density of rational points on a family of diagonal quartic ... thesis presentation.pdf · a family...

Density of Rational Points on a Familyof Diagonal Quartic Surfaces

Thesis advisor CandidateProf. Ronald M. van Luijk Dino Festi

Universiteit Leiden - Universita di Padova

June 21, 2012

Dino Festi Leiden - June 21, 2012

Table of Contents

I A family of diagonal quartic surfaces.

I Elliptic fibers.

I Torsion points on the fibers:

I 2-torsion points,

I 4-torsion points,

I 5-torsion points,

I 3-torsion points,

I Full torsion subgroup.

I Proof of the main theorem.

A family of diagonal quartic surfaces

For any c1, c2 ∈ Q×, let Wc1,c2 =W be the surface given by

(x2 − 2c1y2)(x2 + 2c1y

2) = c2(z2 + 2zw + 2w2)(z2 − 2zw + 2w2).

TheoremLet c1, c2 and W be as above.Let P = (x0 : y0 : z0 : w0) be a rational point on W with x0 and y0 both nonzero.If |2c1| is a square in Q×, then also assume that z0, w0 are not both zero.Then the set of rational points on the surface is Zariski dense.

Fibrations from W to P1:

ψ1 : (x : y : z : w) 7→ (x2 − 2c1y2 : z2 + 2zw + 2w2) = (c2(z

2 − 2zw + 2w2) : x2 + 2c1y2),

ψ2 : (x : y : z : w) 7→ (x2 − 2c1y2 : z2 − 2zw + 2w2) = (c2(z

2 + 2zw + 2w2) : x2 + 2c1y2).

(x2 − 2c1y2)(x2 + 2c1y

2) = c2(z2 + 2zw + 2w2)(z2 − 2zw + 2w2).

ψ1 : (x : y : z : w) 7→ (x2 − 2c1y2 : z2 + 2zw + 2w2) = (c2(z

2 − 2zw + 2w2) : x2 + 2c1y2),

ψ2 : (x : y : z : w) 7→ (x2 − 2c1y2 : z2 − 2zw + 2w2) = (c2(z

2 + 2zw + 2w2) : x2 + 2c1y2).

(x2 − 2c1y2)(x2 + 2c1y

2) = c2(z2 + 2zw + 2w2)(z2 − 2zw + 2w2).

ψ1 : (x : y : z : w) 7→ (x2 − 2c1y2 : z2 + 2zw + 2w2) = (c2(z

2 − 2zw + 2w2) : x2 + 2c1y2),

ψ2 : (x : y : z : w) 7→ (x2 − 2c1y2 : z2 − 2zw + 2w2) = (c2(z

2 + 2zw + 2w2) : x2 + 2c1y2).

Elliptic fibers

ψ1 : (x : y : z : w) 7→ (x2−2c1y2 : z2+2zw+2w2) = (c2(z2−2zw+2w2) : x2+2c1y

The fiber F of ψ1 above (s : 1), with s ∈ Q, is given by:{x2 − 2c1y

2 = s(z2 + 2zw + 2w2)

c2(z2 − 2zw + 2w2) = s(x2 + 2c1y2)

Most fibers have genus 1: intersection of two smooth quadrics in P3.The Jacobian of the fiber is isomorphic over Q to the elliptic curve given by:

y2 = x3 + 4c1(c22 − s4)x2 + 4c21(c42 − 34s4c22 + s8)x.

Its j-invariant and discriminant are

j =2(s8 + 94s4c22 + c42)3

c22s4(s4 − 6s2c2 + c22)2(s4 + 6s2c2 + c22)2

d = 217s4c61c42(s4 − 6s2c2 + c22)2(s4 + 6s2c2 + c22)2.

Remark: If we assume that on the fiber there is at least one rational pointthen the fiber is isomorphic to its Jacobian.

Elliptic fibers

2 = s(z2 + 2zw + 2w2)

c2(z2 − 2zw + 2w2) = s(x2 + 2c1y2)

y2 = x3 + 4c1(c22 − s4)x2 + 4c21(c42 − 34s4c22 + s8)x.

j =2(s8 + 94s4c22 + c42)3

c22s4(s4 − 6s2c2 + c22)2(s4 + 6s2c2 + c22)2

d = 217s4c61c42(s4 − 6s2c2 + c22)2(s4 + 6s2c2 + c22)2.

