Math. Z. DOI 10.1007/s00209-014-1402-7

Mathematische Zeitschrift

Zero-cycles on diagonal cubic surfaces over p-adic fields Tetsuya Uematsu

Received: 3 February 2014 / Accepted: 23 June 2014 © Springer-Verlag Berlin Heidelberg 2014

Abstract By using symbolic generators of the Brauer group of diagonal cubic surfaces, we compute the degree-zero part of the Chow group of zero-cycles on some diagonal cubic surfaces over p-adic fields. We also find that the Chow group of zero-cycles on them are generated by the classes of rational points. Keywords

Zero-cycles · Diagonal cubic surfaces · Brauer groups

Mathematics Subject Classification

Primary 14C15 · 14J26; Secondary 11G25 · 14F22

1 Introduction Let k be a p-adic field and X be a variety over k. Explicit computation of the Chow group CH0 (X ) of zero-cycles on X has been studied by many authors. For example, zero-cycles on some Châtelet surfaces was studied in [1,2]. In this paper, we are concerned with zero-cycles on diagonal cubic surfaces. A diagonal cubic surface X is a smooth projective surface defined by the following equation: x 3 + by 3 + cz 3 + dt 3 = 0, b, c, d ∈ k ∗ .

This work was supported by MEXT Grant-in-Aid for the Global COE Program ‘The research and training center for new development in mathematics’ at Graduate School of Mathematical Sciences, the University of Tokyo, and by JSPS Grant-in-Aid for Scientific Research (B) Grant Number 23340003 (Kanetomo Sato). T. Uematsu (B) Faculty of Science and Engineering, Chuo University, 1-13-27 Kasuga, Bunkyo-ku, Tokyo 112-8551, Japan e-mail: [email protected] Present Address: T. Uematsu Department of General Education, Toyota National College of Technology, 2-1 Eisei-cho, Toyota, Aichi 471-8525, Japan

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Put A0 (X ) := Ker(deg : CH0 (X ) → Z). When the equation of X is of the form x 3 + y 3 + z 3 + dt 3 = 0, we have the following: Theorem 1 [3], [4, Theorem 4.1.1] Let X be the projective surface over k defined by x 3 + y 3 + z 3 + dt 3 = 0, where d ∈ / (k ∗ )3 . (1) If p  = 3,we have ⎧ ⎪ if ord(d) ≡ 0 mod 3, ⎨0 A0 (X ) ∼ if ord(d)  ≡ 0 mod 3 and ζ ∈ / k, = Z /3 Z ⎪ ⎩ 2 (Z /3 Z) if ord(d)  ≡ 0 mod 3 and ζ ∈ k. (2) If p = 3, ζ ∈ k and ord(d) ≡ 1 mod 3, then A0 (X ) ∼ = (Z /3 Z)2 . However, here are some remaining problems: – This list does not cover all the cases. For example, the structure of A0 (X ) is not known yet in the case of p = 3 and ord(d) ≡ 2, or in the case of x 3 + y 3 + cz 3 + dt 3 = 0. – These results tell us only the structure of A0 (X ) and no information about zero-cycles generating A0 (X ). In this article, we give some answers to these problems. First, we compute the structure of A0 (X ). The results are as follows: Theorem 2 Let k be a 3-adic field containing a primitive cubic root of unity. Let X be the projective surface over k defined by x 3 + y 3 + z 3 + dt 3 = 0, where d ∈ / (k ∗ )3 . If ord(d) ≡ 2 mod 3 and the absolute ramification index of k is greater than 3, then we have A0 (X ) ∼ = (Z /3 Z)2 . As is the case of Theorem 1, we prove Theorem 2 by using the same method developed in [4] and symbolic generators of Br(X ) found by Manin [5]. In these proofs, we have to take a regular model X / Ok of X/k, where Ok is the integer ring of k. In the proof of Theorem 1, we can take a model defined by the same equation, that is, x 3 + y 3 + z 3 + dt 3 = 0. On the other hand, since this model is not regular in our situation, we cannot take this model to prove Theorem 2 and need some modifications. For surfaces of the form x 3 + y 3 + cz 3 + dt 3 = 0, we have the following theorem. Theorem 3 Let X be the projective surface over k defined by a homogeneous equation x 3 + y 3 + cz 3 + dt 3 = 0. Assume that c, d, cd, c/d ∈ / (k ∗ )3 . (1) If p  = 3, we have  0 if ord(c) ≡ ord(d) ≡ 0 mod 3, A0 (X ) = Z /3 Z otherwise. (2) If p = 3, ζ ∈ k, ord(c − 1) is greater than the absolute ramification index of k and ord(d) ≡ 1 mod 3, then A0 (X ) ∼ = Z /3 Z. This theorem is proved by using the same method in [4] and a symbolic generator of Br(X ) appeared in [6]. The result of Theorem 3 (2) was announced in [7]. As stated before, these theorems only give us the structure of A0 (X ). By calculating the Brauer–Manin pairing explicitly, we obtain some results for generators of A0 (X ) in Theorem 1. This is a partial answer to the question stated in [4, Section 4.1].

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Theorem 4 Let p  = 3 be a prime number, k be a p-adic field containing a primitive cubic root of unity and X be the projective surface defined by x 3 + y 3 +z 3 +dt 3 = 0 with d ∈ / (k ∗ )3 . Then the Chow group CH0 (X ) of zero-cycles on X is generated by the classes of rational points. In some cases, we find explicit rational points generating A0 (X ). In general, for a variety X over a field, it would be an interesting problem when CH0 (X ) is generated by only the classes of rational points. This paper is organized as follows. In Sect. 2, we first recall a method of computing zerocycles studied in [4]. Secondly, we recall the structure and generators of the Brauer group of diagonal cubic surfaces, which were studied in [5,6,8]. In Sect. 3, we give a proof of Theorem 2. We prove Theorem 3 in Sect. 4. Finally, in Sect. 5, Theorem 4 will be proved. Notation Throughout this paper, let k be a p-adic field, F be its residue field, and Ok be the integer ring of k. We denote a uniformizer of Ok by π. The map ord : k ∗ → Z be the normalized valuation on k. The absolute ramification index of k is denoted by e. Put e := pe/( p − 1). Note that e is integer when k contains a primitive pth root ζ of unity. For a field K , the map {·, ·} p means the second norm residue map K 2M (K ) → 2 H (K , μ⊗2 p ). When K contains a primitive pth root of unity, we fix an isomorphism M 2 ⊗2 ∼ ∼ μ μ⊗2 = p and this map is often considered as the composite K 2 (K ) → H (K , μ p ) = p H 2 (K , μ p ) = p Br(K ). we often write it simply by {·, ·}. The Brauer group Br(X ) of a scheme X always means the cohomological one, that is, He´2t (X, Gm ). A diagonal cubic surface X over k is a smooth projective surface defined by a homogeneous equation x 3 + by 3 + cz 3 + dt 3 = 0, where b, c, d ∈ k ∗ . Note that diagonal cubic surfaces are (geometrically) rational.

