ON THE CHARACTERISTIC POLYNOMIAL OF THE FROBENIUS ON ÉTALE COHOMOLOGY

ANDREAS-STEPHAN ELSENHANS AND JÖRG JAHNEL

Abstract. Let X be a smooth proper variety of even dimension d over a finite field. We establish a restriction on the value at (−1) of the characteristic polynomial of the Frobenius on the middle-dimensional étale cohomology of X with coefficients in ℚ_{l}(d/2).

1. Introduction

Let X be a smooth proper variety over a finite field 𝔽_{q} of characteristic p > 0. Then the geometric Frobenius Frob operates linearly on the l-adic cohomology vector spaces H_{ét}^{i}(X_{\overline{𝔽}_{q}}, ℚ_{l}(j)) . The characteristic polynomial of Frob has rational coefficients [11, Théorème (1.6)].

By far not every polynomial Φ ∈ ℚ[T] may occur as the characteristic polynomial of the Frobenius on a smooth proper variety. The following conditions were established essentially in the 70s of the last century, in the course of working out A. Grothendieck’s ideas in Algebraic Geometry.

Theorem 1.1 (Deligne, Mazur, Ogus). Let X be a smooth proper variety over a finite field 𝔽_{q} of characteristic p. For i, j ∈ ℤ, denote by

Φ_{j}^{(i)} = T^{N} + a_{1}^{(i)}T^{N−1} + ⋯ + a_{N−1}^{(i)}T + a_{N}^{(i)}

the characteristic polynomial of Frob on H_{ét}^{i}(X_{\overline{𝔽}_{q}}, ℚ_{l}(j)) . Then, for every r, one has a_{r}^{(i)} ∈ ℚ and this value is independent of l ≠ p. Moreover,

a) every complex zero of Φ_{j}^{(i)} is of absolute value q^{i/2−j}.

b) If i is odd then all real zeroes of Φ_{j}^{(i)} are of even multiplicity.

c) For every l ≠ p, the zeroes of Φ_{j}^{(i)} are l-adic units, i.e. units in a suitable extension of ℤ_{l}.

d) Put h_{i−m,m}:=dim H^{i−m}(X, Ω_{X}^{m}) and define the step-function G^{(i)} : [0, N] → ℝ by

G^{(i)}(t) = \{ \begin{matrix}0 & for t ≤ h_{i,0} , \\ n & for h_{i,0} + ⋯ + h_{i−n+1,n−1} < t ≤ h_{i,0} + ⋯ + h_{i−n,n} . \end{matrix}

2010 Mathematics Subject Classification. Primary: 14F20, Secondary: 14G15, 14J20, 14J35.

Key words and phrases. Proper variety over finite field, Middle cohomology, Frobenius, Characteristic polynomial, Artin-Tate formula.

Then, for r = 1, …, N, one has a_{r}^{(i)} = 0 or

ν_{q}(a_{r}^{(i)}) ≥ ∫_{0}^{r}[G^{(i)}(t) − j] dt .

Here, ν_{q} is the non-archimedean valuation such that ν_{q}(q) = 1.

Remarks 1.2. i) We will provide references and proof sketches at the beginning of Section 2. ii) Assertion a) immediately implies that Φ_{j}^{(i)} ∈ ℚ[T] fulfills the functional equation

T^{N}Φ(q^{i−2j}/T) = ±q^{\frac{N}{2} (i−2j)}Φ(T)

for N := dim H_{ét}^{i}(X_{\overline{𝔽}_{q}}, ℚ_{l}(j)) . Indeed, on both sides, there are polynomials with leading term ±q^{\frac{N}{2} (i−2j)}T^{N}. They have the same zeroes as, with z, the number \overline{z} = \frac{q^{i−2j}}{z} is a zero of Φ, too. Furthermore, by b), the plus sign always holds when i is odd. The statements c) and d) together show Φ_{j}^{(i)} ∈ ℤ[T] when j ≤ 0. iii) For X projective and i even, we also have

Φ_{j}^{(i)}(q^{i/2−j}) = 0 .  (1)

Indeed, there is the Gal(\overline{𝔽}_{q}/𝔽_{q})-invariant cycle given by the intersection of i/2 hyperplanes. The cycle map [10, Cycle, Théorème 2.3.8.iii)] yields a non-trivial Galois invariant element of H_{ét}^{i}(X_{\overline{𝔽}_{q}}, ℚ_{l}(i/2)).

Remark 1.3. Consider the case that i = 1. Then N = dim H_{ét}^{1}(X_{\overline{𝔽}_{q}}, ℚ_{l}) is always even [12, Corollaire (4.1.5)]. On the other hand, let a polynomial Φ ∈ ℤ[T] be given that is of even degree and fulfills assertions a), b), and c). Then, by the main theorem of T. Honda [23], there exists an abelian variety A such that the eigenvalues of Frob on H_{ét}^{1}(A_{\overline{𝔽}_{q}}, ℚ_{l}) are exactly the zeroes of Φ. One may enforce that the characteristic polynomial is a power of Φ, and, typically, Φ itself may be realized.

1.4. We will show in this note that the same is not true in general for i > 1. In fact, for the characteristic polynomial of Frob on the middle cohomology of a variety of even dimension, we will establish a further condition, which is arithmetic in nature and independent of Theorem 1.1, as well as of formula (1).

Theorem 1.5. Let X be a smooth proper variety of even dimension d over a finite field 𝔽_{q} of characteristic p and Φ = Φ_{d/2}^{(d)} ∈ ℚ[T] be the characteristic polynomial of Frob on H_{ét}^{d}(X_{\overline{𝔽}_{q}}, ℚ_{l}(d/2)) . Put N := deg Φ. Then (−2)^{N}Φ(−1) is a square or p times a square in ℚ.

Remark 1.6. For X a surface, this result may be deduced from the Tate conjecture, via the Artin-Tate formula. Cf. Proposition 3.11, below.

1.7. The correct exponent of p, may, at least for p ≠ 2, be described as follows. Definition. We put

a(X) := ∑_{m=0}^{\frac{d}{2} −1}(\frac{d}{2} − m)h′_{d−m,m} ,

for (h′_{d−m,m})_{m} the abstract Hodge numbers of X in degree d [30, Section 4].

Remarks 1.8. a) Recall that the abstract Hodge numbers are defined as follows. The crystalline cohomology groups H^{i}(X/W) are finitely generated W-modules, for W := W (𝔽_{q}) the Witt ring. They are acted upon by the absolute Frobenius F, the corresponding map is only F-semilinear [8, Exposé I, 2.3.5]. Put H := H^{i}(X/W)/tors. Then, as F: H → H is injective, H/FH is a W-mod-ule of finite length. By the classical invariant factor theorem, there is a unique sequence of integers such that H/FH ≅ ⨁_{m>0}(W/p^{m}W)^{h′_{i−m,m}}. Finally, one defines

h′_{i,0} := rk_{W} H − ∑_{m>0}h′_{i−m,m}.

Observe that a(X) is a geometric quantity. It depends only on the base extension X_{\overline{𝔽}_{q}}.

b) Suppose that X is such that all H^{i}(X/W) are torsion-free and that the conjugate spectral sequence E_{2}^{jm} := H^{j}(X, ℋ ^{m}(Ω_{X/𝔽_{q}}^{•} )) ⟹ H_{dR}^{i}(X) degenerates at E_{2}. Then h′_{i−m,m} = h_{i−m,m} (= dim H^{i−m}(X, Ω_{X}^{m})) [5, Lemma 8.32]. For complete intersections, both assumptions hold ([15, Exposé XI, Théorème 1.5] together with [5, Lemma 8.27.2]). Moreover, the second assumption is automatically fulfilled when dim X ≤ p and X lifts to W ([14, Corollaire 2.4] and [5, Lemma 8.27.2]).

c) Suppose that X is of Hodge-Witt type in degree d, i.e., that the Serre cohomology groups H^{j}(X, WΩ_{X}^{m}) [36] are finitely generated W-modules for j + m = d. The assertion of Theorem 1.9 below may then be formulated entirely in terms of the characteristic polynomial Φ. In fact, denote the zeroes of Φ by z_{1}, …, z_{N}. Then

a(X) = − ∑_{ν_{q}(z_{i})<0} ν_{q}(z_{i})

[Corollary 2.16]. This case includes all varieties that are ordinary in degree d [25, Définition IV.4.12].

