The Picard group of a K3 surface and its reduction modulo p

Andreas-Stephan Elsenhans and Jörg Jahnel

Abstract

We present a method to compute the geometric Picard rank of a K3 surface over ℚ. Contrary to a widely held belief, we show it is possible to verify Picard rank 1 using reduction only at a single prime. Our method is based on deformation theory for invertible sheaves.

1 Introduction

1.1. –––– For K3 surfaces, the Picard group is a highly interesting invariant. In general, it is isomorphic to ℤ^{n} for some n = 1, …, 20. A generic K3 surface over ℂ has Picard rank 1. Nevertheless, the first explicit examples of K3 surfaces over ℚ with geometric Picard rank 1 were constructed by R. van Luijk [vL] as late as 2004. Van Luijk’s method is based on reduction modulo p. It works as follows.

1.2. Approach (van Luijk). —– Let S be a K3 surface.

i) At a place p of good reduction, the Picard group Pic(S\overline{ℚ} ) of the surface injects into the Picard group Pic(S_{\overline{𝔽}_{p}}) of its reduction modulo p. ii) On its part, Pic(S_{\overline{𝔽}_{p}}) injects into the second étale cohomology group H_{ét}^{2}(S_{\overline{𝔽}_{p}}, ℚ_{l}(1)) . iii) Only roots of unity can arise as eigenvalues of the Frobenius on the image of Pic(S_{\overline{𝔽}_{p}}) in H_{ét}^{2}(S_{\overline{𝔽}_{p}}, ℚ_{l}(1)) . The number of eigenvalues of this form is therefore an upper bound for the Picard rank of S_{\overline{𝔽}_{p}}. One may compute the eigenvalues of Frob by counting the points on S, defined over 𝔽_{p} and some finite extensions.

Doing this for one prime, one obtains an upper bound for rk Pic(S_{\overline{𝔽}_{p}}) which is always even. The Tate conjecture asserts that this bound is actually sharp. For proving that the Picard rank over \overline{ℚ} is equal to 1, the best that could happen is to find a prime which yields an upper bound of 2.

iv) In this case, the assumption that the surface would have Picard rank 2 over \overline{ℚ} implies that the discriminants of both Picard groups, Pic(S\overline{ℚ} ) and Pic(S_{\overline{𝔽}_{p}}), are in the same square class. Note here that reduction modulo p respects the intersection product.

v) To obtain a contradiction, one combines information from two primes. It may happen that one has a rank bound of 2 at both places but different square classes for the discriminant do arise. Then, these data are incompatible with Picard rank 2 over \overline{ℚ} .

1.3. The improvement. –––– Approach 1.2 accepts the possibility that Pic(S\overline{ℚ} ) ⊂ Pic(S_{\overline{𝔽}_{p}}) might be a proper sublattice of full rank. If that occurred then one knows at least that the two discriminants differ by a perfect square. This is a standard observation from the theory of lattices.

We will show in this article that such provisions need not be made. From the technical point of view, our main result states that, at least for p ≠ 2, the quotient Pic(S_{\overline{𝔽}_{p}})/ Pic(S\overline{ℚ} ) is always torsion-free. This is true actually in much more generality than just for K3 surfaces. It follows in a rather straightforward manner from deformation theory, a tool developed by A. Grothendieck and M. Artin in the sixties of the last century. To be precise, our result is as follows.

1.4. Theorem. –––– Let p ≠ 2 be a prime number and X be a scheme proper and flat over ℤ. Suppose that the special fiber X_{p} is non-singular and satisfies H^{1}(X_{p}, 𝒪_{X_{p}}) = 0. Then, the specialization homomorphism Pic(X\overline{ℚ} ) → Pic(X_{\overline{𝔽}_{p}}) has a torsion-free cokernel.

1.5. Remarks. –––– a) Recall that, for a K3 surface S, one has H^{1}(S, 𝒪_{S}) = 0 [BPV, Chap. VI, Table 10].

b) We will prove this theorem in 3.4. As an application, one may prove rk Pic(S\overline{ℚ} ) = 1 for a K3 surface S using its reduction only at a single prime. This works as follows.

Approach. Let a K3 surface S be given.

i) For a prime p ≠ 2 of good reduction, perform steps i), ii) and iii) as in 1.2. Thereby, the hope is to prove rk Pic(S_{\overline{𝔽}_{p}}) ≤ 2. Further, compute the discriminant giving two explicit generators. Alternatively, one might use the Artin-Tate formula. ii) Assume rk Pic(S\overline{ℚ} ) = 2. Then, according to Theorem 1.4, every invertible sheaf on S_{\overline{𝔽}_{p}} lifts to S\overline{ℚ} . Use reduction theory of binary quadratic forms or explicit arguments to estimate the degree of a hypothetical effective divisor. Finally, use Gröbner bases to verify that such a divisor does not exist.

1.6. Example. –––– Consider the K3 surface S over ℚ, 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} .

Then, rk Pic(S\overline{ℚ} ) = 1.

