Tate-Shafarevich groups and K3 surfaces

Following (and elaborating on) a method of Logan and van Luijk, we exhibit explicit genus-2 curves whose Jacobians have nontrivial 2-torsion in their Tate-Shafarevich groups.


Introduction
Let C be a curve of genus 2 over a number field k, with Jacobian J.In an effort to describe the (finite) set C(k), we are led to the study of J(k).To determine its rank, we refer to a well-known exact sequence (1) 0 → J(k)/2J(k) → Sel (2) (k, J) → X(k, J) [2] → 0 where the middle group is effectively computable.See [17] for a comprehensive description of the computation, which has been implemented in the computer algebra system MAGMA ( [2]).So computing the group on the left is more or less equivalent to computing the rather mysterious group on the right.In this paper, we find examples of curves C over Q such that X(Q, J) [2] is nonzero, by finding explicit elements of this group.Such elements can be represented by 2-coverings X of J which have points everywhere locally but no k-points.The strategy, as in [12], is to prove that the Hasse principle fails for X by exhibiting a Brauer-Manin obstruction to rational points on the desingularization of the quotient X/ι, where ι is the involution corresponding to multiplication by −1.This desingularization is a K3 surface which can be given explicitly as the smooth complete intersection of 3 quadrics in P 5 .
The result is the following theorem.
Theorem 1.1.Let S be the set of primes splitting completely in a certain finite extension K/Q (this extension is given explicitly in the statement of Proposition 5.3).For all n equal to the product of primes in S, the genus-2 curve satisfies X(Q, Jac(C)) [2] = 0.
The curve with n = 1 was originally obtained by a computer search over products of polynomials of degree 2, 3, 1 with small coefficients.I thank Ronald van Luijk for introducing me to this topic and for many helpful conversations.I would also like to thank Bjorn Poonen for several enlightening comments, and in particular for the method of computing problematic smooth places outlined in the proof of the main theorem.
2. The Brauer-Manin obstruction 2.1.Generalities.First we briefly review the Brauer-Manin obstruction to the Hasse principle.Let V be a smooth proper k-variety, k a number field.Then the map V (A k ) → v V (k v ) is a bijection ( [16], pp.98-99).We will suppose that this set V (A k ) is nonempty.
For any scheme V we can define the Brauer group Br V = H 2 (V et , G m ).For an element A ∈ Br V , define the set where the sum is over all places v of k, and define (2) Here inv v : Br k v → Q/Z is an isomorphism if v is nonarchimedian, or the injection so we say that V has a Brauer-Manin obstruction to the Hasse principle if V (A k ) Br is empty, so that V (k) is as well.For many classes of varieties, it is believed that the Brauer-Manin obstruction to the Hasse principle is "the only one"; that is, if V (A k ) Br is nonempty, then so is V (k).It is not known (even conjecturally) whether or not the Brauer-Manin obstruction to the Hasse principle is the only one for K3 surfaces.

2.2.
The key isomorphism.The map V → Spec k induces a map Br k → Br V , which is injective if V (A k ) is nonempty.Elements in the image of this map are called constant algebras.Two elements of Br V which differ by a constant algebra cut out the same subset of V (A k ), so the intersection (2) defining V (A k ) Br need only be taken over a set of representatives of Br V Br k .
Proposition 2.1.For V a smooth projective geometrically integral k-variety, k a number field, with V (A k ) = ∅, there is an isomorphism Proof: This is a standard consequence of the Hochschild-Serre spectral sequence; see for instance [8], Proposition 1.3.7.
It is, unfortunately, very difficult in general to compute the inverse of the isomorphism (3) explicitly.The crux of the computation is an explicit use of the fact (due to Tate) that H 3 (k, k * ) = 0; that is, one must express an arbitrary 3-cocycle with values in k * as a coboundary.In the next section, we discuss one way around this problem.
