Local Galois Symbols on E E

Transcription

1 Loca Gaois Symbos on E E Jacob Murre and Dinakar Ramakrishnan To Spencer Boch, with admiration Introduction Let E be an eiptic curve over a fied F, F a separabe agebraic cosure of F, and a prime different from the characteristic of F. Denote by E[] the group of -division points of E in E(F ). To any F -rationa point P in E(F ) one associates by Kummer theory a cass [P ] in the Gaois cohomoogy group H 1 (F, E[]), represented by the 1-cocyce β : Ga(F /F ) E[], σ σ ( P ) P. Here P denotes any point in E(F ) with ( ) P = P. Given a pair (P, Q) of F -rationa points, one then has the cup product cass [P, Q] := [P ] [Q] H 2 (F, E[] 2 ). Any such pair (P, Q) aso defines a F -rationa agebraic cyce on the surface E E given by P, Q := [(P, Q) (P, 0) (0, Q) + (0, 0)], where [ ] denotes the cass taken moduo rationa equivaence. It is cear that this zero cyce of degree zero defines, by the paraeogram aw, the trivia cass in the Abanese variety Ab(E E). So P, Q ies in the Abanese kerne T F (E E). It is a known, but non-obvious, fact that the association (P, Q) [P, Q] depends ony on P, Q, and thus resuts in the Gaois symbo map s : T F (E E)/ H 2 (F, E[] 2 ). It is a conjecture of Somekawa and Kato that this map is aways injective ([So]). It is easy to verify this when F is C or R. In the atter case, the image of s is non-trivia iff = 2 and a the 2-torsion points are R-rationa, which can be used to exhibit a non-trivia goba 2-torsion cass in T Q (E E), whenever E is defined by y 2 = (x a)(x b)(x c) with a, b, c Q. It shoud aso be noted that injectivity of the anaog of s fais for certain surfaces occurring as quadric fibrations (cf. [ParS]). However, the genera expectation is that such pathoogies do not occur for abeian surfaces. Partiay supported by the NSF 1

2 Let F denote a non-archimedean oca fied, with ring of integers O F and residua characteristic p. Let E be a semistabe eiptic curve over F with Néron mode E over S =Spec(O F ). Let E[] denote the kerne of mutipication by on E, which defines a finite fat groupscheme E[] over S. Let F (E[]) denotes the smaest Gaois extension of F over which a the -division points of E are rationa. It is easy to see that the image of s is zero if (i) E has good reduction and (ii) p, the reason being that the absoute Gaois group G F acts via its maxima unramified quotient Ga(F nr /F ) Ẑ, which has cohomoogica dimension 1. So we wi concentrate on the more subte = p case. Theorem A Let F be a non-archimedean oca fied of characteristic zero with residue fied F q, q = p r, p odd. Suppose E/F is an eiptic curve over F, which has good, ordinary reduction. Then the foowing hod: (a) s p is injective, with image of F p -dimension 1. (b) The foowing are equivaent: (bi) dim Im(s p ) = 1 (bii) [F (E[p]) : F ] 2 and µ p F (c) Suppose that [F (E[p]) : F ] 2 and µ p F. Then T F (E E)/p Z/p. If E[p] F, then T F (E E)/p even consists of symbos P, Q with P, Q in E(F )/p. Note that [F (E[p]) : F ] is prime to p iff the Ga(F /F )-representation ρ p on E[p] is semisimpe. We obtain: Coroary B T F (E E) is p-divisibe when E[p] is not semisimpe. When a the p-division points of E are F -rationa, the injectivity part of part (a) has aready been asserted, without proof, in [R-S], where the authors show the interesting resut that T F (E E)/p is a quotient of K 2 (F )/p. Our techniques are competey disjoint from theirs, and besides, the deicate part of our proof of injectivity is exacty when [F (E[p]) : F ] is divisibe by p, which is equivaent to the Gaois modue E[p] being non-semisimpe. Our resuts prove in fact that when T F (A)/p is non-zero, i.e., when F (E[p])/F is unramified of degree 2, it is isomorphic to the p-part Br F [p] of the Brauer group of F, which is known ([Ta1]) to be isomorphic to K 2 (F )/p. A competey anaogous resut concerning s hods when E/F has mutipicative reduction, for both = p and p, and this can be proved by arguments simiar to the ones we use in the ordinary case. However, there is aready a paper of Yamazaki ([Y]) giving essentiay the same resut (in the mutipicative case), and we content ourseves to a very brief discussion in section 7 on how to deduce this anaogue from [Y]. The reevant preiminary materia for the paper is assembed in the first two sections and in the Appendix. We have, primariy (but not totay) for the convenience of the reader, suppied proofs of various statements for which we coud not find pubished references, even if they are apparenty known to experts or in the fokore. In a seque we wi use these resuts in conjunction with others (incuding a treatment of the case of supersinguar reduction, p-adic approximation, and a oca-goba emma) and prove two goba theorems about the Gaois symbos on E E moduo p, for any odd prime p. 2

3 The second author woud ike to acknowedge support from the Nationa Science Foundation through the grants DMS and DMS , as we as the hospitaity of the DePrima Mathematics House at Sea Ranch, CA. The first author woud ike to thank Catech for the invitation to visit Pasadena, CA, for severa visits during the past eight years. This paper was finaized in May We are gratefu to various mathematicians who pointed out errors in a previous version, which we have fixed. It gives us great peasure to dedicate this artice to Spencer Boch, whose work has inspired us over the years, as it has so many others. The first author (J.M.) has been a friend and a feow cycist of Boch for many years, whie the second author (D.R.) was Spencer s post-doctora mentee at the University of Chicago during , a period which he remembers with peasure. 1 Preiminaries on Symbos 1.1 Cyces Let F be a fied and X a smooth projective variety of dimension d and defined over F. Let 0 i d and Z i (X) be the group of agebraic cyces on X defined over F and of dimension i, i.e. the free group generated by the subvarieties of X of dimension i and defined over F (see [F], section 1.2). A cyce of dimension i is caed rationay equivaent to zero over F if there exists T Z i+1 (P 1 X) and two points P and Q on P 1, each rationa over F, such that T (P ) = Z and T (Q) = 0, where for R P 1 we put T (R) = pr 2 (T.(R X)) in the usua sense of cacuation with cyces (see [F]). Let Z rat i (X) be the subgroup of Z i (X) generated by the cyces rationay equivaent to zero and CH i (X) = Z i (X)/Z rat i (X) the Chow group of rationa equivaence casses of i-dimensiona cyces on X defined on F. If K F is an extension then we write X K = X F K, Z i (X K ) and CH i (X K ) for the corresponding groups defined over K and if F is the agebraic cosure of F then we write X = X F, Z i ( X) and CH i ( X). 1.2 The Abanese kerne In this paper we are ony concerned with the case when X is of dimension 2, i.e. X is a surface (and in fact we sha ony consider very specia ones, see beow) and with groups of 0-dimensiona cyces contained in CH 0 (X). In that case note that if Z Z 0 (X) then we can write Z = n i P i when P i X( F ), i.e. the P i are points on X but themseves ony defined (in genera) over an extension fied K of F. Put A 0 (X) = Ker(CH 0 (X) Z) deg where deg(z) = n i deg(p i ), i.e. A 0 (X) are the cyce casses of degree zero and put further T (X) := Ker(A 0 (X) Ab(X)), where Ab(X) is the Abanese variety of X (see [B]). Again if K F, write A 0 (X K ), T (X K ) or aso T K (X) for the corresponding groups over K. It is known that both A 0 ( X) and T ( X) are divisibe groups ([B]). Moreover if the group of transcendenta cyces H 2 ( X) trans is non-zero, and if F is a universa domain (ex. F = C), the group T ( X) is huge (infinite dimensiona) by Mumford (in characteristic zero) and Boch (in genera) ([B], 1.22). 1.3 Symbos 3

