Download Solution to 18.782 Homework #10, due December 9

Solution to 18.782 Homework #10, due December 9
Problem 1. Show that the curve E ⊂ P2Q given by t30 + 3t31 + 9t32 is a nice curve of genus one
which has no rational points over Q (so it can not be given the structure of an elliptic curve).
Solution
Let us first show that E has no rational points. If that was false it would have a nontrivial
integer point (since the defining equation is homogeneous). Let us choose x, y, z s.t. x3 + 3y 3 +
9z 3 = 0 and gcd(x, y, z) = 1. Clearly x must be divisible by 3 and so we can write x = 3x′ , x′ ∈ Z.
Then substituting this into the equation and dividing by 3 we see that y should be divisible by 3
and so y = 3y ′ , y ′ ∈ Z. Finally the same argument yields 3 | z. This contradicts our assumption
on gcd(x, y, z) = 1. So E has no rational points.
The curve E is smooth since all affine patches x3 + 3y 3 + 9 = 0, x3 + 3 + 9z 3 = 0, 1 +
3
3y + 9z 3 = 0 are smooth. Moreover, this curve is irreducible: if not, then x3 + 3y 3 + 9z 3 would
be divisible by a linear function ax + by + cz, which is clearly not the case. Finally, as E is
defined by a cubic equation it’s genus equals g = (3 − 1)(3 − 2)/2 = 1. Problem 2. Let b be an integer which is fourth power free. Let E be the elliptic curve which is
projective closure of the curve y 2 = x3 + bx. Show that the group of points of finite order in
E(Q) is isomorphic to:
(a) Z/4Z if b = 4.
(b) Z/2Z × Z/2Z if −b is square.
(c) Z/2Z otherwise.
Solution
Let G ⊆ E(Q) be the subgroup consisting of all points of finite order. Let Gp = Ẽ(Fp ),
where Ẽ is the reduction of E modulo p. Since the discriminant of our curve is −4b3 the
Reduction modulo p theorem (see [Silverman-Tate], p.123) guarantees that for p - 8b3 we have
|G| | |Gp |. The same argument as in Problem 2a) from homework #9 shows that |Gp | = p + 1
for p ≡ 3. Let S = {p + 1|p ≡ 3, p > 8b3 }. We claim that in fact gcd(S) = 4, this implies
4
4
that |G| | 4. This claim easily follows from the Dirichlet’s theorem about primes in arithmetic
progression. Indeed, if the claim was false we would either have 8 | gcd(S) or q | gcd(S) for
an odd prime q. Applying Dirichlet’s theorem to the arithmetic progression 8n + 7 in the first
case and to the arithmetic progression 4q · n + r (where r is a unique residue modulo 4q, s.t.
r ≡ 3, r ≡ 1) we get a contradiction.
4
q
2
Solution to 18.782 Homework #10, due December 9
Since |G| | 4 and G contains both (0, 0) and a point O at infinity there are two possibilities
for |G|, namely 2 or 4. In the first case G ≃ Z/2Z, while in the second case it might be Z/4Z
or Z/2Z × Z/2Z. Let us now deal with each case of the problem separately.
(a) If b = 4 we obviously have |G| = 4 as besides of the two points O, (0, 0), G also
contains (2, ±4). Those latter points are not of order two, since the y coordinate is nonzero, so
they have to be of order 4, hence G ∼
= Z/4Z in this case.
(b) If b = −n2 , then {O, (0, 0), (±n, 0)} ⊆ G and as |G| | 4 the above inclusion is in fact
an equality. Moreover they all satisfy 2P = O and so G ≃ Z/2Z × Z/2Z.
(c) For the rest of the cases let us show that in fact G ≃ Z/2Z. If P = (x, y) ∈ G is an
order 2 point then y = 0 and so x3 + bx = 0 implying x = 0 as −b ∈
/ Z2 . Hence, the only order
2 point of E is (0, 0) and it remains to show there are no order 4 points. If P ∈ G was an order
4 point we would have 2P = O. This is equivalent to the line y = mx being tangent to E at the
point P for some m ∈ Q× , or in other words x2 − m2 x + b being a full square. This happens iff
m4 = 4b and so b = 4N 4 for some N ∈ Z. However, if N = ±1 this would yield case a), while
for the rest of N we get a contradiction with b being fourth power free. Problem 3. Let p be a prime and E be as in the previous problem with b = p. Show that the rank
of E is either 0, 1 or 2. If p = 7 mod 16, show that the rank of E is zero. If p = 3 mod 16,
show that the rank of E is either 0 or 1.
Solution
We use the same notations as in [Silverman-Tate] III.6. According to a result in the book
ᾱ(Γ̄)
we have an equality 2r = #α(Γ)·#
, where r = rk(E), the curve Ē is the projective closure
4
2
3
of y = x − 4bx and homomorphisms α : Γ = E(Q) → Q× /(Q× )2 , ᾱ : Γ̄ = Ē(Q) → Q× /(Q× )2
are those constructed in the proof of Mordell’s theorem.