Of course, quaternion algebras are quite general objects; the reason we study the special quaternion algebras defined above is the following standard lemma: Note that the quaternion algebra (c, g) will always split over the quadratic extension k( √ c), so its image in There is also a well-known formula for the local invariant of such a quaternion algebra in Br V : for any point Algorithm 2.4.Given a class of varieties over k, here is how we look for varieties V in that class with quaternion algebras of the above type generating nonconstant elements in Br 1 (V ): (1) Find a G k -invariant generating set Γ for Pic V (possibly a subgroup of Pic V will work as well; see the comments).(2) The action of the Galois group on Γ induces a map G k → Aut(Γ).By inflationrestriction, H 1 (k, Pic V ) will be isomorphic to H 1 (H, Pic V ), where H is the image of G k inside Aut(Γ).List the cohomology groups H 1 (H, Pic V ) for every subgroup H of Aut(Γ); these are the possibilities for Comments on the algorithm: Carrying out step (1) requires that we know quite a lot about the geometry of V .Even for general K3 surfaces, this is too hard.Some previous attempts to carry out this algorithm explicitly have restricted themselves to Del Pezzo surfaces (e.g.[8], [10], [9]), or special K3 surfaces such as diagonal quartic hypersurfaces ( [3]) or Kummer surfaces ( [1]).As we will see, the K3 surfaces we examine in this paper come equipped with a G k -invariant set Γ of 32 lines generating a subgroup of Pic V which is free of rank 17.
In Step (3), we usually search for a subgroup H maximal with respect to the given properties, so that we can study the broadest possible class of varieties V admitting a nonconstant quaternion algebra of the above type in Br 1 V .
We carry out Step (5) using the following procedure: let P = Pic V .Then there is an isomorphism sending (the class of) a divisor class e on the left to the cocycle c τ = 1 2 (τ e − e) (cf.[9], Lemma 3.1).Since we have computed H 1 (k, P ) [2], we can find a nontrivial c τ and hence a nontrivial e.If this cocycle trivializes upon restriction to G L , then we can find a e on the left defined over L, and then the divisor class d = 1 2 (σe − e) satisfies d + σd = 0 in Pic V L .Moreover, in this case the class in Br 1 (V )/ Br k of the quaternion algebra (c, g) we obtain at the end of the algorithm corresponds via the isomorphism 3 to the class of the cocycle c τ , so if we know that this cocycle is not a coboundary, we are guaranteed that the quaternion algebra we have found is nonconstant.
In practice, we usually write down conditions on V in Step (4) in such a way that generically H 1 (k, P ) [2] is nontrivial, because the image of G k → Aut(Γ) is the subgroup H we found in Step (3); but if the image of G k → Aut(Γ) is strictly contained in H, it might happen that the quaternion algebra (c, g) we construct is constant.
Similarly, if Γ generates a subgroup Q of P , this algorithm constructs a quaternion algebra whose corresponding cocycle in H 1 (k, Q) is not a coboundary.If Q is a proper subgroup of P , the restriction of this quaternion algebra to H 1 (k, P ) might be zero.So in this case (and the problem case in the above paragraph), we will get a bona fide element of Br V from the above algorithm, but it will not furnish a Brauer-Manin obstruction to rational points on V .

Genus-2 curves and K3 surfaces: a review of the theory
The material in this section is taken largely from [12].
3.1.Construction of the K3 surface.Let C be a genus-2 curve over a number field k and J its Jacobian.We wish to associate, to an element δ ∈ Sel (2) (k, J), a K3 surface V δ with points everywhere locally, such that if δ is in the image of J(k)/2J(k), then V δ has a rational point.
If we write q(x) = 5 i=0 a i x i , then q lies in V f,δ if and only if the coefficients of x 3 , x 4 , and x 5 vanish.Considered as polynomials in the a i , these coefficients C 3 , C 4 , C 5 are homogeneous of degree 2. Definition 3.3.For δ ∈ A * f , define V δ to be the variety in P 5 consisting of points (a 0 : • • • : a 5 ) such that 5 i=0 a i x i ∈ V f,δ .One shows ([12], Proposition 2.1.2) that V δ is a smooth complete intersection of the three quadrics C 3 , C 4 , C 5 , which makes it a K3 surface of degree 8.
Let H f be the kernel of the norm map (which is well-defined because the degree of f is even).Then lemma 5.1 of [17] describes a map ∆ k : J(k)/2J(k) → H f whose kernel has order 1 or 2; as usual, it is induced from the homomorphism Div 0 ⊥ (C)(k) → A * f defined by P → x(P ) − x, where Div 0 ⊥ (C) consists of divisors whose support is disjoint from that of div(y).In addition, the lemma gives necessary and sufficient conditions under which the kernel will have order 1.It suffices, for instance, for f to have a factor of odd degree.Cf.Theorem 13.2 of [15]; following section 5 of [17], we say that "k satisfies condition ( ‡)" (thinking of f as fixed) if the kernel has order 1.
