9.8
The formula for rank(E)
Since we now know that E is finitely-generated, it follows from the Structure
Theorem for Finitely Generated Abelian Groups that
E = Z ⊕ · · · ⊕ Z ⊕ Z/(pe11 ) ⊕ · · · Z/(pess ),
where there are r = rank(E) copies of Z.
Proposition 9.6 Let


0 if there are 0 points of order 2 on E,
d = 1 if there is 1 point of order 2 on E,


2 if there are 3 points of order 2 on E.
Then
kE/2Ek = 2s ,
where
s = r + d.
Proof I If
A = A1 ⊕ · · · ⊕ Am
then
2A = 2A1 ⊕ · · · ⊕ 2Am
and so
A/2A = A1 /2A1 ⊕ · · · ⊕ Am /2Am .
Thus it is sufficient to consider the factors of E.
Evidently the r copies of Z will give rise to r copies of Z/(2).
Lemma If A = Z/(2e ) then
A/2A = Z/(2).
Proof of Lemma B Let g be a generator of A, so that
A = {0, g, 2g, . . . , (2e − 1)g}
Then
2A = {0, 2g, 4g, . . . , (2e − 2)g}.
Thus half the elements of A are in 2A, and so A/2A is of order 2, ie A/2A =
Z/(2). C
428–99 9–27
Lemma If A = Z/(pe ), where p is odd,then
A/2A = 0.
Proof of Lemma B Consider the map
θ : A → A : a 7→ 2a.
Then
ker θ = {a ∈ A : 2a = 0} = 0,
since by Lagrange’s Theorem there are no elements of order 2 in A. Hence θ
is injective, and so surjective, ie 2A = A, and A/2A = 0. C
From the two Lemmas it follows that the number of copies of Z/(2) in
E/2E = Z/(2) + · · · + Z/(2)
is equal to r + f , where f is the number of factors of the form Z/(2e ). It
remains to show that f = d.
Lemma The number of elements of order 2 in A is 2f − 1, where f is the
number of factors of the form Z/(2e ).
Proof of Lemma B An element of a direct sum
A = A 1 ⊕ A2 ⊕ · · · ⊕ Am
is of order 1 or 2 if and only if that is true of each component:
2(a1 , a2 , . . . , am ) = 0 ⇐⇒ 2a1 = 0, 2a2 = 0, . . . , 2am = 0.
But there is no element of order 2 in Z/(pe ) if p is odd, by Lagrange’s
Theorem; while there is just one element of order 2 in Z/(2e ), nameley
2e−1 mod 2e .
Thus we have two choices in each factor Z/(2e ), and one choice in each
factor Z/(pe ) (p odd).
It follows that the number of elements of order 1 or 2 is 2f where f is
the number of factors of the form Z/(2e ); and so the number of elements of
order 2 is 2f − 1. C
J
428–99 9–28
9.9
The square-free part
Each rational x ∈ Q× is uniquely expressible in the form
x = dy 2 ,
where y ∈ Q× and d is a square-free integer. Explicitly, if
x = ±2e2 3e3 5e5 · · ·
then
x = ±22 33 55 · · ·
where each p ∈ {0, 1} is given by
p ≡ ep mod 2.
For example,
x = 2/3 7→ d = 6,
x = −3/4 7→ −3.
We may call d the square-free part of x.
Thus each x̄ ∈ Q× /Q×2 is represented by a unique square-free integer d,
establishing an isomorphism
Q× /Q×2 ←→ D,
where D is the group formed by the square-free integers under multiplication
modulo squares, eg
2 · 6 = 3, −3 · 6 = −2.
Let us see how to use this to compute the rank. Recall that
E/E 2 ∼
= im Θ
where
Θ = θα × θβ × θγ ,
with θα , for example, given by
(
x − α if x 6= α
P = (x, y) 7→
p0 (α) if x = α
If P = (x, y) is on the elliptic curve
E(Q) : y 2 = x3 + ax2 + bx + c
then
(a, b, c ∈ Z)
M
m
,
y
=
t2
t3
where m, M, t ∈ Z with gcd(m, t) = 1 = gcd(M, t) and t > 0.
x=
428–99 9–29
9.10
An example
Consider the elliptic curve
E(Q) : y 2 = x3 − x = x(x − 1)(x + 1).
