10.11
Computing the rank — II
Recall that we associate to the elliptic curve
E : y 2 = x3 + ax2 + bx
a second elliptic curve
E1 : y 2 = x3 + a1 x2 + b1 x,
where
a1 = −2a, b1 = a2 − 4b.
The map E → E : P 7→ 2P factorises into two homomorphisms
Θ : E → E1 , Φ : E1 → E,
defined by
2
2
x + ax + b x2 − b
x1 + a1 x1 + b1 x21 − b1
Θ(x, y) =
,
y , Φ(x1 , y1 ) =
,
y1 ,
x
x2
4x1
8x21
except that in each case the point (0, 0) of order 2 maps to 0. (Thus each
homomorphism has kernel {0, (0, 0)}, since every affine point apart from (0, 0)
maps to an affine point.)
It follows (by a little elementary group theory) that
[E : 2E] = [E : im Φ] [im Φ : im ΦΘ]
[E : im Φ] [E1 : im Θ]
=
[ker Φ : ker Φ ∩ im Θ]
Our basic Lemma (corresponding to Mordell’s Lemma in the earlier approach) states that P1 = (x1 , y1 ) ∈ E1 lies in im Θ if and only if x1 is a perfect
square; and similarly P = (x, y) ∈ E lies in im Φ if and only if x is a perfect
square.
Thus if we introduce the auxiliary homomorphisms
χ : E → Q× /Q×2 ,
χ1 : E1 → Q× /Q×2
defined by
χ(x, y) = x̄ (x 6= 0), χ(0, 0) = b̄
χ1 (x1 , y1 ) = x1 (x1 6= 0), χ1 (0, 0) = b¯1 .
428–99 10–26
then
im Θ = ker χ1 ,
im Φ = ker χ.
It follows that
[E : 2E] =
where
k im χk k im χ1 k
,
e
(
1 if b1 is a perfect square,
e=
2 otherwise.
Since r = rank E is given by
2r+d = [E : 2E],
where d = 1 or 2 according as x3 + ax2 + bx has 1 rational root or 3, the rank
is completely determined once we know k im χk and k im χ1 k.
Recall that if
P = (x, y) ∈ E : y 2 = x3 + ax2 + bx + c,
where a, b, c ∈ Z then x, y take the forms
x=
m
M
, y= 3,
2
t
t
with gcd(m, t) = 1 = gcd(M, t).
As in the earlier method, we represent each rational x ∈ Q× /Q×2 by its
square-free part d. Thus if
m = du2
where d is square-free then we may take d as the representative of x̄ ∈
Q× /Q×2 .
Proposition 10.15 Suppose
E(Q) : y 2 = x3 + ax2 + bx
is an elliptic curve with a, b ∈ Z. If d ∈ im χ (where d is square-free) then
d | b. Moreover, if b = dd0 then d ∈ im χ if and only if there exist u, v, t with
gcd(u, t) = 1 = gcd(v, t) such that
du4 + au2 t2 + d0 t4 = v 2 .
Conversely, any solution u, v, t of this equation with gcd(u, t) = 1 arises in
this way from a point on E.
428–99 10–27
Proof I Suppose
P =
Then
M2
du2
=
t6
t2
du2 M
,
t2 t3
∈ E.
d2 u4
du2
+a 2 +b .
t4
t
Thus
M 2 = du2 (d2 u4 + adu2 t2 + bt4 )
= d2 u2 (du4 + au2 t2 + d0 t4 ).
. It follows that du4 + au2 t2 + d0 t4 is a perfect square, say
du4 + au2 t2 + d0 t4 = v 2 .
Conversely, if u, v satisfy this equation then
2
du duv
P =
, 3
∈ E.
t2
t
Finally, gcd(v, t) = 1, since
p | v, t =⇒ p2 | du2 =⇒ p | u,
contradicting gcd(u, t) = 1.
10.12
J
Example
Consider the elliptic curve
y 2 = x3 + 1.
over the rationals. There is one point of order 2 on the curve, namely D =
(−1, 0).
(The point P = (2, 3) is also on the curve. Since
dy
3x2
=
dx
2y
12
=
=2
6
at this point, the tangent at P cuts E again at (X, Y ), where
2 + 2 + X = 22 ,
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ie
X = 0.
It follows that 2P = −D = D, so that P is of order 4.)
The transformation x0 = x + 1, ie x = x0 − 1 (taking the point of order 2
to (0, 0)) brings the curve to our preferred form
E : x3 − 3x2 + 3x
Thus
a = −3, b = 3,
and so
a1 = 6, b1 = −3,
ie the associated curve is
E1 : y 2 = x3 + 6x2 − 3x.
