HIDA THEORY NOTES TAKEN BY PAK-HIN LEE

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1 HIDA THEORY NOTES TAKEN BY PAK-HIN LEE Abstract. These are notes from the (ongoing) Student Number Theory Seminar on Hida theory at Columbia University in Fall 2017, which is organized by David Hansen and Samuel Mundy. Contents 1. Lecture 1 (September 7, 2017): Samuel Mundy Ribet s Converse to Herbrand s Theorem Introduction Modular forms Proof of Ribet s theorem 3 2. Lecture 2 (September 14, 2017): Samuel Mundy Basic Iwasawa Theory Z p -extensions and Iwasawa s theorem Λ-modules Proof of Iwasawa s theorem 7 3. Lecture 3 (September 21, 2017): Samuel Mundy p-adic Properties of L-functions Class number formulas p-adic L-functions Main Conjecture Lecture 4 (September 28, 2017): Pak-Hin Lee Λ-adic modular forms Lecture 5 (October 5, 2017): Pak-Hin Lee Ordinary Λ-adic forms Lecture 6 (October 12, 2017): Pak-Hin Lee Hida Theory and Λ-adic Galois Representations Lecture 7 (October 19, 2017): Pak-Hin Lee Λ-adic Representations via Pseudo-representations Lecture 8 (October 26, 2017): Samuel Mundy Proof of the Main Conjecture, I Reduction to one inclusion Proof of Main Conjecture 14 Last updated: November 10, Please send corrections and comments to phlee@math.columbia.edu. We thank Samuel Mundy for proofreading and editting these notes. 1

2 1. Lecture 1 (September 7, 2017): Samuel Mundy Ribet s Converse to Herbrand s Theorem 1.1. Introduction. Let p be an odd prime, and ω : G Q µ p 1 Z p be the Teichmüller character (so that σζ p = ζp ω(σ) for all σ G Q ). For any Z p [G Q ]-module M and ϕ Gal(Q(ζ p )/Q), write M ϕ = {m M σm = ϕ(σ)m} to be the ϕ-eigenspace of M. Let B n be the n-th Bernoulli number. Today we are going to prove Ribet s converse to Herbrand s theorem: Theorem 1.1 (Ribet). Let m be an odd integer with 3 m p 2. If p B p m, then Cl Q(ζp)[p ] ωm 0. Remark. We have v p (B p 1 ) = 1 and Cl Q(ζp)[p ] ω = 0. The idea is to deduce properties of Galois representations attached to automorphic representations of GL 1 using those for GL Modular forms. Fix embeddings Q Q p C. For N 1 an integer, ψ : (Z/NZ) C a Dirichlet character, and A C a subring containing Z[ψ], write { } S k (N, ψ, A) = cusp forms f = a n (f)q n of weight k, level N, nebentypus ψ a n (f) A n=1 where q = e 2πiz. Denote by T k (N, ψ, A) the sub-a-algebra of End A (S k (N, ψ, A)) generated by the Hecke operators T (n) for all n. We define a pairing In particular, f, T (n) = a n (f). Theorem 1.2. This pairing is perfect. S k (N, ψ, A) T k (N, ψ, A) A f, h a 1 (f h). Theorem 1.3. Let B be an A-algebra with φ : A B. Under the isomorphism Hom A (T k (N, ψ, A), B) S k (N, ψ, A) A B, the A-algebra homomorphisms Hom A-alg (T k (N, ψ, A), B) correspond to the normalized eigenforms, i.e., f = a n (f)q n S k (N, ψ, A) A B B[[q]] with a 1 (f) = 1 such that for all h T k (N, ψ, A), there exists c B such that φ(f h) = cφ(f), where the corresponding homomorphism on the LHS sends an operator to its eigenvalue. Theorem 1.4. Let f = n=1 a n(f)q n S k (N, ψ, A) be a normalized eigenform. Then there exists a continuous Galois representation ρ f : G Q Aut(V ) with dim Qp (V ) = 2 such that (1) ρ f is irreducible; (2) for all l pn, ρ f is unramified at l; (3) for all l pn, tr ρ f (Frob l ) = a l (f); (4) for all l pn, det ρ f (Frob l ) = ψ(l)l k 1. Definition 1.5. A normalized eigenform f S k (1, Q p ) is ordinary if a p (f) p = 1. 2