Here
α = 0, β = 1, γ = −1,
so that
p0 (0) = (0−1)(0+1) = −1,
p0 (1) = (1−0)(1+1) = 2,
p0 (−1) = (−1−0)(−1−1) = 2.
Thus, from above,
im Θ ⊂ S = {(d, e, f ) : d | 1, e | 2, f | 2}.
This gives 32 choices:
d = ±1,
e = ±1, ±2,
f = ±1, ±2.
It follows (since 32 = 25 ) that
kE/2Ek ≤ 5,
and so
rank E ≤ 3.
However, we can restrict the range of im Θ much more than this. In the
first place, since
x(x − 1)(x + 1) = y 2 ,
it follows that def is a perfect square, say
def = g 2 .
This implies firstly that def > 0, and secondly that each prime p dividing
any of d, e, f must in fact divide just two of them. This reduces the number
of cases to 8:
(d, e, f ) = (1, 1, 1), (1, −1, −1), (−1, 1, −1), (−1, −1, 1), (1, 2, 2), (1, −2, −2), (−1, 2, −2), (−1, −
We can reduce the number still further by observing that since
m = du2 , m − t2 = ev 2 , m + t2 = f w2 ,
428–99 9–30
it follows that
d < 0 =⇒ m < 0 =⇒ m − t2 < 0 =⇒ e < 0,
while
d > 0 =⇒ m > 0 =⇒ m + t2 > 0 =⇒ f > 0.
This leaves just 4 choices for d, e, f :
(d, e, f ) = (1, 1, 1), (−1, −1, 1), (1, 2, 2), (−1, −2, 2).
Thus
kE/2Ek ≤ 4
Since d = 2 (as there are 3 points of order 2),
kE/2Ek = 2r+d ≥ 4.
We conclude that
rank E = 0.
9.11
Another example
Now let us consider the elliptic curve
y 2 = x3 − x = x(x − 2)(x + 2).
Here
p0 (0) = −4, p0 (2) = 8, p0 (−2) = 8,
and so
E/2E = im Θ ⊂ {(d, e, f ) : d, e, f | 2}
The group on the right contains 26 elements, since each of d, e, f can take
the values ±1, ±2.
But as before, the condition
def = g 2
restricts the choice considerably. Firstly,
d < 0 =⇒ e < 0. d > 0 =⇒ f > 0.
428–99 9–31
Secondly, the factor 2 occurs in none, or just two, of d, e, f . This reduces the
choice to
(d, e, f ) = (1, 1, 1), (−1, −1, 1), (1, 2, 2), (−1, −2, 2), (2, 1, 2), (−2, −1, 2), (2, 2, 1), (−2, −2, 1).
Thus the rank is either 0 or 1. Can we reduce the choice further, and
reduce the rank to 0? or conversely, can we find a point of infinite order on
the curve, and so show that the rank is 1?
Note that it only necessary to eliminate one case; for we know that
kE/2Ek = 2s ≥ 4, since there are 3 points of order 2 (and so d = 2).
Suppose
(d, e, f ) = (−1, −1, 1).
In this case,
m = −u2 , m − 2t2 = −v 2 , m + 2t2 = w2 .
Thus
u2 − v 2 = 2t2 = u2 + w2 .
Now a2 ≡ 0 or 1 mod 4 according as a is even or odd. Since u2 − v 2 is even
it followu, v are both even or both odd; and in either case u2 − v 2 ≡ 0 mod 4.
So t is even, and therefore u, v must both be odd, since gcd(m, t) = 1 =
gcd(m − 2t2 , t).
9.12
Third example
Consider the elliptic curve
E(Q) : y 2 = x(x − 2)(x + 4) = x3 + 2x2 − 8x.
The point
P = (−1, 3) ∈ E.
(We chose α, β, γ to give this result.)
The slope at P is
dx
3x2 + 4x − 8
=
dy
2y
3
=−
2
at P . It follows that P is of infinite order (since 2P has non-integral coordinates). Thus
r = rank(E) ≥ 1.
