Introduction to Iwasawa Theory

Transcription

1 Introduction to Iwasawa Theory Yi Ouyang Department of Mathematical Sciences Tsinghua University Beijing, China

2 Contents 1 Modules up to pseudo-isomorphism 2 2 Iwasawa modules 8 3 Z p -extensions 15 4 Iwasawa theory of elliptic curves 23 1

3 Chapter 1 Modules up to pseudo-isomorphism Let A be a commutative noetherian integrally closed domain. Let K be the quotient field of A. Let P (A) = { ht = 1} be the set of prime ideals of A of height 1. Then for every P (A), A is a discrete valuation ring. Theorem 1.1. A = A. P (A) Proof. See [7, Theorem 11.5, page 81]. normal noetherian domains. This is a well known theorem about For an A-module M, we let Thus there is a pairing M + := Hom A (M, A). M + M A, (α, m) α(m) which induces a homomorphism of A-modules ϕ M : M M ++. Definition 1.2. An A-module M is called reflexive if the canonical map is an isomorphism. ϕ M : M M ++ = Hom A (Hom A (M, A), A) m (ϕ M (m) : α α(m)) Remark. M + is always torsion free, thus M is reflexive implies that M is torsion free. 2

4 and Assume M is a finitely generated torsion free A-module, then M M M A K = M A K := V M + (M + ) (M + ) A K = M + A K = V where V = Hom K (V, K) is the dual of V. One has M + = {λ V λ(m) A for all m M} (M + ) = {λ V λ(m) A for all m M } = (M ) +. Lemma 1.3. Let M be a finitely generated torsion free A-module, then (1) M + = M +. P (A) (2) M ++ = M. P (A) (3) M is reflexive if and only if M = M. Proof. (1) is trivial. If λ P (A) P (A) M +, then for all m M, λ(m) A for P (A), hence λ(m) A and λ M +. (2) since M = M ++ for ht = 1 (A is a discrete valuation ring). (3) follows from (2). Corollary 1.4. If M is a finitely generated A-module, then M + is reflexive. Definition 1.5. A finitely generated A-module M is called pseudo-null if the following equivalent conditions are fulfilled: (1) M = 0 for all prime ideals in A of height ht( ) 1; (2) If is a prime ideal with ann A (M), then ht( ) 2. Recall that ann A (M) = {a A am = 0}. Remark. (1) For the equivalence of the two conditions: M = 0 if and only if there exists s A\, such that sm = 0, which is equivalent to ann A (M). (2) A pseudo-null module is torsion since M (0) = M A K = 0. (3) If A is a Dedekind domain, then M is pseudo-null if and only if M = 0. (4) If A is a 2-dimensional noetherian integrally closed local ring with finite residue field, then M is pseudo-null if and only if M is finite. Indeed, let m be the maximal ideal of A, if M is finite, there exists r N such that m r M = 0, thus Supp(M) {m}. On the other hand, if Supp(M) {m},then there exists r N such that m r M = 0, thus m r ann A (M), therefore M is a finitely generated A/m r -module, hence finite. Definition 1.6. Let f : M N be a homomorphism of finitely generated A- modules M and N. f is called a pseudo-isomorphism if both ker f and coker f are pseudo-null, equivalently, if the induced homomorphisms f : M N are isomorphisms for all P (A). We write it as f : M N or f : M N. 3

5 Lemma 1.7. Let M be a finitely generated A-torsion module, if 0 α A such that Supp(A/αA) is disjoint to Supp(M) P (A), then α : M M, m αm is a pseudo-isomorphism. Proof. This is clear since α is a unit for every Supp(M) P (A). From now on, we set T A (M) : the torsion submodule of M; F A (M) = M/T A (M) : the maximal torsion free quotient of M. Proposition 1.8. Let M be a finitely generated A-module. Then (1) There exists a pseudo-isomorphism f : M T A (M) F A (M). (2) There exists { i } i I P (A), n i N, and a pseudo-isomorphism g : T A (M) i I A/ ni i where { i, n i } are uniquely determined by T A (M) up to re-numbering. Proof. (1) Let { 1,, h } = Supp(M) P (A). pseudo-null, the homomorphism If h = 0, then T A (M) is is a pseudo-isomorphism. If h > 0, let f : M (0,can) T A (M) F A (M) S = h h A\ i = A\ i. i=1 then S 1 A is a semi-local Dedekind domain with maximal ideals S 1 i, i = 1,, h. Thus S 1 A is a principal ideal domain by the approximation theorem, and S 1 T A (M) is a direct summand of S 1 M and is the torsion module of S 1 M. Since M is finitely generated, then i=1 Hom S 1 A(S 1 M, S 1 T A (M)) = S 1 Hom A (M, T A (M)). Thus there exists f 0 : M T A (M) and s 0 S such that f 0 s 0 : S 1 M S 1 T A (M) 4

6 is the projection of S 1 M onto S 1 T A (M), hence f 0 s 0 S 1 T A (M) = Id S 1 T A (M). Thus there exists s 1 S, f 1 = s 1 f 0 such that f 1 TA (M) = s 1 s 0 Id TA (M). Let by the commutative diagram f = (f 1, can) : M T A (M) F A (M) 0 T A (M) M F A (M) 0 f1 T A (M) f 0 T A (M) T A (M) F A (M) F A (M) 0 By Lemma 1.7, f 1 TA (M) is a pseudo-isomorphism, the snake lemma implies that f is also a pseudo-isomorphism. (2) By the structure theorem of finitely generated modules over a principal ideal domain, there exists an isomorphism h g 0 : S 1 T A (M) = r i S 1 E = S 1 for some uniquely determined Using E = h r i A/ nij i. i=1 j=1 i=1 j=1 A/ nij i Hom S 1 A(S 1 T A (M), S 1 E) = S 1 Hom A (T A (M), E), again we obtain g : T A (M) E and s S, such that g = sg 0. Again using the previous lemma, g is a pseudo-isomorphism. Remark. (1) If M, N are torsion modules, then f : M N implies that there exists g : N M. In general, this is not true. For example, let A = Z p [[T ]] = Λ, N = Λ and M = ker (N Z/pZ). Then M N but N M. (2) If the exact sequence of finitely generated A-torsion modules 0 M M M 0 satisfies that the associated sets of prime ideals of height 1 of M and M are disjoint. Then M M M. Proposition 1.9. Let M be a finitely generated torsion free A-module. Then there exists an injective pseudo-isomorphism of M onto a reflexive A-module M. 5

