Nontrivial Sha for curves of genus 2 arising from K3 surfaces

Nontrivial Sha for curves of genus 2

arising from K3 surfaces

Ronald van Luijk MSRI, Berkeley Joint work with

Adam Logan Waterloo, Canada

May 4, 2006 San Diego

The entire talk on one page

Theorem 1. Let S be the set of primes that split completely in F = Q µ√ −1,√2, √5, q −3(1 + √2), q 6(1 + √5) ¶ .

Then for all n that are products of elements in S, the 2-part of the Tate-Shafarevich group of the Jacobian of the curve

Cn: y2 = −6n(x2 + 1)(x2 − 2x − 1)(x2 + x − 1) is nontrivial.

Proof in short:

We take a specific principal homogeneous space of the Jacobian of C_n that is everywhere locally solvable and show that it does not have any

Goal of this talk:

Understand previous page and why we care.

Main ingredients:

(a) 2-descent on Jacobians

Let C be a smooth, geom. irred. curve of genus 2 over a #-field K.

Let C be a smooth, geom. irred. curve of genus 2 over a #-field K.

Faltings: _{The set C(K) of rational points is finite.}

Many known methods to find C(K) (in special cases) require knowing the Mordell-Weil group J(K) of rational points on the Jacobian J of C.

Mordell-Weil Theorem:

Let C be a smooth, geom. irred. curve of genus 2 over a #-field K.

Faltings: _{The set C(K) of rational points is finite.}

Many known methods to find C(K) (in special cases) require knowing the Mordell-Weil group J(K) of rational points on the Jacobian J of C.

Mordell-Weil Theorem:

J(K) is finitely generated, i.e., J(K) ∼= J(K)_{tors ⊕} Zr. J(K)_tors J(K)/2J(K) =⇒ J(K)_tors r = rank J(K) =⇒ J(K) computable

There are cohomologically defined finite groups

Sel(2)(K, J), the 2-Selmer group,

X(K, J), the Shafarevich-Tate group, with

0 _{→ J(K)/2J(K) → Sel}(2)(K, J) _→ X(K, J)[2] _{→ 0.}

2-descent: compute Sel(2)(K, J) and decide which

There are cohomologically defined finite groups

Sel(2)(K, J), the 2-Selmer group,

X(K, J), the Shafarevich-Tate group, with

0 _{→ J(K)/2J(K) → Sel}(2)(K, J) _→ X(K, J)[2] _{→ 0.}

2-descent: compute Sel(2)(K, J) and decide which

of its elements come from J(K)/2J(K) (i.e., map to 0).

Assumption: We can compute Sel(2)(K, J).

Element of Sel(2)(K, J): a twist π : Y _{→ J} of the map [2] : J _{→ J}

(over K there is an isomorphism σ such that Y_K =∼_σ

π

J_K [2] J_K J_K

commutes), where Y is locally soluble everywhere.

Element of Sel(2)(K, J): a twist π : Y _{→ J} of the map [2] : J _{→ J}

(over K there is an isomorphism σ such that Y_K =∼_σ

π

J_K [2] J_K J_K

commutes), where Y is locally soluble everywhere.

The element Y _{→ J maps to 0 in} X(K, J)[2] iff Y (K) _{6= ∅.}

Solution: A quotient of _{Y .} Y_K σ₁ σ₂ π J_K [2] J_K J_K

Two isomorphisms σ₁ and σ₂ differ by translation by a 2-torsion point, as the morphism

Y_K (σ1,σ2) _J

K × JK

(P,Q)_{7→P −Q J} K has connected domain and finite image.

[_{−1] on J commutes with translation by 2-torsion points ⇒}

Solution: A quotient of _Y

[_{−1] on J commutes with translation by 2-torsion points ⇒}

it induces a unique involution ι of Y_K, defined over K. Set X = Y /ι.

Advantages:

• X is a complete intersection of 3 quadrics in P5. • X(K) = ∅ ⇒ Y (K) = ∅

Disadvantage:

Solution: A quotient of _Y

[_{−1] on J commutes with translation by 2-torsion points ⇒}

it induces a unique involution ι of Y_K, defined over K. Set X = Y /ι.

Advantages:

• X is a complete intersection of 3 quadrics in P5. • X(K) = ∅ ⇒ Y (K) = ∅

Disadvantage:

• This only gives sufficient conditions for Y (K) = ∅.

Situation: Such K3 surfaces are everywhere locally soluble, but may still satisfy X(K) = _{∅. Do they?}

Tool: Brauer-Manin obstruction.

For any scheme Z we set Br Z = H_´et2 (Z,G_m).

For any K-algebra S and any S-point x : Spec S _{→ X}, we get a homo-morphism x∗: Br X _{→ Br S}, yielding a map

Tool: Brauer-Manin obstruction.

For any scheme Z we set Br Z = H_´et2 (Z,G_m).