Elliptic fibers

2 = s(z2 + 2zw + 2w2)

c2(z2 − 2zw + 2w2) = s(x2 + 2c1y2)

y2 = x3 + 4c1(c22 − s4)x2 + 4c21(c42 − 34s4c22 + s8)x.

j =2(s8 + 94s4c22 + c42)3

c22s4(s4 − 6s2c2 + c22)2(s4 + 6s2c2 + c22)2

d = 217s4c61c42(s4 − 6s2c2 + c22)2(s4 + 6s2c2 + c22)2.

Elliptic fibers

2 = s(z2 + 2zw + 2w2)

c2(z2 − 2zw + 2w2) = s(x2 + 2c1y2)

y2 = x3 + 4c1(c22 − s4)x2 + 4c21(c42 − 34s4c22 + s8)x.

j =2(s8 + 94s4c22 + c42)3

c22s4(s4 − 6s2c2 + c22)2(s4 + 6s2c2 + c22)2

d = 217s4c61c42(s4 − 6s2c2 + c22)2(s4 + 6s2c2 + c22)2.

Elliptic fibers

Question 1: Are there singular fibers?

We have exactly ten singular fibers, for both ψ1 and ψ2. Namely the fibersabove (1 : 0), (0 : 1), (s : 1) with

s ∈ S := {(±1±√

2)γ22 , i(±1±

√2)γ2

where γ2 is such that γ42 = c2.

Question 2: Are there rational points on the singular fibers?

Not above the points (s : 1), with s in S. There may be rational points onthe fibers of ψ1 and ψ2 above the points (0 : 1) and (1 : 0), whose union isgiven by: {

x4 − 4c21y4 = 0

z4 + 4w4 = 0

Elliptic fibers

s ∈ S := {(±1±√

2)γ22 , i(±1±

√2)γ2

x4 − 4c21y4 = 0

z4 + 4w4 = 0

Elliptic fibers

s ∈ S := {(±1±√

2)γ22 , i(±1±

√2)γ2

x4 − 4c21y4 = 0

z4 + 4w4 = 0

Elliptic fibers

s ∈ S := {(±1±√

2)γ22 , i(±1±

√2)γ2

x4 − 4c21y4 = 0

z4 + 4w4 = 0

Elliptic fibers

Proposition 2.2.3

Let W and ψ1 defined as before.If |2c1| is not a square in Q then there are no rational points on anysingular fiber.If 2c1 is a square in Q then there are exactly two rational points on thefiber above (0 : 1) and this is the only singular fiber with rational points.If −2c1 is a square in Q then there are exactly two rational points on thefiber above (1 : 0) and this is the only singular fiber with rational points.

The same holds for ψ2.

Lemma 2.2.4

The intersection of the fibers above (0 : 1) and (1 : 0) are the following:

F0 ∩G0 = {(±√

2c1 : 1 : 0 : 0)},

F0 ∩G∞ = {(0 : 0 :√

2ζ8 : 1), (0 : 0 : −√

2ζ38 : 1)},

F∞ ∩G0 = {(0 : 0 : −√

2ζ8 : 1), (0 : 0 :√

2ζ38 : 1)},

F∞ ∩G∞ = {(±√−2c1 : 1 : 0 : 0)}.

Elliptic fibers

Proposition 2.2.3

Let W and ψ1 defined as before.If |2c1| is not a square in Q then there are no rational points on anysingular fiber.If 2c1 is a square in Q then there are exactly two rational points on thefiber above (0 : 1) and this is the only singular fiber with rational points.If −2c1 is a square in Q then there are exactly two rational points on thefiber above (1 : 0) and this is the only singular fiber with rational points.

The same holds for ψ2.

Lemma 2.2.4

The intersection of the fibers above (0 : 1) and (1 : 0) are the following:

F0 ∩G0 = {(±√

2c1 : 1 : 0 : 0)},

F0 ∩G∞ = {(0 : 0 :√

2ζ8 : 1), (0 : 0 : −√

2ζ38 : 1)},

F∞ ∩G0 = {(0 : 0 : −√

2ζ8 : 1), (0 : 0 :√

2ζ38 : 1)},

F∞ ∩G∞ = {(±√−2c1 : 1 : 0 : 0)}.

2-torsion points

Any smooth fiber with at least one rational point, sayP = (x0 : y0 : z0 : w0), is isomorphic over the rationals to the elliptic curve

y2 = x3 + 4c1(c22 − s4)x2 + 4c21(c42 − 34s4c22 + s8)x.