7 Proof. Consider the homomorphism ϕ M : M M ++. One notes that: (1) M = M ++ for ht 1. In particular, ker ϕ M A K = 0, hence ker ϕ M is torsion. As M is torsion free, ker ϕ M = 0; (2) M ++ is reflexive. Proposition Let A be an n-dimensional regular local ring, 2 n <. Let {p 1,, p n } be a regular system of parameters generating the maximal ideal of A. Let p 0 := 0. Then for a finitely generated A-module M, the following assertions are equivalent: (1) For every i = 0,, n 2, the A/(p 0,, p i )-module M/(p 0,, p i )M is reflexive. (2) M is a free A-module. In particular, a reflexive A-module M over a 2-dimensional regular local ring A is free. Proof. We only need to show (1) (2). From (1), M is reflexive, hence torsion free. Let ϕ : A r M be a minimal free presentation of M. Consider the diagram 0 A r p 1 A r (A/p 1 ) r 0 ϕ ϕ ϕ 0 M p1 M M/p 1 M 0 Assume M/p 1 is a free A/p 1 -module, then by Nakayama Lemma and the minimality of r, ϕ is an isomorphism. Hence p 1 : ker ϕ ker ϕ is an isomorphism. By Nakayama again, ker ϕ = 0. We get M is a free A-module. Thus we only need to show M/p 1 is a free A/p 1 -module Note that (i) A/p 1 is a regular local ring of dimension n 1; (ii) Let p i = p i + p 1 A, then { p 2,, p n } is a regular system of parameter of A/p 1. Thus (1) holds for (A/p 1, M/p 1 ). By induction, we only need to check (1) (2) for n = 2. In this case, A/p 1 is regular of dimension 1, hence a discrete valuation ring and an integral domain. Thus Hom A (M +, A/p 1 ) is torsion free, therefore M/p 1 = M ++ /p 1 = Hom A (M +, A) A/p 1 Hom A (M +, A/p 1 ) is also torsion free over d.v.r. A/p 1, which must be free. Theorem 1.11 (Structure Theorem). Let A be a 2-dimensional regular local ring and M be a finitely generated A-module. Then there exists finitely many primes i of height 1, natural numbers n i for each i, nonnegative integer r and a pseudo-isomorphism f : M A r (A/ ni i ), i I 6

8 i, n i and r are uniquely determined by r = dim K M A K, { i i I} = Supp M P (A). 7

9 Chapter 2 Iwasawa modules In this chapter, we let K be a finite extension of Q p and let O be the ring of integers of K, let π be a uniformizing parameter of O. Let k = O/(π) be the residue field of O, which is a finite extension of F p. As a convention, we write Λ = Z p [[T ]]. Lemma 2.1 (Division Lemma). Suppose f = a 0 + a 1 T + O[[T ]] but π f(i.e. f / πo[[t ]]). Let n = min{i a i / (π)}. Then any g O[[T ]] may be uniquely written as g = qf + r where q O[[T ]] and r O[T ] is a polynomial of degree at most n 1. Proof. First we show the uniqueness. If qf + r = 0, we need to show that q = r = 0. If not, we may assume that π q or π r. But 0 = qf + r mod π implies that π r and therefore π qf. Since π f, we have π q, contradiction! For the existence, we have two proofs. First proof: We let τ n = τ be the O-linear map b i T i b i T i n. i=0 Note that (i) τ(t n h) = h for h O[[T ]]. (ii) τ(h) = 0 if and only if h is a polynomial of degree n 1. Write f = πp (T ) + T n U(T ), where P (T ) is a polynomial of degree at most n 1 and U(T ) is a unit in O[[T ]]. For any g O[[T ]], let q(t ) = 1 U i=n ( ( 1) j π j τ P ) j τ(g) O[[T ]]. U j=0 Then τ(qf) = τ(πqp ) + τ(t n qu) = πτ(qp ) + qu 8

10 and πτ(qp ) =τ πp U = ( ( 1) j π j τ P ) j τ(g) U j=0 ( ( 1) j 1 π j τ P ) j τ(g) U j=1 =τ(g) qu. Thus τ(qf) = τ(g). Second proof: Note that k[[t ]] is a discrete valuation ring, it has a simple division algorithm. We let ḡ(t ) be the reduction of g(t ) modulo π. Since f(t ) = T n (unit) in k[[t ]], we have ḡ(t ) = q(t ) f(t ) + r(t ) for suitable q(t ) k[[t ]] and r(t ) k[t ] of degree n 1. Let q 1 (T ) O[[T ]], r 1 (T ) O[T ] (of the same degree of r(t )) be liftings of q(t ) and r(t ) respectively. Then g(t ) = f(t )q 1 (T ) + r 1 (T ) + πg 1 (T ) for some g 1 (T ) O[[T ]]. Apply the same procedure for g 1, we get g(t ) = f(t )q 1 (T ) + r 1 (T ) + π(f(t )q 2(T ) + r 2(T ) + πg 2 (T )) = f(t )q 2 (T ) + r 2 (T ) + π 2 g 2 (T ) where q 2 = q 1 mod π, r 2 = r 1 mod π. Repeat the procedure, we get g(t ) = f(t )q n (T )+r n (T )+π n g n (T ), q n+1 = q n mod π n, r n+1 = r n mod π n. Taking the limits, we get the desired result. Corollary 2.2. If π f O[[T ]], then O[[T ]]/(f) is a free O-module of rank n = inf{i a i / (π)} with basis {T i : i < n}. Definition 2.3. A distinguished polynomial (or Weierstrass polynomial) F (T ) O[T ] is a polynomial of the form F (T ) = T n + a n 1 T n a 0, a i (π). Corollary 2.4. Let F be a distinguished polynomial, then O[T ]/F O[T ] = O[[T ]]/F O[[T ]]. 9