4 We sha ony be concerned with surfaces X which are abeian, even of the form A = E E, where E is an eiptic curve defined over F. If P, Q E(F ), put P, Q := (P, Q) (P, O) (O, Q) + (O, O) where O is the origin on E, and the addition is the addition of cyces (or better: cyce casses). Ceary P, Q T F (A) and P, Q is caed the symbo of P and Q. Definition. ST F (A) denotes the subgroup of T F (A) generated by symbos P, Q with P E(F ), Q E(F ). It is caed the symbo group of A = E E. 1.4 Properties and remarks 1. The symbo is biinear in P and Q. For instance P 1 + P 2, Q = P 1, Q + P 2, Q (where now the first + sign is addition on E!). 2. The symbo is the Pontryagin product P, Q = {(P, 0) (0, 0)} {(0, Q) (0, 0)} 3. If F is finitey generated over the prime fied, then it foows from the theorem of Morde-Wei that ST F (A) is finitey generated. 4. Note the simiarity with the group K 2 (F ) of a fied. Aso the symbo group ST F (A) is reated to, but different from, the group K 2 (E E) defined by Somekawa ([So]); compare aso with [R-S]. 1.5 Restriction and norm/corestriction Let K F. Consider the morphism ϕ K/F : A K A F. a. This induces homomorphisms, caed the restriction homomorphisms: res K/F = ϕ K/F : CH i(a F ) CH i (A K ), T F (A) T K (A) and ST F (A) ST K (A) (see [F], section 1.7). b. Aso we have the norm or corestriction homomorphisms N K/F = (ϕ K/F ) : CH i (A K ) CH i (A F ) and T (A K ) T F (A) (see [F], section 1.4, page 11 and 12). Remarks. 1. If [K : F ] = n we have ϕ ϕ = n 2. It is not cear if N K/F = ϕ induces a homomorphism on the symbo groups themseves. Note however that there is such norm map for cohomoogy ([Se2], p. 127) and for K 2 -theory ([M], p. 137). 4

5 1.6 Some preiminary emmas Lemma Let Z T F (A). Suppose we can write Z = N (P i, Q i ) N (P i, Q i ) with P i, P i, Q i and Q i a in E(F ) for i = 1,..., N. Then Z ST F (A). Proof: Since Z T F (A) we have that N P i = N i=0 i=0 i=1 i=1 P i and N Q i = N Q i i=1 i=1 as sum of points on E. From this it foows immediatey that we can rewrite Z = N {(P i, Q i ) (P i, 0) (0, Q i ) + (0, 0)} N {(P i, Q i ) (P i, 0) (0, Q i ) + (0, 0)} = N P i, Q i N i=1 Coroary i=1 Over the agebraic cosure we have i=1 i=1 T (Ā) = im {res F /K ST K (A)} K where the imit is over a finite extensions K F. P i, Q i ; hence Z ST F (A). Proof: Immediate from Lemma Norms of symbos Next we turn our attention again to T (A) = T F (A) itsef. If K F is a finite extension and P, Q E(K) then consider N K/F ( P, Q ) T F (A). Let ST K/F (A) T F (A) be the subgroup of T F (A) generated by such eements (i.e., coming as norms of the symbos from finite fied extensions K F ). Note that ceary ST F/F (A) = ST F (A) and aso that ST K/F (A) consists of the norms of eements of ST K (A). Lemma With the above notations et T F (A) be the subgroup of T F (A) generated by a the subgroups of type ST K/F (A) of T F (A) for K F finite (with K F ). Then T F (A) = T F (A). Proof: For simpicity we sha (first) assume char(f ) = 0. If c(z) T F (A), then c(z) is the (rationa equivaence) cass of a cyce Z Z 0 (A F ) and moreover Z = Z Z with Z and Z positive (i.e. effective ), of the same degree and both Z Z 0 (A F ) and Z Z 0 (A F ). Fixing our attention on Z Z 0 (A F ) we can, by definition, write Z = Z α where the Z α are (0-dimensiona) α subvarieties of A and irreducibe over F (Remark: in the terminoogy of Wei s Foundations [W] the Z is a rationa chain over F and the Z α are the prime rationa parts of it, see [W], p. 207). For each α we have Z α = n α (P αi, Q αi ) where the (P αi, Q αi ) A( F ) is a set of points which form a compete set of conjugates over F (see [W], 207). i=1 Note that aso n α (P αi, 0) Z 0(A F ) and simiary nα (0, Q αi ) Z 0(A F ) (Note: these cyces are rationa, but not necessariy prime i=1 rationa.) For each α fix an arbitrary i(α) and put K αi(α) = F (P αi(α), Q αi(α) ) and consider now i=1 5

6 the cyce (P αi(α), Q αi(α) ) Z 0(A Kαi(α) ). We have by definition (see [F], section 1.4, p. 11), Z α = N Kαi(α) /F (P αi(α), Q αi(α) ). Furthermore if we put then in T F (A) we have Zαi(α) P = αi(α), Q αi(α) ST Kαi(α) (A) n α N Kαi(α) /F (Zαi(α) ) = P αi, Q αi (again by the definition of the N /F ) and ceary the cyce is in ST Kαi(α) /F (A), i.e. in T F (A). Now doing this for every α and treating simiary Z = Z β, we have, since Z T F (A), that β i=1 Z = Z Z = α Z α β Z β = α P αi, Q αi P βj, βj Q. i β j Hence Z T F (A), which competes the proof (in char 0). Remark. If char(f ) = p > 0, then we have that Z α = p m α i (P αi, Q αi ) where the pm α is the degree of inseparabiity of the fied extension K αi(α) over F (see again [W], p.207). Note that p mα does not depend on the choice of the index i(α), because the fied K αi(α) is determined by α up to conjugation over F. From that point onwards the proof is the same (note in particuar that we sha have Z α = N Kαi(α) /F (P αi, Q αi )). Lemma For P, Q E(F ) we have Q, P = P, Q, i.e., the symbo is skew-symmetric. Proof: Since the symbo is biinear we have a we-defined homomorphism λ: E(F ) Z E(F ) ST F (A) with λ(p Q) = P, Q. Then λ((p + Q) (P + Q)) = (P + Q, P + Q) (P + Q, 0) (0, P + Q) + (0, 0). On the other hand by the biinearity, it aso equas P, P + P, Q + Q, P + Q, Q. On the diagona we have (P + Q, P + Q) + (0, 0) = (P, P ) + (Q, Q), on E 0 we have (P + Q, 0) + (0, 0) = (P, 0) + (Q, 0), and on 0 E we have (0, P + Q) + (0, 0) = (0, P ) + (0, Q). Putting these facts together we get P, Q + Q, P = A usefu emma We wi often have occasion to use the foowing simpe observation: Lemma Let F be any fied and a prime. Let P, Q be points in E(F ), Q E( F ) s.t. Q = Q, and put K = F (Q ). Consider the statements (a) P N K/F E(K) mod E(F ) (b) P, Q T F (A) 6

