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f(ÃÄ) = f(Ã) · Ã(f(Ä)), Ã, Ä " G,
and we have to find a ³ " L× such that f(Ã) = ó/³ for all à " G. Because the f(Ä) are
nonzero, Dedekind s theorem on the independence of characters (see Math 594) implies that
f(Ä)Ä : L ’! L
is not the zero map, i.e., that there exists an ± " L such that
² = f(Ä)ı = 0.
Ä"G
But then, for à " G,
ò = Ã(f(Ä)) · ÃÄ(±) = f(Ã)-1 · f(ÃÄ) · ÃÄ(±) = f(Ã)-1 f(ÃÄ)ÃÄ(±) = f(Ã)-1²,
Ä"G Ä "G Ä "G
which shows that f(Ã) = ²/ò = Ã(²-1)/²-1.
Proposition 12.4. For any exact sequence of G-modules
0 ’! M ’! N ’! P ’! 0,
there is a canonical exact sequence
´
0 ’! H0(G, M) ’! H0(G, N) ’! H0(G, P ) - H1(G, M) ’! H1(G, N) ’! H1(G, P )
’!
ELLIPTIC CURVES 57
G
Proof. The map ´ is defined as follows. Let p " P . There exists an n " N mapping
to p, and Ãn - n " M for all à " G. The map à ’! Ãn - n : G ’! M is a crossed
homomorphism, whose class we define to be ´(p). Another n mapping to p gives rise to a
crossed homomorphism differing from the first by a principal crossed homomorphism, and
so ´(p) is well-defined. The rest of the proof is routine.
Let H be a subgroup of G. The restriction map f ’! f|H defines a homomorphism
Res : H1(G, M) ’! H1(H, M).
Proposition 12.5. If G has order m, then m kills H1(G, M).
Proof. In general, if H is a subgroup of G of index m, then there exists a homomorphism
Cor : Hi(H, M) ’! Hi(G, M) such that the composite Res æ% Cor is multiplication by m. The
proposition is proved by taking H = 1.
Remark 12.6. Let H be a normal subgroup of a group G, and let M be a G-module. Then
MH is a G/H-module, and a crossed homomorphism f : G/H ’! MH defines a crossed
homomorphism G ’! M by composition:
G · · · ’! M
“! *"
f
G/H - ’! MH.
-
In this way we obtain an  inflation homomorphism
Inf : H1(G/H, MH) ’! H1(G, M),
and one verifies easily that the sequence
Inf Res
0 ’! H1(G/H, MH) - H1(G, M) -’! H1(H, M)
’! -
is exact.
Cohomology of infinite Galois groups. Let k be a perfect field, and let kal be an algebraic
closure of k. The automorphisms of kal fixing the elements of k form a group G, which
when endowed with the topology for which the open subgroups are those fixing some finite
extension of k, is called the Galois group of kal over k. The group G is compact, and so any
open subgroup of G is of finite index. Infinite Galois theory says that the intermediate fields
K, k ‚" K ‚" kal, are in natural one-to-one correspondence with the closed subgroups of
G. Under the correspondence intermediate fields of finite degree over k correspond to open
subgroups of G.
A G-module M is said to be discrete if the map G × M ’! M is continuous when M
is given the discrete topology and G is given its natural topology. This is equivalent to
requiring that
M = *"HMH, H open in G,
i.e., to requiring that every element of M is fixed by the subgroup of G fixing some finite
extension of k. For example, M = kal, M = kal×, and M = E(kal) are all discrete G-modules
because
kal = *"K, kal = *"K×, and E(kal) = *"E(K)
where, in each case, the union runs over the finite extensions K of k contained in kal.
58 J.S. MILNE
For an infinite Galois group G, we define H1(G, M) to be the group of continuous crossed
homomorphisms f : G ’! M modulo the subgroup of principal crossed homomorphisms.
With this definition
H1(G, M) = lim H1(G/H, MH)
H
-
’!
where H runs through the open normal subgroups of G. Explicitly, this means that:
(a) H1(G, M) is the union of the images of the inflation maps Inf : H1(G/H, MH) ’!
H1(G, M), H an open normal subgroup of G;
(b) an element ³ " H1(G/H, MH) maps to zero in H1(G, M) if and only if it maps to
zero H1(G/H , MH ) for some open normal subgroup H of G contained in H.
In particular, the group H1(G, M) is torsion.
