K3 surfaces of Picard rank one which are double covers of the projective plane

Andreas-Stephan ELSENHANS^{a} and Jörg JAHNEL^{a} aUniversität Göttingen, Mathematisches Institut, Bunsenstraße 3–5, D-37073 Göttingen, Germany^{1}

Abstract. We construct explicit examples of K3 surfaces over ℚ which are of degree 2 and the geometric Picard rank of which is equal to 1. We construct, particularly, examples in the form w^{2} = det M where M is a symmetric ( 3 × 3) -matrix of ternary quadratic forms or a symmetric ( 6 × 6) -matrix of ternary linear forms. Our method is based on reduction modulo p for p = 3 and 5. Keywords. K3 surface, Picard group, plane sextic, tritangent, conic tangent in six points, Frobenius, Tate conjecture

Introduction

In the projective plane, let a smooth curve B of degree 6 be given by f_{6}( x, y, z) = 0. Then, w^{2} = f_{6}( x, y, z) de�nes an algebraic surface S in a weighted projective space. We have a double cover π: S → P^{2} rami�ed at π^{− 1}( B) .

According to the Enriques classi�cation of algebraic surfaces, S is an example of a K3 surface (of degree two). In general, a K3 surface is a simply connected, projective algebraic surface the canonical class of which is trivial.

Examples 1.1 K3 surfaces embedded into P^{n} are automatically of even degree.

K3 surfaces of small degree may be described, explicitly. A K3 surface of degree two is a double cover of P^{2}, rami�ed in a smooth sextic. K3 surfaces of degree four are smooth quartics in P^{3}. A K3 surface of degree six is a smooth complete intersection of a quadric and a cubic in P^{4}. And, �nally, K3 surfaces of degree eight are smooth complete intersections of three quadrics in P^{5}.

The Picard group of a K3 surface is known to be isomorphic to ℤ^{n} where n may range from 1 to 20. It is generally known that a generic K3 surface over ℂ is of Picard rank one.

Nevertheless, it seems that the �rst explicit examples of K3 surfaces of geometric Picard rank one have been constructed as late as in 2005 [5]. All these examples are of degree four.

The goal of the work described in this article is to provide explicit examples of K3 surfaces de�ned over ℚ which are of degree two and geometric Picard rank one.

1The computer part of this work was executed on the Sun Fire V20z Servers of the Gauß Laboratory for Scienti�c Computing at the Göttingen Mathematisches Institut. Both authors are grateful to Prof. Y. Tschinkel for the permission to use these machines as well as to the system administrators for their support.

Let 𝒮 be a K3 surface over a �nite �eld 𝔽_{q}. Then, we have the �rst Chern class

c_{1}: Pic( 𝒮) − → H _{ét}^{2}( 𝒮_{\overline{𝔽}_{q}} , \overline{ℚ}_{l}( 1) )

into l-adic cohomology at our disposal. There is a natural operation of the Frobenius on H _{ét}^{2}( 𝒮_{\overline{𝔽}_{q}} , \overline{ℚ}_{l}( 1) ) . All eigenvalues are of absolute value 1. The Frobenius operation on the Picard group is compatible with the operation on cohomology.

Every divisor is de�ned over a �nite extension of the ground �eld. Consequently, on the subspace Pic( 𝒮_{\overline{𝔽}_{q}}) ⊗ _{ℤ} \overline{ℚ}_{l} ↪ H _{ét}^{2}( 𝒮_{\overline{𝔽}_{q}}, \overline{ℚ}_{l}( 1) ) , all eigenvalues are roots of unity. Those correspond to eigenvalues of the Frobenius operation on H _{ét}^{2}( 𝒮_{\overline{𝔽}_{q}}, \overline{ℚ}_{l}) which are of the form qζ for ζ a root of unity.

We may therefore estimate the rank of the Picard group Pic( 𝒮_{\overline{𝔽}_{q}}) from above by counting how many eigenvalues are of this particular form.

Estimates from below may be obtained by explicitly constructing divisors. Under certain circumstances, it is possible, that way, to calculate rkPic( 𝒮_{\overline{𝔽}_{q}}) , exactly.

Our general strategy is to use reduction modulo p. If S is a K3 surface over ℚ then there is the inequality

rkPic( S\overline{ℚ} ) ≤ rkPic( S_{\overline{𝔽}_{p}})

which is true for every prime p of good reduction.

Remark 1.2 Consider a complex K3 surface S. Since H ^{1}( S, 𝒪_{S}) = 0, the Picard group of S is discrete and the �rst Chern class

c_{1}: Pic( S) → H ^{2}( S, ℤ) ⊂ H ^{2}( S, ℂ)

is an injection. For divisors, numerical and homological equivalence are known to coincide [4, Corollary 1]. This shows, Pic( S) equals the group of divisors modulo numerical equivalence.

2. Explicit divisors - Geometric constructions over \overline{𝔽}_{p}

In order to estimate the rank of the Picard group from below, one needs to explicitly construct divisors. Calculating discriminants, it is possible to show that the corresponding divisor classes are linearly independent.

Assumption 2.1 For the algebro-geometric considerations described in this section, we assume that we work over a ground �eld which is algebraically closed of characteristic ≠ 2.

Construction 2.2 i) One possible construction is to start with a branch curve “f_{6} = 0” which allows a tritangent line G. The pull-back of G to the K3 surface 𝒮 is a divisor splitting into two irreducible components. The corresponding divisor classes are linearly independent. ii) A second possibility is to use a conic which is tangent to the branch sextic in six points.

Both constructions yield a lower bound of 2 for the rank of the Picard group.

Tritangent. Assume, the line G is a tritangent to “f_{6} = 0”. This means, the restriction of f_{6} to G ≅ P^{1} is a section of 𝒪( 6) , the divisor of which is divisible by 2 in Div( G) . As G is of genus 0, this implies f_{6}| _{G} is the square of a section f ∈ Γ ( G, 𝒪( 3) ) . The form f_{6} may, therefore, be written as f_{6} = f̃^{2}+ lq_{5} for l a linear form de�ning G, f̃ a cubic form lifting f, and a quintic form q_{5}.

Consequently, the restriction of π to π^{− 1}( G) is given by an equation of the form w^{2} = f^{2}( s, t) . We, therefore, have π^{∗}( G) = D_{1} + D_{2} where D_{1} and D_{2} are the two irreducible divisors given by w = ± f( s, t) . Both curves are isomorphic to G. In particular, they are projective lines.