7 Then (a) impies (b). Proof: Let P = N K/F (P ) + P 1, with P E(K), P 1 E(F ). Then P, Q P 1, Q = NK/F (P ), Q, which equas, by the projection formua, N K/F ( P, Res K/F Q ) = N K/F ( P, Q ) = N K/F ( P, Q ) T F (A). 2 Symbos, cup products, and H 2 s (F, E 2 ) 2.1 Degeneration of the spectra sequence Notations and assumption are as before. Let be a prime number with char(f ). We work here with Z coefficients, but the resuts are aso true for Z/ s coefficients (any s 1) and Q -coefficients. Lemma The Hochschid-Serre spectra sequence E pq 2 = Hp (F, H q et (Ā, Z (s)) = H p+q et (A, Z (s)) degenerates at d 2 -eve (and at a d t -eves, t 2) for a r. Remark. We coud take here any abeian variety A instead of E E. Proof: This foows from weight considerations. Consider on A mutipication by n, i.e. n: A A is the map x nx. Then we have a commutative diagram H p (F, H q et (Ā, )) d 2 H p+2 (F, H q 1 et (Ā, )) n n H p (F, H q et (Ā, )) d 2 H p+2 (F, H q 1 (Ā, )) et On the eft we have mutipication by n q, on the right by n q 1. this being true for any n > 0, we must have d 2 = Kummer sequence (and some notations) If E is an eiptic curve defined over F, we write (by abuse of notation) and we have the (eiptic) Kummer sequence E[ n ] := ker{e( F ) E( n F )}, 0 E[ n ] E( F ) n E( F ) 0. This is an exact sequence of Ga( F /F )-modues and gives us a short exact sequence of (cohomoogy) groups 0 E(F )/ n H 1 (F, E[ n ]) H 1 (F, Ē)[n ] 0 7

8 Taking the imit over n, one gets the homomorphism where T (E) = im E[ n ] is the Tate group. n This aows us to define by δ (1) : E(F ) H 1 (F, T (E)) [, ] : E(F ) E(F ) H 2 (F, T (E) 2 ) [P, Q] = δ (1) (P ) δ (1) (Q) We have simiar maps (and we use the same notations) if we take E(F )/ n and E[ n ] 2. Expicity the map δ (1) is given by the foowing: For P E(F ) the cohomoogy cass δ 1 (P ) is represented by the 1-cocyce Ga( F /F ) T (E), ( ( ) 1 σ σ P 1 ( ) 1 P, σ 2 P 1 ) 2 P, Comparison with the usua cyce cass map For every smooth projective variety X defined over F there is the cyce cass map to continuous cohomoogy as defined by Jannsen ([J], emma 6.14) c (i) : CH i (X) Hcont(X, 2i Z (i)) Taking now X = E, resp. X = A, and using the degeneration of the Hochschid-Serre spectra sequence we get c (1) : CH(0) 1 (E) H1 (F, H et(ē, 1 Z (1))) where CH 1 (0)(E) is the Chow group of 0-cyces on E of degree 0, resp. c (2) : T F (A) H 2 (F, H et(ā, 2 Z (2))). Lemma There is a commutative diagram E(F ) δ (1) H 1 (F, T (E)) = CH 1 (0) (E) c (1) H 1 (F, H et(ē, 1 Z (1))) where the vertica map on the eft is P (P ) (0) and the one on the right comes from the we-known isomorphism T (E) H et(ē, 1 Z (1)). Proof: See [R], proof of the emma in the appendix. 8

9 2.4 The symboic part of cohomoogy Let K/F be any finite extension. Given a pair of points P, Q E(K)/, with associated casses δ (1) (P ), δ (1) (Q) in H 1 (K, E[]). We have seen in section 2.2 that by taking cup product in Gaois cohomoogy, we get a cass [P, Q] in H 2 (K, E[] 2 ). By taking the norm (corestriction) from H 2 (K, E[] 2 ) to H 2 (F, E[] 2 ), we then get a cass N K/F ([P, Q] ) H 2 (F, E[] 2 ). We define the symboic part of H 2 (F, E[] 2 ) to be the F -subspace generated by such norms of symbos N K/F ([P, Q] ), where K runs over a possibe finite extensions of F and P, Q run over a pairs of points in E(K)/. Note the simiarity of this definition with the description of T F (E E)/ via Lemma Summary The remarks and maps of the previous sections can be subsumed in the foowing: Proposition There exist maps and a commutative diagram E(F ) 2 δ (1) δ (1) CH 1 c (1) (0) (E) 2 c (1) p 1 p 2 H 1 (F, T (E)) 2 = H 1 (F, H et(ē, 1 Z (1))) 2 Hcont(E, 2 Z (1)) 2 p 1 p 2, CH 1 (A) 2 c (1) c (1) Hcont(A, 2 Z (1)) CH 2 (A) c (2) H 4 cont(a, Z (2)) T F (A) c (2) c H 2 (F, H 2 et(ā, Z (1)))).. H 2 (F, H et(ē, 1 Z (2) 2 ) where c is defined via the Künneth formua H et(ā, 2 Z (2)) = H et(ē, 2 Z (2)) H et(ē, 1 Z (1) 2 H et(ē, 2 Z (2)) and the projection on the reevant (= midde) term. In particuar [P, Q] := δ (1) (P ) δ (1) (Q) = c ( P, Q ) for P, Q E(F ). Moreover the same hods for Z/ n -coefficients (instead of Z ) and aso for Q -coefficients. Remark: Note that the map, on the eft from E(F ) 2 to T F (A) is the symbo map, and that the map on the right from H 1 (F, Het(E, 1 Z (1))) 2 to H 2 (F, Het(E, 1 Z (1)) 2 ) is the one given by taking the cup product in group cohomoogy. 9

10 Proof (and further expanation of the maps) The map p 1 p 2 is induced by the projections p i : A = E E E. The commutativity in upper rectange comes from Lemma The other rectanges are a natura. Coroary With respect to the decomposition H 1 et(ē, Z (1)) 2 = T (E) 2 Sym 2 T (E) Λ 2 T (E) we have that [P, Q] = c ( P, Q ) H 2 (F, Sym 2 T (E)) if 2 (and simiary for Z/ s -coefficients). Proof Step 1. For the sake of carity of the proof we sha first take two different eiptic curves E 1 and E 2 (or, if one prefers, two different copies E 1 and E 2 of E). Put A 12 = E 1 E 2. There exist obvious anaogs of the maps and the commutation diagram of Proposition 2.5.1; ony of course one shoud now write E 1 (F ) E 2 (F ), etc. In particuar we have again the cup product on the right: H 1 (F, T (E 1 )) H 1 (F, T (E 2 )) H 2 (F, T (E 1 ) T (E 2 )) where we write T (E 1 ) = Het(Ēi, 1 Z (1)), i = 1, 2. We get c ( P, Q ) = δ (1) (P ) δ (1) (Q). Now consider aso A 21 = E 2 E 1 and the corresponding diagram for A 21. Consider the natura isomorphism t: A 12 A 21 given by t(x, y) = (y, x) for x E 1 and y E 2 and aso the corresponding isomorphism t : H 1 (Ē1) H 1 (Ē2) H 1 (Ē2) H 1 (Ē1). Caim. If P E 1 (F ) and Q E 2 (F ) then c,a21 ( Q, P ) = t (δ (1) (P ) δ (1) (Q)) Here we have written in order to avoid confusion c,a21 for c in the diagram reative to A 21. Proof of the caim: c,a21 ( Q, P ) = δ (1) (Q) δ (1) (P ) = t ( δ (1) (P ) δ (1) ) (Q) where the first equaity is from the diagram (for A 21 ) in Proposition and the second equaity is a we-known property in cohomoogy (see for instance [Br], p. 111, (3.6), or [Sp], chap.5, 6, p. 250). Step 2. Returning to the case E 1 = E 2 = E, we have Q, P = P, Q in T F (A) by Lemma Step 3. Write c ( P, Q ) = α + β with α H 2 (F, Sym 2 T (E)) and β H 2 (F, Λ 2 T (E)). We get c ( P, Q ) = c Q, P ) from Step 2, and next from Caim in Step 1 we have c ( Q, P ) = t (δ (1) (P ) δ (1) (Q)), hence α + β = t (α + β) = α β, hence β = 0 if 2. Definition Let s : T F (E E)/ H 2 (F, E[] 2 ) 10