2 Preliminaries 2.1 Method of computing the group of zero-cycles Here, we recall a method of computing the group of zero-cycles. Let X be a diagonal cubic surface over k. A fundamental ingredient for calculating A0 (X ) is the following homomorphism, which is induced by the Brauer–Manin pairing due to Manin [9]: Φ X : A0 (X ) → Hom(Br(X )/ Br(k), Q / Z),    n i coresk(Pi )/k invk(Pi ) (α(Pi )) . n i Pi → α → Here Br(X )/ Br(k) = Br(X )/ Br 0 (X ), where Br 0 (X ) = Im(Br(k) → Br(X )). For the injectivity of Φ X , we have the following result due to Colliot-Thélène [10, Proposition 5 and Proposition 7]: Proposition 5 The map Φ X is injective. In fact, he proved the injectivity for rational surfaces. To state the sufficient condition for the surjectivity of Φ X , we prepare some notation. Let X be a scheme which is regular in codimension one and faithfully flat over Spec Ok such that its generic fiber X ×Ok k/k is isomorphic to X/k. Let Y / F be the special fiber of X / Ok . For a generic point η of Y , put Aη := OhX ,η , the henselization of OX ,η at its maximal ideal

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and K η := Frac(Aη ). We note that Aη is a discrete valuation ring and denote a uniformizer of Aη by  . Let eη be the absolute ramification index of  . Note that the residue field of Aη is isomorphic to κ(η). Put ι : Br(X ) → Br(K η ). By the regularity of Aη , we always consider ι Br(Aη ) as a subgroup of Br(K η ). We put ι : Br(X ) → Br(K η ) → Br(K η )/ Br(Aη ). In the paper [4], Saito and Sato gave the following sufficient condition for the surjectivity of Φ X : Proposition 6 [4, Proposition 4.1.3] Assume that there exists a generic point η ∈ Y such that Ker ι ⊂ Br 0 (X ). Then the map Φ X is surjective. When ch F = 3, in order to check this condition, we use some filtrations defined just below. We first define the descending filtration U m K η∗ (m ≥ 0) on K η∗ as  K η∗ m = 0, m ∗ U K η := m 1 +  Aη m ≥ 1. m Put H := H 2 (K η , μ⊗2 3 ). We also define the descending filtration U H (m ≥ 0) on H as the m ∗ 0 ∗ image of U K η ⊗ U K η under a norm residue map {·, ·} = {·, ·}3 : (K η∗ )⊗2 → H . We also m H and put e = 3e /2. The structures denote its mth graded module U m H/U m+1 H by grU η η m of grU H were originally studied by Bloch and Kato in the strictly henselian case (see [11, Corollary 1.4.1]). In the henselian case, a similar result holds:

Proposition 7 [4, equation (2.2.1)] We have the following isomorphisms: ⎧ 2 1 if m = 0, ⎪ ⎨Ωκ(η),log ⊕ Ωκ(η),log m 1 ∼ ρm : grU H = Ωκ(η) if 0 < m < eη , 3  m, ⎪ ⎩ 1 1 ⊕ κ(η)/κ(η)3 if 0 < m < eη , 3 | m, Ωκ(η) /Z κ(η) which are defined by {x1 , x2 } + {, x3 } → (d log x1 ∧ d log x2 , d log x3 ), {1 +  m y1 , x1 } + {1 +  m y2 ,  } → y1 d log x1 − m −1 d y2 , {1 +  m y1 , x1 } + {1 +  m y2 ,  } → (y1 d log x1 , y2 ) for x1 , x2 , x3 ∈ A∗η and y1 , y2 ∈ Aη , where a is the residue class of a in κ(η). In the sequel of the proof, we freely use the following relations (see [4, Sublemma 4.3.9]): For a, b ∈ Aη , i, j ≥ 1, k, l ≥ 0 and h ∈ U k K η∗ , we have {1 +  i a +  j b, h} ≡ {1 +  i a, h} + {1 +  j b, h} ±l

mod U i+ j+k H,

H, {1 +  a(1 +  b) , h} ≡ {1 +  a, h} mod U

 ab , − i a mod U i+2 j H. {1 +  i a, 1 +  j b} ≡ − 1 +  i+ j 1 + ia i

j

i

i+ j+k

(1) (2) (3)

In particular, as a consequence of (3), we have that {·, ·} maps U i K η∗ ⊗ U j K η∗ into U i+ j H . At the end of this subsection, we note that: Lemma 8 [4, Lemma 2.2.1 (2)] If K η contains a primitive cubic root of unity, the image of the map 3 Br(Aη ) 

is contained in U eη H .

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→ 3 Br(K η ) ∼ = H 2 (K η , μ⊗2 3 )

Zero-cycles on diagonal cubic surfaces

2.2 Brauer group of diagonal cubic surfaces We recall the structure and symbolic generators of the Brauer group of diagonal cubic surfaces. Note that by the regularity of any diagonal cubic surface X , the natural inclusion j : Spec k(X ) → X induces the injection j ∗ : Br(X ) → Br(k(X )). In the sequel of this article, we always identify Br(X ) with its image j ∗ Br(X ) in Br(k(X )). Proposition 9 [5, Example 45.3], [8, Proposition 1], [6, Theorem 1.1] Let k be a p-adic field containing a primitive cubic root ζ of unity, b, c, d ∈ k ∗ and X : x 3 + by 3 + cz 3 + dt 3 = 0 be the diagonal cubic surface with coefficients b, c, d. Put f :=

x + ζy x+z , g := ∈ k(X )∗ . x+y x+y

(1) If b = c = d = 1, then Br(X )/ Br(k) = 0. (2) If b = c = 1 and d is not a cube in k ∗ . Then the elements e1 = {d, f }, e2 = {d, g} ∈ 3 Br(k(X )) are contained in Br(X ) and Br(X )/ Br(k) ∼ = (Z /3 Z)e1 ⊕ (Z /3 Z)e2 . (3) If b = 1 and neither c, d, cd nor c/d is a cube in k ∗ . Then the element e3 = {d/c, f } ∈ 3 Br(k(X )) is contained in Br(X ) and Br(X )/ Br(k) ∼ = (Z /3 Z)e3 . Remark 10 For the case when X is of the form x 3 + by 3 + cz 3 + dt 3 = 0 with “general” b, c, d, we know the structure Br(X )/ Br(k) ∼ = Z /3 Z, but there is no uniform generator, that is, a generator algebraically expressed by coefficients b, c, d. For a precise statement, see [6, Section 5]. When k does not contain primitive cubic roots of unity, we have the following proposition. The claim (1) is due to Colliot-Thélène, see [4, Proposition 4.2.6]. The claim (2) can be proved in a similar way and we omit its proof. Proposition 11 Let k be a p-adic field without containing a primitive cubic root ζ of unity, b, c, d ∈ k ∗ and X : x 3 + by 3 + cz 3 + dt 3 = 0 be the diagonal cubic surface with coefficients b, c, d. Put L = k(ζ ). Let ei be the elements of Br(X L ) in Proposition 9. The map cores L/k denotes the corestriction map from Br(X L ) to Br(X ). (1) If b = c = 1 and d ∈ / (k ∗ )3 , then Br(X )/ Br(k) ∼ = (Z /3 Z) cores L/k (e1 ). (2) If b = 1 and c, d, cd, c/d ∈ / (k ∗ )3 , then Br(X )/ Br(k) ∼ = (Z /3 Z) cores L/k (e3 ).

