- Email: [email protected]
Galois cohomology in degree 3 of function fields of curves over number fields
ARTICLE IN PRESS
Journal of Number Theory 107 (2004) 80–94
http://www.elsevier.com/locate/jnt
Galois cohomology in degree 3 of function fields of curves over number fields V. Suresh Department of Mathematics and Statistics, University of Hyderabad, GachiBowli, P.O. Central University, Hyderabad 500 046, India Received 10 February 2003 Communicated by J.-L. Colliot-The´le`ne
Abstract Let k be a field of characteristic not equal to 2. For nZ1; let H n ðk; Z=2Þ denote the nth Galois Cohomology group. The classical Tate’s lemma asserts that if k is a number field then given finitely many elements a1 ; ?; an AH 2 ðk; Z=2Þ; there exist a; b1 ; ?; bn Ak such that ai ¼ ðaÞ,ðbi Þ; where for any lAk ; ðlÞ denotes the image of k in H 1 ðk; Z=2Þ: In this paper we prove a higher dimensional analogue of the Tate’s lemma. r 2004 Elsevier Inc. All rights reserved. Keywords: Galois cohomology; Number fields; Function fields of curves
Introduction Let k be a field of characteristic not equal to 2. For nX1; let H n ðk; Z=2Þ denote the nth Galois cohomology group. We have H 1 ðk; Z=2Þ ¼ k =k2 ; where k is the group of non-zero elements of k: The structure of these groups is well understood when k is a number field or a local field. For example, if k is a number field or a local field then every element in H n ðk; Z=2Þ is of the form a1 ,?,an with ai AH 1 ðk; Z=2Þ: Further, if k is a totally imaginary number field or a p-adic field, then H n ðk; Z=2Þ ¼ 0 for nX3 (cf. [Se, p. 90, 108]). If k is a p-adic field, it is well known that H 2 ðk; Z=2Þ ¼ Z=2 and the non-trivial element of H 2 ðk; Z=2Þ becomes trivial over every quadratic field extension of k (cf. [CF, p. 131]). The classical Tate’s lemma asserts that if k is a
Fax: +91-40-2301-1469. E-mail address: [email protected].
0022-314X/$ - see front matter r 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jnt.2004.01.002
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
81
number field then given finitely many elements a1 ; y; an AH 2 ðk; Z=2Þ; there exist a; b1 ; y; bn Ak such that ai ¼ ðaÞ,ðbi Þ; where for any lAk ; ðlÞ denotes the image of k in H 1 ðk; Z=2Þ: Saltman [S1] proved that if K is a function field of a curve over a p-adic field, pa2; then every element in H 2 ðK; Z=2Þ is a sum of two symbols, i.e., of the form ða1 Þ,ðb1 Þ þ ða2 Þ,ðb2 Þ; where a1 ; a2 ; b1 ; b2 AK (cf. [S2]). Using this result, in [PS], it was proved that every element in H 3 ðK; Z=2Þ is a symbol, i.e. of the form ðaÞ,ðbÞ,ðcÞ for some a; b; cAK : Let K be a function field of a curve over a dyadic field or a number field. It is not known whether there exists a number N such that every element of H 2 ðK; Z=2Þ or H 3 ðK; Z=2Þ is a sum of at most N symbols. The aim of this paper is to prove the following two-dimensional analogue of the classical Tate’s lemma. Theorem. Let k be a p-adic field or a number field and K a function field of a curve over k: Let a1 ; y; an AH 3 ðK; Z=2Þ: Then there exist lAK and bi AH 2 ðK; Z=2Þ such that ai ¼ ðlÞ,bi : In fact we prove the following stronger results, which are used by Parimala and Preeti [PP] in the study of Hasse principle for classical groups over function fields of curves over number fields. Theorem 1. Let k be a dyadic field and K a function field of a curve over k: Let a1 ; y; an AH 3 ðK; Z=2Þ and a; bAK : Then there exist f AK and 2 b1 ; y; bn AH ðK; Z=2Þ such that f is a value of the quadratic form /a; b; abS over K and ai ¼ ðf Þ,bi : Theorem 2. Let k be a number field and K a function field of a curve over k: For a1 ; y; an AH 3 ðK; Z=2Þ and a; b; dAK ; there exists f AK such that (1) f is a value of the quadratic form /a; b; abdS; (2) for every finite non-dyadic place v of k; there exist b1v ; y; bnv AH 3 ðKv ; Z=2Þ such that aiv ¼ ðf Þ,biv ; where for any place v of k; kv is the completion of k at v; Kv ¼ K#k kv and aiv ¼ ai #Kv ; (3) for every finite dyadic place v of k; if d is a square in Kv ; there exist b1v ; y; bnv AH 3 ðKv ; Z=2Þ such that aiv ¼ ðf Þ,biv :
1. Some preliminaries Let F be a field and F the group of non-zero elements of F : By a quaternion algebra over F we mean a central simple algebra of dimension 4 over F : For a; bAF ; let ½a; bÞ denote the quaternion algebra over F generated by i and j satisfying i2 þ i þ a ¼ 0; ij þ ji ¼ j and j 2 ¼ b: It is well known that every quaternion algebra over F is of the form ½a; bÞ for some a; bAF (cf. [Sc, p. 313]). Suppose that the
ARTICLE IN PRESS 82
V. Suresh / Journal of Number Theory 107 (2004) 80–94
characteristic of F is not equal to 2. For a; bAF ; let ða; bÞ denote the quaternion algebra generated by x and y satisfying x2 ¼ a; y2 ¼ b and xy þ 1yx ¼ 0: For a; bAF ; we have isomorphisms of quaternion algebras ða; bÞC 4 a; b and ½a; bÞC 14 a; b : Let BrðF Þ denote the Brauer group of F and 2 BrðF Þ the subgroup of BrðF Þ consisting of 2-torsion elements. Let R be a discrete valuation ring with quotient field F : Let aABrðF Þ: We say that a is unramified at R if a comes from an Azumaya algebra over R: For iX0 and nX0; let H n ðF ; Z=2ðiÞÞ be the cohomology groups defined as in [K, Section 0]. For any field we have [K, Section 0] H 1 ðF ; Z=2ÞCHomcont ðGalðF ab =F Þ; Z=2Þ; H 2 ðF ; Z=2ð1ÞÞC2 BrðF Þ: We use these isomorphisms and identify H 1 ðF ; Z=2Þ with Homcont ðGalðF ab =F Þ; Z=2Þ and H 2 ðF ; Z=2ð1ÞÞ with 2 BrðF Þ: Let k be a dyadic field (a finite extension of Q2 ) or a number field and K a function field of a curve over k: Let R be a discrete valuation ring with the quotient field K and the residue field F : Then either F is a number field or a dyadic field or a function field of a curve over a finite field. If F is a function field of a curve over a finite field of characteristic 2, then we have ½F : F 2 ¼ 2: Let S be a discrete valuation ring with the quotient field F and residue field k: Then k is a finite field. If the characteristic of k is 2, then ½k : k2 ¼ 1: Therefore, we have residue maps [K, Section 1] @ : H 3 ðK; Z=2ð2ÞÞ-H 2 ðF ; Z=2ð1ÞÞ; @ : H 2 ðF ; Z=2ð1ÞÞ-H 1 ðk; Z=2Þ: Note that if F is a local field, then @ : H 2 ðF ; Z=2ð1ÞÞ-H 1 ðk; Z=2Þ is the classical invariant for local fields. We say that an element a in H 3 ðK; Z=2ð2ÞÞ (resp. H 2 ðF ; Z=2ð1ÞÞ) is unramified at R (resp. S) if @ðaÞ ¼ 0: Lemma 1.1. Let k; K; R and F as above with F a function field of a curve. Let u; v; aAK be such that u; v are units in R: Then we have @ð½u; vÞ,ðaÞÞ ¼ nðaÞ½u; % v%Þ; where bar denotes the image in F and n is the valuation at R: Proof. We can replace K by its completion and assume that R and K are complete. Let m be the maximal ideal of R: Suppose that vðaÞ is even. Then, without loss of generality, we assume that a is a unit in R: We need to show that @ð½u; vÞ ðaÞÞ ¼ 0: If v% is a square in F ; then ½u; % v%Þ ¼ 0: Since R is complete, ½u; vÞ ¼ 0 and hence ½u; vÞ ðaÞ ¼ 0: Assume that v% is
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
83
pffiffiffi not a square in F : Since every element of F is a square in F ð v%Þ; there exist a0 ; a2 AR such that a20 a22 v ¼ a modulo m: If a0 Am; then ½u; vÞ and ½u; aÞ are isomorphic modulo m: Since R is complete ½u; vÞ and ½u; aÞ are isomorphic. In particular a is a reduced norm from ½u; vÞ and hence ½u; vÞ ðaÞ ¼ 0: Assume that a0 em: Let f ðX Þ ¼ uX 2 a0 X þ a20 a22 v aAR½X : Then f au0 ¼ a20 a22 v aAm and f 0 au0 D / 0 modulo m; where f 0 ðX Þ is the derivative of f ðX Þ: Since R is complete, by a0 D Hensel’s Lemma [G, p. 47], there exists a1 AR such that f ða1 Þ ¼ 0: Let x; yA½u; vÞ be such that x2 þ x þ u ¼ 0; y2 ¼ v and yx ¼ xy y: Then it is easy to see that the reduced norm of z ¼ a0 þ a1 x þ a2 yA½u; vÞ is a20 þ ua21 va22 a0 a1 : Since f ða1 Þ ¼ 0; we see that the reduced norm of z is a: Hence ½u; vÞ ðaÞ ¼ 0: In particular @ð½u; vÞ ðaÞÞ ¼ 0: Suppose that vðaÞ is odd. Then, without loss of generality, we assume that a is a parameter in R: Let @ð½u; vÞ ðaÞÞ ¼ bAH 2 ðF ; Z=2ð1ÞÞ: Since F is a function field of a curve over a finite field, b is represented by a quaternion algebra Q over F : We need to show that QC½u; % v%Þ: Let E be a finite extension of F which splits ½u; % v%Þ: Let L be a finite extension of K such that E is the residue field of the discrete valuation of L extending the discrete valuation of K (if ECF ½X =ðf ðX Þ for some monic polynomial f ðX ÞAK½X ; let L ¼ K½X =ðf ðX ÞÞ and use induction). Since ½u; % v%Þ splits over E and L is complete, it follows that ½u; vÞ splits over L: In particular bE ¼ @ðð½u; vÞ ðaÞÞL Þ ¼ 0: Hence Q is split