K, S) of OK,S to be the cokernel of the natural map φ : K I(K, S). Note that ker φ : K× I(K, S) is precisely the unit group O . K,S → Next time we prove the two basic finiteness properties: → Theorem 4.3. Let K be a global field and S a nonempty finite set of places of K including all archimedean places. Then

• Cl(K, S) is ﬁnite.

• OK,S is ﬁnitely generated of rank equal to |S| − 1.

5 January 31

Today we’ll deduce the ﬁniteness of the class group and Dirichlet’s units theorem from 1 × the compactness of AK/K . × Let AK,S denote the subgroup

× × × × AK,S = {x ∈ AK | |x|v = 1all v∈ / S} = Kv × Ov . v∈S Y Yv∈/S 1 × 1 1 and let AK,S denote AK,S ∩ AK: then AK,S is an open subgroup of AK,S. The key exact sequence we’ll use here is

1 × × 1 × 1 1 × 1 (AK,S · K )/K AK/K AK/(AK,S · K ) 1

1 × × Here (AK,S · K )/K is an open subgroup, hence also closed. As a consequence, the last 1 × 1 1 × 1 1 map AK/K AK/(AK,S · K ) is continuous using the discrete topology on AK/(AK,S · K×). 1 × 1 1 × Compactness→ of AK/K then implies that the quotient AK/(AK,S · K ) is compact in 1 × × the discrete topology, hence finite. It also implies that the subgroup (AK,S · K )/K is compact. From the first of these facts we’ll get the finiteness of class group, and from the second we’ll get Dirichlet’s units theorem. To get the finiteness of class group, we now just need

Lemma 5.1. ∼ 1 × 1 ClS(K) = AK/K AK,S.

7 Proof. Deﬁne a map 1 φ : AK ClS(K)

vp(ap) by φ(a) = p p , where p ranges over the primes of OK,S (which we know are in bijection with the places of K not in S). → Q × 1 Then check that it is a surjection and that the kernel is K AK,S.

Corollary 5.2. ClS(K) is ﬁnite. Now we look at the units group. This is a bit trickier.

Lemma 5.3. 1 × × ∼ 1 × (AK,S · K )/K = AK,S/OK,S Proof. (This is a special case of the second isomorphism theorem: (A + B)/B =∼ A/(A ∩ B).) 1 × By our previous exact sequence, we know that AK,S/OK,S is compact. This will ultimately allow us to show that OK,S has enough units, but we need some technical lemmas ﬁrst:

Lemma 5.4. The set {x ∈ OK,S | v(x) ∈ [1/2, 2] for all v ∈ S} is ﬁnite. × Proof. This is the intersection inside AK of the discrete set K with the compact set v∈S Rv × v∈/S Kv, where Rv = {x ∈ Kv | v(x) ∈ [1/2, 2]. × QPropositionQ 5.5. a ∈ K is a root of unity iff |a|v = 1 for all v, and the set of all such a is ﬁnite.

Proof. Finiteness of {a | |a|v = 1 for all v} follows from the previous lemma. For the equivalence, is clear. To show observe that {a | |a|v = 1 for all v is a ﬁnite group, hence is torsion. ⇐ 1 ⇒ + + S Now, consider the homomorphism L : AK,S v∈S R = (R ) given by (L(a)) = (|aQ| ) v log→ v v . × n It follows from the lemma that L(OK,S) is a discrete subset of R . 1 The image L(AK,S) is contained in the hyperplane H = { v xv = 0}. In the case where K is a number ﬁeld and S is precisely the archimedean places, in fact L(A1 ) = H. In P K,S class we only handled this case: to do the general case, you need the following fact, 1 which is easy to check: H/L(AK,S) is compact. Now, L induces a map

1 × 1 × AK, S /OK,S L(AK,S)/L(OK,S)

→ 8 since the left hand side is compact, so is the right hand side. Combining with the fact × × above, we hate that H/L(OK,S) compact. Since we also know L(OK,S) is discrete, we × conclude that L(OK,S) is a free abelian group of rank |S| − 1. Finally, Proposition 5.5 tells us that the kernel of L contains only ﬁnitely many ele- × × × ments of OK,S, so OK,S is a ﬁnitely generated abelian group of the same rank as L(OK,S), and we conclude Dirichlet’s units theorem.

6 February 2

1 × Recently we’ve been talking about AK/K . Now we’re going to be moving on to class × × ﬁeld theory, where CK = AK /K is more important. Brief remarks on the relationship between the two. If K is a number ﬁeld, have short exact sequence of topological groups.

0 × × >0 1 AK/K AK/K R 1.

v K× >0 K K× If is any archimedean place→ of AK/ → , then the→ map R→ , v , AK/ gives a × × ∼ 0 × >0 splitting of the short exact sequence, so AK /K = AK/K × R . In the case that K is a function ﬁeld with constant ﬁeld Fq, have→ short→ exact sequence

0 × × Z 1 AK/K AK/K q 1. Again this sequence splits since qZ is free, but we have to be more ad hoc about it. → → → → 6.1 Statement of results of global class ﬁeld theory Let K be a global ﬁeld. Then

Theorem 6.1 (Main theorem of global class ﬁeld theory, Take 1, Part 1). There is a continuous map, the global reciprocity map

× × ab θ/K : CK = AK /K Gal(K /K) (C )0 C which is surjective, and has kernel equal to the connected→ component K of K at the identity. 0 × Remark. The connected component (CK) is equal to the closure of the image of v infinite Kv × × 0 inside AK /K . If K is a function field, this means that CK = {1}. If K is a number field, × × × Q then (by current HW) the map v infinite Kv , AK /K is a closed embedding if and only if K has a unique infinite place (K is Q or an imaginary quadratic field). In that 0 ∼ × 0 Q case CK = K : otherwise CK is a product of some→ number of solenoids, some number of circles, and one copy of R>0. This is a correction from what I said in class. See the exercises on page 367 of∞ Neukirch ANT for an outline of the proof.

9 The global reciprocity map is determined by compatibility properties: speciﬁcally, that the square θ × /Kv ab Kv Gal(Kv /Kv)

× × θK ab AK /K Gal(K /K).

commutes for all v. Here θ/Kv is the reciprocity map defined last semester. More specifically, if L is a finite abelian extension of K and w is any valuation extending L, we have θ × Lw/Kv Kv Gal(Lw/Kv)

× × θL/K AK /K Gal(L/K). × and recall from last semester that, whenever Lw/Kv is unramiﬁed, then Ov ⊂ θLw/Kv In particular this means we may deﬁne

θL/K(a) = θLw/Kv (av) v Y as the right hand side is a finite product, and take the inverse limit over all L/K finite abelian to define θ/K(a). The map θ/K induces a bijection between finite index open subgroups of C − K and finite index open subgroups of Gal(Kab/K). The Galois correspondence gives a bijection between the latter and finite abelian extensions of K. This bijection (finite abelian extensions of K) to (finite index open subgroups of CK) can be described explicitly. × × Note that we can define a norm map NL/K : AL AK by (NL/Ka)v = w extends v NLw/Kv (aw). This norm map extends the norm map N : L× K× (this follows from a HW problem L/K Q from last semester), so induces a norm map : NL/K→: CL CK. × × Can show that NL/K(AL ) is an open subgroup→ of AK (this requires using local class × × field theory to show that NLw/Kv Lw is open in Kv ). → Theorem 6.2 (Global Class Field theory, Take 1, Part 2). If L is a finite abelian extension of K, then θ−1( (Kab K)) = N ( ×) /K Gal / L/K AL .

6.2 Ray class fields and ray class groups Let K be a number field now. 0 We’ll now define a family of open subgroups of CK containing CK, which by the global class field theory correspondence will give us a family of finite abelian extensions of K.

10 Deﬁnition. A modulus m of K is a function from (places of K) to Z≥0 such that • m(v) = 0 for all but ﬁnitely many v

• if v is complex then m(v) = 0

• if v is real then m(v) = 0 or 1. m(v) As a matter of notation we write m = v v , e.g. m = p1p2 1.

For a modulus m, deﬁne the congruenceQ subgroup Um of CK by ∞  

 × × >0 × Um = K · Up m(p) K R  /K  , v  p finite v infinite v finite Y mY(v)=0 mY(v)=1

× mp where Up,mp = {a ∈ Op | a ≡ 1 (mod (pOp) )}. The subgroup Um is open in CK and contains the connected component of the identity. ab Definition. The ray class field Lm of modulus m is the subfield of K fixed by θ/K(Um). ∼ Then Lm is an abelian extension of K with Gal(Lm/K) = CK/Um. This Galois group can be defined non-adelically as

Deﬁnition. The ray class group Clm(OK) is the quotient Im/Pm, where

• Im is the group of fractional ideals of OK relatively prime to m.

• Pm ⊂ Im is the subgroup of principal ideals (a) such that a ∈ Up,m(p) for all ﬁnite places p contained in the modulus m and a is positive at all real places contained in the modulus m. ∼ Proposition 6.3. CK/Um = Clm(OK) Proof. HW.

7 February 5

Correction to last time: θ/K is only surjective when K is a number ﬁeld. The correct statement of the Main Theorem is Theorem 7.1 (Main theorem of global class ﬁeld theory, Take 1, Part 1). There is a continuous map, the global reciprocity map × × ab θ/K : CK = AK /K Gal(K /K) 0 which has dense image, and has kernel equal to the connected component (CK) of CK at the → identity.

11 Because Gal(Kab/K) = lim Gal(L/K), the statement that θ has dense −L/K ﬁnite abelian /K image is equivalent to saying that the composite map

← ab θL/K : CK Gal(K /K) Gal(L/K) is surjective for all ﬁnite L/K. 0 → → 1 × For K a number ﬁeld, CK/CK is compact, as the compact set AK/K surjects onto 0 CK/CK. 0 (Proof: if [(av)] ∈ CK/CK, and v0 is any archimedean place of K, choose x ∈ Kv0 such

that x is in the connected component of 1 in Kv0 (so x > 0 if Kv0 is real), and |x|v0 = c(av0 ). × 1 Then define bv ∈ AK by bv = av for v 6= v0 and bv = av/x for v = v0. Then bv ∈ AK and 0 [bv] ≡ [av] modulo CK.) 0 Since CK is contained in ker θK and the continuous image of a compact set is closed, ab the image θ/K(CK) is closed in Gal(K /K). But we already know that θ/K(CK), and so θ/K is dense in the number field case. If K is a function field with field of constants k = Fq, then CK is totally disconnected (we leave the proof as an exercise). Hence θ/K : CK Gal(K¯ /K) is injective. On the other hand, we have the short exact sequence of topological groups 1 × c Z 1 AK/K CK − q 1 → which means that CK is not compact, so is not profinite, and θ/K can’t be an isomorphism→ of topological→ groups.→ → What’s going on here is that Kab contains K · k, so there’s a group homomorphism Gal(Kab/K) Gal(k/k) =∼ Z^ , and we have the commutative rectangle Z CK → q 1

Gal(Kab/K) Z^ 1. This extends to a morphism of exact sequences

0 × Z 1 AK/K CK q 1

ab 1 IK Gal(K /K) Z^ 1.

ab where IK is deﬁned as the kernel of the map Gal(K /K) Z^ . In this diagram, one can 0 × show that both exact sequences split and the map AK/K IK is an isomorphism. → 7.1 Narrow class ﬁeld take 2 → I’m going to change notation a bit here from Friday.

12 m(v) × For a modulus m = v v , deﬁne the congruence subgroup Um of AK by

Q × >0 Um = Up,m(p) Kv R p finite v infinite v finite Y mY(v)=0 mY(v)=1

m m × × Then let CK be the congruence subgroup of CK given by CK = Um ·K /K . × Any open subgroup of AK must contain some Um, hence any open subgroup of CK m must contain some CK . We restate our deﬁnition from last time in new notation

m Definition. The ray class field of modulus m Lm is the fixed field of θ/K(CK ). m Recall from last time that CK/CK is isomorphic to the ray class group Clm(OK). It follows from definitions of the ray class field and the reciprocity map that, for 0 × any prime p not dividing m, and any prime p of Lm extending p, θLm/Kp Op fixes the completion (Lm)p0 . By local class field theory, this means that (Lm)p0 /Kp is unramified, and hence that p is unramified in the extension Lm/K. Then we have isomorphisms

m Clm(OK) CK/CK Gal(Lm/K). (O ) This isomorphism can be described→ explcitly.→ Note that Clm K is generated by elements of the form [p] where p is a fractional ideal relatively prime to m. The isomor- m × phism Clm(OK) CK/CK sends such a [p] to class [a] of the idele a = (av) ∈ AK with × ap = π is a uniformizer of Kp , and av = 1 v 6= p. Then → θ (a) = θ (π) | Lm/K (Lm)p0 /Kp Lm is the Frobenius element Frobp in Gal(Lm/K) since the local extension is unramiﬁed. It follows that the Frobenius element Frobp depends only on the class of p in Clm(OK).

8 February 7

m × × Recall from last time that the ray class field Lm is the fixed field of θL/K(CK ) = θL/K(UmK /K .

Proposition 8.1. a) Lm is unramiﬁed at all primes not dividing m.

0 b) if m | m then Lm ⊂ Lm0 .

c) Lm1 ∩ Lm2 = Lgcd m1,m2

ab d) ∪mLm = K ,

13 × m Proposition 8.2. Stated a) last time but didn’t prove in class: follows from Op , CK , so × (L ) 0 is fixed by θ (O ), equivalent by local class field theory to saying that (L ) 0 /K m p (Lm)p0 /Kp p m p p is unramified. → (This also works with the convention that a real place v is unramified if and only if every v0 extending v stays real.) b) and c) follow from definitions. × for d): any open subgroup of AK contains some Um, so any open subgroup of CK contains m some CK , and d) follows by Galois correspondence. ∼ × ∼ × Example. Let K = Q. Clm(K) = (Z/mZ) , Clm = (Z/mZ) /{±1}. Then Lm = Q(ζm) −1 and Lm = Q(ζm + ζm ). ∞ ∞ Example. m = 1, Clm(K) = Cl(K), Lm = H is the Hilbert Class Field, the maximal field extension unramified at any places (including infinite places). + + m = vreal v, Clm(K) = Cl (K) is narrow class group, Lm = H is narrow Hilbert class field. Q Definition. If L is a finite abelian extension of m, then the minimal m such that L ⊂ Lm is referred to as the conductor of L, and written f = fL/K.

This means that m is minimal such that the reciprocity map θ/K : CK Gal(L/K) ∼ m factors through Clm(K) = CK/CK . Two interpretations of this: first, this tells us that for any prime p of OK →not dividing L/K f, L is unramified at p and Frobp = p depends only on the image of p in Clf(OK) Secondly, we’ve stated in the main theorem of class field theory that ker θ/K : CK f Gal(L/K) is NL/KCL, so this says that f is minimal such that NL/KCL ⊂ CK. Exercise: the places that ramify in L are exactly those dividing fL/K. → √ K = Q L = Q( p) p > 0 p ≡ 1 ( 4) f = p Example. , , , mod .√ Here, L/K meaning that −1 Q( p)/Q p L ⊂ Q(ζp + ζp ), and for any prime q not equal to p, q = q depends only on the image of q in (Z/pZ)×/ ± 1.

9 February 9

9.1 General strategy Recall that for local fields, we proved main theorem by getting an isomorphism L×/NK× =∼ H^ 0(L/K, L×) =∼ H^ −2(L/K, L×) =∼ Gal(L/K)ab. for each finite Galois extension L/K. Then we took the projective limit, as both sides run through over all finite extensions of K, to obtain an isomorphism L^× =∼ Gal(Ksep/K)ab =∼ Gal(Kab/K). Fundamental information here: for each finite Galois extension L of K, have a Gal(L/K)- module L×, such that

14 if E ⊃ L, then have injection L× , E× and if E/L is Galois, then L× = (E×)Gal(E/L). Can also take the union (inductive limit) of all of the finite extensions to get a × × Gal(Ksep/K) module (Ksep) . From→ this big module we can, for each L, recover L as sep the fixed submodule ((Ksep)×)Gal(K /L, on which Gal(Ksep/K)/ Gal(Ksep/L) =∼ Gal(L/K) acts. × We want to do the same thing but now replace L with CL. In order to do this, we’ll first check that . After that we’ll explain the general formalism of class formations.

