Course 428 Part III — Sample Paper
Timothy Murphy
Length of exam: 1 hour
Attempt 3 questions. (If you attempt more, only the best 3 will
be counted.) All questions carry the same number of marks.
1. State (without proof) Hasse’s Theorem on the number of points on an
elliptic curve over the finite field Fq .
Determine the number of points on the curve
E(F9 ) : y 2 = x3 + x
over the field F9 . What is the group on this curve?
Answer:
(a) Hasse’s Theorem: Suppose E(Fq ) is an elliptic curve over the finite
field Fq . If there are N points on the curve then
√
|N − (q + 1)| ≤ 2 q.
(b) The curve
E(F9 ) : y 2 = x3 + x
is elliptic, since
∆ = −4 · 13 6= 0.
To construct F9 = F32 we need an irreducible polynomial p(x) of
degree 2 over F3 . We take
p(x) = x2 + 1.
This is irreducible over F3 since −1 is a quadratic non-residue
mod3.
Now
F9 = F3 [x]/(x2 + 1).
If we set α = x mod (x2 + 1) then the 9 elements of F9 are
c0 + c1 α,
where c0 , c1 ∈ {0, ±1}; and we multiply these elements using α2 =
−1.
Although not essential, it saves time if we use the Frobenius automorphism
φ : x 7→ x3 : F9 → F9 .
Since α3 = −α it follows that
φ(c0 + c1 α) = c0 − c1 α.
Just half, ie four, of the elements of F×
9 are squares, since the
homomorphism
×
x 7→ x2 : F×
9 → F9
has kernel {±1}.
We have
(±1)2 = 1, ±α)2 = −1;
while
(1 ± α)2 = ±2α = ∓α.
Thus
2
F× = {±1, ±α}.
To find the elements in E(F9 ) we run through the 9 values of x,
and see for which x3 + x is a square.
Note that if
x = c0 + c1 α =⇒ x3 = φ(x) = c0 − c1 α
=⇒ x3 + x = 2c0 = −c0 .
x
x3 + x
0
0
1
−1
−1
1
±α
0
1±α
−1
−1 ± α
1
y
0
±α
±1
0
±α
±1
Adding the point O = [0, 1, 0],
N = |E(F9 )| = 1 + 2 + 2 + 2 + 4 + 4 + 1 = 16.
This satisfies
√
|N − 10| ≤ 2 9 = 6
(just!).
[Although this wasn’t asked, since there are 3 points of order 2 on
the curve it follows that either
E(F9 ) ∼
= Z/(8) ⊕ Z/(2)
or else
E(F9 ) ∼
= Z/(4) ⊕ Z/(4)
I leave it as an exercise to determine which.]
2. Reduce the hyper-elliptic curve
y 2 = x4 + 1
by a birational transformation to Weierstrass form.
Hence prove Fermat’s Theorem for n = 4.
Answer:
(a) We can re-write the equation as
y 2 − x4 = 1,
ie
(y + x2 )(y − x2 ) = 1.
Set
y + x2 = s.
Then
1
y − x2 = .
s
Hence
1
2x2 = s − .
s
Multiplying through by s2 ,
2x2 s2 = s3 − s.
Set
xs = t.
The equation becomes
2t2 = s3 − s.
Finally, on setting
s = u/2, t = v/4
the equation takes the Weierstrass form
v 2 = u3 − 4u.
The birational transformation is
x=
t
v
u
v2
2u3 − v 2
=
, y = s − x2 = − 2 =
,
s
2u
2 4u
4u2
with inverse
u = 2s = 2(x2 + y), v = 4t = 4xs = 2xu = 4x(x2 + y).
(b) To show that
x4 + y 4 = z 4
has no solution in integers with xyz 6= 0, it is sufficient to show
that
x4 + y 4 = z 2
has no solution in integers with xyz 6= 0. On dividing by y 4 ,
4
2
x
z
+1=
.
y
y2
Thus it is sufficient to show that the hyper-elliptic equation
y 2 = x4 + 1
has no rational solution with x 6= 0.