This elliptic curve has only two rational 2-torsion points: O and (0, 0).

Rational 2-torsion Non rational 2-torsionpoints on the fiber points on the fiber

P = (x0 : y0 : z0 : w0) T1 = (−√

2x0 :√

2y0 : 2w0 : z0)

T0 = (−x0 : −y0 : z0 : w0) T2 = (√

2x0 : −√

2y0 : 2w0 : z0)

2-torsion points

y2 = x3 + 4c1(c22 − s4)x2 + 4c21(c42 − 34s4c22 + s8)x.

P = (x0 : y0 : z0 : w0) T1 = (−√

2x0 :√

2y0 : 2w0 : z0)

T0 = (−x0 : −y0 : z0 : w0) T2 = (√

2x0 : −√

2y0 : 2w0 : z0)

2-torsion points

y2 = x3 + 4c1(c22 − s4)x2 + 4c21(c42 − 34s4c22 + s8)x.

P = (x0 : y0 : z0 : w0) T1 = (−√

2x0 :√

2y0 : 2w0 : z0)

T0 = (−x0 : −y0 : z0 : w0) T2 = (√

2x0 : −√

2y0 : 2w0 : z0)

4-torsion points

Claim (Theorem 3.1.4):

Let F be a smooth fiber of ψ1 with a rational point P on it. Then (F,P),viewed as an elliptic curve, has no rational nontrivial 4-torsion points.

Recall that (F, P ) is isomorphic over the rationals to the elliptic curve

y2 = x3 + 4c1(c22 − s4)x2 + 4c21(c42 − 34s4c22 + s8)x.

The 4-division polynomial of the elliptic curve above is

f4(x) = 8xh1(x)h2(x)h3(x),

h1(x) =x2 + 4c1(c22 − s4)x+ 4c21(c42 − 34s4c22 + s8),

h2(x) =x2 − 4c21(s8 − 34s4c22 + c42),

h3(x) =x4 + 8c1(c22 − s4)x3 + 24c21(s8 − 34s4c22 + c42)x2+

+ 32c31(−s12 + 35s8c22 − 35s4c42 + 32c62)x+

+ 16c41(s16 − 68s12c22 + 1158s8c42 − 68s4c62 − c82).

4-torsion points

y2 = x3 + 4c1(c22 − s4)x2 + 4c21(c42 − 34s4c22 + s8)x.

f4(x) = 8xh1(x)h2(x)h3(x),

h1(x) =x2 + 4c1(c22 − s4)x+ 4c21(c42 − 34s4c22 + s8),

h2(x) =x2 − 4c21(s8 − 34s4c22 + c42),

h3(x) =x4 + 8c1(c22 − s4)x3 + 24c21(s8 − 34s4c22 + c42)x2+

+ 32c31(−s12 + 35s8c22 − 35s4c42 + 32c62)x+

+ 16c41(s16 − 68s12c22 + 1158s8c42 − 68s4c62 − c82).

4-torsion points

y2 = x3 + 4c1(c22 − s4)x2 + 4c21(c42 − 34s4c22 + s8)x.

f4(x) = 8xh1(x)h2(x)h3(x),

h1(x) =x2 + 4c1(c22 − s4)x+ 4c21(c42 − 34s4c22 + s8),

h2(x) =x2 − 4c21(s8 − 34s4c22 + c42),

h3(x) =x4 + 8c1(c22 − s4)x3 + 24c21(s8 − 34s4c22 + c42)x2+

+ 32c31(−s12 + 35s8c22 − 35s4c42 + 32c62)x+

+ 16c41(s16 − 68s12c22 + 1158s8c42 − 68s4c62 − c82).

4-torsion points

Claim (Lemma 3.1.3):

For any s ∈ Q× the polynomial h2(x) = x2 − 4c21(s8 − 34s4c22 + c42) has norational roots.

Finding a rational s such that h2 admits a rational root is equivalent tofinding a rationals point on the surface D ⊂ U := A3(z, s, c2) ∩ {sc2 6= 0}given by:

D : z2 − (s8 − 34s4c22 + c42) = 0.

Using the map from D to P2 sending

(z, s, c2) 7→ (z/c2 : s2 : c2),

one can see that the existence of rational points on D implies the existenceof rational points on the curve C ⊂ P2(X,Y, Z) defined by:

C : X2Z2 = Y 4 − 34Y 2Z2 + Z4.