Proof. For the reduction of S at the prime 5, one sees that the branch locus has a tritangent line given by z − 2y = 0. It meets the branch locus at (1 : 0 : 0), (1 : 3 : 1), and (0 : 1 : 2).

The numbers of points over 𝔽_{5^{d}} are, in this order, 41, 751, 15 626, 392 251, 9 759 376, 244 134 376, 6 103 312 501, 152 589 156 251, 3 814 704 296 876, and 95 367 474 609 376. Thus, the traces of the Frobenius on H_{ét}^{2}(S_{\overline{𝔽} _{5}}, \overline{ℚ} _{l}) are 15, 125, 0, 1 625, −6 250, −6 250, −203 125, 1 265 625, 7 031 250, and 42 968 750. Algorithm 23 of [EJ1] shows that the sign in the functional equation is positive. The characteristic polynomial of the Frobenius is therefore completely determined. For its decomposition into prime polynomials, we find (after scaling)

(t − 5)^{2}(t^{20} − 5 t^{19} − 25 t^{18} + 250 t^{17} − 250 t^{16} − 1 875 t^{15} + 12 500 t^{14} − 31 250 t^{13} − 156 250 t^{12} + 390 625 t^{11} + 5 859 375 t^{10} + 9 765 625 t^{9} − 97 656 250 t^{8} − 488 281 250 t^{7} + 4 882 812 500 t^{6} − 18 310 546 875 t^{5} − 61 035 156 250 t^{4}

+ 1 525 878 906 250 t^{3} − 3 814 697 265 625 t^{2} − 19 073 486 328 125 t

+ 95 367 431 640 625) .

This shows rk Pic(S_{\overline{𝔽}_{5}}) ≤ 2.

The splits of the pull-back of the tritangent line are explicit generators for Pic(S_{\overline{𝔽}_{5}}). Such a split l, being a projective line, has self-intersection number l^{2} = −2. Further, lh = 1 for h the pull-back of a line. If we had rk Pic(S\overline{ℚ} ) = 2 then the invertible sheaf 𝒪(l) would lift to S\overline{ℚ} . We had a divisor L on S\overline{ℚ} such that HL = 1 and L^{2} = −2. By [BPV, Ch. VIII, Proposition 3.6.i], such a divisor is automatically effective.

HL = 1 shows that L is obtained from a line on P^{2}, the pull-back of which splits into two components. This is possible only for a tritangent line of the branch locus. [EJ1, Algorithm 8] shows, however, that such a tritangent line does not exist. □

2 The sequence of the Picard lattices

2.1. Remark. –––– The proof of Theorem 1.4 relies on deformation-theoretic methods [Ar, Kl]. For K3 surfaces and prime-to-p torsion, one could have used étale cohomology which appears to be more natural.

In fact, to show Pic(X_{\overline{𝔽}_{p}})/ Pic(X\overline{ℚ} ) has no l-torsion, it is sufficient to consider Pic(X_{\overline{𝔽}_{p}})⊗_{ℤ} ℤ_{l}/ Pic(X\overline{ℚ} )⊗_{ℤ} ℤ_{l} . But Pic(X_{\overline{𝔽}_{p}})⊗_{ℤ} ℤ_{l} ⊆ H_{ét}^{2}(X_{\overline{𝔽}_{p}}, ℤ_{l}(1)) which, by standard comparison theorems, is isomorphic to H_{sing}^{2}(X(ℂ), ℤ)⊗_{ℤ} ℤ_{l}. On the other hand, Pic(X\overline{ℚ} ) ≅ Pic(X_{ℂ} ). Finally, H_{sing}^{2}(X(ℂ), ℤ)/ Pic(X_{ℂ} ) is torsion-free according to the Lefschetz (1, 1)-theorem.

2.2. Notation. –––– Let X be a ℤ_{p}-scheme. Then, we will write X_{p} for the special fiber and, more generally, X_{p^{n}} := X ×_{Spec ℤ_{p}} Spec ℤ/p^{n}ℤ. Finally, let X̂ be the formal scheme obtained by completing X along (p).

2.3. Lemma. –––– Let p ≠ 2 be a prime number and X a ℤ_{p}-scheme which is Noetherian,separated, and fulfills H^{1}(X_{p}, 𝒪_{X_{p}}) = 0. Denote by P ⊆ Pic(X_{p}) the subset of all invertible sheaves allowing a lift as an invertible sheaf on X̂. Then, Pic(X_{p})/P is torsion-free.

Proof. First step. Preliminaries. Assume, to the contrary, that Pic(X_{p})/P has torsion. Then, there are a prime number l and an invertible sheaf ℒ ∈ Pic(X_{p})\P such that ℒ^{⊗l} ∈ P. This means that ℒ^{⊗l} lifts to X̂ but ℒ does not. We have to show that this situation is impossible.

By [Ha, Proposition II.9.6], an invertible sheaf on X̂ is the same as an inverse system (ℐ_{n})_{n} of invertible sheaves ℐ_{n} ∈ Pic(X_{p^{n}}) such that ℐ_{n+1}|_{X_{p^{n}}} = ℐ_{n} for all n. By assumption, we have such a system for ℐ_{0} = ℒ^{⊗l}. It has to be shown that the invertible sheaf ℒ, too, lifts to X_{p^{n}} for all n.

