On the characteristic polynomial of the Frobenius on étale cohomology
aa r X i v : . [ m a t h . AG ] O c t ON THE CHARACTERISTIC POLYNOMIAL OF THEFROBENIUS ON ´ETALE COHOMOLOGY
ANDREAS-STEPHAN ELSENHANS AND J ¨ORG JAHNEL
Abstract.
Let X be a smooth proper variety of even dimension d over a fi-nite field. We establish a restriction on the value at ( −
1) of the characteristicpolynomial of the Frobenius on the middle-dimensional ´etale cohomology of X with coefficients in Q l ( d/ Introduction
Let X be a smooth proper variety over a finite field F q of characteristic p > l -adic cohomology vec-tor spaces H i ´et ( X F q , Q l ( j )). The characteristic polynomial of Frob has rational coef-ficients [11, Th´eor`eme (1.6)].By far not every polynomial Φ ∈ Q [ T ] may occur as the characteristic polyno-mial of the Frobenius on a smooth proper variety. The following conditions wereestablished essentially in the 70s of the last century, in the course of working outA. Grothendieck’s ideas in Algebraic Geometry. Theorem 1.1 (Deligne, Mazur, Ogus) . Let X be a smooth proper variety over afinite field F q of characteristic p . For i, j ∈ Z , denote by Φ ( i ) j = T N + a ( i )1 T N − + · · · + a ( i ) N − T + a ( i ) N the characteristic polynomial of Frob on H i ´et ( X F q , Q l ( j )) . Then, for every r , onehas a ( i ) r ∈ Q and this value is independent of l = p . Moreover, a) every complex zero of Φ ( i ) j is of absolute value q i/ − j . b) If i is odd then all real zeroes of Φ ( i ) j are of even multiplicity. c) For every l = p , the zeroes of Φ ( i ) j are l -adic units, i.e. units in a suitable exten-sion of Z l . d) Put h i − m,m := dim H i − m ( X, Ω mX ) and define the step-function G ( i ) : [0 , N ] → R by G ( i ) ( t ) = (cid:26) t ≤ h i, ,n for h i, + · · · + h i − n +1 ,n − < t ≤ h i, + · · · + h i − n,n . Mathematics Subject Classification.
Primary: 14F20, Secondary: 14G15, 14J20, 14J35.
Key words and phrases.
Proper variety over finite field, Middle cohomology, Frobenius, Char-acteristic polynomial, Artin-Tate formula.
Then, for r = 1 , . . . , N , one has a ( i ) r = 0 or ν q ( a ( i ) r ) ≥ r Z [ G ( i ) ( t ) − j ] dt . Here, ν q is the non-archimedean valuation such that ν q ( q ) = 1 .Remarks . i) We will provide references and proof sketches at the beginning ofSection 2.ii) Assertion a) immediately implies that Φ ( i ) j ∈ Q [ T ] fulfills the functional equation T N Φ( q i − j /T ) = ± q N ( i − j ) Φ( T )for N := dim H i ´et ( X F q , Q l ( j )). Indeed, on both sides, there are polynomials withleading term ± q N ( i − j ) T N . They have the same zeroes as, with z , the number z = q i − j 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 ( i ) j ∈ Z [ T ] when j ≤ X projective and i even, we also haveΦ ( i ) j ( q i/ − j ) = 0 . (1)Indeed, there is the Gal( F q / F q )-invariant cycle given by the intersection of i/ H i ´et ( X F q , Q l ( i/ Remark . Consider the case that i = 1. Then N = dim H ( X F q , Q l ) is alwayseven [12, Corollaire (4.1.5)]. On the other hand, let a polynomial Φ ∈ Z [ T ] be giventhat is of even degree and fulfills assertions a), b), and c). Then, by the main theoremof T. Honda [23], there exists an abelian variety A such that the eigenvalues of Frobon H ( A F q , Q l ) are exactly the zeroes of Φ. One may enforce that the characteristicpolynomial is a power of Φ, and, typically, Φ itself may be realized. 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 ofeven dimension, we will establish a further condition, which is arithmetic in natureand 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 finitefield F q of characteristic p and Φ = Φ ( d ) d/ ∈ Q [ T ] be the characteristic polynomialof Frob on H d ´et ( X F q , Q l ( d/ . Put N := deg Φ .Then ( − N Φ( − is a square or p times a square in Q .Remark . For X a surface, this result may be deduced from the Tate conjecture,via the Artin-Tate formula. Cf. Proposition 3.11, below. N THE CHARACTERISTIC POLYNOMIAL OF FROBENIUS 3
The correct exponent of p , may, at least for p = 2, be described as follows. Definition.