11 Theorem 2.5 (Weierstrass Preparation Theorem). Let f O[[T ]]. Then f can be uniquely written as f = π µ P (T )U(T ) where µ = inf(ord π (a i )), P (T ) is a distinguished polynomial of degree n = inf{i ord π a i = µ}, U(T ) is a unit in O[[T ]]. As a consequence, O[[T ]] is a factorial domain. Proof. One may assume π f. Write f = a 0 + a 1 T + + a n T n + with π a n and π a i for i < n. By the division lemma, T n = q(t )f(t ) + r(t ) with deg r < n and q(t ) O[[T ]]. One has r(t ) = 0 mod π. Therefore f(t )q(t ) = T n r(t ) := P (T ) = T n mod π, we have q(t )a n = 1 mod π and q(t ) := 1 U(T ) (O[[T ]]). Thus in this case f(t ) = U(T )P (T ). The uniqueness follows from the division lemma, since T n = U(T ) 1 f(t ) + (P (T ) T n ). Remark. For π f, then O[[T ]]/(f(t )) = O[T ]/(P (T )). Thus P (T ) is the characteristic polynomial of the linear transformation T : O[[T ]]/(f) O[[T ]]/(f). Corollary 2.6. There are only finitely many x C p, x < 1 such that f(x) = 0. Proof. This is an easy exercise. Lemma 2.7. Let P be a distinguished polynomial. If O[T ], then g(t ) P (T ) O[T ]. g(t ) P (T ) O[[T ]], g(t ) Proof. Let g(t ) = P (T )f(t ), f O[[T ]]. For any root x C p of P (T ), 0 = P (x) = x n + multiple of π, one has x < 1, hence f(x) converges and g(x) = 0. Continue this process, we get P (T ) g(t ) as polynomials, hence f(t ) O[T ]. Let Γ = Z p = lim Z/p n Z. As a profinite group, Γ is compact and procyclic. Let γ be a topological generator of Γ, i.e., Γ = γ. Let Γ n = γ pn be n the unique closed subgroup of index p n of Γ, then Γ/Γ n is cyclic of order p n generated by γ + Γ n. One has an isomorphism ( ) O[Γ/Γ n ] O[T ]/ (1 + T ) pn 1 γ mod Γ n (1 + T ) mod (1 + T ) pn 1 Moreover, if m n 0, the natural map Γ/Γ m Γ/Γ n induces a natural map φ m,n : O[Γ/Γ m ] O[Γ/Γ n ], which is compatible with the isomorphism. We let ( ) O[[Γ]] = lim O[Γ/Γ n ] = lim O[T ]/ (1 + T ) pn 1. n n Note that O is a topological ring, compact and complete with π-adic topology, so are the rings O[Γ/Γ n ], thus O[[Γ]] is endowed with the product topology of π-adic topology, it is also compact and π-complete. The ring O[[Γ]] is called the Iwasawa algebra and its modules are called Iwasawa modules. 10

12 Theorem 2.8. One has a topological isomorphism O[[T ]] O[[Γ]], T γ 1 where O[[T ]] is a compact topological ring complete with (π, T )-topology. Proof. Write ω n (T ) = (1+T ) pn 1. ω n is a distinguished polynomial. Moreover, ω n+1 (T ) ω n (T ) = (1 + T ) pn (p 1) + + (1 + T ) pn + 1 (p, T ) (π, T ), thus ω n (T ) (p, T ) n+1 for n 0. By Corollary 2.4, for every n N, we have a projection O[[T ]] O[[T ]]/(ω n ) O[T ]/(ω n ) O[Γ/Γ n ] which is compatible with the transition map. By the universal property of projective limits, then we have a continuous homomorphism ɛ : O[[T ]] O[[Γ]], T γ 1. On one hand ker ɛ n (ω n) n (p, T )n+1 = 0, thus ɛ is injective. On the other hand, O[[T ]] is compact, hence the image is closed, it is also dense since at every level the map is surjective, hence ɛ is also surjective. From now on let O = Z p and Λ = Z p [[T ]]. Let m = (p, T ) be the maximal ideal of Λ. We identify Z p [[Γ]] and Λ by the above Theorem, though we should keep in mind that this isomorphism depends on the choice of the topological generator γ of Γ. Write ω n (T ) = (1 + T ) pn 1 and ν n,e (T ) = ω n (T )/ω e (T ). Lemma 2.9. If f and g are relatively prime to each other, then Λ/(f, g) <. Proof. Let h (f, g) be of minimal degree. we show that h = p s (up to Z p). If not, h = p s H for deg H 1. By the division algorithm, f = Hq + r, thus p s r (f, g), contradiction! Proposition The prime ideals of Λ are (0), m = (p, T ), (p), (P ) where P are irreducible distinguished polynomials in Λ. Proof. First all in the list are prime ideals. Let be a prime ideal of Λ and h be of minimal degree. Then h = p s H with H = 1 or distinguished (up to Z p). If H 1, then it must be irreducible by minimality. Then (f) where f = p or an irreducible distinguished polynomial. If (f) =, we are done. If not, there exists g such that f, g are relatively prime. By the above lemma, then p N for N 0, which implies p since is prime; also there exists a pair i < j, such that T i T j, as 1 T j i is a unit, T i, hence T. Thus (p, T ). 11

13 Theorem 2.11 (Structure Theorem for Iwasawa modules). For any finitely generated Λ-module M, M Λ r s Λ/p mi i=1 t j=1 Λ/F nj j where r = rank M, m i (i = 1, s), F j and n j (j = 1,, t) are uniquely determined by M. Definition F M = t j=1 F nj j is called the characteristic polynomial of M. If M is a torsion module, we define the Iwasawa invariants of M by λ(m) = s i=1 m i, µ(m) = j n j deg F j = deg F M. Remark. The isomorphism of Z p [[Γ]] and Z p [[T ]] depends on the choice of γ. Therefore if a finitely generated Iwasawa module M considered as a Λ-module, F j and F M depend on the choice of γ, but λ(m) and µ(m) are independent invariants. Lemma 2.13 (Nakayama s Lemma). Let M be a compact Λ-module. Then the following are equivalent: (1) M is finitely generated over Λ; (2) M/T M is a finitely generated Z p -module; (3) M/(p, T )M is a finitely dimensional F p -vector space. Proof. (1) (2) (3) are easy. Assuming (3), let x 1,, x n generate M/(p, T )M as F p -vector space. Let N = Λx Λx n M, then M N = N + (p, T )M N = (p, T ) M N. Thus M/N = (p, T ) n M/N for all n > 0. Consider a small neighborhood U of 0 in M/N. Since (p, T ) n 0 in Λ, for any z M/N, there exists a neighborhood U z of z and some n z such that (p, T ) nz U z U. But M/N is compact, then (p, T ) n M/N U for n 0, hence M/N = (p, T ) n M/N = 0 and M = N is finitely generated over Λ. Theorem Let X be a compact Λ-module. Then (1) X = 0 X/T X = 0 X/mX = 0. (2) X is a finitely generated Λ-module X/T X is a finitely generated Z p -module X/mX is a finite dimensional F p -vector space. Moreover, for a finitely generated Λ-module X, the minimal number of generators of X is dim Fp (X/mX). (3) If X/T X is finite, then X is a torsion Λ-module. 12