The adjunction formula shows − 2 = D_{1}( D_{1} + K ) = D_{1}^{2}. Analogously, D_{2}^{2} = − 2. Finally, we have G^{2} = 1. It follows that ( D_{1}+ D_{2}) ^{2} = 2 which yields D_{1}D_{2} = 3. For the discriminant, we �nd

| \begin{matrix}− 2 & 3 \\ 3 & − 2 \end{matrix} | = − 5 ≠ 0

guaranteeing rkPic( 𝒮) ≥ 2.

Remark 2.3 We note explicitly that this argument works without modi�cation if two or all three points of tangency coincide.

Conic tangent in six points. If C is a conic tangent to the branch curve “f_{6} = 0” in six points then, for the same reasons as above, we have π^{∗}( C) = C_{1} + C_{2} where C_{1} and C_{2} are irreducible divisors. Again, C_{1} and C_{2} are isomorphic to C and, therefore, of genus 0. This shows C_{1}^{2} = C_{2}^{2} = − 2. Further, C^{2} = 4 which implies ( C_{1}+ C_{2}) ^{2} = 8 and C_{1}C_{2} = 6. Here, for the discriminant, we obtain

| \begin{matrix}− 2 & 6 \\ 6 & − 2 \end{matrix} | = − 32 ≠ 0.

Thus, rkPic( 𝒮) ≥ 2 in this case, too.

Remark 2.4 Further tritangents or further conics which are tangent in six points lead to even larger Picard groups.

3. Explicit divisors – Practical tests over 𝔽_{q}

A test for tritangents. The property of a line of being a tritangent may easily be written down as an algebraic condition. Therefore, tritangents may be searched for, in practice, by investigating a Gröbner basis.

More precisely, a general line in P^{2} can be described by a parametrization

g_{a, b} : t ↦ [ 1 : t : ( a + bt) ] .

g_{a, b} is a (possibly degenerate) tritangent of the sextic “f_{6} = 0” if and only if f_{6}∘ g_{a, b} is a perfect square in \overline{𝔽}_{q}[ t] . This means,

f_{6}( g_{a, b}( t) ) = ( c_{0}+ c_{1}t + c_{2}t^{2}+ c_{3}t^{3}) ^{2}

is an equation which encodes the tritangent property of g_{a, b}. Comparing coef�cients, this yields a system of seven equations in c_{0}, c_{1}, c_{2}, and c_{3} which is solvable if and only if g_{a, b} is a tritangent. The latter may be understood as well as a system of equations in a, b, c_{0}, c_{1}, c_{2}, and c_{3} encoding the existence of a tritangent of the form above.

Using Magma, we compute the length of 𝔽_{q}[ a, b, c_{0}, c_{1}, c_{2}, c_{3}] modulo the corresponding ideal I . This is twice the number of the tritangents detected.

The remaining one dimensional family of lines may be tested analogously using the parametrizations g_{a} : t ↦ [ 1 : a : t] and g: t ↦ [ 0 : 1 : t] .

Remarks 3.1 a) To compute the length of 𝔽_{q}[ a, b, c_{0}, c_{1}, c_{2}, c_{3}] / I , a Gröbner basis of I is needed. The time required to compute such a basis over a �nite �eld is usually a few seconds. From the Gröbner basis, the tritangents may be read off, explicitly.

b) Since the existence of a tritangent is a codimension 1 condition, one occasionally �nds tritangents on randomly chosen examples.

A test for conics tangent in six points. A non-degenerate conic in P^{2} allows a parametrization of the form

c: t ↦ [ ( c_{0}+ c_{1}t + c_{2}t^{2}) : ( d_{0}+ d_{1}t + d_{2}t^{2}) : ( e_{0}+ e_{1}t + e_{2}t^{2}) ] .

With the sextic “f_{6} = 0”, all intersection multiplicities are even if and only if f_{6} ∘ c is a perfect square in \overline{𝔽}_{q}[ t] . This may easily be checked by factoring f_{6}∘ c.

For small q, that allows, at least, to search for conics which are de�ned over 𝔽_{q} and tangent in six points. To achieve this, we listed all q^{2}( q^{3}− 1) non-degenerate conics over 𝔽_{q} for q = 3 and 5.

Remark 3.2 A general approach, analogous to the one described above, which would be able to �nd conics de�ned over \overline{𝔽}_{q} does not succeed. The Gröbner basis required becomes too large.

4. Upper bounds – The Frobenius operation on l-adic cohomology

The Lefschetz trace formula. A method to understand the Frobenius operation on H _{ét}^{2}( 𝒮_{\overline{𝔽}_{p}} , \overline{ℚ}_{l}) works as follows.

Count the points on 𝒮 over 𝔽_{p^{d}} and apply the Lefschetz trace formula [6] to compute the trace of the Frobenius φ_{𝔽_{p^{d}}} = φ^{d}. In our situation, it yields

Tr( φ^{d}) = #𝒮 ( 𝔽_{p^{d}}) − p^{2d} − 1.

We have Tr( φ^{d}) = λ_{1}^{d} + ⋯ + λ_{22}^{d} = : σ_{d}( λ_{1}, …, λ_{22}) when we denote the eigenvalues of φ by λ_{1}, …, λ_{22}. Newton’s identity [8]

s_{k}( λ_{1}, …, λ_{22}) = \frac{1}{k} ∑_{r= 0}^{k− 1}( − 1) ^{k+ r+ 1}σ_{k− r}( λ_{1}, …, λ_{22}) s_{r}( λ_{1}, …, λ_{22})

shows that, doing this for d = 1, …, k, one obtains enough information to determine the coef�cient ( − 1) ^{k}s_{k} of t^{22− k} of the characteristic polynomial f_{p} of φ.

Observe that we also have the functional equation

( ∗ )  p^{22}f_{p}( t) = ± t^{22}f_{p}( p^{2}/ t)

at our disposal. It may be used to convert the coef�cient of t^{i} into the one of t^{22− i}.

Methods for counting points. The number #𝒮 ( 𝔽_{q}) of the points may be determined as the sum

∑_{[ x:y:z] ∈ P^{2}( 𝔽_{q})}[ 1 + χ( f_{6}( x, y, z) ) ] .

Here, χ is the quadratic character. The sum is well-de�ned since f_{6}( x, y, z) is uniquely determined up to a sixth-power residue. To count the points naively, one would need q^{2}+ q + 1 evaluations of f_{6} and χ.