3 Theorem 1.6 (Mazur Wiles). If f is ordinary, then there exists a basis v 1, v 2 of V such that α ρ f GQp = 1 χ k 1, α where χ : G Q Z p is the cyclotomic character (i.e., σζ = ζ χ(σ) for all ζ µ p ), and α : G Qp Q p is the unramified character such that α(frobp ) is the unit root of X 2 a p (f)x + p k Proof of Ribet s theorem. The proof consists of the following steps. Step 0: Let κ by any finite field of characteristic p. Then Cl Q(ζp)[p] ωm Fp κ = Hom GQ (G ab,unr Q(ζ p), κ(ωm )) = H 1 unr (G Q(ζp), κ(ω m )) G Q Step 1 (Construction of an eigenform): Let E k (z) = ζ(1 k) 2 Res 1 = H 1 unr (G Q, κ(ω m )) = Ext unr κ[g Q ](κ, κ(ω m )) = Ext unr κ[g Q ](κ(ω p 1 m ), κ). + d n be the Eisenstein series of weight k, where k 4 is even. The following facts are classical: (1) ζ(1 k) = B k /k. (2) v p (B k ) < 0 p 1 k v p (B k ) = 1. (3) E k M k (1, C). If p 1 k, then E k M k (1, Z p ) by (2). (4) If m n (mod p 1), then B m B n (mod p). Note that E 4 (z) = q +, E 6 (z) = q +. Let k 4 and k p m (mod p 1). Write k = 4a + 6b where a, b 0. Then define G k = (240E 4 ) a ( 504E 6 ) b = 1 + a n q n where a n Z. Let F k = E k n=1 ζ(1 k) G k = 2 d k 1 n=1 q n b n q n where b n = B k 2k a n + d n dk 1. Now assume p B p m. Then by (4), p B k and so b n d n dk 1 (mod p). Thus the map n=1 T k (1, Z p ) F p 3

4 T (n) d n d k 1 (mod p) is a Z p -algebra homomorphism, corresponding to F k (mod p). Let m T k (1, Z p ) be the kernel, and p m be a minimal prime. Let f be the eigenform corresponding to the homomorphism T k (1, Z p ) A := T k (1, Z p )/p O L, where L := Frac A. Then f S k (1, O L ) and f F k (mod λ), where λ O L is a uniformizer. Step 2 (Analysis of ρ f ): Write f = n=1 c nq n. Then c p d p d k p k 1 1 so f is ordinary. Let v 1, v 2 V ρf such that α ρ f GQp = 1 χ k 1. α (mod λ), Let σ 0 G Qp be an element such that χ k 1 (σ 0 ) 1 (mod λ) (possible since k 1 is odd, so p 1 k 1). Write β := χ k 1 (σ 0 ). Replacing v 1 by α(σ 0 )v 1 and v 2 by α 1 (σ 0 )v 2 + (something) v 1, we may assume β 0 ρ f (σ 0 ) =. 0 1 aσ b For σ O L [G Q ], let ρ f (σ) = σ. c σ d σ Lemma 1.7. (1) For all σ, τ O L [G Q ], we have a σ, d σ, b σ c τ O L and a σ ω k 1 (σ), d σ 1(σ), b σ c τ 0 (mod λ), where 1 is the trivial G Q -character, and ω k 1, 1 are extended by linearity to O L [G Q ]. (2) C = {c σ σ O L [G Q ]} is a nonzero fractional ideal. (3) c σ = 0 for all σ I l and primes l. Proof. Omitted; play with matrix coefficients using information about the trace. Step 3 (Construction of the lattice): Let M 1 = O L v 1, M 2 = Cv 2 and M = M 1 M 2 V ρf. This is G Q -stable, generated over O L [G Q ] by v 1. Let κ = O L /λ and M = M/λM. Claim. M 2 := M 2 /λm 2 M is a G Q -stable line with trivial action. Let m 2 M 2, so m 2 = c τ v 2 for some c τ C. If σ G Q, then ρ f (σ)m 2 = c τ b σ v 1 + c τ d σ v 2 d σ m 2 = 1(σ)m 2 = m 2 (mod λ) which proves the claim. Let M 1 = M 1 /λm 1. Then M 1 = κ(ω k 1 ) by a similar argument, and there is a short exact sequence 0 κ M κ(ω k 1 ) 0 which is: 4