over E: Now let E be a finite extension of F which splits Q and L a finite extension of K lifting E as above with ½L : K ¼ ½E : F : Then @ðð½u; vÞ ðaÞÞL Þ ¼ bL ¼ 0: Since L is complete, @ is an isomorphism [K, Lemma 1.4] and hence ð½u; vÞ ðaÞÞL ¼ 0: This implies that a is a reduced norm from ½u; vÞL : Since ½L : K ¼ ½E : F ; a remains a parameter in L: Since a parameter is a reduced norm, it follows that ½u; % v%Þ splits over E: Thus a finite extension of F splits Q if and only if it splits ½u; % v%Þ: Since F is a global field, by the class field theory, it is easy to see that QC½u; % v%Þ (cf. Lemma 1.2 below). & Lemma 1.2. Let F be a global field of any characteristic and A and B be two quaternion algebras over F : Suppose that a finite extension L of F splits A if and only if it splits B; then ACB: Proof. Let a be a valuation of F and Fa the completion of F at a: By the local global principal, it is enough to show that A#Fa CB#Fa (cf. [CF, p. 185]). Suppose there is a valuation a of F such that A#Fa is non-trivial and B#Fa is trivial. Let dAF be such that d is a square in Fa and d is a non-square in Fb at all those finitely many pffiffiffi valuations of F where B is non-trivial (cf. [CF, p. 67]). Let L ¼ F ð d Þ: Then, by the local global principal, it follows that B#L is trivial. Since d is a square in Fa ; for any valuation b of L lying over a; we have Lb CFa : Since A#Fa is non-trivial, A#L is non-trivial. This is a contradiction. Therefore, A#Fa is non-trivial if and only if B#Fa is non-trivial. Since there is only one non-trivial quaternion algebra over all the completions of F (cf. [Sc, p. 353]), it follows that A and B are isomorphic over all the completions of F : &
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
84
Lemma 1.3. Let k; K and R be as above. Let aAH 3 ðK; Z=2ð2ÞÞ: Then there exist u; v; pAR such that u; v are units in R and a ½u; vÞ ðpÞ is unramified at R: Proof. Since F is either a number field or a dyadic field or a function field of a curve over a finite field, every element of H 2 ðF ; Z=2ð1ÞÞC2 BrðF Þ is represented by a symbol ½u; % v%Þ for some units u; vAR: In particular @ðaÞ ¼ ½u; % v%Þ for some units u and v in R: Let p be a parameter in R: Then by (1.1), we have @ða ½u; vÞ ðpÞÞ ¼ @ðaÞ ½u; % v%Þ ¼ 0: Hence a ½u; vÞ ðpÞ is unramified at R: & Let k and K be as above. Let X be a regular integral scheme with quotient field K: For iX0; let Xi denote the set of points of X of codimension i: For any xAX; let kðxÞ denote the residue field at x: For xAX1 ; let OX;x denote the discrete valuation ring at x and @x : H 3 ðK; Z=2ð2ÞÞ-H 2 ðkðxÞ; Z=2ð1ÞÞ be the residue homomorphism defined with respect to the discrete valuation ring OX;x : Let Y 3 Hnr ðK=SpecðXÞ; Z=2ð2ÞÞ ¼ kerðH 3 ðK; Z=2ð2ÞÞH 2 ðkðxÞ; Z=2ð1ÞÞÞ: xAX1
An element aAH ðK; Z=2ð2ÞÞ is called unramified at a point xAX1 ; if @x ðaÞ ¼ 0; otherwise it is called ramified at x: We say that a is unramified on X if it is unramified at 3 all points of X1 ; i.e., aAHnr ðK=X; Z=2ð2ÞÞ: We define the ramification divisor of a by X x: ramX a ¼ 3
@x ðaÞa0
For f AK ; we denote by SuppX ðf Þ the support of the principal divisor divX ðf Þ: Let Ok be the ring of integers of k (k a number field or a dyadic field). Let X be a smooth, integral, projective curve over k with K ¼ kðX Þ: Let ai AH 3 ðK; Z=2ð2ÞÞ and fj AK ; 1pipn; 1pjpm: By a result of Lipman on the resolution of singularities (cf. [S1, Proof of 2.1]), there exists a regular, projective model X of X over Ok and two regular curves C and E on X with only normal crossings (i.e., for every xAC-E the maximal ideal of the local ring OX;x is generated by local equations of C and E; such that [ [ SuppðramX ðai ÞÞ, SuppX ðfj ÞCSuppðC þ EÞ: 1pipn
1pjpm
We use this result through out this paper without further reference. Let K be any field of characteristic not equal to 2. Let W ðKÞ be the Witt group of quadratic forms over K and I n ðKÞ be the nth power of the fundamental ideal IðKÞ of W ðKÞ [Sc, Chapter 2].
2. Dyadic fields Lemma 2.1. Let F be a finite field of characteristic equal to 2 and Y a smooth, projective curve over F : Let bAH 2 ðF ðY Þ; Z=2ð1ÞÞC2 BrðF ðY ÞÞ and P1 ; y; Pn be the closed points of Y where b is ramified. Let E be a quadratic extension of F ðY Þ with
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
85
either E=F ðY Þ is inseparable or E=F ðY Þ is separable and for any discrete valuation ring S of E lying over the discrete valuation ring Ri at a point Pi ; S=Ri is unramified with the residue field of S is the quadratic extension of the residue field of Ri : Then b#E ¼ 0: Proof. Suppose that E=F ðY Þ is inseparable. Since E=F ðY Þ is a quadratic extension, pffiffiffi we have E ¼ F ðY Þð f Þ; for some f AF ðY Þ: Since every element in H 2 ðF ðY Þ; Z=2ð1ÞÞ is a symbol, we can write b ¼ ½a; bÞ; for some a; bAF ðY Þ: Since the characteristic of F ðY Þ is 2 and ½F ðY Þ : F ðY Þ2 ¼ 2; every element in F ðY Þ is a square in E: Therefore ½a; bÞ#E is split. Suppose that E is separable over F ðY Þ: By class field theory, it is enough to prove that b#E is unramified at each discrete valuation ring of E: Let S be a discrete valuation ring with E as its quotient field. Let R be the discrete valuation ring of F ðY Þ such that RCS: If b is unramified at R; then b is unramified at S: Suppose that b is ramified at R: Then R ¼ OY ;Pi for some i and the residue of b at Pi is the unique quadratic extension of the residue field at Pi : Since S is unramified over R; we have % where S% is the residue field of S: Since S% is the unique quadratic @S ðbÞ ¼ @R ðbÞ#S; % it follows that @S ðbÞ ¼ 0: Therefore, b is unramified at S: & extension of R; Lemma 2.2. Let R be a discrete valuation ring, F its quotient field and k its residue field. Assume that if charðkÞ ¼ 2; then k is perfect. If a quaternion algebra ½a; bÞ is unramified at R; then ½a; bÞ ¼ ½u; vÞ for some u; vAR : Proof. If charðkÞa2; then we know that if ½a; bÞ is unramified, then ½a; bÞ ¼ ðu; vÞ for 1 h14uw2
2 2 some u; vAR : We have ðu; vÞ ¼ ðuw ; vÞ ¼ 4 uw ; v ¼ 4 ; v for any wAR : 2
Since 2 is a unit in R; we can choose wAR such that 14uw is a unit in R: 4 Assume that charðkÞ ¼ 2: Suppose that ½a; bÞ is unramified at R: Suppose that ½a; bÞ#F Fˆ is split, where Fˆ is the completion of F : Then there exists lA½a; bÞ#F Fˆ with reduced trace and reduced norm equal to 1. Assume that½a; bÞ#Fˆ non-trivial. ˆ ˆ Since ½a; bÞ is unramified at R; ˆ D#k is a Let D be the maximal R-order in ½a; bÞ#F: quaternion algebra over k: Therefore, there exists a separable quadratic extension of k contained in D#k: By lifting this to D; we get an element lAD with reduced norm and trace units. Thus in both the cases there exists an element lA½a; bÞ#Fˆ whose reduced norm and trace are units. Since the reduced trace and reduced norm are polynomial functions, we can find mA½a; bÞ (which is close to l) such that the reduced trace and reduced norm are units in R: Then dividing m by its reduced trace, we can assume that m satisfies a quadratic equation X 2 þ X þ u for some unit u in R: Therefore ½a; bÞ ¼ ½u; cÞ for some cAR: Since ½u; cÞ ¼ ½u; cd 2 Þ for any dAK ; we can assume that the valuation of c is 0 or 1. If the valuation of c is 1, then ½a; bÞ is unramified if and only if the quadratic equation X 2 þ X þ u has solution over the residue field of R: From this it follows that a parameter p in R is a norm from the quadratic extension given by X 2 þ X þ u and hence ½u; cÞ ¼ ½u; pcÞ: Thus ½u; cÞ ¼ ½u; vÞ for some u; vAR : &
ARTICLE IN PRESS 86
V. Suresh / Journal of Number Theory 107 (2004) 80–94
Proposition 2.3. Let A be a regular local ring of dimension 2, K its quotient field and k its residue field. Suppose that k is finite and for every regular parameter p of A (i.e., A=ðpÞ is regular) the residue field kðpÞ at p is either a dyadic field or a function field of a curve over a finite field. Let aAH 3 ðK; Z=2ð2ÞÞ: (i) Suppose a is ramified only at p among the prime elements of A: If p is a regular parameter in A; then a ¼ a0 þ ½u; vÞ ðpÞ 3 ðK=SpecðAÞ; Z=2ð2ÞÞ and u; vAA : for some a0 AHnr (ii) Suppose a is ramified only at p and d among the prime elements of A: If p and d generate the maximal ideal m of A; then
a ¼ a1 þ a 2 ; 3 ðK=SpecðAÞ; Z=2ð2ÞÞ and a2 is a sum of symbols of the type where a1 AHnr
½u; vÞ ðpÞ;
½u; vÞ ðdÞ;
½u; dÞ ðpÞ;
u; v running over the units of A: Proof. Let a and p be as in (i). Since p is a regular parameter of A; there exists a prime element d in A such that the maximal ideal m of A is generated by p and d: We have a complex [K, Proposition 1.7] @
H 3 ðK; Z=2ð2ÞÞ !