3 Proof of Theorem 2 We can assume d = π 2 , where π is a uniformizer of k. Define 1 := 3π −e ∈ O∗k . In the proof of Theorem 1(1), the projective surface X over Ok defined by the same equation as X can be taken as a regular model appearing in Sect. 2.1. But in our case, we can easily see that this equation does not define a regular model over Ok . So we have to take another model. Let X  be the proper model over Ok defined to be Proj Ok [x, y, z, t]/(x 3 + y 3 + z 3 + dt 3 )

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and Z be the closed subscheme of X  defined by the homogeneous ideal (π, x + y + z). X

X 

Put p : → be the blowing-up of X  along Z . Put X t ⊂ X  be the affine open subscheme defined by t  = 0. Now consider the defining equation of p −1 (X t ). Put I = (π, x + y + z) ⊂ A := (X t , OX  ), u = − 1 ((x + y + z)(x y + yz + zx) − x yz) ∈ A.

(4)

Consider the following surjective homomorphism of graded A-algebras In φ : A[r, s] → n≥0

r → π, s → x + y + z. Under the relation (x + y + z)r = πs we have the following equations in A[r, s]: (x 3 + y 3 + z 3 + π 2 )s 2 = (x 3 + y 3 + z 3 )s 2 + (x + y + z)2 r 2 = {(x + y + z)3 − 3(x + y + z)(x y + yz + zx) + 3x yz}s 2 + (x + y + z)2 r 2 = (x + y + z)3 s 2 + ε1 π e us 2 + (x + y + z)2 r 2 = (x + y + z)3 s 2 + ε1 π e−2 u(x + y + z)2 r 2 + (x + y + z)2 r 2 = (x + y + z)2 {(x + y + z)s 2 + (1 + π e−2 u)r 2 }. Thus p −1 (X t ) is explicitly written as a closed subscheme in X t ×Ok P1Ok defined by (x + y + z)r − πs = (x + y + z)s 2 + (1 + π e−2 u)r 2 = 0, where [r : s] is the homogeneous coordinate of P1Ok . Let U be the affine open subscheme of p −1 (X t ) defined by s  = 0. U is written explicitly as follows: U = Spec Ok [x, y, z][r ]/(x + y + z + (1 + π e−2 u)r 2 , r (x + y + z) − π). We see that its special fiber has only one generic point η and its local ring OU,η is a discrete valuation ring with uniformizer r . Now we define X to be U ∪ p −1 (X ). This is a scheme which is regular in codimension one, faithfully flat over Ok and whose generic fiber is isomorphic to X . So we can take this model X to compute the Chow group of zero-cycles on X . Let Aη be the henselization of OX ,η at the maximal ideal (r ) and K η be its fractional field. Note that the residue field of Aη is isomorphic to κ(η) = F(y, z). We now introduce some notation. Let v be the normalized valuation on Aη . Recall e = 3e/2. We have v(ζ − 1) = e , eη = v(3) = 3e, v(π) = 3 and v(x + y + z) = 2. We define

2 , 3 and 4 ∈ A∗η as: 

ζ − 1 = 2 r e , π e−2 = 3 r 3(e−2) , x + y + z = 4 r 2 .

(5)

Note that we have π = −(1 + π e−2 u)r 3 , 4 = −(1 + π e−2 u), 3 = 4e−2

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(6)

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by the defining equation of U . We can determine which filtrations the generators of Br(X )/ Br(k) belong to: 



3e −4 H is nonzero. Proposition 12 (1) {π, f } ∈ U 3e −4 H and its image in grU 3e−4 3e−4 H and its image in grU H is nonzero. (2) {π, g} ∈ U

This proposition is proved by very technical and long calculations. Before proving this proposition, we assert the following implication: Proposition 13 Proposition 12 implies Theorem 2. In [4], the proof of such an implication is omitted. However for the reader’s convenience we include the proof of this proposition, which is given by Kanetomo Sato. Proof Let S be the subgroup of Br(X ) generated by {π, f } and {π, g}. Recall that ι is the natural map Br(X ) → Br(K η )/ Br(Aη ). We need the following lemmas. Lemma 14 The restriction map of ι on S is injective. Proof Let ω = a{π, f } + b{π, g} be an arbitrary element in Ker ι. Since S is a 3-torsion  group, ι(ω) ∈ 3 Br(Aη ) by the definition of ι. Thus by Lemma 8, ι(ω) ∈ U 3e H . This means 3e−4 3e−4 that the image of ι(ω) in grU H is zero. On the other hand, in grU H , the image of {π, f } is zero and the image of {π, g} is nonzero by Proposition 12. This implies that b ≡ 0 3e −4 H , we have a ≡ 0 mod 3 and that ω = a{π, f }. If we consider the image of ι(ω) in grU mod 3 by Proposition 12(2). Therefore ω = 0 and this completes the proof of Lemma 14.   Lemma 15 ι(Br 0 (X )) ∩ ι(S) = {0} Proof By the same argument as in Lemma 14, for any unramified extension k  over k, the map ι

S → Br(K η )/ Br(Aη ) → Br(K η ⊗k k  )/ Br(Aη ⊗Ok Ok  ) is injective. On the other hand, for k  = k ur , the maximal unramified extension of k, the map Br(k) ∼ = Br 0 (X ) → Br(X ) → Br(K η ⊗k k ur )/ Br(Aη ⊗Ok Ok ur )  is trivial since it factors through Br(k ur ) = 0. The claim easily follows from these facts.  We now go back to the proof of Proposition 13. Note that we have the following decomposition: Br(X ) = Br 0 (X ) ⊕ S since X (k)  = ∅ and hence the natural map Br(k) → Br(X ) is injective. Proposition 13 is an immediate consequence of the above two lemmas and Proposition 6.   Now we prove Proposition 12:

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Proof (of Proposition 12) (1) By the defining equation of U , Eqs. (4), (5) and (6), we have 1 + π e−2 u = 1 + 3 r 3(e−2) (− 1 ( 4 r 2 (x(y + z) + yz) − ( 4 r 2 − (y + z))yz))



4 x = 1 + r 3(e−2) (− 1 3 (y + z)yz) 1 + r 2 , yz y f = 1 + (ζ − 1) x+y  2 y = 1 + re x+y 