Here, a number of possibilities arise for optimization. We use two of them which we describe below.

i) Symmetry: If f_{6} is de�ned over 𝔽_{p} then the summands for [ x : y : z] and φ( [ x : y : z] ) are equal. This means, over 𝔽_{p^{d}}, we may save a factor of d if, on the af�ne chart “x = 1”, we put in for y only values from a fundamental domain of the Frobenius.

ii) Decoupling: Suppose, f_{6} contains only monomials of the form x^{i}y^{6− i} or x^{i}z^{6− i}. Then, on the af�ne chart “x = 1”, the form f_{6} may be written as a sum of a function in y and a function in z.

In O( qlogq) steps, for each of the two functions, we build up a table stating how many times it adopts each of its values. Again, we may restrict one of the tables to a fundamental domain of the Frobenius. We tabulate the quadratic character, too. After these preparations, less than q^{2} additions suf�ce to determine the number of points.

The advantage of a decoupled situation is, therefore, that an evaluation of a polynomial in 𝔽_{p^{d}} gets replaced by an addition.

Remark 4.1 Having implemented the point counting in C, these optimizations allow to determine the number of 𝔽_{3^{10}}-rational points on a K3 surface 𝒮 within half an hour (without decoupling) on an AMD Opteron processor.

In a decoupled situation, the number of 𝔽_{5^{9}}-rational points may be counted within two hours. In a few cases, we determined the numbers of points over 𝔽_{5^{10}}. This took around two days. Without decoupling, the same counts would have taken around one day or 25 days, respectively.

This shows, using the methods above, we may effectively compute the traces of φ_{𝔽_{p^{d}}} = φ^{d} for d = 1, …, 9, ( 10) .

An upper bound for rk Pic( 𝒮_{\overline{𝔽}_{p}} ) having counted till d = 10. We know that f_{p}, the characteristic polynomial of the Frobenius, has a zero at p since the pull-back of a line in P^{2} is a divisor de�ned over 𝔽_{p}. Suppose, we determined Tr( φ^{d}) for d = 1, …, 10. We may achieve an upper bound for rkPic( 𝒮_{\overline{𝔽}_{p}} ) as follows.

i) First, assume the minus sign in the functional equation ( ∗ ) . Then, f_{p} automatically has coef�cient 0 at t^{11}. Therefore, the numbers of points counted suf�ce in this case to determine f_{p}, completely.

ii) Then, assume that, on the other hand, the plus sign is present in ( ∗ ) . In this case, the data collected immediately allow to compute all coef�cients of f_{p} except that at t^{11}. Use the known zero at p to determine that �nal coef�cient.

iii) Use the numerical test, described below, to decide which sign is actually present.

iv) Factor f_{p}( pt) into irreducible polynomials. Check which of the factors are cyclotomic polynomials and add their degrees. That sum is an upper bound for rkPic( 𝒮_{\overline{𝔽}_{p}}) . If step iii) had failed then one has to work with both candidates for f_{p} and deal with the maximum.

Verifying rk Pic( 𝒮_{\overline{𝔽}_{p}} ) = 2 having counted till d = 9, only. Assume, 𝒮 is a K3 surface over 𝔽_{p} given by Construction 2.2.i) or ii). We, therefore, know that the rank of the Picard group is at least equal to 2. We suppose that the divisor constructed by pull-back splits already over 𝔽_{p}. This ensures p is a double zero of f_{p}. There is the following method to verify rkPic( 𝒮_{\overline{𝔽}_{p}} ) = 2.

i) First, assume the minus sign in the functional equation ( ∗ ) . This forces another zero of f_{p} at ( − p) . The data collected are then suf�cient to determine f_{p}, completely. The numerical test, described below, may indicate a contradiction.

Otherwise, the veri�cation fails. (In that case, we could still �nd an upper bound for rkPic( 𝒮_{\overline{𝔽}_{p}} ) which is, however, at least equal to 4.)

ii) As we have the plus sign in ( ∗ ) , the data immediately suf�ce to compute all coef�- cients of f_{p} with the exception of those at t^{10}, t^{11}, and t^{12}. The functional equation yields a linear relation for the three remaining coef�cients of f_{p}. From the known double zero at p, one computes another linear condition.

iii) Let n run through all natural numbers such that ϕ( n) ≤ 20. (The largest such n is 66.) Assume, in addition, that there is another zero of the form pζ_{n}. This yields further linear relations. Inspecting this system of linear equations, one either achieves a contradiction or determines all three remaining coef�cients. In the latter case, the numerical test may indicate a contradiction.

If each value of n turned out to be contradictory then we found that rkPic( 𝒮_{\overline{𝔽}_{p}} ) = 2.

Consequently, the equality rkPic( 𝒮_{\overline{𝔽}_{p}} ) = 2 may be effectively provable having determined Tr( φ^{d}) for d = 1, …, 9, ( 10) .

A numerical test. Given a polynomial f of degree 22, we calculate all its zeroes as �oating point numbers. If at least one of them is clearly not of absolute value p then f can not be the characteristic polynomial of the Frobenius for any K3 surface over 𝔽_{p}.

Remarks 4.2 i) This approach will always yield an even number for the upper bound of the Picard rank. Indeed, the bound is

rkPic( 𝒮_{\overline{𝔽}_{p}}) ≤ dim( H _{ét}^{2}( 𝒮_{\overline{𝔽}_{p}}, \overline{ℚ}_{l}) )

− #{ zeroes of f_{p} which are not of the form ζ_{n}p} .

The relevant zeroes come in pairs of complex conjugate numbers. Hence, for a K3 surface the bound is always even.

ii) There is a famous conjecture due to John Tate [7] which implies that the canonical injection c_{1}: Pic( 𝒮_{\overline{𝔽}_{p}}) → H _{ét}^{2}( 𝒮_{\overline{𝔽}_{p}}, \overline{ℚ}_{l}( 1) ) maps actually onto the sum of all eigenspaces for the eigenvalues which are roots of unity. Together with the conjecture of J.-P. Serre claiming that the Frobenius operation on étale cohomology is always semisimple, this would imply that the bound above is actually sharp.

It is a somewhat surprising consequence of the Tate conjecture that the Picard rank of a K3 surface over \overline{𝔽}_{p} is always even. For us, this is bad news. The obvious strategy to prove rkPic( S\overline{ℚ} ) = 1for a K3 surface S over ℚ would be to verify rkPic( S_{\overline{𝔽}_{p}}) = 1for a suitable place p of good reduction. The Tate conjecture indicates that there is no hope for such an approach.