M
@
H 2 ðkðxÞ; Z=2ð1ÞÞ ! H 1 ðk; Z=2Þ:
xASpecðAÞ1
Since kðpÞ is either a dyadic field or a function field of a curve over a finite field, there % bar denoting the image in A=ðpÞ: Since m is exist a; bAA such that @p ðaÞ ¼ ½a; % bÞ; generated by p and d; A=ðpÞ is a discrete valuation ring with d% as a parameter. Since a has residue only at p; @@ðaÞ ¼ @ð@p ðaÞÞ: Since @@ ¼ 0; @p ðaÞ is unramified at the discrete valuation ring A=ðpÞ: Thus, by (2.2), we have @p ðaÞ ¼ ½u; % v%Þ for some units u; vAA: Let a0 ¼ a ½u; vÞ ðpÞ: Then, by (1.7), a0 is unramified on A: Hence 3 ðK=SpecðAÞ; Z=2ð2ÞÞ and a ¼ a0 þ ½u; vÞ ðpÞ: a0 AHnr Now let a; p and d be as in (ii). As above, there exist a; bAkðpÞ ; such that @p ðaÞ is equal to ½a; bÞ: Suppose that ½a; bÞ is unramified at the discrete valuation ring A=ðpÞ: By (2.2), ½a; bÞ ¼ ½u; % v%Þ; for some u; vAA : Then a ½u; vÞ ðpÞ is as in the case (i) and we are done. Assume that ½a; bÞ is ramified at A=ðpÞ: Since A=m is a finite field, one can show that ½a; bÞ ¼ ½u; % vdÞ for some u; vAA : By (1.1), a ½u; vdÞ ðpÞ is ramified at most at ðdÞ: So by the case (i), we are done. & Remark 2.4. Suppose that in the above proposition, K is a function field in one variable over a dyadic field or a number field k; X a regular two-dimensional scheme over the ring of integers Ok and A the local ring at a codimension 2 point
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
87
of X: Then for every prime pAA; the residue field kðpÞ at p is either a dyadic field or a number field or a function field in one variable over a finite field. Therefore, every element in H 2 ðkðpÞ; Z=2ð1ÞÞ is represented by a symbol. Thus A satisfies the hypothesis of (2.3). Let k be a dyadic field, X a smooth, projective, integral curve over k and K ¼ kðX Þ: Let aAH 3 ðK; Z=2ð2ÞÞ: Then there exists X a regular, projective model of X over the ring Ok of integers in k; such that ramX ðaÞCC þ E; where C and E are regular curves on X such that C and E have only normal crossings. Let T ¼ C-E and B be the semi-local ring at T: Since X is regular, B is a regular semi-local ring and hence a unique factorization domain. We use this situation and notation in the rest of this section without further reference. Lemma 2.5. Let k; K and X be as above. Let x be a codimension 2 point of X and A ¼ OX;x : Let S be a discrete valuation ring which dominates A: Then every symbol of the type ½u; vÞ ðaÞ; with u; vAA and aAK ; is unramified at S: Proof. Let u; vAA ; aAK : By (1.1), we have @S ð½u; vÞ ðaÞÞ ¼ nS ðaÞ½u; % v%; bar denoting the image in the residue field of S and nS denoting the valuation of S: Since u; vAA and kðxÞ is a finite field, it follows that ½u; % v%Þ ¼ 0: Hence ½u; vÞ ðaÞ is unramified at S: & Lemma 2.6. Let k; K be as above. Let aAH 3 ðK; Z=2ð2ÞÞ: Let X; C and E be as above. Let L be an extension of K and S a discrete valuation ring with quotient field L: Suppose that there exists xAC-E such that S dominates OX;x : Suppose one of the following conditions holds: (i) the residue field of S contains the quadratic extension of kðxÞ; (ii) there exist local equations px ; dx for C and E; respectively, at x such that either px or dx or px dx is of the form wy2 ; yAS; wAS ; with the image of w in the residue field of S has its square class coming from kðxÞ ; (iii) there exist local equations px ; dx for C and E; respectively, at x such that the valuations of px and dx at S are even. Then aL is unramified at S: Proof. Let A ¼ OX;x : By (2.3), a ¼ a0 þ a00 ; where a0 is unramified on A and a00 is a sum of the symbols of the type ½u; vÞ ðpx Þ; ½u; vÞ ðdx Þ; ½u; px Þ ðdx Þ; with u; vAA : By Lemma 2.7 (below), the extension of a0 to L is unramified at S: Let nS denote the discrete valuation of S; @S denote the residue homomorphism at S and mS denote the maximal ideal of S: By (2.5), ½u; vÞ ðpx Þ; ½u; vÞ ðdx Þ are unramified at S: Suppose that the residue field of S contains a quadratic extension of kðxÞ: Then @S ð½u; px Þ ðdx ÞÞ contains the quadratic extension given by X 2 þ X þ u: % Since the
ARTICLE IN PRESS 88
V. Suresh / Journal of Number Theory 107 (2004) 80–94
unique quadratic extension of kðxÞ is contained in the residue field of S; X 2 þ X þ u% is reducible over the residue field of S: Therefore @S ðaL Þ ¼ 0: Suppose that px ¼ wy2 for some wAS such that w% ¼ ll21 with lAkðxÞ ; and yAS: Then ð½u; px Þ ðdx ÞÞL ¼ ð½u; wÞ ðdx ÞÞL : We have @S ð½u; px Þ ðdx ÞÞ ¼ nS ðdx Þ½u; % wÞ % ¼ nS ðdx Þ½u; % lÞ: Since u; % lAkðxÞ ; as before, it follows that ½u; % lÞ ¼ 0: Similarly, one can prove that if dx ¼ wy2 ; with w; y as above, then @S ð½u; px Þ ðdx ÞÞ ¼ 0: Suppose that px dx ¼ wy2 ; with w; y as above. Since ½u; px Þ ðdx Þ ¼ ½u; px dx Þ ðdx Þ; we have ð½u; px Þ ðdx ÞÞL ¼ ð½u; wÞ ðdx ÞÞL and @S ðð½u; px Þ ðdx ÞÞL Þ ¼ nS ðpx Þ½u; % wÞ % ¼ 0: Therefore a is unramified at S: Suppose that the valuation of px and dx are even. Then by (2.3), it is easy to see that a is unramified at S: & Lemma 2.7. Let k and K be as in (2.6). Let A be a regular local ring of dimension 2 with K as its quotient field. Let L be a finite extension of K and S a discrete valuation ring with quotient field L: Suppose S dominates A: Then the map H 3 ðK; Z=2ð2ÞÞ-H 3 ðL; Z=2ð2ÞÞ restricts to a map 3 3 Hnr ðK=SpecðAÞ; Z=2ð2ÞÞ-Hnr ðL=SpecðSÞ; Z=2ð2ÞÞ:
Proof (Parimala). Let aAH 3 ðKÞ: Since K is a function field in one variable over a padic field, we have I 4 ðKÞ ¼ 0 and e3 : I 3 ðKÞ-H 3 ðKÞ is an isomorphism [AEJ, Corollary 4 and Theorem 2]. Let q0 AI 3 ðKÞ be such that e3 ðq0 Þ ¼ a: Let q be an anisotropic quadratic form over K which represents q0 in I 3 ðKÞ: Let P be any height one prime ideal of A and R ¼ AP : Since a is unramified at R; by [K, Lemma 1.4 (3)], ˆ ¼ 0: Since e3 is an isomorphism, q0 #Kˆ ¼ 0: Hence q#Kˆ is hyperbolic. e3 ðq0 #KÞ ˆ the two forms q over K ¼ ðAP Þ (p a uniformizing Since q is hyperbolic over K; p P xi yi over A#P patch over the Kˆ and give a quadratic form QP parameter in AP ) and over AP such that QP #KCq [CO]. Now, by the general non-sense of the torsors over two-dimensional regular rings [CS], q comes from a quadratic form Q over A: Therefore aL ¼ e3 ðQ#S#L) and hence unramified at S: & Lemma 2.8. With the notation as above, let L be a quadratic extension of K: Let S be a discrete valuation ring with L as its quotient field. Assume that S-K ¼ BðpÞ ; where p is a prime element in B giving a local equation for a component C1 of C: If C1 -Ea|; let C1 -E ¼ fx1 ; y; xr g and dxi a local equation of E at xi ; 1pipr: Suppose that either pffiffiffi C1 -E ¼ | or L ¼ Kð f Þ with f AB satisfying one of the following conditions: (i) f is a parameter in BðpÞ ; (ii) f is a unit in BðpÞ and f% is not a square in B=ðpÞ; (iii) Let Bi be the local ring at ðp; dxi Þ and B* i is the integral closure of Bi in L: Then B* i =ðp; dxi Þ contains the quadratic extension of Bi =ðp; dxi Þ: Then aL is unramified at S:
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
89
Proof. Let A ¼ BðpÞ : Then the residue field kðpÞ of A is the quotient field of B=ðpÞ: Since ramX aCC þ E and C1 is regular curve on X; it follows from the complex [K, 1.7] @
H 3 ðK; Z=2ð2ÞÞ !