4

2 y  = 1 + re − 1 + r2 . z z − 4 r 2

(7)

(8)

Therefore we have {π, f } = {(−r )3 (1 + π e−2 u), f } = {1 + π e−2 u, f }   (7),(1) ≡ 1 + r 3(e−2) (− 1 3 (y + z)yz), f    + 1 + r 3e−4 (− 1 3 4 (y + z)x), f mod U 5e −10 H 



2 y (8),(1)  ≡ 1 + r 3(e−2) (− 1 3 (y + z)yz), 1 + r e − z



2 4 y  3(e−2) e +2 + 1+r (− 1 3 (y + z)yz), 1 + r − z(z − 4 r 2 )



2 y  + 1 + r 3e−4 (− 1 3 4 (y + z)x), 1 + r e − z



2 4 y  + 1 + r 3e−4 (− 1 3 4 (y + z)x), 1 + r e +2 − z(z − 4 r 2 ) 

mod U 4e −4 H. The last congruence follows from  1 + r 3(e−2) (− 1 3 (y + z)yz),  1 + r 3e−4 (− 1 3 4 (y + z)x),

(9) (10) (11)

(12)  f

≡ (9) + (10)

 f ≡ (11) + (12)



mod U 4e −4 H, 

mod U 4e −2 H.

Note that (5e − 10) − (4e − 4) ≥ 0 under our assumption e > 3. In the last expression, (10)    and (11) are both in U 3e −4 H , and (12) is in U 3e −2 H . So in order to prove {π, f } ∈ U 3e −4 H ,  −4 3e it suffices to show the first term (9) is in U H . By Eq. (1), We rewrite (9) as follows: 



2 y (1)   ≡ A + B mod U 5e −12 H, 1 + r 3(e−2) (− 1 3 (y + z)yz), 1 + r e − z 



2 y  , A := 1 + r 3(e−2) (− 1 3 )y 2 z, 1 + r e − z



2 y  B := 1 + r 3(e−2) (− 1 3 )yz 2 , 1 + r e − . z For A and B, we can see:

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Lemma 16 (1) A ∈ U 3e −2 H .  (2) B ∈ U 3e −3 H . Proof (1) we have 

  r 3(e −2) 1 2 3 y 3  3(e−2) 2 ,r

1 3 y z mod U 4e −6 H A ≡ − 1+ 1 + r 3(e−2) (− 1 3 y 2 z)   (2)   ≡ − 1 + r 3(e −2) 1 2 3 y 3 , r 3(e−2) 1 3 y 2 z mod U 5e −12 H. (3)

Using the equation 

r 3(e −2) 1 2 3 = −(ζ − 1)3 r −6 (1 + (ζ − 1))2 (1 + π e−2 u)−2 , we have

   − 1 + r 3(e −2) 1 2 3 y 3 , r 3(e−2) 1 3 y 2 z (2)



≡ −{1 − (ζ − 1)3 r −6 y 3 , r 3(e−2) 1 3 y 2 z} mod U 4e −6 H   = −{1 − ((ζ − 1)r −2 y)3 , 1 3 y 2 z} + (1 − (ζ − 1)r −2 y)3 , 1 3 y 2 z 

3(ζ − 1)r −2 y(1 − (ζ − 1)r −2 y) 2 ,

y z = 1− 1 3 1 − ((ζ − 1)r −2 y)3 

∈ U 3e −2 H. 

Noting that 4e − 6 and 5e − 12 is greater than 3e − 2, this proves A ∈ U 3e −2 H . (2) we have    r 3(e −2) 1 2 3 y 2 z (3)  3(e−2) 2 B ≡ − 1+

1 3 yz ,r mod U 4e −6 H 1 + r 3(e−2) (− 1 3 yz 2 ) (2)





≡ −{1 + r 3(e −2) 1 2 3 y 2 z, r 3(e−2) 1 3 yz 2 } mod U 5e −12 H 

= −{1 + r 3(e −2) 1 2 3 y 2 z, r 3(e−2) 1 3 yz 2 } 



− {1 + r 3(e −2) 1 2 3 y 2 z, −r 3(e −2) 1 2 3 y 2 z} (Steinberg relation) 



= −{1 + r 3(e −2) 1 2 3 y 2 z, −r 5e −12 12 2 32 y 3 z 3 }. Using the equation 

r 5e −12 12 2 32 = (ζ − 1)5 ζ 4 r −12 (1 + π e−2 u)−4 , we get 



− {1 + r 3(e −2) 1 2 3 y 2 z, −r 5e −12 12 2 32 y 3 z 3 } 

= −{1 + r 3(e −2) 1 2 3 y 2 z, −(ζ − 1)5 ζ 4 (1 + π e−2 u)−4 y 3 z 3 } 

= {1 + r 3(e −2) 1 2 3 y 2 z, (ζ − 1)ζ 2 (1 + π e−2 u)} 

= {1 + r 3(e −2) 1 2 3 y 2 z, (ζ − 1)ζ 2 π −e/2 } 

+ {1 + r 3(e −2) 1 2 3 y 2 z, π e/2 (1 + π e−2 u)} (6)



= {1 + r 3(e −2) 1 2 3 y 2 z, (ζ − 1)ζ 2 π −e/2 }

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+ {1 + r 3(e −2) 1 2 3 y 2 z, r e (1 + π e−2 u)e/2+1 } (7)





≡ {1 + r 3(e −2) 1 2 3 y 2 z, (ζ − 1)ζ 2 π −e/2 } mod U 3(e −2)+3(e−2) H.

Now we consider the element 5 := (ζ − 1)ζ 2 π −e/2 ∈ O∗k . For its image 5 ∈ F∗ , there exists a unit a ∈ O∗k such that a 3 = 5 . This means that there exists an element b ∈ Ok such that 5 = a 3 (1 + πb). Using this, we have 



{1 + r 3(e −2) 1 2 3 y 2 z, 5 } = {1 + r 3(e −2) 1 2 3 y 2 z, 1 + πb} 

∈ U 3e −3 H 

Noting that 4e − 6 and 5e − 12 is greater than 3e − 3, this proves B ∈ U 3e −3 , which completes the proof of Lemma 16.   