If k satisfies condition ( ‡), we can identify Sel (2) (k, J) with the set of elements of H f whose images in H f ⊗ k v lie in the image of ∆ kv for all places v of k. 1Henceforth, we will assume that k satisfies ( ‡), and we will think of Sel (2) (k, J) as the subset of H f described in the above paragraph.The goal is then to compute that set and to look for elements in it which are not in the image of ∆ k .This is where the set V f,δ comes in: Proposition 3.2.6 of [12] shows that if δ ∈ Sel (2) (k, J) is in the image of ∆ k , then there is a polynomial q ∈ V f,δ .For the sake of concreteness, we sketch the proof: δ will be congruent mod (A * f )2 k * to (x(P 1 ) − x)(x(P 2 ) − x) for some pair of points P 1 , P 2 ∈ C(k) which are defined and conjugate over a quadratic extension of k.We restate this fact: of degree 6, and let J be its Jacobian.Suppose that k satisfies ( ‡).If δ ∈ Sel (2) (k, J) is in the image of the map J(k)/2J(k) → Sel (2) (k, J) from ( 1), the K3 surface V δ has a rational point.
We can view this result from another perspective as well: elements δ ∈ Sel (2) (k, J) correspond to 2-coverings X δ of J, which are twists of J equipped with a map π : X → J defined over k which is a twist of multiplication by 2 on J k .Such 2-coverings inherit an involution defined over k descending from multiplication by −1 on J. Then the surface V δ is the minimal nonsingular model of the quotient of X δ by this involution.(It is a twist of the desingularized Kummer surface of J; cf.Chapter 16 of [7].) The element δ comes from J(k)/2J(k) if and only if X δ is the trivial twist of J, which happens if and only if X δ (k) is nonempty.But since δ ∈ Sel (2) (k, J), we must have that X δ (k v ) is nonempty for all places v of k.Thus X δ is a counterexample to the Hasse principle.Certainly if X δ has points in some field, then so does V δ ; so V δ has points everywhere locally, and V δ (k) = ∅ will imply that X δ (k) = ∅ and hence that δ maps to a nontrivial element of X(k, J) [2].Remark 3.5.It may be true that X δ (k) is empty while V δ (k) is nonempty; since X δ is really the surface in whose rational points we are interested, one might hope to work directly with it instead of V δ .Of course, the reason we do not do this is that the usual explicit description of X δ is as an intersection of 72 quadrics in P 15 (just as it is for the Jacobian itself-see [7], Chapter 2 for the construction).
Remark 3.6.If f (x) has a k-rational root, we can find a model C ′ for C of the form w 2 = g(u), deg g = 5.If we carry through the above constructions in A g instead of A f , we get a smooth complete intersection of two quadrics in P 4 , which is a Del Pezzo surface W β of degree 4, where β is the element of Sel (2) (k, Jac C ′ ) corresponding to δ (see [6] and [5]).We will see later that V δ is a double cover of W β .Then, just as in the previous remark, it may be true that W β (k) is nonempty while V δ (k) is empty; in fact, this happens for the curve given in Theorem 1.1 (assuming that the Brauer-Manin obstruction to the Hasse principle is the only one for Del Pezzo surfaces).
3.2.The 32 lines and Pic V δ .Here we review the construction of the 32 lines on Pic V δ and analyze the structure of Pic V δ as a G k -module.The ideas are taken from [12], but the notation will be different.
For δ ∈ A * f , let r 1 , . . ., r 6 be the roots of f in k.Fix a choice of square roots z i of δ(r i ) in k.
For an element s = (s 1 , . . ., s 6 ) ∈ (Z/2) 6 , define γ s to be the unique degree-5 polynomial Thus we have 32 lines L s indexed by elements of (Z/2) 6 / ( It is not hard to determine the intersection pairing as applied to the subgroup generated by these lines in Pic V δ ; one obtains The first line follows by the adjunction formula for K3 surfaces, and the second and third lines by direct calculations.From these intersection numbers we obtain Proposition 3.7.( [12], Proposition 2.1.21)The classes of the 32 lines on V δ generate a subgroup of Pic V δ isomorphic to Z 17 .Remark 3.8.In the generic case this subgroup equals all of Pic V δ (Proposition 2.1.30 of [12]).As noted above, even if V δ is not generic, we can still use the subgroup generated by the classes of the lines to give us an element of H 1 (k, Pic V δ ) via restriction.

The algorithm for V δ
Viewed in terms of Algorithm 2.4, the end of the previous section carries out Step (1) for the class of varieties of the form V δ ; then Γ is in our case the set of 32 lines on V δ defined earlier.We proceed to carry out the remaining steps of the algorithm.