M ZAX
1
@
H 2 ðkðZÞ; Z=2ð1ÞÞ !
M yAX
H 1 ðkðyÞ; Z=2Þ
2
that @ðpÞ ðaÞ is possibly ramified only at the discrete valuations of kðpÞ corresponding to C1 -E: Suppose that C1 -E ¼ |: Then it follows that @ðpÞ ðaÞ is unramified at every discrete valuation ring of kðpÞ: Since kðpÞ is either a global field of positive characteristic (so that there are no archimedean primes) or a local field, by class field theory, we have @ðpÞ ðaÞ ¼ 0 and hence aL is unramified at S: Suppose that C1 -Ea|: Suppose that f is a parameter in A: By (1.3), a ½u; vÞ ðf Þ is unramified at A for some units u; vAA: Since f is a square in L; ½u; vÞ ðf Þ is trivial over L and hence aL is unramified at S: Assume that f is as in (ii) or (iii). Suppose kðpÞ is a p-adic field. Then the residue field k of S is a quadratic extension of kðpÞ and @S ðaL Þ ¼ @p ðaÞ#k: Since the unique quaternion division algebra over kðpÞ is split over any quadratic extension of kðpÞ (cf. [Sc, p. 353]), we have @S ðaS Þ ¼ @p ðaÞ#k ¼ 0: If kðpÞ is a function field of a curve over a finite field, then by (2.1), it follows that a is unramified at S: & Lemma 2.9. Let A be regular local ring of dimension 2 with quotient field K and with a finite residue field k: Let m be the maximal ideal of A and a; bAA units. Then there exist p; dAm; positive integer i and u0 ; v0 ; w0 AA=ðp2iþ1 ; dÞ such that m ¼ ðp; dÞ and pffiffiffi % 2 abw2 modulo ðp2iþ1 ; dÞÞ; if L ¼ Kð f Þ and A˜ for any f AA with f% ¼ au % 2 þ bv ˜ the integral closure of A in L; then A=m contains the unique quadratic extension of A=m ¼ k: Proof. Since only finitely many prime elements of A divide 2 and there are infinitely many prime elements in m\m2 ; we can write m ¼ ðp; dÞ; where d does not divide 2: pffiffiffi Then A1 ¼ A=ðdÞ is a dyadic ring with p% a parameter. Let lAA1 be such that A1 ½ l is the unramified extension of A1 : The square class of l is uniquely determined by its image modulo 4p: % Since the discriminant l of the quadratic form /a; b; ab; lS is not a square in A1 ; /a; b; ab; lS is isotropic (cf. [Sc, p. 187]). Thus /a; b; abS represents l over A1 : In particular there exists u0 ; v0 ; w0 AA1 =ð4pÞ such that l ¼ au20 þ bv20 abw20 in A1 =ð4pÞ: Let i be the valuation of 2 in A1 : Then ð4p; dÞ ¼ ðp2iþ1 ; dÞ: Let f AA be any unit such that f ¼ au20 þ bv20 abw20 modulo ðp2iþ1 ; dÞ: pffiffiffi Then it is easy to see that the quadratic extension Kð f Þ has the required property mentioned in the lemma. & Theorem 2.10. Let k be a dyadic field, X a smooth, projective, irreducible curve over k: Let K ¼ kðX Þ and ai AH 3 ðK; Z=2ð2ÞÞ; 1pipn: Let a; bAK : Then there exists f AK pffiffiffi which is a value of the quadratic form /a; b; abS such that ai #Kð f Þ ¼ 0 for 1pipn:
ARTICLE IN PRESS 90
V. Suresh / Journal of Number Theory 107 (2004) 80–94
Proof. Let X be a regular, projective model of X over Ok such that [ SuppðramX ðai ÞÞCSuppðC þ EÞ; Suppð2Þ,SuppðaÞ,SuppðbÞ, 1pipn
where C and E are regular curves on X with only normal crossings. Let T ¼ C-E: Let B be the semi-local ring at T: For xAT; let px ; dx AB be local equations for C and E at x; respectively. Since B is a unique factorization domain with quotient field K; without loss of generality, we assume that a; b are square free in B and SuppSpecðBÞ ðabÞCSuppðC þ EÞ: Let cAB be the greatest common divisor of a and b; so that a ¼ ca0 ; b ¼ cb0 ; with a0 ; b0 AB: Since a and b are square free, c; a0 ; b0 are pairwise coprime. For xAT; choose ux ; vx ; wx as follows: (i) Suppose cðxÞ ¼ 0: Let mx denote the maximal ideal of B at x: Since c; a0 ; b0 are pairwise coprime and the only prime elements of Bmx which divide ca0 b0 are px ; dx ; at least one of a0 and b0 is coprime with px and dx ; and hence is a unit at x: Thus a0 ðxÞa0 or b0 ðxÞa0: Let ux ; vx AkðxÞ be such that a0 ðxÞu2x þ b0 ðxÞv2x a0: Let wx ¼ 0: (ii) Suppose that cðxÞa0 and a0 b0 ðxÞ ¼ 0: Let ux ¼ vx ¼ 1 and wx ¼ 0: (iii) Suppose that cðxÞa0 ðxÞb0 ðxÞa0: Then a; b are units in Bmx : Let ix ¼ i; p0x ¼ 2
2
p; d0x ¼ d and ux ¼ v0 =a0 ; vx ¼ u0 =b0 ; wx ¼ w0 ABmx =ðp2iþ1 ; dÞ be as in (2.9). Let u; v; wAB be such that for any xAT; if x as in (i) or (ii) uðxÞ ¼ ux ; vðxÞ ¼ vx and wðxÞ ¼ wx ; and if x is as in (iii), u ¼ ux ; v ¼ vx ; w ¼ wx modulo ðpx2ix þ1 ; d0x Þ: Let f ¼ ca0 b0 ða0 u2 þ b0 v2 cw2 Þ: Clearly f is a value of ca0 b0 /a0 ; b0 ; cS ¼ /a; b; abS: pffiffiffi pffiffiffi We now show that a#Kð f Þ ¼ 0: Let L ¼ Kð f Þ and k0 be the field of constants of L: Let X 0 be a smooth, projective, irreducible curve over k0 with k0 ðX 0 Þ ¼ L: Let X0 be a regular proper model of X 0 over Ok0 and yAX0 be a point of codimension one. Let S ¼ OX0 ;y be the discrete valuation ring at y: Since X is projective over Ok ; there exists a point zAX of codimension 1 or 2, such that S dominates the local ring A ¼ OX;z : Fix i; 1pipn and let a ¼ ai : Suppose dimðAÞ ¼ 1: Then A is a discrete valuation ring. Suppose that z does not correspond to a component of C or E: Then a is unramified at A and hence aL is unramified at S: Let z correspond to a component C1 of C: The case where z corresponds to a component of E is similar. Suppose that C1 -E ¼ |: Then by (2.8), aL is unramified at S: Suppose that C1 -Ea|: Let p be a prime element of B corresponding to the component C1 : Then A ¼ BðpÞ : Since c; a0 ; b0 are pairwise coprime in B; it follows that at most one of c; a0 ; b0 is divisible by p: Suppose p divides c: Let xAC1 -E: Then by (i), a0 u2 þ b0 v2 cw2 is a unit in OX;x : Since A is a localization of OX;x ; a0 u2 þ b0 v2 cw2 is a unit in A: Further, since p divides c; both a0 and b0 are units in A: Therefore f is a parameter in A and hence by (2.8), aL is unramified at S: Suppose p does not divide c and divides a0 or b0 : Let xAC1 -E: If cðxÞ ¼ 0; then by (i), a0 u2 þ b0 v2 cw2 is a unit at x and hence it is a unit in A: Assume that cðxÞa0:
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
91
Suppose p divides a0 (the other case is similar). If b0 ðxÞa0; then by (ii), ux ¼ 1 and b0 ðxÞv2x cwx ¼ b0 ðxÞa and hence a0 u2 þ b0 v2 cw2 is a unit in A: If b0 ðxÞ ¼ 0; since a0 ; b0 ; c0 are pairwise coprime, we have a0 ¼ a00 p and b0 ¼ b00 dx for some units a00 ; b00 in Bmx : Since wx ¼ 0; write w ¼ w1 p þ w2 dx for some w1 ; w2 in Bmx : We have b0 v2 cw2 ¼ dx ððb00 v2 Þ cw22 dx Þ pðw21 p þ 2w1 w2 dx Þ: Since vx ¼ 1; it follows that b00 v2 cw22 dx is a unit in Bmx and hence a0 u2 þ b0 v2 cw2 is a unit in BðpÞ ¼ A: Therefore, as above, f is a parameter in A and aL is unramified at S: Suppose that p does not divide ca0 b0 : Let xAC1 -E: Suppose that dx divides ab: If dx divides a and b; then dx divides c and does not divide a0 b0 : If dx divides only one of a and b; then dx divides only one of a0 ; b0 and does not divide c: By (i), a0 u2 þ b0 v2 cw2 is a unit. In particular f is a unit at px and not a square modulo px : Therefore by (2.8), a is unramified at S: Suppose that dx does not divide ab: Then a and b are units at x: Let C1 -E ¼ fx ¼ x1 ; y; xr g: Suppose that dxi divides ab for some xi ; then as above, it follows that f is a unit at px and not a square modulo px : Thus, by (2.8), a is unramified at S: Suppose that dxi does not divide ab for all i: Then, by the choice of u; v; w; it follows that f is as in (2.8, (iii)) and