First we compute #α(Γ). We will use the following result from the book: α(Γ) contains b, 1
and all the remaining elements are of the form {b1 mod (Q× )2 }, s.t. b1 ∈ Z, b1 | b and moreover
N 2 = b1 M 4 + bb1 e4 has a nontrivial integer solution with gcd(e, M ) = gcd(e, N ) = 1. Since b = p
is a prime we see that any element in α(Γ) should be a factor of b = p and so α(Γ) ⊆ {±1̄, ±p̄}
(here and in what follows k̄ denotes the image of k ∈ Q× under the projection to Q× /(Q× )2 .
However, when all representatives of the set have different images in this quotient we will omit
¯). Since N 2 = −M 4 −pe4 and N 2 = −pM 4 −e4 don’t have any non-trivial solutions in integers,
−1, −p ∈
/ α(Γ), while 1, p ∈ α(Γ) we get #α(Γ) = 2.
Similarly, we have ᾱ(Γ̄) ⊆ {±1̄, ±2̄, ±4̄, ±p̄, ±2p, ±4p}={±1̄, ±2̄, ±p̄, ±2p}. So #ᾱ(Γ̄) ≤ 8
and substituting into aforementioned formula for 2r we get r = 0, 1 or 2 as desired.
It remains to give a better estimate for ᾱ(Γ̄) in the two special cases. In what follows it will
be important to know that the only quartic residues mod 16 are 0, 1, while the only quadratic
residues are 0, 1, 4, 9.
4
(a) Suppose p ≡ 7. Let us reduce the equality N 2 = b1 M 4 − 4p
b1 e modulo 16. If b1 = ±2,
16
then we get 2M 4 + 2e4 ≡ ∓N 2 . From the list of quadratic and quartic residues mod 16 we
16
see that all e, M, N should be even, contradicting gcd(e, M ) = gcd(e, N ) = 1. Similarly, if
b1 = −1 we would have −M 4 − 4e4 ≡ N 2 and the same argument gives a contradiction. So
16
{−1, 2, −2} ∈
/ ᾱ(Γ̄). However, since ᾱ is a homomorphism and ᾱ((0, 0)) = −4p = −p̄, we see
that p, 2p, −2p ∈
/ ᾱ(Γ̄). So #ᾱ(Γ̄) ≤ 2 and hence 2r ≤ 1 proving r = 0.
Solution to 18.782 Homework #10, due December 9
3
(b) Suppose p ≡ 3. Then by the same argument −1, 2 ∈
/ ᾱ(Γ̄). Similarly to (a) this
16
automatically yields p, −2p ∈
/ ᾱ(Γ̄) as well and hence #ᾱ(Γ̄) ≤ 4 proving r = 0 or 1. Problem 4. Let C be the singular cubic which is the projective closure of y 2 = x3 . Check that
the set of nonsingular points of C (over any field) has a group structure defined similarly to
the case of nonsingular curves (in particular, show that a line passing through two nonsingular
points of C doesn’t pass through its singular point). Show that the map (x, y) 7→ xy extends to
an isomorphism of this group with the additive group of the field.
Solution Let Cns (k) denote the set of nonsingular points of C and S = (0, 0) be the singular
point. The key point is to observe that a line L intersecting C at two points P1 , P2 ∈ Cns (k)
(which may coincide, in which case we understand that L should be tangent to C at P1 = P2 )
intersects it at another point P3 ∈ Cns (k) (again, we don’t exclude the possibility that P3
coincides with P1 or P2 , i.e. what we mean is that the divisor of intersection is P1 + P2 + P3 ). To
see this recall that the degree of this divisor is three by (the easy case of) Bezout Theorem; but
if L ∋ S, then the multiplicity of intersection of L with C at S is at least two. Since 2S +P1 +P2
has degree 4 > 3 we get a contradiction.
Thus we can define the group law on the set of nonsingular points by the same chordtangent rule.
Since we have not discussed Picard groups of singular curves (although they can be defined), the axioms of an abelian group must be checked directly. We will define a bijection
between Cns (k) and A1 (k) sending the chord-tangent composition law to addition, this implies
the group axioms together with the rest of the problem.
In homogeneous coordinates the curve is given by t0 t22 = t31 , and the function ϕ = xy =
t1
2
t2 = t0 t2 t1 is well defined on C \ {S}.
Consider an arbitrary line L : at1 + bt2 + ct0 = 0 not passing through S, i.e. with c ̸= 0.
Let us note that the only point of C with a zero t2 –coordinate is S. So it’s natural to consider
C in the affine chart t2 ̸= 0, where it is defined by the equation z = x3 . In these coordinates L
is given by ax + cz + b = 0. Substituting z = x3 (i.e. looking at the points of L ∩ C = {P, Q, R})
we get an equation b + ax + cx3 = 0, whose solutions sum up to 0 (as there is no x2 term).
This is equivalent to ϕ(P ) + ϕ(Q) + ϕ(R) = 0 proving that ϕ sends our chord-tangent operation
to addition. Finally, ϕ is easily seen to be a bijection. In fact, the inverse map ψ is given by
k ∋ u 7→ (u : 1 : u3 ) ∈ C\{S} 1 . This finishes the proof. Problem 5. Show that for a nice curve X over Q of genus greater than one the set X(Q) is
finite.
Solution
This is a famous Mordell conjecture, proved by G. Faltings in 1983 and now known as
Faltings’ theorem. 1 Another
way to show that ϕ is a homomorphism is to show that ψ is a homomorphism, which can translated
into checking that the determinants of some particular 3 × 3 matrices are zero.