3e −4 We now have {π, f } ∈ U 3e −4 H . Next we consider its image in grU H . The terms  3e −3 H , so we have to prove the sum of (10) and (11) is nonzero in (9) and (12) are in U 3e −4 H. grU 1 . For the term (10), we have We use Proposition 7 and consider their images in Ωκ(η)



2 4 y 3(e−2) e +2 (− 1 3 (y + z)yz), 1 + r 1+r − z(z − 4 r 2 ) ⎧ ⎫ 2 y (y + z)  −4 ⎪ ⎪ 3e ⎪ ⎪ r

1 2 3 4 ⎨ ⎬ (3)  z − 4 r 2 ,

mod U 4e −2 H. ≡ − 1+

(y + z)yz 1 3 3(e−2) (− (y + z)yz) ⎪ ⎪ 1 + r 1 3 ⎪ ⎪ ⎩ ⎭ 1 , since they are in F ⊂ F(y, z) = Note that all i ’s commute with the differential d in Ωκ(η) 1 . Thus the image of (10) in Ω 1 κ(η). We also remark that we have 2 = −1 in Ωκ(η) κ(η) is:

y 2 (y + z) · d log( 1 3 (y + z)yz) z y(y − z) = 1 2 3 4 · (zdy − ydz). z2

− 1 2 3 4 ·

Similarly, for the term (11), we have



2 y 3e−4 e 1+r − (− 1 3 4 (y + z)x), 1 + r z ⎧ ⎫ y  ⎪ ⎪ r 3e −4 1 2 3 4 (y + z)x · ⎨ ⎬ (3) z , r 3e−4 (y + z)x ≡ − 1+ 1 3 4 3e−4 ⎪ ⎪ 1+r (− 1 3 4 (y + z)x) ⎩ ⎭ 1 is: Noting that x = 4 r 2 − (y + z), the image of (11) in Ωκ(η)



− 1 2 3 4 ·



  3e − 4 (y + z)2 y −(y + z)2 y · d log − 1 3 4 (y + z)2 −  · d − 1 2 3 4 z 3e − 4 z

= − 1 2 3 4

123



mod U 4e −4 H.

(y + z)2 (zdy − ydz). z2

Zero-cycles on diagonal cubic surfaces 1 Hence the image of the sum of (10) and (11) in Ωκ(η) is

− 1 2 3 4 (zdy − ydz)  = 0, which completes the proof of Proposition 12(1). (2) We prove this in the same way as (1). Using Eq. (6) and the following expression x+z x+y 

4 (z − y) y , = · 1 + r2 z y( 4 r 2 − z)

g=

we have {π, g} = {1 + π e−2 u, g} 



y e−2 e−2 2 4 (z − y) + 1 + π u, 1 + r = 1 + π u, z y( 4 r 2 − z)

 y (7),(1) ≡ 1 + r 3(e−2) (− 1 3 (y + z)yz), z 

y + 1 + r 3e−4 (− 1 3 4 (y + z)x), z 

(z − y) 4 e−2 2 mod U 6e−10 H. + 1 + π u, 1 + r y( 4 r 2 − z)

(13) (14) (15)

Since the terms (14) and (15) are in U 3e−4 H , in order to prove {π, g} ∈ U 3e−4 H , we have to show that the term (13) is in U 3e−4 H . In fact, we show a stronger result. Rewrite (13) as follows:

 y (1) ≡ C + D mod U 6e−12 H, 1 + r 3(e−2) (− 1 3 (y + z)yz), z 

y , C := 1 + r 3(e−2) (− 1 3 y 2 z), z

 y D := 1 + r 3(e−2) (− 1 3 yz 2 ), . z Here we have the following: Lemma 17 (1) C is in U 3e−3 H . (2) D is in U 3e−3 H . Proof (1) We have 

y C = 1 + r 3(e−2) (− 1 3 y 2 z), z   3(e−2) 2 + 1+r (− 1 3 y z), r 3(e−2) 1 3 y 2 z (Steinberg relation)   = 1 + r 3(e−2) (− 1 3 y 2 z), r 3(e−2) 1 3 y 3   (6) = 1 + r 3(e−2) (− 1 3 y 2 z), ((−r )3 )e−2 · 1 · (1 + π e−2 u)e−2   = 1 + r 3(e−2) (− 1 3 y 2 z), 1

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  + (e − 2) 1 + r 3(e−2) (− 1 3 y 2 z), 1 + π e−2 u (7)

≡ {1 + r 3(e−2) (− 1 3 y 2 z), 1 }

mod U 6e−12 H.

Using the expression 1 = a 3 (1 + πb) similar to 5 in the proof of the Lemma 16(1), we get   {1 + r 3(e−2) (− 1 3 y 2 z), 1 } = 1 + r 3(e−2) (− 1 3 y 2 z), 1 + πb ∈ U 3e−3 H, which completes the proof of (1). (2) As 2D = −D, it suffices to prove 2D ∈ U 3e−3 H . By the similar way to (1), we get

 y2 2D = 1 + r 3(e−2) (− 1 3 yz 2 ), 2 z

2 y = 1 + r 3(e−2) (− 1 3 yz 2 ), 2 z   3(e−2) 2 (− 1 3 yz ), r 3(e−2) 1 3 yz 2 (Steinberg relation) + 1+r   = 1 + r 3(e−2) (− 1 3 yz 2 ), r 3(e−2) 1 3 y 3 . Now we can easily prove 2D ∈ U 3e−3 H in the same way as (1).   Hence we prove {π, g} ∈ U 3e−4 H . To prove the nontriviality of the image of {π, g} in 3e−4 grU H , we only have to show that of the image of the sum of (14) and (15). 1 The image of (14) in Ωκ(η) is:

1 3 4 (y + z)2 · d log

 y (y + z)2 = 1 3 4 · (zdy − ydz). z yz

On the other hand, for the term (15), we have: 

4 (z − y) 1 + π e−2 u, 1 + r 2 y( 4 r 2 − z)



4 (z − y) (7),(2) ≡ 1 + r 3(e−2) (− 1 3 (y + z)yz), 1 + r 2 mod U 3e−2 H y( 4 r 2 − z) ⎧ ⎫ ⎪ ⎪ 3e−4 − 1 3 4 (y + z)(z − y)z ⎪ ⎪ r ⎨ ⎬ (3)

4 r 2 − z ,

mod U 3e−2 H, ≡ − 1+

(y + z)yz 1 3 3(e−2) ⎪ ⎪ 1 + r (−

(y + z)yz) 1 3 ⎪ ⎪ ⎩ ⎭

1 is: and its image in Ωκ(η)

1 3 4 (y + z)(y − z) · d log( 1 3 (y + z)yz) = − 1 3 4 ·

(y − z)2 (zdy − ydz). yz

1 is: Therefore the image of the sum of (14) and (15) in Ωκ(η)

1 3 4 (zdy − ydz)  = 0. This completes the proof of Proposition 12(2) and Theorem 2.