hence by (2.8), a is unramified at S: Suppose dimðAÞ ¼ 2: Then z is a closed point of X: If zeC,E; then a is unramified on A and hence unramified at S (2.7). Assume that zAC,E: If zeC-E; then by (2.3 and 2.5), aL is unramified at S: Suppose that zAC-E: Then A ¼ Bmz ; where mz is the maximal ideal of B at z: Suppose that cðzÞ ¼ 0: Then, by the choice of u; v; w; a0 u2 þ b0 v2 cw2 is a unit at z: Since the only prime elements of A which divide ca0 b0 are pz ; dz and c; a0 ; b0 are pairwise coprime, f ¼ ca0 b0 ða0 u2 þ b0 v2 cw2 Þ is of the form w0 pz or w0 dz or w0 pz dz ; with w0 AA : Since f AL2 ; by (2.6), aL is unramified at S: Suppose that cðzÞa0 and a0 ðzÞb0 ðzÞ ¼ 0: If a0 ðzÞ or b0 ðzÞ is not zero, then, as above, one shows that either pz or dz or pz dz is as in (2.6, (ii)) and hence, by (2.6), aL is unramified at S: Suppose that a0 ðzÞ ¼ b0 ðzÞ ¼ 0: Since the only prime elements of A which divide a0 ; b0 are pz ; dz and a0 ; b0 are coprime and non-units at z; we have a0 ¼ w0 pz and b0 ¼ w00 dz or a0 ¼ w0 dz and b0 ¼ w00 pz for some w0 ; w00 AA : Consider the case where a0 ¼ w0 pz and b0 ¼ w00 dz ; with w0 ; w00 AA (the other case being similar). Let nS denote the valuation at S: Since S dominates A; we have nS ða0 ÞX1 and nS ðb0 ÞX1: We assume without loss of generality that nS ða0 ÞpnS ðb0 Þ: Then, b0 =a0 AS: Since wAm and a0 em2 ; w2 =a0 is in the maximal ideal of S: We have
0
2 b w f ¼ cb0 ðu2 Þ þ 0 v2 c 0 ða0 Þ2 : a a 0
2
Suppose that nS ða0 ÞonS ðb0 Þ: Then u2 þ ba0 v2 c wa0 AS : Since b0 ¼ w00 dz ; w00 AA ; cAA and f AL2 ; it follows that dz is as in (2.6, (ii)) and aL is unramified at S: 0 2 Suppose that nS ða0 Þ ¼ nS ðb0 Þ: If u2 þ ba0 v2 c wa0 AS ; then nS ðf Þ ¼ nS ðb0 Þ þ 2nS ða0 Þ ¼ 3nS ðb0 Þ: Since nS ðf Þ is even, it follows that nS ða0 Þ ¼ nS ðb0 Þ is even. In particular nS ðpz Þ ¼ nS ðdz Þ is even. Therefore by (2.6, (iii)), a is unramified at S: Assume that 0 2 u2 þ ba0 v2 c wa0 is not a unit in S: Let n ¼ nS ða0 Þ ¼ nS ðb0 Þ: Let y be a parameter in S and write a0 ¼ w1 yn ; b0 ¼ w2 yn ; with w1 ; w2 AS : By (ii), u; vAA and wAmx : Since
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
92 0
2
2
2
u2 þ ba0 v2 c wa0 ¼ u2 þ ww21 v2 c wa0 is not a unit in S and c wa0 is also not a unit in S; we have u% 2 w2 ¼ : v%2 w1 In particular, the image of w1 w2 in the residue field of S has its square class coming from kðzÞ: We have ww0 pz dz ¼ a0 b0 ¼ w1 w2 y2n : Therefore pz dz as in (2.6. (ii)). Therefore by (2.6), a is unramified at S: Suppose that cðzÞa0 ðzÞb0 ðzÞa0: Then by the choice of u; v; w it follows that f is as in (2.8. (iii)). Therefore, by (2.8), a is unramified at S: This implies 3 3 that a#LAHnr ðL=X0 ; Z=2ð2ÞÞ: Since Hnr ðL=X0 ; Z=2ð2ÞÞ ¼ 0 by a result of Kato [K, Proposition 5.2], we have a#L ¼ 0: This completes the proof of the theorem. & 3. Number fields Theorem 3.1. Let k be a number field and K a function field of a curve over k: Let a1 ; y; an AH 3 ðK; Z=2ð2ÞÞ and a; b; dAK : Then there exists f AK such that (1) f is a value of /a; b; abdS; (2) for every finite non-dyadic place v of k; ðai Þv ¼ ðf Þ,bi for some bi AH 2 ðKv ; Z=2ð1ÞÞ; (3) for every dyadic place v of k; where d is a square in Kv ; ðai Þv ¼ ðf Þ,bi for some bi AH 2 ðKv ; Z=2ð1ÞÞ: Proof. Let Ok be the ring of integers of k: Let X be a smooth, integral, projective curve over k and K ¼ kðX Þ: Let X be a regular, projective model of X over Ok with SuppX ð2Þ,SuppX ðaÞ,SuppX ðbÞ,SuppX ðdÞ,
n [
SuppðramX ai ÞCSuppðC þ EÞ;
i¼1
where C and E are regular curves on X with only normal crossings. Let T be the finite set of closed points of X containing C-E and one closed points from each component of C and E: Let B be the semi-local ring at T: For xAT; let px ; dx AB be local equations for C and E at x; respectively. Since B is a unique factorization domain with quotient field K; without loss of generality, we assume that a; b; d are square free in B and SuppSpecðBÞ ðabdÞCSuppðC þ EÞ: Let cAB be the greatest common divisor of a and b; so that a ¼ ca0 ; b ¼ cb0 ; with a0 ; b0 AB: Since a and b are square free, c; a0 ; b0 are pairwise coprime. For xAT; choose ux ; vx ; wx as follows: (i) Suppose cðxÞ ¼ 0: Let mx denote the maximal ideal of B at x: Since c; a0 ; b0 are pairwise coprime and the only prime elements of Bmx which divide ca0 b0 are px ; dx ; at least one of a0 and b0 is coprime with px and dx ; and hence is a unit at x: Thus a0 ðxÞa0 or b0 ðxÞa0: Let ux ; vx AkðxÞ be such that a0 ðxÞu2x þ b0 ðxÞv2x a0: Let wx ¼ 0:
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
93
(ii) Suppose that cðxÞa0 and a0 b0 ðxÞ ¼ 0: Let ux ¼ vx ¼ 1 and wx ¼ 0: (iii) Suppose that cðxÞa0 ðxÞb0 ðxÞa0 and charðkðxÞÞa2: Then every element of kðxÞ is represented by the quadratic form /a0 ðxÞ; b0 ðxÞS: Let ux ; vx AkðxÞ be such that cðxÞa0 ðxÞb0 ðxÞða0 ðxÞu2x þ b0 ðxÞv2x ÞekðxÞ2 : Let wx ¼ 0: (iv) Suppose that cðxÞa0 ðxÞb0 ðxÞa0 and charðkðxÞÞ ¼ 2: Let A ¼ Bmx : As in (2.9), let p0x ; d0x Amx be such that p0x does not divide 2. Let A1 ¼ A=ðp0x Þ: Then A1 is a ring of integers in a number field. For each dyadic place n of A1 with d a square in pffiffiffiffiffi A1n ; let ln AA1n be such that A1n ½ ln is unramified. For each such dyadic place % 2xn abdw2xn (cf. proof of 2.9). Let n; let uxn ; vxn ; wxn be such that ln ¼ au % 2xn þ bv ux ; vx ; wx AA1 which are very close to uxn ; vxn ; wxn : Let ix be the maximum positive integer among the positive integers given in (2.9) corresponding to the completions at dyadic places n: Let u; v; wAB be such that for any xAT; if x as in (i) or (ii) or (iii), uðxÞ ¼ ux ; vðxÞ ¼ vx and wðxÞ ¼ wx and if x is as in (iv), u ¼ ux =a% 2 ; v ¼ vx =b%2 ; w ¼ wx modulo ðpx02ix þ1 ; d0x Þ: Let f ¼ ca0 b0 ða0 u2 þ b0 v2 cdw2 Þ: Clearly f is a value of ca0 b0 /a0 ; b0 ; cdS ¼ /a; b; abdS: Let n be any finite place of k: Xn ¼ X kn ; Xn ¼ X On ; Cn ¼ C On and En ¼ E On ; where kn is the completion of k at n and On is the ring of integers of kn : If kn is non-dyadic, as in the proof of Proposition 4.4 in [PS] pffiffiffi and (2.9), it follows that ðai Þn #kn ðXn Þð f Þ ¼ 0: Therefore for every finite nondyadic place v of k; we have ðai Þn ¼ ðf Þ,ðbi Þ for some bi AH 2 ðKn ; Z=2ð1ÞÞ: Let n be any dyadic place with d a square in Kv : Then we have /a; b; abdS ¼ /a; b; abS: pffiffiffi As in the proof of (2.10), it follows that ðai Þn #kn ðXn Þð f Þ ¼ 0: This completes the proof. & Theorem 3.2. Let k be a number field and K a function field of a curve over k: Let a1 ; y; an AH 3 ðK; Z=2ð2ÞÞ: Then there exist f AK and bi AH 2 ðK; Z=2ð1ÞÞ such that ai ¼ ðf Þ,bi : Proof. Let a ¼ b ¼ d ¼ 1: Let f be a value of / 1; 1; 1S as in (3.1). Let n pffiffiffi be a real place of k: Since f is a sum of squares, by [Se], cd2 ðKn ð f ÞÞ ¼ 1: pffiffiffi pffiffiffi Therefore H 3 ðKn ð f Þ; Z=2ð2ÞÞ ¼ 0: In particular, ðai Þn #Kn ð f Þ ¼ 0: Since d ¼ 1 is pffiffiffi a square, by (3.1), if n is a finite place of k; then ai #Kv ð f Þ ¼ 0: Thus, by [K, pffiffiffi Theorem 0.8], ai #Kð f Þ ¼ 0 and hence there exist bi AH 2 ðK; Z=2ð1ÞÞ such that ai ¼ ðf Þ,bi : &
Acknowledgments I thank R. Parimala, A. Wadsworth and J.-L. Colliot-The´le`ne for various helpful discussions during the preparation of this paper. I also thank the referee for many suggestions.