9.2 Inclusion maps for adeles × × If L/K is a finite extension of global fields then have injection AL , AK and AL , AK . This inclusion is a topological embedding. × × → → Proposition 9.1. AK is closed in AL . × Gal(L/K) × if L/K is Galois, then (AL ) = AK . × × × AK ∩ L = K . ∼ Proof. This all follows from AK ⊗K L = AL, both as topological rings, and as Galois modules when L/K Galois. × × × × As a consequence, get inclusion CK = AK /K , CL = AL /L . Proposition 9.2. If L/K is a finite Galois extension, then C = (C )Gal(L/K). → K L × × Proof. Short exact sequence of Gal(L/K)-modules 1 L AL CL 1 gives the long exact sequence

× × Gal(L/K) → 1 → × → → 1 K AK (CL) H (L/K, L ) H1(L K L×) = 1 but / , by→ Hilbert→ 90, so→ we’re done. →

(Can ﬁnd examples to show that this is not true for the ideal class group ClL; the Gal L/K natural map ClK (ClL) is usually neither injective nor surjective.)

9.3 Class formations→ Class formations: G is a proﬁnite group. We consider the set of open subgroups of G (necessarily of ﬁnite index), which we write as {GK | K ∈ X}, and refer to these indices K

as “fields”. The field K0 with GK0 = G is called the “base field.” Write K ⊂ L if GL ⊂ GK. A pair K, L with K ⊂ L is called a “layer” L/K. The degree of the layer is [GK : GL]. We say that L/K is normal if GL is normal in GK, write GL/K = GK/GL. Can formally define intersection and union of subgroups. Define gK by GgK = −1 gGKg .

15 A formation A is a G-module on which G acts continuously when A is given the discrete topology. U Equivalently: A = ∪U⊂G openA G Write A K = AK, so A = ∪U⊂GAK. E.g. K local, G = Gal(Ksep/K), A = (Ksep)×. Or K global, G = Gal(Ksep/K), A = lim C . − L/K ﬁnite L

10 February 12 →

Continuing from last time: (G, (GK)K∈X, A) is a formation. G Write AL = A L . q q q q Write H (L/K) = H (GL/K, AL). (Also write H (/K) = H (GK, A).) If E/L/K with E/K and L/K normal layers then get inflation map q q GL/K/GE/K q q H (L/K) = H (GL/K, AE ) H (GE/K, AE) = H (E/K). Also: restriction map Res : Hq(E/K) Hq(E/L) also: corestriction Cor : Hq(E/L) q H (E/K). → Suppose that L/K is a normal layer→ and g ∈ G is arbitrary. Then there are natural→ ∼ isomorphisms GL/K = GgL/gK and AL AgL. The isomorphisms induce a natural ∗ q q homomorphism g : H (L/K) H (gL/gK). If g ∈ GK then gK = K but also gL = L as ∗ L/K is normal. Exercise: g is the identity.→ Definition. A is a field formation→if for every layer L/K, H1(L/K) = 0.

Example. If K0 is any field, G = Gal(K0/K), If that’s the case then have inflation restriction exact sequence in deg 2: 1 H2(L/K) H2(E/K) H2(E/L). 2 S 2 In particular, this means that H (/K) = L H (L/K). → → → Definition. A is a class formation if for every normal layer L/K there is an isomorphism 2 1 invL/K : H (L/K) [L:K] Z/Z such that the diagrams

2 invL/K 1 → H (L/K) [L:K] Z/Z

inf

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

Res ×[L:K]

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

16 commute (for the ﬁrst diagram assume E/K and L/K both normal, for the second just need E/K normal).

11 February 14

Example. The simplest example of a class formation: ^ ^ G = Z, A = Z with trivial action. GKn = nZ, AKn = Z. 1 1 Then, for m | n, Kn/Km is a layer and H (Kn/Km) = H (GKm /GKn , Z) = 0 because

GKm /GKn is cyclic of order n/m. 2 ∼ ^ 0 ∼ Likewise, H (Kn/Km) = H (Kn/Km) = Z/(n/m)Z, and we can define invKn/Km by 1 : (n m) 1 composing with × n/m Z/ / Z n/m Z/Z. sep sep × Example. K0 local field: G = Gal(K0 /K0), A = (K0 ) is a class formation by last semester. → K G = (Ksep K ) A = S C 0 global field: Gal 0 / 0 , K/K0 finite K: we haven’t proved this is a class formation, but we will.

Proposition 11.1. Suppose that A is a class formation. Then

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

Cor

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

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

g∗ ∼

2 invgL/gK 1 H (gL/gK) [L:K] Z/Z commutes

17 Proof. For a), we make a big commutative diagram:

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

Res ×[L:K]

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

Cor

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

The top square commutes by class formation axiom, and the entire rectangle commutes because Cor ◦ Res = [L : K]. Since the two top vertical arrows are surjective, we can chase the diagram to find that the bottom square commutes. Part b) will be done on HW, but here’s a sketch: first of all, we’ve observed that if g ∈ K, then gL = L, gK = L, and g∗ : H2(L/K) H2(L/K) is the identity map, so we already have the commutative diagram. 0 0 For the general case, find a field L extending→L such that L /K0 is normal. Construct 2 2 0 0 injection H (L/K) H (L /K0) and use that to transfer the result from L /K0 to L/K>

→ 2 1 Definition. Fundamental class uL/K ∈ H (L/K) defined by inv(uL/K) = [L:K] . Proposition 11.2. Let E/L/K be fields with E/K normal. Then

a) ResL uE/K = uE/L

b) uL/K = [E : L] inf uE/K if [L : K] normal

c) CorK uE/L = [L : K]uE/K ∗ d) g (uE/K) = ugE/gK proof: exercise.

Theorem 11.3. For any normal layer L/K, cup product with uL/K gives an isomorphism ^ q ^ q+2 H (GL/K, Z) H (L/K) ab ∼ −2 0 In particular, do the case q = −2: get isomorphism G = H^ (GL K, Z) H^ (L/K) = → L/K / AK/NAL. ^ −2 ab As before, we denote the inverse isomorphism H (GL/K, Z) GL/K by→θL/K. Our goals now: show that GK, CK is a class formation. (will focus on number ﬁelds case) → Need to show two things: H1(L/K) is trivial, and construct invariant map H2(L/K) 1 [L:K] Z/Z. → 18 Intermediate steps: First inequality: show that if L/K is cyclic then |H2(L/K)| ≥ [L : K] (We’ll do this by showing that the Herbrand quotient H2(L/K)/H1(L/K) = [L : K].) Second inequality: show that if L/K is cyclic of prime order, then |H2(L/K)| ≤ [L : K]. (Combining Second inequality with Herbrand quotient, get |H2(L/K)| = [L : K] and H1(L/K) = 0 for cyclic layers. For non-cyclic layers, can deduce that H1(L/K) vanishes 2 and H (L/K) ≤ [L : K]. ) Then: construct invL/K from local reciprocity maps.

× 11.1 Cohomology of AL Let L/K be a finite extension of global fields, with galois group G = Gal(L/K). We’ll × want to compute cohomology of AL as a Gal(L/K)-module. Let S be any finite set of places of K including all infinite ones, and let S¯ be the set of all primes of L lying above some prime of S. A× = A× For simplicity, we’ll write L,S L,S¯ . Then × × × AL,S = Lv0 × Ov0 . v∈S 0 0 Y Yv |v Yv∈/S Yv |v For every v, choose w above v, and let Dw be the decomposition group of w: g ∈ G | ∼ gw = w. Have natural isomorphism Dw = Gal(Lw/Kv). Also, the G-module × × Lv0 = Lgw 0 Yv |v g∈YG/Dw × is induced/co-induced from the Dw-module Lw. q × ∼ q × By Shapiro’s lemma, get H (G, v0|v Lv0 ) = H (Gw, Lw), and this isomorphism is induced by Q q × Res q × π∗ q × H (G, Lv0 ) −− H (Gw, Lv0 ) − H (Gw, Lw) 0 0 Yv |v Yv |v (this works also for Tate cohomology).→ → q × ∼ q × q × Hence H (G, AL,S) = v∈S H (Gw, Lw) × v∈/S H (Gw, Ow). Note that if v is unramified in L, then Hq(G , O×) =∼ Hq(L /K , O×) = 1: so if S Q Q w w w v w contains all unramified primes, we have

q × ∼ q × H (G, AL,S) = H (Gw, Lw) v∈S Y Now, take the direct limit over all sets S containing the ramiﬁed primes, and get

q × q × ∼ M q × H (G, A ) = lim H (G, A ) = H (Lw/Kv, Lw) L S L,S v

19 Consequences:

1 × • H (L/K, AL ) = 0 : the ideles form a ﬁeld formation. • H2(L/K, A×) = L 1 Z/Z. L v [Lw/Kv] H2(L K C ) ∼ 1 Z (We’ll later show / , L = [L/K]Z / . ^ 0 × ∼ ^ 0 × • H (L/K, AL ) = v H (Lw/Kv, Lw). As a consequence we get the Norm theorem × × × for ideles: (av) ∈ A is a norm from A if and only if av ∈ NL for all v. Q K L w (Contrast this with the Hasse Norm theorem: a ∈ K× is a norm from L× if and only if a is a norm locally. This is harder and will need the second inequality. )

× • The Herbrand quotient of AL,S is equal to v∈S[Lw : Kv]. (Note that the module A× does not have a ﬁnite Herbrand quotient, which is why we’ll use A× instead.) L Q L,S

Now we’re going to set ourselves up to compute the Herbrand quotient for CK. First, choose a set of places S of K large enough that:

(a) S contains all inﬁnite places.

(b) S contains all places that ramify in L. × ¯ (c) AL,S surjects onto CL (can do this by making sure that S contains a set of generators for Cl(OK)) × (d) AK,S surjects onto CK (likewise)

Then we have a short exact sequence

× × 1 OL,S AL,S CL 1, × h(AL,S) so the Herbrand quotient h(→CL) = → × . We’ve→ → computed the numerator, will get h(OL,S) the denominator next time.

12 February 21

Lemma 12.1. G is a finite group. L/K fields, K infinite. Two K[G]-modules V1, V2 are isomorphic iff V1 ⊗K L and V2 ⊗K L are isomorphic L[G]- modules.

20 Proof. (In class we reduced to the case where L/K is ﬁnite, but we don’t actually need that for this proof.) ∼ We have an isomorphism: HomL[G](V1 ⊗K L, V2 ⊗K L) = HomK[G](V1, V2) ⊗ L. We choose K-bases for V1 and V2. We have then have determinant maps det : HomL[G](V1 ⊗K L, V2 ⊗K L) L which restricts to det : HomK[G](V1, V2) K. Observe that det is a polynomial map. Be- ∼ cause of the assumption that V1 ⊗K L = V2 ⊗K L, we know that det not identically→ 0 on HomL[G](V1 ⊗K L, V2 ⊗K L). Hence→ its restriction to HomK[G](V1, V2) is not the zero polynomial either, and there must be some φ ∈ HomK[G](V1, V2) with det φ 6= 0.

Remark. Can also prove this via Galois descent: IsomL[G](V1, V2) is a torsor for the group AutL[G](V1). Can show that AutL[G](V1) is the group of units in a central simple algebra over L (in class, I said this was a product of matrix algebras: that’s not actually correct un- 1 less L is sufficiently large), and that H (G, AutL[G](V1)) is trivial. Hence HomL[G](V1, V2) G is the trivial torsor, which implies that IsomL[G] V1, V2) = IsomL[G](V1, V2) is nonempty. ∼ Proposition 12.2. If A, B are G-modules that are f.g. as abelian groups, and A ⊗Z R = B ⊗Z R then h(A) = h(B). ∼ Proof. By Lemma, A ⊗Z Q = B ⊗ ZQ. exercise: this means that there is a finite index subgroup A0 of A/T(A) isomorphic (as G-module) to a finite index subgroup B0 of B/T(B). Hence h(A) = h(A0) = h(B0) = h(B). × ∼ + Have isomorphism (OL,S) ⊗Z R = H ⊂ v0∈S¯ R given by Q (a) ⊗ 1 7 (log |a|v0 )v0∈S¯

Extend to isomorphism → × ∼ + (OL,S ⊕ Z) ⊗Z R = R 0 ¯ vY∈S So we can apply Proposition 12.2 for the following G-modules: × × Let A = (OL,S ⊕ Z). Then h(A) = [L : K]h(OL,S). ∼ G Let B = v0∈S¯ Z: can decompose as B = v∈S Bv, where Bv = v0|v Z = coIndGw Z. By Shapiro’s lemma, the Herbrand quotient of the G-module B = coIndG Z is equal Q Q vQ Gw to the Herbrand quotient of the Gw-module Z (with trivial action), but we know that the latter is equal to |Gw|. Hence h(B) = v h(Bv) = v∈S |Gw| = v∈S[Lw : Kv]. h(A) = [L : K ] h(O× ) = v∈S[Lw:Kv] Conclude that Q v∈QS w v , andQ so L,S [L:K] . Now can compute Q Q

21 h(A× ) L,S v∈S[Lw : Kv] h(CL) = = = [L : K], h(O× ) [L : K ]/[L : K] K,S v∈QS w v completing our proof of the second inequality.Q 0 We’ve now proved that |H^ (L/K, CL)| ≥ [L : K], that is, |CL/NCK| ≥ [L : K]. Note ∼ × × × CL/NCK = AL /L NL/KAK .

13 February 23

13.1 Corollaries of First Inequality × × Corollary 13.1. L/K ﬁnite abelian (or more generally solvable), D ⊂ AK with D ⊂ NL/KAL , × × K D is dense in AK , then L = K. Proof. Without loss of generality we may assume L/K cyclic: otherwise replace L with L0 ⊂ L such that L0/K is cyclic. × × × × × The subgroup NL/KAL is open in AK , so K NL/KAL is an open subgroup of AK . On × × × × × × the other hand, K NL/KAL contains K D, so is also dense. Hence AK = K NL/KAL , so 0 |H^ (L/K, CL)| = 1 implying L = K by ﬁrst inequality.

Corollary 13.2. 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. We prove the contrapositive: suppose all but ﬁnitely many primes split completely in L Let S contain the inﬁnite primes and all the primes that don’t split completely in L. × × Let D = {x | xv = 1 for all v ∈ S}. Then D is contained in NAL and K D is dense in × × × AK . (This is weak approximation: K is dense in v∈S Kv .) Applying the previous corollary, then get L = K. Q

Corollary 13.3. If S is a finite set of places of K and L/K is finite abelian, then Gal(L/K) is L/K generated by v = Frobv ∈ Gal(L/K) for v ∈/ S. Proof. (Wlog S contains all ramified and infinite primes.) Let H be the subgroup of Gal(L/K) generated by {Frobv | v ∈/ S}. Let M be the subfield of L fixed by H. Then all primes of K that do not lie above primes in S must split completely in M. By the contrapositive of previous corollary, must have M = K. By Galois correspondence, H = Gal(L/K).

22 13.2 The Second Inequality We’ll next work towards proving

0 Theorem 13.4 (Second Inequality). K a number ﬁeld, L/K cyclic of degree p, |H^ (L/K, CL)| = |CL/NL/KCK| ≤ p Claim: it’s enough to show this when K contains the pth roots of unity.