Example 12.7. (a) Proposition 12.3 shows that
H1(G, kal×) = lim H1(Gal(K/k), K×) = 0.
H
-
’!
(b) For a field L, let µn(L) = {¶ " L× | ¶n = 1}. From the exact sequence
n
1 ’! µn(kal) ’! kal× - kal× ’! 1
’!
we obtain an exact sequence of cohomology groups
n
1 ’! µn(k) ’! k× - k× ’! H1(G, µn(kal) ’! 1,
’!
H"
and hence a canonical isomorphism H1(G, µn(kal)) - k×/k×n. Note that for k = Q,
’!
Q×/Q×n is infinite if n > 1. For example, the numbers
(-1)µ(") pµ(p),
p prime
where µ(p) = 0 or 1 and all but finitely many are zero, form a set of representatives for the
elements of Q×/Q×2, which is therefore an infinite-dimensional vector space over F2.
(c) If G acts trivially on M, then H1(G, M) is the set of continuous homomorphisms
G ’! M. This set can be identified with the set of pairs (K, ±) consisting of a finite Galois
extension K of k contained in kal and an injective homomorphism ± : Gal(K/k) ’! M.
For an elliptic curve E over k, we abbreviate Hi(Gal(kal/k), E(kal)) to Hi(k, E).
Now consider an elliptic curve E over Q. Let Qal be the algebraic closure of Q in C, and
choose an algebraic closure Qal for Qp. The embedding Q ’! Qp extends to an embedding
p
Qal ’! Qal,
p
Qal ’! Qal
p
‘! ‘!
Q ’! Qp.
The action of Gal(Qal/Qp) on Qal ‚" Qal defines an inclusion
p p
Gal(Qal/Qp) ’! Gal(Qal/Q).
p
Hence any crossed homomorphism
Gal(Qal/Q) ’! E(Qal)
induces (by composition) a crossed homomorphism
Gal(Qal/Q) ’! E(Qal).
p p
ELLIPTIC CURVES 59
In this way, we obtain a homomorphism
H1(Q, E) ’! H1(Qp, E)
that (slightly surprisingly) is independent of the choice of the embedding Qal ’! Qal. A
p
similar remark applies to the cohomology groups of µn and En. Later we ll give a more
natural interpretation of these  localization homomorphisms.
13. The Selmer and Tate-Shafarevich groups
Lemma 13.1. For any elliptic curve E over an algebraically closed field k and any integer
n, the map n : E(k) ’! E(k) is surjective.
Proof. The simplest proof uses algebraic geometry. The map of varieties n : E ’! E has
finite fibres (because E(k)n is finite and it is a homomorphism) and has Zariski-closed image
(because E is complete) of dimension one (because its fibres have dimension 0). Hence it is
surjective as a morphism of algebraic varieties.
Alternatively, let P = (x : y : 1) " E(k). To find a point Q = (x : y : 1) such that nQ = P
one has to solve a pair of polynomial equations in the variables X, Y . In characteristic zero,
these equations can t be inconsistent, because n : E(C) ’! E(C) is surjective, and so, by the
Hilbert Nullstellensatz, they have a solution in k. In characteristic p one has to work a little
harder.
From the lemma we obtain an exact sequence
n
0 ’! En(Qal) ’! E(Qal) - E(Qal) ’! 0
’!
and a cohomology sequence
n n
0 ’! En(Q) ’! E(Q) - E(Q) ’! H1(Q, En) ’! H1(Q, E) - H1(Q, E),
’! ’!
from which we extract the sequence
0 ’! E(Q)/nE(Q) ’! H1(Q, En) ’! H1(Q, E)n ’! 0.
Here, as usual, H1(Q, E)n is the group of elements in H1(Q, E)n killed by n. If H1(Q, En)
were finite, then we could deduce that E(Q)/nE(Q) is finite, but unfortunately, it isn t.
Instead, we proceed as follows. When we consider E as an elliptic curve over Qp we obtain
a similar exact sequence, and there is a commutative diagram:
0 ’! E(Q)/nE(Q) ’! H1(Q, En) ’! H1(Q, E)n ’! 0
“! “! “!
0 ’! E(Qp)/nE(Qp) ’! H1(Qp, En) ’! H1(Qp, E)n ’! 0.
We want replace H1(Q, En) by a subset that contains the image of E(Q)/nE(Q) but which [ Pobierz całość w formacie PDF ]