ARTICLE IN PRESS 94
V. Suresh / Journal of Number Theory 107 (2004) 80–94
References [AEJ] J.K. Arason, R. Elman, B. Jacob, Fields of cohomological 2-dimension three, Math. Ann. 274 (1986) 649–657. [CF] J.W.S. Cassels, A. FrO¨hlich, Algebraic Number Theory, Academic Press, London, 1967. [CO] J.-L. Colliot-The´le`ne, M. Ojanguren, Espaces principaux homoge`nes localement triviaux, Inst. Hautes Etudes Sci. Publ. Math. 75 (1992) 97–122. [CS] J.-L. Colliot-The´le`ne, J.-J. Sansuc, Fibre´s quadratiques et composantes connexes re´elles, Math. Ann. 244 (1979) 105–134. [G] M.J. Greenburg, Lectures on Forms in Many Variables, W.A. Benjamin, Inc., New York, Amsterdam, 1969. [K] K. Kato, A Hasse principle for two-dimensional global fields, J. Reine Angew. Math. 366 (1986) 142–181. [PP] R. Parimala, R. Preeti, Hasse principle for classical groups over function field of curves over number fields, J. Number Theory 101 (2003) 151–184. [PS] R. Parimala, V. Suresh, Isotropy of quadratic forms over function fields of p-adic curves, Inst. Hautes Etudes Sci. Publ. Math. 88 (1998) 129–150. [S1] D.J. Saltman, Division algebras over p-adic curves, J. Ramanujan Math. Soc. 12 (1) (1997) 25–47. [S2] D.J. Saltman, Correction to: division algebras over p-adic curves, J. Ramanujan Math. Soc. 13 (2) (1998) 125–129. [Sc] W. Scharlau, Quadratic and Hermitian Forms, Grundlehren Mathematischen Wissenschaften, Vol. 270, Springer-Verlag, Berlin–Heidelberg–New York, 1985. [Se] J.-P. Serre, Cohomologie Galoisienne, in: Lecture Notes in Mathematics, Vol. 5, Springer, Berlin, 1964, 1994.
Journal of Number Theory 107 (2004) 80–94
http://www.elsevier.com/locate/jnt
Galois cohomology in degree 3 of function fields of curves over number fields V. Suresh Department of Mathematics and Statistics, University of Hyderabad, GachiBowli, P.O. Central University, Hyderabad 500 046, India Received 10 February 2003 Communicated by J.-L. Colliot-The´le`ne
Abstract Let k be a field of characteristic not equal to 2. For nZ1; let H n ðk; Z=2Þ denote the nth Galois Cohomology group. The classical Tate’s lemma asserts that if k is a number field then given finitely many elements a1 ; ?; an AH 2 ðk; Z=2Þ; there exist a; b1 ; ?; bn Ak such that ai ¼ ðaÞ,ðbi Þ; where for any lAk ; ðlÞ denotes the image of k in H 1 ðk; Z=2Þ: In this paper we prove a higher dimensional analogue of the Tate’s lemma. r 2004 Elsevier Inc. All rights reserved. Keywords: Galois cohomology; Number fields; Function fields of curves
Introduction Let k be a field of characteristic not equal to 2. For nX1; let H n ðk; Z=2Þ denote the nth Galois cohomology group. We have H 1 ðk; Z=2Þ ¼ k =k2 ; where k is the group of non-zero elements of k: The structure of these groups is well understood when k is a number field or a local field. For example, if k is a number field or a local field then every element in H n ðk; Z=2Þ is of the form a1 ,?,an with ai AH 1 ðk; Z=2Þ: Further, if k is a totally imaginary number field or a p-adic field, then H n ðk; Z=2Þ ¼ 0 for nX3 (cf. [Se, p. 90, 108]). If k is a p-adic field, it is well known that H 2 ðk; Z=2Þ ¼ Z=2 and the non-trivial element of H 2 ðk; Z=2Þ becomes trivial over every quadratic field extension of k (cf. [CF, p. 131]). The classical Tate’s lemma asserts that if k is a
Fax: +91-40-2301-1469. E-mail address: [email protected].
0022-314X/$ - see front matter r 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jnt.2004.01.002
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
81
number field then given finitely many elements a1 ; y; an AH 2 ðk; Z=2Þ; there exist a; b1 ; y; bn Ak such that ai ¼ ðaÞ,ðbi Þ; where for any lAk ; ðlÞ denotes the image of k in H 1 ðk; Z=2Þ: Saltman [S1] proved that if K is a function field of a curve over a p-adic field, pa2; then every element in H 2 ðK; Z=2Þ is a sum of two symbols, i.e., of the form ða1 Þ,ðb1 Þ þ ða2 Þ,ðb2 Þ; where a1 ; a2 ; b1 ; b2 AK (cf. [S2]). Using this result, in [PS], it was proved that every element in H 3 ðK; Z=2Þ is a symbol, i.e. of the form ðaÞ,ðbÞ,ðcÞ for some a; b; cAK : Let K be a function field of a curve over a dyadic field or a number field. It is not known whether there exists a number N such that every element of H 2 ðK; Z=2Þ or H 3 ðK; Z=2Þ is a sum of at most N symbols. The aim of this paper is to prove the following two-dimensional analogue of the classical Tate’s lemma. Theorem. Let k be a p-adic field or a number field and K a function field of a curve over k: Let a1 ; y; an AH 3 ðK; Z=2Þ: Then there exist lAK and bi AH 2 ðK; Z=2Þ such that ai ¼ ðlÞ,bi : In fact we prove the following stronger results, which are used by Parimala and Preeti [PP] in the study of Hasse principle for classical groups over function fields of curves over number fields. Theorem 1. Let k be a dyadic field and K a function field of a curve over k: Let a1 ; y; an AH 3 ðK; Z=2Þ and a; bAK : Then there exist f AK and 2 b1 ; y; bn AH ðK; Z=2Þ such that f is a value of the quadratic form /a; b; abS over K and ai ¼ ðf Þ,bi : Theorem 2. Let k be a number field and K a function field of a curve over k: For a1 ; y; an AH 3 ðK; Z=2Þ and a; b; dAK ; there exists f AK such that (1) f is a value of the quadratic form /a; b; abdS; (2) for every finite non-dyadic place v of k; there exist b1v ; y; bnv AH 3 ðKv ; Z=2Þ such that aiv ¼ ðf Þ,biv ; where for any place v of k; kv is the completion of k at v; Kv ¼ K#k kv and aiv ¼ ai #Kv ; (3) for every finite dyadic place v of k; if d is a square in Kv ; there exist b1v ; y; bnv AH 3 ðKv ; Z=2Þ such that aiv ¼ ðf Þ,biv :
1. Some preliminaries Let F be a field and F the group of non-zero elements of F : By a quaternion algebra over F we mean a central simple algebra of dimension 4 over F : For a; bAF ; let ½a; bÞ denote the quaternion algebra over F generated by i and j satisfying i2 þ i þ a ¼ 0; ij þ ji ¼ j and j 2 ¼ b: It is well known that every quaternion algebra over F is of the form ½a; bÞ for some a; bAF (cf. [Sc, p. 313]). Suppose that the
ARTICLE IN PRESS 82
V. Suresh / Journal of Number Theory 107 (2004) 80–94