Notice that on C there are at least three rational points:

(1 : 0 : 1), (1 : 0 : −1), (1 : 0 : 0).

The desingularization of the curve C is isomorphic over Q to the ellipticcurve

H : y2 = x3 − x2 − 24x− 36.

H(Q) ∼= Z/4Z

4-torsion points

D : z2 − (s8 − 34s4c22 + c42) = 0.

(z, s, c2) 7→ (z/c2 : s2 : c2),

C : X2Z2 = Y 4 − 34Y 2Z2 + Z4.

(1 : 0 : 1), (1 : 0 : −1), (1 : 0 : 0).

H : y2 = x3 − x2 − 24x− 36.

H(Q) ∼= Z/4Z

4-torsion points

D : z2 − (s8 − 34s4c22 + c42) = 0.

(z, s, c2) 7→ (z/c2 : s2 : c2),

C : X2Z2 = Y 4 − 34Y 2Z2 + Z4.

(1 : 0 : 1), (1 : 0 : −1), (1 : 0 : 0).

H : y2 = x3 − x2 − 24x− 36.

H(Q) ∼= Z/4Z

4-torsion points

D : z2 − (s8 − 34s4c22 + c42) = 0.

(z, s, c2) 7→ (z/c2 : s2 : c2),

C : X2Z2 = Y 4 − 34Y 2Z2 + Z4.

(1 : 0 : 1), (1 : 0 : −1), (1 : 0 : 0).

H : y2 = x3 − x2 − 24x− 36.

H(Q) ∼= Z/4Z

4-torsion points

D : z2 − (s8 − 34s4c22 + c42) = 0.

(z, s, c2) 7→ (z/c2 : s2 : c2),

C : X2Z2 = Y 4 − 34Y 2Z2 + Z4.

(1 : 0 : 1), (1 : 0 : −1), (1 : 0 : 0).

H : y2 = x3 − x2 − 24x− 36.

H(Q) ∼= Z/4ZDino Festi Leiden - June 21, 2012

5-torsion points

Fact (Lemma 3.3.1):

Let E/Q be an elliptic curve defined over Q, and P ∈ E(Q) a 5-torsionpoint. Then there is a number b ∈ Q and an isomorphism ϕ : E → E′,where E′ is the curve defined by

E′ : y2 + (b+ 1)xy + by = x3 + bx2,

such that ϕ(P ) = (0, 0).

5-torsion points

Fact (Lemma 3.3.1):

Let E/Q be an elliptic curve defined over Q, and P ∈ E(Q) a 5-torsionpoint. Then there is a number b ∈ Q and an isomorphism ϕ : E → E′,where E′ is the curve defined by

E′ : y2 + (b+ 1)xy + by = x3 + bx2,

such that ϕ(P ) = (0, 0).

5-torsion pointsSo if we assume that there is a nontrivial rational 5-torsion point on (F, P ),it follows that there is a b ∈ Q× such that the j-invariant of (F, P ) is

− b2 + 12b+ 14− 12b−1 + b−2

b+ 11 + b−1︸︷︷︸=:g(b)

=2(c−2

2 s4 + 94 + c22s−4)3

(c−22 s4 − 34 + c22s

−4)2︸︷︷︸=:f(s)

This means that we have a rational point on the affine curveC′ ⊂ U := A2(s, b) ∩ {sb 6= 0} defined by

C′ : g(b) = f(s).

Claim (Proposition 3.3.3):

The curve C′ has no rational points.

Using the map from C′ to A2(u, c)

θ : (s, b) 7→ ((c−12 s2 − c2s−2)2, b− b−1),

one can see that the existence of rational points on C′ would imply theexistence of rational points on the genus 0 curve C ⊂ A2(u, c), where C isgiven by

C : 2(u+ 96)3(c+ 11) = −(c2 + 12c+ 16)3(u− 32)2.

− b2 + 12b+ 14− 12b−1 + b−2

b+ 11 + b−1︸︷︷︸=:g(b)

=2(c−2

2 s4 + 94 + c22s−4)3

(c−22 s4 − 34 + c22s

−4)2︸︷︷︸=:f(s)

C′ : g(b) = f(s).

θ : (s, b) 7→ ((c−12 s2 − c2s−2)2, b− b−1),

C : 2(u+ 96)3(c+ 11) = −(c2 + 12c+ 16)3(u− 32)2.