Let G Γ be the group (Z/2) 6 / (1, 1, 1, 1, 1, 1) indexing the 32 lines.Then, by a straightforward analysis of the intersection pairing on Γ (which is Proposition 2.2.11 of [12]), Aut(Γ) is isomorphic to G Γ ⋊ S 6 , where the symmetric group acts on G Γ by permuting indices (which corresponds to permuting the square roots z i of δ(r i )), and G Γ acts on itself by addition.
For δ ∈ Sel (2) (k, J), the image of G k → Aut(Γ) must lie in a subgroup of index 2 inside Aut(Γ), because the norm of δ is required to be a square in k.Indeed, this index-2 subgroup is the semi-direct product of S 6 with the index-2 subgroup of G Γ of elements whose sum is 0. This is a group G of order 11520 which acts on Z 17 in a prescribed way.MAGMA can enumerate its subgroups H and compute H 1 (H, Z 17 ) for each one.This is Step (2).
We look in Step (3) for maximal subgroups H such that H 1 (H, Z 17 ) = Z/2 and H 1 (H ′ , Z 17 ) = 0 for some index-2 subgroup H ′ ⊂ H. MAGMA finds two conjugacy classes of such subgroups.One class consists of subgroups of order 96; the other consists of subgroups of order 128.Here we exhibit a Brauer-Manin obstruction coming from the first conjugacy class; presumably we could use the same techniques to try to find one coming from the second class.

4.1.
The subgroup H 96 and the shape of f (x).Consider polynomials f (x) of the shape where f i is an irreducible polynomial of degree i.Let C be the genus-2 curve y 2 = f (x).Pick the ordering of the roots of f that lists the two roots of f 2 , then the three roots of f 3 , then r 6 .Suppose also that δ ∈ Sel (2) (k, J) satisfies the condition that δ(r 1 ) is a square in Q(r 1 ), where r 1 is a root of f 2 (x).Then the image of G k → Aut(Γ) will sit inside the subgroup of G Γ ⋊ S 6 generated by the elements (0, 0, 1, 0, 0, 1), (0, 0, 1, 1, 1, 1), ( 12), (345), (34) where the first two elements are in G Γ and the last three elements are permutations in S 6 .
Of course we will have to use the fact that V δ has points everywhere locally.The easiest way to solve this computational problem, generally speaking, is to reduce it to solving a certain norm equation which we are guaranteed has a solution by the Hasse Norm Theorem.We now show how this can be accomplished for V δ .4.2.The divisor D and the quaternion algebra.The strategy is to take a divisor E in the class of d defined over a higher-degree extension of k, and then to subtract the divisor of a judiciously chosen rational function to E in order to obtain a divisor defined over L. That is, we want τ (E − (h)) = E − (h) for all τ ∈ G L .Here we begin with E = L (0,0,0,1,1,0) + L (0,0,0,1,1,1) − L (0,0,1,0,0,0) + L (0,0,1,0,0,1) , The stabilizer H 12 of the divisor E (not its class) in H 96 is generated by (0, 0, 1, 1, 0, 0) • (34), ( 12), (45).It has order 12, and index 4 in H 48 .We wish to find a function (h) such that E − (h) is fixed by H 48 .That is, for all τ ∈ H 48 , we want E − τ E = div(h/τ h).
There are four left cosets of H 12 in H 48 , and we first identify what the divisor of h/τ h must be for τ in each coset.For ease of notation, we identify an element s ∈ G Γ with the binary number 6 i=1 2 6−i s i .So in this notation . Let E I,B be the sum of the eight lines L s , s ∈ G Γ , where s i j = b j .Then E I,B is a hyperplane section.
Proof: This is a restatement of Lemma 2.1.24in [12].The point of this proposition is that we can write the above divisors as differences of hyperplane sections.For 3 ≤ i ≤ 6, define p i to be a linear polynomial cutting out E {1,2,i},{0,0,0} ; and define q i to be a linear polynomial cutting out E {1,2,i},{0,0,1} .Then we see that The proof of Hilbert Theorem 90 suggests the following choice of h: Subject to some conditions on the p i and q i , which have so far only been defined up to scalar multiples, we will show that this h gives us the rational function we want.Here we outline the conditions we will need.
Condition 1: We require p i and q i to have coefficients in k(r 1 , z i , z 6 ), and to be defined so that (0, 0, 1, 1, 1, 1)p i = q i .This condition will be immediate from the construction of the p i and q i we will outline below.