Since there is just one point of order 2 on E, and b1 is not a perfect square,
2r+1 =
k im χk k im χ1 k
,
2
We start by computing k im χk. Since d | 3,
im χ ⊂ {±1, ±3}.
Since (0, 0) 7→ 3,
im χ = {1, 3} or {±1, ±3}.
Suppose d = −1. Then d0 = −3, and we are looking for solutions of
−u4 − 3u2 t2 − 3t4 = v 2 .
Since the left-hand side is negative while the right-hand side is positive, there
is no such solution. Hence
im χ = {1, 3}.
Turning to im χ1 , we again have d | 3, and so
im χ1 ⊂ {±1, ±3}.
But now (0, 0) 7→ −3. Thus
im χ = {1, −3} or {±1, ±3}.
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Again, consider d = −1. Now d0 = 3, and we are looking for solutions of
−u4 + 6u2 t2 + 3t4 = v 2 .
This implies that
−u4 ≡ v 2 mod 3.
and therefore
3 | u, v
since the quadratic residues mod3 are {0, 1}. But then
32 | u4 , u2 t2 , v 2 =⇒ 32 | 3t4
=⇒ 3 | t,
contradicting the condition gcd(u, t) = 1.
We conclude that
im χ1 = {1, −3}.
Hence
2r+1 =
2·2
,
2
ie
rank E = r = 0.
10.13
Another example
Let us re-visit the curve
E : y 2 = x3 − x,
which we already saw has rank 0 (in the last chapter).
The associated curve is
E1 : y 2 = x3 + 4x,
Since b1 = 4 is a perfect square, while the original equation has three
points of order 2,
2r+2 = k im χk k im χ1 k.
If d ∈ im χ then d | b = −1. Thus
im χ ⊂ {±1}.
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In fact, since (0, 0) 7→ −1,
im χ = {±1}.
Turning to im χ1 , since d | 4 =⇒ d | 2 (as d is square-free),
im χ1 ⊂ {±1, ±2}.
We observe that (2, 4) ∈ E1 . Thus 2 ∈ im χ1 , and so
im χ1 = {1, 2} or {±1, ±2}.
Suppose d = −1. Then d0 = −4, and we are looking for solutions of
−u4 − 4t4 = v 2 ,
which is impossible, since the left-hand side is negative, while the right-hand
side positive. Thus
im χ1 = {1, 2}.
We conclude that
2r+2 = 2 · 2,
whence
rank E = r = 0.
10.14
A third example
Finally, let us look again at the curve
E(Q) : y 2 = x(x − 2)(x + 4) = x3 + 2x2 − 8x,
which we already saw (in the last Chapter) has rank 1, with the point P =
(−1, 3) having infinite order.
Since
a1 = −2a = −4, b1 = a2 − 4b = 36,
the associated curve is
E1 : y 2 = x3 − 4x2 + 36x.
Since b1 = 36 is a perfect square,
2r+2 = k im χk k im χ1 k.
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If d ∈ im χ then d | −8. Thus
im χ ⊂ {±1, ±2}.
Since (2, 0) 7→ 2, while (−4, 0) 7→ −1, we deduce that
im χ = {±1, ±2}.
Turning to im χ1 , we have d | 36. Thus
im χ1 ⊂ {±1, ±2, ±3, ±6}.
The point (0, 0) 7→ 1 (since 36 ≡ 1 modulo squares), which is not much help.
Consider d = −1. In this case d0 = −36, and we have to solve the equation
−u4 − 4u2 t2 − 36t4 = v 2 .
Since the left-hand side is < 0, we conclude that −1 ∈
/ im χ1 .
In fact, any d < 0 will lead to a contradiction in the same way. We
conclude that
im χ1 ⊂ {1, 2, 3, 6}.
Suppose d = 3. Then d0 = 12, and the equation reads
3u4 − 4u2 t2 + 12t4 = v 2 .
But this implies that
−u2 t2 ≡ v 2 mod 3.
Thus 3 | v and 3 | u or t. But
3 | u, v =⇒ 32 | 12t4 =⇒ 3 | t
while
3 | v, t =⇒ 32 | 3u4 =⇒ 3 | u,
and in either case gcd(u, t) > 1, contrary to assumption.
We conclude that 3 ∈
/ im χ1 ; and therefore
2r+2 ≤
4·2
r ≤ 1.
=⇒
However, we recall that the point (−1, 3) ∈ E is of infinite order, and so
rank E = 1.
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