Theorem 1.9. Let X be a smooth proper variety of even dimension d over a finite field 𝔽_{q} of characteristic p ≠ 2 and Φ = Φ_{d/2}^{(d)} ∈ ℚ[T] be the characteristic polynomial of Frob on H_{ét}^{d}(X_{\overline{𝔽}_{q}}, ℚ_{l}(d/2)) . Put N := deg Φ. Then (−2)^{N}q^{a(X)}Φ(−1) is a square in ℚ.

Remarks 1.10. i) It seems not unlikely that an analogous result is true in characteristic 2, too. The difficulties that arise may well be of purely technical nature. Cf. Remark 2.14, below. ii) The assertions may easily be formulated for an arbitrary Tate twist. But this does not lead to anything new. Also, the twist by d/2 appears to be natural as the operation of Frob on H_{ét}^{d}(X_{\overline{𝔽}_{q}}, ℚ_{l}(d/2)) is orthogonal with respect to a ℚ_{l}-valued symmetric, bilinear pairing.

Remarks 1.11. i) One reason for our interest in these conditions is that they simplify the actual computation of the characteristic polynomial Φ for a given variety X. Such computations usually involve point counting on X over several extensions of the base field, cf. Examples 3.13 and 3.15. The conditions given redundantise the most expensive counting steps. ii) Applications of the characteristic polynomials Φ to varieties in characteristic zero include the determination of the Néron-Severi rank. The interested reader might consult the articles [28], [19], and [20].

Acknowledgements. We wish to thank Christian Liedtke (München), who read an earlier version of this article, for many valuable suggestions.

2. The proofs

2.1. Proof of Theorem 1.1. a) This was first proven by P. Deligne in [11, Théorème (1.6)] for the projective case and later in [12, Corollaire (3.3.9)], in general. The assertion was formulated by A. Weil as a part of his famous conjectures.

b) If X is projective then, by the Hard Lefschetz theorem [12, Théorème (4.1.1)] and Poincaré duality, there is a non-degenerate pairing

H_{ét}^{i}(X_{\overline{𝔽}_{q}}, ℚ_{l}(j)) × H_{ét}^{i}(X_{\overline{𝔽}_{q}}, ℚ_{l}(j)) → ℚ_{l}(2j − i)

that is compatible with the operation of Frob. It is alternating as i is odd. The assertion follows directly from this. Cf. the remarks after [12, Corollaire (4.1.5)]. The proper non-projective case has only recently been settled by J. Suh [40, Corollary 2.2.3].

c) As Frob operates on H_{ét}^{i}(X_{\overline{𝔽}_{q}}, ℤ_{l}(j)), the l-adic valuations of the eigenvalues are clearly non-negative. Poincaré duality implies the assertion, cf. [11, (2.4)].

d) This statement was originally known as Katz’s conjecture. The usual formulation is that the Newton polygon of Φ_{0}^{(i)} lies above the Hodge polygon of weight i. Proofs are due to B. Mazur [31] and A. Ogus [5, Theorem 8.39]. □

Notation 2.2. For R an integral domain, K its field of fractions, and H an R-module, we will write H_{K} := H ⊗_{R}K.

Lemma 2.3. Let R be a principal ideal domain, K its field of fractions, and H a free R-module of finite rank, equipped with a perfect, R-bilinear, symmetric pairing H × H → R. Denote its K-bilinear extension by ⟨. , .⟩: H_{K} × H_{K} → K. Furthermore, let a K-linear map σ: H_{K} → H_{K} be given that is orthogonal with respect to the pairing, i.e., ⟨σ(x), σ(y)⟩ = ⟨x, y⟩ for every x, y ∈ H_{K}.

Put

B_{0}(H) := [H/[H ∩ (1 − σ)H]]_{tors} .

a) Then there is a non-degenerate, skew-symmetric R-bilinear pairing

(. , .): B_{0}(H) × B_{0}(H) → K/R .

b) Suppose char K ≠ 2 and ⟨x, x⟩ ∈ 2R for every x ∈ H ∩ (1 − σ)H_{K}. Then (. , .) is alternating. In particular, the length of B_{0}(H) is even.

Remarks 2.4. i) Observe that x ∈ H represents an element of B_{0}(H) if and only if x ∈ H ∩ (1 − σ)H_{K}. ii) For x, y ∈ H_{K} arbitrary, one has ⟨(1 − σ)x, σy⟩ = −⟨x, (1 − σ)y⟩. In particular, as σ: H_{K} → H_{K} is bijective, x ∈ ker(1 − σ) if and only if x ∈ ((1 − σ)H_{K})^{⊥}. I.e., (1−σ)H_{K} is exactly the set of all elements perpendicular to the eigenspace H_{K,1}. This fact is rather obvious, let us nevertheless emphasize that it is true whether σ is semisimple or not.

2.5. Proof of Lemma 2.3. a) Definition. The pairing is defined as follows. For a, b ∈ B_{0}(H), choose representatives x, y ∈ H. Let y′ ∈ H_{K} be such that y = (1 − σ)y′. Then (a, b) := ⟨x, y′⟩ mod R. Well-definedness. For two representatives x_{1}, x_{2} ∈ H, we have x_{1} − x_{2} = (1 − σ)v for some v ∈ H. Thus,

⟨x_{1} − x_{2}, y′⟩ = ⟨(1 − σ)v, y′⟩ = ⟨(1 − σ)σv, σy′⟩ = −⟨σv, (1 − σ)y′⟩
= −⟨v − x_{1} + x_{2}, y⟩ ∈ R ,

as both sides are in H. On the other hand, for two representatives y′_{1}, y′_{2} ∈ H_{K}, we have (1 − σ)(y′_{1} − y′_{2}) = (1 − σ)w ∈ H for a suitable w ∈ H. Therefore, ⟨x, y′_{1} − y′_{2}⟩ = ⟨x, w⟩ + ⟨x, k⟩ for some k ∈ H_{K,1}. The first summand is in R, as both sides are elements of H. The second summand vanishes, since x ∈ (1 − σ)H_{K}. It is clear that (. , .) is R-bilinear. Non-degeneracy. For 0 ≠ b ∈ B_{0}(H), one has a representative y and some y′ ∈ H_{K} such that y = (1 − σ)y′. As y ∉ (1 − σ)H, we see y′ ∉ H + H_{K,1}. The goal is to find some x ∈ H ∩ (1 − σ)H_{K} such that ⟨x, y′⟩ ∉ R.

For this, we observe that the perfect pairing induces an isomorphism

H_{K} ⟶^{≅} Hom(H, K) .

Under this map, H_{K,1} ≅ Hom(H/[H ∩ (1 − σ)H_{K}], K). Furthermore,

H + H_{K,1} ≅ {α ∈ Hom(H, K) | α(H ∩ (1 − σ)H_{K}) ⊆ R} .

Indeed, as H ≅ Hom(H, R), the inclusion “⊆” is obvious. The other inclusion follows from the fact that H ∩ (1 − σ)H_{K} is a direct summand of H. The homomorphism α|_{H∩(1−σ)H_{K}}: H ∩ (1 − σ)H_{K} → R may thus be extended to a homomorphism α′: H → R, corresponding to an element of H. The difference α − α′ vanishes on H ∩ (1 − σ)H_{K}, hence is defined by an element of H_{K,1}.

