Math 160c Spring 2013 Caltech Applications of Global Class Field Theory Course Notes
Andrei Jorza
Contents
1 Local Class Field Theory 2 1.1 Main results ...... 2 1.2 Application ...... 2
2 Glocal Class Field Theory 3 2.1 Adeles ...... 3 2.2 The Dirichlet unit theorem using adeles ...... 3 2.3 Main results ...... 4 2.4 Conductors ...... 4 2.5 Hilbert, ray and ring class elds ...... 5
3 Selmer groups and applications 7 3.1 Selmer systems and Selmer groups ...... 7 3.2 Global duality and dual Selmer groups ...... 8 3.3 Euler characteristics and sizes of Selmer groups ...... 9
4 Rational points on elliptic curves 12 4.1 Facts about elliptic curves ...... 12 4.2 Cohomology of elliptic curves over nite elds ...... 13 4.3 Descent for rational points ...... 14 4.4 Selmer groups for elliptic curves ...... 14 4.5 Mordell-Weil ...... 16 4.6 Standard proof of Mordell-Weil ...... 16
5 Characters with prescribed behavior 17 5.1 Grunwald-Wang ...... 17 5.2 Characters with prescribed nite order local behavior ...... 19 5.3 Characters with prescribed local behavior at in nite places ...... 19
6 Projective Galois representations 22 6.1 A theorem of Tate ...... 22 6.2 Lifting projective Galois representations ...... 24 6.3 Local Galois representations in the tame case ...... 25
1 7 Iwasawa theory for -extensions 28 Zp 7.1 -extensions and Leopoldt's conjecture ...... 28 Zp 7.2 Class groups and Galois modules ...... 34 7.3 The Iwasawa algebra ...... 36 7.4 Modules over the Iwasawa algebra ...... 37 7.5 Class numbers in -extensions ...... 38 Zp 8 Hecke theory for GL(1) 40 8.1 Fourier analysis ...... 40 8.2 Local zeta integrals ...... 41 8.3 Local functional equation and local ε-factors ...... 42 8.4 Global zeta integrals ...... 44 8.5 Global L-functions and ε-factors ...... 45 8.6 Applications ...... 46
9 Hecke theory for Galois representations 48 9.1 Global theory ...... 48 9.2 Deligne's local ε-factors ...... 50 Lecture 1 2013-04-01
1 Local Class Field Theory 1.1 Main results (1.1.1) Have an upper rami cation ltration u for such that −1 , (−1,0] , (0,1] GK u ≥ −1 GK = GK GK = IK GK = . If is an algebraic extension then u u u . PK L/K GL/K = GK /(GK ∩ GL) (1.1.2) The invariant map is an isomorphism ∼ 2 × ∼ such that if then invK : Br(K) = H (GK , K ) = Q/Z L/K invK ◦ cor = invL and invL ◦ res = [L : K] invK . (1.1.3) Tate duality. If is a nite -module let ∗ . The Galois group acts on M GK M = Hom(M, Q/Z) GK ∗ ∗ via ∗ ∗ −1 . Write ∼ . Write i i IK . m ∈ M (gm )(m) = m (g m) M(1) = M ⊗Q/Z µ∞ Hur(GK ,M) = H (GK /IK ,M ) Then there exists a perfect pairing
i 2−i ∗ 2 ∗ 2 2 × H (GK ,M) ⊗ H (GK ,M (1)) → H (GK ,M ⊗ M (1)) → H (GK , µ∞) → H (GK , K ) → Q/Z such that i ⊥ 2−i ∗ . Hur(GK ,M) = Hur (GK ,M (1)) (1.1.4) The Artin map. For nite write −1 ×, 0 × and n n for . K/Qp UK = K UK = OK UK = 1 + ($) n ≥ 1 There exists a homomorphism × ∼ ab such that n n,ab. It has the property that rK : K = WK rK (UK ) = GK × ∨ × × rK (NL/K x) = rL(x) for x ∈ L and rK (x) = cor (rL(x)) for x ∈ K ⊂ L .
1.2 Application We will prove Kronecker-Weber for local elds. Theorem 1.1. The maximal abelian extension of is . Qp Qp(µ∞)
Zb Z ab ∼ ab Zb ab ∼ × Proof. Recall that GK = IK Frob and WK = IK Frob . Thus G = (I )Frob × Frob and I = o K o K Qp Qp p p Qp Zp and so Gab is a quotient of × which we will show to be equal to G under the reciprocity map. Qp Zp o Zb Qp(µ∞)/Qp First, recall ur × and since ϕ(n) if it follows that ur Qp = Qp(ω(α)|α ∈ Fp ) p ≡ 1 (mod n) p - n Qp = Qp(ζn|p - . Next is rami ed over and so ur and therefore ur . This gives n) Qp(ζp) Qp ζp -Qp Qp ∩ Qp(µp∞ ) = Qp G =∼ G × G ur . Qp(µ∞)/Qp Qp(µp∞ )/Qp Qp /Qp
2 2 Glocal Class Field Theory 2.1 Adeles (2.1.1) If is a nite extension then , ×,0 is the connected component of in × . For K/Q K∞ = K ⊗Q R K∞ 1 K∞ an embedding write and write . Then Q , × Q × v : K,→ R v | R v : K,→ C v | C K∞ = v|∞ Kv K∞ = v|∞ Kv and K×,0 = Q (0, ∞) Q ×. ∞ v|R v|C C (2.1.2) Write = Q0 K with the restricted product topology. For a nite set of places S write AK {Ov } v K = Q K and S = Q0 K in which case = K × S . Then the ring = Q0 K has S v∈S v AK v∈ /S,{Ov } v AK S AK AK {Ov } v the product topology, is a discrete subgroup and is compact. K ⊂ AK AK /K × Q0 × × × S,× (2.1.3) Write = × K with the restricted product topology. As above = K × . The AK {Ov } v AK S AK natural inclusion × is not continuous and in fact the topology on × is the subset topology induced AK ⊂ AK AK by the map × which takes to −1 . Write × . De ne 1 × as the AK ,→ AK × AK x (x, x ) | · |AK : AK → (0, ∞) AK ⊂ AK kernel of . Then × 1 is a discrete subgroup, 1 is continuous and 1 × is compact. | · |AK K ⊂ AK AK ⊂ AK AK /K (2.1.4) Strong approximation states that if then S is dense. S 6= ∅ K ⊂ AK 2.2 The Dirichlet unit theorem using adeles Theorem 2.1. Let be a number eld with real embeddings and complex embeddings. Then × is K/Q r s OK a nitely generated abelian group of rank r + s − 1.
Proof. For a nite set S of places which include the in nite places write OK [1/S] = {x ∈ K|v(x) ≥ 0, v∈ / S}. Note that if then × × . We will show that × is a nitely generated abelian S = {v | ∞} O[1/S] = OK OK [1/S] group of rank |S| − 1. Let ClS(K) be the class group of OK [1/S], i.e., the set of ideals modulo the set of principal ideals. An element of × gives the fractional ideal Q v(av ) of and the set of fractional ideals is a = (av) AK v∈ /S($v) OK [1/S] isomorphic to × × Q ×. Write 1 Q in which case × × ∼ 1 1 . AK /KS v∈ /S Ov KS = {x ∈ KS| v∈S |xv|v = 1} AK /KS = AK /KS It is easy to see that 1 × 1 Q × which gives the exact sequence ClS(K) = AK /K KS v∈ /S Ov
1 Y × × 1 × 1 → KS Ov /OK [1/S] → AK /K → ClS(K) → 1 v∈ /S
Immediately one sees ClS(K) as a quotient of a compact group by an open subgroup and so ClS(K) is nite. De ne as the kernel of the summation map and write the kernel of . Then ∆ ⊕vR → R ∆S ⊕v∈SR → R one has the map 1 × given by . Clearly Q × 1 is surjective log : AK /K → ∆ (av) 7→ (log |av|v) v∈ /S Ov KS → ∆S and so get an exact sequence
1 Y × × × KS Ov /OK [1/S] → ∆S/ log OK [1/S] → 0 v∈ /S
× which exhibits ∆S/ log OK [1/S] as the image via a continuous map of an open subgroup of a compact group, × × therefore ∆S/ log OK [1/S] is compact. In particular log OK [1/S] is a lattice in the (|S| − 1)-dimensional ∆S. × We would like to prove that OK [1/S] is a nitely generated abelian group of rank |S| − 1. To do this × it is enough to show that the intersection of the kernel of log with OK [1/S] consists only of torsion. What 1 1 Q × Q Q 1 is the kernel? The kernel of log on is {(av) ∈ ||av|v = 1}, i.e., O × {±1} × S . AK AK v-∞ v v|R v|C This is compact and its intersection with K× is compact and discrete, therefore nite. Since it is nite this intersection must be µ∞(K), i.e., torsion.
3 2.3 Main results (2.3.1) The global Brauer sequence
0 → Br(K) → ⊕v Br(Kv) → Q/Z → 0 is exact where the rightmost map is the sum of the local invariant maps. Application: a global quaternion algebra over any number eld must be nonsplit at an even number of places. (Used to study Hilbert modular forms.) (2.3.2) The Artin map × ab has the property that where × rK : AK → GK rK (1,..., 1, x, 1,...) = rKv (x) x ∈ Kv is placed in position and is the local Artin map. It gives v rKv
× × ×,0 ∼ ab rK : AK /K K∞ = GK such that and induces an isomorphism ab ∼ × × ×. rL(x) = rK (NL/K (x)) rK GL/K = AK /K NL/K AL
Lecture 2 2013-04-03
2.4 Conductors (2.4.1) If is a nite extension and is a continuous complex nite dimensional representation of K/Qp V GK Z ∞ Gu de ne the conductor cond(V ) = codimV (V K )du. Since the Galois action is continuous, V is xed by −1 an open subgroup of and so the integral is nite. For any , and if and GK V cond(V ) ∈ Z≥0 cond(V ) = 0 only if V is unrami ed and cond(V ) ≤ 1 if and only if V is tamely rami ed.
(2.4.2) Let K/Q be a nite extension and V be a continuous complex nite dimensional representation of GK . Proposition 2.2. The representation V is almost everywhere unrami ed. Proof. By the Brauer induction theorem V = P n IndK χ for nite Galois extensions L and characters i Li i i ×. It su ces to show the proposition for a character since then will be unrami ed at χi : GLi → C V χ V all nite places where no nor is rami ed. Now ab × is continuous and the kernel will Li χi χ : GK → C ker χ be open in Gab ∼ × /K×K×,0 and so will contain an open set of the form U Q O× where U ⊂ K× K = AK ∞ S v∈ /S Kv S S is an open set. Then χ will be unrami ed at v∈ / S.
Q cond(V |G ) De ne the conductor cond(V ) = ($v) Kv as an ideal of OK , the product being nite by v-∞ Proposition 2.2. The representation V is unrami ed if and only if cond(V ) = OK and it is tamely rami ed if and only if cond(V ) is square free. Since V has a continuous action of GK , one can nd a nite Galois extension L/K such that GL acts trivially on V . Thus the action of GK factors through the discrete action of GL/K . For every nite place v of K choose a place w of L. Then
Y cond(V |GL /K ) cond(V ) = ($v) w v v-∞
where Kv is chosen as containing Lw. Proposition 2.3. Let L/K be a nite extension of number elds. Then L as a vector space over K has a linear action of GL/K and the discriminant DL/K is equal to cond(L) as ideals of OK .
4 Proof. For each place of x an arbitrary place of . Recall that Q where v K w L DL/K = NL/K ( v DLw/Kv ) is the dierent and R ∞ 1 . Therefore DLw/Kv v(DLw/Kv ) = −1 1 − |Gu | du Lw/Kv
v(cond(L)) = cond(L|GLw/Kv ) Z ∞ Gu = codimL(L Lw/Kv )du −1 Z ∞ = (|G | − |G /Gu |)du L/K L/K Lw/Kv −1 Z ∞ = [L : K] ((1 − 1/|Gu |)du Lw/Kv −1
= [L : K]vKv (DLw/Kv )
= v(NL/K (DLw/Kv ))
2.5 Hilbert, ray and ring class elds
A little classical notation. The class group Cl(K), also known as the wide or weak class group, is the set of fractional ideals of OK modulo principal ideals, equal to ClS(K) as de ned above when S = {v | ∞}. The narrow or strict class group Cl+(K) is the set of fractional ideals modulo principal ideals generated by totally positive elements, i.e., x such that for every real emberdding τ : K,→ R, τ(x) > 0. Let be a number eld. For any open subgroup of × , is an open subgroup of ab and so K/Q U AK rK (U) GK is of the form Gab for a nite extension L /K. In fact L = (Kab)rK (U) is an abelian extension and LU U U ∼ × × ×,0 GLU /K = AK /K K∞ U
2.5.1 The Hilbert class eld
× Q × ∼ Suppose U = K O . Then LU is called the (wide/weak) Hilbert class eld HK of K and G = ∞ v-∞ v HK /K × × × Q × ∼ . AK /K K∞ Ov = Cl(K) + ×,0 Q × + Suppose U = K O . Then L + is called the (narrow/strict) Hilbert class eld H of K and ∞ v-∞ v U K ∼ × × ×,0 Q × ∼ + . Note that G + = /K K∞ Ov = Cl (K) HK /K AK + × × × ×,0 Y HK /HK = K K∞/K K∞ ⊂ {±1} v|R
Two facts about Hilbert class elds: HK /K is the maximal abelian extension which is unrami ed at every place while + is the maximal abelian extension unrami ed at every nite place, and every ideal of HK becomes principal in +. K HK We now give an example application. Example 2.4. If m | n then h | h . Q(µm) Q(µn) Proof. Let be the Hilbert class eld of and be the Hilbert class eld of . The extension Hm Q(µm) Hn Q(µn) (which is an extension since ) is totally rami ed at every and so Q(µn)/Q(µm) m | n p | n/m Q(µn) ∩ Hm = . Now is unrami ed and abelian and so . Therefore Q(ζm) Q(µn)Hm/Q(µn) Q(µn)Hm ⊂ Hn h = [H : (µ )] Q(µn) n Q n = [H : H (µ )]|G | n mQ n HmQ(µn)/Q(µn) = [H : H (µ )]|G | n mQ n Hm/Q(µm) = [H : H (µ )]h n mQ n Q(µm) as desired.
5 2.5.2 Ray class elds Suppose is an ideal of . Let and + consist of totally positive . m OK Km = {x ∈ K|x ≡ 1 (mod m)} Km x ∈ Km Let be the set of ideals coprime to modulo the principal ideals generated by and let + Clm(K) m Km Clm(K) be the set of ideals coprime to modulo the principal ideals generated by +. m Km + ×,0 Q v(m) × Q v(m) Let U = K U and Um = K U . Then HK,m = LU is called the weak ray class m ∞ v-∞ K ∞ v-∞ K m + eld and H = L + is the strict ray class eld of conductor m. K,m Um Proposition 2.5. + × × + and × × . Clm = AK /K Um Clm = AK /K Um Proof. Let × . The Chinese remainder theorem implies that there exists × such that x = (xv) ∈ AK y ∈ K has the property that for , v(m) v(m) and is positive for every ; xy = (xvy) v | m xvy ∈ 1 + ($v) = UK xvy v | R elements of K with this property are ≡ 1 (mod m) (and so in Km) and totally positive. Attach to x the ideal Q v(xv y) × × + ($v) in which case /K U becomes the set of ideals coprime to m modulo principal ideals v-m AK m generated by totally positive x ≡ 1 (mod m). Similarly for Um.
+ Then as before and + . GHK,m/K = Clm(K) G = Clm(K) HK,m/K Example 2.6. 1. If and then + ∼ × and so the strict ray class eld of K = Q m = nZ Cln (K) = (Z/nZ) conductor is . n Q(µn) 2. If then and + + . m = OK HK = HK,m HK = HK,m Proposition 2.7. Let be an ideal of . Then + is the maximal abelian extension of such that m OK HK,m H K Gn = 0 if n ≥ v(m), where w | v is an arbitrary place of H. In particular, the strict Hilbert class eld Hw/Kv + is the maximal abelian extension of which is unrami ed at every nite place. HK K Proof. Let L/K be an abelian extension such that if v is a place of K and w | v is an arbitrary place of L then Gn = 0. Recall Herbrandt's theorem that Gn = Gn,ab/Gn,ab ∩ W ab and so this is equivalent Lw/Kv Lw/Kv Kv Kv Lw to W ab ⊃ Gn,ab. Via the inverse of the Artin map this is equivalent to r−1 (W ab ) ⊃ U n . Equivalently, Lw Kv Kv Lw Kv using the global Artin map, the component at v of the open subgroup r−1(Gab) of × should contain U n . K L AK Kv That L is the largest abelian extension of K such that Gn = 0 for n ≥ v(m) is equivalent to the fact Lw/Kv that U = r−1(Gab) is the largest open subgroup of Gab such that the component in v is included in U v(m). K L K Kv The largest such is +. U Um Lecture 3 2013-04-05
Proposition 2.8. Let m be an ideal of OK . Then the discriminant of H = HK,m over K is
[H:K] v(m)− 1 Y ( qv −1 ) mv v|m
v(D ) Proof. For simplicity write +. Since Q Hw/Kv it su ces to show that for each H = Hm (DL/K ) = NH/K ( mv ) nite place v of K (and an arbitrary choice w | v of H) we have v(D ) = v(m) − 1 . For this we Hw/Kv qv −1 need to know the cardinality of Gn = Gn,ab/Gn,ab ∩ W ab and via the inverse of the local Artin map Hw/Kv Kv Kv Hw this is n n ×. This is trivial if and only if which implies that n × if Uv /(Uv ∩ NHw/Kv Hw n ≥ v(m) Uv ⊂ NHw/Kv Hw and only if . But is maximal among such abelian extensions and so × v(m) giving n ≥ v(m) H NHw/Kv Hw = Uv n n v(m) × × i i+1 G =∼ U /Uv . Since O /(1 + ($ )) =∼ k and (1 + ($ ) )/(1 + ($ ) ) =∼ k it follows that Hw/Kv v Kv v Kv v v Kv
6 n v(m)−n 0 v(m)−1 |G | = qv if n ≥ 1 and |G | = qv (q − 1) as long as v | m. We compute the conductor Hw/Kv Hw/Kv v of as a -representation (see the proof of Proposition 2.3: H GHw/Kv Z ∞ cond(H) = [H : K] (1 − 1/|Gu |)du Hw/Kv −1 v(m) X = [H : K] (1 − 1/|Gn |) Hw/Kv n=0 v(m) 1 X 1 = [H : K] v(m) + 1 − − v(m)−1 v(m)−n qv (qv − 1) n=1 qv 1 = [H : K] v(m) − qv − 1
The following section was not covered in lecture
2.5.3 Ring class elds An order in a number eld is a nitely generated -submodule of such that . O ⊂ K K Z K O ⊗Z Q = K Every order is contained in . The conductor of the order is the cardinality of . O√ OK O OK /O If then every order is of the form which has conductor . An ideal of K = Q( −d) O = Z + OK f f I O is said to be proper if O = {x ∈ K|xI ⊂ I}; I is said to be coprime to f if I + (f) = O. An ideal is proper if and only if it is equivalent to an ideal coprime to f. The ring class group Cl(O) is de ned to be the set of proper ideals of O modulo principal ideals. There is an isomorphism between ideals of OK coprime to the conductor f and proper ideals of O, the map from the rst to the second being I 7→ I ∩ O and the inverse being I 7→ IOK . As such Cl(O) is the set of ideals of O coprime to f modulo principal ideals generated by x such that x ≡ n (mod fOK ) for some integer n coprime to f. Finally, and so there is an open set such that ∼ × × × Cl(O) ⊂ ClfOK (K) UO ⊃ UfOK Cl(O) = AK /K K∞UO in which case where is the ring class eld of . It shows up in the study of Heegner Cl(O) = GLO /K LO O points. √ Proposition 2.9. Let with (wide) Hilbert class eld . For consider K = Q( −d) HK f ≥ 2 O = Z + fOK the order of conductor . Let be the ring class eld of . Then is Galois over and ∼ f Kf O Kf HK GKf /HK = × ×. (OK /fOK ) /(Z/fZ) Proof. The Galois group is isomorphic to the group of principal ideals of coprime to modulo GHK,f /HK OK f those principal ideals generated by x ≡ c (mod fOK ) with c coprime to f. End of section not covered in lecture
3 Selmer groups and applications 3.1 Selmer systems and Selmer groups (3.1.1) Let be a number eld and a nite discrete -module. Then has trivial action of K/Q M GK M GL for some nite extension L/K and so M will be unrami ed at all places v where L/K is unrami ed. A Selmer system is a collection of 1 such that 1 for L = {Lv} Lv ⊂ H (GKv ,M) Lv = Hur(GKv ,M) almost all v.
7 Proposition 3.1. De ne ⊥ 1 ∗ as the annihilator of under the local Tate pairing. Lv ⊂ H (GKv ,M (1)) Lv Then ⊥ ⊥ is also a Selmer system, called the dual Selmer system of . L = {Lv } L Proof. This follows from the fact that 1 ⊥ 1 ∗ . Hur(GKv ,M) = Hur(GKv ,M (1)) De nition 3.2. The Selmer group is 1 1 . HL(K,M) = {x ∈ H (GK ,M)| resv x ∈ Lv} Proposition 3.3. Let L be a Selmer system and let S be a set of places containing the in nite places, nite places such that 1 and nite places such that does not act trivially on . Let v Lv 6= Hur(GKv ,M) v IKv M be the largest extension which is unrami ed outside and let . Then KS/K S GK,S = GKS /K
1 1 1 0 → HL(K,M) → H (GK,S,M) → ⊕v∈SH (GKv ,M)/Lv is exact.
Proof. By de nition 1 is the kernel of the map 1 Q 1 . If then HL(K,M) H (GK ,M) → v H (GKv ,M)/Lv v∈ / S 1 is the kernel of the restriction map 1 1 ∼ . Therefore Lv = Hur(Kv,M) H (Kv,M) → H (Iv,M) = Hom(Iv,M) Q 1 1 Q . v H (GKv ,M)/Lv ⊂ ⊕v∈SH (GKv ,M)/Lv ⊕ v∈ /S Hom(Iv,M) Suppose 1 is the image of some class in 1 . Then is the trivial map c ∈ H (K,M) HL(K,M) c|Iv Iv → M when v∈ / S. The Galois group G is generated by I for v∈ / S and so c| is the trivial class. But then KS v GKS by in ation-restriction 1 since acts trivially on . c ∈ H (GK,S,M) GKS M
3.2 Global duality and dual Selmer groups This section is about the Tate-Poitou nine-term exact sequence, which is a global version of local Tate duality, and its application to Selmer groups.
3.2.1 Tate-Poitou duality
Let K be a number eld and let S be a nite set of places containing the in nite places. Let M be a nite GK -module and suppose that S contains all the nite places v such that Iv acts nontrivially on M and all the nite places v such that v(|M|) > 0. Then M is naturally a GK,S-module. De ne G M Kv v ∞ - 0 GKv Hb (Kv,M) = M /NK /K M v | R v v 0 v | C If is an abelian group write ∨ . If is a nite -module write ∗ A A = Hom(A, Q/Z) M GK M = Hom(M, Q/Z) with the action (gφ)(m) = φ(g−1m).