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4 Proof of Theorem 3 4.1 Proof of Theorem 3(1) We may assume that ord(c) and ord(d) are both in {0, 1, 2}. Moreover, by symmetry, we may assume ord(c) ≤ ord(d). Thus there are the following 12 cases: / k, ord(c) = 0, ord(d) = 0 (1) ζ ∈ k, ord(c) = 0, ord(d) = 0, (1) ζ ∈ (2) ζ ∈ k, ord(c) = 0, ord(d) = 1, (2) ζ ∈ / k, ord(c) = 0, ord(d) = 1 (3) ζ ∈ k, ord(c) = 0, ord(d) = 2, (3) ζ ∈ / k, ord(c) = 0, ord(d) = 2 (4) ζ ∈ k, ord(c) = 1, ord(d) = 1, (4) ζ ∈ / k, ord(c) = 1, ord(d) = 1 (5) ζ ∈ k, ord(c) = 1, ord(d) = 2, (5) ζ ∈ / k, ord(c) = 1, ord(d) = 2 / k, ord(c) = 2, ord(d) = 2. (6) ζ ∈ k, ord(c) = 2, ord(d) = 2, (6) ζ ∈ We take a proper model X of X as follows: X = Proj Ok [x, y, z, t]/(x 3 + y 3 + cz 3 + dt 3 ).

Put Y := X ⊗Ok F. First we consider the case when a primitive cubic root of unity ζ is in k. Case (1) Since X is proper, the Brauer–Manin pairing of X induces the following homomorphism A0 (X ) → Hom(Br(X )/(Br(X ) + Br(k)), Q / Z), which is injective by Proposition 5. Therefore it suffices to prove Br(X )/(Br(X ) + Br(k)) = 0. Noting that X is smooth over Ok in this case, we have the following exact sequence by the purity of Brauer groups (see [4, Remark 3.1.2(3)]): ι

0 → Br(X ) → Br(X ) → Br(K η )/ Br(Aη ),

(16)

where η is the generic point of the special fiber Y . Thus it suffices to show the image of e3 = {d/c, f } in Br(K η ) is contained in Br(Aη ). we prove this claim by using the following residual exact sequence: δ

0 → 3 Br(Aη ) → 3 Br(K η ) → κ(η)∗ /(κ(η)∗ )3 . we can easily show v( f ) = 0, where v is the normalized valuation on Aη , which proves δ(e3 ) = 0 and completes the proof of Case (1). Case (2) Let η be the generic point of Y . Then OX ,η is a regular local ring. By Proposition 5, 6 and 9, we only have to show Ker ι ⊂ Br 0 (X ). Let S be the subgroup of Br(X ) generated by e3 = {π, f }. Now we have the following lemma corresponding to Lemma 14: Lemma 18 the restriction of ι on S is injective.

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Proof Using the residual exact sequence (16) again, we have to prove δ(e3 )  = 0. Since we know that v(c) = 0, v(d) = 1 and f ∈ A∗η , we have δ(e3 ) = f ∈ κ(η)∗ /(κ(η)∗ )3 . Thus we have to show that f is not a cube in κ(η)∗ . Consider the following elliptic curve E c := Proj(F[x, y, z]/(x 3 + y 3 + cz 3 )) instead of E in [4, Section 4.2, Case (iii)]. Note that κ(η) is isomorphic to F(E c )(t). Now suppose that f would be a cube in κ(η). If we consider f as an element of F(E c ), we have f ∈ (F(E c )∗ )3 by the equation F(E c )∗ ∩ (κ(η)∗ )3 = (F(E c )∗ )3 . Thus there exists h ∈ F(E c )∗ such that f = h 3 . In Div(E c ), we have div( f ) = 3[P] − 3[O],

O = [1 : −1 : 0],

P = [1 : −ζ : 0]

and hence div(h) = [P] − [O]. This means the residue class of [P] − [O] is zero in   A0 (E c ) ∼ = E c (F), which is a contradiction to the nontriviality of P − O in E c (F). By the same way as in Lemma 15, we can prove ι(S) ∩ ι(Br 0 (X )) = {0} and hence Ker ι ⊂ Br 0 (X ). Case (3) Let η be the generic point of Y . Then we also prove that OX ,η is a regular local ring. The same argument as in (2) completes the proof of this case. Case (4) In this case, the special fiber Y consists of three irreducible components Yi , i = 0, 1, 2, where Yi = Proj(F[x, y, z, t]/(x + ζ i y)). Let ηi be the generic point of Yi . We can easily see that each OX ,ηi is regular. Thus we have the following residual exact sequence for each ηi : δi

0 → 3 Br(Aηi ) → 3 Br(K ηi ) → κ(ηi )∗ /(κ(ηi )∗ )3 . As in the case (2), it suffices to prove Ker(ι) ⊂ Br 0 (X ). To prove the claim, we only have to show δ0 (e3 )  = 0. Since v( f ) = −1, where v is the normalized valuation on Aη0 , we have δ0 (e3 ) = d/c ∈ κ(η0 )∗ /(κ(η0 )∗ )3 . Noting that κ(η0 ) = F(y, z), if d/c would be in (κ(η0 )∗ )3 , we had d/c ∈ (κ(η0 )∗ )3 ∩ F∗ = (F∗ )3 . Then by Hensel’s lemma, d/c is a cube in Ok and this is in contradiction to the assumption d/c ∈ / (k ∗ )3 . Therefore we have δ0 (e3 )  = 0. Case (5) Let Yi and ηi be as in (4). Each OX ,ηi is also regular in this case. Take d/c as the uniformizer π ∈ Ok and Put c = c/π and d  = d/π. We divide the proof into two cases. (i) the case ζ ∈ (k ∗ )3 . For the same reason as in (4), it suffices to prove δ0 (e3 )  = 0. In Aη0 , we have x + y = (−π) · and therefore

 (x 2 − x y + y 2 )(x + ζ y) c z 3 + d  t 3

 (x 2 − x y + y 2 )(x + ζ y) π, (Steinberg relation) = δ0 c z 3 + d  t 3

δ0 (e3 ) = δ0

123



c z 3 + d  t 3 , x 2 − x y + y2

π, −π −1 ·

Zero-cycles on diagonal cubic surfaces

= =

(x 2 − x y + y 2 )(x + ζ y) c z 3 3(ζ − 1) (x + y = 0 ∈ κ(η)) c

= c

−1

(3 = −ζ 2 (ζ − 1)2 , ζ ∈ (k ∗ )3 )

∈ κ(η0 )∗ /(κ(η0 )∗ )3 . By the same argument as in (4), if c would be in (κ(η0 )∗ )3 , we had c ∈ (k ∗ )3 and cd = c3 /c ∈ (k ∗ )3 , which is a contradiction to the assumption cd ∈ / (k ∗ )3 . Thus we get δ0 (e3 )  = 0. ∗ 3 (ii) the case ζ ∈ / (k ) . In this case, we consider in Aη2 . We have 