Second step. Obstructions. We will construct sheaves ℒ_{n} ∈ Pic(X_{p^{n}}) lifting ℒ, inductively. These will satisfy, in addition, the relation ℒ_{n}^{⊗l} ≅ ℐ_{n}. First, we put ℒ_{0} := ℒ.

For the induction step, consider the short exact sequence

0 ⟶ 𝒦 ⟶ 𝒪_{X_{p^{n+1}}}^{∗} ⟶ 𝒪_{X_{p^{n}}}^{∗} ⟶ 0 .

Here, we have 𝒪_{X_{p}} ≅ 𝒦 via the exponential map x ↦ 1 + p^{n}x (mod p^{n+1}). This yields the commutative diagram with exact rows,

\xymatrix{0 \ar[r] & Pic(X_{p^{n+1}}) \ar[r] \ar[d]^{(.)^{⊗l}} & Pic(X_{p^{n}}) \ar[r] \ar[d]^{(.)^{⊗l}} & H^{2}(X_{p}, 𝒪_{X_{p}}) \ar[d]^{·l} \\ 0 \ar[r] & Pic(X_{p^{n+1}}) \ar[r] & Pic(X_{p^{n}}) \ar[r] & H^{2}(X_{p}, 𝒪_{X_{p}}) .}

The group H^{2}(X_{p}, 𝒪_{X_{p}}) is p-torsion as the sheaf 𝒪_{X_{p}} is annihilated by p. In particular, it is uniquely l-divisible. Further, ℐ_{n} ∈ Pic(X_{p^{n}}) is the image of ℐ_{n+1} ∈ Pic(X_{p^{n+1}}) and ℒ_{n} ∈ Pic(X_{p^{n}}). A standard diagram argument yields some invertible sheaf ℒ_{n+1} ∈ Pic(X_{p^{n+1}}) which is mapped to ℒ_{n} and ℐ_{n+1}. This completes the proof for l ≠ p.

Third step. The case l = p. Here, we first observe the congruence

(1 + p^{n}c)^{p} ≡ 1 + p^{n+1}c (mod p^{n+2})

which, as p > 2, is valid for every n ≥ 1. This has the striking consequence that, for s ∈ Γ(U, 𝒪_{X_{p^{n}}}^{∗}), the power s^{p} automatically defines a section of 𝒪_{X_{p^{n+1}}}^{∗}. Further, we have the commutative diagram

[figure: 0; 𝒪_{X_{p}}; 𝒪_{X_{p^{n+1}}}^{∗}; (.)^{p}; 𝒪_{X_{p^{n}}}^{∗}; (.)^{p}; 0; 0; 𝒪_{X_{p}}; 𝒪_{X_{p^{n+2}}}^{∗}; 𝒪_{X_{p^{n+1}}}^{∗}; 0]

with exact rows. Taking cohomology, this yields the commutative diagram with exact rows,

[figure: 0; Pic(X_{p^{n+1}}); (.)^{⊗p}; Pic(X_{p^{n}}); (.)^{⊗p}; H^{2}(X_{p}, 𝒪_{X_{p}}); 0; Pic(X_{p^{n+2}}); Pic(X_{p^{n+1}}); H^{2}(X_{p}, 𝒪_{X_{p}}) .]

We see, in particular, that the lift of an invertible sheaf, if possible, is unique up to isomorphism.

We will inductively construct a sequence of sheaves ℒ_{n} ∈ Pic(X_{p^{n}}) lifting ℒ such that ℒ_{n}^{⊗p} ≅ ℐ_{n}. To start, simply put ℒ_{0} := ℒ. For the induction step, we observe that ℒ_{n}^{⊗p} ≅ ℐ_{n} implies that ℒ_{n} is mapped to ℐ_{n+1} under the middle vertical arrow in the diagram. Indeed, the lifting of an invertible sheaf is unique. The same diagram argument as in the second step completes the proof. □

2.4. Remark. –––– For p = 2, the same argument shows that Pic(X_{2})/P may only have 2-power torsion.

2.5. –––– To illustrate the effect of the obstructions, suppose that Pic(X_{p}) = ℤ^{n} and H^{2}(X_{p}, 𝒪_{X_{p}}) ≅ 𝔽_{p}. Then, the lattices Λ_{i} := Pic(X_{p^{i}}) form a system {Λ_{i}}_{i∈ℕ} such that

… ⊆ Λ_{i} ⊆ … ⊆ Λ_{2} ⊆ Λ_{1} ,

Λ_{i} ⊆ Λ_{i−1} is always of index 1 or p, and ℒ^{⊗p} ∈ Λ_{i} if and only if ℒ ∈ Λ_{i−1}.