Theorem 3.4. Suppose M is a nite GK,S module as above. Then 1. There is an exact sequence
0 0 2 ∗ ∨ 0 → H (GK,S,M) → ⊕v∈SHb (Kv,M) → H (GK,S,M (1)) →
1 1 1 ∗ ∨ → H (GK,S,M) → ⊕v∈SH (Kv,M) → H (GK,S,M (1)) → 2 2 0 ∗ ∨ → H (GK,S,M) → ⊕v∈SH (Kv,M) → H (GK,S,M (1)) → 0
i ∼ i 2. If i ≥ 3 then H (GK,S,M) = ⊕v∈SH (Kv,M). Lecture 4 2013-04-08
8 3.2.2 Dual Selmer groups
⊥ Suppose L is a Selmer system and L is the dual Selmer system for the cohomology of a nite GK -module M. Proposition 3.5. There is an exact sequence
0 0 2 ∗ ∨ 0 → H (GK,S,M) → ⊕v∈SHb (Kv,M) → H (GK,S,M (1)) →
1 1 ∗ ∨ 1 ∗ ∨ → HL(K,M) → ⊕v∈SLv → H (GK,S,M (1)) → HL⊥ (K,M (1)) → 0 Proof. Tate-Poitou implies that
0 0 2 ∗ ∨ 0 → H (GK,S,M) → ⊕v∈SHb (Kv,M) → H (GK,S,M (1)) →
1 1 1 ∗ ∨ → H (GK,S,M) → ⊕v∈SH (Kv,M) → H (GK,S,M (1)) 1 is exact. Since Lv ⊂ H (Kv,M) is sequence gives the exact sequence
0 0 2 ∗ ∨ 1 ∗ ∨ 0 → H (GK,S,M) → ⊕v∈SHb (Kv,M) → H (GK,S,M (1)) → K → ⊕v∈SLv → H (GK,S,M (1))
1 1 where K is the preimage of ⊕v∈SLv under the restriction map H (GK,S,M) → ⊕v∈SH (Kv,M). By de nition 1 and so get an exact sequence K = HL(K,M)
0 0 2 ∗ ∨ 1 1 ∗ ∨ 0 → H (GK,S,M) → ⊕v∈SHb (Kv,M) → H (GK,S,M (1)) → HL(K,M) → ⊕v∈SLv → H (GK,S,M (1)) Proposition 3.3 replacing M by M ∗(1) and L by L⊥ gives
1 ∗ 1 ∗ 1 ∗ ⊥ 0 → HL⊥ (K,M (1)) → H (GK,S,M (1)) → ⊕v∈SH (Kv,M (1))/Lv which after dualizing and using 1 ∗ ⊥ ∨ ∼ gives (H (Kv,M (1))/Lv ) = Lv
1 ∗ ∨ 1 ∗ ∨ ⊕v∈SLv → H (GK,S,M (1)) → HL⊥ (K,M (1)) → 0 Putting everything together gives the proposition.
3.3 Euler characteristics and sizes of Selmer groups 3.3.1 The Euler characteristic formulae Proposition 3.6. If is a nite extension and is a nite -module then i is nite for K/Qp M GK H (K,M) i ≥ 0. Moreover, Hi(K,M) = 0 for i ≥ 3. Theorem 3.7. Let be a nite extension and a nite -module. Then K/Qp M GK |H0(K,M)||H2(K,M)| χ(K,M) = = |#M| |H1(K,M)| K
i Proposition 3.8. If M is a nite GK,S-module then H (GK,S,M) is nite for i ≥ 0. 0 i ∼ i Proof. This is clear when i = 0 as H (GK,S,M) ⊂ M. When i ≥ 3 then H (GK,S,M) = ⊕v∈SH (Kv,M) by Theorem 3.4 and niteness follows from Proposition 3.6. Again by Theorem 3.4 we get exactness for 1 ∗ ∨ 2 2 2 H (GK,S,M (1)) → H (GK,S,M) → ⊕v∈SH (Kv,M) and niteness of H (GK,S,M) follows from Propo- 1 ∗ ∨ sition 3.6 if we assume niteness of H (GK,S,M (1)) . Therefore it su ces to treat the case of i = 1 and 1 show that H (GK,S,M) is nite for M nite. Since is a -module there exists a nite Galois subextension of such that acts M GK,S L/K KS/K GKS /L trivially on . In ation-restriction gives 1 1 1 . Let M 0 → H (L/K, M) → H (GK,S,M) → H (GKS /L,M) SL for the set of places of lying above places of in in which case and so . L K S LSL = KS GKS /L = GL,SL
9 1 Therefore to show niteness of H (GK,S,M) it su ces to treat the case when GK,S acts trivially on M in which case 1 ∼ ∼ ab H (GK,S,M) = Hom(GK,S,M) = Hom(GK,S,M)
It is unfortunate that we use KS as both the maximal extension of K which is unrami ed outside S and as Q but it should be clear which one we mean from context. So ab ∼ × × ×,0 Q ×. v∈S Kv GK,S = AK /K K∞ v∈ /S Ov We have already seen in the proof of Theorem 2.1 that we have an exact sequence
× × × × Y × 0 → KS /OK [1/S] → AK /K Ov → ClS(K) → 0 v∈ /S
Since is nite to show that ab is nite is su ces to show that × × ClS(K) Hom(GK,S,M) Hom(KS /OK [1/S] ,M) is nite. But × × Q × and × ∼ ∼ Hom(KS /OK [1/S] ,M) ,→ v∈S Hom(Kv ,M) Hom(Kv ,M) = Hom(GKv ,M) = 1 H (Kv,M) is nite by Proposition 3.6. Theorem 3.9. Let be a number eld, a nite -module and as in Theorem 3.4. Then K/Q M GK S Q 0 |H0(G ,M)||H2(G ,M)| |H (Kv,M)| χ(K,M) = K,S K,S = v|∞ 1 [K: ] |H (GK,S,M)| |M| Q
3.3.2 Sizes of Selmer groups and a theorem of Wiles Lemma 3.10. Let be a number eld and nite, and as in Proposition 3.3. Then 1 is K/Q M L S HL(K,M) nite.
Proof. Proposition 3.3 implies that 1 1 which we know is nite by Theorem 3.4. HL(K,M) ⊂ H (GK,S,M) Theorem 3.11. Let K/Q be a number eld and M, L and S as in Proposition 3.3. Then
1 0 |H (K,M)| |H (K,M)| Y |Lv| L = |H1 (K,M ∗(1))| |H0(K,M ∗(1))| |H0(K ,M)| L⊥ v v
Example 3.12. Let p > 2 be a prime and S be a nite set of places of Q containing p and ∞. What is the number of abelian extensions of Q of degree pn which are unrami ed at all places `∈ / S? Proof. Recall that ∨ ∼ ab,∨ and so every continuous homomorphism has open kernel G = G φ : GQ → Q/Z for a niteQ abelianQ extension . If then for every , m and ker φ = GL L/Q [L : Q] = m g ∈ GQ g ∈ GL so m and so . The map taking to is not injective, as mφ(g) = φ(g ) ∈ φ(GL) = 0 φ : GQ → Z/mZ φ L acts on the set of . Reciprocally, every cyclic of degree produces Aut GL/Q φ L/Q m ϕ(m) = | Aut(Z/mZ)| homomorphisms φ. Every n (with trivial -action produces a cyclic extension of degree dividing n. φ : GQ → M = Z/p Z GQ p Therefore we need to study ∼ 1 . If is as above then is unrami ed at if Hom(GQ,M) = H (Q,M) L/Q L `∈ / S 1 and only if φ(I`) = 0 if and only if φ|G ∈ H ( `,M). Therefore L/ is unrami ed outside S if and only Q` ur Q Q if 1 n where 1 for and 1 when . The problem φ ∈ HL(Q, Z/p Z) L` = Hur(Q`,M) `∈ / S Lv = H (Qv,M) v ∈ S now becomes to compute 1 n . The number of with image n is exactly hn = |HL(Q, Z/p Z)| φ Z/p Z hn − hn−1 and n acts on this giving hn−hn−1 extensions cyclic of order n unrami ed outside . Aut(Z/p Z) ϕ(pn) L/Q p S Note that ⊥ for and ⊥ 1 for . Suppose 1 ∗ . Here Lv = 0 v ∈ S L` = Hur(Q`,M) `∈ / S c ∈ HL⊥ (Q,M (1)) ∗ 1 ⊥ 1 M (1) = µ n and so c ∈ H ( , µ n ) such that c| ∈ L ⊂ H ( , µ n ) for all `. p Q p √ Q` ` ur Q` p pn g( α) n n By Kummer theory where × × p , since 1 ∼ × × p . Since c(g) = √n α ∈ /( ) H ( , µpn ) = /( ) p α Q Q Q Q Q 1 ur,× ur,× pn is unrami ed at all primes it follows that is trivial in n ∼ . Therefore c ` c H (IQ` , µp ) = Q` /(Q` ) ur,× n ur,× p . In particular, n n . Let Q a` in which case we just α ∈ (Q` ) v`(α) ∈ p v`(Q` ) = p Z α = ± ` ` −n n n showed n . Therefore Q a`p p × p since . This shows that is trivial and p | a` α = (± ` ` ) ∈ (Q ) p > 2 c so 1 ∗ . HL⊥ (Q,M (1)) = 0
10 Lecture 5 2013-04-10 By Theorem 3.11 it follows that
0 n 1 n |H (Q, Z/p Z)| Y |Lv| |HL(Q, Z/p Z)| = 0 0 n |H ( , µ n )| |H ( , /p )| Q p v Qv Z Z 1 n n Y |H (Qv, Z/p Z)| = p 0 n |H ( v, /p )| v∈S Q Z Z since 0 n n and 0 . The second line in the equation above comes from the H (Q, Z/p Z) = Z/p Z H (Q, µpn ) = 0 rst paragraph of the proof of Theorem 3.11. Now H1(R, Z/pnZ) =∼ {0} and H0(R, Z/pnZ) =∼ Z/pnZ while for ` a prime, Theorem 3.7 implies that
0 n 2 n n |H (Q`, Z/p Z)||H (Q`, Z/p Z)| n n χ(Q`, Z/p Z) = 1 n = |Z/p Z|Q` = |p |Q` |H (Q`, Z/p Z)| Therefore 1 n 2 n |H (Qv, Z/p Z)| |H (Q`, Z/p Z)| 0 n = n |H (Qv, Z/p Z)| |p |Q` But by local Tate duality 2 n 0 ∨ 0 |H (Q`, Z/p Z)| = |H (Q`, µpn ) | = |H (Q`, µpn )| = |µpn (Q`)| = |µpn (F`)| = (pn, ` − 1). Finally,
Y (pn, ` − 1) |H1 ( , /pn )| = L Q Z Z |pn| `∈S−∞ Q` n Y n hn = p (p , ` − 1) `∈S−∞
Finally the number of L/Q cyclic of order pn unrami ed outside of S is then |{`∈S−∞|pn|`−1}| hn − hn−1 Y Y p − 1 = (pn, ` − 1) + (pn−1, ` − 1) pn−1(p − 1) p − 1 `∈S−∞ `∈S−∞
Proof of Theorem 3.11. First note that if then is an unrami ed -module and so 0 ur v∈ / S M GKv H (Kv /Kv,M) → Frobv −1 1 is exact. Therefore 1 0 ur 0 . In M −→ M → Hur(Kv,M) → 0 |Hur(Kv,M)| = |H (Kv /Kv,M)| = |H (Kv,M)| particular the product in the theorem is nite. Proposition 3.5 then implies that
1 2 ∗ ∨ 0 |HL(K,M)| |H (GK,S,M (1)) ||H (GK,S,M)| Y |Lv| 1 ∗ = 1 ∗ ∨ |H ⊥ (K,M (1))| |H (GK,S,M (1)) | 0 L v∈S |Hb (Kv,M)|
∨ G If A is a nite abelian group then |A| = |A |. Note that M is unrami ed outside S and so M KS = M. 0 GK GK GK,S 0 Thus H (K,M) = M = (M S ) = H (GK,S,M). This gives
1 0 |H (K,M)| |H (K,M)| Y |Lv| L = χ(K,M ∗(1)) |H1 (K,M ∗(1))| |H0(K,M ∗(1))| 0 L⊥ v∈S |Hb (Kv,M)| But Theorem 3.9 implies that
Q 0 ∗ |H (Kv,M (1))| χ(K,M ∗(1)) = v|∞ |M ∗(1)|[K:Q]
11 and so, since |M| = |M ∗(1)|,
1 Q 0 ∗ 0 |H (K,M)| v|∞ |H (Kv,M (1))| |H (K,M)| Y |Lv| L = |H1 (K,M ∗(1))| |M|[K:Q] |H0(K,M ∗(1))| 0 L⊥ v∈S |Hb (Kv,M)| so we only need to show that
Q 0 ∗ 0 v|∞ |H (Kv,M (1))| Y |Hb (Kv,M)| = [K: ] 0 |M| Q |H (Kv,M)| v∈S But
0 0 Y |Hb (Kv,M)| Y |Hb (Kv,M)| 0 = 0 |H (Kv,M)| |H (Kv,M)| v∈S v|∞ 0 0 Y |Hb (Kv,M)| Y |Hb (Kv,M)| = 0 0 |H (Kv,M)| |H (Kv,M)| v|R v|C Gal( / ) Y |M C R /N / M| Y = C R |M|−1 |M Gal(C/R)| v|R v|C and
Q |H0(K ,M ∗(1))| v|∞ v Y −1 0 ∗ Y −2 0 ∗ = |M| |H (Kv,M (1))| |M| |H (Kv,M (1))| |M|[K:Q] v|R v|C Y −1 0 ∗ Y −1 = |M| |H (Kv,M (1))| |M| v|R v|C
Write Gal(C/R) = {1, c}. Then it su ces to show that |M c−1/(c + 1)M| |(M ∗(1))c−1| = |M c−1| |M| and this follows from
|(M ∗(1))c−1| = | ker(c − 1 : M ∗(1) → M ∗(1))| = | ker(c + 1 : M ∗ → M ∗)| = | coker(c + 1 : M → M)| = |M|/| Im(c + 1 : M → M)|
Lecture 6 2013-04-12
4 Rational points on elliptic curves 4.1 Facts about elliptic curves De nition 4.1. Let be a eld. An elliptic curve over is a smooth curve 2 3 2 3 in 2 K K y z = x + axz + bz PK whose solutions E(A) = {(x : y : z) ∈ P2(A)|y2z = x3 +axz2 +bz3} form an abelian group with 0 = (0 : 1 : 0) the point at in nity for any K-algebra A.
12 Theorem 4.2. If K/Q is a number eld and E/K is an elliptic curve then E(K) is a nitely generated abelian group. The rank of E is de ned as the rank of E(K). To prove this theorem we need to collect some facts about elliptic curves.
(4.1.1) First if M/L/K are eld extensions then E(M) has an action of GM/L which commuted with the group law. Moreover, E(M)GM/L = E(L). (4.1.2) If n > 1 is an integer there is a map [n]: E(A) → E(A) given by P 7→ P + ··· + P exactly n times. Write E(A)[n] = ker([n]). If E is de ned over a eld K and L/K is an extension then E(L)[n] is a nite group. Note that GL/K acts on E(L)[n]. The map [n] on E(K) is surjective. (4.1.3) Let K be a number eld. If one clears denominators in the equation of E and v is a place of K such 2 3 2 3 that the equation y z = x + axz + bz is still smooth over the residue eld kv then E is said to have good reduction at v. If n is an integer such that v - n then
E(K)[n] ,→ E(kv) (4.1.4) Let K be a number eld. Consider the function H : P2(K) → [1, ∞) de ned by
!1/[K:Q] Y H(x : y : z) = max(|x|v, |y|v, |z|v) v
This is well-de ned because if × then Q . Moreover, does not depend on λ ∈ K v |λ|v = 1 H(x : y : z) the number eld K over which (x : y : z) is de ned. That H(x : y : z) ≥ 1 follows from the fact that Q 1/[K:Q] . H(x : y : z) ≥ ( v |x|v) = 1 Consider h : E(K) → [0, ∞) de ned by h(P ) = log H(P ). Then
1. for any C > 0, {P ∈ E(K)|h(P ) ≤ C} is a nite set,
2. for any Q ∈ E(K) there exists a constant CE,Q such that h(P + Q) ≤ 2h(P ) + CE,Q for all P ∈ E(K) and
2 3. there exists a constant CE,n such that h([n]P ) ≥ n h(P ) − CE,n.
4.2 Cohomology of elliptic curves over nite elds
In this section k represents a nite eld with q elements. Lemma 4.3. Let C be a smooth projective curve of genus 1 over k. If C has a rational point over some nite extension of k then it has a rational point over k. Proof. Riemann-Roch for curves shows that the zeta function of C has the form
1 − aX + qX2 Z(C,X) = (1 − X)(1 − qX) where X r Z(C,X) = exp |C(Fqr )|X /r r≥1 If then 2 by de nition. The formula above then gives in which C(Fq) = 0 Z(C,X) ≡ 1 (mod X ) a = q + 1 case . But the for all contradicting the hypothesis. Z(C,X) = 1 C(Fqr ) = 0 r Proposition 4.4 (Lang). Let k be a nite eld and let E/k be an elliptic curve. Then H1(k, E(k)) = 0.
13 1 Proof. Let c ∈ H (k, E(k)). Embed E(k) ,→ Aut(E(k)) of automorphisms over k by sending P to tp(Q) = P + Q. The Galois cohomology group H1(k, Aut(E(k))) represents forms of E over k. Let C be the form of E over k represented by the cohomology class −c. In other words C is a smooth projective curve of genus 1 ∼ g −1 g −1 over k with a k-isomorphism φ : C/k = E/k such that φ φ = t−c(g) where φ (x) = g(φ(g x)). Lemma 4.3 implies that C has a point P ∈ C(k) and let Q = φ(P ). Then φgφ−1Q = φg(P ) = g(φ(P )) = g −1 g(Q) since P is de ned over k. But φ φ Q = t−c(g)Q = Q − c(g) which gives c(g) = Q − g(Q). Thus c is trivial in H1(k, E(k)).
4.3 Descent for rational points
Lemma 4.5 (Descent). Let K be a number eld and E/K an elliptic curve. If for some integer m, E(K)/mE(K) is nite, then E(K) is nitely generated.
Proof. Let E(K)/mE(K) be represented by Q1,...,Qr with Q1,...,Qr ∈ E(K). We will show by descent that there exists a constant C such that Q1,...,Qr and the nitely many points {P ∈ E(K)|h(P ) ≤ C} generate E(K). Indeed, let and . Let . We would like to express C1 = max CE,−Qi C = 1 + (CE,m + C1)/2 P ∈ E(K) P as a sum of and points of height at most . Write , and for let . Qi C P = mP1 + Qi1 n ≥ 1 Pn = mPn+1 + Qin Then 1 h(P ) ≤ (h([m]P ) + C ) n m2 n E,m 1 = (h(P − Q ) + C ) m2 n−1 in E,m 1 ≤ (2h(Pn−1) + CE,m + CE,−Q ) m2 in 1 ≤ (2h(P ) + C + C ) m2 n−1 E,m 1 Inductively we get
n n ! 2 X 2i−1 h(P ) ≤ h(P ) + (C + C ) n m2 m2i E,m 1 i=1 2 n C + C ≤ h(P ) + E,m 1 m2 m2 − 2 ≤ 2−nh(P ) + C − 1
Now if n it follows that and n Pn j−1 . 2 > h(P ) h(Pn) < C P = [m ]Pn + j=1[m ]Qij 4.4 Selmer groups for elliptic curves
Lemma 4.6. Let n > 1 be an integer, K a eld and E an elliptic curve over E. Then
0 → E(K)/nE(K) −→δK H1(K,E(K)[n]) −→iK H1(K,E(K))[n] → 0
Proof. This is Kummer theory for elliptic curves. Take the GK -cohomology of the short exact sequence [n] 0 → E(K)[m] → E(K) −→ E(K) → 0 and get the long exact sequence
E(K) −→n E(K) → H1(K,E(K)[n]) → H1(K,E(K)) −→n H1(K,E(K)) giving 0 → E(K)/nE(K) → H1(K,E(K)[n]) → H1(K,E(K))[n] → 0
14 De nition 4.7. Let K be a number eld and E an elliptic curve over K. The n-Selmer group of E is ! n 1 Y 1 Sel (E/K) = ker H (K,E(K)[n]) → H (K,E(Kv))[n] v
where the map to 1 is restriction to followed by . H (K,E(Kv))[n] GKv jKv The purpose of this section is to express n as the Selmer group 1 for a suitable Sel (E/K) HL(K,E(K)[n]) Selmer system L. Proposition 4.8. Let be the image of via in 1 . Then Lv E(Kv)/nE(Kv) δKv H (Kv,E(Kv)[n]) L = {Lv} is a Selmer system and n 1 Sel (E/K) = HL(K,E(K)[n]) Lecture 7 2013-04-15
Proof. The equality n 1 follows from the fact that Sel (E/K) = HL(K,E(K)[n])
n 1 Sel (E/K) = {c ∈ H (K,E(K)[n])|iKv ◦ resKv c = 0} 1 = {c ∈ H (K,E(K)[n])| resKv c ∈ ker iKv } 1 = {c ∈ H (K,E(K)[n])| resKv c ∈ Im δKv } 1 = {c ∈ H (K,E(K)[n])| resKv c ∈ Lv} 1 = HL(K,E(K)[n])
To show that L is a Selmer system we only need to check that for v∈ / S for some nite set S one has 1 . We will do this by showing that each group is contained in the other whenever Lv = Hur(Kv,E(K)[n]) v∈ / S where S is an explicit nite set of places. De ne S to be the set of places consisting of in nite places, of nite places where E has bad reduction and nite places above n. Let . Consider with for some . Kummer theory gives that for any v∈ / S c ∈ Lv c = δKv (P ) P ∈ E(Kv) such that then is the cochain . Suppose Q ∈ E(Kv) [n]Q = P δKv P (δP )(g) = g(Q) − Q ∈ E(Kv)[n] M/Kv is the smallest nite extension such that Q ∈ E(M). × There exists a reduction map E(M) → E(kM ) as follows: if x = (a : b : c) ∈ E(M) there exists λ ∈ M such that and at least one of them is in × . Then . The λa, λb, λc ∈ OM OM x = (λa : λb : λc) ∈ E(kM ) projection map is Galois equivariant where the Galois group acts on via the E(M) → E(kM ) GM/Kv E(kM ) projection to G /I ∼ G . M/Kv M/Kv = kM /kKv Let be the reduction of . Let . Then because Q ∈ E(kM ) Q σ ∈ IM/Kv σ(Q)−Q ∈ E(M)[n] [n](g(Q)−Q) = . Moreover, as is trivial in . But injects into ( has g(P ) − P = 0 σ(Q) − Q = 0 σ GkM /kv E(M)[n] E(kM ) E good reduction at v and v - n as v∈ / S) and so, since σ(Q) − Q projects to σ(Q) − Q = 0, it follows that . But then IM/Kv ur and since is the smallest extension of such that σ(Q) = Q Q ∈ E(M) = E(M ∩ Kv ) M Kv it follows that is unrami ed and so . Finally for , Q ∈ E(M) M/Kv IKv = IM σ ∈ IKv = IM ⊂ GM σ(Q) = Q and so which implies that 1 . Therefore 1 . c(σ) = 0 c = δvP ∈ Hur(Kv,E(Kv)[n]) Lv ⊂ Hur(Kv,E(Kv)[n]) Reciprocally, suppose 1 . By de nition 1 ∼ 1 c ∈ Hur(Kv,E(Kv)[n]) Hur(Kv,E(Kv)[n]) = H (kKv ,E(kKv )[n]) which, by Lemma 4.6, stays in the exact sequence
1 1 0 → E(kKv )/nE(kKv ) → H (kKv ,E(kKv )[n]) → H (kKv ,E(kKv ))[n] → 0 But Proposition 4.4 implies that 1 and therefore ∼ 1 . H (kKv ,E(kKv )) = 0 E(kKv )/nE(kKv ) = H (kKv ,E(kKv )[n]) Thus there exists such that . Now Hensel's lemma allows one to nd such P ∈ E(kKv ) c = δP P ∈ E(Kv) that its image in is . Then and so 1 . E(kKv ) P c = δP Hur(Kv,E(Kv)[n]) ⊂ Lv
15 4.5 Mordell-Weil We are now ready to prove the Mordell-Weil Theorem
Proof of Theorem 4.2. Proposition prop:selmer elliptic shows that n ∼ 1 . We Sel (E/K) = HL(K,E(K)[n]) know that E(K)[n] is nite and therefore Lemma 3.10 implies that Seln(E/K) is nite. We have a commu- tative diagram
0 / E(K)/nE(K) / H1(K,E(K)[n]) SSS SSS SSS SSS SS) 1 1 0 / E(Kv)/nE(Kv) / H (Kv,E(Kv)[n]) / H (Kv,E(Kv))[n]
1 and therefore the image of E(K)/nE(K) in H (Kv,E(Kv))[n] is trivial for every v. By de nition, the image of E(K)/nE(K) in H1(K,E(K)[n]) is included in Seln(E/K) and therefore E(K)/nE(K) is nite. Now Lemma 4.5 implies that E(K) is nitely generated.
The following section was not covered in lecture
4.6 Standard proof of Mordell-Weil I include here the proof from Silverman's book, for comparison.
Lemma 4.9. Suppose E is an elliptic curve over a number eld K and L/K is a nite Galois extension. If E(L)/mE(L) is nite then E(K)/mE(K) is nite.