characteristic of F is not equal to 2. For a; bAF ; let ða; bÞ denote the quaternion algebra generated by x and y satisfying x2 ¼ a; y2 ¼ b and xy þ 1yx ¼ 0: For a; bAF ; we have isomorphisms of quaternion algebras ða; bÞC 4 a; b and ½a; bÞC 14 a; b : Let BrðF Þ denote the Brauer group of F and 2 BrðF Þ the subgroup of BrðF Þ consisting of 2-torsion elements. Let R be a discrete valuation ring with quotient field F : Let aABrðF Þ: We say that a is unramified at R if a comes from an Azumaya algebra over R: For iX0 and nX0; let H n ðF ; Z=2ðiÞÞ be the cohomology groups defined as in [K, Section 0]. For any field we have [K, Section 0] H 1 ðF ; Z=2ÞCHomcont ðGalðF ab =F Þ; Z=2Þ; H 2 ðF ; Z=2ð1ÞÞC2 BrðF Þ: We use these isomorphisms and identify H 1 ðF ; Z=2Þ with Homcont ðGalðF ab =F Þ; Z=2Þ and H 2 ðF ; Z=2ð1ÞÞ with 2 BrðF Þ: Let k be a dyadic field (a finite extension of Q2 ) or a number field and K a function field of a curve over k: Let R be a discrete valuation ring with the quotient field K and the residue field F : Then either F is a number field or a dyadic field or a function field of a curve over a finite field. If F is a function field of a curve over a finite field of characteristic 2, then we have ½F : F 2 ¼ 2: Let S be a discrete valuation ring with the quotient field F and residue field k: Then k is a finite field. If the characteristic of k is 2, then ½k : k2 ¼ 1: Therefore, we have residue maps [K, Section 1] @ : H 3 ðK; Z=2ð2ÞÞ-H 2 ðF ; Z=2ð1ÞÞ; @ : H 2 ðF ; Z=2ð1ÞÞ-H 1 ðk; Z=2Þ: Note that if F is a local field, then @ : H 2 ðF ; Z=2ð1ÞÞ-H 1 ðk; Z=2Þ is the classical invariant for local fields. We say that an element a in H 3 ðK; Z=2ð2ÞÞ (resp. H 2 ðF ; Z=2ð1ÞÞ) is unramified at R (resp. S) if @ðaÞ ¼ 0: Lemma 1.1. Let k; K; R and F as above with F a function field of a curve. Let u; v; aAK be such that u; v are units in R: Then we have @ð½u; vÞ,ðaÞÞ ¼ nðaÞ½u; % v%Þ; where bar denotes the image in F and n is the valuation at R: Proof. We can replace K by its completion and assume that R and K are complete. Let m be the maximal ideal of R: Suppose that vðaÞ is even. Then, without loss of generality, we assume that a is a unit in R: We need to show that @ð½u; vÞ ðaÞÞ ¼ 0: If v% is a square in F ; then ½u; % v%Þ ¼ 0: Since R is complete, ½u; vÞ ¼ 0 and hence ½u; vÞ ðaÞ ¼ 0: Assume that v% is
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
83
pffiffiffi not a square in F : Since every element of F is a square in F ð v%Þ; there exist a0 ; a2 AR such that a20 a22 v ¼ a modulo m: If a0 Am; then ½u; vÞ and ½u; aÞ are isomorphic modulo m: Since R is complete ½u; vÞ and ½u; aÞ are isomorphic. In particular a is a reduced norm from ½u; vÞ and hence ½u; vÞ ðaÞ ¼ 0: Assume that a0 em: Let f ðX Þ ¼ uX 2 a0 X þ a20 a22 v aAR½X : Then f au0 ¼ a20 a22 v aAm and f 0 au0 D / 0 modulo m; where f 0 ðX Þ is the derivative of f ðX Þ: Since R is complete, by a0 D Hensel’s Lemma [G, p. 47], there exists a1 AR such that f ða1 Þ ¼ 0: Let x; yA½u; vÞ be such that x2 þ x þ u ¼ 0; y2 ¼ v and yx ¼ xy y: Then it is easy to see that the reduced norm of z ¼ a0 þ a1 x þ a2 yA½u; vÞ is a20 þ ua21 va22 a0 a1 : Since f ða1 Þ ¼ 0; we see that the reduced norm of z is a: Hence ½u; vÞ ðaÞ ¼ 0: In particular @ð½u; vÞ ðaÞÞ ¼ 0: Suppose that vðaÞ is odd. Then, without loss of generality, we assume that a is a parameter in R: Let @ð½u; vÞ ðaÞÞ ¼ bAH 2 ðF ; Z=2ð1ÞÞ: Since F is a function field of a curve over a finite field, b is represented by a quaternion algebra Q over F : We need to show that QC½u; % v%Þ: Let E be a finite extension of F which splits ½u; % v%Þ: Let L be a finite extension of K such that E is the residue field of the discrete valuation of L extending the discrete valuation of K (if ECF ½X =ðf ðX Þ for some monic polynomial f ðX ÞAK½X ; let L ¼ K½X =ðf ðX ÞÞ and use induction). Since ½u; % v%Þ splits over E and L is complete, it follows that ½u; vÞ splits over L: In particular bE ¼ @ðð½u; vÞ ðaÞÞL Þ ¼ 0: Hence Q is split over E: Now let E be a finite extension of F which splits Q and L a finite extension of K lifting E as above with ½L : K ¼ ½E : F : Then @ðð½u; vÞ ðaÞÞL Þ ¼ bL ¼ 0: Since L is complete, @ is an isomorphism [K, Lemma 1.4] and hence ð½u; vÞ ðaÞÞL ¼ 0: This implies that a is a reduced norm from ½u; vÞL : Since ½L : K ¼ ½E : F ; a remains a parameter in L: Since a parameter is a reduced norm, it follows that ½u; % v%Þ splits over E: Thus a finite extension of F splits Q if and only if it splits ½u; % v%Þ: Since F is a global field, by the class field theory, it is easy to see that QC½u; % v%Þ (cf. Lemma 1.2 below). & Lemma 1.2. Let F be a global field of any characteristic and A and B be two quaternion algebras over F : Suppose that a finite extension L of F splits A if and only if it splits B; then ACB: Proof. Let a be a valuation of F and Fa the completion of F at a: By the local global principal, it is enough to show that A#Fa CB#Fa (cf. [CF, p. 185]). Suppose there is a valuation a of F such that A#Fa is non-trivial and B#Fa is trivial. Let dAF be such that d is a square in Fa and d is a non-square in Fb at all those finitely many pffiffiffi valuations of F where B is non-trivial (cf. [CF, p. 67]). Let L ¼ F ð d Þ: Then, by the local global principal, it follows that B#L is trivial. Since d is a square in Fa ; for any valuation b of L lying over a; we have Lb CFa : Since A#Fa is non-trivial, A#L is non-trivial. This is a contradiction. Therefore, A#Fa is non-trivial if and only if B#Fa is non-trivial. Since there is only one non-trivial quaternion algebra over all the completions of F (cf. [Sc, p. 353]), it follows that A and B are isomorphic over all the completions of F : &
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
84
Lemma 1.3. Let k; K and R be as above. Let aAH 3 ðK; Z=2ð2ÞÞ: Then there exist u; v; pAR such that u; v are units in R and a ½u; vÞ ðpÞ is unramified at R: Proof. Since F is either a number field or a dyadic field or a function field of a curve over a finite field, every element of H 2 ðF ; Z=2ð1ÞÞC2 BrðF Þ is represented by a symbol ½u; % v%Þ for some units u; vAR: In particular @ðaÞ ¼ ½u; % v%Þ for some units u and v in R: Let p be a parameter in R: Then by (1.1), we have @ða ½u; vÞ ðpÞÞ ¼ @ðaÞ ½u; % v%Þ ¼ 0: Hence a ½u; vÞ ðpÞ is unramified at R: & Let k and K be as above. Let X be a regular integral scheme with quotient field K: For iX0; let Xi denote the set of points of X of codimension i: For any xAX; let kðxÞ denote the residue field at x: For xAX1 ; let OX;x denote the discrete valuation ring at x and @x : H 3 ðK; Z=2ð2ÞÞ-H 2 ðkðxÞ; Z=2ð1ÞÞ be the residue homomorphism defined with respect to the discrete valuation ring OX;x : Let Y 3 Hnr ðK=SpecðXÞ; Z=2ð2ÞÞ ¼ kerðH 3 ðK; Z=2ð2ÞÞH 2 ðkðxÞ; Z=2ð1ÞÞÞ: xAX1
An element aAH ðK; Z=2ð2ÞÞ is called unramified at a point xAX1 ; if @x ðaÞ ¼ 0; otherwise it is called ramified at x: We say that a is unramified on X if it is unramified at 3 all points of X1 ; i.e., aAHnr ðK=X; Z=2ð2ÞÞ: We define the ramification divisor of a by X x: ramX a ¼ 3
@x ðaÞa0
For f AK ; we denote by SuppX ðf Þ the support of the principal divisor divX ðf Þ: Let Ok be the ring of integers of k (k a number field or a dyadic field). Let X be a smooth, integral, projective curve over k with K ¼ kðX Þ: Let ai AH 3 ðK; Z=2ð2ÞÞ and fj AK ; 1pipn; 1pjpm: By a result of Lipman on the resolution of singularities (cf. [S1, Proof of 2.1]), there exists a regular, projective model X of X over Ok and two regular curves C and E on X with only normal crossings (i.e., for every xAC-E the maximal ideal of the local ring OX;x is generated by local equations of C and E; such that [ [ SuppðramX ðai ÞÞ, SuppX ðfj ÞCSuppðC þ EÞ: 1pipn
1pjpm
We use this result through out this paper without further reference. Let K be any field of characteristic not equal to 2. Let W ðKÞ be the Witt group of quadratic forms over K and I n ðKÞ be the nth power of the fundamental ideal IðKÞ of W ðKÞ [Sc, Chapter 2].