We put a ( X ) := d − X m =0 ( d − m ) h ′ d − m,m , for ( h ′ d − m,m ) m the abstract Hodge numbers of X in degree d [30, Section 4]. Remarks . 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 , thecorresponding map is only F -semilinear [8, Expos´e I, 2.3.5].Put H := H i ( X/W ) / tors. Then, as F : H → H is injective, H/ F H is a W -mod-ule of finite length. By the classical invariant factor theorem, there is a uniquesequence of integers such that H/ F H ∼ = L m> ( W/p m W ) h ′ i − m,m . Finally, one defines h ′ i, := rk W H − P m> h ′ i − m,m .Observe that a ( X ) is a geometric quantity. It depends only on the base exten-sion X F q .b) Suppose that X is such that all H i ( X/W ) are torsion-free and that the conjugatespectral sequence E jm := H j ( X, H m (Ω • X/ F q )) = ⇒ H i dR ( X ) degenerates at E . Then h ′ i − m,m = h i − m,m (= dim H i − m ( X, Ω mX )) [5, Lemma 8.32].For complete intersections, both assumptions hold ([15, Expos´e XI, Th´eor`eme 1.5]together with [5, Lemma 8.27.2]). Moreover, the second assumption is automaticallyfulfilled when dim X ≤ p and X lifts to W ([14, Corollaire 2.4] and [5, Lemma8.27.2]).c) Suppose that X is of Hodge-Witt type in degree d , i.e., that the Serre coho-mology groups H j ( X, W Ω mX ) [36] are finitely generated W -modules for j + m = d .The assertion of Theorem 1.9 below may then be formulated entirely in terms ofthe characteristic polynomial Φ. In fact, denote the zeroes of Φ by z , . . . , z N . Then a ( X ) = − X ν q ( z i ) < ν q ( z i )[Corollary 2.16]. This case includes all varieties that are ordinary in degree d [25,D´efinition IV.4.12]. Theorem 1.9.
Let X be a smooth proper variety of even dimension d over a finitefield F q of characteristic p = 2 and Φ = Φ ( d ) d/ ∈ Q [ T ] be the characteristic polynomialof Frob on H d ´et ( X F q , Q l ( d/ . Put N := deg Φ .Then ( − N q a ( X ) Φ( − is a square in Q .Remarks . i) It seems not unlikely that an analogous result is true in charac-teristic 2, too. The difficulties that arise may well be of purely technical nature.Cf. Remark 2.14, below. ANDREAS-STEPHAN ELSENHANS AND J ¨ORG JAHNEL ii) The assertions may easily be formulated for an arbitrary Tate twist. But thisdoes not lead to anything new. Also, the twist by d/ H d ´et ( X F q , Q l ( d/ Q l -valuedsymmetric, bilinear pairing. Remarks . i) One reason for our interest in these conditions is that they simplifythe actual computation of the characteristic polynomial Φ for a given variety X .Such computations usually involve point counting on X over several extensions ofthe base field, cf. Examples 3.13 and 3.15. The conditions given redundantise themost expensive counting steps.ii) Applications of the characteristic polynomials Φ to varieties in characteristic zeroinclude the determination of the N´eron-Severi rank. The interested reader mightconsult the articles [28], [19], and [20]. Acknowledgements.
We wish to thank Christian Liedtke (M¨unchen), who readan earlier version of this article, for many valuable suggestions.2.
The proofs a) This was first proven by P. Deligne in [11, Th´e-or`eme (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, Th´eor`eme (4.1.1)]and Poincar´e duality, there is a non-degenerate pairing H i ´et ( X F q , Q l ( j )) × H i ´et ( X F q , Q l ( j )) → Q l (2 j − i )that is compatible with the operation of Frob. It is alternating as i is odd. The as-sertion 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, Corol-lary 2.2.3].c) As Frob operates on H i ´et ( X F q , Z l ( j )), the l -adic valuations of the eigenvalues areclearly non-negative. Poincar´e duality implies the assertion, cf. [11, (2.4)].d) This statement was originally known as Katz’s conjecture. The usual formula-tion is that the Newton polygon of Φ ( i )0 lies above the Hodge polygon of weight i .Proofs are due to B. Mazur [31] and A. Ogus [5, Theorem 8.39]. (cid:3) 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 afree R -module of finite rank, equipped with a perfect, R -bilinear, symmetric pairing H × H → R . Denote its K -bilinear extension by h . , . i : H K × H K → K .Furthermore, let a K -linear map σ : H K → H K be given that is orthogonal withrespect to the pairing, i.e., h σ ( x ) , σ ( y ) i = h x, y i for every x, y ∈ H K . N THE CHARACTERISTIC POLYNOMIAL OF FROBENIUS 5
Put B ( H ) := [ H/ [ H ∩ (1 − σ ) H ]] tors . a) Then there is a non-degenerate, skew-symmetric R -bilinear pairing ( . , . ) : B ( H ) × B ( H ) → K/R . b) Suppose char K = 2 and h x, x i ∈ R for every x ∈ H ∩ (1 − σ ) H K . Then ( . , . ) is alternating. In particular, the length of B ( H ) is even.Remarks . i) Observe that x ∈ H represents an element of B ( H ) if and onlyif x ∈ H ∩ (1 − σ ) H K .ii) For x, y ∈ H K arbitrary, one has h (1 − σ ) x, σy i = −h x, (1 − σ ) y i . 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, .This fact is rather obvious, let us nevertheless emphasize that it is true whether σ is semisimple or not. a) Definition.
The pairing is defined as follows.For a, b ∈ B ( H ), choose representatives x, y ∈ H . Let y ′ ∈ H K be such that y = (1 − σ ) y ′ . Then ( a, b ) := h x, y ′ i mod R . Well-definedness.