[m] Proof. Consider the GK -cohomology sequence attached to 1 → E(K)[m] → E(K) −→ E(K) → 1:
[m] [m] 1 → E(K)[m] → E(K) −→ E(K) → H1(K,E(K)[m]) → H1(K,E(K)) −→ H1(K,E(K)) which gives the Kummer isomorphism
1 → E(K)/mE(K) → H1(K,E(K[m])) → H1(K,E(K))[m] Kummer and in ation restriction gives the following diagram
0 / E(L)/mE(L) / H1(L, E(K)[m]) / H1(L, E(K))[m] O O O
0 / E(K)/mE(K) / H1(K,E(K)[m]) / H1(K,E(K))[m] O O O
1 1 ker / H (GL/K ,E(L)[m]) / H (GL/K ,E(L))[m] O O O
0 0 0
1 From the diagram it is clear that the map ker → H (GL/K ,E(L)[m]) is injective. We know that E(L)[m] is nite and GL/K is a nite group and so ker injects into a nite group and thus it is nite. Now E(K)/mE(K) → E(L)/mE(L) is a map with nite kernel and image and so E(K)/mE(K) is nite. We are not ready to prove Theorem 4.2.
16 Proof of Theorem 4.2. By Lemma 4.5 it su ces to show that E(K)/mE(K) is nite. By Lemma 4.9 to show this we may replace K by any nite extension. In particular we may assume that E(K)[m]) ⊂ E(K), which we may do since E(K)[m] is nite. This implies that E(K)[m] = E(K)[m] has trivial GK -action and 1 ∼ so H (K,E(K)[m]) = Hom(GK ,E(K)[m]). If P ∈ E(K) let K(P ) be the nite extension of K generated by the coordinates of P . The Kummer sequence now implies that E(K)/mE(K) ,→ Hom(GK ,E(K)[m]). This map is described explicitly as attaching to P ∈ E(K) the map φP : GK → E(K)[m] given by φP (g) = g(Q) − Q for any Q ∈ E(K) such that [m]Q = P . What is the kernel of φP ? It is K(Q) for the chosen Q with [m]Q = P . Let L be the compositum of all K(Q) such that [m]Q ∈ E(K). Thus φP vanishes on GL ⊂ GK . Moreover, E(K)/mE(K) × GL/K → E(K)[m] de ned by (P, g) 7→ φP (g) is a perfect pairing. Since E(K)[m] is nite to show that E(K)/mE(K) is nite is is enough to show that L/K is a nite extension. m m First, note that φP (g ) = mφP (g) = 0 and so g = 1 in GL/K and so L/K has exponent m. Second, let S be a nite set of places containing the places where E has bad reduction, places v dividing m and the in nite places of K. We now show that L/K is unrami ed outside S. Since the compositum of unrami ed extensions is unrami ed it su ces to show that if [m]Q ∈ E(K) then K(Q) is unrami ed at v∈ / S. We know that as and has good reduction at . We need to show that acts E(K)[m] ,→ E(kv) v - m E v IKv trivially on . Let . Recall that and so . Therefore Q σ ∈ IKv [m]Q = p ∈ E(K) [m](σ(Q) − Q) = σ(P ) − P = 0 . But acts on via the projection ∼ and σ(Q) − Q ∈ E(K)[m] = E(K)[m] ,→ E(kv) GKv E(kv) Gkv = GKv /IKv so in . Therefore in and so acts trivially on as desired. σ(Q) − Q = 0 E(kv) σ(Q) = Q E(K) IKv K(Q) Finally, is an abelian extension because appears as the kernel of the map K(Q)/Q GL φP : GK → E(K)[m] and E(K)[m] is abelian. Therefore L/K is an abelian extension of exponent m which is unrami ed outside S. We want to show that L/K is nite. The inverse of the global Artin map gives an injection
× × ×,0 Y × m Y × GL/K ,→ AK /K K∞ (Kv ) Ov v∈S−∞ v∈ /S
and × × ×,0 Q × m Q × Q × × m. This is so because × is dense in AK /K K∞ v∈S−∞(Kv ) v∈ /S Ov ⊂ v∈S−∞ Kv /(Kv ) K ×,S and so for every × there exists × such that −1 × when and −1 ×,0 AK (av) ∈ AK x ∈ K avx ∈ Ov v∈ / S avx ∈ Kv when v | ∞. Therefore it su ces to show that × × m is nite. But by Kummer theory × × m ∼ 1 Kv /(Kv ) Kv /(Kv ) = H (Kv, µm(Kv)) which is nite because µm(Kv) is nite. End of section not covered in lecture
5 Characters with prescribed behavior 5.1 Grunwald-Wang
Let K be a number eld. The Grunwald-Wang problem asks whether if x ∈ K× is an n-th power in almost all × then is also an -th power in ×. Another way of putting this problem is to study the kernel of Kv x n K the map × × n Y × × n K /(K ) → Kv /(Kv ) v∈ /S × × n 1 where S is some nite set of places of K. Since k /(k ) = H (k, µn) for any eld k this kernel is ! 1 ∼ 1 Y 1 XS(K, µn) = ker H (K, µn) → H (Kv, µn) v∈ /S
Lemma 5.1. Let L/K be a Galois extension of number elds. If all but nitely many places v of K split completely in L then L = K.
17 Proof. Suppose S is a nite set of places of K, containing the in nite places, such that if v∈ / S then splits completely in . For each and a place of let be an integer v = w1 ··· wn L v ∈ S − ∞ w | v L nw ∈ Z≥0 such that U nw ⊂ N L× which is an open subgroup of K×. Let n = max(n ). Kv Lw/Kv w v v w For each a = (a ) ∈ × one may nd x ∈ K× such that a x−1 ∈ U nv for all v ∈ S −∞ and a x−1 ∈ K×,0 v AK v Kv v v for . Indeed, choosing strong approximation states that × is dense in {u},×. Therefore there v | ∞ u∈ / S K AK exists × whose image in {u},× is in the open subset x ∈ K AK Y Y Y a K×,0 a U nv O× v ∞ v Kv v v|∞ v∈S−∞ v∈ /S∪{u}
Then if v | ∞ one has a x−1 ∈ N L× = K×,0, if v ∈ S−∞ one has a x−1 ∈ U nv ⊂ N L× by choice v Lw/Kv w v v Kv Lw/Kv w of and if and then so trivially −1 ×. Therefore −1 × and nv v∈ / S w | v Lw = Kv avx ∈ NLw/Kv Lw ax ∈ NL/K AL so × ×. We deduce that a ∈ K NL/K AL
ab ∼ × × × ∼ GL/K = AK /K NL/K AL = 0 If L/K were abelian then L = K. Suppose L/K is not abelian. Let L/M/K be a subextension such that L/M is abelian and nontrivial. This is always possible as there exists a cyclic subgroup of GL/K . Then let SM be the set of places of M lying above places v ∈ S. Thus for every w∈ / SM , w splits completely in L. Since L/M is abelian it follows that L = M contradicting the fact that L/M is nontrivial. Therefore L = K.
Lecture 8 2013-04-17
Theorem 5.2 (Weak Grunwald-Wang). Let K be a number eld and t = v2(n). 1. If is a cyclic extension then 1 . In particular this is so when . K(ζ2t )/K XS(K, µn) = 1 8 - n 2. If is not cyclic then 1 is a nite 2-torsion group, i.e., if 1 then K(ζ2t )/K XS(K, µn) α ∈ XS(K, µn) α ∈ (K×)n/2. Proof. If (m, n) = 1 and α = am = bn then for pm + qn = 1 one obtains α = (aqbp)mn. Therefore it is enough to show the result when r for a prime . n = p √ p If and 1 then n is a Galois extension. For every one knows √ζn ∈ K α ∈ XS(K, µn) K( α)/K √ v∈ / S that n and therefore the prime of splits completely in n . Lemma 5.1 now shows that √ α ∈ Kv v K K( α) n and therefore in 1 . K( α) = K α = 1 XS(K, µn) n n Q If ζn ∈/ K consider K(ζn). The above shows that α = β for some β ∈ K(ζn). Let X − α = fi(X) the factorization into irreducibles over . Over we know that n Q i and therefore K K(ζn) X − α = (X − βζn) has a root of the form j . The extension is therefore abelian Galois. fi(X) βi = βζn ∈ K(ζn) K(βi)/K For let n for in which case Q and so for some . Since v∈ / S α = αv αv ∈ Kv fi(αv) = 0 fi(αv) = 0 i K(βi)/K is abelian the Galois group acts transitively on the roots of fi and so fi splits completely in Kv and so v splits completely in K(βi). If K(ζn)/K is cyclic of prime power degree then its sub elds are ordered linearly and we may assume that K(β1) ⊂ K(β2) ⊂ .... Since v splits completely in some K(βi) it must also split completely in and so by Lemma 5.1. Finally, n for . K(β1) K(β1) = K α = β1 β1 ∈ K r r r If n = p with p > 2 then K(ζn)/K is cyclic of degree p . If n = 2 and K(ζ2t )/K is assumed cyclic then 1 . XS(K, µn) = 1 √ √ √ If is not cyclic then as is cyclic. Therefore × n. K(ζ2t )√/K K( −1) 6= K K(ζ2t )/K( −1) α ∈ (K( −1) ) Let such that n in which case taking norms one gets 2 √ n and the β ∈ K( −1) α = β α = (NK( −1)/K β) conclusion follows.
The following section was not covered in lecture
18 I include, without proof, the full Grunwald-Wang theorem.
Theorem 5.3 (Strong Grunwald-Wang). For let −1. Let be a number eld and a r ≥ 1 ηr = ζr + ζr K S nite set of places of . Let be the largest integer such that . Then 1 unless: K r ηr ∈ K XS(K, µn) = 1
1. −1, 2 + ηr and −(2 + ηr) are non-squares in K and
2. v2(n) > r and
3. S contains the set SK of all the places v | 2 such that −1, 2 + ηr and −(2 + ηr) are non-squares in Kv. End of section not covered in lecture
5.2 Characters with prescribed nite order local behavior
Let K be a number eld and S a nite set of places of K. Lemma 5.4. For every nite index open subgroup of Q × there exists an open subgroup of P0 P = v∈S Kv U × × such that . AK /K U ∩ P = P0
Proof. This proof has some missing topological details. See [AT09, Chapter 10, Lemma 4]. Let CK = × ×. First, for , n and n are closed subgroups of and n is open in n . AK /K n > 1 P0CK PCK CK P0CK PCK Therefore there exists open such that n n . Let n . Then V ⊂ CK PCK ∩ V ⊂ P0CK U = VP0CK P ∩ U = n n n n n n . P ∩ PCK ∩ P0CK V = P ∩ P0CK (PCK ∩ V ) = P ∩ P0CK = P0(P ∩ CK ) n Now P0 is nite index in P and so for some integer n one has P ⊂ P0 in which case we take the open of such that 2n . Now suppose 2n. Then there exists × U CK P ∩ U = P0(P ∩ CK ) α ∈ P ∩ CK (αv) ∈ P ⊂ AK and × such that × 2n. Since if it follows that × 2n for . Now x ∈ K (αv)x ∈ (AK ) αv = 1 v∈ / S x ∈ (Kv ) v∈ / S Theorem 5.2 implies that × n and so × n so n. Thus 2n n and therefore x ∈ (K ) (αv) ∈ (AK ) (αv) ∈ P P ∩ CK ⊂ P 2n as desired. P ∩ U = P0(P ∩ CK ) = P0 Theorem 5.5. Let K be a number eld and S a nite set of (not necessarily nite) places. For each v ∈ S let × × be a continuous character of nite order . Then there exists a continuous nite order χv : Kv → C nv × × × character χ : K \ → such that χ| × = χ when v ∈ S. AK C Kv v Proof. Let Q ×. Have a character × and let be a nite index P = v∈S Kv ⊗v∈Sχv : P → C P0 = ker ⊗χv subgroup of . Lemma 5.4 provides an open subgroup of × such that . But then ∼ P U AK P ∩ U = P0 P U/U = ∼ 1 P/P ∩ U = P/P0 and therefore the character χS = ⊗v∈Sχv : P/P0 → S extends to a character χS : 1. Finally, is nite index and so extends to a character × × 1 which P U/U → S PU ⊂ CK χS χS : AK /K U → S is a global nite order Hecke character.
5.3 Characters with prescribed local behavior at in nite places Write for the usual absolute value on × and for its square. | · | C | · |C Lemma 5.6. Continuous characters of × are of the form ε t for and . R x 7→ sign(x) |x| ε ∈ {0, 1} tv ∈ C Continuous characters of × are of the form x 7→ (x/|x|)m|x|t for some m ∈ and t ∈ . C C Z C Proof. All continuous homomorphisms from R to C are obtained by scalar multiplication and so all continuous homomorphisms from (0, ∞) to C× are of the form x 7→ xt for t ∈ C. The result then follows from the fact that R× =∼ {−1, 1} × (0, ∞) and C× =∼ S1 × (0, ∞).
De nition 5.7. A Hecke character × × is unitary if for each one has mv itv ψ : AK /K v | ∞ ψv(x) = (x/|x|) |x|v where . The character is said to be algebraic of type if for all . The character is tv ∈ R ψ A0 tv = 0 v | ∞ ψ algebraic of type A if
19 Proposition 5.8. Let be a number eld and for each let ( or if ) and . K v | ∞ mv ∈ Z 0 1 v | R tv ∈ R There exists a Hecke character × × × such that for , mv itv if and only χ : AK /K → C v | ∞ χv(x) = (x/|x|v) |x| if C mv Y ιv(α) itv |ιv(α)| = 1 |ιv(α)|v C v|∞ for in a nite index subgroup of × . α OK Proof. If χ is a global Hecke character let U be the nite index open subgroup of Q O× such that v-∞ v × Q χ|U = 1, i.e., U is the conductor of χ. Then for every α ∈ O ∩ U one has χv(α) = 1 and therefore K v-∞ since Q one also has Q . v χv(α) = 1 v|∞ χv(α) = 1 Reciprocally, let V ⊂ O× be a nite index subgroup. There exists a nite index subgroup U ⊂ Q O× K v-∞ v such that × . There is an exact sequence V ⊃ OK ∩ U
× × × × 1 → K∞U/(OK ∩ U) → AK /K → Cl(U) → 1 where Cl(U) is a nite group. De ne on by letting as required when and when . The hypothesis implies χ K∞U χv v | ∞ χv = 1 v - ∞ that factors through × × . Choosing a section to the exact sequence above, since is χ K∞U/(OK ∩ U) Cl(U) nite, one may extend to a Hecke character χ. (Such a section exists because to an ideal I ∈ Cl(U) one may Q v(I) attach $v which is a homomorphism.) v-∞ Lecture 9 2013-04-19
Lemma 5.9. Let K be a number eld. A continuous Hecke character has nite order if and only if it is trivial on ×,0. K∞ Proof. If then m t for and . If then on ×,0 this v | ∞ φv(x) = (x/|x|) |x| m ∈ Z t ∈ C v = R Kv = (0, ∞) t C ×,0 × is φv(x) = |x| and if this is nite order then t = 0 and so φv| ×,0 = 1. If v = then on K = C Kv C v C iθ imθ t this is φv(re ) = e r . If φv is nite order then necessarily t = 0 and, since for θ irrational the set {mθ mod 2π} ⊂ [0, 2π) is dense, also m = 0 which gives again φ | ×,0 = 1. v Kv f,× Reciprocally, suppose φ| ×,0 = 0. Since φ is continuous there exists an open subgroup U ⊂ such K∞ AK that . But then factors through × × ×,0 which is nite, thereby showing that has nite U ⊂ ker φ φ AK /K K∞ U φ order.
Theorem 5.10 (Weil-Artin). Let K be a number eld and KCM ⊂ K be the maximal CM sub eld (where if has no CM sub elds). Let × × × be an algebraic Hecke character of type . KCM = Q K ψ : AK /K → C A0 Then there exists an algebraic Hecke character ψCM of KCM of type A0 and a nite order character χ of K such that . In particular every algebraic Hecke character of type of a totally real ψ = χ · ψCM ◦ NK/KCM A eld is the product of a nite order character and | · |m for some integer m. AK Proof. By Lemma 5.9 a continuous Hecke character has nite order if and only if it is trivial on ×,0. K∞ Therefore it su ces to show that and agree on × for some algebraic character . ψ ψCM ◦ NK/KCM K∞ ψCM First, assume is Galois with Galois group . Since acts transitively on places of K/Q G G = GK/Q K above a place of Q, the in nite places of K are either all real or all complex. Suppose that is totally real. Then for , mv mv which is trivial on ×,0 K v | ∞ ψv(x) = (x/|x|) = (sign x) Kv and so ψ readily has nite order and the theorem follows. Suppose that K is totally complex. The character ψ is algebraic of type A0 and so for an in nite place , mv and Proposition 5.8 implies the existence of an open subgroup of × such that for v ψv(x) = (x/|x|) U OK
20 one has Q mv . Squaring we get α ∈ U v|∞(ιv(α)/|ιv(α)|) = 1
mv 2mv ! Y ιv(α) Y ιv(α) = |ιv(α)| v|∞ v|∞ ιv(α) = 1 Fixing an embedding , we have and for we write if τ : K,→ C {ιv, ιv} = {τ ◦ g|g ∈ G} g ∈ G mg = mτ◦g τ ◦ g = ιv and mg = −mτ◦g if τ ◦ g = ιv. Then the equation above becomes Y τ ◦ g(α)mg = 1 g∈G and since τ is injective Y g(α)mg = 1 g∈G For any complex embedding let be the (necessarily nontrivial) element of induced by ι : K,→ C cι G complex conjugation on , i.e., . Suppose . Then and so by C ι(x) = ι(cι(x)) ιv = τ ◦ g ιv = ιv ◦ cιv = τ ◦ (cιv g) de nition m = −m . cιv g g The eld is the xed eld of 0 . The equation above becomes KCM {cιcι0 |ι, ι : K,→ C}
Y mg mc g g(α) cιg(α) ι = 1
G/cι for × . α ∈ U ⊂ OK Note that so taking logarithms get |ι(g(α))|C = |ι(cιg(α))|C X (mg + mcιg) log |ι(g(α))|C = 0 G/cι Theorem 2.1 implies that × and therefore the nite index subgroup of × have rank OK U OK #∞−1 = |G|/2−1 which is also the rank of #∞ and so is independent of . But ∆∞ = ker R → R wι = mg + mcιg g wι|G|/2 = P (m + m ) = P m is also independent of ι and so w is independent of ι. Therefore for any g∈G/cι g cιg g∈G g ι 0 other ι : K,→ one has m + m 0 = w 0 = w = m + m and so m = m 0 . Thus m is constant C cιg cιcιg ι ι g cιg g cιcιg g on -orbits and so GK/KCM Y mg Y mg g(α) = g(NK/KCM α) = 1 g∈G g∈G KCM/Q and if ranges over a nite index subgroup of × then ranges over a nite index subgroup 0 of α OK NK/KCM α U O× . What we get is that for α ∈ U 0 ⊂ O× have KCM KCM Y g(α)mg = 1 g∈G KCM/Q and m = −m from the fact that over K have m = −m . Suppose τ : K ,→ is a complex cιg g cιv g g CM C embedding. Then writing c for the unique complex conjugation on KCM we have
!mg 2m Y τ(g(α)) Y τ(g(α)) g = = 1 |τ(g(α))| g∈G /c τ(g(α)) g∈G /c KCM/Q KCM/Q and by restricting the nite index subgroup U 0 a little more we conclude by Proposition 5.8 the existence of mv an algebraic Hecke character ψCM of KCM such that at v | ∞ in KCM, ψCM,v(x) = (x/|x|) where mv = mg −1 × if ιv = τ ◦ g. In that case ψ · ψCM ◦ N is trivial on K and so has nite order. K/KCM ∞
21 Now if is not Galois let be its Galois closure and the maximal CM sub eld of . Let K/Q L/K LCM L and . By de nition and so . Now is H1 = GL/K H2 = GL/LCM K ∩ LCM = KCM GL/KCM = H1H2 ψ ◦ NL/K a character on which is Galois over and so where is an algebraic character L Q ψ ◦ NL/K = χ · η ◦ NL/LCM η of LCM. As before we need to show that the integers {mv} where v | ∞ runs over places of L are constant on orbits of . But the integers are constant along -orbits since they are attached to GL/KCM = H1H2 H1 and they are constant along orbits because . ψ ◦ NL/K H2 ψ ◦ NL/K = χ · η ◦ NL/LCM Lecture 10 2013-04-22
Proposition 5.11. Let be an integer not divisible by 8 and 1 be a continuous n > 1 ω : µn(K)\µn(AK ) → S character such that for . Then there exists a nite order character × × 1 whose ωv = 1 v | C ωe : K \AK → S restriction to is . Here 1 . µn(AK ) ω S = {z ∈ C||z| = 1} Proof. We have an exact sequence
× × 1 µn(K)\µn(AK ) → (AK /K )[n] → X∅(K, µn) → 1 as follows: if then n for some ×. Let be the image of in 1 . What (αv) ∈ CK [n] (αv ) = x x ∈ K y x X∅(K, µn) × n −1 −1 is the kernel of this map? If y = 1 then x ∈ (K ) . But (αv)x = (αv) in CK and αvx ∈ µn(Kv) and so −1 . By Theorem 5.2 since , 1 and so is a character of 1. (αv)x ∈ µn(AK ) 8 - n X∅(K, µn) = 1 ω CK [n] → S For an abelian topological group G denote by G∨ = Hom(G, S1) the Pontryagin dual consisting of continuous homomorphisms. Then (G/H)∨ =∼ H⊥, where H⊥ = {φ ∈ G∨|φ(H) = 1}, and G∨∨ =∼ G. Thus ∨ ∨ ∨ ∼ ∨ ⊥ ∼ n ∨ . By duality get ∨ ∼ ∨ ∨ and (CK /nCK ) = (nCK ) = {x ∈ CK |φ(x ) = 1, ∀φ ∈ CK } = CK [n] CK [n] = CK /nCK so we get an extension 1 which is well-de ned up to -th power characters. We only need to ωe : CK → S n show that the lift can be chosen to have nite order. Theorem 5.10 implies that there exists an algebraic character η of K such that ω and η ◦ N CM e K/KCM dier by a nite order character. Thus it su ces to show that η can be chosen to have nite order, i.e., to be trivial on × . But we know that at , on and so which implies that KCM,∞ v | C ωv = 1 µn n | mv n | mηv for every (necessarily complex) place of KCM. Proposition 5.8 implies that there exists a Hecke character µ mη /n of KCM such that µv(x) = (x/|x|) v (the condition of Proposition 5.8 is automatically satis ed because it is satis ed for η) and in that case ηµ−n is nite order and thus ω(µ ◦ N )−n is nite order with the e K/KCM same restriction to as . µn(AK ) ωe 6 Projective Galois representations 6.1 A theorem of Tate The setup of the rst lemma is that is a nite extension such that and we may choose K/Q` µp(K) ⊂ K a primitive root of unity. Then the -cohomology sequence of p ζp ∈ µp(K) GK 1 → Z/pZ → Qp/Zp −→ Qp/Zp gives 1 p 1 δ 2 H (K, Qp/Zp) −→ H (K, Qp/Zp) −→ H (K, Z/pZ) ∼ where 2 ∼ 2 2 × ∼ = 1 where the rst isomorphism H (K, Z/pZ) = H (K, µp(K)) = H (K, K )[p] = Br(K)[p] −→ p Z/Z is via the identi cation of -modules ∼ via x and the last map is the invariant map GK Z/pZ = µp(K) x 7→ ζp . The local Artin map × ∼ ab ab permits the identi cation of continuous homomorphisms invK rK : K = WK ⊂ GK with continuous homomorphisms × . φ : GK → Qp/Zp φ ◦ rK : K → Qp/Zp Lemma 6.1. Let be as above. K/Q` 1. If φ ∈ H1(K, / ) =∼ Hom(G , / ) then δ(φ) only depends on (φ ◦ r )| . In fact there Qp Zp K Qp Zp K µp(K) exists , independent of , such that . a ∈ Z/pZ φ invK (δ(φ)) = aφ(rK (ζp))
22 2. The connecting homomorphism δ is an isomorphism, i.e., a 6= 0. 3. The constant is independent of . a K/Q` 4. If F is a number eld such that ζ ∈ F and u and u are two nite places of F then a = a . p 1 2 Fu1 Fu2
Proof. Note that 1 p 1 δ 2 is exact and so factors through H (K, Qp/Zp) −→ H (K, Qp/Zp) −→ H (K, Z/pZ)
1 1 δ 2 0 → H (K, Qp/Zp)/pH (K, Qp/Zp) −→ H (K, Z/pZ)