5 nonsplit over G Q because, if it were, M 2 = κ would be a quotient and hence the image of v 1 would belong to the kernel by this quotient. This is because σ 0 acts as β 1 (mod λ) on v 1, but v 1 generates M over G Q. Hence the kernel is all of M, a contradiction. split over I l for all l because O L v 1 (mod λ) is a stable line under I l, since ρ(σ)v 1 = a σ v 1 + c σ v 2 a σ v 1 (mod λ). Since ω k 1 = ω p 1 m, this extension gives a nontrivial class in Ext unr κ[g Q ](κ(ω p 1 m ), κ). This finishes the proof of Theorem Lecture 2 (September 14, 2017): Samuel Mundy Basic Iwasawa Theory 2.1. Z p -extensions and Iwasawa s theorem. Last time we understood better the structure of Cl Q(ζp) as a G Q -module. A natural question is: What about Cl Q(ζp n)? Theorem 2.1 (Iwasawa). There exist integers n 0 and µ, λ, κ such that for all n > n 0, #Cl Q(ζp n)[p ] = p µpn +λn+κ. This statement a priori does not have anything to do with the Galois structure of class groups, but the proof relies heavily on that. Recall that Gal(Q(ζ p n)/q) (Z/p n Z). Let Q(ζ p ) = n 0 Q(ζ pn). Then Gal(Q(ζ p )/Q) Z p, Gal(Q(ζ p )/Q(ζ p )) Z p. The element 1 ζ p n O Q(ζp n) = Z[ζ p n] generates the (totally ramified) prime above p Z. With these facts we are ready to study basic Iwasawa theory. Definition 2.2. Let K 0 be a number field. A Z p -extension of K 0 is an algebraic extension K /K 0 such that Gal(K /K 0 ) Z p topologically; equivalently, it is a tower of finite extensions such that Gal(K n /K 0 ) Z/p n Z. K 0 K 1 K = n 0 K n Proposition 2.3. Let K /K 0 be a Z p -extension. Then: Proof. (1) Some prime is ramified in K. (2) For any ramified prime p O K0, there exists n 0 such that if q O Kn0 is such that q p, then q is totally ramified in K /K n0. (3) Every ramified prime lies over p Z. (1) The maximal unramified abelian extension of K 0 is finite over K 0. (2) Let p be ramified in K /K 0, and I p Gal(K /K 0 ) Z p be the inertia group at p. This is nontrivial and closed. Thus I p p n 0 Z p for some n 0, so I p = Gal(K /K 0 ). 5

6 (3) Let p O K0 be ramified in K. By local class field theory, Gal(K ab 0,p/K 0,p ) = K 0,p πẑ O K 0,p corresponding to the unramified and ramified parts respectively, so by (2) we get an infinite subgroup of O K 0,p isomorphic to Z p, forcing K 0,p to have residue characteristic p. Proposition 2.4. Let L/K be an extension of number fields such that some prime in K is totally ramified in L. Then the norm map is surjective. Nm : Cl L [p ] Cl K [p ] Remark. This is true without taking p-primary parts and essentially the same proof will work. It is also true if the ramified prime is at infinity. Proof. Let M 0 /K be the maximal unramified abelian p-extension of K, and M 1 that for L. Then M 0 and L are linearly disjoint over K (i.e., M 0 L = K in any K). M 1 LM 0 L M 0 K Since LM 0 /L is an unramified abelian p-extension, it is contained in M 1. Then by class field theory, the diagram Cl L [p ] Gal(M 1 /L) commutes. Nm Cl K [p ] This suggests that we should look at Art Art Gal(M 0 /K) C := lim n 0 Cl Kn [p ], restriction to M 0 where the inverse limit is taken with respect to Nm. Write Γ n = Gal(K n /K 0 ). Since Cl Kn [p ] is a Z p [Γ n ]-module, C is a module over Λ := lim n 0 Z p [Γ n ], where the inverse limit is taken with respect to restriction maps Γ m Γ n if m n. Theorem 2.5 (Iwasawa). C is a finitely generated torsion Λ-module. 6

