Cubic points on cubic curves and the Brauer-Manin obstruction on K3 surfaces

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Ronald van Luijk PIMS, Vancouver

University of British Columbia Simon Fraser University

July 10, 2007 Bristol

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Two problems:

(1) Are there cubic curves without cubic points?

(2) Is the Brauer-Manin obstruction the only one on K3 surfaces?

Goal:

(a) Explain the problems (b) Relate them

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Let X be a variety over Q.

If X has no points over R then X has no points over Q. If X has no points over Q_p then X has no points over Q.

Conics satisfy the Hasse principle:

If a conic C has a point over R and over Q_p for every p, then C has a point over Q.

If a variety X over a number field k has points over every completion of k, then we say that X is locally solvable everywhere (LSE).

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Cubic curves in general do not satisfy the Hasse principle. The curve C given by 3x3 + 4y3 + 5z3 = 0 in P2 is LSE, but has no points over Q (Selmer).

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The curve C given by 3x3 + 4y3 + 5z3 = 0 in P2 is LSE, but has no points over Q (Selmer).

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Question 1_{: Over what fields does C acquire points?} Question 2: Over cubic fields that are galois?

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The curve C given by 3x3 + 4y3 + 5z3 = 0 in P2 is LSE, but has no points over Q (Selmer).

Definition: A cubic point is a point over a cubic galois extension.

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Definition: A cubic point is a point over a cubic galois extension.

The line L : 711x + 172y + 785z = 0 intersects C in three cubic points.

Question 3: Does every cubic curve that is LSE have cubic points? (unknown)

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Brauer-Manin obstruction.

Let K be a number field with ring of ad`eles

A_K = Y

v_∈M_K

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Let K be a number field with ring of ad`eles

A_K = Y

v_∈M_K

′ _K_v _{(almost all coordinates are integral).}

Let X be a smooth, absolutely irreducible, projective variety over K. Then the set of ad`elic points is

X(A_K) = Y

v_∈M_K

X(Kv)

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For any scheme Z we set Br Z = H_´_et2 (Z,G_m). For any ring R we set Br R = Br Spec R.

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

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

ρ_S: X(S) _{→ Hom(Br X, Br S).}

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0 _→ Br K → Br A_K → Q/Z

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

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X(K) X(A_K)

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X(K) X(A_K)

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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) = _∅

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0 _{→ Hom(Br}₁X,Br K ) → Hom(Br₁X,Br A_K) → Hom(Br₁X,Q/Z) X(K) X(A_K) ρ_K ρA_K X(A_K)Br1_{= ψ}−1 1 (0) ψ₁

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X(A_K)Br(1) ₌ _∅ _⇒ _{X(K) =} _∅_.

There is a Brauer-Manin obstruction to the Hasse principle if X(A_K) _{6= ∅} and X(A_K)Br = _∅.

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For a class _{S of varieties over K the Brauer-Manin obstruction is}

the only obstruction _{to the Hasse principle if for every X} _{∈ S we have}

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X(A_K)Br(1) ₌ _∅ _⇒ _{X(K) =} _∅_.

For a class _{S of varieties over K the Brauer-Manin obstruction is}

the only obstruction _{to the Hasse principle if for every X} _{∈ S we have}

X(A_K)Br = _∅ _⇔ X(K) = _∅.

Conjecture: The Brauer-Manin obstruction is the only obstruction to the Hasse principle for rationally connected varieties.

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Examples of K3’s:

smooth surfaces of degree 4 in P3, Kummer surfaces.

Question 4: Is the Brauer-Manin obstruction the only obstruction to the Hasse principle for K3 surfaces?

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Let C be a smooth cubic curve over K in P2 and ρ the automorphism ρ : C _{× C → C × C,} (P, Q) _{7→ (Q, R),}

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Relating the two problems

Let C be a smooth cubic curve over K in P2 and ρ the automorphism ρ : C _{× C → C × C,} (P, Q) _{7→ (Q, R),}

with R the third intersection point of C with the line through P and Q. Let X_C be the minimal desingularization of the quotient (C _{× C)/ρ.}

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(i) C is LSE,

(ii) abc _{∈ K}∗ is not a cube,

(iii) C has no cubic points (with K as ground field). Then

X_C(A_K)Br1 _{6= ∅} _and _X

C(K) = ∅

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Theorem (vL)

Let C be the cubic curve in P2_K given by ax3+ by3+ cz3 = 0 and suppose (i) C is LSE,

(iii) C has no cubic points (with K as ground field). Then

X_C(A_K)Br1 _{6= ∅} _and _X

C(K) = ∅

sketch of proof:

(iii) implies X_C(K) = _∅.

Indeed, T _{∈ X}_C(K) corresponds to a galois-invariant orbit

{(P, Q), (Q, R), (R, P )} of ρ on C ×C, so galois acts by even permutations only and P, Q, R are defined over some cubic extension that is galois.

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(i) C is LSE,

(iii) C has no cubic points (with K as ground field). Then X_C(A_K)Br1 _{6= ∅} _and _X C(K) = ∅ sketch of proof: (iii) implies X_C(K) = _∅. (ii) implies Br₁ X_C = Br K.

Indeed, Br₁ X_C/ Br K ∼= H1(K, Pic X_C), and Pic X_C is defined over K(ζ₃, q3 a/c, q3 b/c), with galois group contained in (Z/3Z_×Z/3Z)⋊ Z/2Z.

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Theorem (vL)

Let C be the cubic curve in P2_K given by ax3+ by3+ cz3 = 0 and suppose (i) C is LSE,

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(i) C is LSE,