But 1 ∼ ∼ × . The sequence × p × H (K, Qp/Zp) = Hom(GK , Qp/Zp) = Hom(K , Qp/Zp) 1 → µp(K) → K −→ K is exact and so × × is exact which means that Hom(K , Qp/Zp) → Hom(K , Qp/Zp) → Hom(µp, Qp/Zp) 1 1 . Therefore only depends on the restriction H (K, Qp/Zp)/pH (K, Qp/Zp) ,→ Hom(µp, Qp/Zp) δ(φ) δ ◦ . rK |µp(K) Since it follows that × for a maximal . Let × such that µp ⊂ K µpn ⊂ K n > 0 φ : K → Qp/Zp −n; such a exists by the decomposition × ∼ Z × TF where stands for torsion φ(ζpn ) = p φ K = $K ×µ∞(K)×(OK ) TF free. If there exists × such that then −n and so a for ×. ψ : K → Qp/Zp φ = pψ pψ(ζpn ) = p ψ(ζpn ) = pn+1 a ∈ Zp But then n a in whereas n . We conclude that × p ψ(ζpn ) = p 6= 0 Qp/Zp p ψ(ζpn ) = ψ(1) = 0 Hom(K , Qp/Zp) 6= × and so × × ∼ and so p Hom(K , Qp/Zp) 0 6= Hom(K , Qp/Zp)/p Hom(K , Qp/Zp) ,→ Hom(µp, Qp/Zp) = Z/pZ × × ∼ . Hom(K , Qp/Zp)/p Hom(K , Qp/Zp) = Z/pZ Then is a homomorphism and it follows that there exists such that invK ◦δ Z/pZ → Z/pZ a ∈ Z/pZ invK (δ(φ)) = aφ(rk(ζp)) for all φ. To check that is an isomorphism simply note that injects × × ∼ δ δ Hom(K , Qp/Zp)/p Hom(K , Qp/Zp) = 2 and so is injective which implies that and so is an isomorphism. Z/pZ ,→ H (K, µp) δ a 6= 0 δ Let be a nite extension. Under × ×, ∨ and for × L/K µp ⊂ K ⊂ L rL(ζp) = cor ◦rK (ζp) φ : L → Qp/Zp one has
invK (δK (cor φ)) = aK cor φ(rK (ζp))
invL(δL(φ)) = aLφ(rL(ζp)) ∨ = aLφ(cor rK (ζp))
= aL cor φ(rK (ζp))
But invK (δK (cor φ)) = invL(δL(φ)) and so aK = aL. Consider the number eld . It su ces to show that if and are nite places of then K = Q(ζp) u1 u2 K . Let ∼ 1 δ 2 ∼ 2 × ∼ . aKu = aKu φ ∈ Hom(GK , Qp/Zp) = H (K, Qp/Zp) −→ H (K, µp) = H (K, K )[p] = Br(K)[p] Then1 P 2 because . Therefore P . v invv δ(φ) = 0 0 → Br(K) → ⊕ Br(Kv) → Q/Z → 0 aKv φ(rKv (ζp)) = 0 Since × ab is trivial on × it follows that P . Suppose one rK : AK → GK K φ(rK (ζp)) = v φ(rKv (ζp)) = 0 may choose such that is nonzero at exactly and . Then the equations φ φ(rKv (ζp)) u1 u2 φ(rKu (ζp)) + and give as desired. 1 φ(rKu (ζp)) = 0 aKu φ(rKu (ζp)) + aKu φ(rKu (ζp)) = 0 aKu = aKu We2 now show the existence1 1 of such a global2 character2 . Let 1 2 be a nontrivial character. φ η : µp → Z/pZ Since for nite have × × ∼ one may extend to a v Hom(Kv , Qp/Zp)/p Hom(Kv , Qp/Zp) = Hom(µp(Kv), Qp/Zp) η character × and be restriction get a character × . Now × × is dense and η : Kv → Qp/Zp η : K → Qp/Zp K ⊂ Kv × −1 so there exist φ : K → /p such that φ | × = η and φ | × = η . Let S = {u , u }. Recall that ui ui Z Z u1 K u2 K 1 2 is generated by the inertia groups and and so ab ∼ × × × Q × . Consider GK/K IKu IKu GK,S = AK /K K∞ v∈ /S OK the exactS sequence 1 2 v × × × × × ab 1 → O O /K ∩ O O → GK,S → Cl(K) → 1 Ku1 Ku2 Ku1 Ku2 The character on × × will then be trivial on × × × and, since the cokernel φ = φu1 ⊗ φu2 O O K ∩ O O Ku1 Ku2 Ku1 Ku2 is nite, will extend to a character of ab . Let's check that satis es the requirements. For Cl(K) GK,S φ , and so . If then ±1 . v 6= u1, u2 rKv (ζp) ∈ IKv φ(rKv (ζp)) ∈ φ(IKv ) = 0 v ∈ S φ(rKv (ζp)) = η (ζp) 6= 0
23 Lecture 11 2013-04-24
Theorem 6.2. Let K be a number eld. Then H2(K, Q/Z) = 0. Proof. Since it is enough to show that 2 . In ation-restriction gives Q/Z = ⊕Qp/Zp H (K, Qp/Zp) = 0
2 2 2 1 → H (GK(µp)/K , Qp/Zp) → H (K, Qp/Zp) → H (K(µp), Qp/Zp) Since × and is pro- it follows that 2 . Therefore to show GK(µp)/K ⊂ (Z/pZ) Qp/Zp p H (GK(µp)/K , Qp/Zp) = 0 that 2 it is enough to do the same for . We now assume that . H (K, Qp/Zp) = 0 K(µp) µp ⊂ K Consider the exact sequence
1 δ 2 2 p 2 H (K, Qp/Zp) −→ H (K, Z/pZ) → H (K, Qp/Zp) −→ H (K, Qp/Zp) The group 2 is -power torsion so to show it is trivial it is enough to show that multiplication H (K, Qp/Zp) p by is injective, i.e., the connecting homomorphism is surjective. Note that 2 ∼ 2 p δ H (K, Z/pZ) = H (K, µp) = 2 × ∼ so we need to show that 1 is surjective. H (K, K )[p] = Br(K)[p] H (K, Qp/Zp) → Br(K)[p] In other words we need to show that for every torsion Brauer class there exists a ab α ∈ Br(K)[p] φ : GK → such that . We know that and so projects Qp/Zp δφ = α 0 → Br(K) → ⊕ Br(Kv) → Q/Z → 0 α ∈ Br(K)[p] injectively to a collection with . By Lemma 6.1 there exists such (αv) αv ∈ Br(Kv)[p] φv : GKv → Qp/Zp that and some such that . For all but nitely many , and δφv = αv a ∈ Z/pZ invv αv = aφ(rKv (ζp)) v αv = 0 so and therefore we get a character P . Also for × we φv(µp(Kv)) = 0 φ = φv : µp(AK ) → Qp/Zp ζp ∈ µp(K ) have P −1 P and so factors through . φ(ζp) = φv(rKv (ζp)) = a invv αv = 0 φ φ : µp(K)\µp(AK ) → Qp/Zp Since , for one has . Moreover, as it follows that and so Br(C) = 0 v | C φv = 0 pαv = 0 pφ = 0 1 . Writing we get × having nite order Im φ ⊂ p Z/Z ψ(x) = exp(2πix) Φ = ψ ◦ φ : µp(K)\µp(AK ) → C p and such that for . Proposition 5.11 implies the existence of a nite order extension of from Φv = 1 v | C Φ to × ×. As has nite order it is necessarily trivial on ×,0 and so factors through µp(K)\µp(AK ) AK /K Φ K∞ Φ × × ×,0 ∼ ab. As has nite order, its image lies in and so composing with logarithm we AK /K K∞ = GK Φ ψ(Q/Z) get an extension ab and composing again with projection to gives ab . By φ : GK → Q/Z Qp/Zp φ : GK → Qp/Zp construction δφ = α as desired.
6.2 Lifting projective Galois representations
Lemma 6.3. Let Γ be a pro nite group and let H ⊂ G be topological groups such that H ⊂ Z(G). Let ρ :Γ → G/H be a homomorphism. For each g ∈ Γ let ag be an arbitrary lift of ρ(g) to G. Then c(g, h) = −1 −1 is well-de ned in 2 and there exists a homomorphism such that the image of aghah ag H (Γ,H) ρ :Γ → G ρ(g) in G/H is ρ(g) if and only if c is cohomologically trivial. Proof. Since the image of c(g, h) in G/H is ρ(gh)ρ(h)−1ρ(g)−1 = 1 it follows that c(g, h) ∈ H. Moreover, as it follows that we also have −1 −1 −1 −1. Using these and the fact that c(g, h) ∈ Z(G) c(g, h) = ah ag agh = ag aghah c(g, h) ∈ Z(G) one may check that (dc)(g, h, i) = c(h, i)c(gh, i)−1c(g, hi)c(g, h)−1 = 1 and so c ∈ Z2(Γ,H), where H has trivial Γ action. Also note that if 0 is any other lift of to then writing 0 −1 one has . ag ρ(g) G φ(g) = agag φ(g) ∈ H Writing 0 0 0 −1 0 −1 one gets 0 −1 −1 and so 0 and are equal c (g, h) = agh(ah) (ag) c (g, h) = φ(gh)φ(g) φ(h) c(g, h) c c in 2 . Certainly the lift 0 comes from a homomorphism if and only if 0 if and only H (Γ,H) ag ρ :Γ → G c = 1 if c = c0 is cohomologically trivial in H2(Γ,H). Theorem 6.4. Let be a number eld and be a continuous homomorphism. Then K ρ : GK → PGL(n, Qp) there exists a continuous representation such that the image of in is ρ : GK → GL(n, Qp) ρ(g) PGL(n, Qp) ρ(g) for any g.
24 Proof. Note that via the natural projection and that the kernel is Hm = SL(n, Qp)µnm(Qp) →→ PGL(n, Qp) . 1 → µnm(Qp) → Hm →→ PGL(n, Qp) 2 Let cm ∈ H (GK , µmn) be the cohomology class associated by Lemma 6.3 to an arbitrary lift of ρ to 2 2 0 Hm. The cohomology classes are compatible under the maps H (GK , µnm) → H (GK , µnm0 ) if m | m and so we get a cohomology class
c = lim c ∈ lim H2(G , µ ( )) = H2(G , / ) = 0 −→ m −→ K mn Qp K Q Z
where the last equality if the content of Theorem 6.2. But then c = 0 implies that cm = 0 for some m and therefore lifts to a homomorphism . ρ ρ : GK → Hm ⊂ GL(n, Qp) Lecture 12 2013-04-26
Proposition 6.5. Suppose is a number eld and is a projective Galois represen- K ρ : GK → PGL(n, Qp) tation which is unrami ed almost everywhere. Let be a lift of . Then is unrami ed ρ : GK → GL(n, Qp) ρ ρ almost everywhere. Proof. Let be a nite extension such that (standard Baire category theory F/Qp ρ : GK → GL(n, F ) argument). The representation ρ is continuous and so there exists L/K nite such that ρ(GL) ⊂ 1 + 2 2 p Mn×n(OF ) (the composition of ρ with projection to the discrete group GL(n, F )/(1 + p Mn×n(OF )) has 2 2 2 2 open kernel). Note that log : 1 + p Mn×n(OF ) → p Mn×n(OF ) and exp : p Mn×n(OF ) → 1 + p Mn×n(OF ) are inverses to each other and satisfy the Baker-Campbell-Hausdor formula. Thus the log map is injective 2 n 2 and has the property that log((1 + p X) ) = n log(1 + p X). Therefore log ρ(GL) is a pro-p torsion-free group. Let be a place such that , is unrami ed at and . This implies that v v - p Lw/Kv w | v ρ(IKv ) = 1 and so it su ces to show that . But × IKv = ILw ρ(ILw ) = 1 ρ(ILw ) ⊂ F = ker(GL(n, F ) → PGL(n, F )) ab ab × × TF is abelian and so ρ(I ) = ρ(I ). Now I ∼ O ∼ k × µ ∞ (L ) × (1 + ($ )) where TF stands Lw Lw Lw = Lw = Lw p w w for torsion-free and ∞ is nite because the rami cation of n grows with . Since µp (Lw) Lw(µp )/Lw n ρ(ILw ) is torsion-free it follows that and so TF . But TF, being a ρ(µ∞(Lw)) = 1 ρ(ILw ) = ρ((1 + ($w)) ) (1 + ($w)) subgroup of , is pro- -group whereas its image is in 2 which is pro- with . 1 + ($Lw ) qw 1 + p Mn×n(OF ) p p - qw All subgroups of pro- groups must be pro- and so TF and it follows that . p p ρ((1 + ($w)) ) = 1 ρ(ILw ) = 1
6.3 Local Galois representations in the tame case Theorem 6.6. Let be a nite extension and be an irreducible continuous repre- K/Qp ρ : GK → GL(n, C) sentation. If p - n (the tame case) then there exists an order n extension L/K and a continuous character × × such that ∼ K . χ : L → C ρ = IndL χ Before proving the theorem we give two results of Cliord.
Proposition 6.7. Let G be a pro nite group and let N C G be an open normal subgroup. Let (ρ, V ) be an irreducible representation of G over a eld K and let (ρ, W ) ⊂ V |N be an irreducible component. Then 1. P and there exist such that as representations of . V = g∈G/N ρ(g)W g1, . . . , gn V = ⊕ρ(gi)W N 2. If U is an irreducible representation of N write V [U] be the U-isotypical component in V , i.e., the set of v ∈ V lying in the image of some N-equivariant map U → V . Then the set of U such that V [U] 6= 0 is nite, G acts transitively on it and V = ⊕V [U]. 3. Let U be an irreducible representation of N such that V [U] 6= 0 and write H = {g ∈ G|gV [U] = V [U]}. Then ∼ G as representations of . V = IndH V [U] G
25 Proof. Clearly P is -invariant and must be equal to since is an irreducible - g∈G/N ρ(g)W ⊂ V G V V G representation. Now since N is normal in G, ρ(g)W is also a (necessarily irreducible) representation of N 0 and thus ρ(g)W ∩ ρ(g )W is either 0 or equal to ρ(g)W and therefore V becomes a direct sum of ρ(gi)W for nitely many gi. That the set of U with V [U] 6= 0 is clear from the fact that V is nite dimensional. Next, let U ⊂ V |N irreducible, then U ⊂ ⊕ρ(gi)W and since each ρ(gi)W is irreducible it follows that U = ρ(gi)W for some 0 0 ∼ 0 gi. Finally V [U] = ⊕U where U ⊂ V |N such that U = U as N-representations and so V = ⊕V [U] as N-representations. Finally, let U = U1,U2,...,Um be the nitely many irreducible representations of N such that V = ⊕V [Ui] and let gi such that Ui = ρ(gi)U. It follows that G/H = {g1, . . . , gm}. Any v ∈ V can be represented P P uniquely as v = ρ(gi)vi where vi ∈ U. The map V → K[G] ⊗K[H] V [U] given by v 7→ [gi] ⊗ vi is an isomorphism of vector spaces which can easily be checked to be G-equivariant and therefore is an isomorphism of -representations. Finally, ∼ G via the -equivariant map sending to G K[G] ⊗K[H] V [U] = IndH V [U] G [g] ⊗ v the function sending g to v and every coset other than gH to 0. Lecture 13 2013-04-29
Proposition 6.8. Let G, N, V and W as in Proposition 6.7 such that K is algebraically closed and V = V [W ] (in which case immediately dim W | dim V ). Then there exist irreducible projective representations σ : G → PGL(dim W, K) and τ : G → PGL(dim V/ dim W, K) with τ trivial on N, such that ρ = σ ⊗ τ in PGL(dim V,K).
Proof. The group N acts on ρ(g)W via ρ(n)ρ(g)w = ρ(g)ρg(n)w which makes sense since N is normal in . Recall that as -representations where has the action g −1 . G V = ⊕ρ(g)W N ρ(g)W ρ|W (n)w = ρ|W (g ng)w Since ∼ as is isotypic there exist matrices such that g −1 for all , ρ(g)W = W V Ag ρ|W (n) = Agρ|W (n)Ag n ∈ N where Ag is de ned up to scalars. Let σ(g) = Ag : G → PGL(W ) be the rst projective representation; the fact that this is a homomorphism is straightforward to check.
Proposition 6.7 shows that V = ⊕ρ(gi)W for nitely many gi in which case dim W | dim V as desired. Let r = dim V/ dim W and let U be a vector space spanned by u1, . . . , ur, let w1, . . . , wm be a basis of P W and let wij = ρ(gi)wj be a basis of ρ(gi)W . For g ∈ G write ρ(gig)wj = βijkl(g)wkl (the order really is gig!) which can be done since wkl is a basis for V . De ne U ⊗K W as a G-representation by P which exhibits ∼ as -representations. It su ces to construct µ(g)(ui ⊗wj) = k,l βijkl(g)uk ⊗wl U ⊗W = V G the projective representation τ on PGL(U) such that ρ =∼ σ ⊗ τ. Note that for n ∈ N one has ρ(gin)wj = ρ(gi)ρ|W (n)wj ∈ ρ(gj)W and so µ(n) = 1 ⊗ ρ|W (n). For every one has that g −1 and therefore get that g ∈ G ρ|W (n) = Agρ|W (n)Ag
g g µ (n) = 1 ⊗ ρ |W (n) −1 = (1 ⊗ Ag )(1 ⊗ ρ|W (n))(1 ⊗ Ag) µg(n) = µ(g)−1µ(n)µ(g) −1 = µ(g) (1 ⊗ ρ|W (n))µ(g)
So µ(g)(1 ⊗ Ag) commutes with 1 ⊗ ρ|W (n) for all g ∈ G. For g ∈ G let F (g) = µ(g)(1 ⊗ Ag) ∈ End(U ⊗ W ). Since every element of U ⊗K W can be written uniquely as a linear combination of ui ⊗vi for some vi ∈ W and P the basis vectors ui of U, one can write F (g)(ui ⊗ w) as uj ⊗ Fij(g)(w) for linear maps Fij(g) ∈ End(W ). Since F (g) commutes with 1 ⊗ ρ|W (n) it follows that Fij(g) commutes with ρ|W (n). But (ρ|N ,W ) is ∼ × irreducible and so Fij(g) ∈ Z(EndN (W )) = K by Schur's lemma. Therefore Fij(g) = αij(g) for scalars × P αij(g) ∈ K . Writing τ(g)ui = αij(g)uj it follows that τ(g) ∈ End(U) giving a projective representation τ : G → PGL(U) (since Ag is de ned up to scalars only). That τ is trivial on N follows from the fact that for n ∈ N, µ(n) = 1 ⊗ ρ|W (n). Finally, σ and τ are irreducible because otherwise V would be reducible, which it is not.
26 De nition 6.9. A supersolvable nite group is a nite group G with a descending ltration G = G0 ⊃ such that and the successive quotients are cyclic. G1 ⊃ · · · ⊃ Gn = 1 Gi C G Gi/Gi+1 Proposition 6.10. If (ρ, V ) is an irreducible representation of the supersolvable group G over an alge- braically closed eld K then there exists a subgroup H ⊂ G and a character χ : H → K× such that ∼ G . ρ = IndH χ Proof. If G is abelian then ρ is a character to begin with. We will prove the result by induction on |G|. If ρ : G → GL(V ) is not faithful, i.e., if ker ρ = H C G then ρ factors through ρ : G/H → GL(V ) where |G/H| < |G| and so by induction ρ is induced from a character. Suppose therefore that ρ is faithful. The group is supersolvable with ltration with G/Z(G) G/Z(G) = H0 ⊃ H1 ⊃ · · · ⊃ Hm = 1 Hi C G/Z(G) and Hi/Hi+1 cyclic. Let H = Hm−1Z(G) which will be proper in G if G is not abelian. Since is a cyclic normal subgroup of , is abelian. Since and is Hm−1 = Hm−1/Hm G/Z(G) H C G H 6⊂ Z(G) ρ injective it follows that ρ(H) 6⊂ Z(ρ(G)) = K× where the last equality follows from the irreducibility of ρ. Now Proposition 6.7 implies, since H is normal in G that if U ⊂ V |H is a (necessarily one dimensional) ∼ irreducible then V [U] is irreducible as a representation of HU = {g ∈ G|gV [U] = V [U]} and that V = IndG V [U]. If V 6= V [U] then by the inductive hypothesis V [U] ∼ IndHU χ for T ⊂ H and χ a character HU = T U in which case ∼ G . If then as representations of , ∼ ⊕d where is one-dimensional V = IndT χ V = V [U] H V = U U and so ρ(H) consists of scalar matrices contradicting the construction of H. Lemma 6.11. Let G → H be a surjection of nite groups with abelian kernel and let ρ : H → GL(V ) be a nite dimensional representation. Suppose F ⊂ H is a subgroup and E ⊂ G is its preimage in G. Suppose is the composition of and that there exists such that ρe : G → GL(V ) G → H → GL(V ) τ : E → GL(W ) ∼ G . Then ∼ H for a representation . ρe = IndE τ ρ = IndF σ σ Proof. Let . If then so . But at the same time π : G → H k ∈ ker π k ∈ ker ρ ρe(k) = 1 ker π ⊂ E and so g which implies that is trivial on . Thus descends to a representation ρe(k) = ⊕g∈G/Eτ (k) τ ker π τ ∼ . The map gives ∼ and so G ∼ H which, since is σ : E/ ker π = F → GL(W ) π G/H = E/F IndE τ = (IndF σ) ◦ π π surjective, gives ∼ H as desired. ρ = IndF σ
Lemma 6.12. Let G be a nite group and H C G a normal subgroup such that G/H is supersolvable. If (ρ, V ) is an irreducible representation of G that cannot be written as the induction from a subgroup of G, then V |H is irreducible.
Proof. Suppose V |H is reducible and let W an irreducible H-subrepresentation. Proposition 6.7 implies that unless V is isotypic, i.e., V = V [W ], V can be written as an induction. Therefore V = V [W ] and so V = ⊕ρ(gi)W where all the ρ(gi)W are isomorphic to W . We may therefore apply Proposition 6.8. We obtain σ : G → PGL(W ) and τ : G → PGL(U) where U has dimension dim V/ dim W > 1, τ|H = 1 and ∼ σ|H = ρ|H . Since τ|H = 1, the projective representation τ factors through G/H → PGL(U). For simplicity denote G0 = G/H supersolvable. 0 2 0 Recall from 6.2 that attached to τ : G → PGL(U) is a cohomology class c ∈ H (G , µN ) for some integer in which case one get a genuine representation 0 as follows: let be a xed N τe : G o µN → GL(U) τe lifting of to 0 (not necessarily a homomorphism) and de ne 0 by letting multiplication τ G → GL(U) G o µN be given by in which case setting is in fact a homomorphism. (g, α)(h, β) = (gh, αβc(g, h)) τe(g, α) = τe(g)α Indeed, . Now 0 is also supersolvable because if τe(g, α)(h, β) = τe(gh)αβc(g, h) = τe(g, α)τe(h, β) G o µN 0 0 0 is such that 0 0 and 0 0 is cyclic then 0 0 G = G0 ⊃ ... ⊃ Gi ⊃ ... ⊃ 1 Gi C G Gi/Gi+1 G o µn = G0 o µN ⊃ ... ⊃ 0 is such that 0 0 and 0 0 ∼ 0 0 is cyclic Gi o µN ⊃ ... ⊃ 1 o µN ⊃ 1 Gi o µN C G o µN Gi o µN /Gi+1 o µN = Gi/Gi+1 0 while is also cyclic. As is irreducible and 0 is supersolvable it follows that ∼ G oµN µN /1 τe G o µN τe = IndH0 χ for a character 0 × by Proposition 6.10. The group 0 and so Mackey1 gives χ : H → K 1 o µN C G o µN 1oµN g τ|1 µ = ⊕g∈G0 µ /H(1 µ )(Ind 0 χ) e o N o N o N H ∩1oµN 1 If H ⊂ G, V is a representation of G and N C G then G N g (IndH V )|N = ⊕g∈G/HN (IndH∩N V )
27 0 But evaluating G oµN we get a scalar matrix and so is in fact a character of τe(1, α) = (IndH0 χ)(1, α) χ 0 as all 0 conjugates of are equal. If 0 then this would imply H (1 o µN ) 1 o µN /H ∩ (1 o µN ) χ 1 o µN 6⊂ H 0 that ∼ G oµN would be reducible. Thus 0. Writing 00 0 gives τe = IndH0 χ 1 o µN ⊂ H H = {(g, 1)|(g, α) ∈ H } 0 00 GoµN H = H µN . Note that composing τ with the projection G → G/H gives τ = Ind 00 χ where χ o e e H HoµN extends to H00H by its action on H00. For some integer M a lift to GL(dim W ) of σ(g, α) de ned as σ(g) will give an actual homomorphism whereas lifts to by sending to σe : GoµN oµM → GL(dim W ) τe GoµN oµM → GL(dim V/ dim W ) (g, α, β) ∼ GoµN oµM 00 τ(g, α). Then as a representation of G µN µM have τ = Ind 00 χ where χ on H H µN µM e o o e H H µN µM o o is de ned via the projection to 00 . o o H H o µN ∼ GoµN oµM 0 Let ρ = σ ⊗ τ = Ind 00 (σ ⊗ χ). Let ρ be the composition of the representation ρ with the e e e H H µN µM e e projection o. Theno and 0 are representations of whose projectivisations G o µN o µM → G ρe ρe G o µN o µM agree. Therefore they dier by a character of . Therefore 0 is an induced representations ψ G o µN o µM ρe which implies that ρ is induced by Lemma 6.11. Lecture 14 2013-05-01
Proof of Theorem 6.6. First, ρ is a continuous representation and thus factors through GL/K for some G nite Galois extension L/K. Next, write ρ =∼ Ind L/K ρ0 such that ρ0 cannot be written as an induction. GL/M Since ρ is irreducible it follows that ρ0 is irreducible and p - dim ρ0 | dim ρ. Therefore it su ces to show that if ρ cannot be written as an induction then ρ has dimension 1. Let 1 be the wild inertia. Note that contains as a PL/K = GL/K G0 = GL/K /PL/K G1 = IL/K /PL/K normal subgroup and ∼ which is cyclic. Moreover, and G0/G1 = GkL/kK IL/K = IK /IK ∩GL PL/K = PK /PK ∩ by the Herbrandt quotient theorem and so ∼ . But ∼ Q and GL IL/K /PL/K = IK /PK (IK ∩ GL) IK /PK = Zp(1) so IL/K /PL/K is a nite quotient of an abelian group which must therefore be supersolvable. Lemma 6.12 then shows that is irreducible. ρ|PL/K But is a -group and so must be a power of . But and so as PL/K p dim ρ | |PL/K | p p - dim ρ dim ρ = 1 desired.