5. How to prove rk Pic( S\overline{ℚ} ) = 1

Using the methods described above, on one hand, we can construct even upper bounds for the Picard rank. On the other hand, we can generate lower bounds by explicitly stating divisors. In an optimal situation, this may establish an equality rkPic( 𝒮_{\overline{𝔽}_{p}}) = 2 . How is it possible that way to reach Picard rank 1 for a surface de�ned over ℚ?

For this, a trick due to R. van Luijk [5, Remark 2] is helpful.

Fact 5.1 (van Luijk) Assume, we are given a K3 surface 𝒮 ^{( 3)} over 𝔽_{3} and a K3 surface 𝒮 ^{( 5)} over 𝔽_{5} which are both of geometric Picard rank 2. Suppose further that the discriminants of the intersection forms on Pic( 𝒮_{\overline{𝔽}_{3}}^{( 3)} ) and Pic( 𝒮_{\overline{𝔽}_{5}}^{( 5)} ) are essentially different, i.e. their quotient is not a perfect square in ℚ.

Then, every K3 surface S such that its reduction at 3 is isomorphic to 𝒮 ^{( 3)} and its reduction at 5 is isomorphic to 𝒮 ^{( 5)} is of geometric Picard rank one.

Proof. The reduction maps ι_{p}: Pic( S\overline{ℚ} ) → Pic( S_{\overline{𝔽}_{p}}) = Pic( 𝒮_{\overline{𝔽}_{p}}^{( p)} ) are injective [3, Example 20.3.6]. Observe here, Pic( S\overline{ℚ} ) is equal to the group of divisors on S\overline{ℚ} modulo numerical equivalence.

This immediately leads to the bound rkPic( S\overline{ℚ} ) ≤ 2. Assume, by contradiction, that equality holds. Then, the reductions of Pic( S\overline{ℚ} ) are sublattices of maximal rank in both, Pic( S_{\overline{𝔽}_{3}}) = Pic( 𝒮_{\overline{𝔽}_{3}}^{( 3)} ) and Pic( S_{\overline{𝔽}_{5}}) = Pic( 𝒮_{\overline{𝔽}_{5}}^{( 5)} ) .

The intersection product is compatible with reduction. Therefore, the quotients DiscPic( S\overline{ℚ} ) / DiscPic( 𝒮_{\overline{𝔽}_{3}}^{( 3)} ) and DiscPic( S\overline{ℚ} ) / DiscPic( 𝒮_{\overline{𝔽}_{5}}^{( 5)} ) are perfect squares. This is a contradiction to the assumption. □

Remark 5.2 Suppose that 𝒮 ^{( 3)} and 𝒮 ^{( 5)} are K3 surfaces of degree two given by explicit branch sextics in P^{2}. Then, using the Chinese Remainder Theorem, they can easily be combined to a K3 surface S over ℚ.

If one of them allows a conic tangent in six points and the other a tritangent then the discriminants of the intersection forms on Pic( 𝒮_{\overline{𝔽}_{3}}^{( 3)} ) and Pic( 𝒮_{\overline{𝔽}_{5}}^{( 5)} ) are essentially different as shown in section 2.

Remark 5.3 Suppose S is a K3 surface over ℚ constructed that way. Then, S cannot be isomorphic, not even over \overline{ℚ}, to a K3 surface S′ ⊂ P^{3} of degree 4. In particular, the explicit examples, which we will describe in the next sections, are different from those of R. van Luijk [5].

Indeed, Pic( S\overline{ℚ} ) = ℤ· ⟨ ℒ ⟩ and degS = 2 mean that the intersection form on Pic( S\overline{ℚ} ) is given by ⟨ ℒ^{⊗ n}, ℒ^{⊗ m}⟩ = 2nm. All self-intersection numbers of invertible sheaves on S\overline{ℚ} are of the form 2n^{2} which is always different from 4.

6. An explicit K3 surface of degree two

Examples 6.1 We consider two particular K3 surfaces over �nite �elds.

i) By 𝒳 ^{0}, we denote the surface over 𝔽_{3} given by the equation

w^{2} = ( y^{3}− x^{2}y) ^{2}+ ( x^{2}+ y^{2}+ z^{2}) ( 2x^{3}y + x^{3}z + 2x^{2}yz + x^{2}z^{2}+ 2xy^{3}+ 2y^{4}+ z^{4})
= 2x^{5}y + x^{5}z + x^{4}y^{2}+ 2x^{4}yz + x^{4}z^{2}+ x^{3}y^{3}+ x^{3}y^{2}z + 2x^{3}yz^{2}+ x^{3}z^{3}

+ 2x^{2}y^{3}z+ x^{2}y^{2}z^{2}+ 2x^{2}yz^{3}+ 2x^{2}z^{4}+ 2xy^{5}+ 2xy^{3}z^{2}+ 2y^{4}z^{2}+ y^{2}z^{4}+ z^{6}.

ii) Further, let 𝒴 ^{0} be the K3 surface over 𝔽_{5} given by

w^{2} = x^{5}y + x^{4}y^{2}+ 2x^{3}y^{3}+ x^{2}y^{4}+ xy^{5}+ 4y^{6}+ 2x^{5}z + 2x^{4}z^{2}+ 4x^{3}z^{3}+ 2xz^{5}+ 4z^{6}.

Theorem 6.2 Let S be any K3 surface over ℚ such that its reduction modulo 3 is isomorphic to 𝒳 ^{0} and its reduction modulo 5 is isomorphic to 𝒴 ^{0}. Then, rkPic( S\overline{ℚ} ) = 1.

Proof. We follow the strategy described in Remark 5.2. For the branch locus of 𝒳 ^{0}, the conic given by x^{2} + y^{2} + z^{2} = 0 is tangent in six points. The branch locus of 𝒴_{0} has a tritangent given by z − 2y = 0. It meets the branch locus at [ 1 : 0 : 0] , [ 1 : 3 : 1] , and [ 0 : 1 : 2] .

It remains necessary to show that rkPic( 𝒳_{\overline{𝔽}_{3}}^{0}) ≤ 2 and rkPic( 𝒴_{\overline{𝔽}_{5}}^{0}) ≤ 2 . To verify these assertions, we used the methods described in section 4. We counted points over 𝔽_{3^{d}} and 𝔽_{5^{d}}, respectively, for d ≤ 10. For 𝒴 ^{0}, we could use the faster method since the sextic form on the right hand side is decoupled. □

Corollary 6.3 Let S be the K3 surface over ℚ given by

w^{2} = − 4x^{5}y + 7x^{5}z + x^{4}y^{2}+ 5x^{4}yz + 7x^{4}z^{2}+ 7x^{3}y^{3}− 5x^{3}y^{2}z + 5x^{3}yz^{2}+ 4x^{3}z^{3}
+ 6x^{2}y^{4}+ 5x^{2}y^{3}z − 5x^{2}y^{2}z^{2}+ 5x^{2}yz^{3}+ 5x^{2}z^{4}− 4xy^{5}+ 5xy^{3}z^{2}− 3xz^{5}
− 6y^{6}+ 5y^{4}z^{2}− 5y^{2}z^{4}+ 4z^{6}.

i) Then, rkPic( S\overline{ℚ} ) = 1.

ii) Further, S( ℚ) ≠ / 0. For example, [ 2 ; 0 : 0 : 1] ∈ S( ℚ) .