− b2 + 12b+ 14− 12b−1 + b−2

b+ 11 + b−1︸︷︷︸=:g(b)

=2(c−2

2 s4 + 94 + c22s−4)3

(c−22 s4 − 34 + c22s

−4)2︸︷︷︸=:f(s)

C′ : g(b) = f(s).

θ : (s, b) 7→ ((c−12 s2 − c2s−2)2, b− b−1),

C : 2(u+ 96)3(c+ 11) = −(c2 + 12c+ 16)3(u− 32)2.

− b2 + 12b+ 14− 12b−1 + b−2

b+ 11 + b−1︸︷︷︸=:g(b)

=2(c−2

2 s4 + 94 + c22s−4)3

(c−22 s4 − 34 + c22s

−4)2︸︷︷︸=:f(s)

C′ : g(b) = f(s).

θ : (s, b) 7→ ((c−12 s2 − c2s−2)2, b− b−1),

C : 2(u+ 96)3(c+ 11) = −(c2 + 12c+ 16)3(u− 32)2.

− b2 + 12b+ 14− 12b−1 + b−2

b+ 11 + b−1︸︷︷︸=:g(b)

=2(c−2

2 s4 + 94 + c22s−4)3

(c−22 s4 − 34 + c22s

−4)2︸︷︷︸=:f(s)

C′ : g(b) = f(s).

θ : (s, b) 7→ ((c−12 s2 − c2s−2)2, b− b−1),

C : 2(u+ 96)3(c+ 11) = −(c2 + 12c+ 16)3(u− 32)2.

5-torsion points

Let C ⊂ P2(X,Y, Z) be the projective closure of the curve C, whereu = X/Z and c = Y/Z. Then there exists a birational morphism

ϕ : P1 → C, (p : q) 7→ (X(p, q) : Y (p, q) : Z(p, q))

X(p, q) =25(p− 7168q)2(p2 − 10240pq + 20971520q2)2·

(p2 − 73728

5pq + 54525952q2),

Y (p, q) =− 11(p− 8192q)4(p− 36864

(p3 − 22528p2q +1862270976

11pq2 − 4668629450752

11q3),

Z(p, q) =(p− 8192q)5(p− 7168q)2(p− 36864

5-torsion points

From the definition of the map θ and recalling the parametrization ϕ, theexistence of a rational point on C coming from C′ would imply theexistence of a rational point on the curve M ⊂ P2(p, q, r) defined by

M : q2r2 = 2(p2 − 73728

5pq + 54525952q2)(p− 8192q)(p− 36864

Notice that on this curve there are at least two rational points, namely:(1 : 0 : 8192) and (5 : 0 : 36864), which correspond to the points(8192 : 1), (36864 : 5) ∈ P1(p, q); both the points are sent to (1 : 0 : 0) ∈ Cvia ϕ.We claim that these two are the only rational points on M .M is isomorphic over the rational to the elliptic curve

G : y2z = x3 − 4x2z + 20xz2.

But for the elliptic curve G one can see that

G(Q) ∼= Z/2Z.

5-torsion points

M : q2r2 = 2(p2 − 73728

5pq + 54525952q2)(p− 8192q)(p− 36864

Notice that on this curve there are at least two rational points, namely:(1 : 0 : 8192) and (5 : 0 : 36864), which correspond to the points(8192 : 1), (36864 : 5) ∈ P1(p, q); both the points are sent to (1 : 0 : 0) ∈ Cvia ϕ.We claim that these two are the only rational points on M .

M is isomorphic over the rational to the elliptic curve

G : y2z = x3 − 4x2z + 20xz2.

G(Q) ∼= Z/2Z.

5-torsion points

M : q2r2 = 2(p2 − 73728

5pq + 54525952q2)(p− 8192q)(p− 36864

G : y2z = x3 − 4x2z + 20xz2.

G(Q) ∼= Z/2Z.

5-torsion points

M : q2r2 = 2(p2 − 73728

5pq + 54525952q2)(p− 8192q)(p− 36864

G : y2z = x3 − 4x2z + 20xz2.

G(Q) ∼= Z/2Z.

3-torsion points

Fact (Lemma 3.4.1):

Let E be an elliptic curve defined over Q and P a rational 3-torsion point;assume E has nonzero j-invariant. Then there exist an element b ∈ Q×such that the pair (E,P ) is isomorphic to the pair (Eb, (0, 0)), where Eb isthe elliptic curve defined by

y2 + xy +b

27y = x3.