Now, as y′ ∉ H + H_{K,1}, the corresponding homomorphism does not send H ∩ (1 − σ)H_{K} to R. I.e., there is some x ∈ H ∩ (1 − σ)H_{K}, not mapped to R. This is exactly our claim. Skew-symmetry. Let a, b ∈ B_{0}(H) and choose representatives x, y ∈ H. There are x′, y′ ∈ H_{K} such that x = (1 − σ)x′ and y = (1 − σ)y′. Then

⟨x, y′⟩ + ⟨x′, y⟩ = ⟨x′ − σx′, y′⟩ + ⟨x′, y′ − σy′⟩ = ⟨x′ − σx′, y′ − σy′⟩ = ⟨x, y⟩ ∈ R .

b) For a ∈ B_{0}(H) choose a representative x ∈ H and x′ ∈ H_{K} such that x = (1−σ)x′. Then

2(a, a) = 2⟨x, x′⟩ = [⟨x′, x′⟩ − ⟨σx′, x′⟩] + [⟨σx′, σx′⟩ − ⟨x′, σx′⟩]
= ⟨x′ − σx′, x′ − σx′⟩ = ⟨x, x⟩ ∈ 2R ,

hence (a, a) = 0. Evenness of ℓ(B_{0}(H)). First note that B_{0}(H) is a finitely generated torsion module over a principal ideal domain. This implies that ℓ(B_{0}(H)) is finite.

It is known [37, Exercise 10.20] that, for such a module M, the existence of a non-degenerate alternating pairing implies that ℓ(M) is even. The argument is essentially as follows.

Assume the contrary and let M be a counterexample of minimal length. Choose x ∈ M such that Rx ⊆ M is of length one. I.e., 𝔪 := Ann_{R}(x) is a maximal ideal. Then L_{x} := ⟨x, . ⟩: M → K/R is a nonzero linear form, the image of which is 𝔪^{−1}/R ≅ R/𝔪. Clearly, one has x ∈ ker L_{x}.

There is an alternating form induced on [ker L_{x}]/(x), which is again non-degen-erate. Moreover, ℓ(ker L_{x}) = ℓ(M) − 1 and ℓ([ker L_{x}]/(x)) = ℓ(M) − 2. This is a contradiction. □

Remarks 2.6. i) If 2 is a unit in R then the assumption of b) is automatically fulfilled. ii) When R = ℤ_{l}, σ(H) ⊆ H, and σ is semisimple at 1, this result was proven by Yu. G. Zarhin in [43, 3.3 and Lemma 3.4.1]. It is implicitly contained in the work of J. W. S. Cassels [6]. iii) Most of our applications will be based on the following corollary.

Corollary 2.7. Let (R, ν) be a normalized discrete valuation ring, k its residue field, which we assume to be finite of characteristic ≠2, K the field of fractions, and H a free R-module of finite rank. Suppose there is a perfect, R-bilinear, symmetric pairing H × H → R and denote its K-bilinear extension by ⟨. , .⟩: H_{K} × H_{K} → K. Furthermore, let a K-linear map σ: H_{K} → H_{K} be given that is orthogonal with respect to the pairing and such that 1 is not among its eigenvalues.

a) Then ℓ(σ(H) + H/H) + ν(det(1 − σ)) is even.

b) In particular, if σ(H) ⊆ H then ν(det(1 − σ)) is even.

Proof. a) We write q := #k. According to the definition, the modulus [16, section 14.3] of the map (1 − σ): H_{K} → H_{K} may be computed as q^{ℓ(M/H)−ℓ(M/(1−σ)H)} for every R-module M ⊇ H, (1 − σ)H that is chosen such that the two lengths are finite. Moreover, by [16, Exercise 14.3.6], that modulus is equal to q^{−ν(det(1−σ))}. Consequently,

ν(det(1 − σ)) = ℓ([σ(H) + H]/(1 − σ)H) − ℓ(σ(H) + H/H) .

On the other hand, all the assumptions of Lemma 2.3.b) are fulfilled. As 1 is not an eigenvalue of σ, H/[H ∩ (1 − σ)H] is purely torsion. Thus, we have that ℓ(H/[H ∩ (1 − σ)H]) is even. Furthermore,

H/[H ∩ (1 − σ)H] ≅ [H + (1 − σ)H]/(1 − σ)H = [σ(H) + H]/(1 − σ)H .

I.e., ℓ([σ(H) + H]/(1 − σ)H) is even, too. The assertion follows.

b) is an immediate consequence of a). □

2.8. In order to illustrate the strength of Corollary 2.7, let us show an application to modules of rank two, the smallest non-trivial case. The fact obtained belongs to the not-so-well-known results on real quadratic number fields. Cf. [44, p. 118]. Corollary. Let K = ℚ(\sqrt{d}) be a quadratic number field and ε ∈ K a unit of norm (+1). Then N(1 − ε) = 2 − tr(ε) is a product of some primes dividing the discriminant, a perfect square, and, possibly, a factor 2 and a minus sign. Proof. As K is a quadratic number field, the norm N: 𝒪_{K} → ℤ is a quadratic form. The multiplication map ·ε: 𝒪_{K} → 𝒪_{K} is compatible with this form and, therefore, orthogonal with respect to the symmetric, bilinear form ⟨. , .⟩: 𝒪_{K} × 𝒪_{K} → ℤ associated to N.

The same is true for the corresponding ℤ_{l}-valued pairings between the l-adic completions of 𝒪_{K}, for l any prime number. As these pairings are perfect as long as l does not divide the discriminant of K, the assertion follows from Corollary 2.7.b).

□

2.9. Proof of Theorem 1.5. We clearly have that (−2)^{N}Φ(−1) ∈ ℚ. Furthermore, we may assume that (−1) is not among the zeroes of Φ as, otherwise, the assertion is true, trivially.

Then (−2)^{N}Φ(−1) > 0. Indeed, as Φ ∈ ℝ[T] and there is no real zero different from 1,

(−2)^{N}Φ(−1) = 2^{N}(1 + z_{1}) · . . . · (1 + z_{N})

is the product of several factors of the form z\overline{z} = |z|^{2} for z ∈ ℂ and some factors that are equal to 2. To prove the assertion, we will show that (−2)^{N}Φ(−1) is of even l-adic valuation for every prime number l ≠ p. Put H := H_{ét}^{d}(X_{\overline{𝔽}_{q}}, ℤ_{l}(d/2))/tors. By Poincaré duality [2, Exp. XVIII, formule (3.2.6.2)], cf. [38, Chap. 6, Sec. 2, Theorem 18 and Chap. 5, Sec. 5, Theorem 3], the bilinear pairing

⟨. , .⟩: H × H ⟶ H_{ét}^{2d}(X_{\overline{𝔽}_{q}}, ℤ_{l}(d)) ⟶^{≅} ℤ_{l} ,

given by cup product and trace map, is perfect. As d is even, it is symmetric, too. The operation of Frob on H_{ℚ_{l}} is orthogonal with respect to the ℚ_{l}-linear extension of this pairing.

First case. l ≠ 2. The operation of (− Frob) is orthogonal with respect to the pairing, too. As 1 is not among its eigenvalues, Corollary 2.7.b) shows that ν_{l}(det(1 + Frob)) = ν_{l}(Φ(−1)) is even.