11 be the homomorphism induced by the reduction of c moduo and by using the isomorphism of H 1 (E F, Z (1)) with the -adic Tate modue of E, which is the inverse imit of E[ n ] over n. Thanks to the discussion above, the definition of the symboic part of H 2 (F, E[] 2 ) in section 2.4, and Lemma 1.7.1, we obtain the foowing: Proposition The image of T F (E E)/ under s is H 2 s (F, E[] 2 ). 3 A key Proposition Let E/F be an eiptic curve over a oca fied F with good ordinary reduction. We wi henceforth take = p, the residua characteristic of F. A basic fact is that the representation ρ p of G F on E[p] is reducibe, so the matrix of this representation is trianguar. Since the determinant is the mod p cycotomic character χ p, we may write ( χp ν (3.1) ρ p = 1 ), 0 ν where ν is an unramified character of finite order, such that E[p] is semisimpe (as a G F -modue) iff = 0. Note that ν is necessariy of order at most 2 when E has mutipicative reduction, with ν = 1 iff E has spit mutipicative reduction. On the other hand, ν can have arbitrary order (dividing p 1) when E has good, ordinary reduction. In any case, there is a natura G F -submodue C F of E[p] of dimension 1, such that we have a short exact sequence of G F -modues: (3.2) 0 C F E[p] C F 0, with G F acting on C F by χ p ν 1 and on C F by ν. Ceary, E[p] is semisimpe iff the sequence (3.2) spits. The natura G F -map C 2 F E[p] 2 induces a homomorphism (3.3) γ F : H 2 (F, C 2 F ) H2 F, E[p] 2 ). The key resut we prove in this section is the foowing: Proposition C Let F be a non-archimedean oca fied with odd residua characteristic p, and E an eiptic curve over F with good, ordinary reduction. Denote by Im(s p ) the image of T F (E E)/p under the Gaois symbo map s p into H 2 F, E[p] 2 ). Then we have (a) Im(s p ) Im(γ F ). (b) The dimension of Im(s p ) is at most 1, and it is zero dimensiona if either µ p F or ν 2 1. Remark: If E/F has mutipicative reduction, with an arbitrary odd prime (possiby equa to p), then again the Gaois representation ρ on E[] has a simiar shape, and in fact, ν is at most quadratic, refecting the fact that E attains spit mutipicative reduction over at east a quadratic extension of F, over which the two tangent directions at the node are rationa. An anaogue of 11

12 Proposition C hods in that case, thanks to Proposition A.3.3 in the Appendix, once we assume that does not divide the order of the component group of the specia fibre E s of the Néron mode E. For the sake of brevity, we are not treating this case here. The buk of this section wi be invoved in proving the foowing resut, which at first seems weaker than Proposition C: Proposition 3.4 Let F, E, p be as in Proposition C. Then, for a points P, Q in E(F )/p, we have s p ( P, Q ) Im(γ F ). Caim 3.5 Proposition 3.4 = Part (a) of Proposition C: Proof of this caim goes via some emmas. Lemma 3.6 (Behavior of γ under finite extensions) Let K/F be finite. We then have two commutative diagrams, one for the norm map N = N K/F and the other for the restriction map Res=Res K/F : H 2 (K, C 2 K ) γ K H 2 (K, E[p] 2 ) N Res N Res H 2 (F, C 2 F ) H 2 (F, E[p] 2 ) γ F This Lemma foows from the compatibiity of the exact sequence (3.2) (of Gaois modues) with base extension. Lemma 3.7 In order to prove that Im(s p ) Im(γ F ), it suffices to prove it for the image of symbos, i.e., that Im (s p (ST F,p (A))) Im(γ F ), where Proof. ST F,p (A) := Im (ST F (A) T F (A)/p). Use the commutativity of the diagram in Lemma 3.6 for the norm. Lemma 3.8 In order to prove that Im(s p ) Im(γ F ), we may assume that µ p F and that ν = 1, i.e., that we have the foowing exact sequence for groupschemes over S: ( ) 0 µ p,s E[p] (Z/p) S 0. Proof. There is a finite extension K/F such that ( ) hods over K, with p [K : F ]. Now use the diagram(s) in Lemma 3.6 as foows: Let P, Q E(F ) E(K), then Res(P ) = P, Res(Q) = Q, and we have [K : F ]s F,p ( P, Q ) = N{Res(s F,p ( P, Q } = N{s K,p ( Res(P ), Res(Q) )} = N{s K,p ( P, Q )} Therefore, if s K,p ( P, Q ) Im(γ K ), then (again by using Lemma 3.6 for norm) we see that N{s K,p ( P, Q )} is contained in Im(γ F ). Hence [K : F ]s F,p ( P, Q ) ies in Im(γ F ). Finay, since p [K : F ], s F,p ( P, Q ) itsef beongs to Im(γ F ). This proves Caim

13 3.9 Proof of Proposition 3.4 Since we may take µ p F and ν = 1, the exact sequence ( ) in Lemma 3.8 hods, compatiby with the corresponding one over F of Ga(F /F )-modues. Taking cohomoogy, we get a commutative diagram of F p -vector spaces with exact rows: O F /p E(F )/p Z/p H 1 (F, µ p ) H 1 (F, E[p]) H 1 (F, Z/p) 0 Here the vertica maps are isomorphisms (see the Appendix), which induce the horizonta maps on the top row. Definition 3.10 Put and choose a non-canonica decomposition U F := Im (O F /p E(F )/p), E(F )/p U F W F, with W F Z/p. Notation 3.11 If S 1, S 2 are subsets of E(F )/p, we denotes by S 1, S 2 the subgroup of ST F,p (A) generated by the symbos s 1, s 2, with s 1 S 1 and s 2 S 2. Lemma 3.12 ST F,p (A) is generated by the two vector subspaces Σ 1 := U F, E(F )/p and Σ 2 := E(F )/p, U F. Proof. ST F,p (A) is ceary generated by Σ 1, Σ 2 and by W F, W F. However, W F is onedimensiona and the pairing, is skew-symmetric, so W F, W F = 0. Now we have a commutative diagram (3.13-i) where the top map α 1 factors as O F /p E(F )/p α 1 σ 1 H 2 (F, µ p E[p]) E(F )/p E(F )/p sp β 1 H 2 (F, E[p] 2 ), O F /p E(F )/p U F E(F )/p E(F )/p E(F )/p. We aso get a simiar diagram (3.13-ii) by repacing O F /p (E(F )/p) (resp. U F (E(F )/p)) by (E(F )/p) O F /p (resp. (E(F )/p) U F ). The maps α j and their factoring are obvious, s p is the map constructed in section 2, and the vertica maps σ j are defined entirey anaogousy. Lemma 3.14 The image of s p : ST F,p (A) H 2 (F, E[p] 2 ) is generated by β 1 (Im(σ 1 )) and β 2 (Im(σ 2 )). Proof. (3.13-ii). Immediate by Lemma 3.12 together with the commutative diagrams (3.13-i) and 13