According to Lemma 2.7 below, the system {Λ_{i} ⊗_{ℤ} ℤ_{p}}_{i∈ℕ} is isomorphic to {ℤ_{p}⊕ ⋯ ⊕ℤ_{p}⊕p^{i}ℤ_{p}}_{i∈ℕ} . I.e., there is a linear functional

H: Pic(X_{p}) = ℤ^{n} → ℤ_{p}, (x_{1}, …, x_{n}) ↦ a_{1}x_{1} + ⋯ + a_{n}x_{n}

with coefficients a_{1}, …, a_{n} ∈ ℤ_{p} such that, for ℒ ∈ Pic(X_{p}) arbitrary, p^{i}|H(ℒ) if and only if ℒ lifts to Pic(X_{p^{i}}).

H somehow collects all the obstruction maps into a single homomorphism. Further, H(ℒ) = 0 if and only if ℒ ∈ P. This shows again that Pic(X_{p})/P ↪ ℤ_{p} is torsion-free.

2.6. Remark. –––– This formulation also indicates that it is difficult to show rk P ≤ rk Pic(X_{p}) − 2. For this, one had to ensure that the ℤ-rank of im H is at least 2. But this is impossible knowing only p-adic approximations of a_{1}, …, a_{n}.

2.7. Lemma. –––– Let {Λ_{i}}_{i∈ℕ} be a sequence of p-adic lattices such that

i) Λ_{i+1} ⊂ Λ_{i}, ii) Λ_{i}/Λ_{i+1} = ℤ/pℤ, iii) x ∈ Λ_{i}\Λ_{i+1} ⟹ px ∈ Λ_{i+1}\Λ_{i+2}. Then, there exists a basis (b_{1}, …, b_{n}) of Λ_{1} such that Λ_{i} = ⟨b_{1}, …, b_{n−1}, p^{i−1}b_{n}⟩. Proof (cf. [We]). We first observe that Λ_{1}/Λ_{i} ≅ ℤ/p^{i−1}ℤ. Indeed, the quotient Λ_{1}/Λ_{i} is precisely of order p^{i−1}. Further, for x ∈ Λ_{1} \ Λ_{2}, we find px ∈ Λ_{2} \ Λ_{3} and, finally, p^{i−2}x ∈ Λ_{i−1} \Λ_{i}. In particular, we see that Λ_{0}/Λ_{i} has an element of order p^{i−1}.

Let now i be fixed. By the elementary divisor theorem, there exists a basis (b_{1}, …, b_{n}) of Λ_{0} such that (p^{e_{1}}b_{1}, …, p^{e_{n}}b_{n}) is a basis of Λ_{i}. As this yields Λ_{1}/Λ_{i} ≅ ℤ/p^{e_{1}}ℤ×⋯×ℤ/p^{e_{n}}ℤ, we may conclude e_{1} = ⋯ = e_{n−1} = 0 and e_{n} = i−1. The only lattices between Λ_{0} and Λ_{i} are ⟨b_{1}, …, b_{n−1}, p^{j}b_{n}⟩ for j = 1, …, i − 2. Thus, we have shown the assertion for a finite chain of lattices.

To prove it for the infinite sequence, we observe that the space of all bases of Λ_{1} is compact in the p-adic topology. For every i ∈ ℕ, there is a basis B^{(i)} = (b_{1}^{(i)}, …, b_{n}^{(i)}) of Λ_{1} having the desired property for the finite subsequence Λ_{1}, …, Λ_{i}. Consider the limit (b_{1}, …, b_{n}) of a convergent subsequence of {B^{(i)}}_{i∈ℕ} .

We claim that (b_{1}, …, b_{n−1}, p^{i−1}b_{n}) is a basis for Λ_{i}. Indeed, (b_{1}, …, b_{n−1}, p^{i−1}b_{n}) is arbitrarily close to a basis which completes the proof. □

3 The quotient Pic(X_{\overline{�}_{p}})/ Pic(X\overline{�} )

3.1. Sublemma. –––– Let p be a prime number and X be a ℤ_{p}-scheme which is proper and flat. Suppose that the generic fiber X_{η} is connected and the special fiber X_{p} is non-singular. Then, X_{p} is irreducible. Proof. The function field K := Γ(X_{η}, 𝒪_{X_{η}}) is a finite extension of ℚ_{p}. Further, O := Γ(X, 𝒪_{X}) is a finite ℤ_{p}-algebra being an integral domain with quotient field K. Clearly, O/pO is contained in Γ(X_{p}, 𝒪_{X_{p}}). But, according to the assumption, the latter does not have nilpotent elements other than zero. Hence, p generates the maximal ideal of O. This means, K/ℚ_{p} is necessarily unramified and O = 𝒪_{K} is its ring of integers. Stein factorization provides us with a morphism X → Spec 𝒪_{K} with connected fibers. From this, we immediately see that X_{p} is connected. As X_{p} is non-singular, this is enough for irreducibility. □

3.2. Lemma. –––– Let p ≠ 2 be a prime number and X be a ℤ_{p}-scheme which is proper and flat. Suppose that the special fiber X_{p} is non-singular and satisfies H^{1}(X_{p}, 𝒪_{X_{p}}) = 0. Then, the specialization homomorphism sp: Pic(X_{η}) → Pic(X_{p}) from the generic fiber has a torsion-free cokernel. Proof. As each connected component may be treated separately, we assume without restriction that X is connected. Further, the assumption implies that X is nonsingular. Hence, X is actually irreducible. This implies that X_{η} is irreducible, too. Finally, we conclude irreducibility of X_{p} from Sublemma 3.1.