7 We will see how Theorem 2.1 follows from this Λ-modules. We need to first understand the structure of Λ. Let γ Γ := Gal(K /K 0 ) Λ be a topological generator. An observation due to Serre is that: Proposition 2.6 (Serre). There is an isomorphism of topological rings, induced by γ 1 T. Λ Z p [[T ]] Proof. Omitted; see Lang s Cyclotomic Fields I and II, Chapter 5. Corollary 2.7 (Nakayama s lemma). If M is a compact Λ-module and M/T M = M/(γ 1)M is finitely generated over Z p, then any set of generators in M/T M lifts to a set of generators in M. In particular, M is finitely generated. Definition 2.8. A quasi-isomorphism of Λ-modules M, N is a map ϕ : M N such that there exists an exact sequence with A, B of finite cardinality. 0 A M ϕ N B 0 Quasi-isomorphism is an equivalence relation. Theorem 2.9 (Weierstrass preparation). Let f Z p [[T ]] be a nonconstant power series. Then there exist a unique integer n 0 and unique g, u Z p [[T ]] such that g is distinguished and u Z p [[T ]], and f = p n ug. (A monic polynomial g = T n + a n 1 T n a 0 is distinguished if p a i for all i = 0,, n 1.) Theorem Let M be a finitely generated Λ-module. Then there exist integers r, n i and distinguished f j (where i I and j J with I and J finite) such that M q-iso Λ r i I Λ/(p n i ) j J Λ/(f j ) Proof of Iwasawa s theorem. We will prove Theorem 2.1 and Theorem 2.5 in the case when there is only one ramified prime p O K0 above p. Proof of Theorem 2.5. For simplicity, we further assume that p is totally ramified in K /K 0. Then we make the main Claim. C/(γ 1)C Cl K0. Let M n /K n be the maximal unramified abelian p-extension, and M := n 0 M n, so M /K is the maximal unramified pro-p abelian extension. Let G C := Gal(M /K ) = C, G := Gal(M /K 0 ) and Γ = Gal(K /K 0 ). Claim (Subclaim 1). If I G is the inertia at p, then G G C I. 7

8 There is an exact sequence 1 Gal(M /K ) Gal(M /K 0 ) Gal(K /K 0 ) 1. G C G Γ Since p is totally ramified in K, I surjects onto Γ and has empty intersection with G C. This proves Γ I and subclaim 1. Let Γ I act on G C by conjugation. Then this action is the same as that on C: Frob σq = σ Frob q σ 1. Claim (Subclaim 2). Let G be the commutator subgroup of G. Then G = (γ 1)G. Let g G. Then (γ 1)g = γg γ 1 g 1 G, where γ is a lift of γ to I. Conversely, G/(γ 1)G is the largest quotient where Γ acts trivially. So conjugation by elements in I is trivial, and since G C is abelian and G = G C I, we have that G/(γ 1)G is abelian and hence (γ 1)G G. This proves subclaim 2. To prove the main claim that C/(γ 1)C Cl K0 [p ], it is the same to prove G C /(γ 1)G C = Gal(M0 /K 0 ). Note (γ 1)G C = (γ 1)G (because if g G C, ι I and γ Γ, then picking a lift γ I, we have (γ 1)gι = γgι γ 1 ι 1 g 1 = γgιι 1 γ 1 g 1 = (γ 1)g; the reverse inclusion is trivial). Then G C /(γ 1)G C = G C /(γ 1)G = G/I (γ 1)G = G/I G = Gal(M 0 /K 0 ). This proves the main claim. Since Cl K0 [p ] = C/(γ 1)C is finite, by Nakayama s lemma C is finitely generated. Write C q-iso Λ r Λ/(p n i ) Λ/(f j ). Then since Λ r /(γ 1)Λ r Z r p, we have r = 0. This proves Theorem 2.5. Remark. Since C/(γ pn 1)C is finite, we see that f j is coprime to (1 + T ) pn 1 for all j, n. Finally, let us prove Theorem 2.1: There exist n 0 and µ, λ, κ such that for all n > n 0, #Cl Kn [p ] = p µpn +λn+κ. Exercise. (1) #Λ/(p m, γ pn 1) = p mpn. (2) Let f be a distinguished polynomial coprime to (1 + T ) pn 1 for all n. Then there exists n 0 such that if n > n 0, then #Λ/(f, γ pn 1) = p deg(f)n+c for some c = c(f). Proof of Theorem 2.1. Replacing K 0 by K n where n 0, we may assume the ramified prime is totally ramified. Then setting µ = i n i and λ = j deg f j, we get (by the previous exercise and remark) κ such that #Cl Kn [p ] = #C/(γ pn 1)C = p µpn +λn+κ 8