Proof of Claim: Suppose L/K is any cyclic extension of global ﬁelds of degree p. let K0 = 0 0 K(ζp), L = L(ζp) = LK . Observe that K0/K has degree d | p − 1, so K0 and L are linearly disjoint, and L0/L = d also We get a commutative diagram

NL0/L CL CL0 CL

×d NL/K NL0/K0 NL/K

NK0/K CK CK0 CK

×d

0 CK/NL/KCL CK/NL0/K0 CL CK/NL/KCL

×d

Observe that CK/NCL is a group of exponent p relatively prime to d, so multiplication by d is an automorphism of CK/NCL. Looking at the bottom row, we see that the induced 0 0 map CK/NL/KCL CK/NL0/K0 CL must be injective, and CK/NL0/K0 CL CK/NL/KCL must be surjective. 0 0 Hence |CK/NL/→KCL| ≤ |CK/NL0/K0 CL|, and so it’s enough to bound→ the size of the latter. √ n Hence we can assume that K contains µp, and by Kummer L = K( a) for some a ∈ K×. × × × Need to show that [AK : K NAL ] ≤ p. × × × We’ll actually show that [AK : K F] ≤ p, where F ⊂ NAL is an appropriately chosen subgroup. Let S be an appropriately chosen set of places of K. We’ll choose S later: for now we just need that S contains all primes ramiﬁed in L/K.

23 Let v1, ... , vk be additional places of K, with the property that each vi splits completely in L. We’ll choose them later: this will be the hard part of the proof. ∗ Let S = S ∩ {v1, ... , vk}. Now take × p × F = (Kv ) Kv Ov . v∈S v ,...,v ∗ Y 1Y k vY∈/S × We check that F ⊂ NAL : we know that we can do this locally. × p × For v ∈ S, (Kv ) ⊂ NLw/Kv Lw as [Lw : Kv] divides p. × × For each vi, Lw = Kv as vi splits completely. So N L = K as needed. i i Lwi /Kvi wi vi ∗ × × For v ∈/ S , we know that Lw/Kv is unramiﬁed, so NLw/Kv Lw includes Ov as needed.

14 February 26

We choose S to: (a) contain all inﬁnite places and all primes dividing p,

(b) contain all v with |a|v 6= 1. (so a is an S-unit) × (c) Be large enough that AK,S CK is surjective Lemma 14.1. Let K be a global field, v a finite place√ of K such that Kv does not have residue × → p × p characteristic p, and b ∈ K . Then is ramified in K( b)/K iff b ∈ Op · (Kv) , and v splits p completely iff b ∈ Kv . Proof. exercise. Corollary: L/K is unramified outside S, as needed in the proof above that F ⊂ × NL/KAL . × × Now we break up [AK : K F] into a local and global component. × × × × Observe that F ⊂ AK,S∗ : since AK,S∗ surjects onto AK /K , we have that

× × × AK,S∗ AK /K F × with kernel F × OK,S, so × × ∼ × × AK /K F = AK,S∗ /F × OK,S∗ Now, have SES:

× × ∗ ∗ 0 F × OK,S∗ /F AK,S /F AK,S /F × OK,S∗ 1 × × F × O F =∼ O ∗ F ∩ O where the ﬁrst term→ K,S∗ / → K,S / → K,S∗ by the second→ isomorphism theorem.

24 So × × × ∗ ∗ [AK : K F] = [AK,S : F]/[OK,S : F ∩ OK,S∗ ]. (1) × ∗ ∗ and we just need to ﬁnd [AK,S : F] and [OK,S : F ∩ OK,S∗ ]. The ﬁrst one is a product of local factors:

× × p [AK,S∗ : F] = Kv /(Kv ) v∈S Y × × p 2 −1 By HW: if Kv is a local ﬁeld containing µp, then Kv /(Kv ) = p · |p|v . 2 −1 Hence [AK,S∗ : F] = v∈S p · v∈S |p|v . The second term vanishes by the product formula (using |p|v = 1 for v ∈/ S), so. Q Q 2|S| 2|S| [AK,S∗ : F] = p = p For the global part: × × p observe that F ∩ OK,S∗ contains (OK,S∗ ) . So × × p × × p ∗ ∗ [OK,S : F ∩ OK,S∗ ] = [OK,S :(OK,S∗ ) ]/[F ∩ OK,S∗ :(OK,S∗ ) ]

× p ∗ ∗ [OK,S :(OK,S∗ ) ] = |S | = |S| + k by Dirichlet’s units theorem. So, plugging into equation 1, get |S|−k p AK = p [F : OK,S∗ ]

We’ll show we can choose places v1, ... , vk, split in L/K with k = |S| − 1 and F ∩ OK,S∗ = p OK,S∗ .

√p Proposition 14.2. Suppose that v1, ... , vk are places of K such that if K( b) is unramiﬁed p ∗ × p ∗ outside S and completely split at all primes of S, then b ∈ (K ) . Then F ∩ OK,S = OK,S∗ .

√p Proof. Proof: suppose b ∈ F ∩ OK,S∗ . Then by lemma 14.1, K( b) has the appropriate × p ∗ p behavior, hence b ∈ (K ) , but since b is an S -unit, in fact b ∈ OK,S∗ .

Now we get to choose v1, ... , vk. pp Let T = K( OK,S). Then T/K is an extension of exponent p with ∼ p ∼ | Gal(T/K) = Hom(OK,S/(OK,S) , µp) = (Z/p) S|. Note that L ⊂ T, and Gal(T/L) is an extension of exponent p with Gal(T/L) =∼ (Z/p)k where k = |S| − 1. Choose places w1, ... , wk of L, not lying above any places of S, such that Frobw1 , ... , Frobwk ∈ Gal(T/L) form a basis for Gal(T/L) as a Fp-vector space: this here is where we are using the ﬁrst inequality. Let v1, ... , vk be the places of K below w1, ... , wk.

25 15 February 28

Let v1, ... , vk be the places of K below w1, ... , wk.

We now look at their splitting behavior in L. Observe that the elements Frobwi ∈ e wi/vi Gal(T/L) and Frobvi ∈ Gal(T/K) are related by Frobwi = Frobvi where e = ewi/vi

is the inertia degree, which can be either 1 or p. But if e = p, then have Frobwi = 0,

contradicting that Frobwi generates, so e = 1 and vi splits completely in L.

Choose one more place vk+1 so that Frobv1 , ... , Frobvk+1 generate Gal(T/K). Next, we show that

Lemma 15.1. k+1 O× (O× )p O× (O×)p K,S/ K,S vi / vi i=1 Y is bijective. →

Proof. Injectivity is Kummer theory: Suppose c ∈ O× , is such that [c] ∈ ker φ. Then √ K,S p consider the extension K( c) ⊂ T. √ For each i, c is a pth power in O× , so v splits completely in K( p c), hence Frob √ v,i √ vi p p ﬁxes K( c) for 1 ≤ i ≤ k + 1. Since the Frobvi generate, we have K( c) = K, so c is a pth power and [c] = 1. Surjectivity then follows by counting orders: RHS has order pk+1, LHS has order p|S| = pk+1

Corollary 15.2. The map

k O× (O× )p O× (O×)p K,S/ K,S vi / vi i=1 Y is surjective. →

Now we can show that√ the condition on the vi is satisﬁed. p × n × M = K( b) D = K × (K ) × ∗ O For this: Let . Then let v∈S v vi vi v∈/S v . Then for × the same reasons as previously, D ⊂ NAM. × × Q Q × Q On the other hand, DK contains DOK,S, which equals AK,S by the previous corollary. × × × × By assumption on S, AK = AK,SK , so DK contains all of AK. By our corollary to the ﬁrst inequality, M = K as required. 0 2 1 Hence we have |H^ (L/K, CL)| = |H (L/K), CL| = p, and |H (L/K, CL)| = 1. By HW, this 1 2 implies that H (L/K, CL) = 1 for any abelian extension L/K, and also |H (L/K), CL| ≤ [L : K]. Some consequences:

26 2 × L 2 × Corollary 15.3. for L/K abelian, the map H (L/K, L ) v H (Lw/Kv, Lw) is injective. Proof. Use short exact sequence 0 K× A× C 0 K K→ Corollary 15.4 (Hasse Norm Theorem). If L/K is a cyclic extension and a ∈ K×, then × × → → × → → a ∈ NLw for all primes w of L implies a ∈ NL . ^ 0 × ^ 0 × Proof. This is just the statement that H (L/K), L H (Lw/Kv, Lw) is injective. Note that the statement that L/K is cyclic is critical here: there are counterexamples √ √ → without, e.g. L = Q( 13, 17), a = −1. 2 What we need now is an invariant map (equiv, a fundamental class) H (L/K, CL) 1 [L:K]Z /Z. We’ll work on building this next time, as a sum of local invariants. → 16 March 2

General comment: for proof of second inequality we really needed characteristic 0, because Kummer theory. At this point, all ﬁelds should be assumed to be number ﬁelds unless stated otherwise. . Today we’re going to work towards proving the following reciprocity laws:

2 × (a) inv-reciprocity: for L/K if α ∈ H (L/K, L ) then v invv α = 0. × P (b) θ-reciprocity: for L/K if α ∈ K then v θLw/Kv α = 0. We’ll prove inv-reciprocity for all L/K Galois,Q and θ-reciprocity for all L/K abelian. Before this, we need some technical lemmas which will let us reduce to the case of cyclotomic extensions.

Proposition 16.1. If K is a number ﬁeld, and α ∈ Br(K) = H2(K, K¯ ×), there exists some cyclic cyclotomic extension L/K such that Res α = 0 ∈ Br(L). (e.g. α is split by L) L Proof. From last time, we know that Br(L) injects into w Br(Lw). By the commutative diagram Br(K) Res Br(L)

Res α ∈ (L ) [L : K ] Br(Kv) Br(Lw) the image of Br w is zero iff w v is a multiple of

invv invw ×[Lw:Kv] Q/Z Q/Z the order of the element invv(α) ∈ Q/Z. So it’ll be enough to prove

27 Lemma 16.2. Let S be a finite set of finite primes of K. Then there exists a totally complex cyclic cyclotomic extension L/K such that m | [Lw : Kv] for all v ∈ S proof of lemma: reduce to case K = Q, by replacing m with m[K : Q]. rest is HW. 2 × 1 OK, now we define invL/K : H (L/K, AL ) [L:K] Z/Z by invL/K(c) = v invLw/Kv c. Compatibilities: P invN/K(inf c) =→invL/K(c)

invN/L(ResL/K c) = [L : K] invN/K(c)

invN/L(Cor c) = invN/K(c)

(eg. this is almost a class formation, except that invL/K is not an isomorphism: in general it’s neither injective nor surjective) We check invN/L(ResL/K c) = [L : K] invN/K(c) which is the hardest of these. Indeed:

( c) = ( c) invN/L ResL/K invNw0 /Lw ResLw/Kv w X = [L : K ] c w v invNw0 /Lv w X = [L : K ] c w v invNw0 /Lv v X Xw|v = [L : K] invN/K(c).

Let’s also deﬁne θL/K : AK Gal(L/K) by θL/K(a) = v θLw/Kv av. These maps are related by Q → Proposition 16.3. χ(θL/K(a)) = invL/K([a] ∪ δχ). for any χ ∈ Hom(Gal(L/K), Q/Z) =∼ H1(L/K, Q/Z), and a ∈ K×.

Proof. Follows from the same statement for local θ and inv, which we proved last semester. There are two related reciprocity properties we want to prove: θ-reciprocity if a ∈ L× 2 × then θL/K(a) = 1. inv-reciprocity invL/K vanishes on H (L/K, L ) For any L/K, inv-reciprocity implies θ-reciprocity. Observe that θ-reciprocity implies inv-reciprocity if L/K cyclic. Indeed choose χ to generate H1(L/K, Q/Z), so δχ generates H2(L/K, Z). For any [a] ∈ H^ 0(L/K, L×) we ^ 0 × have, 0 = χ(θL/K(a)) = invL/K([a] ∪ δχ). But − ∪ δχ is an isomorphism H (L/K, L ) 2 × 2 × H (L/K, L ), so invL/K is 0 on all of H (L/K, L ). → 28 17 March 5

17.1 Checking reciprocity laws We start out by checking θ-reciprocity for K = Q and L a cyclotomic extension: note that for us this will mean that L is any subfield of Q(ζn): but it’s enough to check when L = Q(ζn) In fact, it’s enough to check when L = Q(ζ`r ), since any cyclotomic extension is a compositum of such. We just need to check θ-reciprocity for a generating set of Q×: we check it for a = `, a = p 6= `, and a = −1. a = p 6= ` p−1 • θ`(p) sends ζ 7 ζ (Lubin-tate) p • θp(p) = Frobp (unramified theory) which sends ζ 7 ζ → • θ 0 (p) = 1 (unramified) p → • θ (p) = 1 a = ` ∞ • θ`(`) = 1 (Lubin-Tate) 0 • θp0 (`) = 1 (unramified theory) when p 6= ` • θ (`) = 1 0 • θ∞p0 (−1) all p 6= ` −1 • θ`(−1) sends ζ 7 ζ (Lubin-Tate) • θ (−1) is complex conjugation. → So everything∞ checks out and θ-reciprocity holds for L/Q cyclotomic. It follows that inv-reciprocity holds for L/Q cyclic cyclotomic. Since any element of Br(Q) is represented by some H2(L/Q, L×) where L is cyclic cyclotomic, get that inv-reciprocity holds for K = Q, L arbitrary. Now, suppose that K and L are arbitrary. Let E/Q be a Galois extension with L ⊂ E. Cor inf Then have H2(L/K, L×) −− H2(E/K, E×) − H2(E/Q, Q×), and both maps are compatible with inv, so inv-reciprocity holds for arbitrary L/K. Hence also θ-reciprocity holds. → → × × × Observe: now we have a map θ : AL /L NAK Gal(L/K). One route would be to stop here and show that this map is an isomorphism, but instead we’ll push on and work on showing that the ideles form a class formation,→ which will automatically give us everything.

29 2 × 2 17.2 Invariants on H (L/K, AL ) and H (L/K, CL). From inv-reciprocity, we get maps

inv 1 H2(L/K, L×) , H2(L/K, A×) −− Z/Z L [L : K] and the composition of these two→ maps is 0. Natural→ question, does this form a short exact sequence? Proposition 17.1.

i∗ inv 1 1 H2(L/K, L×) − H2(L/K, A×) −− Z/Z 0 L [L : K] is exact when L/K is→ cyclic. → → → Proof. Surjectivity: need to show that [L : K] is the least common multiple of the local degrees [Lw : Kv]. This follows from the fact that elements Frobv, of order [Lw : Kv] generate the cylic group Gal(L/K) of order [L : K]. ker inv = im i∗: We already have ker inv ⊃ im i∗. To get equality, we count orders, and compare with the following exact sequence:

2 × 2 × j∗ 2 δ 3 × 1 H (L/K, L ) H (L/K, AL ) − H (L/K, CL) − H (L/K, K ) = 1 where the ﬁnal term is 1 by Hilbert 90 and periodicity of cohomology for cyclic groups. → → → 2 → × Surjectivity of inv tells us that the index of ker inv in H (L/K, AL ) is precisely [L : K]. On the other hand, the cohomology exact sequence tells us that index of im i∗ in 2 × 2 H (L/K, AL ) is equal to |H (L/K, CL)|, which is at most [L : K] by the second inequality. 2 × Since we know already ker inv ⊃ im i∗, the index of the former in H (L/K, AL ) is at most that of the latter: comparing with the previous two observations, equality must 2 × hold and ker inv = im i∗ (and both have index [L : K] in H (L/K, AL ). It follows from the previous proof that if L/K is cyclic, the invariant map inv : 2 1 2 H (L/K, AL) [L:K] Z/Z factors through H (L/K, CL) and the induced map, which we denote by 1 → inv : H2(L/K, C ) Z/Z, L [L : K] is an isomorphism. → 2 1 Unfortunately this approach doesn’t work to deﬁne inv : H (L/K, CL) [L:K] Z/Z when L is not cyclic. We’ll have to do the same thing we did last semester, which is to diagram-chase to move inv from the extensions we understand to the ones we→ don’t. We’ll need to show the following lemma: 0 0 2 Lemma 17.2. L/K any Galois extension, L /K cyclic with [L : K] = [L : K], then H (L/K, CL) 2 0 2 S 2 and H (L /K, CL) have same image inside H (K¯ /K, CK¯ ) = M H (M/K, CM).