7 Iwasawa theory for -extensions Zp The main result of this section is the following theorem of Iwasawa on -extensions. The main reference is Zp [Was97, 13]. Theorem 7.1. Suppose is a prime. Let be any extension of the number eld . Then p > 2 K∞/K Zp K there exist integers and depending only on such that n for . λ, µ ≥ 0 ν K vp(hKn ) = λn + µp + ν n >> 0
7.1 -extensions and Leopoldt's conjecture Zp
Proposition 7.2. Let K be a number eld. There exists a tower of extensions K = K0 ⊂ K1 ⊂ ... such that is Galois over with Galois group ∼ and ∼ n . K∞ = ∪Kn K GK∞/K = Zp GK∞/Kn = p Zp Proof. The extension is abelian with Galois group n+1 × ∼ × n . Let Q(µpn+1 )/Q (Z/p Z) = (Z/pZ) × Z/p Z × n be the sub eld of (µ n+1 ) xed under ( /p ) . Then G ∼ /p . Writing = ∪ gives Qn Q p Z Z Qn/Q = Z Z Q∞ Qn G =∼ lim /pn =∼ . Q∞/Q ←− Z Z Zp Now let K = K with Galois group G ∼ G . But the latter is an open subgroup of ∞ Q∞ K∞/K = Q∞/Q∞∩K G ∼ and so is of the form pk for some k ≥ 0 giving G ∼ pk ∼ additively. This is the Q∞/Q = Zp Zp K∞/K = Zp = Zp cyclotomic -extension. Writing to be the sub eld of xed by n produces the desired tower. Zp Kn K∞ p Zp Having produced a extension of we would like to answer the question of how many such extensions Zp K there exist. To answer such question we need the following result from group theory:
28 Lemma 7.3. Let be a pro- group. Then is generated by 1 elements. The maximal G p G dimFp H (G, Fp) torsion free subgroup of is generated by 1 elements. G rankZp H (G, Zp) Proof. See [NSW08, Proposition 3.9.1].
Lemma 7.4. Let G be a pro nite group and let H be the maximal abelian pro-p torsion-free subquotient of G. Then 1 1 rankZp H (G, Zp) = rankZp H (H, Zp) Proof. As acts trivially on it follows that 1 ∼ ∼ ab ∼ 1 ab . G Zp H (G, Zp) = Hom(G, Zp) = Hom(G , Zp) = H (G , Zp) Let H = Gab/U for U open. Then in ation-restriction gives
1 1 ab 1 1 → H (H, Zp) → H (G , Zp) → H (U, Zp) so it su ces to show that 1 . Now let be the maximal pro- quotient of , in rankZp H (U, Zp) = 0 V = U/N p U which case V is torsion (by choice of U) and every nite quotient of N will have cardinality coprime to p. Again in ation-restriction gives
1 1 1 1 → H (V, Zp) → H (U, Zp) → H (N, Zp) The group is nite so 1 is torsion and so 1 which implies that 1 V H (V, Zp) rankZp H (V, Zp) = 0 rankZp H (U, Zp) = 1 so it su ces to show that 1 . But rankZp H (N, Zp) rankZp H (N, Zp) = 0 ! H1(N, ) = lim lim H1(N/M, /pn ) Zp ←− −→ Z Z n M⊂N p Zp⊂Zp where M is open normal in N. But N/M will be nite with cardinality invertible in Z/pnZ and so 1 . H (N, Zp) = 0 Proposition 7.5. Let be a nitely generated -module. M Zp 1. If is a nite extension and carries an action of then K/Qp M GK
0 1 2 χ(GK ,M) = rankZp H (GK ,M) − rankZp H (GK ,M) + rankZp H (GK ,M) = −[K : Qp] rankZp M Moreover, i 2−i ∗ . rankZp H (GK ,M) = rankZp H (GK ,M (1)) 2. If K/Q is a number eld, S is a nite set of places which includes the in nite places, the places where M is rami ed and the places above p, and M carries an action of GK,S, then
0 1 2 χ(GK,S,M) = rankZp H (GK,S,M) − rankZp H (GK ,M) + rankZp H (GK ,M)
X GKv = rankZp M − [K : Q] rankZp M v|∞
Proof. First note that if is a nitely generated -module and is one of and then Propositions M Zp G GK GK,S 3.6 and 3.8 imply that i is a nitely generated -module. Next one may write H (G, M) Zp M = Mtors ⊕ MTF i where Mtors (the nite torsion) is stable under G and MTF is torsion-free. As H (G, Mtors) is nite it follows that i . Using the exact sequence i i i rankZp H (G, Mtors) = 0 H (G, Mtors) → H (G, M) → H (G, MTF) → i+1 we deduce that i i so for the rest of the argument we H (G, Mtors) rankZp H (G, M) = rankZp H (G, MTF) may assume that M is in fact torsion-free. For any nitely generated module one has that . Applying Zp X rankZp X = dimFp (X/pX) − dimFp X[p] this observation to X = Hi(G, M) get that
i i 1 1 rankZp H (G, M) = dimFp H (G, M)/pH (G, M) − dimFp H (G, M)[p]
29 Since is a free module get an exact sequence p which gives the exact M Zp 0 → M −→ M → M/pM → 0 sequence 0 → Hi(G, M)/pHi(G, M) → Hi(G, M/pM) → Hi+1(G, M)[p] → 0 We deduce that
0 1 2 χ(G, M) = rankZp H (G, M) − rankZp H (G, M) + rankZp H (G, M) 0 0 0 = dimFp H (G, M)/pH (G, M) − dimFp H (G, M)[p] 1 1 1 − (dimFp H (G, M)/pH (G, M) − dimFp H (G, M)[p]) 2 2 2 + dimFp H (G, M)/pH (G, M) − dimFp H (G, M)[p] 0 1 2 = dimFp H (G, M/pM) − dimFp H (G, M/pM) + dimFp H (G, M/pM) 0 3 − dimFp H (G, M)[p] − dimFp H (G, M)[p] 0 3 = dimFp χ(G, M/pM) − dimFp H (G, M)[p] − dimFp H (G, M)[p] 3 = dimFp χ(G, M/pM) − dimFp H (G, M)[p] where for the last equality note that H0(G, M) ⊂ M is torsion-free and so H0(G, M)[p] = 0. If for then by G = GK K/Qp dimFp χ(G, M/pM) = −[K : Qp] dimFp (M/pM) = −[K : Qp] rankZp M 3 Theorem 3.7 and H (GK ,M) = 0. This concludes the proof of the rst part. If G = GK,S then by Theorem 3.9
X GKv dimFp χ(G, M/pM) = dimFp (M/pM) − [K : Q] dimFp (M/pM) v|∞ 3 X 3 dimFp H (GK,S,M)[p] = dimFp H (GKv ,M)[p] v∈∞
It therefore su ces to show that for v | ∞ one has
GKv 3 GKv dimFp (M/pM) − dimFp H (GKv ,M)[p] = rankZp M
If v | R then
3 1 H (GC/R,M)[p] = H (GC/R,M)[p] = M[2][p] = 0
where the last line comes from odd . If then trivially 3 . H (Z/nZ, Z) = 0 v | C H (GKv ,M) = 0 When then is cyclic and so 1 . But from the exact sequence GKv GKv v | ∞ GKv H (GKv ,M) = 0 0 → M /pM → GKv 1 we deduce that GKv GKv GKv GKv (M/pM) → H (GKv ,M)[p] → 0 dimFp (M/pM) = dimFp M /pM = rankZp M as desired. Proposition 7.6. Let be a nite extension and be a -extension. If then is K/Q` K∞/K Zp p 6= ` K∞/K the unique unrami ed extension with Galois group . If there are exactly independent Zp p = ` [K : Q`] + 1 such extensions K∞/K. Proof. Certainly ∼ and so there exists a unique unrami ed -extension. Let p be the GKur/K = Zb →→ Zp Zp K /K composite of all extensions, which will then be the maximal abelian pro- extension of with torsion-free Zp p K Galois group over . The number of independent extensions is equal to the rank of the abelian pro- K Zp Zp p group GKp/K . If then p t the maximal tamely rami ed extension of since is pro- . But recall p 6= ` K ⊂ K K PK = GK/Kt ` from local class eld theory that ab ∼ ab ab ∼ × ∼ Zb × ∼ Zb × . GKt/K = GK /PK = Kd/(1+mK ) = FrobK ×OK /(1+mK ) = FrobK ×kK The largest torsion-free pro- subquotient of this is Zp corresponding to the unique unrami ed p FrobK Zp
30 extension. Another way of seeing this is by recalling from local class eld theory that t : GKt/Kur = ∼ = Q and ∼ . If and then −1 since is IK /PK −→ q6=` Zq(1) GKur/K = Zb σ ∈ GKt/K τ ∈ IK /PK στσ ∈ IK /PK IK normal in and −1 where acts on Q via the Tate twist. Now contains GK t(στσ ) = σ(t(τ)) σ q6=` Zq(1) GKt/Kp −1 −1 the commutant [GKt/K ,GKt/K ]. Let σ ∈ GKt/K and τ ∈ GKt/Kur in which case στσ τ ∈ GKt/Kp . −1 −1 −1 But then t(στσ τ ) = σ(t(τ))t(τ) has to be trivial in GKp/K the maximal abelian torsion-free pro-p subquotient of GKt/K and so σ(t(τ)) = t(τ) in this quotient which, since no nontrivial element of GKt/K acts trivially on nontrivial Tate twists, implies that t(τ) = 0 in the quotient. But then IK /PK projects to 0 and so the maximal abelian torsion-free pro-p subquotient of GKt/K is also the maximal abelian torsion-free pro- subquotient of ∼ , i.e., as desired. p GKur/K = Zb Zp 1 Now suppose that . The number of independent extensions, by Lemma 7.3, is p . p = ` Zp rankZp H (GK /K , Zp) The group GKp/K is the maximal abelian pro-p torsion-free subquotient of GK and so Lemma 7.4 implies 1 1 that p . rankZp H (GK /K , Zp) = rankZp H (GK , Zp) Finally, Proposition 7.5 gives that 2 0 and rankZp H (GK , Zp) = rankZp H (GK , Zp(1)) = 0
0 0 0 rankZp H (GK , Zp) − rankZp H (GK , Zp) + rankZp H (GK , Zp) = −[K : Qp] from where immediately we get that 1 . rankZp H (GK , Zp) = [K : Qp] + 1
Lecture 15 2013-05-03
Lemma 7.7. 1. If K/ is a nite extension then lim K× ⊗ /pn ∼ K× ⊗ . Qp ←− Z Z = Zp 2. If K is a number eld and S is the set of places containing the in nite places and the places above p then lim O× ⊗ /pn ∼ O× ⊗ ←− K,S Z Z = K,S Zp × × TF Proof. For the rst part write K ∼ $Z × k × µ ∞ (K) × (1 + m ) . It su ces to show that lim M ⊗ = K K p K ←− n ∼ for each part separately. This is clear for Z , × and which is a nite group. Z/p Z = M ⊗ Zp $K kK µp∞ (K) The group TF is a nitely generated torsion-free -module ( 2 ∼ 2 under the logarithm (1+mK ) Zp 1+p OK = p OK map and the latter is a nitely generated -module; 2 is nite index in TF and so the Zp 1 + p OK (1 + mK ) latter is also nitely generated) and therefore it is of the form r. Finally the result is true for and thus Zp Zp also for K×. For the second part suppose M is a nitely presented abelian group with presentation Zr → Zs → M → 0. Let K ⊂ Zr be the kernel of Zr → Zs in which case (Zr/K) ⊗ (Z/pnZ) satis es the Mittag-Le er condition and so R1 lim(( r/K) ⊗ ( /pn )) = 0. But then we deduce that lim M ⊗ /pn ∼ s/ r ∼ M ⊗ . From ←− Z Z Z ←− Z Z = Zp Zp = Zp Theorem 2.1 it follows that × is a nitely generated abelian group with rank . The group × OK,S |S| − 1 OK,S is therefore nitely presented and the second part follows.
Proposition 7.8. Let K be a number eld with r1 real places and r2 complex places. Let p be a prime. Then the number of independent extensions is where , called the Leopoldt defect, satis es Zp 1 + r2 + δK δK . If is a -extension then is unrami ed outside of places above . 0 ≤ δK ≤ r1 + r2 − 1 K∞/K Zp K∞/K p
Remark 1. One of the many equivalent formulations of Leopoldt's conjecture is that always δK = 0. In particular, if K = Q or K/Q is quadratic imaginary then Leopoldt's conjecture is true. In the case of K = Q the unique extension is the cyclotomic one while in the case of quadratic imaginary elds one has an Zp additional extension called the anticyclotomic extension. Zp
Proof of Proposition 7.8. First, let v | ` 6= p and let w be a place of K∞ above v. Then K∞,w/Kv is abelian with Galois group a subgroup of and so is a extension as well. Thus is unrami ed Zp Zp K∞,w/Kv by Proposition 7.6.
31 Let p be as before the composite of all the extensions in which case p is the maximal abelian K /K Zp K /K pro-p torsion-free extension of K which is unrami ed outside of p. Let S be the nite set of places containing p the in nite places and the places above p. Then K ⊂ KS and GKp/K is a quotient of GK,S. As before the 1 number of independent extensions is p . Z Zp Z = rankZp H (GK /K , Zp) The group GKp/K is the maximal abelian pro-p torsion-free subquotient of GK,S and so by Lemma 7.4 it follows that 1 1 p Z = rankZp H (GK /K , Zp) = rankZp H (GK,S, Zp) The set contains the in nite places and the places above ( is everywhere unrami ed as it carries S p Zp the trivial GK -action) and so Proposition 7.5 implies that
0 1 2 rankZp H (GK,S, Zp) − rankZp H (GK,S, Zp) + rankZp H (GK,S, Zp) = (r1 + r2) − [K : Q] = −r2 giving 1 2 Z = rankZp H (GK,S, Zp) = rankZp H (GK,S, Zp) + r2 + 1 Let 2 in which case . To prove the inequality δK = rankZp H (GK,S, Zp) ≥ 0 Z = 1 + r2 + δK δK ≤ r1 + r2 − 1 we need to show that 2 × . rankZp H (GK,S, Zp) ≤ r1 + r2 − 1 = rankZ OK The Poitou-Tate sequence (Theorem 3.4) applied to Z/pnZ gives an exact sequence
1 n 1 n 1 ∨ H (GK,S, Z/p Z) → ⊕v∈SH (Kv, Z/p Z) → H (GK,S, µpn ) →
2 n 2 n 0 ∨ → H (GK,S, Z/p Z) → ⊕v∈SH (Kv, Z/p Z) → H (GK,S, µpn ) → 0 Dualizing and using Tate duality to write i n ∨ ∼ 2−i we get H (Kv, Z/p Z) = H (Kv, µpn )
0 2 n ∨ 1 1 ⊕v∈SH (Kv, µpn ) → H (GK,S, Z/p Z) → H (GK,S, µpn ) → ⊕v∈SH (Kv, µpn ) Kummer theory gives 1 × × pn and taking projective limits one gets H (Kv, µpn ) = Kv /(Kv )
2 n ∨ 1 1 ⊕ lim µ n (K ) → lim H (G , /p ) → lim H (G , µ n ) → ⊕ lim H (K , µ n ) v∈S ←− p v ←− K,S Z Z ←− K,S p v∈S ←− v p p The projective maps in ⊕ µ n (K ) are x 7→ x and since µ ∞ (K ) is nite (e is nite whereas v∈S p v p v Kv /Qp n e is in nite) it follows that lim µ n (K ) = 0. Using A/A =∼ A ⊗ /n and Lemma 7.7 we obtain Qp(µp∞ )/Qp ←− p v Z Z 2 n ∨ 1 × 0 → lim H (G , /p ) → lim H (G , µ n ) → ⊕ K ⊗ ←− K,S Z Z ←− K,S p v∈S v Zp First, note that
lim H2(G , /pn )∨ = (lim H2(G , /pn ))∨ = H2(G , lim /pn )∨ = H2(G , / )∨ ←− K,S Z Z −→ K,S Z Z K,S −→ Z Z K,S Qp Zp and so 2 ∨ 1 × 0 → H (G , / ) → lim H (G , µ n ) → ⊕ K ⊗ K,S Qp Zp ←− K,S p v∈S v Zp For the next step, we need a little notation. Let × , Q × and let ES = lim OL,S JS = lim v|S Lv −→L⊂KS −→L⊂KS Q × × . It is a classical computation in global class eld theory (see [NSW08, 8.3.8] CS = lim v|S Lv /OL,S −→L⊂KS or [Mil13, Theorem 5.1]) that H0(G ,C ) = × /K× Q O× . The G cohomology sequence for K,S S AK v∈ /S Kv K,S 1 → ES → JS → CS → 1 gives
0 0 1 1 H (GK,S,JS) → H (GK,S,CS) → H (GK,S,ES) → H (GK,S,JS)
1 Hilbert 90 gives H (GK,S,JS) = 0 and so Y Y H1(G ,E ) ∼ × /K× K× O× ∼ Cl (K) K,S S = AK v Kv = S v∈S v∈ /S
32 Since S contains the places of K above p the sequence
1 → µpn → ES → ES → 1 √ is exact. Indeed, if α ∈ E then α ∈ O× for some L ⊂ K and K( pn α)/K will be unrami ed at places S √ L,S S√ away from and so pn . Necessarily then pn × where . S M = K( α) ⊂ KS α ∈ OM,S M ⊂ KS Now Kummer theory gives
× × pn 1 1 n 0 → OK,S/(OK,S) → H (GK,S, µpn ) → H (GK,S,ES)[p ] → 0
1 n n where H (GK,S,ES)[p ] = ClS(K)[p ]. Taking projective limits as n → ∞ one gets
× n 1 n 0 → lim O ⊗ /p → lim H (G , µ n ) → lim Cl (K)[p ] ←− K,S Z Z ←− K,S p ←− S and since lim A[pn] = 0 for any nite group A it follows that, using Lemma 7.7, ←− × 1 O ⊗ ∼ lim H (G , µ n ) K,S Zp = ←− K,S p Plugging this back into the exact sequence above yields
2 ∨ × × 0 → H (GK,S, Qp/Zp) → OK,S ⊗ Zp → ⊕v∈SKv ⊗ Zp Consider the exact sequence
× × × × 0 → OK → OK,S → ⊕v∈S−∞Kv /Ov → Cl(K) → ClS(K) → 0 Since is at over we get after tensoring Zp Z
× × × × 0 → OK ⊗ Zp → OK,S ⊗ Zp → ⊕v∈S−∞Kv /Ov ⊗ Zp
Now 2 ∨ maps via × to in × and so to in × × . c ∈ H (GK,S, Qp/Zp) OK,S ⊗ Zp 0 ⊕v∈SKv ⊗ Zp 0 ⊕v∈S−∞Kv /Ov ⊗ Zp But then the image of in × lies in fact in × . c OK,S ⊗ Zp OK ⊗ Zp Now
2 ∨ × rankZp H (GK,S, Qp/Zp) ≤ rankZp OK ⊗ Zp × ≤ rankZ OK = r1 + r2 − 1 but at the same time if we write 2 r where is nite torsion and 2 H (GK,S, Zp) = Zp⊕X X r = rankZp H (GK,S, Zp) then
2 ∨ 2 ∨ rankZp H (GK,S, Qp/Zp) = rankZp (H (GK , Zp) ⊗ Qp/Zp) r ∨ = rankZp ((Qp/Zp) ⊕ (X ⊗Zp Qp/Zp)) 2 = rankZp H (GK,S, Zp) since and ∨ ∼ . Therefore 2 as desired. X ⊗Zp Qp/Zp = 0 (Qp/Zp) = Zp rankZp H (GK,S, Zp) ≤ r1 + r2 − 1 Remark 2. Given that 2 another formulation of Leopoldt's conjecture is that δK = rankZp H (GK,S, Zp) 2 . rankZp H (GK,S, Zp) = 0 Lecture 16 2013-05-08
33 7.2 Class groups and Galois modules Lemma 7.9. Let be a extension of a number eld. There exists such that is K∞/K Zp m ≥ 0 K∞/Km totally rami ed at all places of rami cation.
Proof. We already know from Proposition 7.8 that K∞/K can ramify only at the places v1, . . . , vr of K above . Fixing a place of above (since is Galois any will do) let . Now p wi K∞ vi K∞/K Ii = IK∞,w /Kv is an open subgroup of ∼ and so is of the form m ∼ for some . Supposei i that ∩Ii GK∞/K = Zp p Zp = GK∞/Km m are places of . Then . But then Herbrandt's theorem gives w | t | v K∞,Km,K GK∞,w/Km,t ⊂ GK∞/Km ⊂ Iv and so is totally rami ed. IK∞,w/Km,t = Iv ∩ GK∞,w/Km,t = GK∞,w/Km,t K∞,w/Km,t √ √ Remark 3. If and , the -extension, with ∼ then is K0 = Q( −6) K1 ⊂ K∞ Z2 GK1/K0 = Z/2Z K1 = K0( 2) unrami ed over K0. (This is [Was97, Exercise 13.3].)
De nition 7.10. Let Ln/Kn be the maximal abelian extension of Kn which is unrami ed at all nite places and has p-power order. This is a subextension of the Hilbert class eld xed by the prime to p part of Cl(Kn). Let L∞ = ∪Ln. In this case
vp(hKn ) = vp([Ln : Kn]) = logp[Ln : Kn] Lemma 7.11. The Galois group X = G ∼ Cl(K ) carries an action of G by letting g·σ = gσg−1 n Ln/Kn = n Kn/K e e for g ∈ G , σ ∈ X and g any lift of g to G . Kn/K n e Ln/K Proof. First, any other lift of g to G is of the form gh for h ∈ G ⊂ G . Then Ln/K e Ln/Kn Ln/K −1 −1 −1 ghσe (ghe ) = ghσhe ge −1 = gσe ge
since h, σ ∈ Xn which is an abelian group by de nition. Therefore the action is independent of the choice of lift. Finally, if g, h ∈ G then gh is a choice of lift of gh and so Kn/K ee
−1 −1 g · (h · σ) = geehσeh ge −1 = geehσ(geeh) −1 = ghσf ghf = (gh) · σ and so this is indeed a group action.
