Algebraic Number Theory Notes

Home , Abelian extension, Class formation, Global field, Principal ideal theorem

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.