Second case. l = 2. Here, the argument is a bit more involved. First, note that the tangent sheaf 𝒯_{X} of X is defined over the base field 𝔽_{q}. This shows that the Chern classes c_{i}(𝒯_{X}) ∈ H_{ét}^{2i}(X_{\overline{𝔽}_{q}}, ℤ_{2}(i)) are invariant under Frob. We therefore see from Lemma 2.10 that there is a Frob-invariant element ω ∈ H_{ét}^{d}(X_{\overline{𝔽}_{q}}, ℤ_{2}(d/2)) such that ⟨ω, x⟩ + ⟨x, x⟩ ∈ 2ℤ_{2} for every x ∈ H_{ét}^{d}(X_{\overline{𝔽}_{q}}, ℤ_{2}(d/2)).

For x ∈ (1 − Frob)H_{ℚ_{2}} , the fact that ω is Frob-invariant implies ⟨ω, x⟩ = 0. Hence, ⟨x, x⟩ ∈ 2ℤ_{2} for x ∈ H ∩ (1 − Frob)H_{ℚ_{2}} . According to Lemma 2.3.b), [H/(1 − Frob)H]_{tors} is of even length.

An application to X_{𝔽_{q^{2}}} shows that [H/(1 − Frob^{2})H]_{tors} is of even length, too. Lemma 2.11 now yields the assertion. □

Lemma 2.10. For every even d > 0, there exists a polynomial P_{d} ∈ ℤ[T_{0}, …, T_{d/2}], weighted homogeneous of degree d for T_{i} of weight 2i, such that the following is true. For a smooth projective variety X of dimension d over a finite field 𝔽_{q} of characteristic ≠ 2, put ω := P_{d}(c_{0}(𝒯_{X}), … c_{d/2}(𝒯_{X})) ∈ H_{ét}^{d}(X_{\overline{𝔽}_{q}}, ℤ_{2}(d/2)) for 𝒯_{X} the tangent sheaf and c_{i} the i-th Chern class. Then

⟨ω, x⟩ + ⟨x, x⟩ ∈ 2ℤ_{2}

for every x ∈ H_{ét}^{d}(X_{\overline{𝔽}_{q}}, ℤ_{2}(d/2)). Proof. This is a standard result in topology, cf. [33, §§8 and 11], and carries over to the étale site. Denote by ε: H_{ét}^{2∗}(X_{\overline{𝔽}_{q}}, ℤ_{2}(∗)) ⟶ H_{ét}^{2∗}(X_{\overline{𝔽}_{q}}, ℤ/2ℤ) the reduction map. Furthermore, for simplicity, write c_{i} := c_{i}(𝒯_{X}) ∈ H_{ét}^{2i}(X_{\overline{𝔽}_{q}}, ℤ_{2}(i)) for the Chern classes.

For k ∈ ℕ_{0}, let ν_{2k} ∈ H_{ét}^{2k}(X_{\overline{𝔽}_{q}}, ℤ/2ℤ) be the 2k-th Wu class of X [41, p. 578]. If X is of dimension d then, according to the very definition of the Wu class,

ν_{d} ∪ x + x ∪ x = Sq^{d}(x) + Sq^{d}(x) = 0

for every x ∈ H_{ét}^{d}(X_{\overline{𝔽}_{q}}, ℤ/2ℤ) [41, Prop. 2.2.(2)]. We will inductively construct polynomials P_{2k} ∈ ℤ[T_{0}, …, T_{k}] such that ε(P_{2k}(c_{0}(𝒯_{X}), … c_{d/2}(𝒯_{X}))) = ν_{2k}.

For every k ∈ ℕ_{0}, there is the formula of Wu [41, Proposition 0.5],

Sq^{0}(ν_{2k}) + Sq^{2}(ν_{2k−2}) + ⋯ + Sq^{2k}(ν_{0}) = ε(c_{k}) .

Moreover, for the Steenrod squares of the Chern classes, there are the formulas

Sq^{2j}(ε(c_{i}))=ε(c_{j})ε(c_{i}) + \binom{2j − 2i}{2}ε(c_{j−1})ε(c_{i+1}) + ⋯ + \binom{2j − 2i}{2j}ε(c_{0})ε(c_{i+j}).

Indeed, these follow in a purely formal manner from the definitions of Chern classes and Steenrod squares, cf. [33, Problem 8-A].

As Sq^{0} = id, Wu’s formula implies that we may choose P_{0}(T_{0}) := T_{0}. Furthermore, having P_{0}, …, P_{2k−2} already constructed, it shows that

ν_{2k} = ε(c_{k}) − ∑_{i=0}^{k} Sq^{2i}(ν_{2k−2i}) = ε(c_{k}) − ∑_{i=0}^{k} Sq^{2i}(ε(P_{2k−2i}(c_{0}, …, c_{k−i})))

= ε(c_{k}) − ∑_{i=0}^{k} P_{2k−2i}(Sq^{2i}(ε(c_{0}), …, Sq^{2i}(ε(c_{k−i})))) .

Plugging into this the formula for the Steenrod squares of the Chern classes, we see that ν_{2k} is the reduction of a polynomial expression in c_{0}, c_{1}, …, c_{k}. □

Lemma 2.11. Let (R, ν) be a normalized discrete valuation ring of characteristic 0, k its residue field, which we assume to be finite, K the field of fractions, and H a free R-module of finite rank, equipped with a non-degenerate, symmetric K-bilinear pairing ⟨. , .⟩: H_{K} × H_{K} → K. Moreover, let an R-linear map σ: H → H be given that is orthogonal with respect to the pairing and does not have the eigenvalue (−1). Suppose that [H/(1 − σ)H]_{tors} and [H/(1 − σ^{2})H]_{tors} are of even lengths. Then

ν(det(1 + σ)) ≡ ν(2)·rk H (mod 2) .

Proof. We will prove this technical lemma in several steps. First step. (1 − σ)H/(1 − σ^{2})H is of even length. Since (−1) is not an eigenvalue of σ, the map (1 + σ): H_{K} → H_{K} is a bijection. In particular, (1 − σ)H_{K} = (1 − σ^{2})H_{K}. Furthermore, there is the commutative diagram of short exact sequences

[figure: 0; (1−σ)H/(1−σ^{2})H; H/(1−σ^{2})H; H/(1−σ)H; 0; 0; 0; H/H ∩(1−σ^{2})H_{K}; H/H ∩(1−σ)H_{K}; 0 .]

As the vertical arrows are surjective, the 9-lemma yields exactness of

0 ⟶ (1 − σ)H/(1 − σ^{2})H ⟶ [H/(1 − σ^{2})H]_{tors} ⟶ [H/(1 − σ)H]_{tors} ⟶ 0 .

Since ℓ([H/(1 − σ)H]_{tors}) and ℓ([H/(1 − σ^{2})H]_{tors}) are even, we see that the length of (1 − σ)H/(1 − σ^{2})H is even, too. This is the claim.

Second step. ν(det(1 + σ)|_{(1−σ)H_{K}}) is even. Both, (1 − σ)H and (1 − σ^{2})H, are R-submodules of maximal rank in the K-vector space (1 − σ)H_{K}. Moreover, (1 − σ^{2})H = (1 + σ)|_{(1−σ)H_{K}}[(1 − σ)H]. Hence, the modulus of (1 + σ)|_{(1−σ)H_{K}} is q^{−ℓ((1−σ)H/(1−σ^{2})H)} for q := #k.

On the other hand, by [16, Exercise 14.3.6], this modulus is q^{−ν(det(1+σ)|_{(1−σ)H_{K}})}. The result established in the first step implies the claim. Third step. ν(det(1 + σ)) ≡ [dim ker(1 − σ)]ν(2) (mod 2). According to C. Jordan, we have H_{K} = ker(1 − σ)^{r} + (1 − σ)H_{K} for r large. On ker(1 − σ)^{r}, the homomorphism σ has only the eigenvalue 1. In particular, (1−σ)H_{K} ⊆ H_{K} has a complement V such that 2 is the only eigenvalue of (1+σ)|_{V}. Hence,

ν(det(1 + σ)) = ν(det(1 + σ))|_{V} + ν(det(1 + σ))|_{(1−σ)H_{K}}
≡ ν(det(1 + σ))|_{V} (mod 2)
= ν(2)[dim V]
= ν(2)[rk H − dim(1 − σ)H_{K}]
= ν(2)[dim ker(1 − σ)] .