Condition 2: We need that σp i = p j and σq i = q j if σ ∈ H 96 ∩ S 6 sends i to j, and so that This condition is easy to satisfy, as we will shortly see.
Condition 3: We also want that p i q i = p j q j for 3 ≤ i, j ≤ 5.This condition is more difficult to satisfy; we will need to use the fact that V δ has points everywhere locally.Proposition 4.2.If p i and q i satisfy Conditions 1,2,3, then the divisor E − (h) is defined over the field L = k( δ(r 6 )).
Proof: By condition 1, the rational function h is defined over the field k(r 1 , z 3 , z 4 , z 5 ).Note that z 6 is automatically contained in this field extension because of the requirement that the product of the z i is in k (and the requirement that z 1 ∈ k(r 1 )).The Galois group of this field extension is a subgroup of H 96 .So we must only show that E − τ E = div(h/τ h) for every τ ∈ H 48 .
Proposition 4.3.In the above situation, we can construct functions p i and q i satisfying Conditions 1, 2, 3.
Define q i by replacing z i with −z i everywhere.We must now show that the intersection divisor of the hyperplane cut out by p i with V δ is in fact E {1,2,i},{0,0,0} .
To see this, let q(x) = 5 j=0 a j x j , and suppose that q(r m ) = 1 z m (tr m + u) for m = 1, 2, i. Then Hence the lines whose classes appear in E {1,2,i},{0,0,0} lie on the hyperplane cut out by p i .
Similarly we can show that the intersection divisor of the hyperplane cut out by q i with V δ is E {1,2,i},{0,0,1} .Now Conditions 1 and 2 are immediate from the construction of p i and q i .Note also that p i q i actually has coefficients in k(z 2 i ), as it is invariant under the transposition of the indices 1, 2 as well as the map z i → −z i .Now we must arrange for Condition 3 to hold by multiplying the p's and q's by suitable constants.For 3 ≤ i ≤ 6, consider the rational function p i q i p 6 q 6 on V .Note that its divisor is 0, so it is a nonzero constant u i .To evaluate this constant, let P v ∈ V (k v ) be a point not in the support of this rational function, and note that Let L i = k(z 2 i ).For any place w of L i , the expression u i = a v (1/b v ) exhibits u i ∈ L i as the product of a norm from (L i ) w (z i ) and a norm from (L i ) w (z 6 ).By Lemma 4.4, we see that and if we let d j be the image of d i under an element of H 96 ∩ S 6 transposing i and j, we see that and thus if we replace p i by p i /d i and q i by q i /(σd i ), we have found functions satisfying Condition (3).Once we have constructed h such that E−τ E = div(h/τ h), the rational function g we want will be a function whose divisor is (E −(h))+σ(E −(h)), where σ is any element of H 96 \H 48 .The most convenient σ to choose is certainly (0, 0, 1, 1, 1, 1) ∈ G Γ , because σE = −E.Thus the rational function g we use is a function whose divisor is −(h • (0, 0, 1, 1, 1, 1)h).We can drop the negative sign, since we have a quaternion algebra, i.e. (c, g) = (c, 1/g).So we can take g = h • (0, 0, 1, 1, 1, 1)h times a constant where we know that we will be able to find a constant so that g is invariant under the action of the Galois group.
Remark 4.5.It appears at first glance that g will always be a norm from k(z 6 ) to k, which would make the quaternion algebra we have constructed trivial in Br V .But h is not itself defined over k(z 6 ), so indeed this algebra is nontrivial in general.
Remark 4.6.Here is how we arrange for 3 to hold in practice.We know from the proof above that there are constants d i ∈ k(z i , z 6 ) (which are conjugate to each other under the action of S 3 on the indices 3, 4, 5) such that ( We consider the normal form N i of p i q i with respect to a Gröbner basis for the ideal generated by the defining equations of V δ .The normal form is a uniquely determined representative of the equivalence class of p i q i modulo this ideal; it is clear from the construction of the normal form that the normal forms of p i q i and p j q j are permuted just as p i q i and p j q j are, by the action of S 3 on the indices.Fixing a monomial appearing in N i and calling its coefficient c i , we note that c i ∈ k(z i ) should satisfy Note also that N σ (d i ) c i is independent of i, but since this quantity is in k(z 2 i , z 6 ) * for each i, it follows that it lies in k(z 6 ) * .