Lemma 7.12. Let v1, . . . , vs be the places of Km above p that ramify (necessarily totally) in K∞ and let ∼ wi | vi be any place of L∞. If Im,i = I then G = X∞Im,i for all i. L∞,wi /Km,vi L∞/Km Proof. Since is unrami ed it follows that . Therefore ∼ L∞/K∞ Im,i ∩ X∞ = 1 Im,i ,→ GL∞/Km /X∞ = . Denote by as well the image of in . The subextension was chosen GK∞/Km Im,i Im,i GK∞/Km Km such that where is the place of below . The decomposition group IK∞,u /Km,v = GK∞,u /Km,v ui K∞ wi i i is ani open subgroupi and so it is of the form for some . But GK∞,u /Km,v ⊂ GK∞/Km GK∞/Kn ni ≥ m by choicei of i , and so which impliesi that . But Km GK∞/Km = ∩Im,i GK∞,Km = ∩GK∞/Kn min ni = m acts transitively on the places and so for all i proving that ∼ . Finally GK∞/Km vi ni = m i Im,i = GK∞/Km this implies that ∼ . GL∞/Km = X∞Im,i Lemma 7.13. Let be a topological generator of ∼ in which case pm is a topological γ0 GK∞/K = Zp γm = γ0 generator of ∼ m . Then GK∞/Km = p Zp
[GL∞/Km ,GL∞/Km ] = (γm − 1) · X∞ where the action of G on X = lim X = G is de ned in Lemma 7.11. K∞/Km ∞ ←− n L∞/K∞
34 Proof. Identify ∼ in which case we denote by the lift to as well. Then −1. Im,1 = GK∞/Km γm Im,1 γm·x = γmxγm If then may write for and and it is easy to g1, g2 ∈ GL∞/Km = Im,1X∞ gi = hixi hi ∈ GK∞/Km xi ∈ X∞ check (using that Im,1 and X∞ are abelian) that −1 −1 g1g2g1 g2 = ((1 − h2)h1 · x1)((h1 − 1)h2 · x2) Taking and gives −1 −1 and so . h1 = γm h2 = 1 (γm − 1) · x2 = g1g2g1 g2 (γm − 1) · X∞ ⊂ [GL∞/Km ,GL∞/Km ] Going in the other direction, if then α for some . Then γ ∈ GK∞/Km γ = γm α ∈ Zp X α γ − 1 = ((γ − 1) + 1)α − 1 = (γ − 1)n m n m n≥1 and so . We deduce that (1 − h2)h1 · x1, (1 − h1)h2 · x2 ∈ (γm − 1) · X∞ [GL∞/Km ,GL∞/Km ] ⊂ (γm − 1) · X∞ and equality follows. Lemma 7.14. Let be the image of ∼ . Since there σm,i ∈ Im,i γm ∈ GK∞/Km = Im,i Im,i ⊂ GL∞/Km = X∞Im,i exists such that . Let be the -submodule generated by gm,i ∈ X∞ σm,i = gm,iσm,1 Ym ⊂ X∞ Zp gm,2, . . . , gm,s and (γm − 1)X∞. Then Xn = X∞/νn,m · Ym where 2 pn−m−1. (Here the action of on is that de ned in Lemma 7.11.) νn,m = 1+γm +···+γm νn,m Ym ⊂ X∞
Proof. By de nition Ln is the maximal abelian unrami ed p-extension of Kn while L∞ is some p-extension of Kn. Therefore Ln is the maximal abelian unrami ed subextension of L∞. Translating to Galois groups, is generated by the commutant (to make abelian) and (to make GL∞/Ln [GL∞,Kn ,GL∞/Kn ] Ln/Kn In,i Ln/Kn unrami ed). Note that pn−m so σn,i = σm,i
pn−m σn,i = σm,i pn−m = (gm,iσm,1) pn−m−1 Y k −k pn−m = (σm,1gm,iσm,1)σm,1 k=0 pn−m−1 Y k pn−m = (γm · gm,i)σm,1 k=0
= (νn,m · gm,i)σn,1 where the fourth equality is by de nition of the action since is a lift of to . We γm · − σm,1 γm GL∞/K conclude that gn,i = νn,m · gm,i. Now
Xn = GLn/Kn
= GL∞/Kn /GL∞/Ln
= X∞In,1/h[GL∞/Kn ,GL∞/Kn ],In,1,...,In,si
= X∞In,1/h[GL∞/Kn ,GL∞/Kn ],In,1, gn,2, . . . , gn,si
= X∞/h[GL∞/Kn ,GL∞/Kn ], gn,2, . . . , gn,si
= X∞/h(γn − 1) · X∞, gn,2, . . . , gn,si where the third equality uses Lemma 7.12 and the sixth equality uses Lemma 7.13. Finally, Yn is generated by (γn − 1) · X∞ = νn,m(γm − 1) · X∞ and gn,i = νn,m · gm,i and so Yn = νn,m · Ym giving
Xn = X∞/νn,m · Ym
35 Lecture 17 2013-05-10
7.3 The Iwasawa algebra
Recall that our goal is Theorem 7.1 where we study vp(|Xn|) where Xn is a quotient of X∞. To study the cardinality of as we need to study as a module over . In fact we will show that Xn n → ∞ X∞ GK∞/K X∞ is a module over , we will study nitely generated modules over and we will deduce the theorem Zp[[T ]] Zp[[T ]] from a structure theorem. We begin by collecting some facts, with brief sketches of proofs, about power series rings. Throughout is a nite extension. L/Qp
Lemma 7.15 (Division with remainder). If f, g ∈ OL[[T ]] such that f(T ) = a0 + a1T + ··· with ai ∈ mL for and × then one may uniquely write 0 ≤ i ≤ n − 1 an ∈ OL g = qf + r
for a power series q ∈ OL[[T ]] and a polynomial r ∈ OL[T ] of degree < n. Proof. Write and for a uniformizer of , let −1 n , a polynomial U(T ) = an + an+1T + ··· $L L P = $L (f − UT ) of degree < n. Consider the linear operator
∞ ∞ X i X i−n τ( biT ) = aiT i=0 i=n
and the multiplication by P/U operator mP/U . If
∞ −1 X i i i q(T ) = U(T ) (−1) $L(τ ◦ mP/U ) ◦ τ(g) i=0 then one may check that this series converges and that g = qf + r for a polynomial r of degree < n. For details see [Was97, Proposition 7.2].
n De nition 7.16. A polynomial P ∈ OL[T ] is said to be distinguished if it is of the form P (T ) = T + n−1 an−1T + ··· + a0 with ai ∈ mL.
Lemma 7.17 (Weierstrass preparation). Let f ∈ OL[[T ]] such that f(T ) = a0 + a1T + ··· with ai ∈ mL for and ×. Then one may write uniquely as where × 0 ≤ i ≤ n − 1 an ∈ OL f f(T ) = P (T )U(T ) U ∈ (OL[[T ]]) and P ∈ OL[T ] is a distinguished polynomial of degree n.
n Proof. By Lemma 7.15 it follows that T = q(T )f(T ) + r(T ) for a polynomial r of degree < n. Modulo mL n n+1 have f(T ) ≡ anT + O(T ) and so
n n n+1 T − r(T ) = q(T )f(T ) ≡ q(T )(anT + O(T )) (mod mL)
n which implies that r(T ) ≡ 0 (mod mL). Let P (T ) = T − r(T ) of degree n and distinguished. Reducing n+1 n n modulo the ideal (mL,T ) the above equation becomes T ≡ anq(0)T and so q(0) 6= 0 which means that × −1 q ∈ (OL[[T ]]) and let U(T ) = q(T ) . Finally f(T ) = P (T )U(T ). If f = PU then T n = U(T )−1f(T ) + r(T ) and uniqueness of P and U follows from the uniqueness statement of Lemma 7.15.
Corollary 7.18. If then n for a distinguished polynomial . f ∈ OL[[T ]] (f) = ($LP ) P Proof. Let be the largest exponent such that n . Then −n will satisfy the hypothesis n f ≡ 0 (mod mL) P = f$L of Lemma 7.17 and so n for a unit . f = $LPU U
36 De nition 7.19. If is a pro nite group and a nite extension then the completed group ring G L/Qp OL[[G]] is de ned as O [[G]] = lim O [G/H] L ←− L where H ⊂ G runs through the open normal subgroups of G. The ring OL[[G]] is called the Iwasawa algebra of G. Proposition 7.20. Let ∼ be topologically generated by and let be a nite extension. Then G = Zp γ L/Qp γ 7→ 1 + T yields an isomorphism ∼ OL[[G]] = OL[[T ]] Proof. The open normal subgroups of are of the form n so we have n ∼ G H = p Zp OL[G/H] = OL[Z/p Z] = pn by sending the generator of n to . Therefore OL[T ]/((1 + T ) − 1) Z/p Z 1 + T
n O [[G]] ∼ lim O [T ]/((1 + T )p − 1) L = ←− L It su ces to show that n O [[T ]] ∼ lim O [T ]/((1 + T )p − 1) L = ←− L pn n+1 Let Pn(T ) = (1+T ) −1. It is easy to see that Pn+1/Pn ∈ (mL,T ) and so Pn ∈ (mL,T ) by induction. n Let f ∈ OL[[T ]]. Lemma 7.15 produces a power series qn and a polynomial fn of degree < p such that
f(T ) = qn(T )Pn(T ) + fn(T ) in which case f ≡ f (mod P ) for all m ≥ n. This provides a map f 7→ (f ) from O [[T ]] to lim O [T ]/(P (T )). m n n n L ←− L n n+1 Finally, ∩(Pn(T )) ⊂ ∩(mL,T ) = 0 and so this map is injective. Now for surjectivity, suppose that (f ) ∈ lim O [T ]/(P (T )). Since (P (T )) ⊂ (m ,T )n+1, it follows n ←− L n n L n+1 that for m ≥ n we have fm ≡ fn (mod (mL,T ) ). But OL[[T ]] is complete for the (mL,T )-adic topology n+1 and so there exists f ∈ OL[[T ]] such that f ≡ fn (mod (mL,T ) ). It remains to show that in fact f ≡ fn (mod Pn(T )). By de nition there exists qm,n ∈ OL[T ] such that fm − fn = qm,nPn. In the (mL,T )-adic topology of OL[[T ]] we have f − fn fm − fn = lim = lim qm,n Pn m Pn m which, being a limit of polynomials, must be a power series in OL[[T ]] if the sequence converges. (Here we may use that OL[[T ]] is closed in its fraction eld. Writing qn = limm qm,n get f = qnPn + fn as desired. Lecture 18 2013-05-13
7.4 Modules over the Iwasawa algebra
De nition 7.21. The Iwasawa algebra is Λ = OL[[T ]]. Lemma 7.22 (Nakayama). Suppose X is a compact topological Λ-module. Then X is nitely generated if and only if X/(mL,T )X is nite. Proof. See for instance [Was97, Lemma 13.16].
De nition 7.23. Let M,N be two Λ-modules. Say M ∼ N if there exists a morphism of modules M → N with nite kernel and cokernel. Lemma 7.24. Let be a nite extension. L/Qp 1. If f, g ∈ Λ are coprime then (f, g) is nite index in Λ.
2. The prime ideals of Λ are 0, mL, (mL,T ) and (P (T )) where P ∈ OL[T ] is irreducible and distinguished. The prime ideal (mL,T ) is the unique maximal ideal.
37 3. If f ∈ Λ such that f is not a unit then Λ/(f) is in nite. 4. If M is a nitely generated Λ module then
s ! t r M ni M mj M ∼ Λ ⊕ Λ/($L ) ⊕ Λ/(fj(T ) ) i=1 j=1
for distinguished irreducible polynomials fj.
Proof. The rst part. Corollary 7.18 implies that we may choose f, g to be products of powers of $L and distinguished polynomials or else (f, g) = Λ. Since f, g are coprime, without loss of generality assume . Let be a polynomial of minimal degree. Write n with either or a distinguished $L - f h ∈ (f, g) h = $L` ` 1 polynomial. If then for gives n a polynomial of smaller ` 6= 1 f = q` + r deg r < deg ` = deg h $Lr ∈ (f, g) degree than . Thus n . Now n and so . But h h = $L ∈ (f, g) ($L, f) = (h, f) ⊂ (f, g) Λ/(f, h) →→ Λ/(f, g) n ∼ n which consists of polynomials of degree and coe cients Λ/(f, h) = Λ/($L, f) = (OL/mL)[[T ]]/(f) < deg f in n and therefore is nite. OL/mL The second part. The ideals listed are prime. Suppose p is a proper prime ideal. By Corollary 7.18 every non-unit in p is a polynomial. Let f ∈ p be a polynomial of minimal degree. If $L ∈ p then p/$L is a prime ideal of kL[[T ]] which is a PID with maximal ideal T and so p = ($L) or p = ($L,T ). Suppose $L ∈/ p. If p 6= (f) then there exists g ∈ p − (f) necessarily coprime to f. Then p ⊃ (f, g) will have nite index in Λ by the rst part. But then n for some contradicting the assumption that . $L ∈ p n $L ∈/ p The third part. Since we care about the ideal , by Corollary 7.18, n where or is (f) (f) = ($Lg) g = 1 g a distinguished polynomial. If n > 0 then (f) ⊂ ($L) and so Λ/(f) →→ kL[[T ]] which is in nite. If n = 0 then g 6= 1 is a distinguished polynomial and no two elements in OL can be equal in Λ/(f) so the quotient is in nite. The fourth part is a big exercise in linear algebra in the style of the classi cation of nitely generated modules over PIDs. See for instance [Was97, Theorem 13.12].
Lemma 7.25. Let M ∼ N as Λ-modules and let fn ∈ Λ such that each M/fnM is nite. Then each N/fnN is nite and
vp(|M/fnM|) = vp(|N/fnN|) + C(1) where the notation C(1) is taken to mean constant for n >> 0. Proof. Unenlightening exercise in using the snake lemma. See [Was97, Lemma 13.21].
7.5 Class numbers in -extensions Zp Lemma 7.26. Let be a extension of a number eld . Let be as de ned in Lemma 7.13. K∞/K Zp K X∞ Then and are nitely generated -modules. Ym X∞ Zp[[T ]] Proof. The group carries an action of and thus carries an action of Xn Zp[GKn/K ] X∞ = lim Xn lim Zp[GKn/K ] = ∼ by Proposition 7.20. ←− ←− Zp[[GK∞/K ]] = Zp[[T ]] Recall that ∼ sending . Let be as in Lemma 7.9. By de nition for Zp[[GK∞/K ]] = Zp[[T ]] γ0 7→ T + 1 m n > m we have pn−m−1 γn − 1 X m ν = = (1 + T )ip ∈ (p, T ) ⊂ Λ n,m γ − 1 m i=0 and so by Lemma 7.14 ∼ ∼ Ym/(p, T )Ym = Ym/νn,m · Ym ⊂ X∞/νn,m · Ym = Xn ∼ is nite. By Lemma 7.22 we deduce that Ym is nitely generated. Finally, X∞/Ym = Xm is nite and so X∞ is nitely generated.
38 Proof of Theorem 7.1. We will show that there exist nonnegative integers µ, λ, ν such that for n >> 0
n vp(hKn ) = vp(|Xn|) = λn + µp + ν in other words that n vp(|Xn|) = λn + µp + C(1) ∼ Note that 0 → Ym → X∞ → X∞/Ym → 0 where X∞/Ym = Xm is nite; therefore Ym ∼ X∞ which shows ∼ that Ym/νn,m · Ym ∼ X∞/νn,m · Ym = Xn. By Lemma 7.25 it su ces to show that n vp(|Ym/νn,mYm|) = λn + µp + C(1)
Lemma 7.26 shows that Ym is a nitely generated Λ-module and so by Lemma 7.24 implies that
m ∼ r M ni M j Ym = Λ ⊕ Λ/(p ) ⊕ Λ/(fj )
First, note that Ym/νn,mYm is nite but νn,m ∈ (p, T ) it is not a unit and therefore Λ/νn,m is in nite by Lemma 7.24. This implies that r = 0. Now X ni X mj vp(|Ym/νn,mYm|) = vp(|Λ/(p , νn,m)|) + vp(|Λ/(fj , νn,m)|)
is a nite sum so it is enough to show that for each direct summand M of Ym one has n vp(|M/νn,mM|) = λM n + µM p + C(1) where . λM , µM ∈ Z≥0 k k Suppose M = Λ/(p ). Then M/νn,mM = Λ/(p , νn,m) consists, using the division algorithm of Lemma 7.15 as n m, of polynomials of degree n m with coe cients in k . Therefore deg νn,m = p − p < p − p Z/p Z k(pn−pm) n |M/νn,mM| = p and so vp(|M/νn,mM|) = kp + C(1) as desired. Now suppose that M = Λ/(f r) where f is distinguished and therefore g = f r of degree d is also distinguished. If k ≥ d, the division algorithm gives T k = q(T )g(T ) + r(T ) with deg r < d. Modulo p, d and so k d which implies that and so k . If n g =≡ T T ≡ qT + r r ≡ 0 (mod p) T ≡ pZp[T ] (mod g) p > d then pn and so by induction pn+k k . Let (1 + T ) ≡ 1 + pZp[T ] (mod g) (1 + T ) ≡ 1 + p Zp[T ] (mod g) n0 ≥ m n n such that p > d. If n ≥ n0, p > d and k ≥ 1 then p−1 ! p−1 X pn+ki X k i Pn+k+1 = Pn+k (1 + T ) ≡ Pn+k (1 + p Zp[T ]) (mod g) ≡ Pn+kp(1 + pZp[T ]) (mod g) i=0 i=0 where recall that pk . But is invertible in and so in , Pk(T ) = (1 + T ) − 1 1 + pZp[T ] Λ Λ/(g) νn+k+1,n+k = Pn+k+1/Pn+k acts (up to a unit) by multiplication by p. Now g is distinguished so p - g and therefore multiplication by p is injective on M = Λ/(g). Therefore
|M/νn,mM| = |M/νn,n−1 ··· νn0+2,n0+1νn0+1,mM|
n−n0−1 = |M/p νn0+1,mM|
n−n0−1 n−n0−1 n−n0−1 = |M/p M||p M/p νn0+1,mM|
n−n0−1 = |M/p M||M/νn0+1,mM|
n−n0−1 = |Λ/(p , g)||M/νn0+1,mM|
n−n0−1 = |(Z/p Z)[T ]/(g)||M/νn0+1,mM|
d(n−n0−1) = p |M/νn0+1,mM| This implies that
vp(|M/νn,mM|) = d(n − n0 − 1) + vp(|M/νn0+1,mM|) = dn + C(1)
39 Remark 4. It is a theorem of Ferrero and Washington that if K/Q is abelian Galois then µ = 0. In general, if is the cyclotomic -extension then it is expected that . K∞/K Zp µ = 0 Lecture 19 2013-05-15
8 Hecke theory for GL(1)
Hecke theory refers to the study of L-functions attached to various arithmetic or analytic objects and their functional equations. It is worth spending a little time understanding what the point is, as the results are fairly technical. Suppose is a number eld and be a continuous Galois representation. One de nes K ρ : GK → GL(n, C) the L-function of ρ as
Y −s IKv −1 L(ρ, s) = det(1 − ρ(Frobv)qv |ρ ) v-∞ which is an analytic function for Re s >> 0. However, a priori, it is not known what kind of behavior L has on C. Is it meromorphic? Analytic? Does it have a functional equation? The strategy for tackling these questions is to nd an analytic construction of the L-function in a context where these questions can be answered naturally using Fourier transforms. Hecke theory for GL(1) is the topic of Tate's thesis, whose main results we explain, without detailed proofs.
8.1 Fourier analysis 8.1.1 Measures
1 Let G be a locally compact topological abelian group and let µG be a Haar measure. Let Gb = Hom(G, S ) be the space of continuous characters. Then G is compact if and only if Gb is discrete and Gb =∼ G. If H ⊂ G is a closed subgroup then G/H[ =∼ H⊥ = {φ ∈ Gb|φ(H) = 1} and G/Hb ⊥ =∼ Hb. There exists a unique Haar measure µG/µH on G/H such that for every φ ∈ Cc(G) with compact support Z Z Z φ(g)dµG,g = φ(gh)dµH,h d(µG/µH )g G G/H H Q0 ∼ Q0 If G = Gv is a restricted product with respect to the open subgroups Uv ⊂ Gv then G = ⊥ Gv. {Uv } b {Uv } b If µv is a Haar measure for Uv such that for almost all v, µv(Uv) = 1 then µ = ⊗µv is a Haar measure for Q0 G . {Uv } v
8.1.2 Fourier transforms for abelian groups
1 For a Haar measure µ on G the Fourier transform Fµ : L (G, µ) → C(Gb) de ned by Z Fµ(φ)(χ) = φ(g)χ(g)dµg G extends by continuity to 2 2 for the unique (dual) Haar measure on such that for Fµ : L (G, µ) → L (G,b µb) µb Gb every φ ∈ Cc(G), Z Z 2 2 |φ| dµ = |Fµ(φ)| dµb G Gb Then F F φ = φ under the canonical identi cation Gb ∼ G. µb µ b = Suppose Γ ⊂ G is a discrete subgroup such that G/Γ is compact. Poisson summation states that
X X ⊥ φ(γ) = Fµ(φ)(γ ) γ∈Γ γ⊥∈Γ⊥
40 8.1.3 Fourier transforms for vector spaces If is , , a nite extension of or where is a number eld and is nontrivial then G R C Qp AK /K K ψ ∈ Gb a 7→ (x 7→ ψ(ax)) gives a noncanonical identi cation G =∼ Gb. Write S(G) for the space of Schwarz functions: when or these are functions all of whose derivatives decay faster than polynomials, when G = R C G = K/Qp is a nite extension then these are locally constant functions with compact support. ∼ Choosing φ ∈ Gb as above gives a Fourier transform Fµ,ψ : S(G) → S(G). Via the identi cation G = Gb, the explicit formula is Z Fµ,ψ(φ)(h) = φ(g)ψ(hg)dµg G Write ∗ for the transfer of the dual measure from to using . µψ µb Gb G ψ Lemma 8.1. If K = R let ψ(x) = exp(2πix) and µ([0, 1]) = 1. If K = C let ψ(x) = exp(2πi Re x) and . If is a nite extension let be such that . µ([0, 1] × [0, i]) = 2 K/Qp λ : Qp → Z[1/p] λ(x) + x ∈ Zp Then λ(x) is well-de ned up to and ψ(x) = exp(2πiλ(Tr (x)) is a well-de ned character. Suppose Z K/Qp −1 −1/2 µ(OK ) = [D : OK ] . K/Qp In all three cases, ∗ . µψ = µ Proof. [Tat67, Theorem 2.2.2].