For two representatives x , x ∈ H , we have x − x = (1 − σ ) v for some v ∈ H . Thus, h x − x , y ′ i = h (1 − σ ) v, y ′ i = h (1 − σ ) σv, σy ′ i = −h σv, (1 − σ ) y ′ i = −h v − x + x , y i ∈ R , as both sides are in H . On the other hand, for two representatives y ′ , y ′ ∈ H K ,we have (1 − σ )( y ′ − y ′ ) = (1 − σ ) w ∈ H for a suitable w ∈ H . Therefore, h x, y ′ − y ′ i = h x, w i + h x, k i for some k ∈ H K, . The first summand is in R , as bothsides are elements of H . The second summand vanishes, since x ∈ (1 − σ ) H K . It isclear that ( . , . ) is R -bilinear. Non-degeneracy.
For 0 = b ∈ B ( 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, . The goal is to findsome x ∈ H ∩ (1 − σ ) H K such that h x, y ′ i 6∈ R .For this, we observe that the perfect pairing induces an isomorphism H K ∼ = −→ Hom(
H, K ) . Under this map, H K, ∼ = Hom( H/ [ H ∩ (1 − σ ) H K ] , K ). Furthermore, H + H K, ∼ = { α ∈ Hom(
H, K ) | α ( H ∩ (1 − σ ) H K ) ⊆ R } . Indeed, as H ∼ = Hom( H, R ), the inclusion “ ⊆ ” is obvious. The other inclusion followsfrom 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 ′ vanisheson H ∩ (1 − σ ) H K , hence is defined by an element of H K, . ANDREAS-STEPHAN ELSENHANS AND J ¨ORG JAHNEL
Now, as y ′ H + H K, , 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 ( 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 h x, y ′ i + h x ′ , y i = h x ′ − σx ′ , y ′ i + h x ′ , y ′ − σy ′ i = h x ′ − σx ′ , y ′ − σy ′ i = h x, y i ∈ R . b) For a ∈ B ( H ) choose a representative x ∈ H and x ′ ∈ H K such that x = (1 − σ ) x ′ .Then2( a, a ) = 2 h x, x ′ i = [ h x ′ , x ′ i − h σx ′ , x ′ i ] + [ h σx ′ , σx ′ i − h x ′ , σx ′ i ]= h x ′ − σx ′ , x ′ − σx ′ i = h x, x i ∈ R , hence ( a, a ) = 0.
Evenness of ℓ ( B ( H )) . First note that B ( H ) is a finitely generated torsion moduleover a principal ideal domain. This implies that ℓ ( B ( H )) is finite.It is known [37, Exercise 10.20] that, for such a module M , the existence ofa non-degenerate alternating pairing implies that ℓ ( M ) is even. The argument isessentially 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 amaximal ideal. Then L x := h x, . i : M → K/R is a nonzero linear form, the imageof which is m − /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 ) − ℓ ([ker L x ] / ( x )) = ℓ ( M ) −
2. This is acontradiction. (cid:3)
Remarks . 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 byYu. G. Zarhin in [43, 3.3 and Lemma 3.4.1]. It is implicitly contained in the work ofJ. 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 residuefield, 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, symmetricpairing H × H → R and denote its K -bilinear extension by h . , . i : H K × H K → K .Furthermore, let a K -linear map σ : H K → H K be given that is orthogonal withrespect to the pairing and such that 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. N THE CHARACTERISTIC POLYNOMIAL OF FROBENIUS 7
Proof. a) We write q := k . According to the definition, the modulus [16, sec-tion 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 lengthsare 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 isnot 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). (cid:3) In order to illustrate the strength of Corollary 2.7, let us show an applicationto modules of rank two, the smallest non-trivial case. The fact obtained belongs tothe not-so-well-known results on real quadratic number fields. Cf. [44, p. 118].
Corollary.
Let K = Q ( √ d ) be a quadratic number field and ε ∈ K a unit ofnorm (+1) . Then N(1 − ε ) = 2 − tr( ε ) is a product of some primes dividing thediscriminant, a perfect square, and, possibly, a factor 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 h . , . i : O K × O K → Z asso-ciated to N.The same is true for the corresponding Z l -valued pairings between the l -adiccompletions 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). (cid:3) We clearly have that ( − N Φ( − ∈ Q . Further-more, we may assume that ( −
1) is not among the zeroes of Φ as, otherwise, theassertion is true, trivially.Then ( − N Φ( − >
0. Indeed, as Φ ∈ R [ T ] and there is no real zero differentfrom 1, ( − N Φ( −
1) = 2 N (1 + z ) · . . . · (1 + z N )is the product of several factors of the form zz = | z | for z ∈ C and some factorsthat are equal to 2. To prove the assertion, we will show that ( − N Φ( −
1) is ofeven l -adic valuation for every prime number l = p .Put H := H d ´et ( X F q , Z l ( d/ / tors. By Poincar´e duality [2, Exp. XVIII, formule(3.2.6.2)], cf. [38, Chap. 6, Sec. 2, Theorem 18 and Chap. 5, Sec. 5, Theorem 3], thebilinear pairing h . , . i : H × H −→ H d ´et ( X F q , Z l ( d )) ∼ = −→ Z l , ANDREAS-STEPHAN ELSENHANS AND J ¨ORG JAHNEL 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 extensionof this pairing. First case. l = 2.The operation of ( − Frob) is orthogonal with respect to the pairing, too. As 1 is notamong its eigenvalues, Corollary 2.7.b) shows that ν l (det(1 + Frob)) = ν l (Φ( − 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 i ´et ( X F q , Z ( i )) are invariant under Frob. We therefore see from Lemma2.10 that there is a Frob-invariant element ω ∈ H d ´et ( X F q , Z ( d/ h ω, x i + h x, x i ∈ Z for every x ∈ H d ´et ( X F q , Z ( d/ x ∈ (1 − Frob) H Q , the fact that ω is Frob-invariant implies h ω, x i = 0.Hence, h x, x i ∈ Z for x ∈ H ∩ (1 − Frob) H Q . According to Lemma 2.3.b),[ H/ (1 − Frob) H ] tors is of even length.An application to X F q shows that [ H/ (1 − Frob ) H ] tors is of even length, too.Lemma 2.11 now yields the assertion. (cid:3) Lemma 2.10.