Fourth step. Orthogonality. Finally, σ is orthogonal with respect to a non-degenerate bilinear form. In this situation, it is well-known [42, Example 2.6.C.i.b)] (see also [39, §1.A1, Lemma 4.ii)]) that

dim ker(1 − σ) ≡ dim ker(1 − σ)^{r} (mod 2)

for r ≫ 0.

The right hand side is the algebraic multiplicity of the eigenvalue 1. As the eigenvalues of an orthogonal matrix come in pairs (z, \frac{1}{z}) and, by our assumption, (−1) is not an eigenvalue, it has the same parity as rk H. The assertion follows.

□

Remarks 2.12. i) There is a conjecture of J.-P. Serre that the operation of Frob on l-adic cohomology is always semisimple. Then, as the eigenvalues come in pairs (z, \frac{1}{z}), the congruence dim ker(1 − Frob) ≡ N (mod 2) is clear. Thus, conjecturally, the argument on the sizes of the Jordan blocks is not necessary in the application to l-adic cohomology. ii) Suppose that q = p^{k} for a prime p ≠ 2 and let X be a surface such that the canonical sheaf K ∈ Pic(X_{\overline{𝔽}_{q}}) is divisible by 2. Then the case l = 2 of Theorem 1.5 may be treated directly. Indeed, in this situation, Wu’s formula [41, Proposition 2.1] implies that ⟨x, x⟩ ∈ 2ℤ_{2} for every x ∈ H. Therefore, by Lemma 2.3.b), H/(1 + Frob)H is of even length. Moreover, the assumption enforces that K^{2} is even. Hence, N = dim H_{ét}^{2}(X_{\overline{𝔽}_{q}}, ℚ_{2}(1)) is even, too, by Noether’s formula [3, I.14].

2.13. Proof of Theorem 1.9. Here, according to our assumption, we have p ≠ 2. Again, we may assume without restriction that Φ(−1) ≠ 0. In view of Theorem 1.5, it will suffice to prove that q^{a(X)}Φ(−1) is of even p-adic valuation. Writing q = p^{k}, this means that ka(X) + ν_{p}((1 + z_{1}) ⋯ (1 + z_{N})) is even.

For this, let W := W (𝔽_{q}) be the Witt ring and K its field of fractions. The crystalline cohomology groups H^{d}(X/W) are finitely generated W-modules, acted upon semilinearly by the absolute Frobenius F. The operation of F^{k} is W -linear and such that its characteristic polynomial coincides with Φ_{0}^{(d)} [26].

Again, H := H^{d}(X/W)/tors is equipped with a natural perfect pairing [4, Ch. VII, Théorème 2.1.3]

⟨. , .⟩: H × H ⟶ W .

The Frobenius operation is, however, compatible with this pairing only in the sense that ⟨F(x), F(y)⟩ = p^{d}⟨x, y⟩ for every x, y ∈ H [8, Exposé II, Exemple 1.1.ii)]. The map σ := F/p^{d/2} : H_{K} → H_{K} respects the pairing ⟨. , .⟩. H is, in fact, a Dieudonné module and thus carries a rich structure [29]. We shall use only a small part of it. Let us distinguish two cases. First case. k is odd. Observe that H is a free module as well over the Witt ring W (𝔽_{p}) and σ is W (𝔽_{p})- linear. Clearly, rk_{W (𝔽_{p})} H = k·rk_{W} H. The eigenvalues of σ, as a W (𝔽_{p})-linear map, are all the k-th roots of the zeroes z_{1}, …, z_{N} of Φ.

Furthermore, σ is orthogonal with respect to the Q_{p} := Q(W (𝔽_{p}))-bilinear extension H_{K} × H_{K} → Q_{p} of the perfect pairing tr_{W (𝔽_{p^{k}})/W (𝔽_{p})} ∘ ⟨. , .⟩: H × H → W (𝔽_{p}) . The operation of (−σ) is orthogonal with respect to the pairing, too. Thus, Corollary 2.7 shows that ℓ_{W (𝔽_{p})} (σ(H)+H/H)+ν_{W (𝔽_{p})} (det(1+σ)) is even. By Lemma 2.15, the first summand is equal to ka(X). The second summand is the p-adic valuation of

∏_{i=1}^{N} ∏_{r^{k}=z_{i}}(1 + r) = ∏_{i=1}^{N}(1 + z_{i}) .

Second case. k is even. Here, the argument is slightly more involved. We first observe that H is a free module over the Witt ring W (𝔽_{p^{k/2}}) , too, and that σ^{k/2} is W (𝔽_{p^{k/2}}) -linear. Clearly, rk_{W (𝔽_{p^{k/2}})} H = 2·rk_{W} H. Moreover, σ^{k/2} is orthogonal with respect to the Q_{p^{k/2}} := Q(W (𝔽_{p^{k/2}}))-bilinear extension H_{K} × H_{K} → Q_{p^{k/2}} of the perfect pairing tr_{W (𝔽_{p^{k}})/W (𝔽_{p^{k/2}})} ∘ ⟨. , .⟩: H × H → W (𝔽_{p^{k/2}}) .

Now choose a unit u ∈ W (𝔽_{p^{k}}) such that tr_{W (𝔽_{p^{k}})/W (𝔽_{p^{k/2}})} (u) = 0, i.e., such that its conjugate is (−u). Then σ^{k/2}(ux) = −uσ^{k/2}(x) for all x ∈ H.

We define a W (𝔽_{p^{k/2}}) -linear map T: H_{K} → H_{K} by

T (x) := σ^{k/2}(ux) .

This yields T ∘ T = −u^{2}σ^{k}. Furthermore, T is W (𝔽_{p^{k}}) -semilinear and one has ⟨T x, T y⟩ = ⟨ux, uy⟩ = u^{2}⟨x, y⟩ for all x, y ∈ H.

We see that the eigenvalues of T, as a W (𝔽_{p^{k/2}}) -linear map, are all the square roots of the numbers (−u^{2}z_{1}), …, (−u^{2}z_{N}), for z_{1}, …, z_{N} the zeroes of Φ. Moreover, [T (H) + H]/H is of even W (𝔽_{p^{k/2}}) -length as it is, in fact, a W (𝔽_{p^{k}}) -module.

Finally, put \overline{H} := H ⊗_{W (𝔽_{p^{k/2}})} W (\overline{𝔽}_{p}) and extend T to a W (\overline{𝔽}_{p})-linear map \overline{T}: \overline{H} → \overline{H}. Then \frac{1}{u} \overline{T} is orthogonal. On the other hand,

[\frac{1}{u} \overline{T} (\overline{H}) + \overline{H}]/\overline{H} = [\overline{T} (\overline{H}) + \overline{H}]/\overline{H} ,

as \frac{1}{u} is a unit, and the latter W (\overline{𝔽}_{p})-module is of even length. Corollary 2.7 shows that ν_{W (\overline{𝔽}_{p})}(det(1− \frac{1}{u} \overline{T})) is even. This number is nothing but the p-adic valuation of

∏_{i=1}^{N} ∏_{r^{2}=−z_{i}}(1 − r) = ∏_{i=1}^{N}(1 + z_{i}) .  □

Remark 2.14. One might want to prove Theorem 1.9 in characteristic 2 along the same lines as Theorem 1.5. For this, one would need, at least, the theory of Steenrod squares and Wu classes, as well as Wu’s formula, for crystalline cohomology of varieties in characteristic 2. It seems, however, that such a theory is not yet available in the literature.