For any K-algebra S and any S-point x : Spec S _{→ X}, we get a homo-morphism x∗: Br X _{→ Br S}, yielding a map

ρ_S: X(S) _{→ Hom(Br X, Br S).}

Apply this to K and to the ring of ad`eles

A_K = Y

v_∈MK

From class field theory (and comparison theorems) we have

0 _→ Br K → BrA_K → Q_/Z

0 _{→ Hom(Br X, Br K ) → Hom(Br X, Br} A_K) → Hom(Br X,Q_/Z) X(K)

ρ_K ρA_K

0 _{→ Hom(Br X, Br K ) → Hom(Br X, Br} A_K) → Hom(Br X,Q_/Z)

X(K) X(A_K)

0 _{→ Hom(Br X, Br K ) → Hom(Br X, Br} A_K) → Hom(Br X,Q_/Z)

X(K) X(A_K)

ρ_K ρA_K

X(A_K)Br = ψ−1(0)

0 _{→ Hom(Br X, Br K ) → Hom(Br X, Br} A_K) → Hom(Br X,Q_/Z) X(K) X(A_K) ρ_K ρA_K X(A_K)Br = ψ−1(0) ψ X(A_K)Br = _{∅ ⇒} X(K) = _∅

0 _{→ Hom(Br1}X,Br K ) → Hom(Br₁X,BrA_K) → Hom(Br₁X,Q_/Z) X(K) X(A_K) ρ_K ρA_K X(A_K)Br1_{= ψ}−1 1 (0) ψ₁ X(A_K)Br1_{= ∅ ⇒ X(K) = ∅} Br₁ X = ker(Br X _{→ Br X)}

X(A_K)Br1 _{= ∅ ⇒ X(K) = ∅.} Two steps:

• Compute Br1 Z/ Br K for the desingularization(!) Z of X = Y /ι. The Hochschild-Serre spectral sequence gives

X(A_K)Br1 _{= ∅ ⇒ X(K) = ∅.} Two steps:

• Compute Br1 Z/ Br K for the desingularization(!) Z of X = Y /ι. The Hochschild-Serre spectral sequence gives

Br₁ Z/ Br K ∼= H1(G_K, Pic Z).

X(A_K)Br1 _{= ∅ ⇒ X(K) = ∅.} Two steps:

• Compute Br1 Z/ Br K for the desingularization(!) Z of X = Y /ι. The Hochschild-Serre spectral sequence gives

Br₁ Z/ Br K ∼= H1(G_K, Pic Z).

• Compute Z(A_K)Br1_.

Making the theory and the ideas explicit

Consider C : y2 = f , with f _{∈ K[X] separable of degree 6.} Set A = K[θ] = K[X]/f , a product of fields.

(P, Q) (x_P − θ)(x_Q − θ) J(K)/2J(K) _{φ A}∗_/K∗_A∗2

Making the theory and the ideas explicit

Consider C : y2 = f , with f _{∈ K[X] separable of degree 6.} Set A = K[θ] = K[X]/f , a product of fields.

(P, Q) (x_P − θ)(x_Q − θ) J(K)/2J(K) _{φ A}∗_/K∗_A∗2

J(Kv)/2J(Kv) _φ

v A∗v/Kv∗A∗v 2 We consider the computable fake Selmer group

For δ _{∈ Sel}(2)_f (K, J) _{⊂ A}∗/K∗A∗2 we set

Xδ = {α ∈ A∗ : δα2 = c2θ2 + c1θ + c0} ⊂ A∗. Then the corresponding

X_{δ ⊂} P(A)

For δ _{∈ Sel}(2)_f (K, J) _{⊂ A}∗/K∗A∗2 we set

Xδ = {α ∈ A∗ : δα2 = c2θ2 + c1θ + c0} ⊂ A∗. Then the corresponding

X_{δ ⊂} P(A)

is a complete intersection of three quadrics C₃ = C₄ = C₅ = 0.

Fact: The K3 surface X_δ is the quotient Y_δ/_hιi. For every ξ _{∈ AK} with ξ2 = δ, the set

After fixing one square root ξ of δ in A_K, the remaining square roots of δ are parametrized by µ₂(A_K) and the lines by µ₂(A_K)/_{±1}.

Since we have A_K =∼ L

ω,f (ω)=0 K, the lines are parametrized by parti-tions of the roots of f into two parts.

1 2 12 34 134 156 56 35 36 45 46 135 146 136 145

Proposition: Generically the group Pic Z has rank 17, generated by the set Λ of 32 lines.

Corollary: _{GK acts on Pic Z through a subgroup of Aut}intΛ (which has size 23040).

We can compute H1(G, Pic Z) for all 2455 possible subgroups G of Autint Λ (up to conjugacy).