428–99 9–32
We have
p0 (0) = −8,
p0 (2) = 12,
p0 (−4) = 24.
Thus
im Θ ⊂ S{(d, e, f ) : d | 2, e | 6, f | 6; def = g 2 }.
Note that any two of d, e, f determine the third since eg f = de (modulo
squares).
If
P = (m/t2 , M/t3 ) 7→ (d, e, f )
then
m = du2 ,
m − 2t2 = ev 2 ,
m + 4t2 = f w2 .
Thus
d > 0 =⇒ m > 0 =⇒ f > 0 =⇒ e > 0,
while
d < 0 =⇒ m < 0 =⇒ e < 0 =⇒ f > 0.
(So f > 0 in all cases.)
It follows that
kSk = 16,
with
S = {d = ±1, ±2, f = 1, 2, 3, 6}.
It follows that s ≤ 4, and so
rank(E) = s − d = s − 2 ≤ 2.
Thus rank(E) = 1 or 2.
In order to prove that rank(E) = 1 it is sufficient to show that one of the
16 elements of S does not lie in im Θ. For kSk is a power of 2, so if it is < 16
it must be ≤ 8.
Let us take the element (−1, −1, 1). Suppose this arises from a point
P = (m/t2 , M/t3 ), where for the moment we assume that P is not of order
2. Then
m = −u2 , m − 2t2 = −v 2 , m + 4t2 = w2 .
Thus
2t2 = v 2 − u2 ,
4t2 = u2 + w2 .
From the second equation,
u2 + w2 ≡ 0 mod 4 =⇒ u, w even,
428–99 9–33
since a2 ≡ 0 or 1 mod 4 according as a is even or odd. It follows that t is
odd, since
gcd(m, t) = 1 =⇒ gcd(u, t) = 1.
But then t2 ≡ 1 mod 4, and so
v 2 − u2 ≡ 2 mod 4,
which is impossible.
(Alternatively, adding the two equations,
6t2 = v 2 + w2 .
Thus
v 2 + w2 ≡ 0 mod 3 =⇒ v ≡ w ≡ 0 mod 3
=⇒ t ≡ 0 mod 3
=⇒ u ≡ 0 mod 3,
contradicting gcd(m, t) = 1.)
9.13
Final example
The elliptic curve
E(Q) : y 2 = x(x + 1)(x − 14) = x3 − 13x2 − 14x
is more complicated, but the method is the same.
We have
p0 (0) = −14,
p0 (−1) = 15,
p0 (14) = 14 · 15.
Thus
im Θ ⊂ S = {(d, e, f ) : d | 14, e | 15, f | 14 · 15; def = g 2 }.
if P = (m/t2 , M/t3 ) 7→ (d, e, f ) (M 6= 0) then
m = du2 ,
m + t2 = ev 2 ,
m − 14t2 = f w2 .
In particular,
d > 0 =⇒ e > 0 =⇒ f > 0
428–99 9–34
while
d < 0 =⇒ f < 0 =⇒ e > 0
(giving e > 0 in all cases).
We have
d = ±1, ±2, ±7, ±14,
e = 1, 3, 5, 15.
Thus
kSk = 25 =⇒ s ≤ 5 =⇒ r ≤ 3.
The elements of order 2 give rise to the points
(0, 0) 7→ (p0 (0), 1, −14) = (−14, 1, −14),
(−1, 0) 7→ (−1, p0 (−1), −15) = (−1, 15, −15),
(14, 0) 7→ (14, 15, p0 (14)) = (14, 15, 14 · 15),
while of course
0 = [0, 1, 0] 7→ (1, 1, 1).
Thus the torsion group gives rise the subgroup
D = {(1, 1, 1), (−14, 1, −14), (−1, 15, −15), (14, 15, 14 · 15).
We can regard S as a 5-dimensional vector space over F2 , with 5 coordinates defined by: the sign of d, the factor 2 in d, the factor 7 in d, the factor
3 in e, the factor 5 in e. Thus
(0, 0) 7→ (−14, 1, −14) ←→ (1, 1, 1, 0, 0),
(−1, 0) 7→ (−1, 15, −15) ←→ (1, 0, 0, 1, 1),
(14, 0) 7→ (14, 15, 14 · 15) ←→ (0, 1, 1, 1, 1).