−ζ 2 y + ζ y δ2 (e3 ) = δ2 π, −ζ 2 y + y 

1 π, = δ2 −ζ = ζ ∈ κ(η2 )∗ /(κ(η2 )∗ )3 . 2

By the assumption on ζ and a similar argument in Case (4), we have δ2 (e3 )  = 0 and A0 (X ) = Z /3 Z. Case (6). We can prove the theorem in the same way as in Case (4). Now we complete the proof of the case ζ ∈ k. Next we deal with the case ζ ∈ / k. Put L := k(ζ ). √ √ Case (1) . Using a field extension k( 3 c, 3 d)/k of degree 9, Proposition 9(1) and a norm argument, we know A0 (X ) is 9-torsion. Moreover, using a field extension L/k of degree two, the result of Case (1) and a norm argument, we also find A0 (X ) is 2-torsion. The claim follows from these facts. Cases (2) –(6) . In Cases (4) , (5) and (6) , the special fiber Y consists of the following two irreducible components: Y0 = Proj F[x, y, z, t]/(x + y), Y1 = Proj F[x, y, z, t]/(x 2 − x y + y 2 ). Let ηi be the generic point of Yi . We easily see OX ,ηi are all regular in each cases. We have a finite étale morphism X  := X ⊗Ok O L → X . Let Y  be the special fiber of X  . Choose a generic point η of Y  such that (X  , η ) is the same pair as in Cases (2)– (6) for X L /L. Let η ∈ X be the image of η . Put Aη := OhX  ,η , K η = Frac(Aη ) and ι : Br(X L ) → Br(K η ). Using Ker ι ⊂ Br 0 (X L ), Proposition 11(2) and a norm argument, we see that Ker ι ⊂ Br 0 (X ) for each case. Therefore we obtain A0 (X ) ∼ = Z /3 Z in these cases and complete the proof of Theorem 3(1). 4.2 Proof of Theorem 3(2) We may assume that ord(d) = 1. We take π := d/c as a uniformizer of Ok . If we put n := ord(c − 1), then we can write c = 1 + δπ n for some δ ∈ O∗k . Note that n > e by the

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second assumption. Let X be the projective model over Ok defined by the same equation x 3 + y 3 + cz 3 + dt 3 = 0 as X . The reduced part of its special fiber is: Proj F[x, y, z, t]/(x + y + z). Let η be its generic point. In an affine open subscheme Spec Ok [x, y, z]/(x 3 + y 3 + cz 3 + d) of X , the point η is the one corresponding to the prime ideal (π, x + y + z). Now we consider a local ring OX ,η = (Ok [x, y, z]/(x 3 + y 3 + cz 3 + d))(π,x+y+z) . Put r := x + y + z,

1 := 3π −e ∈ O∗k , F(x, y, z) := (y + z)yz + x(r − x)r, u := δπ

z + δπ

n−e 3

n−e+1

− 1 F(x, y, z).

(17) (18)

Then we have r 3 = x 3 + y 3 + z 3 + 3F(x, y, z) = −(δπ n z 3 + d) + π e 1 F(x, y, z)   = −π 1 + δπ n + δπ n−1 z 3 − π e−1 1 F(x, y, z) = −(1 + uπ e−1 )π.

(19)

Since n > e, we see that u ∈ Ok and the maximal ideal of OX ,η is generated by one element r . In particular, OX ,η is a discrete valuation ring. Let Aη := OhX ,η and v be the normalized valuation on Aη . We have v(π) = 3, eη = v(3) = 3e and v(ζ − 1) = e . Put 

2 := (ζ − 1) · r −e ∈ A∗η , 3 := π e−1 r −3(e−1) ∈ A∗η . By the same argument as in Proposition 13, in order to prove Theorem 3(2), we only have to show the following: 



3e −2 H is nonzero. Proposition 19 {π, f } ∈ U 3e −2 H and its image in grU

Proof Note that (ζ − 1)y x+y − 2 y  − 2 y  + r e +1 . = 1 + re z z(z − r )

f =1+

Using Eqs. (1) and (2), we obtain: − {π, f } (19)

= {1 + uπ e−1 , f }  

δ(z 3 + π) (18) ,f = 1 + π e−1 (− 1 F) 1 + π n−e − 1 F

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Zero-cycles on diagonal cubic surfaces (2)



≡ {1 + π e−1 (− 1 F), f } mod U 3e H (n > e)   (17) = 1 + r 3(e−1) (− 1 3 )(y + z)yz + r 3e−2 (− 1 3 )x(r − x), f 

− 2 y (1)  − 2 y  + r e +1 ≡ 1 + r 3(e−1) (− 1 3 )(y + z)yz, 1 + r e z z(z − r ) 

− y − 2 y    2 + r e +1 mod U 5e −5 H + 1 + r 3e−2 (− 1 3 )x(r − x), 1 + r e z z(z − r ) 

− y (1)  2 (20) ≡ 1 + r 3(e−1) (− 1 3 )(y + z)yz, 1 + r e z

 − 2 y  + 1 + r 3(e−1) (− 1 3 )(y + z)yz, 1 + r e +1 (21) z(z − r )

  − 2 y + 1 + r 3e−2 (− 1 3 )x(r − x), 1 + r e (22) z 

− 2 y   mod U 4e −2 H. (23) + 1 + r 3e−2 (− 1 3 )x(r − x), 1 + r e +1 z(z − r ) 



In the last expression, the terms (21) and (22) are in U 3e −2 H and the term (23) is in U 3e −1 H .   So it suffices to show the first term (20) is in U 3e −2 H in order to prove {π, f } ∈ U 3e −2 H . We rewrite (20) as follows:

 (1)  − 2 y  1 + r 3(e−1) (− 1 3 )(y + z)yz, 1 + r e ≡ A + B mod U 5e −6 H, z

  − 2 y A := 1 + r 3(e−1) (− 1 3 )y 2 z, 1 + r e , z 

 − 2 y . B := 1 + r 3(e−1) (− 1 3 )yz 2 , 1 + r e z For A and B, by a similar calculation as in Lemma 16, we can see the following 

Lemma 20 (1) A ∈ U 3e −1 H .  (2) B ∈ U 3e H . 



3e −2 H . The terms (20) and (23) are in U 3e −1 H , Next we consider the image of {π, f } in grU 3e −2 1 . H∼ so we have to prove the sum of (21) and (22) is nonzero in grU = Ωκ(η) 1 are By a similar calculation as in Proposition 12(1), the images of (21) and (22) in Ωκ(η)

1 2 3

y(y − z) (y + z)2 (zdy − ydz) and − 1 2 3 (zdy − ydz) 2 z z2

1 is respectively, and therefore their sum in Ωκ(η)

− 1 2 3 (zdy − ydz)  = 0. This completes the proof of Proposition 19 and Theorem 3(2).

 

Remark 21 In our proof, the assumption ord(c − 1) > e is essential, since this enable us to exclude the terms appearing δ and reduce to the same situation as in [4]. If ord(c − 1) ≤ e, we cannot ignore such terms, at least in the above way, which seems to make computation of A0 (X ) more difficult.