For every even d > , there exists a polynomial P d ∈ Z [ T , . . . , T d/ ] ,weighted homogeneous of degree d for T i of weight i , such that the following is true.For a smooth projective variety X of dimension d over a finite field F q of character-istic = 2 , put ω := P d ( c ( T X ) , . . . c d/ ( T X )) ∈ H d ´et ( X F q , Z ( d/ for T X the tangentsheaf and c i the i -th Chern class. Then h ω, x i + h x, x i ∈ Z for every x ∈ H d ´et ( X F q , Z ( d/ . Proof.
This is a standard result in topology, cf. [33, §§ ε : H ∗ ´et ( X F q , Z ( ∗ )) −→ H ∗ ´et ( X F q , Z / Z ) the reduc-tion map. Furthermore, for simplicity, write c i := c i ( T X ) ∈ H i ´et ( X F q , Z ( i )) for theChern classes.For k ∈ N , let ν k ∈ H k ´et ( X F q , Z / Z ) be the 2 k -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 ) = 0for every x ∈ H d ´et ( X F q , Z / Z ) [41, Prop. 2.2.(2)]. We will inductively construct poly-nomials P k ∈ Z [ T , . . . , T k ] such that ε ( P k ( c ( T X ) , . . . c d/ ( T X ))) = ν k .For every k ∈ N , there is the formula of Wu [41, Proposition 0.5],Sq ( ν k ) + Sq ( ν k − ) + · · · + Sq k ( ν ) = ε ( c k ) . N THE CHARACTERISTIC POLYNOMIAL OF FROBENIUS 9
Moreover, for the Steenrod squares of the Chern classes, there are the formulasSq j ( ε ( c i )) = ε ( c j ) ε ( c i ) + (cid:18) j − i (cid:19) ε ( c j − ) ε ( c i +1 ) + · · · + (cid:18) j − i j (cid:19) ε ( c ) ε ( c i + j ) . Indeed, these follow in a purely formal manner from the definitions of Chern classesand Steenrod squares, cf. [33, Problem 8-A].As Sq = id, Wu’s formula implies that we may choose P ( T ) := T . Further-more, having P , . . . , P k − already constructed, it shows that ν k = ε ( c k ) − k X i =0 Sq i ( ν k − i ) = ε ( c k ) − k X i =0 Sq i ( ε ( P k − i ( c , . . . , c k − i )))= ε ( c k ) − k X i =0 P k − i (Sq i ( ε ( c ) , . . . , Sq i ( ε ( c k − i )))) . Plugging into this the formula for the Steenrod squares of the Chern classes, we seethat ν k is the reduction of a polynomial expression in c , c , . . . , c k . (cid:3) Lemma 2.11.