Remarks 6.4 i) For the K3 surface 𝒳 ^{0}, our calculations show the following.

The numbers of the points de�ned over 𝔽_{3^{d}} for d = 1, …, 10 are, in this order, 14, 92, 758, 6752, 59834, 532820, 4796120, 43068728, 387421463, and 3487077812. The traces of the Frobenius φ_{𝔽_{p^{d}}} = φ^{d} on H _{ét}^{2}( 𝒳_{\overline{𝔽}_{3}}^{0}, \overline{ℚ}_{l}) are equal to 4, 10, 28, 190, 784, 1378, 13150, 22006, 973, and 293410.

The sign in the functional equation is positive. For the decomposition of the characteristic polynomial f_{p} of the Frobenius, we �nd (after scaling to zeroes of absolute value 1)

( t − 1) ^{2}( 3t^{20}+ 2t^{19}+ 2t^{18}+ 2t^{17}+ t^{16}− 2t^{13}− 2t^{12}− t^{11}− 2t^{10}
− t^{9}− 2t^{8}− 2t^{7}+ t^{4}+ 2t^{3}+ 2t^{2}+ 2t + 3) / 3

with an irreducible polynomial of degree 20. The assumption of the negative sign leads to zeroes the absolute values of which range (without scaling) from 2. 598 to 3. 464.

ii) For the K3 surface 𝒴 ^{0}, our calculations yield the following results.

The numbers of points over 𝔽_{5^{d}} are, in this order, 41, 751, 15626, 392251, 9759376, 244134376, 6103312501, 152589156251, 3814704296876, and 95367474609376. The traces of the Frobenius on H _{ét}^{2}( 𝒴_{\overline{𝔽}_{5}}^{0}, \overline{ℚ}_{l}) are 15, 125, 0, 1625, − 6250, − 6250, − 203125, 1265625, 7031250, and 42968750.

The sign in the functional equation is positive. For the decomposition of the scaled characteristic polynomial of the Frobenius, we �nd

( t − 1) ^{2}( 5t^{20}− 5t^{19}− 5t^{18}+ 10t^{17}− 2t^{16}− 3t^{15}+ 4t^{14}− 2t^{13}− 2t^{12}+ t^{11}
+ 3t^{10}+ t^{9}− 2t^{8}− 2t^{7}+ 4t^{6}− 3t^{5}− 2t^{4}+ 10t^{3}− 5t^{2}− 5t + 5) / 5.

The assumption of the negative sign leads to zeroes the absolute values of which range (without scaling) from 3. 908 to 6. 398.

7. An explicit K3 surface of degree two given by a symmetric ( 3 × 3) -determinant

Examples 7.1 Consider the following two K3 surfaces over �nite �elds.

i) By 𝒳 , we denote the surface over 𝔽_{3} given by the equation w^{2} = f_{6}( x, y, z) for

f_{6}( x, y, z) = det( \begin{matrix}2xy+ 2y^{2}+ yz & 2x^{2}+ xy+ xz+ yz+ 2z^{2} & 2x^{2}+ xz+ yz+ z^{2} \\ 2x^{2}+ xy+ xz+ yz+ 2z^{2} & 2x^{2}+ xy & xy+ y^{2}+ yz+ 2z^{2} \\ 2x^{2}+ xz+ yz+ z^{2} & xy+ y^{2}+ yz+ 2z^{2} & 2x^{2}+ 2xy+ 2y^{2}+ 2yz \end{matrix})

= 2x^{6}+ 2x^{5}y + 2x^{5}z + 2x^{4}y^{2}+ x^{4}yz + x^{3}y^{3}+ x^{3}yz^{2}+ x^{3}z^{3}+ 2x^{2}y^{4}

+ x^{2}y^{3}z + 2x^{2}y^{2}z^{2}+ xy^{5}+ xy^{2}z^{3}+ y^{6}+ y^{5}z + y^{2}z^{4}+ yz^{5}+ 2z^{6}.

ii) Further, let 𝒴 be the K3 surface over 𝔽_{5} given by w^{2} = f_{6}( x, y, z) for

f_{6}( x, y, z) = det( \begin{matrix}4x^{2}+ 4xz+ y^{2} & 2x^{2}+ 3z^{2} & 4x^{2}+ 2xy+ 2xz+ 4y^{2}+ 3yz+ 2z^{2} \\ 2x^{2}+ 3z^{2} & 2x^{2}+ 4xy+ 4y^{2}+ yz+ 3z^{2} & 4xy+ 2xz+ y^{2}+ 4yz+ 4z^{2} \\ 4x^{2}+ 2xy+ 2xz+ 4y^{2}+ 3yz+ 2z^{2} & 4xy+ 2xz+ y^{2}+ 4yz+ 4z^{2} & 4x^{2}+ xz+ 3z^{2} \end{matrix})

= 4x^{6}+ 2x^{5}y + x^{5}z + x^{4}y^{2}+ x^{4}z^{2}+ x^{3}y^{3}+ 4x^{3}z^{3}

+ 2x^{2}y^{4}+ 2x^{2}z^{4}+ 4xy^{5}+ xz^{5}+ 4z^{6}.

Theorem 7.2 Let S be any K3 surface over ℚ such that its reduction modulo 3 is isomorphic to 𝒳 and its reduction modulo 5 is isomorphic to 𝒴 . Then, rkPic( S\overline{ℚ} ) = 1.

Proof. Consider the branch locus of 𝒳 . For the conic C, given by x^{2}+ xy+ 2xz+ z^{2} = 0, there is the parametrization

q: u ↦ [ u^{2} : 2 : ( 2u^{2}+ 2u) ] .

We �nd

f_{6}( q( u) ) = ( u + 1) ^{2}( u^{5}+ u^{4}+ u^{3}+ u + 1) ^{2},

i.e. C admits the property of being tangent in six points and the corresponding divisor on 𝒳 splits already over 𝔽_{3}. The branch sextic of 𝒴 has a degenerate tritangent given by x = 0.

