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’1 g
= φ2 φ2 (φ’1 θφ1 )g (P ) (since θg = θ)
2
’1
= (φ2 θφ1 )g (P ) + Tg (by (8.20))
2

= g(φ’1 θφ1 )(g ’1 P ) + Tg
2
(by (8.19))
2
= g(g ’1 P + P0 ) + Tg
2
(by (8.23))
2
= P + gP0 + Tg .




© 2008 by Taylor & Francis Group, LLC
249
SECTION 8.9 GALOIS COHOMOLOGY

Therefore,
1 2
Tg ’ Tg = „φ1 (g) ’ „φ2 (g) = gP0 ’ P0 .
Conversely, suppose there exists P1 such that

„φ1 (g) ’ „φ2 (g) = gP1 ’ P1 . (8.25)

De¬ne θ : C1 ’ C2 by

θ(Q) = φ2 (φ’1 (Q) + P1 ).
1

Clearly, θ satis¬es (8.23). We need to show that θ is de¬ned over Q. If
Q ∈ C(Q), then

θg (Q) = gθ(g ’1 Q) (by (8.19))
= gφ2 φ’1 (g ’1 Q) + P1
1
= φg ((φg )’1 (Q) + gP1 )
2 1
= φ2 (φ2 φ2 )((φg )’1 (Q) + gP1 )
’1 g
1
g ’1 2
= φ2 (φ1 ) (Q) + gP1 + Tg (by (8.24))
= φ2 φ’1 (Q) ’ Tg (g) + gP1 + Tg
1 2
(by (8.22))
1
= φ2 (φ’1 (Q) + P1 ) (by (8.25))
1
= θ(Q).

Therefore, θ is de¬ned over Q, so the pairs (C1 , φ1 ) and (C2 , φ2 ) are equivalent.


Proposition 8.33 says that we have a map

equivalence classes of pairs (C, φ) ’ H 1 (G, E(Q)).

It can be shown that this is a bijection (see [109]). The most important
property for us is the following.


PROPOSITION 8.34
Let „φ correspond to the pair (C, φ). Then „φ ∈ B(G, E(Q)) (= coboundaries)
if and only if C has a rational point (that is, a point with coordinates in Q).


Let P ∈ E(Q). Then
PROOF

gP + Tg = φ’1 φg (gP ) = φ’1 (gφ(P ))

and
P = φ’1 (φ(P )).
Therefore,
Tg = P ’ gP ⇐’ gφ(P ) = φ(P ).




© 2008 by Taylor & Francis Group, LLC
250 CHAPTER 8 ELLIPTIC CURVES OVER Q

If C has a rational point Q, choose P such that φ(P ) = Q. Then gQ = Q for
all g implies that
Tg = g(’P ) ’ (’P )
for all g ∈ G. Conversely, if Tg = g(’P ) ’ (’P ) for all g then gφ(P ) = φ(P )
for all g ∈ G, so φ(P ) is a rational point.

Propositions 8.33 and 8.34 give us a reinterpretation in terms of cohomol-
ogy groups of the fundamental question of when certain curves have rational
points.


Example 8.12
Consider the curve C1,p,p from Section 8.8. It was given by the equations

x = u2
x ’ 2p = pv 2
x + 2p = pw2 .

These were rewritten as

w2 ’ v 2 = 4, u2 ’ pv 2 = 2p.

The method of Section 2.5.4 changes this to

C : s2 = 2p(t4 + 6t2 + 1).

Finally, the transformation

2t2 (x ’ p) x2 + 4px ’ 4p2
2p (x + 2p)

, s = ’ 2p +
t= = 2p
x(x ’ 2p)
y 2p

(use the formulas of Section 2.5.3, plus a minor change of variables) changes
the equation to
E : y 2 = x(x ’ 2p)(x + 2p).
We want to relate the curve C1,p,p from Section 8.8 to a cohomology class in
H 1 (G, E(Q)). The map

φ: E ’C
(x, y) ’ (t, s)

gives a map from E to C. Since the equations for E and C have coe¬cients
in Q, these curves are de¬ned over Q. However, φ is not de¬ned over Q.
A short computation shows that

(x, y) + (’2p, 0) = (x1 , y1 )




© 2008 by Taylor & Francis Group, LLC
251
SECTION 8.9 GALOIS COHOMOLOGY

on E, where
’8p2 y
2p ’ x
x1 = 2p , y1 = .
(x + 2p)2
2p + x
Another calculation shows that

φ(x1 , y1 ) = (’t, ’s).
√ √
Let g ∈ G be such that g( 2p) = ’ 2p. Then φg is the transformation
√ √
obtained by changing 2p to ’ 2p in the formulas for φ. Therefore,

φg (x, y) = (’t, ’s) = φ(x1 , y1 ).

We obtain
φ’1 φg (x, y) = (x, y) + (’2p, 0).


Now suppose g ∈ G satis¬es g 2p = + 2p. Then φg = φ, so

φ’1 φg (x, y) = (x, y).