Let ( R, ν ) be a normalized discrete valuation ring of characteristic , k its residue field, which we assume to be finite, K the field of fractions, and H afree R -module of finite rank, equipped with a non-degenerate, symmetric K -bilinearpairing h . , . i : H K × H K → K .Moreover, let an R -linear map σ : H → H be given that is orthogonal with respectto the pairing and does not have the eigenvalue ( − . Suppose that [ H/ (1 − σ ) H ] tors and [ H/ (1 − σ ) 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 ) 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 ) H K . Furthermore, there is the commutativediagram of short exact sequences0 / / (1 − σ ) H/ (1 − σ ) H / / (cid:15) (cid:15) (cid:15) (cid:15) H/ (1 − σ ) H / / (cid:15) (cid:15) (cid:15) (cid:15) H/ (1 − σ ) H / / (cid:15) (cid:15) (cid:15) (cid:15) / / / / H/H ∩ (1 − σ ) H K H/H ∩ (1 − σ ) H K / / . As the vertical arrows are surjective, the 9-lemma yields exactness of0 −→ (1 − σ ) H/ (1 − σ ) H −→ [ H/ (1 − σ ) H ] tors −→ [ H/ (1 − σ ) H ] tors −→ . Since ℓ ([ H/ (1 − σ ) H ] tors ) and ℓ ([ H/ (1 − σ ) H ] tors ) are even, we see that the lengthof (1 − σ ) H/ (1 − σ ) 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 ) H , are R -submodules of maximal rank in the K -vectorspace (1 − σ ) H K . Moreover, (1 − σ ) H = (1 + σ ) | (1 − σ ) H K [(1 − σ ) H ]. Hence, themodulus of (1 + σ ) | (1 − σ ) H K is q − ℓ ((1 − σ ) H/ (1 − σ ) H ) for q := k .On the other hand, by [16, Exercise 14.3.6], this modulus is q − ν (det(1+ σ ) | (1 − σ ) HK ) .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 situa-tion, it is well-known [42, Example 2.6.C.i.b)] (see also [39, § − σ ) ≡ dim ker(1 − σ ) r (mod 2)for r ≫ z, z ) and, by our assumption,( −
1) is not an eigenvalue, it has the same parity as rk H . The assertion follows. (cid:3) Remarks . 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, 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 applicationto l -adic cohomology.ii) Suppose that q = p k for a prime p = 2 and let X be a surface such that thecanonical sheaf K ∈ Pic( X F q ) is divisible by 2. Then the case l = 2 of Theorem 1.5may be treated directly.Indeed, in this situation, Wu’s formula [41, Proposition 2.1] implies that h x, x i ∈ Z for every x ∈ H . Therefore, by Lemma 2.3.b), H/ (1 + Frob) H is of even length.Moreover, the assumption enforces that K is even. Hence, N = dim H ( X F q , Q (1))is even, too, by Noether’s formula [3, I.14]. N THE CHARACTERISTIC POLYNOMIAL OF FROBENIUS 11
Here, according to our assumption, we have p = 2.Again, we may assume without restriction that Φ( − = 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 + z N )) is even.For this, let W := W ( F q ) be the Witt ring and K its field of fractions. The crys-talline cohomology groups H d ( X/W ) are finitely generated W -modules, acted uponsemilinearly by the absolute Frobenius F . The operation of F k is W -linear and suchthat its characteristic polynomial coincides with Φ ( d )0 [26].Again, H := H d ( X/W ) / tors is equipped with a natural perfect pairing [4, Ch. VII,Th´eor`eme 2.1.3] h . , . i : H × H −→ W .
The Frobenius operation is, however, compatible with this pairing only in the sensethat h F ( x ) , F ( y ) i = p d h x, y i for every x, y ∈ H [8, Expos´e II, Exemple 1.1.ii)].The map σ := F /p d/ : H K → H K respects the pairing h . , . i . H is, in fact, a Dieu-donn´e module and thus carries a rich structure [29]. We shall use only a small partof 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 , . . . , z N of Φ.Furthermore, σ is orthogonal with respect to the Q p := Q( W ( F p ))-bilinear exten-sion H K × H K → Q p of the perfect pairing tr W ( F pk ) /W ( F p ) ◦ h . , . i : H × H → W ( F p ).The operation of ( − σ ) is orthogonal with respect to the pairing, too. Thus, Corol-lary 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 N Y i =1 Y r k = z i (1 + r ) = N Y i =1 (1 + z i ) . Second case. k is even.Here, the argument is slightly more involved. We first observe that H is a freemodule over the Witt ring W ( F p k/ ), too, and that σ k/ is W ( F p k/ )-linear.Clearly, rk W ( F pk/ ) H = 2 · rk W H . Moreover, σ k/ is orthogonal with respect to theQ p k/ := Q( W ( F p k/ ))-bilinear extension H K × H K → Q p k/ of the perfect pairingtr W ( F pk ) /W ( F pk/ ) ◦ h . , . i : H × H → W ( F p k/ ).Now choose a unit u ∈ W ( F p k ) such that tr W ( F pk ) /W ( F pk/ ) ( u ) = 0, i.e., such thatits conjugate is ( − u ). Then σ k/ ( ux ) = − uσ k/ ( x ) for all x ∈ H .We define a W ( F p k/ )-linear map T : H K → H K by T ( x ) := σ k/ ( ux ) . This yields T ◦ T = − u σ k . Furthermore, T is W ( F p k )-semilinear and one has h T x, T y i = h ux, uy i = u h x, y i for all x, y ∈ H . We see that the eigenvalues of T , as a W ( F p k/ )-linear map, are all the squareroots of the numbers ( − u z ) , . . . , ( − u z N ), for z , . . . , z N the zeroes of Φ. More-over, [ T ( H ) + H ] /H is of even W ( F p k/ )-length as it is, in fact, a W ( F p k )-module.Finally, put H := H ⊗ W ( F pk/ ) W ( F p ) and extend T to a W ( F p )-linear map T : H → H . Then u T is orthogonal. On the other hand,[ u T ( H ) + H ] /H = [ T ( H ) + H ] /H , as u is a unit, and the latter W ( F p )-module is of even length. Corollary 2.7 showsthat ν W ( F p ) (det(1 − u T )) is even. This number is nothing but the p -adic valuation of N Y i =1 Y r = − z i (1 − r ) = N Y i =1 (1 + z i ) . (cid:3) Remark . One might want to prove Theorem 1.9 in characteristic 2 along thesame lines as Theorem 1.5. For this, one would need, at least, the theory of Steenrodsquares and Wu classes, as well as Wu’s formula, for crystalline cohomology ofvarieties in characteristic 2. It seems, however, that such a theory is not yet availablein 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/ . Then a ( X ) = ℓ W ( σ ( H ) + H/H ) . Proof.