14 Proof. (1) and (2) are Nakayama s Lemma. For (3), by (2), X is a finitely generated Λ-module. Let x 1,, x d be a set of generators. Suppose X/T X has exponent p k, then p k x i T X for 1 i d. Write p k x i = d T a ij (T )x j, j=1 and let A = (p k δ ij T a ij (T )) i, j and g(t ) = det A. Then g(t )x i = 0 for all i = 1,, d but g(0) = p dk 0, hence X is torsion. Remark. One can replace T by any distinguished polynomial in the above Theorem. Lemma Let g be a distinguished polynomial of degree d prime to ω n /ω e for every n > e. Then for n 0, Λ/(g, ω n ) = p dn+o(1). Proof. We know Λ/(g, ω n ) is finite for n 0 by Lemma 2.9. Write V = Λ/(g(T )). Then T d = pq(t ) mod g and by induction, for k d, T k = p poly. mod g. Then for p n d, and (1 + T ) pn = 1 + p poly. mod g (1 + T ) pn+1 = (1 + p poly.) p = 1 + p 2 poly. mod g, ω n+2 (T ) ω n+1 (T ) n+2 + T )p 1 =(1 (1 + T ) pn+1 1 =(1 + T ) (p 1)pn (1 + T ) pn =p(1 + p poly.) mod g. Thus ωn+2(t ) ω n+1(t ) acts as p unit on V for pn d. For n 0 > e, p n d and n n 0, then ω n+2 V = ωn+2(t ) ω n+1(t ) (ω n+1v ) = pω n+1 V, and V/ω n+2 V = V/pV pv/pω n+1 V = V/pV V/ω n+1 V This finishes the proof. =p d(n n0+1) V/ω n0+1v = p nd+c. Lemma For a Λ-module M, let M Γ = M/T M and M Γ = M γ=1. If there is exact sequence then there is a long exact sequence 0 M M M 0, 0 M Γ M Γ M Γ M Γ M Γ M Γ 0. 13

15 Proof. Apply the snake lemma to the commutative diagram with exact rows. 0 M M M 0 γ 1 γ 1 γ 1 0 M M M 0 Remark. Replace γ by γ pn, we can have similar results. Proposition Let M be a finitely generated torsion Λ-module such that M/ω n M is finite for all n 0. Then for n 0, M/ω n M = p µ(m)pn +λ(m)+o(1) where λ(m) and µ(m) are Iwasawa invariants. Proof. By the above lemma, we can replace M by a torsion Λ-module of the form s i=1 ( Λ/p k i ) t j=1 (Λ/f j(t ) mj ). Now just apply Lemma

16 Chapter 3 Z p -extensions Definition 3.1. A Z p -extension is a Galois extension K /K whose Galois group is isomorphic to the ring of p-adic integers Z p. Proposition 3.2. There are exactly one sub-extension K n of K inside K with Galois group Gal(K n /K) = Z/p n Z cyclic of order p n. Proof. This follows easily from the fundamental theorem of Galois Theory, as the only closed subgroups of Z p are 0 and p n Z p for n N. Proposition 3.3. Let K be a number field, then Z p -extensions over K are unramified outside p. Proof. Let v be a prime of K not lying above p. We need to show the inertia subgroup I of v is 0. if not, I = p n Z p for some n N. By local class field theory, U Kv I = p n Z p is surjective, but U Kv = finite groups Z a l for a N and l p, this is impossible. Lemma 3.4. Let K be a number field. Then there exists at least one prime ramified in K /K, and there exists n 0 such that every prime which ramifies in K /K is totally ramified in K /K n. Proof. This is an easy exercise. Suppose K is a number field. Let E = O K be the group of global units. Let E 1 = {x E x 1 mod for p}. Let U 1, be the group of local units congruent to 1 mod. Then we have a diagonal map ψ : E U = p U, ɛ (ɛ,, ɛ) such that ψ(e 1 ) U 1 = p U 1,. 15

17 Lemma 3.5. (1) ψ(e 1 ) = U 1 K v p U v. (2) ψ(e) = U K v p U v. Proof. (1). is easy. For, we write U n = v p U n,v, where U n,v is the group of local units congruent to 1 mod v n, then K U v = (K U v U n ), v p n v p ψ(e 1 ) = n ψ(e 1 )U n. It suffices to show that U 1 K v p U vu n ψ(e 1 )U n. For any element xu u n U 1 K v p U vu n where x K, u v p U v and u n U n, we have x E 1 and for v p, (xu ) v = 1. Then xu u n = ψ(x)u n ψ(e 1 )U n. The proof of (2) is similar to (1). Conjecture 3.6 (Leopoldt Conjecture). rank Z E 1 = rank Zp E 1 Z Z p. Leopodlt Conjecture is true for abelian number fields. Let δ = rank Z E 1 rank Zp E 1 Z Z p. Then δ 0 and δ = 0 if Leopoldt Conjecture holds. Example 3.7. Note that 7, 13 are independent over Z, but log 3 13/ log 3 7 Z 3, thus 7, 13 Z3 = 7 Z3. Theorem 3.8. Let K be the composite of all Z p -extensions of K inside K ab. Then Gal( K/K) = Z r2+1+δ p where r 2 is the number of complex embeddings of K and δ is the Leopoldt defection. Proof. Since K/K is unramified outside p, we first consider the maximal abelian extension F of K unramified outside p. Let H be the maximal unramified abelian extension of K inside F, i.e. the Hilbert class field of K. Write J k the group of ideles of K and I K the ideal class group of K. By Class field theory, then Gal(F/K) = J K /K U v, v p Gal(H/K) = I K = J K /K v U v. Write V = K v p U v. We have Gal(F/H) = K v U v /V = UV/V = U/(U V ). 16