14 Tensoring the exact sequence (3.15) 0 µ p E[p] Z/p 0 with µ p from the eft and the right, and taking Gaois cohomoogy, we get two natura homomorphisms (3.16 i) γ 1 : H 2 (F, µ 2 p ) H 2 (F, µ p E[p]) and (3.16 ii) γ 2 : H 2 (F, µ 2 p ) H 2 (F, E[p] µ p ). Lemma 3.17 Im(σ j ) Im(γ j ), for j = 1, 2. Proof of Lemma We give a proof for j = 1 and eave the other (entirey simiar) case to the reader. By tensoring (3.15) by µ p, we obtain the foowing exact sequence of G F -modues: (3.18) 0 µ 2 p µ p E[p] µ p 0 Consider now the foowing commutative diagram, in which the bottom row is exact: (3.19) H 2 (F, µ 2 p ) γ 1 where σ 1 = i 1 σ 1 is the map from (3.13-i). U F E(F )/p σ 1 H 2 (F, µ p E[p]) i 1 H 2 (F, µ p E[p]) ε 0 2 H (F, µp ) ε H 2 (F, µ p ) To begin, the exactness of the bottom row foows immediatey from the exact sequence (3.18). The factorization of σ 1 is the crucia point, and this hods because U F comes from O F /p and is mapped to H 1 (F, µ p ) via H 1 (F, µ p ). Simiary, E(F )/p maps into H 1 (F, E[p]). To prove Lemma 3.17, it suffices, by the exactness of the bottom row, to see that Im(σ 1 ) is contained in Ker(ε), i.e., to see that ε σ 1 = 0. Luckiy for us, this composite map factors through ε 0 σ 1, which vanishes because Hf 2(S, µ p,s), and hence H 2 (F, µ p ), is zero by [Mi2], part III, Lemma 1.1. Putting these Lemmas together, we get the truth of Proposition

15 3.20 Proof of Proposition C As we saw earier, Proposition 3.4, which has now been proved, impies (by Caim 3.5) part (a) of Proposition C. So we need to prove ony part (b). By part (a) of Prop. C, the dimension of the image of s p is at most that of H 2 (F, C 2 ). Since E[p] is sefdua, the short exact sequence (3.2) shows that the Cartier dua of C is C. It foows easiy that (C 2 ) D is C 2 ( 1). By the oca duaity, we then get H 2 (F, C 2 ) H 0 (F, C 2 ( 1)), As C is a ine over F p with G F -action, the dimension of the group on the right is ess than or equa to 1, with equaity hoding iff C 2 µ p. Since G F acts on C by an unramified character ν, for C 2 to be µ p as a G F -modue, it is necessary that µ p F. So µ p F = Im(s p ) = 0. Now suppose µ p F. Then for Im(s p ) to be non-zero, it is necessary that C 2 Z/p, impying that ν 2 = 1. 4 Vanishing of s p in the non-semisimpe case The object of this section is to prove the foowing: Proposition D Suppose E/F is an eiptic curve (with good ordinary reduction) over a nonarchimedean oca fied F of odd residua characteristic p. Assume that the G F -modue E[p] is not semisimpe. Then we have s p (T F (E E)/p) = 0. Combining this with Proposition C, we get the foowing Coroary 4.1 Let F be a non-archimedean oca fied with residua characteristic p > 2, and E an eiptic curve over F with good, ordinary reduction. Then Im(s p ) is zero whenever [F (E[p]) : F ] > 2. Proof of Proposition D. Let us first note a few basic things concerning base change to a finite extension K/F of degree m prime to p. To begin, since E/F has ordinary reduction, the Gaois representation ρ F on E[p] is trianguar, and it is semisimpe iff the image does not contain any eement of order p. It foows that ρ K is semisimpe iff ρ F is semisimpe. Moreover, the functoriaity of the Gaois symbo map reative to the respective norm and restriction homomorphisms, together with the fact that the composition of restriction with norm is mutipication by m, impies, as p m, that for any θ T F (E E)/p, we have s p,k (res K/F (θ)) = 0 = s p (θ) = 0. So it suffices to prove Proposition D, after possiby repacing F by a finite prime-to-p extension, under the assumption that F contains µ p and ν = 1, sti with E[p] non-semisimpe. Thus we have a non-spit, short exact sequence of finite fat groupschemes over S: (4.2) 0 µ p,s E[p] (Z/p) S 0, 15

16 with the representation ρ F of G =Ga(F /F ) on E[p] having the form: (4.3) ( ) 1 α 0 1 reative to a suitabe basis. Here α : G Z/p is a non-zero homomorphism, and F (α), the smaest extension of F over which α becomes trivia, is a ramified p-extension. Using the exact sequence (4.2), both over S and over F, we get the foowing commutative diagram: (4.4) 0 E[p](S) Z/p Hf 1(S, µ ψ S p) H 1 f (S, E[p]) Hf 1 (S, Z/p) E[p](F ) 0 Z/p H 1 (F, µ p ) ψ H 1 (F, E[p]) H 1 (F, Z/p) X + Y Z = ψ(y ) where X H 1 (F, µ p ) H 1 (F, µ p ) is the subspace given by the image of Z/p. Since µ p is in F, we can identify H 2 (F, µ 2 p ) with Br F [p] Z/p. By our assumption, X 0 as the sequence (4.2) does not spit. Now take e X, with e 0. Then e O F /p F /F p = H 1 (F, µ p ). Since K = F (e 1/p ) is ceary the smaest extension of F over which (4.2) spits, K is aso F (α) and hence a ramified p-extension of F. Then, appying [Se2], Prop.5 (iii), we get a v H 1 (F, µ p ) which is not a norm from K, and so the cup product {e, v} is non-zero in Br F [p] (cf. [Ta1], Prop.4.3). Now consider the commutative diagram (4.5) H 1 (F, µ p ) 2 ψ 2 H 1 (F, E[p]) 2 H 2 (F, µ 2 γ p ) H 2 (F, E[p] 2 ) where ψ is the map defined in the previous diagram, and γ = γ F is the map defined in section 3 with C F = µ p and C F = Z/p. By Proposition C, the image of s p is contained in that of γ. On the other hand, ψ(e) ψ(v) = 0 because ψ(e) = 0. Hence γ({e, v}) = 0, and the image of s p is zero as asserted. 5 Non-trivia casses in Im(s p ) when [F (E[p]) : F ] 2 Proposition E Let E be an eiptic curve over a non-archimedean oca fied F of residua characteristic p 2. Assume that E has good ordinary reduction, and that [F (E[p]) : F ] 2, with µ p F. Then Im(s p ) 0. Moreover, if F (E[p]) = F, i.e., if a the p-division points are rationa over F, then there exist points P, Q of E(F )/p such that s p ( P, Q ) 0. 16

17 In this case, up to repacing F by a finite unramified extension, we may choose P to be a p-power torsion point. Proof. First suppose that K := F (E[p]) is quadratic over F. Reca that over F, C F (resp. C F ) is given by the character χν 1 (resp. ν), and since µ p F and ν quadratic, we have H 2 (F, C 2 F ) H2 (F, µ p ) = Br F [p] Z/p. Suppose we have proved the existence of a cass θ K in T K (E E)/p such that s p,k (θ K ) is nonzero, and this image must be, thanks to Proposition C, in the image of a cass t K in Br K [p]. Put θ := N K/F (θ K ) T F (E E)/p. Then s p (θ) equas N K/F (s p,k (θ K )), which is in the image of t := N K/F (t K ) Br F [p], which is non-zero because the norm map on the Brauer group is an isomorphism. So we may, and we wi, assume henceforth in the proof of this Proposition that E[p]) F. Now et us ook at the basic setup carefuy. Since E has good reduction, the Néron mode is an eiptic curve over S. Moreover, since E has ordinary reduction with E[p] F, we aso have E[p] = (µ p ) S Z/p as group schemes over S (and as sheaves in S fat ). By the Appendix A.3.2, (5.1) E(F )/p E(S)/p F H 1 f (S, E[p]) = Hf 1(S, µ p) Hf 1 (S, Z/p) O F /p Z/p, where the boundary map F is an isomorphism, and we have used the identification of Hf 1 (S, Z/p) with Het(k, 1 Z/p) Z/p ([Mi1], p. 114, Thm. 3.9). Therefore we have a 1-1 correspondence (5.2) P (ūp, n p ) with P E(F )/p, ū p O F /p, n p Z/p. The ordered pair on the right of (5.2) can be viewed as an eement of H 1 f (S, E[p]) or of its (isomorphic) image H1 (F, E[p]) in H 1 (F, E[p]). In Gaois cohomoogy, we have the decomposition (5.3) H 2 (F, E[p] 2 ) H 2 (F, µ 2 p ) H 2 (F, µ p ) 2 H 2 (F, Z/p). We have a simiar one for H 2 f(s, E[p] 2 ). It is essentia to note that by Proposition C, (5.4 i) Im(s p ) H 2 (F, µ 2 p ) H 2 (F, E[p] 2 ), and in addition, (5.4 ii) H 2 (F, µ 2 p ) H 2 (F, µ p ) = Br F [p] Z/p, which is impied by the fact that µ p F (since it is the determinant of E[p])). In terms of the decomposition (5.2) from above, we get, for a P, Q E(F ), (5.5) s p ( P, Q ) = ū P ū Q Br F [p] Z/p. One knows that (cf. [L2], chapter II, sec. 3, Prop.6) (5.6) dim Fp O F /p = µ p (F ) p [F :Q p]. 17