To verify that rkPic( 𝒳_{\overline{𝔽}_{3}}) ≤ 2 and rkPic( 𝒴_{\overline{𝔽}_{5}}) ≤ 2 , again, we used the methods described in section 4. We counted points over 𝔽_{3^{d}}, respectively 𝔽_{5^{d}}, for d ≤ 10. Observe that, for 𝒴 , we could use the faster method since the sextic form on the right hand side is decoupled. □

Corollary 7.3 Let S be the K3 surface over ℚ given by w^{2} = f_{6}( x, y, z) for

f_{6}( x, y, z) = det( \begin{matrix}− 6x^{2} + 5xy − 6xz − 4339y^{2} − 5yz & 2x^{2} − 5xy − 5xz − 150y^{2} − 5yz − 7z^{2} & − x^{2} − 3xy + 7xz − 6y^{2} − 2yz + 7z^{2} \\ 2x^{2} − 5xy − 5xz − 150y^{2} − 5yz − 7z^{2} & 2x^{2} + 4xy − 6y^{2} + 6yz + 3z^{2} & 4xy − 3xz + y^{2} + 4yz − z^{2} \\ − x^{2} − 3xy + 7xz − 6y^{2} − 2yz + 7z^{2} & 4xy − 3xz + y^{2} + 4yz − z^{2} & − x^{2} + 5xy + 6xz + 5y^{2} + 5yz + 3z^{2} \end{matrix})

= 14x^{6}− 118x^{5}y − 64x^{5}z + 8021x^{4}y^{2}+ 220x^{4}yz − 114x^{4}z^{2}
− 20249x^{3}y^{3}− 47700x^{3}y^{2}z − 635x^{3}yz^{2}+ 4x^{3}z^{3}

− 64753x^{2}y^{4}− 247925x^{2}y^{3}z + 26045x^{2}y^{2}z^{2}− 2745x^{2}yz^{3}− 153x^{2}z^{4}

− 33821xy^{5}− 107100xy^{4}z − 463245xy^{3}z^{2}− 62450xy^{2}z^{3}

− 3075xyz^{4}− 384xz^{5}

+ 24025y^{6}− 77345y^{5}z − 143880y^{4}z^{2}− 201885y^{3}z^{3}− 39455y^{2}z^{4}

− 1055yz^{5}− 196z^{6}. i) Then, rkPic( S\overline{ℚ} ) = 1.

ii) Further, S( ℚ) ≠ / 0. For example, [ 155 ; 0 : 1 : 0] ∈ S( ℚ) .

Remarks 7.4 i) For the K3 surface 𝒳 , our calculations show the following.

The numbers of the points de�ned over 𝔽_{3^{d}} for d = 1, …, 10 are, in this order, 14, 88, 800, 6664, 59114, 531136, 4782344, 43029952, 387550223 and 3486755578. The traces of the Frobenius φ_{𝔽_{p^{d}}} = φ^{d} on H _{ét}^{2}( 𝒳_{\overline{𝔽}_{3}}, \overline{ℚ}_{l}) are 5, 7, 71, 103, 65, − 305, − 625, − 16769, 129734, and − 28823.

The decomposition of the scaled characteristic polynomial is

( t − 1) ^{2}( 3t^{20}+ t^{19}+ 2t^{18}+ t^{16}+ t^{15}+ 2t^{14}+ 2t^{13}+ 3t^{12}
+ 2t^{10}+ 3t^{8}+ 2t^{7}+ 2t^{6}+ t^{5}+ t^{4}+ 2t^{2}+ t + 3) / 3.

By consequence, the geometric Picard rank is equal to 2.

ii) For the K3 surface 𝒴 , our calculations yield the following results.

The numbers of points over 𝔽_{5^{d}} are, in this order, 33, 669, 15522, 391861, 9768668, 244132734, 6103019942, 152588860821, 3814709624898, and 95367420137974. The traces of the Frobenius on H _{ét}^{2}( 𝒴_{\overline{𝔽}_{5}}, \overline{ℚ}_{l}) are 8, 44, − 103, 1236, 3043, − 7891, − 495683, 970196, 12359273, and − 11502651.

The decomposition of the scaled characteristic polynomial is

( t − 1) ^{2}( 5t^{20}+ 2t^{19}+ t^{18}+ 5t^{17}+ 2t^{16}+ 2t^{15}+ 5t^{14}+ 8t^{13}+ 4t^{12}+ 2t^{11}
+ 8t^{10}+ 2t^{9}+ 4t^{8}+ 8t^{7}+ 5t^{6}+ 2t^{5}+ 2t^{4}+ 5t^{3}+ t^{2}+ 2t + 5) / 5.

Consequently, the geometric Picard rank is equal to 2.

A probabilistic method to construct symmetric ( 3 × 3) -matrices with decoupled determinant. A general ternary sextic has 28 coef�cients. It is decoupled if 15 of these vanish. Thus, a randomly chosen sextic form in 𝔽_{q}[ x, y, z] is decoupled with a probability of q^{− 15}. This is too low for our purposes.

On the other hand, we can think of decoupling as solving an non-linear system of 15 equations in 36 variables. One could try to attack this system by a Gröbner base calculation. We use a mixture of both methods. More precisely, we do the following.

Method 7.5 We construct the matrix M in the particular form

M := ( \begin{matrix}a( x, y, z) & b( x, z) & c_{1}( x, y, z) \\ b( x, z) & c_{2}( x, y, z) & c_{3}( x, y, z) \\ c_{1}( x, y, z) & c_{3}( x, y, z) & d( x, z) \end{matrix} ) .

i) We choose the quadratic forms c_{1}, c_{2}, c_{3}, and d, randomly.

ii) In a second step, we have to �x the nine coef�cients of the quadratic forms a and b. The coef�cients of det M at x^{6− i− j}y^{i}z^{j} for i, j > 0 are linear functions of the coef�cients of a and b. Observe that the summand − b^{2}d does not contribute to these critical coef�- cients.

Thus, we have to solve a system of 15 linear equations in nine variables. Naively, such a system is solvable with a probability of q^{− 6}.

If it is not solvable then we go back to the �rst step.

Remarks 7.6 i) We randomly generated a sample of 30 surfaces over 𝔽_{3}. For each of them, the branch locus was smooth and had passed the two tests described in section 3, to exclude the existence of a tritangent and to ensure there was exactly one conic over 𝔽_{3} tangent in six points.