30 18 March 7

We prove the following lemma which we stated last time. For brevity let H2(L/K) denote 2 H (L/K, CL). Lemma 18.1. L/K any Galois extension, L0/K cyclic with [L0 : K] = [L : K], then H2(L/K) and 2 0 2 H (L /K) have the same image inside H (K¯ /K, CK¯ ). Proof. To show inf H2(L0/K) ⊂ inf H2(L/K): let N = LL0, so N/L is cyclic with [N : L] | [L0 : K]. There’s a short exact sequence: 1 H2(L/K) H2(N/K) H2(N/L). 2 0 2 0 × Suppose c ∈ H (L /K). Then lift c toc ˜ ∈ H (L /K, AL ). 2 We need to show that ResL/K infN→/L0 c = 0 inside→ H (N/L)→: for this enough to check that it has invariant 0.

invN/L(ResL/K infN/L0 c) = invN/L(ResL/K infN/L0 j∗(c˜))

= invN/L j∗(ResL/K infN/L0 (c˜))

= invN/L(ResL/K infN/L0 (c˜))

= [L : K] invL0/K(c˜) = 0. for equality, compare orders and use second inequality.

2 0 1 0 The maps inv : H (L /K) [L0:K]Z /Z for L /K cyclic glue to give an isomorphism

2 2 inv/K : H (K¯ /K) = lim H (L/K) = lim Q/Z. → L/K L/K cyclic

2 1 By the previous lemma, this restricts to an isomorphism inv→L/K : H (L/K) [L:K] > 2 1 We deﬁne the fundamental class uL/K ∈ H (L/K) by inv(uL/K) = [L:K] Observe that the following diagram commutes → 2 × H (L/K, AK ) inv

1 j∗ [L:K] Z/Z inv

2 H (L/K, CK) Now we know that CK is a class formation. (Really we need to check that inv is compatible with inﬂation and restriction, but that’s straightforward.) By Tate’s theorem ^ i ^ i+2 ∪uL/K : H (L/K, Z) H (L/K, CK) is an isomorphism.

→ 31 ab In particular, for i = −2, we get an isomorphism Gal(L/K) CK/NCL. Let θL/K be the inverse of this isomorphism. The cup product is compatible with inﬂation, restriction, corestriction, conjugation, so θL/K is also. → The argument we did last semester for χ(θL/K(a)) = inv(a ∪ δχ) works equally well in any class formation, and this property determines θ. ab 0 To check that θL/K : CK/NCL Gal(L/K) is given by θ([a]) = v θvav, let θ ([a]) = 0 v θvav, and we check χ(θ (a)) = inv(a ∪ δχ): but we’ve previously observed this (using 2 × Q the inv map from H (L/K, A )→, but we’ve seen that the two different inv maps are Q K compatible).

18.1 Existence

Proposition 18.2. θL/K is continuous using the adelic topology on CK and the discrete topology on Gal(L/K)ab

Proof. Two approaches. One is to note that it’s enough to show that ker θL/K = NL/KCL ∼ 0 >0 ∼ 0 >0 0 is closed in CK. This is true because CL = CL × R , CK = CK × R , and NL/KCL is 0 closed in CK because it’s the continuous image of a compact set. Alternatively, use formula for θL/K as a product of local factors and show continuity directly.

The kernel of θL/K is the intersection of all normic subgroups of CK. Where, analogously to last semester, A ⊂ CK is normic if A = NCL for some L/K ﬁnite Galois. Basic facts of normic subgroups move over: normic subgroups correspond 1-1 to ﬁnite abelian extensions of K, A and B normic implies A ∩ B normic, A normic implies A0 supset A normic.

19 March 9

Normic subgroups are finite index (by the main theorem of class field theory) and open (by Proposition refcontinuous). We’ll show that any finite index open subgroup of CK is normic. × For any finite set S of primes of K, let US be the image of v∈S 1 × v∈/S Ov in CK. Any open subgroup of finite index in C must contain (C )nU for some S. K QK S Q Theorem 19.1. Let K be an number field, and S is a finite set of primes of K such that

• S contains the inﬁnite primes and primes dividing n

× • AK,S surjects onto CK.

32 n pn × If K contains µn then (CK) US is the norm group of T = K( (OK,S))/K. n If K does not contain µn then (CK) US is still normic.

Proof. This is going to be a lot like our proof of the second inequality. First we observe that ∼ × × n ∼ |S| Gal(T/K) = Hom(OK,S/(OK,S) , Z/nZ) = (Z/nZ) n n One consequence is that (CK) is in the kernel of θT/K, and so (CK) ⊂ NT/KCT . × × Also T is unramiﬁed outside S, so v∈S 1 × v∈/S Ov ⊂ NT/KAT . This gives one inclusion (C )nU ⊂ NC . K S Q T Q To get the other inclusion, we count orders: By assumption AK,S surjects onto CK. Since

× × AK,S = Kv Ov v∈S Y Yv∈/S × we also have that v∈S Kv surjects onto CK/US, and there is a short exact sequence

Q × × 1 OK,S Kv CK/US 1. v∈§ Y Quotienting out by nth powers→ (or→ equivalently→ tensoring→ with Z/nZ , we get a right exact sequence

× × n × × n n OK,S/(OK,S) (Kv )/(Kv ) CK/(CK) US 1 (2) v∈S Y We can calculate the orders of everything→ here: → → × × n |S| We already know that |OK,S/(OK,S) | = n . × × n 2 −1 For each n, (Kv )/(Kv ) has order n |n|v (We’ve seen this for n prime: exercise to extend this to all n). × × n Multiplying together and using the product formula, we see that v∈S(Kv )/(Kv ) has order n2|S| |n|−1 = n2|S|. v∈S v Q If we show that O× /(O× )n injects into K×/(K×)n, it will follow from (2) that Q K,S K,S v v v n | Q |CK/(CK) US| = p N| = | Gal(T/K)| = |CK/NCT |

N and thus we must have (CK) US = CT . Again, we use Kummer theory: if [a] is in the kernel of the map

× × n × × n OK,S/(OK,S) Kv /(Kv ) , v Y √ n → then L = K( a) is unramiﬁed outside S and totally√ split everywhere in S, meaning that × n × n AK,S ⊂ NAL , and so CK = NCL and K = L = K( a). It follows that a ∈ (OK,S) .

33 That proves the theorem when K contains the nth roots of unity. For the part when K does not contain nth roots of unity, we do the same thing we did for local ﬁelds. 0 0 0 0 n let K = K(µn), S the set of primes above S, then exists L /K such that (CK0 ) US0 = 0 NL0/K0 CL0 . Extend L to L/K Galois. 0 n n n Then NL/KCL ⊂ NK0/KNL0/K0 (CL) = NK0/K(CK0 ) US0 ⊂ (CK) US, so (CK) US contans a normic subgroup, and is itself normic.

19.1 Kernel of θ/K

Now we know that the kernel of θ/K is the intersection of all finite index open subgroups of CK. This must contain the connected component DK of the identity (which we’ve previously described as the closure of vinfinite Kv): on the other hand, CK/DK is compact + totally disconnected, so profinite, and hence the kernel of θ/K is precisely DK. Q ab Also, the image of θ/K is all of Gal(K /K): previously we observed that the image ab of θ/K was dense in Gal(K /K), but since CK/DK is compact, the image must also be closed in Gal(Kab/K), so it must be all of Gal(Kab/K). ab That is, θ/K is an isomorphism CK/DK Gal(K /K). T n Can also describe DK as the set of all divisible elements of CK, that is, n(CK) . (See HW.) → In class, I claimed that all finite index subgroups of CK are open: however, this is false.

20 March 19

We’ve proved global class ﬁeld theory, will be doing applications today + next time. We have an isomorphism

ab θL/K : CK/NL/KCL Gal(L/K) or equivalently, ^ 0 →^ −2 θL/K : H (L/K, CL) H (L/K, Z). This map is compatible with inﬂation, restriction, corestriction. We’ll be particularly interested in the restriction compatibility today:→ let L/M/K be a tower of ﬁelds with L/K Galois. Then we have a commutative square θ L/K ab CK/NL/KCL Gal(L/K)

V where the ﬁrst vertical map is induced by the in- θ L/M ab CM/NL/KCL Gal(L/M) clusion CK , CM, and the second one is the transfer map deﬁned in problem set 6 last

→ 34 semester.

20.1 Hilbert Class Field Let K be a number field. Recall that the Hilbert class field H of K is the maximal extension of K unramified everywhere (including infinite places), e.g. it is the ray class field of ∼ C /NC Gal(H/K) K H θ ∼ modulus 1. We have isomorphisms ∼

Cl(OK). Example. K = Q, H = K = Q. √ √ Example. K = Q( −5), H = K( −5, i). √ 3 Example. K = Q( −23), H = K(α) where α is a root of x − x + 1. (The field H is a S3 extension of Q, the splitting field of the polynomial x3 − x + 1 with discriminant −23.) Basic properties of the Hilbert class field H:

• the element Frobp ∈ Gal(H/K) depends only on [p] ∈ Cl(K)

• p splits completely in OH iff Frobp = 1 iff p ∈ OK is principal.

Corollary 20.1. For p a prime of Q, there exists π ∈ OK with NK/Qπ = p if and only if p splits completely in OH.

Proof. if p does not split completely in OK then neither condition is true. May assume then that p splits completely in OK, write (p) = NK/Qp. If there exists π ∈ OK with NK/Qπ = p we must have p = (gπ) for some g ∈ Gal(K/Q) by unique factorization. Hence such a π exists iff p is princpal. But we know p is principal iff p splits completely in OH, which is the case iff p splits completely in OH, as desired. But the Hilbert class ﬁeld also has another important property, which is that any fractional ideal a of OK becomes principal in OH: that is, aOH is principal. This was the last of Hilbert’s conjectures about the Hilbert class ﬁeld to be settled, in 1929 when Artin reduced it to a group theoretic result (which was in turn proved by Furtwangler). The group theoretic result is

Theorem 20.2 (Principal Ideal theorem of group theory). If G is a ﬁnite group, G0 = [G, G], G00 = [G0, G0], then the transfer map G/G0 G0/G00 is trivial.

We won’t prove it (most class ﬁeld theory books skip the proof, though Artin-Tate → includes a proof). We’ll deduce from this

35 Theorem 20.3 (Principal Ideal theorem of class field theory). The natural map Cl(K) Cl(H) is trivial. → Proof. Let H1 be the class field of H. Observe that H1/K is unramified, and hence that H is the maximal abelian subextension of H. Let G = Gal(H/K). By the previous observation, the subgroup Gal(H1/H) is equal to the commutator subgroup G0 = [G, G], and G00 = {1}. We get a commutative diagram

∼ θ ab 0 Cl(K) CK/NH/KCH Gal(H/K) G/G ∼ ∼

θ ab 0 CK/NH1/KCH1 Gal(H1/K) G/G

V V

∼ θ ab 0 00 Cl(H) CH/NH1/H Gal(H1/H) G /G

By the group-theoretic principal ideal theorem, the transfer map V is trivial, hence the map on the left hand side is also trivial.

21 March 20

21.1 Class field towers Question if K is a number field does it embed in L with | Cl(L)| = 1? one approach: take H0 = H to be the Hilbert class field of K, H1 to be the Hilbert class field of HS0, H3 the Hilbert class field of H2, etc... If this tower eventually stabilizes at some Hn, then Hn must have class number 1. On the other hand, if K has any finite extension L with trivial class group, then HL is an unramified abelian nextension of L, so H ⊂ L. Repeating this argument get that every Hn is contained in L, so the class field tower must stabilize. Hence the answer to this question is yes if and only if the class field tower stabilizes. It was an open question for a while in the middle of the 20th century whether class field tower must always stabilize. As it turns out, the answer is no. (Golod-Shafarevich) The proof looks at the maximal pro-p subextension: inductively, define H0,p to be the maximal everywhere unramified abelian extenison of K with exponent p, and Hi+1,p to be the maximal everywhere unramified abelian extension of Hi,p with exponent p. This tower are called the p-class field tower of K.

Theorem 21.1 (Golod-Shafarevich). If the p-class ﬁeld tower stabilizes, then the p-rank of p Cl(K) is at most 2 + 2 [K : Q] + 1

36 In particular, if p = [K : Q] = 2 this says that if the 2-rank of Cl(K) is ≥ 6, then K has an infinite 2-class field tower. Results of genus theory say that if an imaginary quadratic extension K/Q has k ramified primes, then the 2-rank of Cl(K) is equal to k − 1. (We’ll go into this on the problem set). If a real quadratic extension K/Q has k ramified primes then the 2-rank of Cl(K) is either k − 1 or k − 2. It follows that if K/Q is imaginary quadratic and has at least 7 ramified primes, or real quadratic and has at least 8 ramified primes, then K has an infinite class field tower. Actually, using a slightly sharper form of Golod-Shafarevch, in the imaginary quadratic case can show that 6 ramified primes suffice: see Cassels-Frohlich for details.

21.2 Intro to L-functions Next we’ll be talking about L-functions.

an Deﬁnition. A Dirichlet series : is a series of a form ns .

The most famous example is the Riemann ζ functionP ζ(s) with an = 1 for all n. 1 It’s known for having an Euler product : ζ(s) = p 1−ps , for having meromorphic continuation to the entire half-plane, and for having a functional equation. In general, Q any Dirichlet series with all of those properties is called an L-function, and there are many different types of them, of which we’ll only mention a few. We want to generalize, the notion of Euler product to any number field: if K is a number field, then 1 . (1 − a Np−s) ··· (1 − a Np−s) p 1,p k,p Y where p ranges through all but finitely many primes of OK. Hence Np is the absolute norm of p: this is equal to |OK/p|, and is also the residue field characteristic of the completion Kp. Our first example is the Dedekind zeta function of a number field:

1 s 1 = Na 1 − Nps p aX⊂OK Y 21.3 Digression: zeta functions of function ﬁelds These are in many ways much nicer than number ﬁelds! We can write

1 s 1 ζ (s) = = . Fq[t] Na 1 − Nps p a⊂XFq[t] Y 37 And we can compute this easily, because any ideal of Fq[t] has a unique monic generator, so this is

1 qd 1 = = . qs deg x qds 1 − q1−s d≥1 x∈FqX[t] monic X We observe that the Euler product here can be viewed as a product over all places of K excluding the “place at inﬁnity” (coming from the valuation v(a) = − deg a): 1 ζ (s) = Fq[t] 1 − |k |s v v vY6= Y

and we can add in the place at inﬁnity to get∞ the zeta function of P1(Fq): 1 1 ζ = = . P1(Fq) 1 − |k |s (1 − q1−s)(1 − qs) v v v Y Y If we write t = q−s, this becomes 1 ζ (s) = P1(Fq) (1 − qt)(1 − t)

so we have a rational function, making analytic continuation trivial! We can do the same thing with any curve C over Fq. One can show (one proof uses Riemann-Roch) that you get the same denominator and a different numerator

(1 − α1t) ··· (1 − α2gt) ζ (s) = . C (1 − qt)(1 − t)

The analogue of the Riemann hypothesis for function ﬁelds of curves then says that 1/2 all αi have absolute value q . This was proven by Weil in 1940: he conjectured a generalization to higher-dimensional varieties. Deligne used etale cohomology to proved this generalization in 1974, and won the Fields medal for it.