So we search for n ∈ k(z 6 ) * such that c i n is a norm from k(z i , z 6 ) to k(z 2 i , z i z 6 ); we are guaranteed that such n exist, and in practice we find them quickly (e.g. using the NormEquation function in MAGMA).
In fact, in every explicit example the author has written down, he has been able to find n ∈ k * ; but the author does not know if it is always possible to find n ∈ k * in general.

Proof of the main theorem
Theorem 5.1.Let C be the hyperelliptic curve Proof: Let J = Jac(C).Let f (x) be the sextic polynomial defining C. First note that it is easy to show that J(Q)/2J(Q) is a vector space over Z/2 of dimension at least 2; J(Q) [2] = Z/2 because there is a quadratic factor of f (x), and there is also the image of a point in J(Q) coming from the point (−1, 0) ∈ C(Q).According to Stoll's algorithm from [17], now implemented in MAGMA as TwoSelmerGroup, we can compute Sel (2) (Q, J), and find that it is a vector space over Z/2 of dimension equal to 4.Moreover, since C has a rational point, its Jacobian is even (in the language of [14]); that is, the order of X(Q, J) [2] is a square.Hence it is either 1 or 4. We must therefore only exhibit one nontrivial element of X(Q, J) [2] to complete the proof. Define Then MAGMA shows that δ gives an element of Sel (2) (J).The K3 surface V δ is given by the vanishing of the polynomials . Since δ(r 6 ) = −7, we will be using the Azumaya algebra (−7, g), where g is constructed as in the previous section.Note that the denominator of g, which is a constant multiple of h•(0, 0, 1, 1, 1, 1)h, is p 4 p 5 q 4 q 5 .This is a constant multiple of (p 6 q 6 ) 2 .Plugging in any point in V (k v ) to (p 6 q 6 ) 2 yields a norm from k v (z 6 ) to k v , so when we compute the invariant of g, we can ignore the contribution coming from its denominator.The numerator is a homogeneous polynomial F of degree 4, so we are reduced now to computing the expressions (−7, F (P v )) v for all places v of Q and all points P v ∈ V (k v ).(These expressions are invariant under projective scaling of the coordinates of P v , since F has even degree.) We wish to compute the primes of bad reduction for V δ next, for the purposes of invariant computations.This is a standard computation using the minors of the matrix of partial derivatives; we can also note that the only bad primes should be primes dividing the discriminant of the minimal Galois extension over which all the lines are defined.In our case the bad primes are 2, 3, 7, 83, 739, ∞.Note that if −7 is a square in Q p we know automatically that the quaternion algebra (−7, F/G) will have constant invariant 0 at p.So we do not have to analyze p = 2, 739.
These will not be the only primes to consider when we compute the possibilities for inv p A(P p ), P p ∈ V δ (Q p ).Here we repeat a standard lemma that allows us to restrict our consideration to a finite set of primes.Lemma 5.2.Let V be a smooth projective geometrically integral k-variety, k a number field, g ∈ k(V ) * , and let (L/k, g) be a quaternion algebra in Br V .Let v be a nonarchimedian place of k and suppose that v does not ramify in L and that V has smooth reduction at v. Then inv v (L/k, g)(P v ) is independent of the choice of P v ∈ V (k v ).
Proof: See Lemma 8.4, [9].So if p is not a bad prime for V δ and p does not ramify in k(z 6 ), then the invariants of the quaternion algebra (δ(r 6 ), g)(P p ) will be independent of the choice of P p .When will it be nonzero?
The proof of Lemma 8.4 of [9] implies that, if δ(r 6 ) is not a square mod p, the invariant of (δ(r 6 ), g)(P p ) will equal m p /2 ∈ 1 2 Z/Z, where m p is the integer such that m p V p appears in the divisor of g (here V p denotes the special fiber of the smooth model V of V δ over Spec Z p Pick a point P ∈ V δ (Q).Write g = F/G, where F and G are as above; they are homogeneous polynomials in six variables with integer coefficients.Since G is a norm from k(z 6 )(V δ ) to k(V δ ), we can ignore it for the purposes of invariant computations.If there is a prime p over p not dividing F (P ), then P p ∈ V p (F p ) is not in the support of g (or some rational function obtained from g by multiplying the denominator by a norm from k(z 6 )(V δ )), so the integer m must equal 0. So the set of primes at which inv p (δ(r 6 ), g)(P p ) is not always 0, independent of P p ∈ V δ (Q p ), is contained in the union of the set of bad primes with the set of primes of Z dividing N(F (P )), where N denotes the absolute norm down to Q.This is true for all P , so after combining results for various P we obtain the set {5, 61, 347, 739, 3433, 4337, 6833, 663149189, 69804594311} of primes p at which V δ has smooth reduction but the integer m p is possibly nonzero.After throwing out the primes for which −7 is a square in Q * p , we get {5, 61, 3433, 663149189}.