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5 Proof of Theorem 4 Since X (k)  = ∅, to prove the theorem, we only have to show that A0 (X ) is generated by some differences of classes of rational points. Moreover, we may assume ord(d)  ≡ 0 mod 3, since the claim is trivial if ord(d) ≡ 0 mod 3 by Theorem 1. We divide its proof into 3 parts: (1) ζ ∈ / (k ∗ )3 , (2) ζ ∈ (k ∗ )3 , 1 − ζ ∈ / (k ∗ )3 , (3) ζ, 1 − ζ ∈ (k ∗ )3 . In the case (1), we have the following explicit generators: Proposition 22 Assume ζ ∈ / (k ∗ )3 . We define two zero-cycles C1 and C2 on X as follows: C1 := [1 : 0 : −ζ : 0] − [0 : 1 : −1 : 0], C2 := [1 : 0 : −ζ : 0] − [1 : 0 : −ζ 2 : 0]. Then these cycles generate A0 (X ). Proof We prove the claim in the case ord(d) = 1. The claim in the other cases easily follows from this case. We note that we have the following isomorphism:   ord(a) b ⊗2 ∼ 2 1 ∗ ∗ 3 ord(a) ord(b) ∼ F /(F ) , {a, b} → (−1) H (k, μ3 ) = H (F, μ3 ) = . a ord(b) Thus for the function f ∈ k(X )∗ and a point P ∈ X (k) such that f (P) ∈ O∗k , we can identify the Brauer–Manin pair [P], {d, f } ∈ Br(k) with f (P) ∈ F∗ /(F∗ )3 . The same computation holds for g ∈ k(X )∗ . Under this identification, we can see that 2

C1 , {d, f } = ζ , C1 , {d, g} = 1 − ζ , 2

C2 , {d, f } = 1, C2 , {d, g} = ζ . Under the assumption ζ ∈ / (k ∗ )3 , ζ is not trivial in F∗ /(F∗ )3 . Therefore, no matter which 1 − ζ is trivial or not, the images of C1 and C2 under the isomorphism A0 (X ) ∼ = Hom(Br(X )/ Br(k), Q / Z) ∼ = Z /3 Z ⊕ Z /3 Z generate the whole group. This completes the proof of Proposition 22.   Now we consider the case (2). By the assumption 1 − ζ ∈ / (k ∗ )3 , we can see that C1 in Proposition 22 defines a non-trivial element in A0 (X ). However, unlike the case (1), it seems to be difficult to take another zero-cycle which is linearly independent of C1 explicitly. Instead we can prove the existence of such an element. Let E be the projective smooth curve over Ok defined by the homogeneous equation x 3 + y 3 + z 3 = 0. Put E := E ⊗Ok F. Consider f and g as elements of F(E)∗ . Using a similar construction appeared in [8, Lemma 9] and [12, 2.7], we have the following result on an evaluation map from E(F). Proposition 23 Assume that ζ ∈ (k ∗ )3 . Then there exists a rational point P ∈ E(F) such that the value f (P) is not trivial in F∗ /(F∗ )3 .

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Proof Put O := [1 : −1 : 0] and Q := [1 : −ζ 2 : 0]. Define a curve E  to be a smooth compactification of C = {u 3 − f = 0} ⊂ P1E , where [u : v] is the homogeneous coordinate of P1F . Explicitly, E  is obtained by the blow-up of C = {(x + y)u 3 − (x + ζ y)v 3 = 0} along two points ([x : y : z], [u : v]) = (O, [0 : 1]), (Q, [1 : 0]). Then we can see that the natural map π : E  → E is an unramified cover of degree 3 and the fiber π −1 (O) consists three F-rational points. Fix one point O  ∈ π −1 (O). Then the map π : (E  , O  ) → (E, O) is an F-isogeny of elliptic curves with degree 3. Thus we have the following exact sequence of G F -modules: π

0 → Ker π → E  (F) → E(F) → 0. Taking cohomologies, we obtain the exact sequence δ

E(F) → H 1 (F, Ker π) → H 1 (F, E  ). Since any curve over a finite field with genus one has a rational point by the Hasse–Weil bound, we have H 1 (F, E  ) = 0 and hence the map δ is surjective. We also have an isomorphism ∼ =

w : Ker π → μ3 induced by the Weil pairing E[3] × E[3] → μ3 . Moreover, by Kummer theory, we have a natural isomorphism ∂ : F∗ /(F∗ )3 → H 1 (F, μ3 ). Thus we have the following surjective homomorphism: ∂ −1 ◦ w ◦ δ : E(F) → H 1 (F, Ker π) → H 1 (F, μ3 ) → F∗ /(F∗ )3 . By construction, we can see that this map is written explicitly by  1 if P = O or Q, −1 (∂ ◦ w ◦ δ)(P) = f (P) otherwise. Therefore there exists a rational point P ∈ E(F) such that f (P) is non-trivial in F∗ /(F∗ )3 . This completes the proof of Proposition 23.   Now we go back to the proof of the case (2). By Proposition 23, we have a point P ∈ E(F) such that f (P) is non-trivial in F∗ /(F∗ )3 . By the assumption p  = 3 and Hensel’s lemma,  ∈ E (Ok ) of P. Define a zero-cycle C3 on X to be: there exists a lift P  : 0] − [0 : 1 : −1 : 0]. C3 := [ P  we have the following By the construction of P, C3 , {d, f } = f (P)  ≡ 1. Hence we find that this C3 is a non-trivial element in A0 (X ). We also see that C1 and C3 generate the whole group, which completes the proof of the case (2). Finally we deal with the case (3). By a similar argument to Proposition 23 and changes of coordinates, we can prove the following: Proposition 24 Assume that ζ and 1 − ζ ∈ (k ∗ )3 . Let E be as above. Then there exist rational points P1 , P2 , P3 ∈ E(F) such that the values g(P1 ), ( f /g)(P2 ) and ( f g)(P3 ) are not trivial in F∗ /(F∗ )3 .

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T. Uematsu

By Proposition 24, there exist a rational point P ∈ E(F) such that ( f /g)(P) = θ , a fixed  ∈ E (Ok ). Put a zero-cycle C4 as generator of F∗ /(F∗ )3 . We can take its lift P  : 0] − [0 : 1 : −1 : 0]. C4 := [ P Then we have the following three subcases: (i) If f (P) = θ and g(P) = 1, we take Q ∈ E(F) so that g(Q) = θ . (ii) If f (P) = θ 2 and g(P) = θ , we take Q ∈ E(F) so that f (Q) = θ . (iii) If f (P) = 1 and g(P) = θ 2 , we take Q ∈ E(F) so that ( f g)(Q) = θ .  ∈ E (Ok ) be a lift of Q and define Let Q  : 0] − [0 : 1 : −1 : 0]. C5 := [ Q Then we can easily see that C4 and C5 generate the group A0 (X ), which completes the proof of the case (3) and Theorem 4. Acknowledgments The author is deeply grateful to Professor Kanetomo Sato for suggesting to him problems of constructing zero-cycles and for discussing about the proof of Proposition 13. He would like to express his gratitude to the referee for pointing out many mistakes and giving valuable comments.

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