22 March 23

22.1 Dirichlet L-functions + generalizations

Dirichlet L-functions: Let m be a modulus of OK. Then a Dirichlet characters χ of mod- 1 ulus m is a homomorphism Clm(OK) S , and the associated Dirichlet L-function is χ(a) 1 L(s, χ) = → = (Na)s 1 − (χ(p))−s (aX,m)=1 Yp-m 38 Example: K = Q, m = m , χ is a character Z/mZ× S1. We’ll be spending this section of the class studying Dirichlet L-functions, but before doing that I want to briefly∞ mention a couple generalizations.→ ∼ m 1 Hecke L-functions : instead of taking a character Clm = CK/CK S , use an arbitrary 1 continuous homomorphism ψ : CK S , which can now be surjective rather than having finite image. The Hecke L-function of ψ is → → 1 −s 1 − ψ(πp)Np Yp∈/S where πp ∈ CK is the element represented by the idele which is a uniformizer πp in the p-component, and 1 everywhere else. While Dirichlet L-functions are can be used to show e.g. Dirichlet’s theorem that the residues of primes of Z, are equidistributed in Z/mZ×, Hecke L-functions can be used to show that the arguments of prime ideals of Z[i] are equidistributed. More precisely, any prime ideal of Z[i] can be written as (π) for a unique π in the upper right-hand quadrant, and the arguments of those generators π are equidistributed in the interval [0, π/2]. ∼ Artin L-functions: Recall that Cm = Gal(Lm/K) is the Galois group of the ray class field of modulus m. Hence we can view Dirichlet L-functions as being of the form 1 −s (3) 1 − (χ(Frobp)) Yp-m 1 where now χ is a character Gal(Lm/K) S , and Frobp ∈ Gal(Lm/K). If now we replace Lm by an arbitrary, possibly non-abelian, Galois extension L/K, then 1 (3) still describes an L-function. Of course,→ if χ is just a homomorphism Gal(L/K) S then χ factors through the Galois group of the maximal abelian subextension and we get nothing new. However if we expand our notion of character to include characters→ of irreducible represenations of Gal(L/K), of any dimension, then we get a larger class of L-functions knows an Artin L-functions which play an important role in modern number theory.

22.2 Convergence results for Dirichlet Series

an Given a Dirichlet series ns , we now consider is domain of convergence. b As long as an is O(n ) for some b, the Dirichlet series will converge locally uniformly P to an analytic function in the half-plane Re s > b + 1. This is sharp in the case of the Riemann zeta function, however can often do better e.g. for all non-trivial Dirichlet characters, the Dirichlet series converges locally uniformly for Re s > 1. b Proposition 22.1. Let S(x) = n≤x as be the sequence of partial sums. If S(x) = O(x ), then an s converges locally uniformly to an holomorphic function in the half plane Re s > b. n P P 39 Proof. (Non-rigorous Sketch, see pg 181 of Milne’s Class Field Theory for a full proof). Partial summation: rewrite the sum as 1 1 S(x)( − ) ns (n + 1)s x X 1 1 d −s Then we can approximate the ﬁnite difference ns − (n+1)s by the derivative dx −x |x=n = −s−1 1 1 b−s−1 b.

Proposition 22.2. Can meromorphically continue ζ(s) to the strip Re s > 0, with at most a pole at s = 1.

Proof. We can’t apply the previous proposition to ζ(s) because it’s not holomorphic at

The Dirichlet series for ζ2(s) is holomorphic for Re s > 0 by the previous proposition, so gives a holomorphic. Hence we can meromorphically extend ζ(s) to Re s > 0 by ζ2(s) ζ(s) = 1−2s . This tells us that ζ(s) is holomorphic everywhere in the region except s = 1 + k 2πi k ζ(s) possibly at log 2 for ∈ Z: this is not quite good enough, since we want holomorphic everywhere. So we do the same thing again with 3 in place of 2: let 1 1 2 1 1 2 ζ (s) = ζ (1 − 31−s) = + − + + − + ··· 3 s 1s 2s 3s 4s 5s 6s The same argument as before tells us that ζ(s) is holomorphic at all points of Re s > 0 1 + k 2πi k log 2 except possibly log 3 for ∈ Z. Since log 3 ∈/ Q, combining this our previous result tells us that ζ(s) is holomorphic everywhere in Re s > 0 excepting s = 1, where it may have a pole.

~~1 1 Lemma 22.3. For s real and s > 1, ζ(s) ∈ [ s−1 , 1 + s−1 ], so the Riemann zeta function has a simple pole at s = 1 with residue 1.~~

−s 1 Proof. Consider upper and lower Riemann sums for 1 x = s−1 . ∞ Next time we’ll do the same thing for the DedekindR zeta function ζK(s). We’ll see that it has a pole at s = 1, and that the residue of this pole is related to the class number of K.

~~40 23 March 26~~

b an Proposition 23.1. if S(n) − a0n ≤ Cn , ns can be analytically continued to a meromorphic function on <(s) > b with a simple pole at s = 1 with residue a . P 0 Proof. Use an an − a0 = a ζ (s) + . ns 0 n ns X X

Now want to understand the behavior of general Dirichlet L-functions. Let m be a modulus of K, and let K be a class in Clm(OK). Then we can deﬁne the partial zeta function: 1 ζ(s, K) = . Nas a∈K X This not an L function (no Euler product), but the partial zeta functions of modulus m have the same span as the Dirichlet L-functions of modulus m. Let S(x, K) = #{ ideals a ∈ K | Na ≤ x} Then one can show (though we’ll only sketch)

~~Proposition 23.2. 1−1/d) S(x, K) = gmx + O(x ) where 2r(2π)s reg(m) g = m 1 2 wmNm| Disc(K)| / In particular, it doesn’t depend on K.~~

We haven’t defined all the notation in the definition of gm, so let’s do that now. As usual, r is the number of real places of K and s then number of complex places. rm The norm Nm is equal to N(mfin)2 where mfin is the finite part of m and rm is the number of real places of m. (I forgot the 2rm in class!) The last two factors reg(m) and wm both involve the unit group

~~× × OK,m = {a ∈ OK | a ≡ 1 (mod m) and |a|v > 0for all real v | m}.~~

× First of all, wm is the number of roots of unity in OK,m. × + Then reg(m) = regulator of m: recall we have a map L = OK into v| R given by a 7 | log av1 , ··· log avr+s |: in fact the image is a lattice inside a hyperplane H. The × Q ∞ regulator reg(m) = is the covolume of L(OK,m) inside H. If →m = 1, we write reg(m) = reg(OK), and this is called the regulator of OK.

41 In the case where K is a real quadratic field then reg(m) = log |u| where u ∈ K ⊂ R×, × u > 0 is a generator for OK,m with |u| > 1. We won’t give a full proof of Proposition 23.2, but some comments: To count S(x, K) = #{ ideals a ∈ K | Na ≤ x}, choose any representative c ∈ K. Then any a ∈ K is of the form a = ac where a ∈ c−1 × and a ≡ 1 (mod m), and a is uniquely defined up to multiplication by elements of OK,m. In class we sketched out how this count works when K is imaginary quadratic or real quadratic. × If K is imaginary quadratic: for simplicity assume that wm = 1, so OK,m is trivial. x Then the region of allowable a is the intersection of the circle |Na| < |Nc| with the lattice {a ∈ c−1 | a ≡ 1 (mod m)}, so can use geometry of numbers to count the number of lattice points in a region. Now let K be real quadratic (and again we’ll assume wm = 1 for simplicity). Then OK embeds as a lattice inside R × R, of which {a ∈ c−1 | a ≡ 1 (mod m)} is a sublattice. In this case the set x {a ∈ R × R | |Na| < } |Nc| is a region bounded by hyperbolas, having infinite volume, and containing infinitely many lattice points. However, in this case a is only unique up to multiplication by × elements of OK,m, and one can show that the volume of a fundamental domain is pro- x reg m x log |u| portional to |Nc| = |Nc| .

~~24 March 28~~

Now, we give a sketch proof of Proposition 23.2: Sketch: Let c be any representative of the ray class K. By the argument given last time, S(x, K) counts the number of a ∈ c−1 with a ≡ 1 (mod m) (and a positive at all real x × places of m), such that Na < Nc , modulo the action of OK . × Let F be a fundamental domain for v| Kv/OK . Then we are counting the number of points of a certain lattice that lie in F and have Nf < y. We can estimate this with Q geometry of numbers: ∞ r s {f ∈ F | Nf < x } 2 (2π) reg(m)x x {a ∈ c−1 | a ≡ 1 The volume of Nc is equal to wmNc . The lattice p Nm (mod m)} has covolume | Disc(K)| Nc . Hence 2r(2π)s reg(m) S(x K) = x + O(x1−1/d) , 1 2 wmNm| Disc(K)| / (the error term is proportional to the surface area of F), as desired.

~~42 Combining with Proposition 23.1, we get.~~

~~1 Proposition 24.1. ζ(s, K) has analytic continuation to Re s > 1 − d with a simple pole at s = 1 of residue gm, and no other poles.~~

1 Theorem 24.2. The function L(s, χ) analytically continues to Re s > 1 − d (where d = [K : Q]) If χ is not the trivial character then L(s, χ) is holomorphic at s = 1. If χ is trivial then L(s, χ) has a pole of residue | Clm(OK)|gm.

~~1 Proof. We have L(s, χ) = K χ(K)ζK(s), so it analytically continues to Re s > 1 − d , with only pole at s = 1 with residue χ(K)gm. P K Specializing to m = (1) and χP= 1~~

~~r s 2 (2π) reg(K)hK Theorem 24.3 (Class Number Formula). ζK,s has a simple pole at s = 1 equal to 1/2 wK| Disc(K)| at s = 1.~~

Special cases: when K is imaginary quadratic this is 2πhK : so we actually get a | Disc(K)|1/2 formula for hK. reg(K)hK When K is real quadratic this is 1/2 . wK| Disc(K)| The Class Number Formula is important for a number of reasons: one is that it’s analogous to the Birch and Swinnerton-Dyer conjecture (see Zagier “The Birch-Swinnerton- Dyer Conjecture from a Naive Point of View”, and Lemmermeyer “Conics: A poor man’s elliptic curve”)

~~24.1 Densities for sets of primes~~

~~Let T be a set of primes of OK. There are multiple reasonable notions of density for T.~~

~~1 1 + Deﬁnition. Dirichlet density: if p∈T Nps = δ log 1−s + O(1) as s 1 , then T has Dirichlet density δ. P #pinT|Np≤x → Deﬁnition. The natural density is limx p|Np≤x .~~

→∞1 −1 n Deﬁnition. Polar density: if ( p∈T (1 − Nps ) ) can be analytically continued to a function with a pole of order m at s = 1, then the polar density of T is m/n. Q All three densities have the following properties

~~• They are monotonic where deﬁned.~~

~~• ﬁnite additivity~~

~~• if T has density 0, so does any subset of T.~~

43 • ﬁnite sets have density 0 Veriﬁcations of these are straightforward. Proposition 24.4. If natural density exists, then so does Dirichlet density and the two are equal . If polar density exists, then so does Dirichlet density and the two are equal. Sketches. For natural density, use partial summation. For polar density, use 1 1 log( (1 − )−1) = + O(1) Nps Nps p∈T p∈T Y X as s 1+ Remark. The set of primes of Z with leading digit 1 (in decimal) has a Dirichlet density log 2→ ( log 10 but no natural density.) Remark. Polar density must be a rational number, whereas natural and Dirichlet densities can be irrational.

Proposition 24.5. The set of all primes of OK has density 1 in all three deﬁnitions. Proof. This is clear for natural density, so must also hold for Dirichlet density. 1 −1 ζ (s) = (1 − s ) For polar density, note that we’ve proved that K p⊂OK Np has a simple pole at s = 1. Q

~~25 March 30~~

~~We’ll be working with Dirichlet densities here: the sets we’ll be considering will also have natural density, but that would require more careful error-bounding.~~

Proposition 25.1. The set T of primes p of OK such that Np is not a prime of Z has polar density 0. Proof. Let d = [K : Q]. Write d 1 j (1 − )−1 = (1 − p−js)−#{p|Np=p } Nps p∈T j=2 p Y Y Y Because {p | Np = pj} ≤ d always, this is a sub-product of ζ(2s)dζ(3s)d ··· ζ(ds)d, which converges for ~~1/2. Hence 1 (1 − )−1 Nps p∈T Y also converges to a holomorphic function on Res s > 1/2.~~

44 Proposition 25.2. Let L be a finite Galois extension of K. Then the set of primes of K that split 1 completely in L, denoted Spl(L/K), has polar density [L:K] . Proof. Let S be this set, and let T be the set of primes above S. We observe if pL is a prime of L with NpL prime, and which does not lie above a ramified prime of K, then pL ∈ T. For this note that if we let pK be the prime of K under [L:K]/g L, we have NpL = NpK , where g is the number of primes of L above K. Hence NpL is only prime in the case g = [L : K], which means that pK ∈ Spl(L/K). Then T has polar density 1 by the previous proposition and the fact that there are [ only finitely many ramified primes. But ζL,T (s) = ζK,S(s) L : K], so S must have polar 1 density [L:K] . Now we work towards the proof of the second inequality. Note that we haven’t used any class field theory so far.

Theorem 25.3. Let H be any subgroup of Clm = Clm(OK), and let G = Clm(OK)/H . Then at most one character χ of H can have L(1, χ) = 0, and it must be a simple zero. 1 The Dirichlet density δ(p | [p] ∈ H) is |G| if L(1, χ) 6= 0 for all characters of Clm /H, and 0 if L(1, χ) = 0 for some χ. 1 + Proof. Consider the behavior of |G| χ log(L(s, χ)) as s 1 where χ ranges over all characters of G. P This is → ! 1 1 χ − log(1 − χ(p)Np−s) = χ(p)Np−s + O(1) = O(1) |G| |G| p p χ X X X X (we skip the details of checking convergence). On the other hand, it’s also log s χ ords=1 L(s, χ) + O(1). P To get rid of the annoying possibility that some L(1, χ) might have a zero at s = 1, we’ll have to use class ﬁeld theory. But ﬁrst we’ll give the analytic proof of the second inequality. Theorem 25.4. L/K Galois, m a modulus: × × Then |CK /NL/KCL Um| ≤ [L : K]. × × × Since we can take m such that Um = CL , we get also that CK /NL/KCL ≤ [L : K]. × × ∼ Proof. G = CK /NL/KCL Um = Clm(OK)/H, where H is the subgroup of Clm(OK) generated by ideals which are norms from L. 1 We know that Dirichlet density δ{p | [p] ∈ H} is either |G| or 0. On the other hand, if p splits completely in L, then [p] ∈ H, so the density must be 1 ≥ [L:K] , and we must have |G| ≤ [L : K]. As a corollary, we see that for any nontrivial character χ of G, L(1, χ) 6= 0.

45 Corollary 25.5. If χ is a nontrivial character of Clm(OK) then L(1, χ) 6= 0. × Proof. In the setup of the previous theorem, take L such that NL/KCL ⊂ Um. Then G = Clm(OK) and we see, as in the proof above, that for any nontrivial character χ of G, L(1, χ) 6= 0.

~~26 April 2~~

26.1 Cebotarev Density Theorem Setup: let L/K be a Galois extension of number ﬁelds. Then we have a map (unramiﬁed primes of K) (conjugacy classes of Gal(L/K)) given by p 7 [Frobp]. → Theorem 26.1 (Cebotarev Density Thoerem). For any conjugacy class [g] of G = Gal(L/K), → the set of primes Tg = {p | [Frobp] = [g]} |[g]| has Dirichlet density |G| .