18 Since we have assumed that F contains a the p-th roots of unity, this dimension is at east p 2. Caim 5.7 Let u, v be units in O F which are ineary independent in the F p -vector space V := O F /(O F )p. Then F [u 1/p ] and F [v 1/p ] are disjoint p-extensions of F. This is we known, but we give an argument for competeness. Pick p-th roots α, β of u, v respectivey. If the extensions are not disjoint, we must have α = p 1 j=0 c jβ j in K = F [v 1/p ], with {c j } F. A generator σ of Ga(K/F must send β to wβ for some p-th root of unity w 1 (assuming, as we can, that v is not a p-th power in F ), and moreover, σ wi send α to w i α for some i. Thus α σ can be computed in two different ways, resuting in the identity j c jw i β j = j c jw j β j, from which the Caim foows. Consequenty, since the dimension of V is at east 2 and since F has a unique unramified p- extension, we can find u O F such that K := F [u1/p ] is a ramified p-extension. Fix such a u and et Q E(F ) be given by (ū, 0). By [Se2], Prop. 5 (iii) on page 72 (see aso the Remark on page 95), there exists v O F s.t. v / N K/F (O K ). Then by [Ta1], prop. 4.3, page 266, {v, u} = 0 in Br F [p]. Take P E(F ) such that P ( v, 0) then s p ( P, Q ) = {v, u} H 2 (F, E[p] 2 ) Br F [p] and {v, u} = 0. It remains to show that we may choose P to be a p-power torsion point after possiby repacing F by a finite unramified extension. Since E[p] is in F, µ p is in F ; reca that µ p (F ) E[p](F ). So we may pick a non-trivia p-th root of unity ζ in F. Let m be the smaest positive integer such that ζ := w pm 1 is in F F p. Let F /F be the unramified extension of F such that over the corresponding residua extension, a the p m -torsion points of E s are rationa; note however that E[p m ] need not be in F. This resuts in the foowing short exact sequence: eading to the incusion 0 µ p m E[p m ] Z/p m 0, µ p m(f ) E[p m ](F ). Since F /F is unramified, w cannot beong to F p, and the corresponding point P, say, in E[p m ](F ) is not in pe(f ). Put L = F ( 1 p P ) = F (w 1/p ), which is a ramified p-extension. So there exists a unit u in O F such that {w, u} is not trivia in Br F [p]. Now et Q be a point in E(F ) such that its cass in E(F )/p is given by (u, 0) O F /p Z/p. It is cear that P, Q is non-zero in T F (A)/p. This proves Proposition E. 6 Injectivity of s p Proposition F Let E be an eiptic curve over a non-archimedean oca fied F of odd residua characteristic p, such that E has good, ordinary reduction. Then s p is injective on T F (E E)/p. In view of Proposition E, we have the foowing 18

19 Coroary 6.1 Let F, E, p be as in Proposition F. Then T F (E E)/p is a cycic group of order p. Moreover, if E[p] F, it even consists of symbos P, Q, with P, Q E(F )/p. To prove Proposition F, we wi need to consider separatey the cases when E[p] is semisimpe and non-semisimpe. Proof of Proposition F in the semisimpe case Again, to prove injectivity, we may repace F by any finite extension of prime-to-p degree. Since p does not divide [F (E[p]) : F ] when E[p] is semisimpe, we may assume (in this case) that a the p-torsion points of E are rationa over F. Remark. The injectivity of s p when E[p] F has been announced without proof, and in fact for a more genera situation, by Raskind and Spiess [R-S], but the method of their paper is competey different from ours. There are three steps in our proof of injectivity when E[p] F : Step I: Injectivity of s p on symbos Pick any pair of points P, Q in E(F ). We have to show that if s p (< P, Q >) = 0, then the symbo < P, Q > ies in pt F (A). To achieve Step I, it suffices to prove that the condition (a) of emma hods. In the correspondence (5.2), et P (ū P, n P ) and Q (ū Q, n Q ). Put K 1 = F ( p u Q ) and take K 2 to be the unique unramified extension of F of degree ( p if ) n Q 0; otherwise take K 2 = F. Consider the compositum K 1 K 2 of K 1, K 2, and K := F 1 p Q, a the fieds being viewed as subfieds of F. (6.2) From (5.1) we get the foowing commutative diagram: E(O K )/p Hf 1(O K, E[p]) O K /(O K )p Z/p N K/F Res Res N K/F Res E(O F )/p Hf 1(O F, E[p]) O F /(O F )p Z/p. N K/F =id Here the map Res= Res K/F on E(O F )/p and O F /p induced by the incusion F K is the obvious restriction map. However, on Z/p, Res comes from the residue fieds F of F and F of K via the identifications (6.3) H 1 f (S, Z/p) H1 (F, Z/p) Hom(Ga(F/F), Z/p) Z/p and the corresponding one for K and F. As Ga(F/F) = Ẑ, we see by taking F F F with [F : F] = d dividing p, that Ga(F/F ) is the obvious subgroup of Ga(F/F) corresponding to dẑ; consequenty, the map Res on the Z/p summand is mutipication by d. Finay note that N K/F Res K/F = [K : F ]id. Lemma 6.4 With K, K 1, K 2 as above, we have K = K 1 K 2. 19

20 In other words we have the foowing diagram K = K 1 K 2.. K 1 K 2 p p. F. (If ū Q 1, n Q 0) K 1 F ramified K 2 F unramified Proof: Put L = K 1 K 2. a) K L: Appy the diagram (6.2) with L instead of K. We caim that L (Res(Q)) = 0 and hence K L. If n Q = 0, so that L = K 1, we get L (Res(Q)) = (Res(u Q ), 0) = 0. On the other hand, if n Q 0, then K 2 is the unique unramified p-extension of F and the restriction map on Z/p is, by the remarks above, given by mutipication by p. So we sti have L (Res(Q)) = 0, as caimed. b) K L: In this case we have K (Res(Q)) = 0. On the other hand, K (Res(Q)) = (Res(u Q ), Res(n Q )) = (0, 0); hence K K 1. If n Q = 0, K 2 = F and we are done. So suppose n Q 0. Then, since Res(n Q ) is zero, we see that the residua extension of K/F must be non-trivia and so must contain the unique unramified p-extension K 2 of F. Hence K contains L = K 1 K 2. Lemma 6.5 Let P, Q be in E(F )/p with coordinates in the sense of (5.2), namey P = (u P, n P ) and Q = (u Q, n Q ). Let P 0, Q 0 be the points in E(F )/p with coordinates (u P, 0), (u Q, 0) respectivey. Then P, Q = P 0, Q 0 T F (A)/p. Proof of Lemma 6.5. Put P 1 = (0, n P ) and Q 1 = (0, n Q ). Then by inearity, First note that inearity, P, Q = P 0, Q 0 + P 1, Q 0 + P 0, Q 1 + P 1, Q 1. P 1, Q 1 = n P n Q (0, 1), (0, 1). It is immediate, since p 2, to see that (0, 1), (0, 1) is zero by the skew-symmetry of.,.. Thus we have, by bi-additivity, P 1, Q 1 = 0. Next we show that in T F (A)/p, P 0, Q 1 = 0 = P 1, Q 0. We wi prove the triviaity of P 0, Q 1 ; the triviaity of P 1, Q 0 wi then foow by the symmetry of the argument. There is nothing to prove if n Q = 0, so we ( may (and we wi) assume that n Q 0. Then it corresponds to the unramified p-extension M = F 1 p 1) Q of F. It is known that every unit 20