18 Note that U = U 1 (finite group), then U/U V and U 1 /(U 1 V ) differ by a finite group. Note that U 1 = (finite group) Z [K:Q] p, then by Lemma 3.5 U 1 /U 1 V = U 1 /ψ(e 1 ) = finite Z r2+1+δ p. Thus K v U v /K v p U v = finite Z r 2+1+δ p and hence Gal(F/K) Z r2+1+δ p = finite. Suppose the quotient is of order N. Write J = Gal(F/K) = J k /V. Then NZ r2+1+δ p NJ Z r2+1+δ p, thus NJ = Z r2+1+δ p as Z p -modules. Let J N = {x J Nx = 0}, then J /J N = NJ. J N is a finite group with order N: otherwise, there exist distinct elements x, x J N with the same image at J /Zp r2+1+δ, then x x Z r2+1+δ p and N(x x ) = 0, contradiction! By definition, the fixed field of J N must be K and we get the Theorem. Theorem 3.9 (Iwasawa). Let K = K 0 K n K be a tower of Z p - extensions. Let p en be the exact p-power dividing h(k n ), the order of ideal class group of K n. Then there exist integers λ 0, µ 0 and ν such that for n sufficiently large. e n = λn + µp n + ν Let Gal(K /K) = Γ. We fix a topological generator γ 0 of Γ. For every n N, let L n be the maximal unramified abelian p-extension of K n. By the maximality, L n /K is Galois. Let L = L n. Then L/K is also Galois. Write X n = Gal(L n /K n ), X = Gal(L/K ) and G = Gal(L/K). We n 0 17

19 have the following diagram: Γ K K n Γ n X Xn Γ/Γ n K L 0 L n G L For n 0, then all primes which are ramified in K /K are totally ramified in K /K n. Then for n 0, K n+1 L n = K n and X n = Gal(L n /K n ) = Gal(L n K n+1 /K n+1 ), thus a quotient of X n+1. Moreover X n = Gal(Ln K /K ) and lim X n = Gal( L n K /K ) = Gal(L/K ) = X. Since X n is an abelian p-group, there is an Z p -action on X n, since Gal(L n /K) is Galois, X n is also equipped with an Γ/Γ n -action: let γ Γ/Γ n, let γ be any lifting of γ in Gal(L n /K), then for x X n, x γ = γx γ 1 is independent of the choices of the lifting. Then X n is a Z p [Γ n ]-module. Passing to the limit, X = lim X n is a lim Z p [Γ n ] = Z p [[Γ]] = Λ-module. We make the following assumption at first: All primes ramified in K /K are totally ramified. Let 1,, s be primes of K which ramify in K /K. Fix i of L lying above i, let I i G be the inertia group. Since L/K is unramified, I i X = 1. Since K /K is totally ramified at i, I i = G/X = Γ, thus G = I i X = XI i, i = 1,, s. We identify I 1 with Γ. Let σ i be a topological generator of I i, then σ i = a i σ 1 for some a i X. Lemma With the assumption. Then G = [G, G] = X γ0 1 = T X. 18

20 Proof. Let a = αx, b = βy for α, β Γ and x, y X. Then aba 1 b 1 =αxβyx 1 α 1 y 1 β 1 = x α αβyx 1 β 1 y 1 β 1 =x α αβyx 1 β 1 α 1 βy 1 β 1 = x α (yx 1 ) αβ (y 1 ) β =x α(1 β) y β(α 1). Let β = 1 and α = γ 0, then y γ0 ) G, hence X γ0 1 G. On the other hand, write β = γ c 0 for c Z p, then x α(1 β) = x α(1 γc 0 ) X γ0 1 since 1 γ c 0 = 1 (1 + T ) c = 1 ( c n) T n T Λ. Similarly y β(α 1) X γ0 1, hence G X γ0 1. Lemma With the assumption. Let Y 0 = T X, a 2,, a s. Let ν n = ω n /ω 0 = (1+T )pn 1 T and let Y n = ν n Y 0. Then for n 0. X n = X/Yn Proof. For n = 0, L 0 /K is the maximal unramified abelian p-extension of K, thus the maximal abelian unramified extension inside the Galois extension L/K, by Galois theory, Gal(L/L 0 ) is the closed subgroup generated by I i for 1 i s and G, i.e., Gal(L/L 0 ) = I 1 Y 0 and X 0 = G/I 1 Y 0 = I 1 X/I 1 Y 0 = X/Y 0. For general n, just replace K by K n, γ 0 by γ pn 0 and Y 0 by Y n. Theorem X is a finitely generated torsion Λ-module. Proof. First with the assumption. To show that X is finitely generated is the same as to show that Y 0 is finitely generated. But Y/ν 1 Y is finite and ν 1 (p, T ), by Nakayama s Lemma, Y is a finitely generated Λ-module. Moreover Y and X are torsion by remark of Theorem In general, suppose all primes ramified in K /K are totally ramified in K /K e. Replace K by K e, then for n e, X n = X/ν n,e Y e where ν n,e = ω n /ω e and Y e is the corresponding Y 0 for the extension K /K e. Similarly we can show that Y e is finitely generated and hence X is finitely generated. Lemma Let M 1 M 2 be two finitely generated Λ-module with a given quasi-isomorphism. If M 1 /ν n,e M 1 < for all n e. Then there exist some constant c and some n 0 e, such that for n n 0. M 1 /ν n,e M 1 = p c M 2 /ν n,e M 2 19

21 Proof. Consider the diagram 0 ν n,e M 1 M 1 M 1 /ν n,e M 1 0 φ n φ φ n 0 ν n,e M 2 M 2 M 2 /ν n,e M 2 0 by the snake lemma, we have an exact sequence 0 ker φ n ker φ ker φ n coker φ n coker φ coker φ n 0. We have (1) ker φ n ker φ ; (2) coker φ n coker φ ; (3) coker φ n coker φ ; (4) ker φ n ker φ coker φ. Now for m n, we have (a) ker φ n ker φ m ; (b) coker φ n coker φ m ; (c) coker φ n coker φ m. (3) and (c) needs a little more explanation, others are easy. For (c), let ν m,e y ν m,e M 2, let z ν n,e M 2 be a representative of ν n,e y in coker φ n. Then ν n,e y z = φ(ν n,e x) for ν n,e x ν n,e M 1 and ν m,e y is represented by ν m,n z in coker φ m. The proof of (3) is similar. By (2) and (b), the sizes of coker φ n s are non-decreasing with an upper bound coker φ, when n 0, coker φ n will be stable. Similarly the sizes of ker φ n and coker φ n will be stable when n 0, hence also the size of ker φ n by the long exact sequence. Proof of Theorem 3.9. By Theorem 3.12, X E = s ( Λ/p k i ) t (Λ/f j (T ) mj ) i=1 where f j (T ) s are irreducible distinguished polynomials. By the above lemma 3.13, X n = X/ν n,e Y 0 is equal to E/ν n,e E up to a bounded factor. Note that j=1 Λ/(p ki, ν n,e = p ki(pn p e) = p kipn +c. We have to compute Λ/(ν n,e, f j (T ) mj ). Let g be a distinguished polynomial of degree d. Write V = Λ/(g(T )). Then T d = pq(t ) mod g and by induction, for k d, T k = p poly. mod g. Then for p n d, (1 + T ) pn = 1 + p poly. mod g and (1 + T ) pn+1 = (1 + p poly.) p = 1 + p 2 poly. mod g, 20