We’ve already proved this for the special case where g = 1: in that case T1 = Spl(L/K). We ﬁrst apply the results from last time to get the case when L/K is abelian, and then we’ll bootstrap from there using the same argument that worked when g = 1.

Proposition 26.2. Let C be any subset of Clm = Clm(OK). Then the density of {p | |[p] ∈ C}is |C| | Clm | Proof. Enough to do when C = {K} is a singleton. Then, by the same argument used in the proof of Theorem 25.3

1 1 1 + s = χ(K) log L(s, χ) + O(1) = log(1/(1 − s)) + O(1) as s 1 Np | Clm | | Clm | p∈K χ X X → χ m 1 (1 (1 − s)) where runs over all characters of modulus . Here the | Cl(m)| log / comes from L(s, 1), and all other L-functions are O(1) as s 1+.

~~Corollary 26.3 (Abelian Cebotarev Density). Let L/K be an abelian extension. Then L/K → satisﬁes the Cebotarev density theorem.~~

Proof. Pick m with L ⊂ Lm, and let H ⊂ Clm be the subgroup corresponding to Gal(L/K) ⊂ Gal(Lm/K). Then Frobp = g ∈ Gal(L/K) if and only if [p] ∈ gH, and the result follows from the previous proposition.

46 Proof of Full Cebotarev Density. We’ll ignore all primes of K that ramify in L, since there are only finitely many such. Let L/K = n, and let the order of g be m. Let M be the intermediate field Lhgi fixed by the cyclic subgroup generated by. g, so L/M is cyclic of degree m. Importantly, L/M is abelian, so we’ll be able to use Abelian Cebotarev. On the other hand M/K might not even be Galois, but that will be fine. First we look at what it means for p to belong to the set Tg. If that is the case then [Frobp] = [g], and for some prime pL above p, the Frobenius element of Gal(L/K) at pL is

equal to g. In particular, this means that g generates the decomposition group DpL , and M is the decomposition ﬁeld of pL. Let pM be the unique prime of M under pL. Then the inertia degree fM/K = 1, and

FrobpM = g ∈ Gal(L/M). Let TM,g be the set of pM satisfying these properties. If we have any pM ∈ TM,g, then p = pM ∩ OK belongs to Tg (but the map pM 7 p is not one-to-one). Observe that 1 1 1 → = + O(1) = log(1/(1 − s)) + O(1) (4) Np−s Np−s m p ∈T M M MXM,g FrobX(pM)=g by abelian Cebotarev. Now we need to move down to K. To do this we need to know how many pM ∈ TM,g lie above some p in Tg. Since pL is uniquely determined by pM, it’s enough to count the

~~number of pL lying above p with FrobpL = g.~~

~~For this, ﬁrst choose some pL,0 with FrobpL,0 = g. Any other prime of L lying above p~~

takes the form pL = hpL0 , where h is unique up to right multiplication by powers of g. −1 Then Frobhp = hgh , which equals g iff h is in the centralizer of g. So the number of L0 |C(g)| n possible pL is equal to m = |[g]|m . Combining with (4), we get that 1 |[g]|m 1 |[g]| = = log(1/(1 − s)) + O(1) Np−s n Np−s n p∈T p ∈T M Xg MXM,g 26.2 Splitting sets of an extension Recall that for L/K a ﬁnite Galois extension, Spl(L/K) denotes the set of primes of K that split completely in L. The notation Spl(L/K) still makes sense if L/K is not Galois. Exercise: Spl(L/K) = Spl(L0/K) where L0/K is the Galois closure of L/K. 0 1 Hence if L/K is not Galois the Dirichlet density of Spl(L/K) = Spl(L /K) is [L0:K] . Galois extensions are entirely characterized by their splitting sets. Theorem 26.4. If L, L0 are Galois extensions of K, then Spl(L/K) = Spl(L0/K) implies L = L0. Proof. If Spl(L/K) = Spl(L0/K) then Spl(LL0/K) = Spl(L/K). But Spl(LL0/K) has density 1 1 0 0 [LL0:K] and Spl(L/K) has density [L:K] , so [LL : K] = [L : K] and L ⊂ L. Likewise L ⊂ L and L = L0.

47 Note that this fails if L and L0 are not Galois: in fact, there exist non-isomorphic Galois extensions L, L0 of a number ﬁeld K such that every prime of K has the same splitting behavior in L as in L0. (See pages 362-363 of Cassels-Frohlich for an outline.)

~~27 April 4~~

~~27.1 Intro Complex Multiplication~~

ab S Recall that for Q Kronecker-Weber gives us an explicit description of Q = n Q(ζn). More specifically, the field Q(ζn) is the ray class field of modulus n infinity. The reason this works is that we’re adjoining torsion points of an algebraic group. In this case, the algebraic group is the multiplicative group Gm, which has endomorphism ∼ ring End(Gm) = Z. For every n we have the torsion subgroupGm[n] = µn of points ∼ killed by the nth power map. Then we have an inclusion Gal(Q(ζn)/Q) , Aut(Gm[n]) = (Z/nZ)×, which we’ve seen is an isomorphism. Ideally we’d like to generalize this setup with Q replaced by some other→ number field K. (Ultimately this will only work for K imaginary quadratic.) Hovewer, if K 6= Q, Kab is larger than K(ζ ) (In class I noted that we know that if K is imaginary quadratic, then ab 2 K will contain a Zp extension, but K(ζ ) won’t. I’m not sure what the best way to proceed for general∞ K is, but you can probably just find a quadratic extension of K that isn’t contained in K(ζ ).). ∞ So we’ll want to replace Gm by some other one-dimensional algebraic group. We don’t have many choices:∞ in fact, the only other thing we can really do is take E to be an elliptic curve defined over K. If we just take E to be a general elliptic curve defined over K, then the n-torsion subgroup E[n] is isomorphic to (Z/nZ)2. The extension K(E[n]) then has Galois group ∼ embedding into Aut(E[n]) = GL2(Z/nZ), so this is not a reliable way of producing abelian extensions. Instead, as in the Lubin-Tate theory, we will want to consider elliptic curves with extra endomorphisms. We’ll later see the important theorem:

~~Proposition 27.1. If E is an elliptic curve over C, then the endomorphism algebra End(E) is isomorphic either to Z or to an order O in an imaginary quadratic ﬁeld.~~

In the latter case we say E has CM by O. So we might try the following: let K be an imaginary quadratic field. Suppose we can find an elliptic curve E defined over K with CM by OK, and such that all endomorphisms of E are defined over K. Then for any ideal a of OK let E[a] be the a-torsion subgroup of E, namely E[a] = {x ∈ E(C) | ax = 0 for all a ∈ a}.

~~48 ∼ ∼ × Then one can show E[a] = OK/a as OK/a-modules, and AutOK (E[a]) = (OK/a) . Hence we’d get an injection ∼ × Gal(K(E[a])/K) , AutOK (E[a]) = (OK/a) ,~~

K(E[a]) K meaning that here is legitimately→ an abelian extension of . The bad news here is this setup works out for only finitely many imaginary quadratic fields K: in fact, exactly those with class number 1. The good news is if K is an imaginary quadratic field, then any elliptic curve E with complex multiplication by E is defined, not over K, but over the Hilbert class field H of K. In fact, H = K(j(E)) is generated by 0 j j(E) K 0 (x − j(E )) the -invariant, and the minimal polynomial for over is E CM by OK . So this already gives us a way to get explicit generators for H. Q Note that this means that the number of elliptic curves with CM by OK is finite and equal to hK. We can see this bijectively as follows: if [a] ∈ Cl(K), then the quotient C/a of the complex numbers by the lattice a is an elliptic curve with CM by OK. If we want to go beyond H and construct the ray class field Lm, then we follow the plan sketched above. Let E be any elliptic curve with CM by O. We know that E is defined over H, so the field H(E[m]) generated by adjoining coordinates of the m-torsion points of E is an abelian extension of H, which one can show is unramified only over primes dividing m. By the argument above H(E[m]) is an abelian extension of H, and we × have an injection Gal(H(E[m])/H) , (OK/m) : this should in fact be an isomorphism. It’s not quite true that H(E[m]) = Lm, as claimed in class. The problem is that H(E[m]) might not be an abelian→ extension of K. What is true is that H(E[m]) ⊃ Lm and [H(E[m])/H : Lm] ≤ 6, so we’re not very far off. In support of this, note that Gal(Lm/H) is the kernel of the natural map Clm(OK) × × Cl(OK), which is in turn the quotient of (OK/m) by the image of OK . On the other ∼ × hand, Gal(H(E[m])/H) = (OK/m) , so this is consistent with H(E[m])/H being a sub-→ group of Gal(Lm/H) (and note this means that the degree [H(E[m])/H : Lm] is at most 6). We’ll now develop enough of the theory of elliptic curves to be able to prove all this. We’ll start from the analytic point of view, via elliptic functions. References for this section are Cox Primes of the Form x2 + ny2 and Silverman Advanced Topics in the Arithmetic of Elliptic Curves. I’ll start out following Cox, but eventually we’ll go beyond what Cox does

27.2 Elliptic Functions Let L be a lattice in C. Then an elliptic function is a meromorphic function on C/L. Equiv- alently, if ω1, ω2 are generators for L, an elliptic function is a meromorphic function f on C with f(z) = f(z + ω1) = f(z + ω2).

49 Weierstrass ℘-function. 1 1 1 ℘(z, L) = + − . z2 (z − ω)2 ω2 ω∈XLr{0} Properties: Fix L and write ℘(z) = ℘(z, L). Prop: ℘(z) is an elliptic function for L with double poles at the points of L. L C r > 2 G (L) = 1 Exercise: if is a lattice in and is a positive integer then r ω∈Lr{0} ωr converges absolutely. (This is an Eisenstein series for L.) 1 P Then, assuming z, z−ω bounded, 1 1 ω2 − (z − ω)2) − = = O(ω−3) (z − ω)2 ω2 ω2(z − ω)2

~~so the series for ℘(z) converges locally uniformly and absolutely on C2 − L. ω L z = ω ℘(z) = 1 + Likewise, for any ∈ , in the neighborhood of , we have (z−ω)2 holomorphic.~~

~~28 April 6~~

28.1 More on the Weierstrass ℘-function So far we’ve just deﬁned ℘(z) = ℘(z, L): now we’ll check some properties and see that it is in fact an elliptic function. First, we have ℘(−z) = ℘(z): clear since ω ∈ L iff −ω ∈ L. Now, to check periodicity, observe that for any ω ∈ L, the function ℘(z, L) − ℘(z + ω, L) is entire, so must be constant. But picking z = −ω/2 we have that the constant must be 0. Hence ℘(z, L) is an elliptic function. The Weierstrass ℘-function does not generate the ﬁeld of meromorphic functions on C/L: to do that we must also add in the derivative 1 ℘0(z) = −2 . (z − ω)3 ω∈L X This is a function on C/L with a pole of order 3 at 0: hence it must have three zeroes. 1 Can check that these occur at the three nonzero points of 2 L/L. We’ll be able to get an algebraic relation between these two. For this it will be useful to get a power series for ℘(z) at the origin. Lemma 28.1. 1 ℘(z) = + (2n + 1)G z2n. z2 2n+2 n≥1 X L −n Recall Gn = ω∈L−0 ω .

50 Proof. Sum 1 1 − = (m + 1)ω−m−2zm (z − ω)2 ω2 m≥1 X over all ω ∈ L r 0, and observe that the terms with even exponent vanish. Can check that 0 2 3 (℘ (z)) = 4℘(z) − g2(L)℘(z) − g3

where g2 = 60G4(L) and g3(L) = 120G6(L): check this by checking that the difference between both sides is entire and vanishes at the origin. As a corollary, get (exercise: by induction) that all G2n(L) are polynomials in g2(L) and g3(L). If we put a grading on the ring generated by g2 and g3 by saying that g2 has weight 2 and g3 has weight 3 then G2n has weight n. 2 2 If we let E be the elliptic curve with equation y = 4x − g2x − g3, the map z 7 (℘(z), ℘0(z)) is a map of Riemann surfaces C/L E. We’ll next show that it’s an isomorphism. → → Proposition 28.2. ℘(z) = ℘(w) iff z ≡ ±w (mod L).

Proof. View ℘(z) − ℘(w) as a function of z ∈ C/L. It has its only pole of order 2 at z = 0, so must have exactly two zeroes (with multiplicity). If w ∈/ L/2 then these are two single zeroes at ±w, if w ∈ L/2 then this is one double zero at w.

~~Corollary 28.3. The map C/L E given by z 7 (℘(z), ℘0(z)) is an isomorphism.~~

Proof. First we show injectivity: suppose (℘(z), ℘0(z)) = (℘(w), ℘0(w)) . By the previous → → proposition we have w ≡ ±z (mod L). If z ∈ 1/2L then we’re done. Otherwise, 0 ∼ It then follows that z 7 (℘(z), ℘ (z)) is injective, so gives an isomorphism CL = E.) Consequence: any elliptic function is a rational function in p, p0. → Proposition 28.4. Addition Theorem:

~~1 p0(z) − p0(w)2 p(z + w) = −p(z) − p(w) + 4 p(z) − p(w)~~

~~Sketch. Viewed as a function of z this is holomorphic everywhere and vanishes at the origin.~~

~~Deﬁne discriminant 3 2 ∆(L) = g2 − 27g3 2 3 this is the discriminant of polynomial 4x − g2x − g3 and is never 0. 3 g2 and deﬁne j(L) = 1728 ∆ .~~

~~51 Theorem 28.5. If L and L0 are lattices in C then j(L) = j(L0) iff L0 is homothetic to L.~~

0 2 0 Proof. Only if direction is clear. If j(L) = j(L ) then there exists λ with g2(L) = λ g2(L ), 3 0 n 0 g3(L) = λ g3(L ). As consequence G2n(L) = λ G2n(L ). 0 Hence ℘(λz) = ℘L0 (z). Comparing poles we have λL = L . Now we study which C/L have extra endomorphisms. Terminology: an isogeny from E E0 is a nonzero homomorphism of complex Lie groups. Note that if φ : C/L C/L is an isogeny, then φ lifts to a homomorphism φ˜ : C C, and φ˜ must→ be multiplication by some α. → → Theorem 28.6. L a lattice, ℘(z) = ℘L(z), α ∈ C not an integer. Then TFAE:

~~a) The multiplication by α map C C induces an isogeny C/L C/L.~~

~~b) αL ⊂ L → → c) ℘(αz) is a rational function in ℘(z)~~

~~d) There is an order O in an imaginary quadratic ﬁeld K such that α ∈ O and L is homothetic to a proper fractional O-ideal.~~

~~29 April 9~~

~~We prove the theorem stated last time.~~

Proof. a) b) is clear. For c) implies b), if ℘(αz) is a rational function in ℘(z), then the set of poles of ℘(αz) −1 −1 must be invariant⇔ under translation by L. But that set is α L, so must have α L ⊃ L, equivalent to b) a) implies c): pull back the function ℘(z) by the map α : C/L C/L to get that ℘(αz) is a meromorphic function on C/L, that is, an elliptic function for L. We also have ℘(αz) = ℘(−αz), so ℘(αz) is an even elliptic function for L. Exercise:→℘(z) generates the ﬁeld of even elliptic functions on C/L, and the implication follows. d) implies b) is clear. for b) implies d): wlog 1 ∈ L. Then, since αL ⊂ L, we have Z[α] ⊂ L. Hence Z[α] is a Z-module of rank 2, so α must be an algebraic integer of degree 2. Also, since Z[α] is discrete in C, the ring K(α) must be an imaginary quadratic ﬁeld. Let O = Z[α]: we 1 have L ⊂ O, and both are lattices in C, so we must also have n O ⊂ L for some L. As well, OL ⊂ L, so L is a fractional ideal of O, as desired.