21 in F is the norm of a unit in M /p. This proves that the point P 0 in E(F )/p is a norm from M. Thus P 0, Q 1 is zero by Lemma Putting everything together, we get as asserted in the Lemma. P, Q = P 0, Q 0 Proof of Step I: Let P, Q E(F )/p be such that s p ( P, Q ) = 0. We have to show that P, Q = 0 in T F (A)/p. By Lemma 6.5 we may assume that n Q = 0. So it foows ( that ) the eement {ū P, ū Q } := ū P ū Q is zero in the Brauer group Br F [p]. Then putting K 1 = K, we have by [Ta1], prop. 4.3, we have u P = N K1 /F (u 1 ) with u 1 K 1 and where u P O F with image ū P. It foows that u 1 must aso be a unit in O K1. Now with the notations introduced at the beginning of Step I we have, since n Q = 0, that K = K 1 and so [K : F ] = p. In the diagram (6.2), take in the upper right corner the eement (u, 0). Then we get an eement P E(K)/p such that N K/F P P (mod p). Hence condition (a) of Lemma hods, yieding Step I. u 1/p Q Step II Put Lemma 6.6 s p is injective on ST F,p (A). ST F,p (A) = Im (ST F (A) T F (A)/p) Proof: This foows from Caim: If V is an F p -vector space with an aternating biinear form [, ]: V V W, with W a 1-dimensiona F p -vector space, then the conditions (i) and (ii) of [Ta1], p. 266, are satisfied. Proof: This is Proposition 4.5 of [Ta1], p The proof goes verbay through. In our case we appy this to V = E(F )/p, W = Br F [p] and [ P, Q] = s p ( P, Q ) = ū P ū Q. By ([Ta1], Coroary on p. 266), the s p is injective on ST F,p (A). Step III: Injectivity of s p in the genera case (but sti assuming E[p] F ). For this we sha use the foowing Remark: If K F is any finite extension (of non-archimedean oca fieds), then N K/F : Br K Br F is an isomorphism. Indeed, the invariant map inv F : Br(F ) Q/Z is an isomorphism and moreover (cf. [Se2], chap.xiii, Prop.7), if n = [K : F ], the foowing diagram commutes: Br F inv F Q/Z Res K/F Br K inv K n Q/Z 21

22 As Q/Z is divisibe, every eement in Br K is the restriction of an eement in Br F. Now since Res K/F : Br(F ) Br(K) is aready mutipication by n in Q/Z, the projection formua N K/F Res K/F = [K : F ]Id = nid impies that the norm N K/F on Br K must correspond to the identity on Q/Z. Proof of Step III By Step II, we see from Lemma that it suffices to prove the foowing resut, which may be of independent interest. Proposition 6.7 Let K F be a finite extension, and E[p] F. Then N K/F (ST K,p (A)) is a subset of ST F,p (A) and hence ST F,p (A) = T F (A)/p. In other words, T F (A) is generated by symbos moduo pt F (A). It is an open question as to whether T F (A) is different from ST F (A), though many expect it to be so. Proof of Proposition 6.7. We have to prove that if P, Q E(K), then the norm to F of P, Q is mapped into ST F,p (A). Since we have the assertion is a consequence of the foowing Lemma 6.8 Ims p (ST F (A)) = Ims p (T F (A)/p Z/p, If [K : F ] = n and P, Q E(K), P, Q E(F ) are such that s p,f (N K/F P, Q ) = s p,f ( P, Q ), then, assuming that a the p-torsion in E(F ) is F -rationa,. Proof of Lemma 6.8. Subemma 6.9 N K/F ( P, Q ) P, Q (mod p T F (A)) We start with a few simpe subemmas. If K/F is unramified, then Lemma 6.8 is true. Proof. Indeed, in this case, every point in E(F ) is the norm of a point in E(K) (cf. [Ma], Coroary 4.4). So we can write P N K/F (P 1 )( mod pe(k)) with P 1 E(K), from which it foows that N K/F ( P 1, Q ) P, Q (modpt F (A)). On the other hand, by the remark above (in the beginning of Step III) about the norm map on the Brauer group, we have s p,k ( P 1, Q ) = s p,k ( P, Q ) and appying Step II to P 1, Q and P, Q for K, we are done. Subemma 6.10 If K/F = n with p n, then Lemma 6.8 is true. 22

23 Proof. Indeed, as the cokerne of N K/F is annihiated by n which is prime to p, the norm map on E(K)/p is surjective onto E(F )/p. Let P 1 E(K) satisfy P N K/F (P 1 )(modpe(f )). Then by the projection formua, ( ) P, Q N K/F ( P 1, Q ) (modpt F (A)). Since s p,f N K/F = N K/F s p,k and since N K/F is non-trivia, and hence an isomorphism, on the one-dimensiona F p -space Br K [p], we see that s p,k ( P, Q ) = s p,k ( P 1, Q ). Appying Lemma 6.6 for K we see that P, Q ) equas P 1, Q in T K (A)/p. Now we are done by ( ). Subemma 6.11 If K F 1 F and if 6.8 is true for both the pairs K/F 1 and F 1 /F, then it is true for K/F. Proof. Let P, Q E(F ) and P, Q E(K) be such that (a) s p,f (N K/F P, Q ) = s p,f ( P, Q ). Appying Proposition E with F 1 in the pace of F, we get points P 1, Q 1 in E(F 1 ) such that (b) s p,f1 ( P 1, Q 1 ) = N K/F1 (s p,k ( P, Q )). Since the right hand side is the same as s p,f1 (N K/F1 ( P, Q )), we may appy 6.8 to the pair K/F 1 to concude that (c) P 1, Q 1 = N K/F1 ( P, Q ). Appying N F1 /F to both sides of (b), using the facts that the norm map commutes withs p and that N K/F = N K/F1 N F1 /F, and appeaing to (a), we get (d) s p,f (N F1 /F P 1, Q 1 ) = s p,f ( P, Q ). Appying (6.8) to F 1 /F, we then get (e) P, Q = N F1 /F ( P 1, Q 1 ). The assertion of the Subemma now foows by combining (c) and (e). Subemma 6.12 It suffices to prove 6.8 for a finite Gaois extensions K /F with E[p] F and [F : Q p ] <. Proof. Assume 6.8 for a finite Gaois extensions K /F as above. Let K/F be a finite, nonnorma extension with Gaois cosure L, and et E[p] F. Suppose P, Q E(F ) and P, Q E(K satisfy the hypothesis of 6.8. We may use the surjectivity of N L/K : Br L Br K and Proposition E (over L) to deduce the existence of points P, Q E(L) such that s p,k (N L/K P, Q ) = s p,k ( P, Q ). 23