Z_/2 Z_/4 (Z_/2)2 Z_/4 (Z_/2)2 (Z/2)3 (Z/2)3 Z_{/2 ×} Z_{/4 (}Z_/2)2 Z_{/2 ×} Z_/4 Z_{/2 ×} Z_/4 Z_{/2 Aut}int_Λ 12 15 20 120 24 24 48 48 96 96

Z_/4 (Z_/2)2 Z_/4 (Z_/2)2 Z_{/2 Aut}int_Λ 12 15 20 120 24 24 48 48 96 96 (Z/2)3 (Z/2)3 Z_{/2 ×} Z_{/4 (}Z_/2)2 Z_{/2 ×} Z_/4 Z_{/2 ×} Z_/4

1 2 12 34 134 156 56 35 36 45 46 135 146 136 145 An elliptic fibration

There is a group E of order 384 such that if the Galois action factors through E, then Z has an elliptic fibration over K, where four of the fibers consist of four lines intersecting in a cycle.

H1(G, Pic Z) =∼ Br1 Z/ Br K

H1(G, Pic_vert Z)

We need to express a nonzero element of Br₁ Z/ Br K as an Azumaya algebra...

Definition: A central simple algebra over K is a simple K-algebra with center equal to K.

“Definition”: An Azumaya algebra is to a central simple algebra, as a sheaf is to a vectorspace.

For a, b _{∈ K, let (a, b) denote the central simple algebra over K,} gener-ated by 1, i, j, ij with i2 = a, j2 = b, and ji = _−ij.

Proposition 2. _{Let φ : V →} P1 be an elliptic fibration, and suppose that V has bad fibers of type I₄ over P = (α : 1), where [Q(α) : Q] = 4. Suppose further that the field of definition of the components of the fiber at P is Q(α, √c), where c _∈ Q(α) is of square norm. Then the pullback of the algebra cores_Q(α)/Q(c, t _{− α) ∈ Br}Q(t) to V (where t

Theorem 3. Let f factor as the product of three quadratic polynomials f₁, f₂, f₃, let δ_i be the images of δ in Q[x]/(f_i), and let V be the K3 surface constructed from f, δ. Suppose further that the splitting field of f is of degree 8; that the norm of δ₁ is a square; that the norms of δ₂ and δ₃ multiplied by the discriminant of f₁ are squares; and that the δ_i are otherwise generic, so that the field of definition of the lines of V has degree 32. Then both elliptic fibrations associated to the factorization f = (f₁)(f₂f₃) satisfy the conditions of Proposition 2, and the elements of the vertical Brauer group constructed in that proposition map to the same (nontrivial) element of H1(Q, Pic V ) and hence to the same element of Br V / Br K.

Theorem 4. Let f factor as the product of three quadratic polynomials f₁, f₂, f₃, let δ_i be the images of δ in Q[x]/(f_i), and let V be the K3 surface constructed from f, δ. Suppose further that the splitting field of f is of degree 8; that the norm of δ₁ is a square; that the norms of δ₂ and δ₃ multiplied by the discriminant of f₁ are squares; and that the δ_i are otherwise generic, so that the field of definition of the lines of V has degree 32. Then both elliptic fibrations associated to the factorization f = (f₁)(f₂f₃) satisfy the conditions of Proposition 2, and the elements of the vertical Brauer group constructed in that proposition map to the same (nontrivial) element of H1(Q, Pic V ) and hence to the same element of Br V / Br K.

Cn: y2 = −6nf1f2f3

I will spare you the details of actually evaluating the Azumaya algebra on the rational points.

The local invariants are constant at every prime. It is 1₂ at 2 and it equals 0 everywhere else. This does not sum to 0...

We conclude that X_δ does not have any rational points, so neither does the homogeneous space Y_δ that covers X_δ.

I will spare you the details of actually evaluating the Azumaya algebra on the rational points.

The local invariants are constant at every prime. It is 1₂ at 2 and it equals 0 everywhere else. This does not sum to 0...

We conclude that X_δ does not have any rational points, so neither does the homogeneous space Y_δ that covers X_δ.

S = ½ p : p split in Q µ√ −1,√2,√5, q −3(1 + √2), q 6(1 + √5) ¶¾ . Cn: y2 = −6n(x2 + 1)(x2 − 2x − 1)(x2 + x − 1). δ = ³3,_{−(1 +} √2), (1 + √5)/2´

A standard (but slightly tedious) computation:

S = ½ p : p split in Q µ√ −1,√2,√5, q −3(1 + √2), q 6(1 + √5) ¶¾ . C_n: y2 = _−6n(x2 + 1)(x2 _{− 2x − 1)(x}2 + x _{− 1).} δ = ³3,_{−(1 +} √2), (1 + √5)/2´

A standard (but slightly tedious) computation:

δ is in the Selmer group, i.e., Y_δ is locally solvable everywhere.

We reduce to the case of primes of bad reduction for C_n, namely ∞, 2, 3, 5, and primes dividing n.