There is a specialization map Pic(X_{η}) → Pic(X) given by taking the Zariski closure in X of a Weil divisor on X_{η}. This map is injective as the restriction forms a section to it. It is a surjection, too, as the only vertical divisors are principal, associated to the powers of (p).

Further, by A. Grothendieck’s existence theorem [EGA III, Corollaire (5.1.6)], one has Pic(X) = Pic(X̂). The assertion now follows from Lemma 2.3. □

3.3. Corollary. –––– Let p ≠ 2 be a prime number and X be a ℤ_{p}-scheme which is proper and flat. Suppose that the special fiber X_{p} is non-singular and satisfies H^{1}(X_{p}, 𝒪_{X_{p}}) = 0. Further, let K/ℚ_{p} be an unramified field extension and denote the residue field of K by k. Then, the cokernels of the specialization homomorphisms

i) sp_{K}: Pic(X_{K}) → Pic(X_{k}), ii) sp_{ℚ_{p}^{nr}} : Pic(X_{ℚ_{p}^{nr}} ) → Pic(X_{\overline{𝔽}_{p}}), and

iii) sp_{\overline{ℚ} _{p}}: Pic(X_{\overline{ℚ} _{p}}) → Pic(X_{\overline{𝔽}_{p}}) are torsion-free. Proof. i) Apply Lemma 3.2 to the fiber product X ×_{Spec ℤ_{p}} Spec 𝒪_{K}. ii) As the filtered direct limit functor is exact, the desired cokernel is the same as

\underrightarrow{lim} coker(sp_{K}: Pic(X_{K}) → Pic(X_{k}))

where K is running over the unramified extensions of ℚ_{p} and k denotes the residue field of K. As all the cokernels are torsion-free, the assertion follows. iii) We claim that sp_{\overline{ℚ} _{p}} has the same image in Pic(X_{\overline{𝔽}_{p}}) as sp_{ℚ_{p}^{nr}} . Let ℒ ∈ Pic(X_{\overline{ℚ} _{p}}) . The Galois group Γ := Gal(\overline{ℚ} _{p}/ℚ_{p}^{nr}) sends ℒ to a finite orbit \{ℒ_{1}, …, ℒ_{m}\} . The specializations of ℒ_{1}, …, ℒ_{m} in Pic(X_{\overline{𝔽}_{p}}) are all the same. Therefore,

m·sp_{\overline{ℚ} _{p}}(ℒ) = sp_{\overline{ℚ} _{p}}(ℒ^{⊗m}) = sp_{\overline{ℚ} _{p}}(ℒ_{1}⊗ ⋯ ⊗ℒ_{m}) = sp_{ℚ_{p}^{nr}} (ℒ_{1}⊗ ⋯ ⊗ℒ_{m})

since ℒ_{1}⊗ ⋯ ⊗ℒ_{m} is Γ-invariant. Hence, m·sp_{\overline{ℚ} _{p}}(ℒ) ∈ im sp_{ℚ_{p}^{nr}} . As sp_{ℚ_{p}^{nr}} has a torsion-free cokernel, we see that sp_{\overline{ℚ} _{p}}(ℒ) ∈ im sp_{ℚ_{p}^{nr}} , too. □

3.4. Theorem. –––– Let p ≠ 2 be a prime number and X be a scheme proper and flat over ℤ. Suppose that the special fiber X_{p} is non-singular and satisfies H^{1}(X_{p}, 𝒪_{X_{p}}) = 0.

Then, the specialization homomorphism sp\overline{ℚ} : Pic(X\overline{ℚ} ) → Pic(X_{\overline{𝔽}_{p}}) has a torsionfree cokernel.

Proof. There is a canonical injection Pic(X\overline{ℚ} ) ↪ Pic(X_{\overline{ℚ} _{p}}). We have to show that both Picard groups have the same image under specialization to Pic(X_{\overline{𝔽}_{p}}).

For this, we switch at first to the scheme X_{Z} for Z = ℤ[\frac{1}{m}] where m is an integer divisible by all primes of bad reduction but not by p. Again, we may assume without restriction that X_{Z} is connected. Further, by construction, X_{Z} is nonsingular and, therefore, irreducible. By virtue of Sublemma 3.1, all the special fibers of X_{Z} are irreducible.

According to a theorem of Grothendieck (cf. [Kl, Theorem 4.8]), the Picard scheme Pic_{X_{Z}/Z} exists in this situation as a scheme, locally of finite type over Z. This means, we are given a morphism i: Spec \overline{ℚ} _{p} → Pic_{X_{Z}/Z} and have to show that there is a morphism Spec \overline{ℚ} → Pic_{X_{Z}/Z} such that the specializations modulo p are the same.