We could establish the equality rkPic( 𝒳_{\overline{𝔽}_{3}}) = 2 in three of the examples. Example 7.1.i) reproduces one of them.

ii) Using the probabilistic method described above, we generated a sample of 50 surfaces over 𝔽_{5}. We made sure that, for each of them, the branch sextic was smooth, had exactly one tritangent, and no conic over 𝔽_{5} tangent in six points. Further, it was decoupled by construction. It took Magma approximately one hour to generate that sample.

Having counted points over 𝔽_{5^{d}} for d ≤ 9, we could establish the equality rkPic( 𝒴_{\overline{𝔽}_{5}}) = 2 in two of the examples. For those, we determined, in addition, the numbers of points over 𝔽_{5^{10}}. Example 7.1.ii) reproduces one of the two.

8. An explicit K3 surface of degree two given by a symmetric ( 6 × 6) -determinant

Examples 8.1 Consider the following K3 surfaces over �nite �elds.

i) By 𝒳 ′, we denote the surface over 𝔽_{3} given by the equation w^{2} = f_{6}( x, y, z) for

f_{6}( x, y, z) = det[ x( \begin{matrix}0 & 0 & 0 & 0 & 2 & 2 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 2 & 1 & 1 & 2 \\ 0 & 1 & 1 & 1 & 1 & 1 \\ 2 & 0 & 1 & 1 & 2 & 2 \\ 2 & 1 & 2 & 1 & 2 & 0 \end{matrix} ) + y( \begin{matrix}0 & 2 & 1 & 1 & 2 & 1 \\ 2 & 2 & 2 & 2 & 2 & 2 \\ 1 & 2 & 0 & 1 & 1 & 0 \\ 1 & 2 & 1 & 1 & 1 & 1 \\ 2 & 2 & 1 & 1 & 2 & 2 \\ 1 & 2 & 0 & 1 & 2 & 0 \end{matrix} ) + z( \begin{matrix}2 & 1 & 1 & 1 & 2 & 0 \\ 1 & 1 & 0 & 1 & 2 & 1 \\ 1 & 0 & 2 & 1 & 1 & 2 \\ 1 & 1 & 1 & 0 & 0 & 1 \\ 2 & 2 & 1 & 0 & 2 & 2 \\ 0 & 1 & 2 & 1 & 2 & 0 \end{matrix} ) ]

= x^{6}+ x^{5}y + 2x^{5}z + 2x^{4}y^{2}+ 2x^{4}yz + 2x^{2}y^{3}z + x^{2}z^{4}

+ 2xy^{5}+ 2xy^{4}z + 2y^{6}+ 2y^{5}z + y^{2}z^{4}+ 2yz^{5}.

ii) Further, let 𝒴 ′ be the K3 surface over 𝔽_{5} given by w^{2} = f_{6}( x, y, z) for

f_{6}( x, y, z) = det[ x( \begin{matrix}3 & 4 & 3 & 4 & 4 & 1 \\ 4 & 3 & 0 & 2 & 1 & 0 \\ 3 & 0 & 4 & 0 & 3 & 0 \\ 4 & 2 & 0 & 2 & 1 & 3 \\ 4 & 1 & 3 & 1 & 0 & 2 \\ 1 & 0 & 0 & 3 & 2 & 1 \end{matrix} ) + y( \begin{matrix}0 & 0 & 3 & 3 & 0 & 1 \\ 0 & 2 & 0 & 1 & 3 & 1 \\ 3 & 0 & 5 & 0 & 3 & 5 \\ 3 & 1 & 0 & 3 & 5 & 5 \\ 0 & 3 & 3 & 5 & 0 & 1 \\ 1 & 1 & 5 & 5 & 1 & 3 \end{matrix} ) + z( \begin{matrix}2 & 1 & 1 & 1 & 0 & 5 \\ 1 & 0 & 4 & 4 & 4 & 4 \\ 1 & 4 & 2 & 3 & 0 & 2 \\ 1 & 4 & 3 & 1 & 2 & 2 \\ 0 & 4 & 0 & 2 & 3 & 1 \\ 5 & 4 & 2 & 2 & 1 & 0 \end{matrix} ) ]

= 2x^{6}+ x^{5}y + 2x^{4}y^{2}+ 3x^{4}z^{2}+ x^{3}y^{3}+ 2x^{3}z^{3}+ x^{2}y^{4}+ 3x^{2}z^{4}+ 2xy^{5}+ 4z^{6}.

Theorem 8.2 Let S be any K3 surface over ℚ such that its reduction modulo 3 is isomorphic to 𝒳 ′ and its reduction modulo 5 is isomorphic to 𝒴 ′. Then, rkPic( S\overline{ℚ} ) = 1.

Proof. Consider the branch locus of 𝒳 ′. For the conic C, given by xz+ y^{2}+ 2yz+ 2z^{2} = 0, there is the parametrization

q: u ↦ [ ( 2u^{2}+ u + 1) : u : 1] .

We �nd

f_{6}( q( u) ) = ( u^{2}+ u + 2) ^{2}( u^{4}+ u + 2) ^{2},

i.e. C admits the property of being tangent in six points and the corresponding divisor on 𝒳 ′ splits already over 𝔽_{3}. The branch sextic of 𝒴 ′ has a degenerate tritangent given by x = 0.

To verify that rkPic( 𝒳′_{\overline{𝔽}_{3}}) ≤ 2 and rkPic( 𝒴′_{\overline{𝔽}_{5}}) ≤ 2 , again, we used the methods described in section 4. We counted points over 𝔽_{3^{d}} and 𝔽_{5^{d}}, respectively, for d ≤ 10. Observe, for 𝒴 ′, we could use the faster method since the sextic form on the right hand side is decoupled. □

Corollary 8.3 Let S be the K3 surface over ℚ given by w^{2} = f_{6}( x, y, z) for

f_{6}( x, y, z) = det[ x( \begin{matrix}− 2382 & − 21 & 3 & − 6 & − 1 & − 4 \\ − 21 & 28 & 0 & 7 & 6 & − 5 \\ 3 & 0 & − 1 & − 5 & − 2 & 5 \\ − 6 & 7 & − 5 & 7 & 1 & − 2 \\ − 1 & 6 & − 2 & 1 & 5 & 2 \\ − 4 & − 5 & 5 & − 2 & 2 & 6 \end{matrix} ) + y( \begin{matrix}0 & 5 & − 2 & − 2 & 5 & 1 \\ 5 & 2 & 5 & − 4 & − 7 & − 4 \\ − 2 & 5 & 0 & − 5 & − 2 & 0 \\ − 2 & − 4 & − 5 & − 2 & − 5 & − 5 \\ 5 & − 7 & − 2 & − 5 & 5 & − 4 \\ 1 & − 4 & 0 & − 5 & − 4 & 3 \end{matrix} ) + z( \begin{matrix}2 & 1 & 1 & 1 & 5 & 0 \\ 1 & − 5 & − 6 & 4 & − 1 & 4 \\ 1 & − 6 & 2 & − 2 & − 5 & 2 \\ 1 & 4 & − 2 & 6 & − 3 & 7 \\ 5 & − 1 & − 5 & − 3 & − 7 & − 4 \\ 0 & 4 & 2 & 7 & − 4 & 0 \end{matrix} ) ]