24 As L/K is Gaois, we have by hypothesis, (i) N L/K ( P, Q ) = P, Q. By construction, we aso have s p,f (N L/F P, Q ) = s p,f ( P, Q ). As L/F is Gaois, we have (ii) N L/F ( P, Q ) = P, Q. The assertion now foows by appying N K/F to both sides of (i) and comparing with (ii). Thanks to this ast subemma, we may assume that K/F is Gaois. Appeaing to the previous three subemmas, we may assume that we are in the foowing key case: (K) K/F is a totay ramified, cycic extension of degree p, with E[p] F So it suffices to prove the foowing Lemma hods in the key case (K). Proof. Suppose K/F is a cycic, ramified p-extension with E[p] F, and et P, Q E(F ), P, Q ine(k) satisfy the hypothesis of 6.8. Since µ p = det(e[p]), the p-th roots of unity are in F and so W := O F /p = H 1 (F, µ p ) H 1 (F, Z/p) = Hom(Ga(F /F ), Z/p). In other words, ines in W correspond to cycic p-extensions of F, and we can write K = F ( u 1/p) for some u O F. For every w O F, et w denote its image in W. Put m = dim F p W. Then m 3 by [L2], chapter II, sec. 3, Proposition 6. Let W 1 denote the ine spanned in W by the unique unramified p-extension of F. Since m > 2, we can find some v O F such that v is not in the inear span of W 1 and u. Using the Caim stated in the proof of Proposition E, we see that L := F ( v 1/p) is ineary disjoint from K over F. Both K and L are totay ramified p-extensions of F, and so is the biquadratic extension KL. Then KL/K is a cycic, ramified p-extension, and by oca cass fied theory ([Se2], chap. V, sec 3), there exists y O K not ying in N KL/K((KL) ). Hence v y is non-zero in Br K [p] (see [Ta1], Prop.4.3, p.266, for exampe). Let P 1 E(F ) and Q 1 E(K) be such that in the sense of (5.2) we have Then P 1 (v, 0), Q 1 (y, 0). s p,k ( P 1, Q 1 ) = v y 0. Since Br K [p] is one-dimensiona (over F p ), we may repace Q 1 by a mutipe and assume that s p,k ( P 1, Q 1 ) = s p,k ( P, Q ). Hence P 1, Q 1 equiv P, Q mod pt K (A) 24

25 by Step II for K. So we get by the hypothesis, s p,f (N K/F P 1, Q 1 ) = s p,f ( P, Q ). But P 1 is by construction in E(F ), and so the projection formua says that N K/F P 1, Q 1 = P 1, N K/F (Q 1 ). Now we are done by appying Step II to F. Now we have competed the proof of Proposition F when E[p] is semisimpe. Proof of Proposition F in the non-semisimpe case Here [F (E[p]) : F ] is divisibe by p, so we cannot assume that E[p] F. However, we may, as in section 4, assume that µ p F and that ν = 1, which impies that the p-torsion points of the specia fibre E s are rationa over the residue fied F q of F. We have in effect (4.2) through (4.4). Consider the commutative diagram (4.4). We can write (6.14) O F /p = H 1 f (S, µ p) H 1 (F, µ p ) = X + Y S, with X Z/p and Y S some compement. Furthermore, we have (cf. [Mi1], Theorem 3.9, p.114) Hence H 1 f (S, Z/p) H1 et(k, Z/p) = Z/p. (6.15) H 1 f (S, E[p]) = Z S Z/p, wherez S = ψ S (Y S ). (Note that the surjectivity of the right map on the top row of the diagram (4.4) comes from the fact that Hf 2(S, µ p) is 0 ([Mi2], chapter III, Lemma 1.1). Reca (cf. (6.2)) that E(F )/p is isomorphic to Hf 1 (S, E[p]). So by (6.15), we have a bijective correspondence (6.16) P (ũ P, n P ) where P runs over points in E(F )/p. Compare this with (5.2). Lemma 6.17 Fix any odd prime p. Let K be an arbitrary finite extension of F where E[p] remains non-semisimpe as a G K -modue. Assume that (u, 0), (u, 0) = 0 in T K (A)/p for a u, u O K /p. Then P, Q = 0 for a P, Q E(F )/p. Proof of Lemma. Let P = (u P, n P ), Q = (u Q, n Q ) be in E(F )/p. Then by the biinearity and skew-symmetry of.,., the fact that 1, 1 = 0, and aso the hypothesis of the Lemma (appied to K = F ), it suffices to show that (u P, 0), (0, n Q ) = (u Q, 0), (0, n P ) = 0. It suffices to prove the triviaity of the first one, as the argument is identica for the second. Let L denote the unique unramified p-extension of F. Then there exists u O L /p such that 25

26 u P = N L/F (u ). Since (u P, 0), (0, n Q ) equas N L/F ( (u P, 0), Res L/F ((0, n Q )) ) by the projection formua, it suffices to show that (u, 0), Res L/F ((0, n Q )) = 0. Now reca (cf. the discussion around (6.3)) that the restriction map H 1 (O F, Z/p) H 1 (O L, Z/p) is zero. This impies that Res L/F ((0, n Q )) = (u, 0) for some u O L /p. (We cannot caim that this restriction is (0, 0) in E(L)/p because the spitting of the surjection H 1 (O K, E[p]) H 1 (O K, Z/p) is not canonica when E[p] is non-semisimpe over K. This point is what makes this Lemma deicate.) Thus we have ony to check that (u, 0), (u, 0) = 0 in T L (A)/p. This is a consequence of the hypothesis of the Lemma, which we can appy to K = L because E[p] remains non-semisimpe over any unramified extension of F. Done. Step I As in the semisimpe case, there are three steps in the proof. Injectivity on symbos: We sha use the foowing terminoogy. For u O F, write u and ũ for its respective images in O F /p and Y, seen as a quotient of H1 f (S, µ p)/x. We wi aso denote by ũ the corresponding eement in Z = ψ S (Y S ) Hf 1 (S, E[p]), which shoud not cause any confusion as Y is isomorphic to Z. We use simiar notation in H 1 (F, ). For u in O F /p, we denote by P u the eement in E(F )/p corresponding to the pair (ũ, 0) given by (6.16). As in the proof of Proposition D under the diagram (4.4), pick a non-zero eement e X O F /p. By definition, ẽ = 0, so that P e = (ẽ, 0) is the zero eement of E(F )/p. As we have seen in the proof of Proposition D, there exists a v in O F /p such that [e, v] := e v is non-zero in Br F [p]. Now et x, y O F /p be such that s p( P x, P y ) = 0. We have to show that (6.18) P x, P y = 0 T F (A)/p. There are two cases to consider. Case (a) [x, y] = 0 Br F [p]: Then x is a norm from K = F (y 1/p ). This impies, as in the proof in the semisimpe case, that P x is a norm from E(K)/p, and by Lemma 1.8.1, we then have (6.18). Case (b) [x, y] 0: Since Br F [p] Z/p, we may, after modifying by a scaar, assume that [x, y] = [e, v] in Br F [p]. Then by Tate ([Ta1], page 266, conditions (i), (ii)), there are eements a, b O F /p such that Caim 6.19 T F (A)/p. [x, y] = [x, b] = [a, b] = [a, v] = [e, v]. For each pair of neighbors in this sequence, the corresponding symbos are equa in Proof of Caim From [x, y] = [x, b] we have, by inearity, that [x, yb 1 ] = 0, hence x is a norm from L := F ((yb 1 ) 1/p ). Hence P x is a norm from from E(L)/p, and so we get P x, P yb 1 = 0 in T F (A)/p. Consequenty, now by the inearity of,, P x, P y = P x, P b. The remaining assertions of the Caim are proved in exacty the same way. This Caim finishes the proof of Step I since P e, P v = 0, which hods because P e = 0 in E(F )/p. 26