9 for sufficiently large n. Remark. For the cyclotomic extension Q(ζ p ), it is known that µ = 0; this is the Ferrero Washington theorem. 3. Lecture 3 (September 21, 2017): Samuel Mundy p-adic Properties of L-functions 3.1. Class number formulas. Fix an odd prime p and embeddings Q Q p C. Denote K = Q(ζ p n) throughout. Definition 3.1. (1) A character ψ : Gal(Q(ζ p n)/q) C is odd if ψ(c) = 1 for the complex conjugation c, and even otherwise (so ψ(c) = 1). (2) Given ψ, L(s, ψ) is the complex analytic function continuing n=1 ψ (n) n s, where ψ is the primitive character associated with ψ. (In particular, L(s, 1) = ζ(s) and not (1 p s )ζ(s).) (3) Let K + = Q(ζ p n) + := Q(ζ p n + ζ 1 p n ), so [K : K+ ] = 2. Let h = #Cl K, h + = #Cl K + and h = h/h +. Remark. Nm : Cl K Cl K + is surjective, since K/K + is totally ramified at. This implies h Z. We collect the following standard facts. Theorem 3.2. (1) Dirichlet s class number formula: Let ζ L (s) be the Dedekind zeta function for a number field L. Then Res s=1 ζ L (s) = 2r 1 (2π) r 2 Reg L h L, #µ L L where the regulator is Reg L = det log σ i α j for a Z-basis {α j } of O L /torsion. In particular, we have Res s=1 ζ K (s) = (2π)N/2 Reg K h 2p n K Λ(s, ψ) := and Res s=1 ζ K +(s) = 2N/2 Reg K +h + 2, K + where N = [K : Q] = (p 1)p n 1. (2) Let ψ : Gal(K/Q) C have conductor m(ψ), and write { m(ψ) s/2 π s/2 Γ( s )L(s, ψ) 2 if ψ is even, m(ψ) s/2 π s/2 Γ( s+1 )L(s, ψ) if ψ is odd. 2 Then ψ( 1)m(ψ) Λ(s, ψ) = Λ(1 s, ψ), S(ψ) where S(ψ) is the Gauss sum. (3) S(ψ) = i N/2 K and S(ψ) = K +. ψ 1 (4) O K = µ p no K +. ψ 1 ψ even 9

10 (5) Let B n,ψ be the generalized Bernoulli numbers, defined by m(ψ) 1 a=0 Then for every integer n 1, We have the following formula for h. Theorem 3.3. Proof. Note that Then h = 2p n Res s=1 ζ K (s) = ψ 1 ψ(a)t e at e m(ψ)t 1 = n=0 L(1 n, ψ) = B n,ψ n. B n,ψ T n n!. ( 1 (5) L(0, ψ) = 2p ) n B 1,ψ. L(1, ψ) and Res s=1 ζ K +(s) = ψ 1 ψ even h = h h + (1) = 2N/2 Reg K +/2 K + (2π) N/2 Reg K /2p n K (2) = Reg K + Reg K (4) = p n K L(1, ψ) L(1, ψ). L(0, ψ)γ( 1 )i m(ψ)1/2 π N/2 2 K + S(ψ)m(ψ) 1/2 π 1/2 Γ(1) 1 2 N/2 1 pn π N/2 K L(0, ψ) S(ψ) 1 π N/2 i N/2 K + (3) = 2p n Remark. The p-part of h is 1 L(0, ψ). 2 0 and v p (pb 1,ω 1) = 0. Thus the theorem gives #Cl Q(ζp)[p ] ψ = ψ ω #Cl K [p ] ψ (left as an exercise). Recall that Cl Q(ζp)[p ] ω = ψ ω 1 #Z p /(L(0, ψ)). Therefore, inverting each character on the right hand side, we have #Cl Q(ζp)[p ] ψ = #Z p /(L(0, ψ 1 )). ψ ω ψ ω In fact this is true without taking products, as we will see in Corollary 3.8. This strengthens Herbrand Ribet, because v p (B p m ) = v p (B 1,ω m). 10