22 ω n+2 (T ) ω n+1 (T ) n+2 + T )p 1 =(1 (1 + T ) pn+1 1 =(1 + T ) (p 1)pn (1 + T ) pn =p(1 + p poly.) mod g. Thus ωn+2(t ) ω n+1(t ) acts as p unit on V for pn d. For n 0 > e, p n d and n n 0, then ν n+2,e V = ωn+2(t ) ω n+1(t ) (ν n+1,ev ) = pν n+1,e V, and V/ν n+2,e V = V/pV pv/pν n+1,e V = V/pV V/ν n+1,e V =p d(n n0+1) V/ν n0+1,ev = p nd+c. Plug the above result in the case g = F mj j, we have when n 0, E/ν n,e E = p µpn +λn+c with µ = µ(e) = k i and λ = λ(e) = m j deg F j. We just showed that the module X is a finitely generated Λ-module. There are more examples. We now consider the following special case: K = Q(ζ p ), K n = Q(ζ p n+1) and K = Q(ζ p ). We let = Gal(Q(ζ p )/Q) = (Z/pZ). Then Gal(K /Q) = Γ. Let E n be the group of global units of K n and C n be the subgroup generated by ζ p n+1 and ζ p n+1 1 as Gal(K n /Q)-module, which is called the group of cyclotomic units. We recall the map ψ maps E n into the finitely generated Z p [K n /Q]-module U Kn,. Let E n = ψ(e n ) and C n = ψ(c n ). Let p E = lim E n, n N C = lim C n n N with the transition maps given by the norm map. Then E and C are finitely generated Z p [[Gal(K /Q)]] = Λ[ ]-modules. For any character χ : Z p and a Λ[ ]-module M, let M χ = e χ M be the χ-part of M. Then E χ, C χ and (E/C) χ are finitely generated Λ-modules. Recall E n /C n are finite for all n N, and E n /C n = E/C ω, then E/C is Λ-torsion and so is n(e/c) (E/C)χ. Similarly X is a Λ[ ]-module and X χ is Λ-torsion. Then the Iwasawa Main Conjecture is the following theorem of Mazur-Wiles: Theorem 3.14 (Main Conjecture). If χ is even (i.e., χ( 1) = 1), then (Char X χ ) = (Char(E/C) χ ). The main conjecture has another equivalent form. By the proof of Theorem 3.8, we know for any number field K, the maximal abelian pro-p extension of K unramified outside p has Z p -rank r 2 (K) δ(k). In a Z p -extension 21

23 K /K, let M n (resp. M ) be the maximal abelian pro-p extension of K n (resp. K ) unramified outside p. Then K M n M. Let X n = Gal(M n /K ), X = Gal(M /K ). Then X is a finitely generated Λ-module since X n = X /ω n X is finitely generated as Z p -module. Back to the special case. Then δ(k) = 0 and X is of Λ-rank r 2 (K) + 1, and there is an action of on X. One can show that if χ is even, X χ is a torsion Λ-module. On the other hand, the p-adic L-function L p (s, χ) is given by L p (1 s, χ) = g((1 + T ) s 1) for some g(t ) Λ. Then Theorem 3.15 (Equivalent form of Main Conjecture). For χ even, (Char(X χ )) = (g(t )). 22

24 Chapter 4 Iwasawa theory of elliptic curves Let K be any number field. For an elliptic curve E defined over K, the theorem of Mordell-Weil claims that the set of K-rational points E(K) of E is a finitely generated abelian group, that is E(K) = Z r T for T the torsion group of E(K) and r the rank of E(K). The study of r(e(k)) is a major problem in the arithmetic of elliptic curve. For example, the famous Birch-Swinnerton-Dyer Conjecture claims that this rank equals the order of zeroes of L(E, s), the L-function of E, at s = 0, and gives a conjectural relation about the leading terms of L(E, s). Let F /F be a Z p -extension and F n be the n-th layer. Let E be an elliptic curve defined over F. One can ask how rank E(F n ) varies as n varies. We shall study this question in this chapter. First let us introduce the definitions of Selmer groups and Shafarevich groups. Let L be a field of characteristic 0 and E be an elliptic curve defined over L. Let L be an algebraic closure of L. Let G L = Gal(L/L). We write H i (L, ) for the cohomology group H i (G L, ). For the exact sequence taking the Galois cohomology, one has 0 E[n] E [n] E 0, (4.1) 0 E(L) κ H 1 (L, E[n]) H 1 (L, E)[n] 0, ne(l) where the Kummer map κ is defined as follows: For b E(L), choose a E(L) such that na = b, then κ(b) is the cohomological class associated to the cocycle κ(b)(σ) = a σ a, σ G L. 23

25 Let v be a place of L, then we get a local exact sequence analogue to (4.1). If we regard G Lv as a subgroup of G L, then the restriction maps from H 1 (L, ) to H 1 (L v, ) yield the following commutative diagram: 0 E(L) ne(l) 0 E(Lv) ne(l v) κ H 1 (L, E[n]) H 1 (L, E)[n] 0 κ η H 1 (L v, E[n]) H 1 (L v, E)[n] 0 The n-th Selmer group of E over L is the group Sel E (L)[n] = v ker (H 1 (L, E[n]) H 1 (L v, E(L v ))[n]). The Shafarevich-Tate group of E over L is the group X E (L) = v ker (H 1 (L, E(L)) H 1 (L η, E(L v )). Easily by diagram chasing, these two groups and the Mordell-Weil group are related by the following important fundamental exact sequence (4.2) 0 E(L)/nE(L) Sel E (L)[n] X E (L)[n] 0. For every pair (n, m) such that n m, we have the following commutative diagram 0 E(L) ne(l) H 1 (L, E[n]) H 1 (L, E)[n] 0 0 E(L) me(l) H1 (L, E[m]) H 1 (L, E)[m] 0 where the vertical maps are natural injections. Furthermore, the local analogue of the above diagram also holds and the restriction maps are compatible with the diagrams. Passing to the limit, we have 0 E(L) Q/Z κ H 1 (L, E(L) tors ) H 1 (L, E) 0 0 E(L v ) Q/Z The Selmer group of E over L is the group κ v H 1 (L v, E(L v ) tors ) H 1 (L v, E) 0 Sel E (L) = v ker (H 1 (L, E(L) tors ) H 1 (L v, E(L v ))). One has the exact sequence (4.3) 0 E(L) Q/Z Sel E (L) X E (L) 0. 24