3-torsion points

Fact (Lemma 3.4.1):

Let E be an elliptic curve defined over Q and P a rational 3-torsion point;assume E has nonzero j-invariant. Then there exist an element b ∈ Q×such that the pair (E,P ) is isomorphic to the pair (Eb, (0, 0)), where Eb isthe elliptic curve defined by

y2 + xy +b

27y = x3.

2933 (b− 9/8)3

b4 − b3︸︷︷︸=:g(b)

=2(c−2

2 s4 + 94 + c22s−4)3

(c−22 s4 − 34 + c22s

−4)2︸︷︷︸=:f(s)

This means that we have a rational point on the affine curveC′ ⊂ U := A2(b, s) ∩ {bs 6= 0} defined by

C′ : g(b) = f(s).

Claim:

Using the map from C′ to A2 defined by

ε : (b, s) 7→ (b, c−12 s2 + c2s

−2),

one can see that the existence of rational points on C′ would imply theexistence of rational points on the genus 2 curve C ⊂ A2(p, q), where C isgiven by

C : 2(u+ 96)3(c+ 11) = −(c2 + 12c+ 16)3(u− 32)2.

2933 (b− 9/8)3

b4 − b3︸︷︷︸=:g(b)

=2(c−2

2 s4 + 94 + c22s−4)3

(c−22 s4 − 34 + c22s

−4)2︸︷︷︸=:f(s)

C′ : g(b) = f(s).

Claim:

ε : (b, s) 7→ (b, c−12 s2 + c2s

−2),

C : 2(u+ 96)3(c+ 11) = −(c2 + 12c+ 16)3(u− 32)2.

2933 (b− 9/8)3

b4 − b3︸︷︷︸=:g(b)

=2(c−2

2 s4 + 94 + c22s−4)3

(c−22 s4 − 34 + c22s

−4)2︸︷︷︸=:f(s)

C′ : g(b) = f(s).

Claim:

ε : (b, s) 7→ (b, c−12 s2 + c2s

−2),

C : 2(u+ 96)3(c+ 11) = −(c2 + 12c+ 16)3(u− 32)2.

2933 (b− 9/8)3

b4 − b3︸︷︷︸=:g(b)

=2(c−2

2 s4 + 94 + c22s−4)3

(c−22 s4 − 34 + c22s

−4)2︸︷︷︸=:f(s)

C′ : g(b) = f(s).

Claim:

ε : (b, s) 7→ (b, c−12 s2 + c2s

−2),

C : 2(u+ 96)3(c+ 11) = −(c2 + 12c+ 16)3(u− 32)2.

2933 (b− 9/8)3

b4 − b3︸︷︷︸=:g(b)

=2(c−2

2 s4 + 94 + c22s−4)3

(c−22 s4 − 34 + c22s

−4)2︸︷︷︸=:f(s)

C′ : g(b) = f(s).

Claim:

ε : (b, s) 7→ (b, c−12 s2 + c2s

−2),

C : 2(u+ 96)3(c+ 11) = −(c2 + 12c+ 16)3(u− 32)2.

3-torsion points

Lemma 3.4.2

There are no rational points on C coming from C′ via ε.

Ingredients of the proof:

I There are at least four rational points on C, namely (0,±6), (1,±6).

I The curve C is isomorphic over the rationals to the hyperelliptic curveH : y2 = 4x5 − 14x4 − 8x3 + 36x2 + 36x+ 9.

I Using the Chabauty’s method one can see that H admits only sixrational points.

I These six points correspond on C to the points (0,±6), (1,±6). Thisshows that these four points are the only rational points on C.

I These four points cannot come from rational points of C′ via ε, sincethe equations c−1

2 s2 + c2s−2 = ±6 have no rational solutions.

3-torsion points

Lemma 3.4.2

3-torsion points

Lemma 3.4.2

3-torsion points

Lemma 3.4.2

3-torsion points

Lemma 3.4.2

3-torsion points

Lemma 3.4.2

Full torsion subgroup

Mazur Theorem

Let E/Q be an elliptic curve defined over Q. Then the torsion subgroupETor(Q) of E(Q) is isomorphic to one of the following fifteen groups:

Z/NZ with 1 ≤ N ≤ 10, N = 12

Z/2Z× Z/2NZ with 1 ≤ N ≤ 4.