52 Now we do some facts about orders in imaginary quadratic fields and their ideals. First of all, if O is an order in an imaginary quadratic field K, the index f = [OK : O] is called the conductor of O, and O = Z + fOK. (The ⊂ inclusion is clear, and both rings have index f inside O.) If O is an order in an imaginary quadratic field, then the class group Cl(O) is the quotient of the group of all invertible fractional ideals of O, modded out by all principal fractional ideals. Let h(O) = | Cl(O)|. Exercise: a fractional ideal a of a quadratic order O is invertible if and only if a is not 0 0 a fractional ideal of O for O ⊃ O. Equivalently, the set EndO(a) = {x | xa ⊂ a} is equal to O. This means that if L ⊂ C is a lattice such that C/L has complex multiplication, there is a unique imaginary quadratic order O such that L is homothetic to an ideal of O. This order O is given by {α ∈ C | αO ⊂ O} = End(C/L).

~~Theorem 29.1. Let O be an order in an imaginary quadratic ﬁeld, and let a be a proper fractional O-ideal. Then j(a) is an algebraic number of degree at most h(O).~~

Proof. Consider the set S = {j(E) | End(E) =∼ O} ⊂ C: this set is invariant under Aut(C). We now use the fact that any isomorphism between subfields of C can be extended to all of C (proof: Zorn’s lemma). Hence the set C r Q¯ of all transcendentals forms a single infinite orbit for Aut(C), and so the finite set S can’t contain any transcendentals. Now we know S ⊂ Q¯ is invariant under Gal(Q¯ /Q) and has size equal to h(O). It follows that any element of S is an algebraic number of degree ≤ h(O). In particular, a ∈ S so we’re done. Alternate argument to show that j(a) is algebraic: (don’t read unless you actually care, other argument is better) assume j(a) is a transcendental complex number. The short version is if this were the case we would have a elliptic curve with CM defined over a transcendental extension, and we could specialize to obtain infinitely many distinct elliptic curves with complex multiplication by O. 2 3 The long version: the elliptic curve C/a has a model E : y = 4x − c2x − c3 in Weierstrass form with c2, c3 defined over the extension Q(j(a)) of Q. Write O = Z[α], and let α have minimal polynomial x2 − bx − c = 0. Let α : E E be the isogeny induced by multiplication by α: we have

α2 − bα − c = 0 (E) → ∈ End . Let K be a field over which α is defined. Then K is a finitely generated extension of Q, so is isomorphic to the field K0(x1, ... , xk) for some finite extension K0/Q and transcendental vari- ables x1, ... , xk. The transcendental element j(a) ∈ K will be some non-zero rational function in K0(x1, . . . xk). Now, view E as an elliptic curve over K0(x1, . . . xk). Specializing and setting x1 = a1, ... , xk = k ak for any (a1, ... , ak) in some Zariski-open subset of C , will give a elliptic curve Ea1,...,ak

~~53 2 with an isogeny α = αa1,...,ak : Ea1,...,ak Ea1,...,ak satisfying α − bα − c = 0. By varying~~

(a1, ... , ak), we can get inﬁnitely many non-isomorphic elliptic curves Ea1,...,ak with complex multiplication by O (and possibly also some→ larger order). But we know that any elliptic curve with complex multiplication by O must be isomorphic to C/a for some fractional ideal a of O (in this case a need not be invertible). But there are only ﬁnitely many such up to homothety, a contradiction. Note that this proof also shows that any elliptic curve E with complex multiplication has (E) ∈ Q¯ .

Later we’ll show this is sharp, and that j(E) generates the ring class field of O, which we now define. If O = OK the ring class field of O is the Hilbert class field of O. More generally, if O is an order, then the ring class field LO is the abelian extension of K defined by the condition that × Gal(LO/K) = CK/ Uv vYfinite × × where Uv is the closure of O in the ring of integers Ov of the localization Kv . × In particular O can be written as Z + fOK, and then Uv = (Zp + fOv) . Observe that LO ⊂ Lf where Lm is the ray class field of K with modulus f.

~~30 April 11~~

So far we’ve seen that if L is a lattice with End(L) = O is larger than Z (“L has complex multiplication by O”), then O is an order in an imaginary quadratic ﬁeld, L is homothetic to some fractional ideal a of O (write L ∼ a), and j(L) = (a) is an algebraic number of degree at most equal to h(O). We want to improve this by showing that j(L) is an algebraic integer of degree precisely equal to h(O), and that K(j(L)) is the ring class ﬁeld LO. We’re going to be giving Deuring’s original proof of this fact, roughly following the exposition in Cox’s book. Three key ingredients: First: Characterization of lattices with complex multiplication using cyclic sublattices.

~~Deﬁnition. A sublattice L0 of L with L/L0 =∼ Z/mZ is called a cyclic sublattice of index m.~~

Claim: If L has complex multiplication , then we can always find an integer m (in fact, infinitely many such!) and a cyclic sublattice L0 of L such that L0 is homothetic to L. (Proof: Suppose L ∼ a, where a is an ideal of O = Z + fOK. Let m = p be a prime relatively prime to a and f such that p splits in the field LO. Exercise: this implies

54 that p = ππ¯ for π ∈ O. Then let L0 = πO: certainly L0 is homothetic to L, and also 0 ∼ ∼ L /L = OK/πOK = Z/pZ.) Second: Modular forms and q-expansions: We’ll use these to show that there exists an equation Φm(X, Y) ∈ Z[X, Y] the modular equation of level m such that if L is a lattice in C and L0 a cyclic sublattice of index m, 0 then Φm(j(L), j(L )) = 0. Combining this with the ﬁrst part, we get if L has complex multiplication, then Φm(j(L), j(L)) = 0 for some m. This immediately gives integrality. Third: It still remains to show that K(j(L)) = LO. We’ll do this by checking that the same primes of K split in both ﬁelds. (Actually we’ll have to be a little more careful than that, because we won’t yet know that K(j(L)) is Galois.)

~~30.1 The j-function as modular function~~

~~Theorem 30.1. For any τ in the upper half-plane H, Deﬁne g2(τ) = g2([1, τ]), g3(τ) = g3([1, τ]) and j(τ) = j([1, τ]) as the corresponding functions of the lattice in C generated by 1 and z.~~

aτ+b a b Let SL2(Z) act on H by γ(τ) = cτ+d where γ = c d . Note that if [1, τ] = L then [1, γτ] = (cz + d)−1L is homothetic to L. a b We then observe that for any matrix γ = c d ∈ SL2(Z), we have g2(γ(τ)) = (cτ + 4 3 d) g2(z), g3(γ(τ)) = (cτ + d) g2(z). aτ+b Finally, j(τ) = j( cτ+d ), and so j descends to a function on the modular curve Y(1) = SL2(Z)\H. Draw standard fundamental domain for SL2(Z)\H, and observe that Y(1) is a punctured sphere. We can compactify Y(1) to a compact riemann surface X(1) by adding a point at infinity. The function field of X(1) is then the field of modular functions defined by

Deﬁnition. A meromorphic function f(τ) on the upper half-plane is a modular function for SL2(Z) if f(τ) = f(γτ) if f(γτ) = f(τ) for all γ ∈ SL2(Z), and if f(τ) can be written as n 2πiτ a Laurent series n≥−k cnq in q = e that converges for Re τ sufﬁciently large. P 31 April 13

~~Formulas: let q = e2πiτ. Then~~

~~(2πi)2k G (τ) = G ([1, z]) = 2ζ(2k) + 2 σ (n)qn 2k 2k (2k − 1)! 2k−1 n≥1 X~~

~~where σ2k−1(n) is the sum of the (2k − 1)st powers of the divisors of k. (Proof will be on HW.)~~

55 Specifically, 4 g (τ) = 60G (τ) = π4(1 + 240 σ (n)qn) 2 4 3 3 and X 8 g (τ) = 120G (τ) = π6(1 + 504 σ (n)qn). 3 6 27 5 X ∆(τ) = (2π)12 τ(n)qn n≥1 X has integer coefficients. 1 1 j(z) = + c(n)qn = + 744 + 196884q + ··· q q n≥0 X where all coefficients are integers. Hence j(z) in in fact a modular function – indeed, j is a meromorphic function on X(1) with a simple pole at the cusp. Hence j induces an isomorphism X(1) CP1 In particular, this means that → Theorem 31.1. The field of modular functions for SL2(Z) is equal to C(j(z)).

~~Modular curves of higher level:~~

~~a b Γ (m) = { | c ≡ 0 (mod m)} 0 c d~~

~~Y0(m) = Γ0(m)\H.~~

~~0 0 Proposition 31.2. Y0(m) = Γ0(m)\H parametrizes pairs of lattices L, L ⊂ C where L ⊂ L is cyclic of index m, up to homothety~~

Proof. The parametrization is given by τ 7 [1, τ], [1, mτ]. Easily seen to be well-deﬁned and surjective. We already know that SL2(Z)\H parametrizes→ latties L, so it’s enough to show that a b for γ = c d ∈ SL2(Z) [1, mτ] ∼ [1, mγτ]

(∼ denotes homothety) iff γ ∈ Γ0(m). But m(aτ + b) [1, mγτ] = [1, ] ∼ [cτ + d, m(aτ + b)] cτ + d and the latter is an index m sublattice of L, so it equals [1, mτ] if and only if it is contained in [1, mτ], which happens exactly when c ≡ 0 (mod m).

56 Compactify Y0(m) to X0(m) by adding in points at cusps (there will be multiple cusps now). A modular function for Γ0(m) is a meromorphic function on X0(m). Equivalent condition: f is a function on H with f(γτ) = f(τ) for any γ in Γ0(τ), and 1/m also for any γ ∈ SL2(Z), f(γ(τ)) is a Laurent series in q . Of course any modular function for SL2(Z), e.g. j(τ) is also a modular function for the subgroup Γ0(m). On the other hand, j(mτ) is a modular function for Γ0(m), since we’ve seen that the homothety type of the lattice [1, mτ] is Γ0(m) invariant. Observe that X0(m) is a ramiﬁed cover of X(1) with degree equal to (HW!) 1 [SL (Z): Γ (m)] = m (1 + ). 2 0 p Yp|m

Denote this degree by dm. Hence the field of modular functions for Γ0(m) is a degree dm extension of C(j(z)). In particular, this tell us already that there’s some algebraic relation between j(z) and j(mz), of degree at most dm. Caution: the covering of curves X0(m) X(1) is not normal, and likewise, the extension of function fields is not Galois. → Proposition 31.3. j(τ) and j(mτ) generate the field of modular functions for Γ0(m).

Proof. It’ll be enough to show that [C(j(τ), j(mτ)) : C(j(τ)] ≥ dm. For this, we use the following strategy: if L/K is a separable field extension (not necessarily Galois), to show [L : K] ≥ d it’s enough to exhibit a (possibly infinite) extension field E/K and d distinct embeddings L , E extending the fixed embedding K , E. In our case, K = C(j(τ)), L = C(j(τ), j(mτ)), and E is the field of meromorphic functions on the upper half plane. We→ aready have a natural embedding of →K into E. Then for any [γ] ∈ Γ0(m)\ SL2(Z), define an embedding φγ : L , E by

~~j(τ) j(γ(τ)) = j(τ) j(mτ) j(mγτ) 7 , 7 →. [1 mγτ] These embeddings are all→ distinct, (the lattices →, are generically distinct, so have distinct j-invariants), and the result follows.~~

~~32 April 16~~

Our agenda here is to construct a minimal polynomial Φm(X, Y) such that Φm(j(mτ), j(τ)) = 0. This polynomial is known as the modular polynomial or modular equation. From what we did at the end of last time, will want Φm(j(mγτ), j(τ)) = 0 for every γ ∈ SL2(Z).

~~57 Let~~

0 fm(X, τ) = (X − j(mγτ)) = (X − j(L )) 0 γ∈Γ0(mY)\ SL2(Z) L ⊂[1,τ] cyclicY index m This is a polynomial in X whose coeffs are functions of τ. We’ll show that actually this can be written as fm(X, τ) = Φm(X, j(τ)) with Φm ∈ Z[X, Y]. i First, observe that fm(X, τ) = fm(X, γτ) so each coeff of X of fm(X, τ) is a holomorphic function on SL2(Z)\H. To show that these coefﬁcients are actually modular functions, we need to show that they can be expressed as power series in q = e2πiτ. We’ll work on making our formula for fm(X, τ) more explicit, so that we can show that the coefﬁcients of fm(X, τ) are not just modular function, but elements of C[j].

32.1 Classiﬁcation of cyclic index m sublattices Let L = [1, τ]: then any cyclic index m sublattice L0 ⊂ L must have generators of the form [d, aτ + b] where ad = m, 0 ≤ b < d and (a, b, d) = 1. a b In other words: Let C(m) = { 0 d | ad = m, 0 ≤ b < d, (a, b, d) = 1}. Then any cyclic index m sublattice of [1, τ] is homothetic to [1, γτ] for some σ ∈ Cm. Equivalently, get a classiﬁcation of cosets Γ0(m)\ SL2(Z): these are all of the form −1 m 0 SL2(Z) ∩ γσ0 SL2(Z)σ, where σ0 = 0 1 , and σ ∈ C(m) is arbitrary.

~~32.2 Proof that the modular polynomial exists and has integer coefﬁ- cients~~

~~Theorem 32.1. The function fm(X, τ) deﬁned above can be written as Φm(X, j(τ)) where Φm ∈ Z[X, Y]. Furthermore, Φm is irreducible as an element of C(Y)[X].~~

i Proof. First, we show that the coefﬁcient of X in fm(X, τ) is a modular form holomorphic away from the cusp, hence an element of C[j(τ)]. We’ve already checked everything apart from having a q-series. Note that

0 fm(X, τ) = (X − j(L )) = (X − j(στ)) 0 σ∈C L ⊂[1,τ] cyclicY index m Ym and so its coefﬁcients are, up to sign, elementary symmetric polynomials in the set j(στ). 1/m 2πiτ/m 2πi/m a b 2πiστ ab a2 Let Q = q = e and ζm = e . For any σ = 0 d , e = ζm Q , and so −ab −a2 abk a2k j(στ) = ζm Q + ckζm Q . k≥0 X

58 is a Laurent series in Q = q1/m with integer coefficients. Hence the coefficient of Xi in 1/m fm(X, τ) is an element of C((q )). However, we also have fm(X, τ) = fm(X, τ + 1), so this Laurent series must actually belong to C((q)). Hence the coefficients of fm(X, τ), viewed as a polynomial in X, are modular functions in τ with poles only at the cusp. It follows that fm(X, τ) ∈ C[j(τ), X]. In fact, we get that the Laurent series for this coefficient actually belongs to Z[ζm]((q)). Additionaly, we observe that the set of q-series for {j(στ) | σ ∈ Cm} is invariant under the action of Gal(Q(ζm)/Q), so all coefficients of fm(X, τ) actually belong to Z((q)). Now, we use the fact (exercise!) that an element of C[j(τ)] with q-series in Z((q)) actually belongs to Z[j(τ)]. We conclude that fm(X, τ) = Φm(X, j(τ)). where Φm(X, Y) ∈ Z[X, Y]. Finally, we check irreduciblity: We already know that the minimal polynomial of j(mτ) over the field C(j(τ)) has degree dm, so it must be equal to Φm(X, j(τ)) (up to scaling by elements of the field C(j(τ)). It follows that Φm must be irreducible as a polynomial in C(Y)[X].