26 Let p be a prime number, then the p-primary Selmer group is given by Sel E (L) p = v =ker ker (H 1 (L, E[p ]) H 1 (L v, E(L v ))[p ]) ( and one has an exact sequence Put H 1 (L, E[p ] v ) H 1 (L v, E[p ]) Im κ v 0 E(L) Q p /Z p Sel E (L) p X E (L) p 0. H E (L v ) = H1 (L v, E[p ]) Im κ v, Denote by P E (L) the product of H E (L v ) for all primes v of L. Then Put then one has an exact sequence Sel E (L) p = ker (H 1 (L, E[p ]) P E (L)). G E (L) = Im (H 1 (L, E[p ]) P E (L)), (4.4) 0 Sel E (L) p H 1 (L, E[p ]) G E (L) 0. Suppose furthermore that the extension L/F is a Galois extension. Write G = Gal(L/F ). For every intermediate field F of L/F, write G(L/F ) = Gal(L/F ). One has the following commutative diagram with exact rows 0 Sel E (F ) p H 1 (F, E[p ]) G E (F ) 0 0 s L/F Sel E (L) G(L/F ) p h L/F g L/F H 1 (L, E[φ ]) G(L/F ) G E (L) G(L/F ) where the vertical maps s L/F, h L/F and g L/F are natural restrictions. The snake lemma then gives the exact sequence: (4.5) 0 ker s L/F ker h L/F ker g L/F coker s L/F coker h L/F. Theorem 4.1 (Mazur s Control Theorem). If F /F is a Z p -extension, assuming that E has good ordinary reduction at all primes of F lying over p. Let F n be the n-th layer of the Z p extension. Then the natural maps s n = s F /F n : Sel E (F n ) p Sel E (F ) Γn p have finite kernels and cokernels, whose orders are bounded as n. We first give some consequences of Mazur s Control Theorem: 25

27 Corollary 4.2. Suppose E is an elliptic curve defined over F such that E has good, ordinary reduction at all primes lying above p. If E(F ) and X E (F ) are both finite, then Sel E (F ) p is Λ-cotorison. Consequently, rank Z E(F n ) is bounded as n varies. Proof. Let X = Hom(Sel E (F ) p, Q p /Z p ). Then X is an Λ-module. Moreover, X/T X = Hom(Sel E (F ) Γ p, Q p /Z p ) is finite since Sel E (F ) p is finite, thus X is a finitely generated Λ-torsion module, hence Sel E (F ) p is Λ-cotorison. Now X/X Zp-tors = Z λ p, thus (Sel E (F ) p ) div = (Qp /Z p ) λ and (Sel E (F n ) p ) div = (Q p /Z p ) tn for some t n λ. Since E(F n ) Q p /Z p (Sel E (F n ) p ) div through the Kummer map, we have rank E(F n ) λ. Corollary 4.3. Suppose E is an elliptic curve defined over F such that E has good, ordinary reduction at all primes lying above p. If E(F n ) and X E (F n ) are finite for all n, then there exist λ, µ 0, depending only on E and F /F, such that X E (F n ) p = p λn+µpn +O(1). Proof. From assumptions, Sel E (F n ) p are finite. Let X = Hom(Sel E (F ) p, Q p /Z p ). Then X/ω n X = Sel E (F ) Γn p < for all n, thus X is a finitely generated Λ- module. Apply Proposition 2.17, we get X/ω n X = p λ(x)n+µ(x)pn +O(1). The result then follows. Corollary 4.4. Suppose E is an elliptic curve defined over F such that E has good, ordinary reduction at all primes lying above p. Let r = corank Λ (Sel E (F ) p ) = rank Λ X, then corank Zp Sel E (F n ) p = rp n + O(1). Proof. Let X = Hom(Sel E (F ) p, Q p /Z p ). Then X is a finitely generated Λ- module, say pseudoisomorphic to Λ r Y Z for Y a free Z p -module of finite rank and Z a torsion group of bounded components. Since X/ω n X is the Pontragin dual of Sel E (F ) Γn p, and the size of latter one differs from Sel E (F n ) p by a finite bounded value, then corank Zp Sel E (F n ) p = rank Zp X/ω n X = rp n + O(1). Green- We won t give a complete proof of the control theorem here (cf. berg [3]). One has to use the exact sequence 0 ker s n ker h n ker g n coker s n coker h n, then to study ker s n and coker s n, it suffices to study ker h n, coker h n and ker g n. The first two are easy by the inflation-restrction exact sequence, but the third one needs more analysis. One needs to study the local restriction r v : H1 (F n,v, E[p ]) Im κ v H1 (L η, E[p ]) Im κ η, 26

28 for every place v. For v p, it is easy. For v p, it is more difficult. Here we only prove Theorem 4.6, which will be key to the study of the local maps. One can use Tate s duality theorem for local fields to prove Theorem 4.6, but we give a proof using methods of Iwasawa theory. We first have: Lemma 4.5. Let K be a finite extension over Q p and let F/K be a finite abelian extension with Galois group. Let χ : Z p be a character of. Let M F be a maximal abelian p-extension over F. Then M F /K is Galois and { rank Zp Gal(M F /F ) χ [K : Q p ] + 1, if χ = 1, = [K : Q p ], otherwise. Proof. M F /K is Galois since M F is maximal. By class field theory, the isomorphism lim n F /F pn Gal(M F /F ) is -equivariant. Recall that F = π F U F, the p-completion of π F is a copy of Z p, with a trivial action of, the p- completion of U F is isomorphic to O F = O K [ ] µ p (F ). Thus { rank Zp Gal(M F /F ) χ [K : Q p ] + 1, if χ = 1, = [K : Q p ], otherwise. Theorem 4.6. Let K v be a finite extension of Q p. Suppose that A is a G kv - module and that A = Q p /Z p as a group. Then H 1 (K v, A) is a cofinitely generated Z p -module of Z p -corank { 1, if A = µ p or A = Q p /Z p ; = [K v : Q p ] + 0, otherwise. Proof. G Kv acts on A = Q p /Z p through a character ψ : G Kv Aut(Q p /Z p ) = Z P : for any g G K v and a A, ga = ψ(g)a. If A = Q p /Z p (i.e. ψ = 1) or A = µ p (i.e. ψ is the cyclotomic character), the theorem is easy to check following from Lemma 4.5 and Kummer theory. We suppose now A is not Q p /Z p or µ p, there are two cases: (1) Im ψ is finite. Let H = ker ψ, then G = G Kv /H is finite. Let F = K v H be the field fixed by H, then Gal(F/K v ) = G is a finite abelian group. We consider the inflation-restriction sequence 0 H 1 (G, A) H 1 (K v, A) H 1 (H, A) G H 2 (G, A). 27