After the birational transformation above, it is sufficient to show
that the elliptic equation
E(Q) : v 2 = u3 − 4u
has no rational solutions apart from (0, 0) and (±2, 0). (Any solution with u 6= 0 will give a solution of the hyper-elliptic equation; while the solutions (x, y) = (0, ±1) of the latter give (u, v) =
(±2, 0).)
In effect, we must show that the rank of this curve is 0, and that its
torsion-group consists of the three points of order 2 (with v = 0)
and the point O = [0, 1, 0].
We use our second method for computing the rank. The associated
elliptic curve is
E1 (Q) : v 2 = u3 + 16u
Since the equation has 3 points of order 2, while b1 = 16 is a
perfect square, the rank r is given by
2r+2 = |im χ| · |im χ1 |.
where
2
χ : E → Q× /Q× , χ1 : E1 → Q× /Q×
2
are the associated homomorphisms.
We have
im χ ⊂ {±1, ±2}.
Also (0, 0) 7→ b = −4 ∼ −1 while (2, 0) 7→ 2. Hence
im χ = {±1, ±2}.
Similarly
im χ1 ⊂ {±1, ±2}.
If d < 0 then d0 = b1 /d < 0, and we would be looking for nontrivial solutions of
du4 + d0 t4 = v 2
which is impossible. Hence
im χ1 ⊂ {1, 2}.
Suppose d = 2. Then d0 = 8, and we are looking for solutions of
2u4 + 8t4 = v 2 .
Evidently 2 | v. But then
2 | v =⇒
=⇒
=⇒
=⇒
2|u
8 | v2
4|v
2 | t,
contradicting the condition that gcd(u, t) = 1. Hence im χ1 = {1},
and so
2r+2 = 4 · 1 =⇒ r = 0,
ie there are only a finite number of points on E(Q).
Finally, to determine the points of finite order, recall that the homomorphism Θ : E → E1 is given by
Θ(x, y) =
x2 + ax + b
.
x
for (x, y) 6= (0, 0), ie in this case
x2 − 4
4
Θ(x, y) =
=x− .
x
x
Now if P = (x, y) is of finite order then so is Θ(P ). Hence the
coefficients of both P and Θ(P ) must be integral. Thus
x | 4,
ie x = ±1, ±2, ±4. But
y 2 = x3 − 4x.
It is a trivial matter to see that the only values of x among these
making y a perfect square are x = ±2, with y = 0.
We conclude that the only rational points on the elliptic curve are
the points of order 2 (with y = 0) and the neutral element O.
Thus there are no non-trivial solutions of Fermat’s Theorem for
n = 4.
3. Show that every non-singular cubic curve over k can be brought to
Weierstrass form by a birational transformation over k.
Reduce the curve
X3 + Y 3 + Z3 = 0
to Weierstrass form.
Answer: [The first part of the question should have read “Show that
every non-singular cubic curve over a field k of characteristic 6= 2 containing at least one point over k can be brought . . . ”.
(a) Suppose P ∈ Γ. If P is a point of inflexion then the curve can be
brought to Weierstrass form by a projective transformation, taking
P into O = [0, 1, 0] and the tangent at P into the line at infinity
Z = 0.
If P is not a point of inflexion, then the tangent at P must meet the
curve again at a second point O. By a projective transformation
we can bring O to [0, 0, 1] and the line OP to the line X = 0.
To simplify the discussion we now use affine coordinates (x, y).
Thus O = (0, 0), while P is on the x-axis y = 0.
The equation for the curve can be written
p3 (x, y) + p2 (x, y) + p1 (x, y) = 0,
where p1 , p2 , p3 are homogeneous of degrees 1,2,3, respectively. (There
is no constant term since the origin (0, 0) is on the curve.)