~~33 April 18~~

~~Properties of the modular equation:~~

~~Proposition 33.1. (a) Φm(X, Y) = Φm(Y, X) if m > 1~~

~~(b) If m is not a square, then Φm(X, X) is a polynomial of degree > 1 with leading coefﬁcient ±1~~

~~p p (c) If m is a prime then Φp(X, Y) ≡ (X − Y)(X − Y )(mod p).~~

0 0 Proof. a): the roots of Φm(j(L), X) are the j-invariants of the lattices L such that L has a cyclic index m sublattice homothetic to L. However (argue) this is the case iff L has 0 a cyclic index m sublattice homothetic to L . This shows that Φm(j(L), X) is equal to Φm(X, j(L))) up to a constant factor. Since j(L) can be any complex number, we have that Φm(Y, X) = cΦm(X, Y), and the constant c must be ±1. But if c were equal to 1, we’d have Φm(X, X) = 0 for all X, which is impossible since we know Φm(j(L), j(L)) 6= 0 when L doesn’t have CM. k b) : Enough to show that if Φm(j(τ), j(τ)) = k≥−N ckq then c−N = 1. Each factor here is P 2 j(τ) − j(σ(τ)) = (Q−n − ζabQ−a ) + holomorphic

59 by assumption the leading terms don’t cancel, so leading coefﬁcient c−N is a root of unity. But we also have c−N ∈ Z so c−N = ±1. 1 i c) The set Cp has the following p − 1 elements: σi = 0 p for i = 0, ... , p − 1, and p 0 σ = 0 1 . We’ll work in Z[ζp][X]((Q)), and show that ∞ p p Φm(X, j(τ)) ≡ (j(τ) − X)(X − j(τ)) (mod 1 − ζp)

~~2πiσ τ p2 p p First, since e 0 = Q = q , j(σ0τ) ≡ j(τ) mod1 − ζp by Frobenius. On the other hand, observe that~~

~~−i −1 ik k −1 k j(σiτ) = ζp Q + ckζmQ ≡ Q + ckQ (mod 1 − ζp) k≥0 k≥0 X X We now multiply the factors together, and get~~

p −1 k p p p Φm(X, j(τ)) ≡ (X − j(τ) )(X − Q − ckQ ) ≡ (X − j(τ) )(X − j(τ)) (mod 1 − ζp). k≥0 X Since both sides of this equality have integer coefﬁcients, the equality also holds mod p. Hence the difference of both sides is an element of pZ[X]((q)) ∩ C[X, j(τ)] = Z[X, j(τ)] and the result follows.

~~Corollary 33.2. If L has complex multiplication by O, then j(L) ∈ Z^~~

Proof. We’ve previously seen that there exists a prime p (in fact, infinitely many such) and a cyclic sublattice L0 of L of index p such that L0 is homothetic to L. 0 Hence Φp(j(L), j(L)) = Φp(j(L ), j(L)) = 0, but we’ve just seen that Φp has leading coefficient ±1. Next stop: the main theorem of class field theory. ∼ Exercise: LO is Galois over Q, with Gal(LO/Q) = Cl(O) o Z/2Z. (Here the nontrivial element σ ∈ Z/2Z acts on Cl(O) by σ([a]) = [a−1]). We’ll be showing that if a is an invertible ideal of O then K(j(a)) = LO. Our strategy: Let M = K(j(a) and L = LO. We know that L is Galois over Q, but we know nothing about M. First we’ll show that M ⊂ L by showing that Spl(L/Q) ⊂ Spl(M/Q) (up to finitely many exceptions). Since L is Galois, this gives that the Galois closure M¯ of M over Q is contained in L, and so M ⊂ L. To show that in fact M = L: we’ll show that Spl(M/Q) ⊂ Spl0(L/Q) (again up to a finitely many exceptions), where Spl0(L/Q) is defined as in Problem 1 of Problem Set 10. It then follows from Problem Set 1 of problem set 10 than M = L.

~~60 34 April 20~~

~~We now prove Theorem 34.1 (Main Theorem of Class ﬁeld Theory). If a is an invertible ideal of an order O in a quadratic ﬁeld K, then K((a)) = LO.~~

Proof. Let M = K(j(a)), and L = LO. First we show that M ⊂ L by showing that, with finitely many exceptions, any prime of Q that splits completely in L also splits completely in M. Exclude the primes p that are not relatively prime to the conductor of O. Also exclude any primes p such that p divides the index [OM : OK[j(a)]]. If p splits completely in LO then (by HW) p = Nπ where π ∈ O. Then πa is homothetic to a, so j(a) is a root of the polynomial Φp(X, X). p 2 Note that the mod p reduction of Φp(X, X) is −(X − X) . So if pM is any prime of p M above p, then j(a) ≡ j(a) (mod p)M. Hence the reduction j(a) ∈ OM/pM actually belongs to Fp. Because of our assumption that p - [OM : OK[j(a)]], we know OM/pM, is generated ∼ by j(a), and so OM/pM = Fp. Since we chose pM over p arbitrary, it follows that p splits completely in Km. For the other direction: we’ll show that, with finitely many exceptions, any prime in Spl0(M/Q) is also in Spl(L/Q). Exclude the finite set of primes p dividing the conductor or sharing a factor with i

~~61 In fact, one can explicitly describe how Gal(L/K) =∼ Cl(O) acts on the set {j(a) | a ∈ Cl(O)}:~~

~~Theorem 34.2 (Shimura Reciprocity). For any unramiﬁed prime p of OK relatively prime to the conductor of O,~~

~~Frobp((j(a)) = j((p ∩ O)a).~~

~~(Discussion of possible future topics: plan is Class Number 1 problem next time.)~~

~~35 April 22~~

~~35.1 The Cube Root of the j-function~~

g3 j = 2 γ = g2 ∆ Recall ∆ . We can take the cube root to deﬁne a new function 2 ∆1/3 : note that is never zero on the upper-half plane, so it has a unique cube root ∆1/3 with q-expansion ∆1/3 = q1/3 + ··· . a b For general γ = c d ∈ SL2(Z), γ2(τ) and γ2(γτ) differ by a cube root of unity. Can show that γ2(γτ) = γ2(τ) iff 3 | b, c or 3 | a, d. It follows from this that γ2(3τ) is a modular function for Γ0(9), so γ2(3τ) is a polynomial in j(τ) and j(3τ).

Theorem 35.1. Let O be an order in an imaginary√ quadratic ﬁeld, and 3 - D = Disc(O). √ 3+ −m Let τ0 = −m if D = −4m and τ0 = 2 otherwise (importantly, gcd(τ0, 3) = 1). Then Q(γ2(τ0)) = Q(j(τ0)).

Proof. Applying the HW, we have that γ2(τ0) ∈ Q(j(τ0/3), j(3τ0)). Can check that that τ0 0 both [1, 3τ0] and [1, 3 ] are invertible ideals of O = Z[3τ], so j(τ0/3), j(3τ0) ∈ LO 0 . It follows that γ2(τ0) ∈ LO 0 . The next step is to show γ2(τ0) ∈ LO. First we compute the index LO 0 /ŁO. Assuming O× = ±1 for simplicity, one can show (we skip details), 0 0 ∼ × × [LO 0 : LO] = | Cl(O )|/| Cl(O)| = | ker(Cl(O ) Cl(O))| = ((O/3O) /Z/3Z ).

In particular, this degree is either 2 or 4. → 3 Now, we observe that [LO(γ2(τ0)) : LO] | 3 as γ2(τ0) = j(τ0) ∈ LO. But also [LO(γ2(τ0)) : LO] | [LO 0 : LO] | 4. As 3 and 4 are relatively prime, conclude that γ2(τ) ∈ LO. Now, one can check that γ2(τ) is real, and that Q(j2(τ)) = LO ∩ R. Hence Q(γ2(τ) ⊂ Q(j(τ)), and the other inclusion is clear.

~~62 35.2 The Weber Functions~~

We now introduce three more modular functions: f, f1, f2 can be deﬁned as power series 1/48 48 48 48 lying in Q((q )). One can show that SL2(Z) ﬁxes the set {f , f1 , f2 }, and can describe the SL2(Z) action on the three functions explicitly. 8 8 8 3 Important property: f , −f1, −f2 are√ roots of X − γ2x − 16√. Useful identities: f(τ)f1(τ)f2(τ) = 2 and f1(2τ)f2(τ) = 2. Important property:

√ 2 Proposition√ 35.2. If m ≡ −3 (mod 8) then K(f( −m) ) is the ring class ﬁeld of O = Z[ −m] (note that O is not the full ring of integers). √ Proof. Proof is similar to the previous proposition, but messier: for τ0 = −m, use that 6 f(8τ0) is a modular form for Γ0(64). Ultimately have to do a bit of Galois theory as well as degree counting.

~~Theorem 35.3. Let K be an imaginary quadratic ﬁeld of discriminant dK. Then h(OK) = 1 iff~~

~~dK = −3, −4, −7, −8, −11, −19, −43, −67, −163~~

Proof. Easy cases: if 4 | dK then 2 ramifies in OK, and so OK must contain an ideal of norm 2. This ideal can be principal only if dK = −4, −8. n−1 So dK is odd. By HW, | Cl(K)[2]| = 2 where n is the number of prime factors of dK. Hence we must have dK = −p. May assume p 6= 3. If p ≡ 7 mod 8, then 2 splits in OK, and so again OK must contain a principal ideal of norm 2: only possible if dK = −7. Left with dK = −p = −3 (mod 8). We already know that dK = −3 has class number 1, so will assume that 3 - p. 0 √ Let O = OK and O = Z[ −p]. By a similar argument to the one sketched above, × × (L 0 L ) ∼ (O 2O) F 3 2 O [L 0 : L ] = 3 Gal O / O = / / 2 has order (as√ is inert in K). Hence O O . If K 1 L = K [K(j( −p)) : K] = 3 has class number√ , then O , and √. Taking real subfields, we get that Q(j( −p)) is a degree 3 extension of Q. Hence Q(f( −p)2) is a cubic extension of Q. √ 3+ −p −1 2 Let τ0 = 2 and α = ζ8 f2(τ0) . Using the Weber function identities √ 2 √ −1 √ = f1(2τ0) = f1(3 + −p) = ζ16 f( −p) f2(τ0) √ α = √2 α α4 Q(f( −p)2) One deduces f( −p)2 , so , and also , generate the cubic field . What is the minimal polynomial of α4 over Q? 4 8 On the one hand α = −f2(τ0) , which satisfies

~~3 x − γ2(τ)x − 16 = 0.~~

63 γ (τ) x3 − γ (τ)x − 16 = 0 This means that 2 is an integer such that a solution to √ 2 . is the fourth power of another element of the same cubic ﬁeld Q(f( −p)2)! This is not something that normally happens. 2 2 To see what the speciﬁc constraint is on γ2, let x + ax + bx + c = 0 be the minimal polynomial of α (which is an algebraic integer because α4 is). Then α2 has minimal polynomial x3 + ex2 + fx + g = 0 with e = 2b − a2, f = x2 − 2ac, g = −c2 and α4 has minimal polynomial

~~x3 + (2f − e2)x2 + (f2 − 2eg)x − g2.~~

2 2 2 Setting the minimal polyomials equal, get 2f = e , g = 16, f − 2eg = −γ2(τ0)). Deduce g = −4, c = ±2, wlog c = 2. Plugging in to 2f = e2, we obtain 2(b2 − 4a) = (2b − a2)2. Oberving that a and b must be even, and setting X = −a/2 and Y = (b − a2)/2, we get the equation 2X(X3 + 1) = Y2. Standard methods show that this equation only has the roots (X, Y) = (0, 0), (−1, 0), (1, ±2) and (2, ±6). If we then solve for j(τ0), we get exactly the j-invariants for dK = −3, −19, −67, −11, −163, −43 respectively.

~~36 April 25~~

~~Correction from the end of Friday’s class: If a has CM by OK and p is any prime of OK, then~~

~~−1 Frobp(j(a)) = j((p ∩ O)a) = j((p ∩ O) a). (5)~~

~~(This statement is a special case of Shimura reciprocity, which applies more generally.)~~

36.1 The Elliptic Curve Perspective (Reference for this section: Serre’s article on Complex Multiplication in Cassels and Frohlich) Let O be an imaginary quadratic order. Let Ell(O) be the set of isom classes of elliptic curves E/Q¯ with End(E) =∼ O. We’ve put Ell(O) in bijection with Cl(O), but this requires analytic methods. Can’t do this bijection algebraically, but can do something closely related. Let Cl(O) act on Ell(O) analytically by a ∗ (C/L) = C/a−1L. This action can also be described algebraically: ∼ Claim: a ∗ E = HomO(a, E) as O-modules.

~~64 Proof of Claim: write E =∼ C/b. Then have SES sequence 0 b C E 0, using that a is a projective O-module, we get~~

~~−1 → → → → 0 a b HomO(a, C) HomO(a, E) 0~~

Suppose E is any elliptic curve, over an arbitrary ﬁeld L, with an inclusion O , → → → → EndL(E). How can we put the structure of an elliptic curve on the O-module a ∗ E = HomO(a, E)? → Actually, let’s generalize the question. If M is a f.g. O-module, doesHomO(M, E) have a natural structure of abelian variety? ∼ a ∼ a First of all, if M = O is a free O-module, then HomO(M, E) = E is naturally an abelian variety. Generally, take a presentation Oa φ Ob M 0, and deﬁne

Hom (M, E) = ker(φ∗ : Hom(Ob, E) Hom(Oa, E)) = ker(φt : Eb Ea). O → → → where φ∗ = φt can be described as multiplication by the transpose matrix to that of φ. → → A dimension count gives that a ∗ E = HomO(a, E) is one-dimensional, and can also show that a ∗ E is connected, so a ∗ E is another elliptic curve isogenous to E. Also, a ∗ E also has CM by O (we have an inclusion O , End(a ∗ E)). There are two nice things about this definition of a ∗ E. First, it works over any base field L, including fields of positive characteristic.→ Secondly, it’s functorial in L. As before, Gal(K¯ /K) acts on Ell(O). Using our new definition of a ∗ E, it’s easy to check that the two actions on Ell(O) are compatible

g(a ∗ E) = a ∗ g(E) for any g ∈ Gal(K¯ /K) and a ∈ Cl(O). It follows that there is a group homomorphism φ : Gal(LO/K) Cl(O) determined by g(E) = φ(g) ∗ E for any E ∈ Ell(O). We’ll show this map φ is inverse to the map θ : Cl(O) Gal(L→O/K) induced by the Artin map: note that this statement is equivalent to (5). This will use some theory of elliptic curves: → Let p be a prime of OK, unramiﬁed in LO, and pO = p ∩ O. As a minor abuse of notation, we’ll write p and pO interchangeably where it won’t cause confusion: e.g. p ∗ E = pO ∗ E. Choose a prime pL of L = LO lying above p, and assume that E has good reduction at pL (this excludes only ﬁnitely many primes). Let p be the norm of p p Let E˜ be the mod pL reduction of E. Let E˜ be the elliptic curve obtained by raising coeffs of E˜ to the pth power (in particular j(E˜ p) = j(E˜ )p . Assuming some theory of elliptic curves, we’ll sketch the proof of Claim: E˜ Np =∼ p ∗ E˜ =∼ p ∗˜ E. Np Once we prove the claim, taking j-invariants gives j(p ∗ E) ≡ j(E ) mod pL, so p ∗ E = Frobp(E) as desired.

65 Case 1: p is split in OK. Then have a isogeny of elliptic curves E˜ p ∗ E˜ of degree p. Can show inseparable by testing derivative at origin. One can show (eg by looking at function fields) that, up to isomorphism, the only such isogeny is→ the the Frobenius isogeny E˜ E˜ p. So p ∗ E =∼ E˜ p = E˜ Np as desired. (It’s actually possible to finish at this point, by observing that we’ve checked that θ(Frob(p))→ = [p] for a density 1 subset of primes of K.) Case 2: p is inert in OK. In this case, one can show that the elliptic curve E˜ has no points of order p over Fp (E is supersingular), and that there is a unique isogeny from E˜ 2 having degree p, namely the Frobenius E˜ E˜ p . We then finish as in Case 1.

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