For each p in this set we find that m p is odd.This is not hard to check: as the invariant is constant, we need only lift one point not in the support of F/G in V δ (Q p ) to high enough precision in order to evaluate F at it.Since the invariant at each of these primes is 1/2, we obtain v inv v (−7, F/G)(P v ) = inv 3 (−7, F/G)(P 3 ) + inv 7 (−7, F/G)(P 7 ) + inv 83 (−7, F/G)(P 83 ) + inv ∞ (−7, F/G)(P ∞ ).
To carry out the invariant computation at ∞, we first show that the value of the function F on the points in V δ (R) is either always positive or always negative.To see this, consider the polynomial whose roots are ±z 3 , ±z 4 , ±z 5 ; this polynomial can be written as h(x 2 ), where h(x) = x 3 − 126x 2 + 7938x + 250047.Notice that h(x) has one real root, which is negative.Under any embedding of the splitting field of h(x 2 ) into C such that the image of z 2 3 is the negative real root, the images of z 2 4 and z 2 5 are distinct and conjugate.Choose such an embedding so that z 4 and −z 5 are complex conjugates.Then consider the action of complex conjugation on the p i and q i ; we see that it sends p 3 to q 3 and vice versa, while it sends p 4 to q 5 and p 5 to q 4 .Now F is a constant multiple of (p 3 p 5 + p 3 p 4 + p 4 p 5 + p 3 q 3 )(q 3 q 5 + q 3 q 4 + q 4 q 5 + p 3 q 3 ), and if we plug in a point in V δ (R) to both these factors, we see immediately that we get the product of a complex number and its conjugate.So if we can show that F is never zero on V δ (R), we can deduce that it is either always positive or always negative.It can be easily verified-either by plugging in a specific real-valued point or noticing that F is actually a positive constant multiple of the above function-that F is always positive.This shows that the invariant is actually zero.
To see that F never vanishes on V δ (R), note that if F (P ) = 0, then both factors of F vanish at P .So P lies on the intersection of the K3 surface V δ with the two hyperplanes cut out by the factors of F ; this intersection is zero-dimensional, and MAGMA computes that there are no real points on it.
At 83 we note that there is one singular F 83 -valued point in the special fiber V 83 , but it does not lift to any points mod 83 2 .By a theorem of Bright ([4], Theorem 1), the invariant is constant above any smooth F 83 -point, so we need merely write down all points in V δ (F 83 ), and lift each one to high enough precision in order to evaluate F/G at it.(Hensel's lemma guarantees that any lift of a smooth F 83 -point will lift to a Z 83 -point.)Doing this computation for each of the 6960 smooth points in V δ (F 83 ), we find that each point has a lift P mod 83 5 such that the 83-adic valuation of F (P ) is either 2 or 4. The invariant at P is 0 or 1/2 depending on whether or not F (P ) has even or odd valuation, respectively; so in all cases, we find that the invariant at 83 is zero.
At 7, we compute that V (F 7 ) has 71 points.One of these points, P s = (3 : 6 : 3 : 1 : 2 : 1), is singular, and the set {P ∈ V (F 7 ) : F (P ) = 0} has 15 elements (including the singular point).For the other 56 points, we find that F (P ) = 1, 2, or 4, which are the squares in F * 7 , so the invariant is zero at all these points.Again using Theorem 1 of [4], we find one lift P of each of the 14 remaining nonsingular points in V (F 7 ) to a point in V (Z 7 ) (at least one of whose coordinates is a unit-we will call such P normalized) and compute F (P ) for each such lift.In particular, we see that F (P ) ≡ 7 2 , 2 • 7 2 , or 4 • 7 2 mod 7 3 for each of these points, so F (P ) is a square in Z 7 , so the invariant is zero at these points as well.
Above the point P s , Bright's theorem does not apply, and we must actually consider all possible lifts of P s to P ∈ V (Z 7 ).A lengthy MAGMA computation shows that for any such normalized P , F (P ) ≡ 2 • 7 4 mod 7 5 , so once again the invariant is zero at all these points, and hence the invariant is zero at all points in V (Q 7 ).