Putting everything together, we see that the pair (C, φ) is of the type con-
sidered above. We obtain an element of H 1 (G, E[2]) that can be regarded as
an element of H 1 (G, E(Q)). The cocycle „φ is given by
√ √
∞ if g 2p √ + 2p √
=
„φ (g) = Tg =
(’2p, 0) if g 2p = ’ 2p

The cohomology class of „φ is nontrivial in H 1 (G, E(Q)), and hence also in
H 1 (G, E[2]), because C has no rational points. Note that „φ is a homomor-
phism from G to E[2]. This corresponds to the fact that G acts trivially on
E[2] in the present case, so H 1 (G, E[2]) =√
Hom(G, E[2]). The kernel of „ is
the subgroup of G of index 2 that ¬xes Q( 2p).

In general, if E is given by y 2 = (x ’ e1 )(x ’ e2 )(x ’ e3 ) with e1 , e2 , e3 ∈ Q,
then a 2-descent yields curves Ca,b,c , as in Section 8.2. These curves yield
elements of H 1 (G, E[2]). The curves that have rational points give cocycles
in Z(G, E(Q)) that are coboundaries. We also saw in the descent procedure
that a rational point on a curve Ca,b,c comes from a rational point on E. This
discussion is summarized by the exact sequence

0 ’ E(Q)/2E(Q) ’ H 1 (G, E[2]) ’ H 1 (G, E(Q))[2] ’ 0.

All of the preceding applies when Q is replaced by a p-adic ¬eld Qp with
p ¤ ∞. We have an exact sequence

0 ’ E(Qp )/2E(Qp ) ’ H 1 (Gp , E[2]) ’ H 1 (Gp , E(Qp ))[2] ’ 0,

where
Gp = Gal(Qp /Qp ).




© 2008 by Taylor & Francis Group, LLC
252 CHAPTER 8 ELLIPTIC CURVES OVER Q

The group Gp can be regarded as a subgroup of G. Recall that cocycles in
Z(G, E[2]) are maps from G to E[2] with certain properties. Such maps may
be restricted to Gp to obtain elements of Z(Gp , E[2]). A curve Ca,b,c yields an
element of H 1 (G, E[2]). This yields an element of H 1 (Gp , E[2]) that becomes
trivial in H 1 (Gp , E(Qp )) if and only if Ca,b,c has a p-adic point.
In Section 8.7, we de¬ned S2 to be those triples (a, b, c) such that Ca,b,c
has a p-adic point for all p ¤ ∞. This means that S2 is the set of triples
(a, b, c) such that the corresponding cohomology class in H 1 (G, E[2]) becomes
trivial in H 1 (Gp , E(Qp )) for all p ¤ ∞. Moreover, 2 is S2 modulo those
triples coming from points in E(Q). All of this can be expressed in terms
of cohomology. We can also replace 2 by an arbitrary n ≥ 1. De¬ne the
Shafarevich-Tate group to be
⎛ ⎞

= Ker ⎝H 1 (G, E(Q)) ’ H 1 (Gp , E(Qp ))⎠
p¤∞


and de¬ne the n-Selmer group to be
⎛ ⎞

Sn = Ker ⎝H 1 (Gp , E[n]) ’ H 1 (Gp , E(Qp ))⎠ .
p¤∞


The Shafarevich-Tate group can be thought of as consisting of equivalence
classes of pairs (C, φ) such that C has a p-adic point for all p ¤ ∞. This
group is nontrivial if there exists such a C that has no rational points. In
Section 8.8, we gave an example of such a curve. The n-Selmer group Sn can
be regarded as the generalization to n-descents of the curves Ca,b,c that arise
in 2-descents. It is straightforward to use the de¬nitions to deduce the basic
descent sequence

0 ’ E(Q)/nE(Q) ’ Sn ’ [n] ’ 0,

where [n] is the n-torsion in . When one is doing descent, the goal is to
obtain information about E(Q)/nE(Q). However, the calculations take place
in Sn . The group [n] is the obstruction to transferring information back to
E(Q)/nE(Q).
The group Sn depends on n. It is ¬nite (we proved this in the case where n =
2 and E[2] ⊆ E(Q)). The group is independent of n. Its n-torsion [n] is
¬nite since it is the quotient of the ¬nite group Sn . It was conjectured by Tate
and Shafarevich in the early 1960s that is ¬nite; this is still unproved in
general. The ¬rst examples where was proved ¬nite were given by Rubin
in 1986 (for all CM curves over Q with analytic rank 0; see Section 14.2) and
by Kolyvagin in 1987 (for all elliptic curves over Q with analytic rank 0 or 1).
No other examples over Q are known.