11 3.2. p-adic L-functions. Fix γ Γ Λ a topological generator of Γ. Let χ cyc be the cyclotomic character. Let ζ be a p n -th root of 1 (not necessarily primitive). For k Z, let φ k,ζ : Λ Z p [ζ] send γ to χ k cyc(γ)ζ. Theorem 3.4. Let ψ : Gal(Q(ζ p )/Q) C be an odd character, and { χ cyc (γ)γ 1 if ψ = ω 1, h ψ = 1 if ψ ω 1. Then there exists g ψ Λ such that φ k,ζ (g ψ ) φ k,ζ (h ψ ) = L( k, ψψ ζω k ), where ψ ζ : Gal(Q(ζ p n)/q) C sends the image of γ to ζ. Definition 3.5. g ψ is called a p-adic L-function. The construction of g ψ uses Bernoulli measures, whose existence implies the Kummer congruences Main Conjecture. Definition 3.6. For a finitely generated torsion Λ-module M, write M q-iso i Λ/(p n i ) j Λ/(f j ) as in the classification theorem. Then the characteristic ideal of M is ( Char(M) := p ni ) f j Λ. Recall that last time we constructed a finitely generated torsion Λ-module i C := lim n 1 Cl Q(ζp n)[p ]. Let C ψ be the ψ-eigenspace of C, where ψ : Gal(Q(ζ p )/Q) C is an odd character. Remark. We have C ψ = lim n ( ClQ(ζp n)[p ] ψ). For a Λ-module M, let M ι be the same Λ-module except with g Γ acting as g 1. The goal of the first part of our seminar is to prove the Theorem 3.7 (Main Conjecture). Char(C ψ,ι ) = (g ψ 1). Corollary 3.8. Suppose ψ ω. Then To prove this we need the #Cl Q(ζp)[p ] ψ = #Z p /(L(0, ψ 1 )). Proposition 3.9. C (and hence C ι ) has no nonzero finite cardinality Λ-submodules. Proof. See Washington P j

12 Exercise. If M is a finitely generated torsion Λ-module with no nonzero finite cardinality Λ-submodules, and if M/(γ pn 1)M is finite for all n, then Proof of Corollary 3.8. #M/(γ pn 1)M = #Λ/(Char(M), γ pn 1). #Cl Q(ζp)[p ] ψ = #C ψ /(γ 1)C ψ = #C ψ,ι /(γ 1)C ψ,ι = #Λ/(γ 1, Char(C ψ,ι )) (by Exercise) = #Λ/(γ 1, g ψ 1) (by MC) = #Z p /(φ 0,1 (g ψ 1)) = #Z p /(L(0, ψ 1 )). 4. Lecture 4 (September 28, 2017): Pak-Hin Lee Λ-adic modular forms 5. Lecture 5 (October 5, 2017): Pak-Hin Lee Ordinary Λ-adic forms 6. Lecture 6 (October 12, 2017): Pak-Hin Lee Hida Theory and Λ-adic Galois Representations 7. Lecture 7 (October 19, 2017): Pak-Hin Lee Λ-adic Representations via Pseudo-representations These will be updated soon! 8. Lecture 8 (October 26, 2017): Samuel Mundy Proof of the Main Conjecture, I Let p be an odd prime, and ψ = ω m where m is odd and ω is the Teichmüller character. Decompose Gal(Q(ζ p )/Q) = Γ = Gal(Q(ζ p )/Q) γ. Let Λ = Z p [[Γ]], with the specialization map ν k,ζ sending γ Γ Λ to ζχ k cyc(γ). We identify Γ = 1 + pz p Z p via the cyclotomic character χ cyc. Recall there exists a unique g ψ Λ such that, writing { χ cyc (γ)γ 1 if ψ = ω 1, h ψ := 1 otherwise, we have where ψ ζ sends γ to ζ. ν k,ζ (g ψ ) ν k,ζ (h ψ ) = L( k, ψψ cycω k ) Remark. The g ψ here is Pak-Hin s g ψω, so g ω 1 corresponds to his g 1. We have to switch from even to odd. Recall that we constructed X = lim Cl Q(ζp n n)[p ], where the minus sign means the odd part of the eigenspace for complex conjugation. This is a finitely generated Λ-module. Let X ψ be the ψ-eigenspace, and X ψ,ι be X ψ on which γ acts as γ 1. 12