First, observe that σ ( H ), being the image of an F -semilinear map, is indeeda W -module. Furthermore, we have H/ F H ∼ = L m> ( W/p m W ) h ′ d − m,m . Hence, thereis a basis of H such that, under the corresponding isomorphism H ∼ = W N , one has F H ∼ = L m ≥ p m W h ′ d − m,m . Therefore, σ ( H ) ∼ = L m ≥ p m − d/ W h ′ d − m,m . The assertion fol-lows. (cid:3) 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 cohomologygroups H j ( X, W Ω mX ) are finitely generated W -modules for j + m = d . Then a ( X ) = − X ν q ( z i ) < ν q ( z i ) . Proof.
We have H d ( X/W ) ∼ = L m H d − m,m for H d − m,m := H d − m ( X, W Ω mX ), as isshown in [25, Th´eor`eme IV.4.5]. On H d − m,m , F operates as p m F for F the usualFrobenius on Serre cohomology. Thus, σ acts as p m − d/ F . For m ≥ d/
2, this ensuresthat 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/ − m − 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 | L m An application to the odd-dimensional supersingular case.Lemma 3.1. Let p be a prime and d and d two odd integers. Moreover, let Φ ∈ Q [ T ] and Φ ∈ Q [ T ] be polynomials of even degrees N and N that fulfill thefunctional equations Φ ( p d /T ) = p d N / T N Φ ( T ) a nd Φ ( p d /T ) = p d N / T N Φ ( T ) . For z (1) i the zeroes of Φ and z (2) j the zeroes of Φ , let Φ be the monic polynomialwith the zeroes z (1) i z (2) j /p d d .Then p N N / Φ( − is a square in Q . Proof. The assumption implies that the zeroes come in pairs with products p d and p d , respectively. For two pairs of zeroes, the corresponding four zeroes of Φ are z (1) i p d / · z (2) j p d / , z (1) i p d / · p d / z (2) j , p d / z (1) i · z (2) j p d / , and p d / z (1) i · p d / z (2) j . We now observe the identity( − − p u u )( − − u /u )( − − u /u )( − − p/u u ) = p ( u p + u p +1 /u +1 /u ) , (2)which applies, since the four zeroes may rationally be written as1 p z (1) i p ( d − / · z (2) j p ( d − / , z (1) i p ( d − / · p ( d − / z (2) j , p ( d − / z (1) i · z (2) j p ( d − / , and p p ( d − / z (1) i · p ( d − / z (2) j . It shows that the product Q i,j ( − − z (1) i z (2) j /p d d ) is p N N / times a square. In-deed, the sums occurring on the right hand side of (2) form a Gal( Q / Q )-invariantset of N N / (cid:3) Proposition 3.2. Let X be a smooth proper variety of odd dimension d over afinite field F p k for p a prime and k odd. Suppose that p > H d ´et ( X F p , Q l ) , dim H d ´et ( X F p , Q l ) ≡ , and that all eigenvalues of Frob on the cohomology are of p -adic valuation dk/ .I.e., that the Newton polygon has constant slope d/ .Then ± p − 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, togetherwith (1.10) and (1.11)]. The eigenvalues of Frob on H ( C F p , Q l ) are ± p − p k . In-deed, the assumptions imply p ≥ 5, and hence p > p p . The claim follows fromHasse’s bound. For Φ the characteristic polynomial of Frob on V := [ H d ´et ( X F p , Q l ) ⊗ H ( C F p , Q l )]( d +12 ) , Lemma 3.1 guarantees that p Φ( − 1) is a perfect square. But all eigenvalues of Frobon V are p -adic units. As they are l -adic units for every prime l = p , too, they mustbe roots of unity [7, Sec. 18, Lemma 2].To prove the assertion, we need to show Φ( − 1) = 0. Assuming the contrary, wesee from Lemma 3.6 that, for some e ≥ 1, all primitive 2 p e -th roots of unity mustbe eigenvalues of Frob on V . As with z , ( − z ) is an eigenvalue, too, this enforcesdim V ≥ p − H d ´et ( X F p , Q l ) ≥ p − 1, a contradiction. (cid:3) Remark . 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 theproducts [ H i ( X, Q l ) ⊗ H i ( C, Q l )]( d +12 ) for i + i = d + 1, i = 1. Example . Proposition 3.2 may fail in small characteristic, as is seen from theelliptic curve C over F , given by y = x − x − 1. Then C ( F ) = 1, whichshows that C is supersingular. Moreover, the characteristic polynomial of Frob is T − T + 3. The eigenvalues are the two primitive twelfth roots of unity withpositive real part, multiplied by √ Remark . When trying to carry over the argument from the proof, it turns outthat, on V := [ H ( C F , Q l ) × H ( C F , Q l )](1), there are the primitive sixth roots ofunity occurring as eigenvalues, together with 1, which appears twice. Therefore,Φ( T ) = ( T − T + 1)( T − and Φ( − 1) = 12.Alternatively, one might combine C with C ′ : y = x − x , which is supersingu-lar having