The curve meets the y-axis where
p3 (0, 1)y 3 + p2 (0, 1)y 2 + p1 (0, 1)y = 0,
ie at (0, 0) and where
p3 (0, 1)y 2 + p2 (0, 1)y + p1 (0, 1) = 0,
By hypothesis this has a double root at P . Thus
p2 (0, 1)2 = 4p3 (0, 1)p1 (0, 1).
Now let y = xt. In terms of x, t, the equation becomes
p3 (1, t)x3 + p2 (1, t)x2 + p1 (1, t)x = 0.
Taking out the factor x, we see that the birational transformation
(x, y) → (x, t), where t = y/x, brings the curve to the form
p3 (1, t)x2 + p2 (1, t)x + p1 (1, t) = 0.
On multiplying by 4p3 (1, t), this can be written
(2p3 (1, t)x + p2 (1, t))2 = p2 (1, t)2 − 4p3 (1, t)p1 (1, t).
Set
s = 2p3 (1, t)x + p2 (1, t).
In terms of s, t, the equation becomes
s2 = p2 (1, t)2 − 4p3 (1, t)p1 (1, t).
The coefficient of t4 in the quartic in t on the right is
p2 (0, 1)2 − 4p3 (0, 1)p1 (0, 1) = 0.
Thus the equation takes the form
s2 = a3 t3 + a2 t2 + a1 t + a0 .
[The inverse transformation from s, t to x, y is given by
x=
s − p2 (1, t)
,
2p3 (1, t)
y = tx =
s − p2 (1, t)
t,
2p3 (1, t)
so the transformation is birational.]
The coefficient a3 6= 0, since otherwise the curve would be a conic,
in which case the original cubic would be singular. The final transformation
s = a23 y 0 , t = a3 x0
brings the curve to Weierstrassian form
y 02 = x03 + ax02 + bx0 + c.
(b) The curve
Γ(Q) : X 3 + Y 3 + Z 3 = 0
contains the 3 points
P = [0, 1, −1], Q = [−1, 0, 1], R = [1, −1, 0].
The tangent at P is
(∂F/∂X)P X + (∂F/∂Y )P Y + (∂F/∂Z)P Z = 0,
that is,
(3X 2 )P X + (3Y 2 )P Y + (3Z 2 )P Z = 0,
ie
Y + Z = 0.
This meets the curve where
X 3 = 0.
Thus P is a point of inflexion.
[This is clear from Fermat’s Last Theorem for n = 3, which states
that P, Q, R are the only points on Γ over Q. If P were not a point
of inflexion then the tangent at P would have to meet Γ again at
Q or R. But if the tangent passed throught Q then by symmetry
it would also pass through R, and so would have 4 points on Γ.]
Accordingly, we are in the first case; we can bring Γ to Weierstrassian form by a projective transformation, taking P = [0, 1, −1]
to [0, 1, 0] and the line Y + Z = 0 to Z = 0. If we set
Z0 = Y + Z
the second condition will be fulfilled. Now if we set
X 0 = X, Y 0 = Y
the first condition will also be satisfied. The inverse transformation is
X = X 0, Y = Y 0, Z = Z 0 − Y 0.
This brings the curve to the form
X 03 + Z 03 − 3Z 02 Y 0 + 3Z 0 Y 02 ,
or in affine form
3y 02 − 3y 0 = −x03 − 1.
The transformation
y 0 = y/32 , x0 = −x/3
takes this to the Weierstrassian form
y 2 − 9y = x3 − 27.
[We can bring this to standard form — although we are not asked
to do that — by completing the square on the left, setting y1 =
y − 9/2. This gives
27
y12 = x3 − .
4
We can make the coordinates integral by the transformation y2 =
23 y1 , x2 = 22 x, bringing the curve to the form
y22 = x32 − 24 33 .
However, the question only asked that the curve be brought to
Weierstrassian form.
Note that the cubic has no rational roots, so our techniques for
computing the rank do not apply. This explains why one has to
introduce algebraic numbers to solve Fermat’s Last Theorem in the
case n = 3.]
4. Outline Lenstra’s Elliptic Curve Factorisation method.