13 Theorem 8.1 (Main Conjecture). Char(X ψ,ι ) = (g ψ 1) Reduction to one inclusion. Let us show that it is enough to prove either or. Fix n 0. Then we have ν 0,ζ (g ω k) = ν 0,ζ (h ω k)l(0, ω k ψ ζ ) ζ µ p n 0 k p 2 ζ,k k odd = ν 0,ζ (h ω 1) L(0, ω k ψ ζ ) ζ ζ,k = (χ cyc (γ)ζ 1) L(0, ψ) Hence, v p ( ζ,k ν 0,ζ (g ω k) ) ζ = (χ pn cyc(γ) 1) L(0, ψ). ψ ψ Gal(Q(ζ p n+1 )/Q) ( = v p (χ pn cyc(γ) 1) + v p L(0, ψ) ψ ( ) = n v p #Cl Q(ζp n+1 )[p ] /p n = 1 + v p ( #X ι /(γ pn 1)X ι ) ) (by CNF) ( = v p #Λ/(Char(X ι ), γ pn 1) ) + o(1) ( )),ι = v p (#Λ/ k Char(Xωk ), γ pn 1 + o(1). On the other hand, if ϖ is a uniformizer in Z p [ζ p n], we have 1 v p ν 0,ζ (g ω k) = (p 1)p v n 1 ϖ ν 0,ζ (g ω k) ζ,k ζ,k 1 = (p 1)p v n 1 p #Λ Zp[ζp n]/(g ω k, γ ζ) ζ,k 1 = (p 1)p v n 1 p (#Λ ) Zp[ζpn]/ k,ζ (g ω k, γ ζ) + o(1) 1 = (p 1)p v ( n 1 p #ΛZp[ζ p n]/ ( k g ω k, γpn 1 )) + o(1) = v p ( #Λ/ (k g ω k, γpn 1 )) + o(1), where the second equality is sort of like v ϖ (#Z p /p) = v p (#Z p [ζ p n]/p). Therefore, ( )) ( (,ι v p (#Λ/ k Char(Xωk ), γ pn 1 = v p #Λ/ k g ω k, 1 )) + o(1). γpn 13