29 If p 2, then Z p is pro-cyclic, G is cyclic in this case and H 1 (G, A) = H 2 (G, A). Suppose G is generated by σ and ψ(σ) = a A, then H 1 (G, A) = N A/(a 1)A. Note that a 1 and A/(a 1)A is finite, so H 1 (G, A) and H 2 (G, A) are both finite. If p = 2, then G = Im ψ = Z/2 n Z Z/2Z or Z/2 n Z. In each case one can verify that H 1 (G, A) and H 2 (G, A) are finite. Now H acts trivially on A, then H 1 (H, A) = Hom(H, A) = Hom(Gal(K v /F ) ab, A) = Hom(Gal(M F /F ), A). Thus H 1 (H, A) G = Hom G (Gal(M F /F ), A) = Hom(Gal(M F /F ) χ, Q p /Z p ) where χ is the restriction of ψ at G. The theorem follows from Lemma 4.5. (2) Im ψ is infinite. Let F = K v ker ψ and G = Gal(F /K v ), Note that G = Im ψ Z p, one can write G = Γ, where is a subgroup of Z/(p 1)Z or Z/2Z if p = 2. Let F = F Γ. Again we need to consider the inflationrestriction sequence 0 H 1 (G, A) H 1 (K v, A) H 1 (F, A) G H 2 (G, A). First consider the spectral sequence H p (, H q (Γ, A)) H p+q (G, A). For n = p + q = 2, as Z p has cohomological dimension 1, H 2 (Γ, A) = 0. If prime p 2, the order of is prime to p, H 1 (Γ, A) and A γ are p-groups, hence H 1 (, H 1 (Γ, A)) = 0 and H 2 (, A Γ ) = 0, thus H 2 (G, A) = 0. If p = 2 and trivial, again H 2 (G, A) = 0; if = Z/2Z, one can get H 2 (G, A) = Z/2Z, but easy to see it is finite. For n = p + q = 1, for = 1, easily to see H 1 (G, A) = H 1 (Γ, A) is finite; for prime p 2 or = 1, we have H 1 (, H 0 (Γ, A)) = 0 and H 0 (, H 1 (Γ, A)) = 0; for p = 2 and = Z/2Z 1, both are again finite. Thus H 1 (G, A) is finite. So we have corank Zp H 1 (K v, A) = corank Zp H 1 (F, A) G. Let F n = F Γn. Fix an algebraic closure Q p of Q p. Let M n be the maximal abelian pro-p extension of F n and M be the maximal abelian pro-p extension of F. Let X = Gal(M /F ) and X n = Gal(M n /F n ). By Lemma 4.5, { Gal(M n /F n ) χ = [K v : Q p ]p n 1, χ = 1; + 0, χ 1. Hence Gal(M n /F ) χ = [K v : Q p ]p n. Write ψ and ψ Γ the restrictions of ψ on and Γ. Then corank Zp H 1 (K v, A) = corank Zp H 1 (F, A) G = corank Zp Hom(Gal(Q p /F ), A) G = corank Zp Hom G (X, A) = rank Zp X ψ = rank Zp (X ψ ) ψγ = rank Zp X ψ (γ 0 ψ(γ 0 )) = rank Z p X ψ 28 T b

30 where b = ψ(γ 0 ) 1 pz p. We need to study X, X ψ. Note for p 2, X ψ = e ψ X for e χ the idempotent element of χ. Note that M n is the maximal abelian sub-extension inside M /F n, thus By the exact sequence Gal(M /M n ) = Gal(M /F n ). 1 X Gal(M /F n ) Γ n 1 then any element in Gal(M /F n ) is of the form αx for α = γ pn m 0 and x X. Let αx, βy Gal(M /F n ), then αxβyx 1 α 1 y 1 β 1 = x α(1 β) y (α 1)β, we have Gal(M /F n ) = ω n X. Since X is compact, ω n X is closed and Gal(M /M n ) = ω n X and Gal(M n /F ) = X/ω n X. By Nakayama Lemma, X is a finitely generated Λ-module of rank [K v : Q p ]. Moreover, X/ω n X is -equivariant, Gal(M n /F ) χ = (X/ω n X) χ = X χ /ω n X χ, then X χ is a finitely generated Λ-module of rank [K v : Q p ]. By Class field theoy, since p [F : Q p ], G F has p-adic cohomological dimension 1, hence H 1 (F, Q p /Z p ) is a divisible group. Thus X = H 1 (F, Q p /Z p ) is torsion free as Z p -module. Thus X has no nonzero finite Λ-submodules. Let Y = X Λ tors and W = X/Y. Then W is torsion free and 0 Y ω n Y X ω n X W ω n W 0 is exact by snake lemma. W has Λ-rank [K v : Q p ] and hence W/ω n W has Z p -rank K v : Q p ] p n, the same as the Z p -rank of X/ω n X. Therefore Y/ω n Y is finite and must be isomorphic to a subgroup of (X/ω n X) Zp tors = µ p (F n ). On one hand, if µ p (F ) is finite, then Y = lim Y/ω n Y is finite and hence n Y = 0. On the other hand, if Y is infinite, then Y = lim Y/ω n Y is pro-cyclic n and therefore = Z p as a Z p -module. Suppose W Λ r is a quasi-isomorphism, then 0 W Λ r B 0 is exact and B is a finite Λ-module, by snake lemma again, (W/ω n W ) Zp tors is bounded by ker (ω n : B B), which equals B when n 0. Therefore if µ p (F n ) is unbounded, then Y/ω n Y is also unbounded and Y is infinite. Hence if µ p F, then Y = T p (µ p ). Now we can finish the proof of the Theorem. We have corank Zp H 1 (K v, A) = rank Zp X ψ /(T b) for T b a distinguished polynomial of degree 1. As X is quasi-isomorphic to Λ [Kv:Qp] if µ p F, or T p (µ p ) Λ [Kv:Qp]. In the latter case, µ p F and ψ gives the action of on µ p. As we assume ψ is not the cyclotomic character, T p (µ p ) ψ = 0 and X ψ /(T b) is of Z p -rank [K v : Q p ]. 29

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