= 76139167x^{6}+ 231184081x^{5}y + 210075725x^{5}z
+ 25609337x^{4}y^{2}+ 487337315x^{4}yz − 314154987x^{4}z^{2}

− 141937719x^{3}y^{3}+ 283035180x^{3}y^{2}z − 434149815x^{3}yz^{2}− 5367468x^{3}z^{3}

− 175763034x^{2}y^{4}+ 168686090x^{2}y^{3}z − 421490010x^{2}y^{2}z^{2}
+ 160009155x^{2}yz^{3}− 153566957x^{2}z^{4}
− 90295273xy^{5}+ 175779575xy^{4}z − 285747180xy^{3}z^{2}
+ 327585255xy^{2}z^{3}− 215766345xyz^{4}+ 94479045xz^{5}
+ 133220y^{6}+ 31145y^{5}z + 380715y^{4}z^{2}− 324195y^{3}z^{3}− 476810y^{2}z^{4}
+ 402845yz^{5}− 174261z^{6}.

i) Then, rkPic( S\overline{ℚ} ) = 1.

ii) Further, S( ℚ) ≠ / 0. For example, [ 1286 ; 1 : 1 : 1] ∈ S( ℚ) .

Remarks 8.4 i) For the K3 surface 𝒳 ′, our calculations show the following.

The numbers of the points de�ned over 𝔽_{3^{d}} for d = 1, …, 10 are, in this order, 12, 90, 783, 6534, 59697, 535329, 4793661, 43079526, 387521091, and 3487248045. The traces of the Frobenius φ_{𝔽_{p^{d}}} = φ^{d} on H _{ét}^{2}( 𝒳′_{\overline{𝔽}_{3}}, \overline{ℚ}_{l}) are 3, 9, 54, − 27, 648, 3888, 10 692, 32 805, 100 602, and 463644.

The decomposition of the scaled characteristic polynomial is

( t − 1) ^{2}( 3t^{20}+ 3t^{19}+ 3t^{18}+ 2t^{17}+ 3t^{16}+ 2t^{15}− 2t^{13}− 3t^{12}− 4t^{11}

− 6t^{10}− 4t^{9}− 3t^{8}− 2t^{7}+ 2t^{5}+ 3t^{4}+ 2t^{3}+ 3t^{2}+ 3t + 3) / 3.

Consequently, the geometric Picard rank is equal to 2. ii) For the K3 surface 𝒴 ′, our calculations yield the following results. The numbers of points over 𝔽_{5^{d}} are, in this order, 36, 666, 15711, 391706, 9763601, 244152021, 6103934341, 152589189186, 3814705355181, and 95367412593451. The traces of the Frobenius on H _{ét}^{2}( 𝒴′_{\overline{𝔽}_{5}}, \overline{ℚ}_{l}) are 11, 41, 86, 1081, − 2024, 11396, 418716, 1298561, 8089556, and − 19047174. The decomposition of the scaled characteristic polynomial is

( t − 1) ^{2}( 5t^{20}− t^{19}+ t^{18}+ 2t^{17}+ 3t^{15}+ t^{14}− 2t^{13}+ t^{12}− t^{11}
+ 2t^{10}− t^{9}+ t^{8}− 2t^{7}+ t^{6}+ 3t^{5}+ 2t^{3}+ t^{2}− t + 5) / 5.

By consequence, the geometric Picard rank is equal to 2.

A probabilistic method to construct symmetric ( 6 × 6) -matrices with decoupled determinant. Method 8.5 a) We construct a symmetric ( 6 × 6) -matrix M _{0} the entries of which are linear forms only in y and z. The goal is that its determinant is decoupled, i.e.

det M _{0} = ay^{6}+ bz^{6}

for certain a, b ∈ 𝔽_{q}, not both vanishing. This leads to �ve conditions for the coef�cients.

i) We choose all entries in M _{0} randomly except for ( M _{0}) _{11}. ii) The determinant is linear in the coef�cients of ( M _{0}) _{11}. Therefore, we have a system of �ve linear equations in two variables. Such a system is solvable with a probability of q^{− 3} which is enough for our purposes. If there is no solution then we return to step i).

b) We construct M in the form

M := M _{0}+ xA

for A a symmetric matrix with entries in 𝔽_{q}.

i) First, look at the monomials xy^{i}z^{5− i} for i = 1, …, 4, only. To make their coef�- cients vanish leads to a system of four linear equations. In general, its solutions form a 17-dimensional vector space. ii) For decoupling, there are six further coef�cients which are required to be zero. This means, we are left with 17 parameters and six non-linear equations. We choose the parameters randomly and iterate this procedure until a solution has been found. Naively, the probability to hit a solution is q^{− 6}, each time.

Remarks 8.6 i) We randomly generated a sample of 50 surfaces over 𝔽_{3}. For each of them, the branch sextic was smooth and had passed the two tests described in section 3, to exclude the existence of a tritangent and to ensure there was exactly one conic over 𝔽_{3} tangent in six points.

We established rkPic( 𝒳′_{\overline{𝔽}_{3}}) = 2 in eleven of the examples. Example 8.1.i) is one of them.

ii) Using the probabilistic method described above, we generated a sample of 120 surfaces over 𝔽_{5}. For each of them, the branch sextic was decoupled, by construction. We made sure, in addition, that it was smooth, had exactly one tritangent, and no conic, de�ned over 𝔽_{5}, which was tangent in six points. It should be remarked that it took Magma half a day to generate that sample.

Having counted points over 𝔽_{5^{d}} for d ≤ 9, we could establish the equality rkPic( 𝒴′_{\overline{𝔽}_{5}} ) = 2 in three of the examples. For those, we determined, in addition, the numbers of points over 𝔽_{5^{10}}. Example 8.1.ii) reproduces one of the three.

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