14 By hypothesis,ι k Char(Xωk ) either contains or is contained in k g ω k, so we must have equality because Λ/(these ideals, γ pn 1) have the same order up to some bounded power of p. Thus we get Char(X ωk,ι ) = (g ω k) Proof of Main Conjecture. We will prove that Char(X ψ,ι ) (g ψ 1). proj Let Ψ : G Q Γ χcyc Z p, and define E ψ = g ψ + h ψ n=1 d n l e d (d,p)=1 l primes Remark. My ν k,ζ (E ψ ) is Pak-Hin s ν k+1,ζ (E ψω ), so ψψ(frob e l) qn. v k,ζ (E ψ ) M ord k+1(p N, ψωψ ζ ω k 1 ) = M ord k+1(p N, ψψ ζ ω k ). This is okay if we built the theory oddly, and we will assume that we did. Thus E ψ M ord (1, ψ; Λ). Step 1 (Construction of a Λ-adic cusp form): Exercise. If p = 3, 5, 7, then g ψ Λ for all. Moreover, g ω 1 Λ for all p > 2. If p 11 and k = 3, 5, then ν 0,1 (g ω k) = L(0, ω k ) = B 1,ω k B k+1 (mod p). Note B 4 = 1 and B 30 6 = 1 which are units in Z 42 p, so g ω k Λ for k = 3 or 5. Now let p > 2. Let a, b Z >0 be such that m + 1 4a + 6b (mod p 1). Then let F := E ψ g ψ g a ω 3 g b ω 5 E a ω 3Eb ω 5 Sord (1, ψ; Λ). Step 2 (g ψ and the Eisenstein ideal): Let T = T ord (1, ψ; Λ). Define the Eisenstein ideal by I := T l 1 ψψ(frob l ) for l p, U p 1 T. The structure map Λ T/I is surjective, so denoting the kernel by J we have Λ/J = T/I. Claim. J (g ψ ). Proof. If ψ = ω 1, then (g ψ ) = Λ so there is nothing to prove. Otherwise, if ψ ω 1, we will show that there exists a surjective ϕ I : T/I Λ/(g ψ ) such that the diagram Λ structure commutes, from which the claim follows. Consider ϕ I defined by proj T/I Λ/(g ψ ) ϕ I ϕ I : T l (mod I) a l (F) (mod g ψ ). This is an algebra homomorphism and fits in the diagram above, so T/I Λ/(g ψ ) is surjective. 14

15 Step 3 (Another reduction): The following reduction will be carried out next time. Proposition 8.2. To prove the Main Conjecture, it suffices to find a faithful T-module M 2 and a surjective Λ-module map X ψ 1,ι M 2 /IM 2. Proof. Let R be a Noetherian ring, and M a finitely generated R-module with a finite presentation R m L R n M 0. The Fitting ideal of M, denoted Fitt R (M), is defined to be the ideal in R generated by the n n minors of the matrix representing L. We have the following facts: (0) This is well-defined. (1) If M = R/I, then Fitt R (M) = I. (2) Fitt R (M N) = Fitt R (M) Fitt R (N). (3) If 0 M M M 0 is an exact sequence, then Fitt R (M ) Fitt R (M ) Fitt R (M). Now choose a quasi-isomorphism 0 M Λ/(f i ) Λ/(p n j ) i j X ψ 1,ι N 0, where M and N are of finite cardinality. Then M = 0 because i Λ/(f i) j Λ/(pn j ) does not have finite submodules. By Facts (1), (2), (3), ( ) Fitt Λ Λ/(f i ) Λ/(p n j ) = f i p nj Fitt Λ (N) Fitt Λ (X ψ 1,ι ), i j i.e., Char(X ψ 1,ι ) Fitt Λ (N) Fitt Λ (X ψ 1,ι ). Since #N <, any Λ-module map Λ N has kernel with finite index, so choosing 0 n N (if it exists) we get an exact sequence 0 Λ/I N N 0 where I has finite index in Λ, and #N < #N. By induction and Fact (3), we deduce that Fitt Λ (N) has finite index in Λ. (4) If M N is a surjective R-module map, then Fitt R (M) Fitt R (N). Thus by assumption. (5) Fitt R/I (M/IM) = Fitt R (M) (mod I). Thus Fitt Λ (X ψ 1,ι ) Fitt Λ (M 2 /IM 2 ) Fitt Λ (M 2 /IM 2 ) (mod J) = Fitt Λ/J (M 2 /IM 2 ) (6) If M is a faithful R-module, then Fitt R M = = Fitt T/I (M 2 /IM 2 ) = Fitt T (M 2 ) (mod I).

16 So Fitt T (M 2 ) = 0 and hence Fitt Λ (M 2 /IM 2 ) J. Since J (g ψ ), we get Char(X ψ 1,ι ) Fitt Λ N (g ψ ). Let f ψ = f i p n j, so Char(X ψ 1,ι ) = (f ψ ). Then find two coprime elements α, β Fitt Λ N and we get g ψ f ψ α and g ψ f ψ β. Since Λ is a UFD, this implies g ψ f ψ and therefore Char(X ψ 1,ι ) (g ψ ). 16