At 3, we compute that V (F 7 ) has 40 points.All of them are singular, so Bright's theorem does not apply.A MAGMA computation similar to the one at 7 shows that if P is any normalized point in V (Z 3 ), v 3 (F (P )) = 5.So the invariant is 1/2 at all points in V (Q 3 ).
(cf. [6] and [5]).Then the K3 surface V δ obtained from C 1 is a double cover of W β , which is a smooth Del Pezzo surface of degree 4. It ramifies over the projectivization of {s(u) ∈ A g : β(u)s(u) 2 ≡ linear (mod g)}.
Proof: Note that k[x]/(f 5 (x)) ∼ = k[u]/(g(u)).Now there is a natural isomorphism between A * f /(A * f ) 2 Q * and A g /(A * g ) 2 , as the first group is isomorphic to (k[x]/f 5 (x)) * ⊕ k * squares • k * and the second group is isomorphic to (k[u]/(g(u))) * squares It is not hard to see that this isomorphism restricts to an isomorphism φ : Sel (2) (k, Jac C 1 ) → Sel (2) (k, Jac C 2 ) (where these Selmer groups are realized, as in [17], as subgroups of A * f /(A * f ) 2 k * and A g /(A * g ) 2 , respectively).The fact that W β is a smooth Del Pezzo surface of degree 4 is Lemma 17 of [5].Now suppose q ∈ V f,δ , so that δq 2 is congruent to a quadratic polynomial c mod f .Let c 1 (u) = u 2 c(1/u + a)/δ(a); then c 1 is also quadratic.If β = φ(δ), then β(u) ≡ u 6 δ(1/u + a)/δ(a) mod (A * g ) 2 , and if we choose this representative polynomial for the class of β, we see that the polynomial s(u) defined by the formula (5) s(u) ≡ u −2 q(1/u + a) mod g(u) is in W g,β , because β(u)s(u) 2 ≡ c 1 (u) (mod g(u)).Note that s(u) is well-defined because u is invertible mod g(u).So (5) gives a map V f,δ → W g,β , which induces a map V δ → W β .Now consider s(u) ∈ W g,β , and suppose that βs 2 ≡ c 1 mod g.Let r 1 , . . ., r 6 be the roots of f in k, where r 6 = a.Now q ∈ V f,δ maps to s if and only if Note that the quadratic polynomial associated to q is c(x) = δ(a)(x − a) 2 c 1 (1/(x − a)).So if q is to lie in V f,δ , we must have that q(a) 2 is the leading term of c 1 .Generically there are two choices for the square root of this leading term (these choices coincide when the leading term is 0).The result follows.
As mentioned above, the lemma implies that if W β has a Brauer-Manin obstruction to rational points, then so does V δ , and if W β fails the Hasse principle, then so does V δ .So the "K3 method" of exhibiting explicit elements of X [2] given above can be viewed as a generalization of the "Del Pezzo method" of [6] and [5].(Of course, the K3 method also works more generally; it applies to any genus-2 curve y 2 = f (x), while the Del Pezzo method works only for curves with odd-degree models.The example given in [12] is an example of an application of the K3 method to a curve without an odd-degree model.) For the curve y 2 = (x 2 − 5x + 1)(x 3 − 7x + 10)(x + 1) and specific choice of δ given in (4), this construction produces a Del Pezzo surface W β with points everywhere locally.As it happens, however, Logan's program for Del Pezzo surfaces ( [11]) shows that W β has no Brauer-Manin obstruction to rational points.We expect that it satisfies the Hasse principle as well (although the size of the coefficients in the defining equations for W β precludes a successful search for rational points of small naive height).In any case, this shows that the K3 method does in fact give a generalization of the Del Pezzo method even in the case when both methods apply.
) Given d and L, find a divisor D defined over L whose class is d.(Such a divisor is guaranteed to exist whenever V has points everywhere locally: see[8], Proposition 1.3.4.)Then our quaternion algebra is (c, g), where L = k( √ c) and g is a function whose divisor is D + σD.
and b v is a norm from k v (z 6 ) to k v .Lemma 4.4.([13], VIII.1.11)Let k be a field of characteristic = 2.An element c ∈ k * is the product of a norm from k( √ a) and a norm from k( √ b) if and only if, as an element of k( √ ab), it is a norm from k( √ a, √ b).
s 1 and s 2 differ in 0 or 6 places 1 if s 1 and s 2 differ in 1 or 5 places 0 otherwise