Lemma 2.15. Let X be a smooth proper variety of even dimension d over 𝔽_{q}, H := H^{d}(X/W)/tors, and σ = F/p^{d/2}. Then a(X) = ℓ_{W} (σ(H) + H/H). Proof. First, observe that σ(H), being the image of an F-semilinear map, is indeed a W-module. Furthermore, we have H/FH ≅ ⨁_{m>0}(W/p^{m}W)^{h′_{d−m,m}}. Hence, there is a basis of H such that, under the corresponding isomorphism H ≅ W^{N}, one has

FH ≅ ⨁_{m≥0} p^{m}W^{h′_{d−m,m}}. Therefore, σ(H) ≅ ⨁_{m≥0}p^{m−d/2}W^{h′_{d−m,m}}. The assertion fol-

lows. □

Corollary 2.16. Let X be a smooth proper variety of even dimension d over 𝔽_{q}. Suppose that X is of Hodge-Witt type in degree d, i.e., that the Serre cohomology groups H^{j}(X, WΩ_{X}^{m}) are finitely generated W-modules for j + m = d. Then

a(X) = − ∑_{ν_{q}(z_{i})<0} ν_{q}(z_{i}) .

Proof. We have H^{d}(X/W) ≅ ⨁_{m} H^{d−m,m} for H^{d−m,m} := H^{d−m}(X, WΩ_{X}^{m}), as is shown in [25, Théorème IV.4.5]. On H^{d−m,m}, F operates as p^{m}F for F the usual Frobenius on Serre cohomology. Thus, σ acts as p^{m−d/2}F. For m ≥ d/2, this ensures that the corresponding summand is mapped to H.

Thus, assume that m < d/2. On Serre cohomology, there is a second operator, the Verschiebung V, such that F V = p. Hence σp^{d/2−m−1}V = id, implying

H^{d−m,m}⊗_{W} Q(W) ⊇ σ(H^{d−m,m}) ⊇ H^{d−m,m} .

Lemma 2.15 shows that a(X) = −ν_{q}(det(σ|_{⨁_{m<d/2} H^{d−m,m}})), which is equivalent to the assertion. □

3. An application and examples

3.1. An application to the odd-dimensional supersingular case.

Lemma 3.1. Let p be a prime and d_{1} and d_{2} two odd integers. Moreover, let Φ_{1} ∈ ℚ[T] and Φ_{2} ∈ ℚ[T] be polynomials of even degrees N_{1} and N_{2} that fulfill the functional equations

Φ_{1}(p^{d_{1}}/T) = \frac{p^{d_{1}N_{1}/2}}{T^{N_{1}}} Φ_{1}(T) and Φ_{2}(p^{d_{2}}/T) = \frac{p^{d_{2}N_{2}/2}}{T^{N_{2}}} Φ_{2}(T) .

For z_{i}^{(1)} the zeroes of Φ_{1} and z_{j}^{(2)} the zeroes of Φ_{2}, let Φ be the monic polynomial with the zeroes z_{i}^{(1)}z_{j}^{(2)}/p^{\frac{d_{1}+d_{2}}{2}} . Then p^{N_{1}N_{2}/4}Φ(−1) is a square in ℚ.

Proof. The assumption implies that the zeroes come in pairs with products p^{d_{1}} and p^{d_{2}}, respectively. For two pairs of zeroes, the corresponding four zeroes of Φ are

\frac{z_{i}^{(1)}}{p^{d_{1}/2}} · \frac{z_{j}^{(2)}}{p^{d_{2}/2}} , \frac{z_{i}^{(1)}}{p^{d_{1}/2}} · \frac{p^{d_{2}/2}}{z_{j}^{(2)}} , \frac{p^{d_{1}/2}}{z_{i}^{(1)}} · \frac{z_{j}^{(2)}}{p^{d_{2}/2}} , and \frac{p^{d_{1}/2}}{z_{i}^{(1)}} · \frac{p^{d_{2}/2}}{z_{j}^{(2)}} .

We now observe the identity

(−1− \frac{1}{p}u_{1}u_{2})(−1−u_{1}/u_{2})(−1−u_{2}/u_{1})(−1−p/u_{1}u_{2}) = p(\frac{u_{1}}{p} + \frac{u_{2}}{p} +1/u_{1}+1/u_{2})^{2} , (2)

which applies, since the four zeroes may rationally be written as

\frac{1}{p} \frac{z_{i}^{(1)}}{p^{(d_{1}−1)/2}} ·\frac{z_{j}^{(2)}}{p^{(d_{2}−1)/2}} , \frac{z_{i}^{(1)}}{p^{(d_{1}−1)/2}} ·\frac{p^{(d_{2}−1)/2}}{z_{j}^{(2)}} , \frac{p^{(d_{1}−1)/2}}{z_{i}^{(1)}} ·\frac{z_{j}^{(2)}}{p^{(d_{2}−1)/2}} , and p\frac{p^{(d_{1}−1)/2}}{z_{i}^{(1)}} ·\frac{p^{(d_{2}−1)/2}}{z_{j}^{(2)}} .

It shows that the product ∏_{i,j}(−1 − z_{i}^{(1)}z_{j}^{(2)}/p^{\frac{d_{1}+d_{2}}{2}} ) is p^{N_{1}N_{2}/4} times a square. Indeed, the sums occurring on the right hand side of (2) form a Gal(\overline{ℚ} /ℚ)-invariant set of N_{1}N_{2}/4 elements. □

Proposition 3.2. Let X be a smooth proper variety of odd dimension d over a finite field 𝔽_{p^{k}} for p a prime and k odd. Suppose that p > 1 + dim H_{ét}^{d}(X_{\overline{𝔽}_{p}}, ℚ_{l}),

dim H_{ét}^{d}(X_{\overline{𝔽}_{p}}, ℚ_{l}) ≡ 2 (mod 4) ,

and that all eigenvalues of Frob on the cohomology are of p-adic valuation dk/2. I.e., that the Newton polygon has constant slope d/2.

Then ±\sqrt{−p^{dk}} are among the eigenvalues.

Proof. Let C be the base extension of a supersingular elliptic curve defined over 𝔽_{p}. Such do exist by the work of M. Eichler [17], see also [21, Proposition 2.4, together with (1.10) and (1.11)]. The eigenvalues of Frob on H_{ét}^{1}(C_{\overline{𝔽}_{p}}, ℚ_{l}) are ±\sqrt{−p^{k}}. Indeed, the assumptions imply p ≥ 5, and hence p > 2\sqrt{p}. The claim follows from Hasse’s bound.

For Φ the characteristic polynomial of Frob on

V := [H_{ét}^{d}(X_{\overline{𝔽}_{p}}, ℚ_{l}) ⊗ H_{ét}^{1}(C_{\overline{𝔽}_{p}}, ℚ_{l})](\frac{d+1}{2}) ,

Lemma 3.1 guarantees that pΦ(−1) is a perfect square. But all eigenvalues of Frob on V are p-adic units. As they are l-adic units for every prime l ≠ p, too, they must be roots of unity [7, Sec. 18, Lemma 2].

To prove the assertion, we need to show Φ(−1) = 0. Assuming the contrary, we see from Lemma 3.6 that, for some e ≥ 1, all primitive 2p^{e}-th roots of unity must be eigenvalues of Frob on V. As with z, (−z) is an eigenvalue, too, this enforces dim V ≥ 2(p − 1), that is dim H_{ét}^{d}(X_{\overline{𝔽}_{p}}, ℚ_{l}) ≥ p − 1, a contradiction. □

Remark 3.3. In principle, the idea behind this proof is to apply Theorem 1.9 to X × C. This is, however, not sufficient as there may be eigenvalues (−1) on the products [H^{i_{1}}(X, ℚ_{l}) ⊗ H^{i_{2}}(C, ℚ_{l})](\frac{d+1}{2}) for i_{1} + i_{2} = d + 1, i_{2} ≠ 1.