© 2008 by Taylor & Francis Group, LLC
253
EXERCISES




Exercises
8.1 Show that each of the following elliptic curves has the stated torsion
group.
y2 = x3 ’ 2; 0
(a)
y2 = x3 + 8; Z2
(b)
y2 = x3 + 4; Z3
(c)
y2 = x3 + 4x; Z4
(d)
y2 = x3 ’ 432x + 8208; Z5
(e)
y2 = x3 + 1; Z6
(f)
y2 = x3 ’ 1323x + 6395814; Z7
(g)
y2 = x3 ’ 44091x + 3304854; Z8
(h)
y2 = x3 ’ 219x + 1654; Z9
(i)
y2 = x3 ’ 58347x + 3954150; Z10
(j)
y2 = x3 ’ 33339627x + 73697852646; Z12
(k)
y2 = x3 ’ x; Z2 • Z2
(l)
y2 = x3 ’ 12987x ’ 263466; Z4 • Z2
(m)
y2 = x3 ’ 24003x + 1296702; Z6 • Z2
(n)
y2 = x3 ’ 1386747x + 368636886; Z8 • Z2
(o)
Parameterizations of elliptic curves with given torsion groups can be
found in [67].
8.2 Let E be an elliptic curve over Q given by an equation of the form
y 2 = x3 + Cx2 + Ax + B, with A, B, C ∈ Z.
(a) Modify the proof of Theorem 8.1 to obtain a homomorphism
»r : Er /E3r ’’ Zp2r
(see [68, pp. 51-52]).
(b) Show that (x, y) ∈ E(Q) is a torsion point, then x, y ∈ Z.
8.3 (a) Show that the map »r , applied to the curve y 2 = x3 , is the map of
Theorem 2.30 divided by pr and reduced mod p4r .
(b) Consider the map »r of Exercise 8.2, applied to the curve E : y 2 =
x3 + ax2 . Let ψ be as in Theorem 2.31. The map »r ψ ’1 gives a
map
y + ±x x
’ p’r (mod p2r ).
y ’ ±x y
Use the Taylor series for log((1 + t)/(1 ’ t)) to show that the map
(2±)»r ψ ’1 is p’r times the logarithm map, reduced mod p2r .




© 2008 by Taylor & Francis Group, LLC
254 CHAPTER 8 ELLIPTIC CURVES OVER Q

8.4 Let E be given by y 2 = x3 + Ax + B with A, B ∈ Z. Let P = (x, y) be
a point on E.

(a) Let 2P = (x2 , y2 ). Show that

y 2 4x2 (3x2 + 4A) ’ 3x2 + 5Ax + 27B = 4A3 + 27B 2 .

(b) Show that if both P and 2P have coordinates in Z, then y 2 divides
4A3 + 27B 2 . This gives another way to ¬nish the proof of the
Lutz-Nagell theorem.

8.5 Let E be the elliptic curve over Q given by y 2 + xy = x3 + x2 ’ 11x.
Show that the point
11 11
,’
P=
4 8
is a point of order 2. This shows that the integrality part of Theorem 8.7
(see also Exercise 8.2), which is stated for Weierstrass equations, does
not hold for generalized Weierstrass equations. However, since changing
from generalized Weierstrass form to the form in Exercise 8.2 a¬ects
only powers of 2 in the denominators, only the prime 2 can occur in the
denominators of torsion points in generalized Weierstrass form.

8.6 Show that the Mordell-Weil group E(Q) of the elliptic curve y 2 = x3 ’x
is isomorphic to Z2 • Z2 .

8.7 Suppose E(Q) is generated by one point Q of in¬nite order. Suppose
we take R1 = 3Q, which generates E(Q)/2E(Q). Show that the process
with P0 = Q and
Pi = Rji + 2Pi+1 ,
as in Section 8.3, never terminates. This shows that a set of represen-
tatives of E(Q)/2E(Q) does not necessarily generate E(Q).

8.8 Show that there is a set of representatives of E(Q)/2E(Q) that gener-
ates E(Q). (Hint: This mostly follows from the Mordell-Weil theorem.
However, it does not handle the odd order torsion. Use Corollary 3.13
to show that the odd order torsion in E(Q) is cyclic. In the set of rep-
resentatives, use a generator of this cyclic group for the representative
of the trivial coset.)

8.9 Let E be an elliptic curve de¬ned over Q and let n be a positive integer.
Assume that E[n] ⊆ E(Q). Let P ∈ E(Q) and let Q ∈ E(Q) be such
that nQ = P . De¬ne a map δP : Gal(Q/Q) ’ E[n] by δP (σ) = σQ’Q.

(a) Let σ ∈ Gal(Q/Q). Show that σQ ’ Q ∈ E[n].
(b) Show that δP is a cocycle in Z(G, E[n]).




© 2008 by Taylor & Francis Group, LLC
255
EXERCISES

(c) Suppose we choose Q with nQ = P , and thus obtain a cocycle
δP . Show that δP ’ δP is a coboundary.
(d) Suppose that δP (σ) is a coboundary. Show that there exists Q ∈
E(Q) such that nQ = P .

This shows that we have an injection E(Q)/nE(Q) ’ H 1 (G, E[n]).
This is the map of Equation 8.18.




© 2008 by Taylor & Francis Group, LLC
© 2008 by Taylor & Francis Group, LLC

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