four points. On the product [ H ( C F , Q l ) ⊗ H ( C ′ F , Q l )](1), one findsΦ( T ) = ( T − T + 1)( T + T + 1) and Φ( − 1) = 3. Lemma 3.6. Let Φ ∈ Q [ T ] be a monic polynomial such that all its roots are rootsof unity. Assume that | Φ( − | 6 = 0 is a multiple of a prime number p .Then Φ is divisible by the cyclotomic polynomial φ p e , for some e ≥ . In particular,if p = 2 then Φ is divisible by φ e , for some e ≥ . Proof. Φ is a product of cyclotomic polynomials φ n . For these, it is well known [34,Section 3] that φ ( − 1) = − φ ( − 1) = 0, φ p 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. (cid:3) The even-dimensional supersingular case.Proposition 3.7. Let X be a smooth proper variety of even dimension d over afinite field F p k for a prime p = 2 and k odd. Suppose that a ( X ) ≡ andthat all eigenvalues of Frob on H d ´et ( X F p , Q l ( d/ are p -adic units. I.e., that theNewton polygon has constant slope d/ . N THE CHARACTERISTIC POLYNOMIAL OF FROBENIUS 15 a) Then ( − is an eigenvalue or, for some e ≥ , the primitive p e -th roots ofunity are eigenvalues. b) If p > dim H d ´et ( X F p , Q l ( d/ or p > dim H d ´et ( X F p , Q l ( d/ and X is projec-tive then ( − is an eigenvalue. Proof. The eigenvalues are l -adic units, too, for every prime number l = p , hencethey are roots of unity. Theorem 1.9 ensures that p Φ( − 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 − p e − ≥ p − p e -th roots of unity and, in the projective case, the number 1. Thus, al-together, there are at least p − 1, respectively p , of them. (cid:3) Corollary 3.8. Let X be a supersingular K surface over a finite field F p k for p > a prime and k odd. Then ( − is an eigenvalue of Frob on H ( X F p , Q l (1)) . Proof. For K E [13, Propo-sition 1.1.a)] and, hence, the conjugate spectral sequence degenerates at E [5,Lemma 8.27.2]. Moreover, all H i ( X/W ) are torsion-free ([24, II.7.2] or [13, Proposi-tion 1.1.c)]). Consequently, we have a ( X ) = dim H ( X, O X ) = 1. Cf. Remark 1.8.b).The claim now follows from Proposition 3.7.b). (cid:3) Remarks . i) Corollary 3.8 refines the observation of M. Artin [1, 6.8] that thefield of definition of the rank-22 Picard group always contains F p .ii) More generally, let X be any K F p k for p = 2 aprime. Theorem 1.9 then asserts that, for z , . . . , z the eigenvalues of Frob on H ( X F p , Q l (1)), the expression p k Φ( − 1) = p k (1 + z ) . . . (1 + z ) is always a squarein Q .iii) Corollary 3.8 is clearly false in small characteristic, as may be seen from theexample below. Examples . i) For C the supersingular elliptic curve over F from Example 3.4,put X := Kum( C × C ). ThenΦ( T ) = ( T − T + 1)( T + T + 1) ( T − . In particular, we have Φ( − 1) = 3 · , in agreement with Remark 3.9.ii).Indeed, we saw that the characteristic polynomial of Frob on H (( C × C ) F , Q l (1))is ( T − T + 1)( T − . Furthermore, the 16 two-torsion points form five orbits ofsize three together with the origin.ii) Let X := Kum( C × C ′ ) be the Kummer surface associated to the product ofthe two supersingular elliptic curves over F , considered in Remark 3.5. Then, onceagain, Φ( T ) = ( T − T + 1)( T + T + 1) ( T − .In fact, we saw that the characteristic polynomial of Frob on H (( C × C ′ ) F , Q l (1))is ( T − T + 1)( T + T + 1)( T − . In addition, four two-torsion points are definedover the base field, while the others form four orbits of size three. Surfaces. The Artin-Tate formula. For surfaces, the assertion of Theo-rem 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 ofcharacteristic p and let Φ = Φ (2)1 ∈ Q [ T ] be the characteristic polynomial of Frob on H ( X F q , Q l (1)) . Put N := deg Φ and α ( X ) := dim H ( X, O X ) − dim H ( X, O X ) + dim H ( X F q , Q l ) . Suppose that the Tate conjecture is true for X .Then ( − N q α ( X ) Φ( − is a square in Q . Proof. Denote the zeroes of Φ, i.e. the eigenvalues of Frob, by z , . . . , z ̺ = 1, z ̺ +1 , . . . , z N = 1. If Φ( − 1) = 0 then the assertion is true, trivially. Thus, let ussuppose the contrary from now on. Then the zeroes z i = 1 come in pairs of complexconjugate numbers. In particular, N − ̺ is even.Furthermore, Frob and Frob have the eigenvalue 1 with the same multiplicity.Hence, the Tate conjecture predicts the rank of Pic( X F q ) not to be higher than thatof Pic( X ). This shows that