Example 3.4. Proposition 3.2 may fail in small characteristic, as is seen from the elliptic curve C over 𝔽_{3}, given by y^{2} = x^{3} − x − 1. Then #C(𝔽_{3}) = 1, which shows that C is supersingular. Moreover, the characteristic polynomial of Frob is T^{2} − 3T + 3. The eigenvalues are the two primitive twelfth roots of unity with positive real part, multiplied by \sqrt{3}.

Remark 3.5. When trying to carry over the argument from the proof, it turns out that, on V := [H_{ét}^{1}(C_{\overline{𝔽}_{3}}, ℚ_{l}) × H_{ét}^{1}(C_{\overline{𝔽}_{3}}, ℚ_{l})](1) , there are the primitive sixth roots of unity occurring as eigenvalues, together with 1, which appears twice. Therefore, Φ(T) = (T^{2} − T + 1)(T − 1)^{2} and Φ(−1) = 12.

Alternatively, one might combine C with C′: y^{2} = x^{3} − x, which is supersingular having four points. On the product [H_{ét}^{1}(C_{\overline{𝔽}_{3}}, ℚ_{l}) ⊗ H_{ét}^{1}(C′_{\overline{𝔽}_{3}}, ℚ_{l})](1) , one finds Φ(T) = (T^{2} − T + 1)(T^{2} + T + 1) and Φ(−1) = 3.

Lemma 3.6. Let Φ ∈ ℚ[T] be a monic polynomial such that all its roots are roots of unity. Assume that |Φ(−1)| ≠ 0 is a multiple of a prime number p. Then Φ is divisible by the cyclotomic polynomial ϕ_{2p^{e}}, for some e ≥ 1. In particular, if p = 2 then Φ is divisible by ϕ_{2^{e}}, for some e ≥ 2.

Proof. Φ is a product of cyclotomic polynomials ϕ_{n}. For these, it is well known [34, Section 3] that ϕ_{1}(−1) = −2, ϕ_{2}(−1) = 0, ϕ_{2p^{e}}(−1) = p for p any prime number and e ≥ 1, and ϕ_{n}(−1) = 1 in all other cases. The assertion follows directly from this.

□

3.2. The even-dimensional supersingular case.

Proposition 3.7. Let X be a smooth proper variety of even dimension d over a finite field 𝔽_{p^{k}} for a prime p ≠ 2 and k odd. Suppose that a(X) ≡ 1 (mod 2) and that all eigenvalues of Frob on H_{ét}^{d}(X_{\overline{𝔽}_{p}}, ℚ_{l}(d/2)) are p-adic units. I.e., that the Newton polygon has constant slope d/2. a) Then (−1) is an eigenvalue or, for some e ≥ 1, the primitive 2p^{e}-th roots of unity are eigenvalues.

b) If p > dim H_{ét}^{d}(X_{\overline{𝔽}_{p}}, ℚ_{l}(d/2)) + 1 or p > dim H_{ét}^{d}(X_{\overline{𝔽}_{p}}, ℚ_{l}(d/2)) and X is projective then (−1) is an eigenvalue.

Proof. The eigenvalues are l-adic units, too, for every prime number l ≠ p, hence they are roots of unity. Theorem 1.9 ensures that pΦ(−1) is a perfect square. Applying Lemma 3.6 again, we immediately obtain assertion a).

For b), assume the contrary. Then, as eigenvalues, we have the (p−1)p^{e−1} ≥ p−1 primitive 2p^{e}-th roots of unity and, in the projective case, the number 1. Thus, altogether, there are at least p − 1, respectively p, of them. □

Corollary 3.8. Let X be a supersingular K3 surface over a finite field 𝔽_{p^{k}} for p > 19 a prime and k odd. Then (−1) is an eigenvalue of Frob on H_{ét}^{2}(X_{\overline{𝔽}_{p}}, ℚ_{l}(1)).

Proof. For K3 surfaces, the Hodge spectral sequence degenerates at E_{1} [13, Proposition 1.1.a)] and, hence, the conjugate spectral sequence degenerates at E_{2} [5, Lemma 8.27.2]. Moreover, all H^{i}(X/W) are torsion-free ([24, II.7.2] or [13, Proposition 1.1.c)]). Consequently, we have a(X) = dim H^{2}(X, 𝒪_{X}) = 1. Cf. Remark 1.8.b). The claim now follows from Proposition 3.7.b). □

Remarks 3.9. i) Corollary 3.8 refines the observation of M. Artin [1, 6.8] that the field of definition of the rank-22 Picard group always contains 𝔽_{p^{2}}. ii) More generally, let X be any K3 surface over a finite field 𝔽_{p^{k}} for p ≠ 2 a prime. Theorem 1.9 then asserts that, for z_{1}, …, z_{22} the eigenvalues of Frob on H_{ét}^{2}(X_{\overline{𝔽}_{p}}, ℚ_{l}(1)) , the expression p^{k}Φ(−1) = p^{k}(1 + z_{1}) … (1 + z_{22}) is always a square in ℚ. iii) Corollary 3.8 is clearly false in small characteristic, as may be seen from the example below.

Examples 3.10. i) For C the supersingular elliptic curve over 𝔽_{3} from Example 3.4, put X := Kum(C × C). Then

Φ(T) = (T^{2} − T + 1)(T^{2} + T + 1)^{5}(T − 1)^{10}.

In particular, we have Φ(−1) = 3·2^{10}, in agreement with Remark 3.9.ii). Indeed, we saw that the characteristic polynomial of Frob on H_{ét}^{2}((C × C)_{\overline{𝔽}_{3}}, ℚ_{l}(1)) is (T^{2} − T + 1)(T − 1)^{4}. Furthermore, the 16 two-torsion points form five orbits of size three together with the origin. ii) Let X := Kum(C × C′) be the Kummer surface associated to the product of the two supersingular elliptic curves over 𝔽_{3}, considered in Remark 3.5. Then, once again, Φ(T) = (T^{2} − T + 1)(T^{2} + T + 1)^{5}(T − 1)^{10}. In fact, we saw that the characteristic polynomial of Frob on H_{ét}^{2}((C × C′)_{\overline{𝔽}_{3}}, ℚ_{l}(1)) is (T^{2} − T + 1)(T^{2} + T + 1)(T − 1)^{2}. In addition, four two-torsion points are defined over the base field, while the others form four orbits of size three.

3.3. Surfaces. The Artin-Tate formula. For surfaces, the assertion of Theorem 1.5 is implied by the Tate conjecture. More precisely,

Proposition 3.11. Let X be a smooth projective surface over a finite field 𝔽_{q} of characteristic p and let Φ = Φ_{1}^{(2)} ∈ ℚ[T] be the characteristic polynomial of Frob on H_{ét}^{2}(X_{\overline{𝔽}_{q}}, ℚ_{l}(1)) . Put N := deg Φ and

α(X) := dim H^{2}(X, 𝒪_{X}) − dim H^{1}(X, 𝒪_{X}) + \frac{1}{2} dim H_{ét}^{1}(X_{\overline{𝔽}_{q}}, ℚ_{l}) .

Suppose that the Tate conjecture is true for X. Then (−2)^{N}q^{α(X)}Φ(−1) is a square in ℚ.

Proof. Denote the zeroes of Φ, i.e. the eigenvalues of Frob, by z_{1}, …, z_{ϱ} = 1, z_{ϱ+1}, …, z_{N} ≠ 1. If Φ(−1) = 0 then the assertion is true, trivially. Thus, let us suppose the contrary from now on. Then the zeroes z_{i} ≠ 1 come in pairs of complex conjugate numbers. In particular, N − ϱ is even.