Our aim is to prove that rank(E) = 0 by showing that im Θ = D. At
first sight one might think we would have to apply our congruence technique
to 25 − 22 = 28 cases. However, we can simplify the task by choosing a
complementary subspace to D – that is, a subspace of U ⊂ S of dimension 3
such that
U ∩ D = 0,
in which case
S = D ⊕ U.
If now we can show that no elements of U except for (1, 1, 1) are in im Θ then
it will follow that
S = im Θ ⊕ U ;
428–99 9–35
whence
dim im Θ = dim D =⇒ im Θ = D.
For our subspace U let us take those vectors with 3rd and 5th components
0, ie
U = {(d, e, f ) ∈ S : d = ±1, ±2, e = 1, 3}.
We see at once that U ∩D = {(1, 1, 1)} (the zero element of our vector space),
so U is — as required — complementary to D. It is sufficient therefore to
show that no element of U apart from (1, 1, 1) can be in im Θ. (This reduces
the number of cases to be considered from 28 to 7.)
1. (−1, 1, −1): in this case
m = −u2 ,
m + t2 = v 2 ,
m − 14t2 = −w2 ,
ie
t2 = u2 + v 2 ,
14t2 = w2 − u2 .
From the second equation t must be even, since otherwise w2 − u2 ≡
2 mod 4, which is impossible.
But then from the first equation, u2 + v 2 ≡ 0 mod 4, which implies that
u, v are both even, contradicting gcd(u, t) = 1.
2. (2, 1, 2): in this case
m = 2u2 ,
m + t2 = v 2 ,
m − 14t2 = 2w2 ,
ie
t2 = v 2 − 2u2 ,
7t2 = u2 − w2 .
From the second equation t must be even, since otherwise 7t2 ≡ 3 mod
4, and u2 − w2 cannot be ≡ 3 mod 4.
But then from the first equation, v is even and so u is even, contradicting gcd(u, t) = 1.
3. (−2, 1, −2): in this case
m = −2u2 ,
m + t2 = v 2 ,
428–99 9–36
m − 14t2 = −2w2 ,
ie
t2 = v 2 + 2u2 ,
7t2 = w2 − u2 .
As in the last case, from the second equation t must be even, and then
from the first equation, so must v and u, contradicting gcd(u, t) = 1.
4. (1, 3, 3): in this case
m = u2 ,
m + t2 = 3v 2 ,
m − 14t2 = 3w2 ,
ie
t2 = 3v 2 − u2 ,
14t2 = u2 − 3w2 .
From the second equation
u2 − 3w2 ≡ 0 mod 7.
Since 3 is a quadratic non-residue mod7, it follows that u ≡ w ≡
0 mod 7, which implies (by the second equation) that 7 | t, so again
gcd(t, u) > 1.
5. (−1, 3, −3): in this case
m = −u2 ,
m + t2 = 3v 2 ,
m − 14t2 = −3w2 ,
ie
t2 = 3v 2 + u2 ,
14t2 = 3w2 − u2 .
As in the last case, since 3 is not a quadratic residue mod 7, the second
equation implies that 7 | u, w, t, contradicting gcd(u, t) = 1.
6. (2, 3, 6): in this case
m = 2u2 ,
m + t2 = 3v 2 ,
m − 14t2 = 6w2 ,
ie
t2 = 3v 2 − 2u2 ,
14t2 = u2 − 6w2 .
Again, since 6 is not a quadratic residue mod7, this leads to a contradiction.
428–99 9–37
7. (−2, 3, −6): in this case
m = −2u2 ,
m + t2 = 3v 2 ,
m − 14t2 = −6w2 ,
ie
t2 = 3v 2 + 2u2 ,
7t2 = u2 − 3w2 .
Since 3 is not a quadratic residue mod7, this again leads to a contradiction.
We conclude that
im Θ = D,
ie
rank E = 0.
428–99 9–38