X F q , 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 ), at least upto square factors. The results are( − ̺ − q α ( X ) N Y i = ̺ +1 (1 − z i ) and ( − ̺ − q α ( X ) N Y i = ̺ +1 (1 − z i ) . Moreover, equality of the ranks implies that disc Pic( X ) / disc Pic( X F q ) is a neces-sarily perfect square. This is a standard observation from the theory of lattices.We conclude that q α ( X ) Q Ni = ̺ +1 (1 + z i ) is a square in Q .On the other hand, ( − N q α ( X ) Φ( − 1) = 2 N + ̺ q α ( X ) Q Ni = ̺ +1 (1 + z i ) such that theassertion follows from the fact that ̺ ≡ N (mod 2). (cid:3) Remarks . i) The Artin-Tate formula appears to us as a very natural conse-quence of the Tate conjecture and the cohomological machinery. Thus, we findit very astonishing that it has the potential to produce incompatible results for avariety and its base extension.Of course, this does not happen for polynomials that really occur as the characteristicpolynomial of the Frobenius on a certain variety. But it occurs for polynomialsthat otherwise look plausible. This observation was actually the starting point ofour investigations.ii) One might want to compare the Picard lattice of X with that of X q n for n > H ( X, O X ) = dim H ( X F q , Q l ) and that X fulfills the assump-tions of Remark 1.8.b). Then α ( X ) = a ( X ) = dim H ( X, O X ). We do not know N THE CHARACTERISTIC POLYNOMIAL OF FROBENIUS 17 how closely Milne’s invariant α ( X ) and our invariant a ( X ) are related for “patho-logical” surfaces. Example . Let X be the double cover of P F , given by w = 6 x + 6 x y + 2 x z + 6 x y + 5 x z + 5 x y + x y + 6 xy + 5 xz + 3 y + 5 z . This is a K X over the finite fields F , . . . , F 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 ( X F , Q l (1)), this informationleaves us with two candidates, one for each sign in the functional equation,Φ i ( t ) = 17 (cid:0) t − t + t − t + 6 t − t − t + 4 t − t − t + ( − i ( − t − t + 4 t − t − t + 6 t − t + t − t + 7) (cid:1) for i = 0 , 1. All roots are of absolute value 1.However, Φ ( − 1) = 60 / − N a ( X ) Φ ( − 1) = 2 · · is the characteristic polynomial of Frob on H ( X F , Q l (1)). The minus sign holds in the functional equation. Remark . Alternatively, we may argue as follows. Assume Φ is the charac-teristic polynomial. Then rk Pic( X F ) = rk Pic( X ) = 2. Indeed, the Tate conjec-ture is proven for K ≥ X F ) ∈ ( − Q ∗ ) and disc Pic( X ) ∈ ( − Q ∗ ) . As − − = 15 is a non-square, this is contradictory.3.4. Cubic fourfolds. Example . Let X be the subvariety of P F , given by x + x x + x x + x x + x x x + x x x + x x x + x x x + x x x + x x x + x x + x x x + x x + x + x x + x x x + x x x + x x + x x x + x x + x x x + x + x x + x x + x x + x x + x x + x x + x + x x + x x + x = 0 . This is a smooth cubic fourfold. We have dim H ( X, O X ) = 0, dim H ( X, Ω X ) = 1,and dim H ( X, Ω X ) = 21, According to Remark 1.8.b), this shows N = 23and a ( X ) = 1.The numbers of points on X over the finite fields F , . . . , F 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 ( X F , Q l (2)) isΦ( t ) = 12 ( t − t − t − t + 2 t − t + t + t − t + t + t − t + t + t − t + t + t − t + 2 t − t − t + 2) . It turns out that Φ( − 1) = − 1, in agreement with Theorem 1.5. Observe that, inthis example, the assertion of Theorem 1.9 is true, albeit the characteristic of thebase field is 2. Remark . The degree 22 factor of Φ is irreducible over Q . In particular, X iscertainly not special in the sense of B. Hassett [22]. References [1] M. Artin , Supersingular K (1974), 543–567.[2] M. Artin, A. Grothendieck, et J.-L. Verdier (avec la collaboration de P. Deligne et B. Saint-Donat ), Th´eorie des topos et cohomologie ´etale des sch´emas, S´eminaire de G´eom´etrieAlg´ebrique du Bois Marie 1963–1964 (SGA 4), Lecture Notes in Math. 269, 270, 305, Springer,Berlin, Heidelberg, New York 1972–1973.[3] A. Beauville , Complex algebraic surfaces, LMS Lecture Note Series 68, Cambridge UniversityPress, Cambridge 1983.[4] P. Berthelot , Cohomologie cristalline des sch´emas de caract´eristique p > 0, Lecture Notes inMath. 407, Springer, Berlin, New York 1974.[5] P. Berthelot and A. Ogus , Notes on crystalline cohomology, Princeton University Press,Princeton 1978.[6] J. W. S. Cassels , Arithmetic on curves of genus 1. VIII. 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