Let K be a number eld. For each place v x ψv ∈ Kbv such that for almost all v, ker ψv = Ov. Then Q0 . Using to identify ∼ get ∼ ⊥. Thus ∼ via ψ = ⊗ψv ∈ dK = ⊥ Kv ψv Kv = Kv Ov = O K = bK A {Ov } c b v A A . Under this identi cation ⊥ is simply and ∼ . a 7→ (x 7→ ψ(ax)) K ⊂ AdK K ⊂ AK A\K /K = K If is the Haar measure on inducing the discrete measure on the discrete subgroup and µ AK K ⊂ AK /K inducing on the compact group then ∗ . µ(AK /K) = 1 AK /K µψ = µ Let 0 consist of where for almost all . If S(AK ) = ⊗vS(Kv) φ = ⊗φv φv = charOv v φ = ⊗φv ∈ S(AK ) then
Fµ,ψφ = ⊗Fµv ,ψv φv
with Fourier inversion Fµ,ψ−1 Fµ,ψφ = φ. The Poisson summation formula for states that K ⊂ AK X X −1 |a|AK φ(aα) = Fµ,ψ(φ)(a α) α∈K α∈K 8.2 Local zeta integrals Suppose or a nite extension of . For a continuous character × × we would like to K = R, C Qp χ : K → C de ne analytically an L-function. The idea is to de ne for each test function φ ∈ S(K) and Haar measure ν on K× Z s ζ(φ, χ, ν, s) = φ(x)χ(x)|x|K dνx K× and recover the L-function as a common denominator as the test function φ varies. If K = and χ(x) = (x/|x|)ε|x|t de ne R R L(χ, s) = π−(s+t+ε)/2Γ((s + t + ε)/2) If K = and χ(x) = (x/|x|)m|x|t de ne C C L(χ, s) = 2(2π)−(s+t+|m|/2)Γ(s + t + |m|/2) If then K/Qp ( × 1 χ(OK ) 6= 1 L(χ, s) = −s −1 × (1 − χ($K )qK ) χ(OK ) = 1
41 Proposition 8.2. There exists a test function φχ such that ζ(φ, χχ, ν, s) = L(χ, s). For every test function ζ(φ, χ, ν, s) φ ∈ S(K), is holomorphic. L(χ, s)
2 1. If K = , ν = dx/|x| and χ = (x/|x|)ε|x|t then φ = xεe−πx . R R χ 2dxdy m t n −2π|x| −n −2π|x| 2. If K = , ν = and χ(x) = (x/|x|) |x| then φ = x e C if n ≥ 0 and φ = x e C C π(x2+y2) C χ χ if n < 0. 3. If is a nite extension, × and is unrami ed then ; if is rami ed of K/Qp ν(OK ) = 1 χ φχ = charOK χ f −1 conductor f ≥ 1 then φχ = ν(1 + m ) char f . K 1+mK Proof. See [Tat67, 2.5] the corresponding functions of z on page 316 for K = R, on page 318 for K = C and on page 320 for . K/Qp Lecture 20 2013-05-17
8.3 Local functional equation and local ε-factors Now that we have de ned L-functions analytically we should remark that they do not contain much infor- mation about the characters. In fact, for every rami ed character χ, one has L(χ, s) = 1, and more generally the L-function of a Galois representation does not take into account the rami ed part of the representation. To study the rami ed part one needs the ε-factor which arise naturally in the context of functional equations. Suppose or a nite extension of and × ×. Let nontrivial identifying K = R, C Qp χ : K → C ψ ∈ Kb Kb with K, µ a Haar measure on K and ν a Haar measure on K×. Proposition 8.3. For every φ ∈ S(K)
−1 ζ(φ, χ, ν, s)γ(χ, ψ, µ, s) = ζ(Fµ,ψφ, χ , ν, 1 − s) for γ(χ, ψ, µ, s) not depending on φ and ν. Proof. [Tat67, Theorem 2.4.1]. Theorem 8.4. The function
L(χ, s) ε(χ, ψ, µ, s) = γ(χ, ψ, µ, s) L(χ−1, 1 − s)
is of the form A · Bs where A, B ∈ K. Let ψ and µ as in Lemma 8.1. If K = and χ(x) = (x/|x|)ε|x|t then ε(χ, ψ, µ, s) = iε. If K = and R R C χ(x) = (x/|x|)m|x|t then ε(χ, ψ, µ, s) = i|m|. Finally supposeC is a nite extension and is any nontrivial character of . Let be the conductor K/Qp ψ K f of , i.e., the smallest integer such that f and let be the conductor of , i.e., the smallest integer χ χ(UK ) = 1 −d ψ such that ψ(m−d) = 1. (For example if ψ is as in Lemma 8.1 then d = v (D ).) Then K K K/Qp ! ! −d X x −1 x −(d+f)s ε(χ, ψ, µ, s) = µ(mK ) ψ d+f χ d+f qK × f $K $K x∈OK /UK
Proof. See [CF86, 2.5] explicit expressions for ρ(c) on page 317 for K = R, on page 319 for K = C and on page 322 for . K/Qp
42 Corollary 8.5. Have
s−1 ε(χ, ψ(a · −), µ, s) = χ(a)|a|K ε(χ, ψ, µ, s) ε(χ, ψ, rµ, s) = rε(χ, ψ, µ, s)
Proof. The second equality is immediate. For the rst equality note that the conductor of ψ(a · −) is equal to −d − vK (a) and so ! ! −d−v (a) X ax x −(d+v (a)+f)s ε(χ, µ, ψ(a · −), s) = µ(m K ) ψ χ−1 q K K d+vK (a)+f d+vK (a)+f K × f $K $K x∈OK /UK ! ! X ax ax −(d+v (a)+f)s = χ(a)|a|−1µ(m−d) ψ χ−1 q K K K d+vK (a)+f d+vK (a)+f K × f $K $K x∈OK /UK ! ! s−1 −d X y −1 y −(d+f)s = χ(a)|a|K µ(mK ) ψ d+f χ d+f qK × f $K $K y∈OK /UK s−1 = χ(a)|a|K ε(χ, ψ, µ, s) where we used that −d−vK (a) −d −d−vK (a) −d vK (a) −1 as is a Haar measure µ(mK )/µ(mK ) = [mK : mK ] = qK = |a|K µ and we denoted −vK (a). y = xa$K Proposition 8.6. Let be a nite extension, a nontrivial character of and × × a K/Qp ψ K η : K → C continuous character of conductor f ≥ 1.
1. For 0 ≤ a ≤ f/2 there exists ca ∈ K such that η(1 + x) = ψ(cax) for vK (x) ≥ f − a. 2. If × × are continuous characters of conductors and such that then χ1, χ2 : K → C f1 f2 f1, f2 ≤ a
ε(χ1η, ψ, µ, s)χ1(ca) = ε(χ2η, ψ, µ, s)χ2(ca)
Proof. Let × × be any continuous character of conductor . First, if χ : K → C f ≥ 1 vK (x), vK (y) ≥ f − a then xy where xy 2f−2a f . Therefore (1 + x)(1 + y) = (1 + x + y)(1 + 1+x+y ) 1 + 1+x+y ∈ UK ⊂ UK = ker χ χ((1 + x)(1 + y)) = χ(1 + x + y) and so x 7→ χ(1 + x) is an additive character which case then be recovered as ψ(cax) for some ca ∈ K. Applying this to χ = η yields the rst result. Recall that −1 f as has conductor . But then . In ker(ψ(ca · −)) = ca ker(ψ) = mK χ f vK (ca) = −d − f particular, in the formula of Theorem 8.4 −d X −1 −(d+f)s ε(χ, ψ, µ, s) = µ(mK ) ψ(xca)χ (xca) qK × f OK /UK where we replace the sum over with a sum over d+f . x xca$K Writing x = y(1 + z) gives
X −1 X X −1 ψK (cax)χ (cax) = ψ(cay(z + 1))χ (cay(z + 1)) × f × f−a f−a f x∈OK /UK y∈OK /UK z∈mK /mK X −1 X −1 = ψ(cay)χ (cay) ψ(cayz)χ (1 + z) × f−a f−a f y∈OK /UK z∈mK /mK X −1 X = ψ(cay)χ (cay) ψ(caz(y − 1)) × f−a f−a f y∈OK /UK z∈mK /mK
43 where the last line follows from the fact that χ(1 + z) = ψ(caz) as vK (z) ≥ f − a. If f−a then u = $K X X ψ(caz(y − 1)) = ψ(ca(z + u)(y − 1)) f−a f f−a f z∈℘K /℘K z∈mK /mK X = ψ(cau(y − 1)) ψ(caz(y − 1)) f−a f z∈mK /mK X and therefore ψ(caz(y−1)) = 0 unless ψ(cau(y−1)) = 1, which can only happen if vK (cau(y−1)) ≥ f−a f z∈mK /mK . But and and therefore the sum vanishes unless a . If a then −d vK (c) = −d − f vK (u) = f − a y ∈ UK y ∈ UK ψ(caz(y − 1)) = 1 and so X f−a f a ψ(caz(y − 1)) = |mK /mK | = qK f−a f z∈mK /mK We get X −1 a X −1 ψK (cax)χ (cax) = qK ψ(cay)χ (cay) × f a f−a x∈OK /UK y∈UK /UK which gives −(d+f)s −d a X −1 ε(χ, ψ, µ, s) = qK µK (mK )qK ψ(cay)χ (cay) a f−a y∈UK /UK
We now apply the above to χ = χ1η and χ2η. Suppose for instance that χ = χ1η. Then χ has conductor and for every f−a as . Therefore f χ(1 + x) = χ1(1 + x)η(1 + x) = η(1 + x) = ψ(cax) x ∈ UK f1 ≤ a ≤ f − a a−(d+f)s −d X −1 ε(χ1η, ψ, µ, s) = qK µK (mK ) ψ(cay)(χ1η) (cay) a f−a y∈UK /UK −1 a−(d+f)s −d X −1 = χ1 (ca)qK µK (mK ) ψ(cay)η (cay) a f−a y∈UK /UK because if a then −1 −1 . y ∈ UK ⊂ ker χ1 χ1 (cay) = χ1 (ca) The conclusion follows from the fact that ε(χ1η, ψ, µ, s)χ1(ca) does not depend on χ1. Lecture 21 2013-05-20
8.4 Global zeta integrals Suppose × × × is a continuous Hecke character, is a Haar measure on × and . χ : AK /K → C ν AK φ ∈ S(AK ) De ne Z s ζ(φ, χ, ν, s) = φ(x)χ(x)|x| dνx × AK AK
× ∼ Q0 × × × Since = ×,⊥ Kdv we may write χ = ⊗χv where χv : K → is unrami ed at all but nitely Ac {Ov } v C many . By de nition and write where is a Haar measure on × with the property v φ = ⊗φv ν = ⊗νv νv Kv that × for almost all . Then νx(Ov ) = 1 v Y ζ(φ, χ, ν, s) = ζ(φv, χv, νv, s)
44 which converges for Re s > t + 1 where |χ | = |x|tv for t ≤ t a real number. Such a t can always be found v Kv v if × , i.e., × × × is a continuous Hecke character. χ(K ) = 1 χ : AK /K → C Theorem 8.7. The integral ζ(φ, χ, ν, s) satis es the functional equation
−1 ζ(φ, χ, ν, s) = ζ(Fµ,ψφ, χ , ν, 1 − s)
s0 It has analytic continuation to unless χ = | · | in which case it has a simple pole at s = −s0 with residue C AK 1 1 × and a simple pole at with residue 1 1 . Here 1 on 1 is −ν (AK /K )φ(0) s = 1 − s0 ν (AK /K)Fµ,ψ(φ)(0) ν AK the Haar measure such that the quotient measure on × 1 ∼ is the measure , while the Haar AK /AK = (0, ∞) dt/t measure on 1 × is the quotient measure by the discrete Haar measure on ×. AK /K K Proof. See [Tat67, Main Theorem 4.4.1].
Corollary 8.8. If is a number eld and × × × is a continuous Hecke character then K χ : AK /K → C Y γ(χv, ψv, µv, s) = 1. v Proof. This is immediate from Theorem 8.7 and Proposition 8.3.
8.5 Global L-functions and ε-factors Let × × × be a continuous Hecke character and write . De ne χ : AK /K → C χ = ⊗χv Y L(χ, s) = L(χv, s) v Write Y ε(χ, s) = ε(χv, ψv, µv, s)
dx 2dxdy which does not depend on ψ or µ. Choose ν such that νv = if v | , νv = √ for v | and |x| R π x2+y2 C × for . Choose as in Lemma 8.1. νv(Ov ) = 1 v - ∞ µv Theorem 8.9. The function L(χ, s) has analytic continuation to unless χ = | · |s0 in which case it C AK has a simple pole at with residue 1 1 × and a simple pole at with residue s = −s0 −ν (AK /K ) s = 1 − s0 1 1 × p −1 where is the discriminant of . Moreover ν (AK /K ) |DK | DK K/Q L(χ, s) = ε(χ, s)L(χ−1, 1 − s)
Proof. Let S be the nite set of places such that v | ∞ or χv is rami ed or ker ψv 6= Ov or µv(Ov) 6= 1 or × . For every place choose such that . In particular, for , ν(Ov ) 6= 1 v φv ζ(φv, χv, νv, s) = L(χv, s) v∈ / S by Proposition 8.2 and in this case we compute φv = charOv Z Z
(Fψv ,µv φv)(x) = charOv (y)ψv(xy)dµv,y = ψv(xy)dµv,y Kv Ov ( µ (O ) ψ (x) = 1 = v v v 0 ψv(x) 6= 1 where since is unrami ed for by choice of . Therefore . ker ψv = Ov ψv v∈ / S S Fψv ,µv φv = µv(Ov)φv = φv Then for , , −1 −1 and . v∈ / S Fµv ,ψv φv = charOv ζ(Fµv ,ψv φv, χv , νv, 1 − s) = L(χv , 1 − s) ε(χv, ψv, µv, s) = 1 Thus
ζ(φ, χ, ν, s) Y ζ(φv, χv, νv, s) 1 = = L(χ, s) L(χv, s) v∈S −1 −1 ζ(Fµ,ψφ, χ , ν, 1 − s) Y ζ(Fψ ,µ φv, χ , νv, 1 − s) = v v v L(χ−1, 1 − s) L(χ−1, 1 − s) v∈S v
45 and so
−1 −1 L(χ ,1−s) −1 ζ(F φ, χ , ν, 1 − s) Q v Q ε(χ , ψ , µ , s) L(χ , 1 − s)ε(χ, s) µ,ψ v∈S ζ(F φ ,χ−1,1−s) v∈S v v v = µv ,ψv v v L(χ, s) ζ(φ, χ, ν, s) Q L(χv ,s) v∈S ζ(φv ,χv ,νv ,s) −1 −1 ζ(Fµ,ψφ, χ , ν, 1 − s) Y ζ(φv, χv, νv, s) ε(χv, ψv, µv, s)L(χv , 1 − s) = −1 ζ(φ, χ, ν, s) ζ(F φ , χ , 1 − s) L(χv, s) v∈S µv ,ψv v v = 1
s0 Since L(χ, s) = ζ(φ, χ, ν, s), L(χ, s) is analytic unless χ = | · | for some s0 in which case it has simple poles AK at s = −s0 and s = 1 − s0. It remains to compute the residues. By Theorem 8.7 the residue at is 1 1 × and the −s0 −ν (AK /K )φ(0) residue at is 1 1 × so it su ces to compute and . Recall from 1 − s0 ν (AK /K )Fµ,ψ(φ)(0) φ(0) Fµ,ψ(φ)(0) Proposition 8.2 that for which is unrami ed we can choose for all . When χv = 1 φv = charOv v - ∞ v | R then −πx2 and when then −π(x2+y2). In particular Q as φv(x) = e v | C φv(x + iy) = e φ(0) = φv(0) = 1 desired. It remains to show that p −1. But for we have chosen −πx2 , Fµ,ψ(φ)(0) = |DK | v | R φv(x) = e ψv(x) = 2πix e and µv = dx for which Z −πx2 Fµv ,ψv φv(0) = e dx = 1 R For we have chosen −2π(x2+y2), 4πix and for which v | C φv(x + iy) = e ψv(x + iy) = e µv = 2dxdy Z −2π(x2+y2) Fµv ,ψv φv(0) = e 2dxdy = 1 C
For we only need to look at s0 which is unrami ed and since × we have v - ∞ χv(x) = |x|v ν(Ov ) = 1 φv = charOv and we have already computed the Fourier transform
−1 −1/2 Fµ ,ψ φv(0) = µv(Ov) = [D : Ov] v v Kv /Qp where the last equality follows from Lemma 8.1. Therefore
Y −1 −1/2 Fµ,ψ(φ)(0) = [D : Ov] Kv /Qp v-∞ Y = (N D )−1/2 Kv /Qp Kv /Qp v-∞ p −1 = |DK | as desired.
8.6 Applications
Theorem 8.10 (Analytic class number formula). Let K be a number eld and let 1 denote the trivial Hecke character of K. Show that L(1, s) has a simple pole at s = 1 with residue 2nh R lim(s − 1)L(1, s) = K K s→1 p wK |DK |
where n is the number of in nite places of K, hK = | Cl(K)|, RK is the regulator of K (de ned as the absolute value of the rank of the matrix as ranges through a set of generators of × and (log(|ui|v))i,v ui OK ), and is the discriminant of . v | ∞ wK = |µ∞(K)| DK K/Q
46 Proof. Theorem 8.9 shows that has a simple pole with residue 1 1 × p −1 so we just need L(1, s) ν (AK /K ) |DK | to compute this volume. Recall from Theorem 2.1 the exact sequence
1 Y × × 1 × 0 → K∞ Ov /OK → AK /K → Cl(K) → 0 v-∞ and that 1 is the quotient measure on 1 × induced from the Haar measure on 1 coming from on ν AK /K AK ν × by the discrete measure on × 1 . This gives AK K ⊂ AK
1 1 × 1 1 Y × × 1 1 1 Y × × ν (AK /K ) = ν (K∞ Ov /OK )ν (Cl(K)) = hK ν (K∞ Ov /OK ) v-∞ v-∞ Also recall the exact sequence
Y Y 1 Y × 1 Y × × × 0 → {±1} S Ov /µ∞(K) → K∞ Ov /OK → ∆∞/ log OK → 0 v|R v|C v-∞ v-∞ Writing ν1 for the measure on 1 Y × × 1 × K∞ Ov /OK ⊂ AK /K v-∞ Q Q 1 Q × 1 and for the subset measure on {±1} S O /µ∞(K) we get the quotient measure ν on v|R v|C v-∞ v × which gives ∆∞/ log OK
1 1 Y × × 1 Y Y 1 Y × 1 × ν (K∞ Ov /OK ) = ν ( {±1} S Ov /µ∞(K))ν (∆∞/ log OK ) v-∞ v|R v|C v-∞
What are the measures on the kernel and image? If v | R then we have × 0 → {±1} → R → R → 0 via . The measure on × is and so the measure on the image is . If x 7→ log |x| R νv = dx/|x| = d log |x| R dx v | C then 1 × 0 → S → C → R → 0 via . Recall that 2dxdy which in polar coordinates and becomes z 7→ log |z|C νv = π(x2+y2) x = r cos θ y = r sin θ 2rdrdθ 2drdθ dθd log r2 and so we can put the measure on 1 yielding the measure on the νv = πr2 = πr = π dθ/π S dx quotient R. This produces the standard Lebesgue measure on and so the volume of × is precisely ∆∞ ∆∞/ log OK . The volume of Q Q 1 Q × is n where is the number of in nite places. Putting RK v| {±1} v| S v ∞ Ov 2 n everything together givesR C - n 1 1 × 2 hK RK ν (AK /K ) = wK
The following section was not covered in lecture
Corollary 8.11. Let ζK be the Dedekind ζ-function of K. Show that
r s 2 (2π) hK RK lim(s − 1)ζK (s) = s→1 p wK |DK | where , , and are as in Theorem 8.10, is the number of non-real embeddings , r hK RK wK 2s K,→ C and is the discriminant of . hK = | Cl(K)| DK K/Q
47 Proof. Note that if r1 is the number of real places and r2 is the number of complex places then
r1 −s/2 −s r2 L(1, s) = π Γ(s/2) 2(2π) Γ(s) ζK (s)
and the result follows from Theorem 8.10 and the fact that
−r2 ress=1 L(1, s) = π ress=1 ζK (s)
Theorem 8.12 (Strong multiplicity one for characters). Let be a number eld and × × × K χ1, χ2 : AK /K → C be two continuous Hecke characters such that χ1,v = χ2,v for almost all v. Then χ1 = χ2. Proof. Let −1 such that for where is a nite set of places which, by assumption, χ = χ1χ2 χv = 1 v∈ / S S does not include the in nite places. Then Y Y L(χ, s) = L(χv, s) L(1v, s) v∈s v∈ /S
Y L(χv, s) = L(1, s) L(1v, s) v∈S
If is unrami ed let and otherwise let . Then −s −1 and so χv αv = χv($v) αv = 0 L(χv, s) = (1 − αvqv )
Y 1 − q−s L(χ, s) = L(1, s) v 1 − α q−s v∈S v v
−s 1−qv But each −s is nonzero at s = 0 and s = 1. Thus L(χ, s) has a pole at s = 0 and s = 1 which implies, 1−αv qv by Theorem 8.9, that χ = 1.
End of section not covered in lecture
Lecture 22 2013-05-22
9 Hecke theory for Galois representations 9.1 Global theory Let be a nite extension and be a continuous Galois representation. We have K/Q ρ : GK → GL(n, C) already de ned
∞ Y −s IKv −1 L (ρ, s) = det(1 − ρ(Frobv)qv |ρ ) v-∞ Lemma 9.1. Let K be a number eld and L/K a nite extension. 1. The function L∞(−, s) extends to virtual representations. In particular, if ρ is a virtual representation such that ρ = 0 then L∞(ρ, s) = 1. 2. If then ρ : GL → GL(n, C) ∞ K ∞ L (IndL ρ, s) = L (ρ, s)
48 Proof. For the rst part, note that ρ being continuous will have open kernel of the form GL for L/K nite Galois. Thus . Maschke's theorem then implies that is completely reducible and ρ : GL/K → GL(n, C) ρ ∞ ∞ ∞ so we only need to check that L (ρ1 ⊕ ρ2, s) = L (ρ1, s)L (ρ2, s) which is immediate from the fact that Iv det(1 − ρ1 ⊕ ρ2(Frobv)X|(ρ1 ⊕ ρ2) ) = det(1 − ρ1(Frobv)X|ρ1) det(1 − ρ1(Frobv)X|ρ1). The second part requires some work. See for example [Neu99, Chapter VII, Proposition 10.4 (iv)]. The idea is to use the decomposition (IndK ρ)| = ⊕ IndKv (ρ| ) and then to express the action of Frob L GKv w|v Lw GLw v on the inertial invariants of this space as a matrix in terms of the action of Frobw on the inertial invariants of the ρ| . GLw s ∞ Theorem 9.2. There exists L∞(ρ, s), and ε(ρ, s) of the form A · B such that if L(ρ, s) = L∞(ρ, s)L (ρ, s) then L(ρ, s) = ε(ρ, s)L(ρ∗, 1 − s)
where ρ∗ = Hom(ρ, C) with action (ρ∗(g)f)(v) = f(ρ(g−1)(v)).
Proof. If × is a continuous character then factors through ab ∼ × × ×,0 and therefore χ : GK → C χ GK = AK /K K∞ get × × ×. It is easy to see that ∞ with a Galois character is ∞ with a Hecke χ : AK /K → C L (χ, s) χ L (χ, s) χ character. By Theorem 8.9 L(χ, s) = ε(χ, s)L(χ−1, 1 − s). As ρ is continuous, there exists a nite Galois extension L/K such that GL ⊂ ker ρ and so ρ factors through . Brauer's theorem implies the existence of cyclic extensions and charac- GL/K → GL(n, C) L/Li/K ters χ : G → × such that ρ = P m IndK χ in the Grothendieck group of continuous representations i L/Li C i Li i of GK . Let L (ρ, s) = Q L (IndK χ , s)mi in which case L(ρ, s) = Q L(χ , s)mi . Let ε(ρ, s) = Q ε(χ , s)mi . ∞ ∞ Li i i i Then
Y mi L(ρ, s) = L(χi, s)
Y mi −1 mi = ε(χi, s) L(χi , 1 − s) = ε(ρ, s)L(ρ∗, 1 − s)
as desired. It remains to show that L(ρ, s) (and therefore ε(ρ, s)) is well-de ned, i.e., if P n IndK χ = 0 as a i Li i virtual representation, then Q L(χ , s)ni = 1. By Lemma 9.1, Q L∞(χ , s)ni = L∞(P n IndK χ , s) = 1. i i i Li i Therefore it su ces to show that Q ni . We will, in fact, show that for each place of , i L∞(χi, s) = 1 v | ∞ K Q Q L (χ , s)ni = 1 where w are places of L lying above v. i wi|v ∞ i,wi i i First, note that the characters are ( nite order) characters of and so or where is χi GL/Li χi,wi = 1 σ σ the sign character for real places w . Next, from P n IndK χ = 0 restricting to G we get i i Li i Kv
X X Kv X X Kv ni Ind χi,wi = ni Ind 1 = 0 Li,wi Li,wi i wi|v i wi|v
If v | then w | for all w and so we deduce that P n P 1 = P n [L : K]1 = 0 which is C i C i i i wi|v i i i equivalent to P . At the same time −s and we compute ni[Li : K] = 0 L(χi,wi , s) = L(1, s) = 2(2π) Γ(s)
Y Y ni Y Y −s ni L(χi,wi , s) = (2(2π) Γ(s))
i wi|v i wi|v −s P n [L :K] = (2(2π) Γ(s)) i i i = 1
If denote by the set of with and , by the set of such that v | R I (i, wi) wi | R χi,wi = 1 J (i, wi) wi | R and and by the set of such that and (necessarily) . Then the formula χi,wi = σ H (i, wi) wi | C χi,wi = 1
49 P P Kv ni Ind 1 = 0 becomes i wi|v Li,wi X X X ni · 1 + ni · σ + ni(1 + σ) = 0
(i,wi)∈I (i,wi)∈J (i,wi)∈H
as IndR 1 = 1 ⊕ σ. We deduce that P n + P n = 0 and P n + P n = 0. We compute C I i H i J i H i
Y Y ni Y ni Y ni Y ni L(χi,wi , s) = L(1R, s) L(σ, s) L(1C, s) i wi|v (i,wi)∈I (i,wi)∈J (i,wi)∈H P P P I ni J ni H ni = L(1R, s) L(σ, s) L(1C, s) P P P − H ni − H ni H ni = L(1R, s) L(σ, s) L(1C, s) P n L(1 , s) H i = C L(1R, s)L(σ, s) = 1
where we used the identity π−s/2Γ(s/2) π−(s+1)/2Γ((s + 1)/2) = 2(2π)−sΓ(s) in other words . L(1R, s)L(σ, s) = L(1C, s)
9.2 Deligne's local ε-factors Having settled the issue of the existence of for a global one is left with the natural ε(ρ, s) ρ : GK → GL(n, C) question of de ning for . Such an -factor would encode information about ε(ρv, ψ, µ, s) ρv : GKv → GL(n, C) ε the rami cation of ρv and appears naturally in the statement of the local Langlands correspondence. Theorem 9.3. Let be a nite extension, nontrivial and a Haar measure on . There exist K/Qp ψ ∈ Kb µ K s ε(ρ, ψ, µ, s) (of the form A·B ) attached to nite dimensional continuous representations ρ of GK such that: 1. If × is a character then as de ned for characters of ×. χ : GK → C ε(χ, ψ, µ, s) = ε(χ ◦ rK , ψ, µ, s) K
2. ε(ρ1 ⊕ ρ2, ψ, µ, s) = ε(ρ1, ψ, µ, s)ε(ρ2, ψ, µ, s) so ε(−, ψ, µ, s) is multiplicative on the Grothendieck ring. 3. For r ∈ (0, ∞), ε(ρ, ψ, rµ, s) = rdim ρε(ρ, ψ, µ, s).