Locally, near the image of i, we have an affine open subset U ≅ Spec R ⊆ Pic_{X_{Z}/Z} for R a finitely generated Z-algebra. We are thus given a ring homomorphism ι: R → \overline{ℚ} _{p}. This actually maps R to 𝒪_{K} for a suitable finite extension K/ℚ_{p}. Unfortunately, as a Z-algebra, 𝒪_{K} is not finitely generated. On the other hand, im ι =: S ⊂ 𝒪_{K} is clearly a finitely generated Z-algebra. We fix a set of generators {T_{1}, …, T_{n}} of S.

Preserving the induced homomorphism to 𝔽_{q} := 𝒪_{K}/𝔪_{K}, our goal is to replace ι by a homomorphism to another subring S′ ⊂ 𝒪_{K} such that S is finite as a Z-module. For this, we will construct an algebra homomorphism φ: S → S′ such that ν(x − φ(x)) ≥ 1 for every x ∈ S. Here, ν denotes the discrete valuation on 𝒪_{K}.

To perform this construction, we apply Noether normalization [ZS, Ch. V, §4, Theorem 8] to S ⊗_{ℤ} ℚ. This states that S ⊗_{ℤ} ℚ is an integral extension of a polynomial ring ℚ[X_{1}, …, X_{k}] ⊆ S⊗_{ℤ} ℚ. We send X_{1}, …, X_{k} to elements of 𝒪_{K} algebraic over ℚ such that ν(X_{i} − φ(X_{i})) ≫ 0. Then, this extends to a homomorphism of the whole of S⊗_{ℤ}ℚ. We claim that ν(T_{i} − φ(T_{i})) ≥ 1 for i = 1, …, n. Indeed, as T_{i} is integral over ℚ[X_{1}, …, X_{k}, T_{1}, …, T_{i−1}], this follows from an iterated application of Hensel’s lemma in the form of [Na, Proposition 5.5].

Since S is generated by T_{1}, …, T_{n} as a Z-algebra and ν(z) ≥ 0 for every z ∈ Z, we see that ν(x − φ(x)) ≥ 1 for every x ∈ S. This completes the proof. □

4 An explicit obstruction

4.1. Proposition. –––– Let S be a K3 surface of degree 2 over ℚ, given explicitly by

w^{2} = f_{6}(x, y, z)

for f_{6} ∈ ℤ[x, y, z] of degree 6. Suppose, for a prime p ≠ 2, there is an 𝔽_{p}-rational tritangent line “ℓ = 0” of the ramification locus of S_{p}. Write l for a split of the pull-back of the tritangent. One has f_{6} ≡ f_{3}^{2} + ℓf_{5} (mod p) for homogeneous forms f_{3}, f_{5} ∈ ℤ[x, y, z]. Put

G(x, y, z) := (f_{6} − f_{3}^{2} − ℓf_{5})/p .

Then, the obstruction to lifting 𝒪(l) to S_{p^{2}} is ((−G) mod (p, ℓ, f_{3}, f_{5})).

Proof. First step. An affine open covering of S_{p}. On S_{p}, we have w^{2} = f_{3}^{2} + ℓf_{5} and, for h a quadric, w^{2} = (f_{3} + ℓh)^{2} + ℓf′_{5} where f′_{5} := f_{5} − 2f_{3}h − ℓh^{2}. On “ℓ = 0”, f_{3} and f_{5} have no common zero as this would cause a singularity on S_{p}. Hence, for a suitably chosen h, the three forms ℓ, f_{5}, and f′_{5} do not have a common zero. For this, it may be necessary to extend the ground field. The sets “ℓ ≠ 0”, “f_{5} ≠ 0”, and “f′_{5} ≠ 0”, form an affine open covering of S_{p}. We may extend them in the obvious manner to an affine open covering of S.

Second step. The invertible sheaf 𝒪(5l). We start with 𝒪(5l) instead of 𝒪(l) as this will turn out to be easier. 𝒪(5l) is given by the rational functions 1 on “ℓ ≠ 0”, \frac{f_{5}^{3}}{(w+f_{3})^{5}} on “f_{5} ≠ 0”, and \frac{f′_{5}^{3}}{(w+f_{3}+ℓh)^{5}} on “f′_{5} ≠ 0”. Thus, the transition functions are

\frac{f_{5}^{3}}{(w + f_{3})^{5}} = \frac{(w − f_{3})^{5}}{ℓ^{5}f_{5}^{2}} , \frac{(w + f_{3})^{5}f′_{5}^{3}}{(w + f_{3} + ℓh)^{5}f_{5}^{3}} = \frac{(w + f_{3})^{5}(w − f_{3} − ℓh)^{5}}{ℓ^{5}f_{5}^{3}f′_{5}^{2}} ,

and \frac{(w + f_{3} + ℓh)^{5}}{f′_{5}^{3}} .