Furthermore, Frob and Frob^{2} have the eigenvalue 1 with the same multiplicity. Hence, the Tate conjecture predicts the rank of Pic(X_{𝔽_{q^{2}}} ) not to be higher than that of Pic(X). This shows that X_{𝔽_{q^{2}}} , too, fulfills the Tate conjecture.

We are therefore in a situation where the Artin-Tate formula [32, Theorem 6.1] computes the discriminants of the Picard lattices Pic(X) and Pic(X_{𝔽_{q^{2}}} ), at least up to square factors. The results are

(−1)^{ϱ−1}q^{α(X)}∏_{i=ϱ+1}^{N}(1 − z_{i}) and (−1)^{ϱ−1}q^{2α(X)}∏_{i=ϱ+1}^{N}(1 − z_{i}^{2}) .

Moreover, equality of the ranks implies that disc Pic(X)/ disc Pic(X_{𝔽_{q^{2}}} ) is a necessarily perfect square. This is a standard observation from the theory of lattices. We conclude that q^{α(X)} ∏_{i=ϱ+1}^{N}(1 + z_{i}) is a square in ℚ.

On the other hand, (−2)^{N}q^{α(X)}Φ(−1) = 2^{N+ϱ}q^{α(X)} ∏_{i=ϱ+1}^{N}(1 + z_{i}) such that the assertion follows from the fact that ϱ ≡ N (mod 2). □

Remarks 3.12. i) The Artin-Tate formula appears to us as a very natural consequence of the Tate conjecture and the cohomological machinery. Thus, we find it very astonishing that it has the potential to produce incompatible results for a variety and its base extension. Of course, this does not happen for polynomials that really occur as the characteristic polynomial of the Frobenius on a certain variety. But it occurs for polynomials that otherwise look plausible. This observation was actually the starting point of our investigations. ii) One might want to compare the Picard lattice of X with that of X_{q^{n}} for n > 2. But this leads to nothing new [18, Corollary 18.i)]. iii) Suppose dim H^{1}(X, 𝒪_{X}) = \frac{1}{2} dim H_{ét}^{1}(X_{\overline{𝔽}_{q}}, ℚ_{l}) and that X fulfills the assumptions of Remark 1.8.b). Then α(X) = a(X) = dim H^{2}(X, 𝒪_{X}) . We do not know how closely Milne’s invariant α(X) and our invariant a(X) are related for “pathological” surfaces.

Example 3.13. Let X be the double cover of P_{𝔽_{7}}^{2} , given by

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

This is a K3 surface of degree two.

The numbers of points on X over the finite fields 𝔽_{7}, …, 𝔽_{7^{10}} are 60, 2 488, 118 587, 5 765 828, 282 498 600, 13 841 656 159, 678 225 676 496, 33 232 936 342 644, 1 628 413 665 268 026, and 79 792 266 679 604 918.

For the characteristic polynomial of Frob on H_{ét}^{2}(X_{\overline{𝔽}_{7}}, ℚ_{l}(1)) , this information leaves us with two candidates, one for each sign in the functional equation,

Φ_{i}(t) = \frac{1}{7}(7t^{22} − 10t^{21} + t^{20} − t^{19} + 6t^{18} − 3t^{17} − 2t^{16} + 4t^{14} − t^{13} − t^{12}
+ (−1)^{i}(−t^{10} − t^{9} + 4t^{8} − 2t^{6} − 3t^{5} + 6t^{4} − t^{3} + t^{2} − 10t + 7))

for i = 0, 1. All roots are of absolute value 1.

However, Φ_{0}(−1) = 60/7 and (−2)^{N}7^{a(X)}Φ_{0}(−1) = 2^{24}·3·5 is a non-square, which contradicts Theorem 1.5. Therefore, Φ_{1} is the characteristic polynomial of Frob on H_{ét}^{2}(X_{\overline{𝔽}_{7}}, ℚ_{l}(1)) . The minus sign holds in the functional equation.

Remark 3.14. Alternatively, we may argue as follows. Assume Φ_{0} is the characteristic polynomial. Then rk Pic(X_{𝔽_{49}} ) = rk Pic(X) = 2. Indeed, the Tate conjecture is proven for K3 surfaces in characteristic ≥ 3 [9, Corollary 2], [35, Theorem 1], cf. [27]. The Artin-Tate formula states that disc Pic(X_{𝔽_{49}} ) ∈ (−465)(ℚ^{∗})^{2} and disc Pic(X) ∈ (−31)(ℚ^{∗})^{2}. As \frac{−465}{−31} = 15 is a non-square, this is contradictory.

3.4. Cubic fourfolds.

Example 3.15. Let X be the subvariety of P_{𝔽_{2}}^{5} , given by

x_{0}^{3} + x_{0}^{2}x_{1} + x_{0}^{2}x_{4} + x_{0}^{2}x_{5} + x_{0}x_{1}x_{2} + x_{0}x_{1}x_{3} + x_{0}x_{1}x_{4} + x_{0}x_{2}x_{3} + x_{0}x_{2}x_{4}
+ x_{0}x_{3}x_{4} + x_{0}x_{4}^{2} + x_{0}x_{4}x_{5} + x_{0}x_{5}^{2} + x_{1}^{3} + x_{1}x_{2}^{2} + x_{1}x_{2}x_{4} + x_{1}x_{2}x_{5} + x_{1}x_{3}^{2}
+ x_{1}x_{3}x_{5} + x_{1}x_{4}^{2} + x_{1}x_{4}x_{5} + x_{2}^{3} + x_{2}^{2}x_{5} + x_{2}x_{3}^{2} + x_{3}^{2}x_{4} + x_{3}^{2}x_{5} + x_{3}x_{4}^{2}
+ x_{3}x_{5}^{2} + x_{4}^{3} + x_{4}^{2}x_{5} + x_{4}x_{5}^{2} + x_{5}^{3} = 0 .

This is a smooth cubic fourfold. We have dim H^{4}(X, 𝒪_{X}) = 0, dim H^{3}(X, Ω_{X}^{1}) = 1, and dim H^{2}(X, Ω_{X}^{2}) = 21, According to Remark 1.8.b), this shows N = 23 and a(X) = 1.

The numbers of points on X over the finite fields 𝔽_{2}, …, 𝔽_{2^{11}} are 33, 361, 4 545, 69 665, 1 084 673, 17 044 609, 270 543 873, 4 311 990 785, 68 853 026 817, 1 100 586 076 161, and 17 600 769 409 025. The characteristic polynomial of Frob on H_{ét}^{4}(X_{\overline{𝔽}_{2}}, ℚ_{l}(2)) is

Φ(t) = \frac{1}{2}(t − 1)(2t^{22} − t^{21} − t^{20} + 2t^{19} − 2t^{17} + t^{16} + t^{15} − 2t^{14} + t^{13} + t^{12}
− t^{11} + t^{10} + t^{9} − 2t^{8} + t^{7} + t^{6} − 2t^{5} + 2t^{3} − t^{2} − t + 2) .

It turns out that Φ(−1) = −1, in agreement with Theorem 1.5. Observe that, in this example, the assertion of Theorem 1.9 is true, albeit the characteristic of the base field is 2.

Remark 3.16. The degree 22 factor of Φ is irreducible over ℚ. In particular, X is certainly not special in the sense of B. Hassett [22].

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School of Mathematics and Statistics F07, University of Sydney, NSW 2006, Sydney, Australia

E-mail address: stephan@maths.usyd.edu.au
URL: http://www.maths.usyd.edu.au/ut/people?who=AS Elsenhans

Département Mathematik, Universität Siegen, Walter-Flex-Str. 3, D-57068 Siegen, Germany

E-mail address: ~jjahnel@mathematik.uni-siegen.de
URL: http://wwwuser.gwdg.de/~jjahnel/