Theorem 3.5.2

Let s ∈ Q× and F be the fiber of ψ1 over (s : 1) on W . Assume that F hasat least one rational point, P = (x0 : y0 : z0 : w0). Denote by E = (F, P )the fiber viewed as elliptic curve. Then

ETor(Q) = {(x0 : y0 : z0 : w0), (−x0 : −y0 : z0 : w0)} ' Z/2Z.

Full torsion subgroup

Mazur Theorem

Let E/Q be an elliptic curve defined over Q. Then the torsion subgroupETor(Q) of E(Q) is isomorphic to one of the following fifteen groups:

Z/NZ with 1 ≤ N ≤ 10, N = 12

Z/2Z× Z/2NZ with 1 ≤ N ≤ 4.

Theorem 3.5.2

Let s ∈ Q× and F be the fiber of ψ1 over (s : 1) on W . Assume that F hasat least one rational point, P = (x0 : y0 : z0 : w0). Denote by E = (F, P )the fiber viewed as elliptic curve. Then

ETor(Q) = {(x0 : y0 : z0 : w0), (−x0 : −y0 : z0 : w0)} ' Z/2Z.

Proof of the main theorem

Theorem 4.1.1

Let c1, c2 be two nonzero rationals and W be the surface defined as

W : x4 − 4c21y4 − c2z4 − 4c2w

4 = 0.

Let P = (x0 : y0 : z0 : w0) be a rational point on W with x0, y0 bothnonzero.If |2c1| is a square in Q×, then also assume that z0, w0 are not both zero.Then the set of rational points on the surface is Zariski dense.

Proof. From the hypotheses it follows that P is not on the intersection oftwo singular fibers.So let F be a smooth fiber passing through P .Consider the point P ′ = (−x0 : y0 : z0 : w0) ∈ F : from the hypotheses itfollows that P ′ 6= P, (−x0 : −y0 : z0 : w0), hence it has infinite order.So we have infinitely many rational points on F . To each but finitely manyof them we can apply the same argument with respect to the otherfibration. Then the Zariski density follows.

Theorem 4.1.1

W : x4 − 4c21y4 − c2z4 − 4c2w

4 = 0.

Proof. From the hypotheses it follows that P is not on the intersection oftwo singular fibers.

So let F be a smooth fiber passing through P .Consider the point P ′ = (−x0 : y0 : z0 : w0) ∈ F : from the hypotheses itfollows that P ′ 6= P, (−x0 : −y0 : z0 : w0), hence it has infinite order.So we have infinitely many rational points on F . To each but finitely manyof them we can apply the same argument with respect to the otherfibration. Then the Zariski density follows.

Theorem 4.1.1

W : x4 − 4c21y4 − c2z4 − 4c2w

4 = 0.

Proof. From the hypotheses it follows that P is not on the intersection oftwo singular fibers.So let F be a smooth fiber passing through P .

Consider the point P ′ = (−x0 : y0 : z0 : w0) ∈ F : from the hypotheses itfollows that P ′ 6= P, (−x0 : −y0 : z0 : w0), hence it has infinite order.So we have infinitely many rational points on F . To each but finitely manyof them we can apply the same argument with respect to the otherfibration. Then the Zariski density follows.

Theorem 4.1.1

W : x4 − 4c21y4 − c2z4 − 4c2w

4 = 0.

Proof. From the hypotheses it follows that P is not on the intersection oftwo singular fibers.So let F be a smooth fiber passing through P .Consider the point P ′ = (−x0 : y0 : z0 : w0) ∈ F : from the hypotheses itfollows that P ′ 6= P, (−x0 : −y0 : z0 : w0), hence it has infinite order.

So we have infinitely many rational points on F . To each but finitely manyof them we can apply the same argument with respect to the otherfibration. Then the Zariski density follows.

Theorem 4.1.1

W : x4 − 4c21y4 − c2z4 − 4c2w

4 = 0.

Proof. From the hypotheses it follows that P is not on the intersection oftwo singular fibers.So let F be a smooth fiber passing through P .Consider the point P ′ = (−x0 : y0 : z0 : w0) ∈ F : from the hypotheses itfollows that P ′ 6= P, (−x0 : −y0 : z0 : w0), hence it has infinite order.So we have infinitely many rational points on F . To each but finitely manyof them we can apply the same argument with respect to the otherfibration. Then the Zariski density follows.

THE END

Thank you for the attention.

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