4. For ×, (s−1) dim ρ . a ∈ K ε(ρ, ψ(a · −), µ, s) = det ρ(rK (a))|a|K ε(ρ, ψ, µ, s) P 5. If L/K is a nite extension and ρ is a representation of virtual dimension 0 (i.e., ρ = miρi in the Grothendieck ring with P ) then K 0 for any Haar mi dim ρi = 0 ε(IndL ρ, ψ, µ, s) = ε(ρ, ψ ◦ TrL/K , µ , s) measure µ0 of L. 6. Have −s(cond ρ−dim ρ cond ψ) ε(ρ, ψ, µ, s) = ε(ρ, ψ, µ, 0) · qK
7. There exists fρ such that if χ is a character of conductor f ≥ fρ then
ε(ρ ⊗ χ, ψ, µ, s) = det ρ(c)−1ε(χ, ψ, µ, s)dim ρ
where c is such that χ(1 + x) = ψ(cx) for vK (x) ≥ df/2e.
50 In principle this should follow from the Brauer induction theorem. Indeed ρ trivializes GL for some nite Galois extension L/K and thus factors through GL/K . There exist L/Mi/K subextensions and character × such that in the Grothendieck ring χi : GL/Mi → C X ρ − dim ρ · 1 = n IndL (χ − 1) i Mi i where . Thus ni ∈ Z Y ni ε(ρ − dim ρ · 1, ψ, µ, s) = ε(χi − 1, ψ ◦ TrMi/K , µi, s) giving ε(ρ, ψ, µ, s). However, the challenge is to show that this de nition does not depend on Mi, χi and ni. In fact there is no current local proof of this fact. The actual proof will use the global ε-factors of Theorem 9.2 which are known to exist. To do this we need to go from the local to the global setting and in the process prove results that ensure that choices do not aect the outcome. Lemma 9.4. Let L/K be a nite Galois extension of p-adic elds. There exists a nite Galois extension of number elds , a nite place of and a unique place of such that and . E/F v0 F u0 E Fv0 = K Eu0 = L ∼ Moreover, GE/F = GL/K . Proof. Since one may choose a number eld which is dense in . Let be the composite Q ⊂ Qp E0 ⊂ L L E of {σ(E0)|α ∈ GL/K } and let F = E ∩ K. Since E ⊂ L and F ⊂ K are dense, GE/F = GL/K . The dense embedding de nes a nite place of with and the dense embedding de nes a F ⊂ K v0 F Fv0 = K E ⊂ L nite place of with . Now is xed by and so is the only place of above u0 E Eu0 = L u0 GL/K = GE/F u0 E v0.
Lemma 9.5. Let L/K be a nite Galois extension of p-adic elds and let e = eL/K be the rami cation index. Then for x ∈ OL one has
d2vL(x)/ee NL/K (1 + x) ≡ 1 + TrL/K (x) (mod mK ) −1 Proof. If σ ∈ GL/K then vK (σ(x)) = vK (x) = e vL(x) and therefore if I ⊂ GL/K is a set of cardinality i then Q i . If then vK ( σ∈I σ(x)) = e vL(x) i ≥ 2 X Y i 2vL(x) v ( σ(x)) ≥ v (x) ≥ K e L e I⊂GL/K ,|I|=i σ∈I and so Y NL/K (1 + x) = (1 + σ(x))
σ∈GL/K [L:K] X X Y = 1 + TrL/K (x) + σ(x)
i=2 I⊂GL/K ,|I|=i σ∈I
d2vL(x)/ee ≡ 1 + TrL/K (mod mK )
Lecture 23 2013-05-24
Lemma 9.6. Let be nite extensions. Let be the smallest integer such that `L/K . L/K/Qp `L/K GK ⊂ GL Then for y ≥ `L/K , −1 φL/K (y) = eL/K y − vL(DL/K ) Z ∞ −1 where recall that φL/K is the rami cation function de ned as φL/K (x) = [GL/K,0 : GL/K,u] du when 0 is Galois, and −1 where is the Galois closure of in the non-Galois case. L/K φL/K = φE/K ◦ φE/L E L/K
51 Proof. Start with L/K Galois. The graph of the function φL/K is piece-wise linear with in ection points at the jumps in the rami cation ltration. In particular, for −1 , the slope of the graph of x ≥ φL/K (`L/K ) φL/K is −1 . This implies that for , eL/K y = φL/K (x) ≥ `L/K
−1 −1 φL/K (y) = eL/K y + φL/K (`L/K ) − eL/K `L/K
Let −1 . The we need to show that . But k = φL/K (`L/K ) ∈ Z eL/K φL/K (k) − k = vL(DL/K )
k Z eL/K du eL/K φL/K (k) − k = − k 0 [GL/K,0 : GL/K,u] Z k = |GL/K,u|du − k 0 k X = (|GL/K,i| − 1) i=1
= vL(DL/K )
Now suppose L/K is not Galois and let E be the Galois closure. For y >> 0 one has
−1 y = φE/L(φE/L(y)) = eE/LφE/L − vE(DE/L) and so
−1 −1 φL/K (y) = φE/L(φE/K (y)) −1 φ (y) + vE(DE/L) = E/K eE/L
e y − vE(D ) + vE(D ) = E/K E/K E/L eE/L
= eL/K y − vL(DL/K )
where the last equality follows from . To show that −1 for DE/K = DE/LDL/K φL/K (y) = eL/K y − vL(DL/K ) it su ces to show that −1 0 where 0 means right derivative. Using the chain y ≥ `L/K (φL/K ) (`L/K ) = eL/K () rule we get
−1 0 −1 0 (φL/K ) (`L/K ) = (φE/L ◦ φE/K ) (`L/K ) 0 −1 φ (φ (`L/K )) = E/L E/K 0 −1 φE/K (φE/K (`L/K ))
[IE/K : GE/K,φ−1 (` )] = E/K L/K [IE/L : G −1 ] E/L,φE/K (`L/K )
eL/K |GE/L,φ−1 (` )| = E/K L/K |G −1 | E/K,φE/K (`L/K )
as 0 by de nition. Thus it is enough to show that φL/K (x) = 1/[IL/K : GL/K,x]
|G −1 | = |G −1 | E/K,φE/K (`L/K ) E/L,φE/K (`L/K )
52 −1 −1 `L/K φE/L(φE/K (`L/K )) φL/K (`L/K ) But G −1 = G while G −1 = G = G . But Herbrand E/K,φE/K (`L/K ) E/K E/L,φE/K (`L/K ) E/L E/L implies that −1 −1 φL/K (`L/K ) φL/K (φL/K (`L/K )) `L/K `L/K GE/L = GE/K ∩ GE/L = GE/K ∩ GE/L = GE/K
where the last equality follows from the fact that by de nition of we have `L/K . `L/K GE/K ⊂ GE/L
Lemma 9.7. Let L/K be a nite extension of p-adic local elds and let α : K× → C× be a continuous character of conductor cond(α) > `L/K . Then
−1 cond(α ◦ NL/K ) = φL/K (cond(α))
Proof. Let the smallest integer such that vanishes on m. Via the local Artin m = cond(α◦NL/K ) α◦NL/K UK map is the smallest integer such that −1 vanishes on m. But −1 −1 m α ◦ NL/K ◦ rL GL α ◦ NL/K ◦ rL = α ◦ rK which would then have to vanish on m with smallest with this property. Herbrand's theorem says GL m that u φL/K (u) and so is the smallest integer such that −1 is trivial on φL/K (m) GL = GK ∩ GL m α ◦ rK GK ∩ φL/K (m) `L/K max(φL/K (m),`L/K ). This implies that . Since GL = GK ∩ GK = GK max(φL/K (m), `L/K ) ≥ cond(α) cond(α) > `L/K we get that φL/K (m) ≥ cond(α) and m is minimal with this property. The result follows since −1 from the previous lemma. φL/K (`L/K ) ∈ Z
Lecture 24 2013-06-05
Lemma 9.8. Let L/K be a nite Galois extension of p-adic elds. There exists an integer nL/K which depends only on L/K with the following property. Suppose α : K× → C× is a continuous character of conductor n ≥ nL/K and a = bn/2c − vL(DL/K ). Let ψ be a nontrivial additive character of K and let ca ∈ K from the proof of Proposition 8.6 such that α(1 + x) = ψ(cax) for x ∈ K with vK (x) ≥ n − a. If L/M/K is a subextension and χ is any character of GL/M then
−1 ε(χ · α ◦ NM/K , ψ ◦ TrM/K , µ, s) = χ(ca) ε(α ◦ NM/K , ψ ◦ TrM/K , µ, s)
for any Haar measure µ on M.
Proof. We will apply the stability Proposition 8.6 for χ1 = χ, χ2 = 1 and η = α ◦ NM/K . Denote by ca,η,ψ such that η(1 + x) = ψ(ca,η,ψx) for vK (x) ≥ cond(η) − a where a ≤ cond(η)/2. Let
nL/K ≥ max(`M/K , 2(vL(DL/K ) + vK (DM/K ) + `L/M /eM/K ))
as M varies among the subextensions L/M/K. We will show that if χ is a character of GL/M then for b = eM/K a − vM (DM/K ) we have cond(χ), cond(1) ≤ b ≤ cond(α ◦ NM/K )/2. Then Proposition 8.6 would imply that
−1 ε(χ · α ◦ NM/K , ψ ◦ TrM/K , µ, s) = χ(cb,α◦NM/K ,ψ◦TrM/K ) ε(α ◦ NM/K , ψ ◦ TrM/K , µ, s) and the Lemma would follow if we could check that
cb,α◦NM/K ,ψ◦TrM/K = ca,α,ψ
From n ≥ nL/K ≥ 2(vL(DL/K ) + vK (DM/K ) + `L/M /eM/K ) we deduce that b ≥ `L/M . But if χ is a character of it will be trivial on `L/M and so as desired. To check that GL/M GM ⊂ GL cond(χ) ≤ `L/M ≤ b b ≤ cond(α ◦ NM/K )/2 it su ces to check that
−1 b = eM/K (n/2 − vL(DL/K )) − vM (DM/K ) ≤ cond(α ◦ NM/K )/2 = φM/K (n)/2
53 where the last equality follows from Lemma 9.7 as cond(α) = n ≥ nL/K ≥ `M/K . This inequality is equivalent to
−1 φM/K (n)/2 ≥ eM/K (n/2 − vL(DL/K )) − vM (DM/K )
(eM/K n − vM (DM/K ))/2 ≥ eM/K (n/2 − vL(DL/K )) − vM (DM/K )
eM/K vL(DL/K ) + vM (DM/K )/2 ≥ 0 which is clear. It remains to show that . Let such that . cb,α◦NM/K ,ψ◦TrM/K = ca,α,ψ x ∈ M vM (x) ≥ cond(α ◦ NM/K ) − b Then vM (x) ≥ eM/K (n − a) + vM (DM/K ) and so vK (TrM/K (x)) ≥ vK (x) ≥ n − a > 0. By Lemma 9.5, NM/K (1 + x) = 1 + TrM/K (x) + y where vK (y) ≥ 2vM (x)/eM/K ≥ 2(n − a) ≥ n. Therefore
y α(NM/K (1 + x)) = α(1 + TrM/K (x))α 1 + 1 + TrM/K (x)
= α(1 + TrM/K (x))
= ψ(ca,α,ψ TrM/K (x))
= ψ(TrM/K (ca,α,ψx))
y n since 1+ ∈ U ⊂ ker α and vK (TrM/K (x)) ≥ n−a. By de nition this implies that cb,α◦N ,ψ◦Tr = 1+TrM/K (x) K M/K M/K ca,α,ψ and the result of the lemma follows.
Proof of Theorem 9.3. Any continuous Galois representation ρ will be trivial on GL for some nite Galois extension L/K. For a xed nite Galois extension L/K and any representation ρ of GL/K we will construct ε(ρ, ψ, µ, s). ∼ Let E/F be the nite Galois extension from Lemma 9.4. Since GL/K = GE/F every representation ρ of is also a representation of . GL/K ρe GE/F Let S be the nite set of places of F containing the places where E/F rami es and the place v0. For each choose a nite order character of × of conductor for a (any) place of . v ∈ S − v0 αv Fv nv ≥ nEu/Fv u | v E For let . By Theorem 5.5 there exists a continuous character of × × such that v = v0 αv = 1 α AF /F α|Fv = αv for . Choose a nontrivial character in such that and a Haar measure on v ∈ S ψF AcF /F ψF,v0 = ψ µF AF such that , and for . When let × such that µF (AF /F ) = 1 µF,v0 = µ µF,v(Ov) = 1 v∈ / S v ∈ S − v0 cv ∈ Fv for (see Lemma 9.8). When choose and αv(1 + x) = ψv(cvx) v(x) ≥ dnv/2e − v(DEu/Fv ) v∈ / S − v0 cv = 1 let ×. c = (cv) ∈ AF Suppose L/M/K is a subextension with corresponding global subextension E/H/F and ρ is a represen- tation of giving a representation of . Let be the places of over places in and again we GL/M ρe GE/H SH H S denote by v0 the unique place of H over v0. Let µM be a Haar measure on M and let µH be a Haar measure on giving volume 1 to and such that . We will de ne AH AH /H µH,v0 = µM !− dim ρ Y ε(ρ, ψ◦Tr , µ , s) = ε(ρ·α◦N , s) det ρ(r (c)) ε(α ◦ N , ψ ◦ Tr , µ , s) M/L M e H/F e H v Hw/Fv F,v Hw/Fv H,w w∈SH −v0 where v denotes the place of F under w. A priori ε(ρ, ψ ◦ TrM/K , µM , s) depends on α. We will check the following four facts:
1. if χ is a character of GL/M then ε(χ, ψ ◦ TrM/K , µM , s) = ε(χ ◦ rM , Tr ◦ TrM/K , µM , s) as de ned for characters,
2. ε(−, ψ ◦ TrM/K , µM , s) extends to the Grothendieck group,
3. if ρ has virtual dimension 0 then ε(ρ, ψ ◦ TrM/L, µM , s) does not depend on µM
54 4. if are subextensions and is a virtual representation of of virtual dimension 0 L/M1/M2/K ρ GL/M1 then for any choices of µ and µ have ε(IndM2 ρ, ψ ◦ Tr , µ , s) = ε(ρ, ψ ◦ Tr , µ , s). M1 M2 M1 M2/K M2 M1/K M1 Independence of : Use Brauer induction to nd , characters of and integers such α L/Li/K χi GL/Li ni that ρ − dim ρ · 1 = P n IndK (χ − 1). Using properties 2 and 5 we deduce i Li i
ni ε(ρ, ψ, µ, s) Y ε(χi, ψ ◦ TrLi/K , µi, s) dim ρ = ε(1, ψ, µ, s) ε(1, ψ ◦ TrLi/K , µi, s) and independence of α is clear. Lecture 25 2013-06-07
Property 1 and fact 1: We now check that if L/M/K is a subextension and χ is a character of GL/M then ε(χ, ψ ◦ TrM/K , µM , s) = ε(χ ◦ rM , Tr ◦ TrM/K , µM , s). This implies property 1. By construction Y ε(χ, ψ , µ , s) = ε(χ · α ◦ N , s)χ(r (c)) ε(α ◦ N , ψ , µ , s) M M e H/F e H v Hw/Fv H,w H,w w∈SH −v0 where SH are the places of H over places in S and again we denote by v0 the unique place of H over v0; here v is the place of F under w, and ψH = ψF ◦ TrH/F . Now SH contains all the places of rami cation of and , being a character of , can only ramify at . Therefore by de nition E/H χe GE/H w ∈ SH Y ε(χ · α ◦ N , s) = ε(χ · α ◦ N , ψ , µ , s) e H/F ew v Hw/Fv H,w H,w w∈SH Q where recall that αv = 1 and so αv ◦N = 1 and cv = 1 by choice. Since χ(rH (c)) = χw(rH (cv)) 0 0 Hv0 /Fv0 0 e w∈SH e w we need to check that
ε(χ , ψ , µ , s) = ε(χ , ψ , µ , s)χ (1) ev0 H,v0 H,v0 ev0 H,v0 H,v0 ev0 Y × ε(χ α ◦ N , ψ , µ , s)χ (c )ε(α ◦ N , ψ , µ , s)−1 ew v Hw/Fv H,w H,w ew v v Hw/Fv H,w H,w w∈SH −v0 for which is is enough to show that ε(χ α ◦ N , ψ , µ , s)χ (c ) = ε(α ◦ N , ψ , µ , s). ew v Hw/Fv H,w H,w ew v v Hw/Fv H,w H,w Since this is implied by Lemma 9.8. cond(αv) = nv ≥ nEu/Lv Property 2 and fact 2: The fact that ε(ρ, ψ ◦ TrM/K , µM , s) extends to the Grothendieck group is automatic from the fact that L(−, s), det and dim extend. dim ρ Property 3 and fact 3: We will show that ε(ρ, ψ ◦TrM/K , rµM , s) = r ε(ρ, ψ ◦TrM/K , µM , s) which also implies that if ρ has virtual dimension 0 then ε(ρ, ψ ◦ TrM/K , µM , s) does not depend on µM . Recall that gives volume 1 to . Let and with , µH = ⊗µH,v AH /H r > 0 µH,r = ⊗µH,r,w µH,r,v0 = rµH,v0
55 −1 µH,r,u = r µH,u for some u ∈ SH − v0 above t and µH,r,w = µH,w for w 6= v0, u. Then
!− dim ρ ε(ρ, ψ ◦ TrM/K , rµM , s) Y = ε(αv ◦ NHw/Fv , ψF,v ◦ TrHw/Fv , µH,r,w, s) ε(ρα ◦ NH/F , s) det ρ(rH (c)) e w∈SH −v0 −1 = ε(αt ◦ NHu/Ft , ψF,t ◦ TrHu/Ft , r µH,u, s)× − dim ρ Y × ε(αv ◦ NHw/Fv , ψF,v ◦ TrHw/Fv , µH,w, s)
w∈SH −{v0,u} !− dim ρ dim ρ Y = r ε(αv ◦ NHw/Fv , ψF,v ◦ TrHw/Fv , µH,w, s)
w∈SH −v0 dim ρ r ε(ρ, ψ ◦ Tr , µM , s) = M/K ε(ραe ◦ NH/F , s) det ρ(rH (c)) by Theorem 8.4. Property 4: We have seen that
ni Y ε(χi, ψ ◦ TrLi/K (a · −), µi, s) ε(ρ, ψ(a · −), µ, s) = ε(1, ψ(a · −), µ, s)dim ρ ε(1, ψ ◦ TrLi/K (a · −), µi, s) ε(χ , ψ ◦ Tr , µ , s)ni (s−1) dim ρ Y ni dim ρ Y i Li/K i = (|a|K χi(rLi (a)) )ε(1, ψ, µ, s) ε(1, ψ ◦ TrLi/K , µi, s) (s−1) dim ρ = det ρ(rK (a))|a|K ε(ρ, ψ, µ, s)
by Corollary 8.5 and the fact that if ρ − dim ρ · 1 = P n IndK (χ − 1) then i Li i Y Y det ρ ◦ r = χ (cor∨ ◦r )ni = (χ ◦ r )ni K i Li/K K i Li
Property 5 and fact 4: Suppose L/M1/M2/K corresponds to E/H1/H2/F and ρ is a virtual dimension 0 representation of G giving ρ of G . Then L/M1 e E/H1
M2 H2 H2 ε(Ind ρ, ψ ◦ Tr , µ, s) = ε(Ind ρ · α ◦ N , s) det Ind ρ(rH (c)) M1 M2/K H1 e H2/F H1 e 2 H2 ∨ = ε(Ind (ρ · α ◦ N ), s) det ρ(cor ◦rH (c)) H1 e H1/F e H1/H2 2 = ε(ρ · α ◦ N , s) det ρ(r (c)) e H1/F e H1
= ε(ρ, ψ ◦ TrM1/K , µM1 , s)
M2 H2 ∨ ∨ since dim Ind ρ = [M1 : M2] dim ρ = 0 and det Ind ρ = ρ ◦ cor and cor ◦rH = rH . M1 H1 e e H1/H2 H1/H2 2 1 Certainly global L-functions are inductive in that if ρ is a representation of G then L(IndH2 ρ, s) = E/H1 H1 L(ρ, s) and therefore ε(IndH2 ρ, s) = ε(ρ, s). Moreover, det IndH2 ρ = and dim IndH2 ρ = [H : H ] dim ρ. H1 H1 H1 1 2 Note that while the character ψ does not seem to appear in the formulae, it does as c is de ned in terms of ψ. Property 6: Let be such that −n(ρ,ψ)s. We know from Theorem 8.4 n(ρ, ψ) ε(ρ, ψ, µ, s) = ε(ρ, ψ, µ, 0)qK that n(χ, ψ) = cond(χ) − cond(ψ). Also property 2 implies that n(ρ1 ⊕ ρ2, ψ) = n(ρ1, ψ) + n(ρ2, ψ) and so n(−, ψ) extends to the Grothendieck group. Now suppose L/K is a nite extension and χ is a character of ×. Then property 5 gives K since fL/K . Thus if L n(IndL (χ − 1), ψ) = fL/K n(χ − 1, ψ ◦ TrL/K ) qL = qK
56 ρ − dim ρ · 1 = P n IndK (χ − 1) then i Li i X n(ρ, ψ) = (dim ρ)n(1, ψ) + n n(IndK (χ − 1), ψ) i Li i X = (dim ρ)n(1, ψ) + nifLi/K n(χi − 1, ψ ◦ TrLi/K ) X = −(dim ρ) cond(ψ) + nifLi/K cond(χi) because n(1, ψ) = − cond(ψ)+cond(1) = − cond(ψ) and n(χ−1, ψ ◦TrL/K ) = cond(χ)−cond(1) = cond(χ). The result now follows from the computation
cond(ρ) = cond(ρ − dim ρ · 1) X = n cond(IndK (χ − 1)) i Li i X = nifLi/K (cond(χi) + vLi (DLi/K ) − cond(1) − vLi (DLi/K ) X = nifLi/K cond(χi) using the fact that K . cond(IndL ρ) = fL/K (cond(ρ) + dim ρvL(DL/K )) Property 7: As before if ρ − dim ρ · 1 = P n IndK (χ − 1) then i Li i
ni Y ε(χi · χ ◦ NLi/K , ψ ◦ TrLi/K , µi, s) ε(ρ ⊗ χ, ψ, µ, s) = ε(χ, ψ, µ, s)dim ρ ε(χ ◦ NLi/K , ψ ◦ TrLi/K , µi, s)
If then Lemma 9.8 gives −1 cond(χ) = f ≥ nL/K ε(χi · χ ◦ NLi/K , ψ ◦ TrLi/K , µi, s) = χi(c) ε(χ ◦ NLi/K , ψ ◦ . Thus TrLi/K , µi, s)
dim ρ Y −ni ε(ρ ⊗ χ, ψ, µ, s) = ε(χ, ψ, µ, s) χi(rLi (c)) Y = ε(χ, ψ, µ, s)dim ρ χ (cor∨ ◦r (c))−ni i Li/K K dim ρ −1 = ε(χ, ψ, µ, s) det ρ(rK (c)) as desired.
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