Third step. The obstruction. We may lift the first and third transition functions naively. The middle one is a transition function between “f_{5} ≠ 0” and “f′_{5} ≠ 0” and, thus, must not have a pole at “ℓ ≠ 0”. We lift (w + f_{3})(w − f_{3}) as ℓf_{5} and obtain, in total,

\frac{[f_{5} − h(w + f_{3})]^{5}}{f_{5}^{3}f′_{5}^{2}} .

The product of the three lifts is

\frac{(w − f_{3})^{5}[f_{5} − h(w + f_{3})]^{5}(w + f_{3} + ℓh)^{5}}{ℓ^{5}f_{5}^{5}f′_{5}^{5}} .

Observe that, in the form described, the transition functions may be lifted even to the affine open subsets of S, not just to S_{p^{2}}. Hence, the exponential of the obstruction for 𝒪(l) is \frac{(w−f_{3})[f_{5}−h(w+f_{3})](w+f_{3}+ℓh)}{ℓf_{5}f′_{5}}, also in the case that p = 5.

Evaluating this expression, making use of the identity w^{2}−f_{3}^{2} = ℓf_{5}+pG, we end up with 1 + p \frac{G(f_{5}−hw−hf_{3}−ℓh^{2})}{ℓf_{5}f′_{5}}. Therefore, the obstruction to lifting 𝒪(l) is given by the Čech cocycle

\frac{G(f_{5} − hw − hf_{3} − ℓh^{2})}{ℓf_{5}f′_{5}} .

Fourth step. Simplification. Any rational function having poles in only two of the three divisors considered is a Čech coboundary. Without changing the cohomology class, we may therefore add to the numerator forms being homogeneous of degree 11 and belonging to the ideal (ℓ, f_{5}, f′_{5}).

On the line “ℓ = 0”, f_{5} and f′_{5} have no zeroes in common. Thus, they are coprime in the graded ring 𝔽_{p}[x, y, z]/(ℓ). Consequently, f_{5} and f′_{5} already generate the full 10-dimensional space of forms of degree 9. Even more, they must generate the space of forms of degree 11. This shows that we may simplify the Čech cocycle to \frac{−Ghw}{ℓf_{5}f′_{5}}.

Hence, the obstruction to lifting 𝒪(l) is ((−Gh) mod (p, ℓ, f_{5}, f′_{5})). The ideal is the same as (p, ℓ, hf_{3}, f_{5}). Thus, the question is whether ((−Gh) mod (p, ℓ)) is a combination of hf_{3} and f_{5}. As, on the line “ℓ = 0” on S_{p}, h and f_{5} have no common zeroes, they are coprime. ((−Gh) mod (p, ℓ)) must be a combination of hf_{3} and hf_{5}. We may, as well, consider ((−G) mod (p, ℓ, f_{3}, f_{5})). □

4.2. Example. –––– Let S be a K3 surface over ℚ given by w^{2} = f_{6}(x, y, z). Suppose

f_{6}(x, y, z) ≡ x^{6} + 2x^{5}z + 2x^{4}y^{2} + 2x^{4}z^{2} + 2x^{3}y^{3} + 2x^{3}z^{3}
+ 2x^{2}y^{4} + 2x^{2}y^{3}z + x^{2}z^{4} + xy^{3}z^{2} + 2xz^{5} + y^{6} (mod 3) .

Assume further that the coefficient of y^{2}z^{4} is not divisible by 9. Then, rk Pic(S\overline{ℚ} ) = 1. Proof. A direct calculation shows that, modulo 3, the right hand side is f_{3}^{2} + xf_{5} for f_{3} = 2x^{3} + 2x^{2}z + xz^{2} + 2y^{3} and f_{5} = 2x^{3}y^{2} + x^{2}z^{3} + 2xy^{4} + 2z^{5}. Thus, the branch locus of S_{3} has a tritangent line given by x = 0.

The numbers of points over 𝔽_{3^{d}} are, in this order, 19, 127, 676, 6 751, 58 564, 532 414, 4 791 232, 43 038 703, 387 383 311, and 3 486 675 052. For the decomposition of the characteristic polynomial of the Frobenius, we find

(t − 3)^{2}(t^{20} − 3 t^{19} − 9 t^{18} + 72 t^{17} − 8 1t^{16} − 324 t^{15} + 1 458 t^{14} − 2 916 t^{13} + 4 374 t^{12} + 26 244 t^{11} − 137 781 t^{10} + 236 196 t^{9} + 354 294 t^{8} − 2 125 764 t^{7} + 9 565 938 t^{6} − 19 131 876 t^{5} − 43 046 721 t^{4} + 344 373 768 t^{3} − 387 420 489 t^{2}

− 1 162 261 467 t + 348 6784 401) .

This shows rk Pic(S_{\overline{𝔽}_{3}}) ≤ 2.

Let l be a split of the pull-back of the tritangent line. We have to show that the obstruction to lifting 𝒪(l) is non-zero. For this, we observe that x, f_{3}, and f_{5} do not generate the monomial y^{2}z^{4}. However, G contains this monomial by its very definition. □

4.3. Example. –––– Consider the K3 surface S over ℚ, given by w^{2} = f_{6}(x, y, z) for

f_{6}(x, y, z) = 4x^{6} + 2x^{5}y + 12x^{5}z + 2x^{4}y^{2} + 4x^{4}yz + 12x^{4}z^{2} + 24x^{3}y^{3} − 57x^{3}y^{2}z
− 9x^{3}yz^{2} + 6x^{3}z^{3} + 8x^{2}y^{4} − 5x^{2}y^{3}z − 72x^{2}y^{2}z^{2} + 7x^{2}yz^{3} + 4x^{2}z^{4}
+ 20xy^{4}z − 52xy^{3}z^{2} − 57xy^{2}z^{3} + 7xyz^{4} + 4y^{5}z − 7y^{4}z^{2} − 18y^{3}z^{3}
+ 7y^{2}z^{4} + 12yz^{5} + 2z^{6} .

Then, rk Pic(S\overline{ℚ} ) = 3. Proof. We have

f_{6} = (2x^{3} + 2x^{2}z + 2y^{2}z + yz^{2} + z^{3})^{2}
+ (2x^{2} + 2xz + yz + z^{2})(x^{3}y + 2x^{3}z + x^{2}y^{2} + x^{2}yz + 2x^{2}z^{2} + 12xy^{3}
− 34xy^{2}z − 9xyz^{2} − 2xz^{3} + 4y^{4} − 15y^{3}z − 7y^{2}z^{2} + 9yz^{3} + z^{4})

and

f_{6} = 4(x^{3} + 2x^{2}y + 2x^{2}z + xy^{2} + xyz + xz^{2} + y^{2}z + yz^{2} + z^{3})^{2}
− (x^{2} + xz + yz + z^{2})(14x^{3}y + 4x^{3}z + 22x^{2}y^{2} + 22x^{2}yz + 8x^{2}z^{2} − 8xy^{3}
+ 61xy^{2}z + 9xyz^{2} + 6xz^{3} − 4y^{4} + 15y^{3}z + 11y^{2}z^{2} − 6yz^{3} + 2z^{4}) .

Hence, there are two conics C_{1} and C_{2} each of which is six times tangent to the ramification locus of S. The splits of their pull-backs yield the intersection matrix

( \begin{matrix}−2 & 6 & 1 & 3 \\ 6 & −2 & 3 & 1 \\ 1 & 3 & −2 & 6 \\ 3 & 1 & 6 & −2 \end{matrix} )

which is of rank 3. Hence, rk Pic(S\overline{ℚ} ) ≥ 3.

On the other hand, S has good reduction at the prime p = 3. Point counting over extensions of 𝔽_{3} shows that the characteristic polynomial of the Frobenius operating on S_{3} is

(t − 3)^{4}(t^{18} + 3 t^{17} + 6 t^{16} + 18 t^{15} + 108 t^{14} + 405 t^{13} + 972 t^{12} + 2 187 t^{11} + 13 122 t^{10} + 52 488 t^{9} + 118 098 t^{8} + 177 147 t^{7} + 708 588 t^{6} + 2 657 205 t^{5} + 6 377 292 t^{4} + 9 565 938 t^{3} + 28 697 814 t^{2} + 129 140 163 t + 387 420 489) .

Consequently, we have rk Pic(S_{\overline{𝔽}_{3}}) ≤ 4.

In particular, the assumption rk Pic(S\overline{ℚ} ) > 3 implies rk Pic(S\overline{ℚ} ) = rk Pic(S_{\overline{𝔽}_{3}}). Theorem 3.4 guarantees that the specialization map sp\overline{ℚ} : Pic(S\overline{ℚ} ) → Pic(S_{\overline{𝔽}_{3}}) must be bijective. Giving one line bundle ℒ ∈ Pic(S_{\overline{𝔽}_{3}}) with a non-trivial obstruction will be enough to yield a contradiction.

For this, observe that the ramification locus of S_{3} has a tritangent line given by x + y + z = 0. Indeed,

f_{6}(x, y, z) ≡ (x^{3} + x^{2}y + xy^{2} + y^{3})^{2} + (x + y + z)(2x^{3}y^{2} + x^{3}yz + 2x^{2}yz^{2} + 2xy^{4}

+ xy^{3}z + xy^{2}z^{2} + 2xyz^{3} + xz^{4} + 2y^{5} + 2y^{4}z + yz^{4} + 2z^{5}) (mod 3) .

Modulo the ideal (3, x + y + z), we have f_{3} ≡ x^{3} + x^{2}y + xy^{2} + y^{3}, f_{5} ≡ −(x^{5} + x^{3}y^{2} + x^{2}y^{3} + xy^{4} + y^{5}), and G ≡ x^{6} + 2x^{5}y + x^{4}y^{2} + 2xy^{5} + y^{6}. Trying to generate G by 3, x + y + z, f_{3}, and f_{5} now leads to linear system of seven equations in six unknowns which is easily seen to be unsolvable. □

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