2. Dyadic fields Lemma 2.1. Let F be a finite field of characteristic equal to 2 and Y a smooth, projective curve over F : Let bAH 2 ðF ðY Þ; Z=2ð1ÞÞC2 BrðF ðY ÞÞ and P1 ; y; Pn be the closed points of Y where b is ramified. Let E be a quadratic extension of F ðY Þ with
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
85
either E=F ðY Þ is inseparable or E=F ðY Þ is separable and for any discrete valuation ring S of E lying over the discrete valuation ring Ri at a point Pi ; S=Ri is unramified with the residue field of S is the quadratic extension of the residue field of Ri : Then b#E ¼ 0: Proof. Suppose that E=F ðY Þ is inseparable. Since E=F ðY Þ is a quadratic extension, pffiffiffi we have E ¼ F ðY Þð f Þ; for some f AF ðY Þ: Since every element in H 2 ðF ðY Þ; Z=2ð1ÞÞ is a symbol, we can write b ¼ ½a; bÞ; for some a; bAF ðY Þ: Since the characteristic of F ðY Þ is 2 and ½F ðY Þ : F ðY Þ2 ¼ 2; every element in F ðY Þ is a square in E: Therefore ½a; bÞ#E is split. Suppose that E is separable over F ðY Þ: By class field theory, it is enough to prove that b#E is unramified at each discrete valuation ring of E: Let S be a discrete valuation ring with E as its quotient field. Let R be the discrete valuation ring of F ðY Þ such that RCS: If b is unramified at R; then b is unramified at S: Suppose that b is ramified at R: Then R ¼ OY ;Pi for some i and the residue of b at Pi is the unique quadratic extension of the residue field at Pi : Since S is unramified over R; we have % where S% is the residue field of S: Since S% is the unique quadratic @S ðbÞ ¼ @R ðbÞ#S; % it follows that @S ðbÞ ¼ 0: Therefore, b is unramified at S: & extension of R; Lemma 2.2. Let R be a discrete valuation ring, F its quotient field and k its residue field. Assume that if charðkÞ ¼ 2; then k is perfect. If a quaternion algebra ½a; bÞ is unramified at R; then ½a; bÞ ¼ ½u; vÞ for some u; vAR : Proof. If charðkÞa2; then we know that if ½a; bÞ is unramified, then ½a; bÞ ¼ ðu; vÞ for 1 h14uw2
2 2 some u; vAR : We have ðu; vÞ ¼ ðuw ; vÞ ¼ 4 uw ; v ¼ 4 ; v for any wAR : 2
Since 2 is a unit in R; we can choose wAR such that 14uw is a unit in R: 4 Assume that charðkÞ ¼ 2: Suppose that ½a; bÞ is unramified at R: Suppose that ½a; bÞ#F Fˆ is split, where Fˆ is the completion of F : Then there exists lA½a; bÞ#F Fˆ with reduced trace and reduced norm equal to 1. Assume that½a; bÞ#Fˆ non-trivial. ˆ ˆ Since ½a; bÞ is unramified at R; ˆ D#k is a Let D be the maximal R-order in ½a; bÞ#F: quaternion algebra over k: Therefore, there exists a separable quadratic extension of k contained in D#k: By lifting this to D; we get an element lAD with reduced norm and trace units. Thus in both the cases there exists an element lA½a; bÞ#Fˆ whose reduced norm and trace are units. Since the reduced trace and reduced norm are polynomial functions, we can find mA½a; bÞ (which is close to l) such that the reduced trace and reduced norm are units in R: Then dividing m by its reduced trace, we can assume that m satisfies a quadratic equation X 2 þ X þ u for some unit u in R: Therefore ½a; bÞ ¼ ½u; cÞ for some cAR: Since ½u; cÞ ¼ ½u; cd 2 Þ for any dAK ; we can assume that the valuation of c is 0 or 1. If the valuation of c is 1, then ½a; bÞ is unramified if and only if the quadratic equation X 2 þ X þ u has solution over the residue field of R: From this it follows that a parameter p in R is a norm from the quadratic extension given by X 2 þ X þ u and hence ½u; cÞ ¼ ½u; pcÞ: Thus ½u; cÞ ¼ ½u; vÞ for some u; vAR : &
ARTICLE IN PRESS 86
V. Suresh / Journal of Number Theory 107 (2004) 80–94
Proposition 2.3. Let A be a regular local ring of dimension 2, K its quotient field and k its residue field. Suppose that k is finite and for every regular parameter p of A (i.e., A=ðpÞ is regular) the residue field kðpÞ at p is either a dyadic field or a function field of a curve over a finite field. Let aAH 3 ðK; Z=2ð2ÞÞ: (i) Suppose a is ramified only at p among the prime elements of A: If p is a regular parameter in A; then a ¼ a0 þ ½u; vÞ ðpÞ 3 ðK=SpecðAÞ; Z=2ð2ÞÞ and u; vAA : for some a0 AHnr (ii) Suppose a is ramified only at p and d among the prime elements of A: If p and d generate the maximal ideal m of A; then
a ¼ a1 þ a 2 ; 3 ðK=SpecðAÞ; Z=2ð2ÞÞ and a2 is a sum of symbols of the type where a1 AHnr
½u; vÞ ðpÞ;
½u; vÞ ðdÞ;
½u; dÞ ðpÞ;
u; v running over the units of A: Proof. Let a and p be as in (i). Since p is a regular parameter of A; there exists a prime element d in A such that the maximal ideal m of A is generated by p and d: We have a complex [K, Proposition 1.7] @
H 3 ðK; Z=2ð2ÞÞ !
M
@
H 2 ðkðxÞ; Z=2ð1ÞÞ ! H 1 ðk; Z=2Þ:
xASpecðAÞ1
Since kðpÞ is either a dyadic field or a function field of a curve over a finite field, there % bar denoting the image in A=ðpÞ: Since m is exist a; bAA such that @p ðaÞ ¼ ½a; % bÞ; generated by p and d; A=ðpÞ is a discrete valuation ring with d% as a parameter. Since a has residue only at p; @@ðaÞ ¼ @ð@p ðaÞÞ: Since @@ ¼ 0; @p ðaÞ is unramified at the discrete valuation ring A=ðpÞ: Thus, by (2.2), we have @p ðaÞ ¼ ½u; % v%Þ for some units u; vAA: Let a0 ¼ a ½u; vÞ ðpÞ: Then, by (1.7), a0 is unramified on A: Hence 3 ðK=SpecðAÞ; Z=2ð2ÞÞ and a ¼ a0 þ ½u; vÞ ðpÞ: a0 AHnr Now let a; p and d be as in (ii). As above, there exist a; bAkðpÞ ; such that @p ðaÞ is equal to ½a; bÞ: Suppose that ½a; bÞ is unramified at the discrete valuation ring A=ðpÞ: By (2.2), ½a; bÞ ¼ ½u; % v%Þ; for some u; vAA : Then a ½u; vÞ ðpÞ is as in the case (i) and we are done. Assume that ½a; bÞ is ramified at A=ðpÞ: Since A=m is a finite field, one can show that ½a; bÞ ¼ ½u; % vdÞ for some u; vAA : By (1.1), a ½u; vdÞ ðpÞ is ramified at most at ðdÞ: So by the case (i), we are done. & Remark 2.4. Suppose that in the above proposition, K is a function field in one variable over a dyadic field or a number field k; X a regular two-dimensional scheme over the ring of integers Ok and A the local ring at a codimension 2 point
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
87
of X: Then for every prime pAA; the residue field kðpÞ at p is either a dyadic field or a number field or a function field in one variable over a finite field. Therefore, every element in H 2 ðkðpÞ; Z=2ð1ÞÞ is represented by a symbol. Thus A satisfies the hypothesis of (2.3). Let k be a dyadic field, X a smooth, projective, integral curve over k and K ¼ kðX Þ: Let aAH 3 ðK; Z=2ð2ÞÞ: Then there exists X a regular, projective model of X over the ring Ok of integers in k; such that ramX ðaÞCC þ E; where C and E are regular curves on X such that C and E have only normal crossings. Let T ¼ C-E and B be the semi-local ring at T: Since X is regular, B is a regular semi-local ring and hence a unique factorization domain. We use this situation and notation in the rest of this section without further reference. Lemma 2.5. Let k; K and X be as above. Let x be a codimension 2 point of X and A ¼ OX;x : Let S be a discrete valuation ring which dominates A: Then every symbol of the type ½u; vÞ ðaÞ; with u; vAA and aAK ; is unramified at S: Proof. Let u; vAA ; aAK : By (1.1), we have @S ð½u; vÞ ðaÞÞ ¼ nS ðaÞ½u; % v%; bar denoting the image in the residue field of S and nS denoting the valuation of S: Since u; vAA and kðxÞ is a finite field, it follows that ½u; % v%Þ ¼ 0: Hence ½u; vÞ ðaÞ is unramified at S: & Lemma 2.6. Let k; K be as above. Let aAH 3 ðK; Z=2ð2ÞÞ: Let X; C and E be as above. Let L be an extension of K and S a discrete valuation ring with quotient field L: Suppose that there exists xAC-E such that S dominates OX;x : Suppose one of the following conditions holds: (i) the residue field of S contains the quadratic extension of kðxÞ; (ii) there exist local equations px ; dx for C and E; respectively, at x such that either px or dx or px dx is of the form wy2 ; yAS; wAS ; with the image of w in the residue field of S has its square class coming from kðxÞ ; (iii) there exist local equations px ; dx for C and E; respectively, at x such that the valuations of px and dx at S are even. Then aL is unramified at S: Proof. Let A ¼ OX;x : By (2.3), a ¼ a0 þ a00 ; where a0 is unramified on A and a00 is a sum of the symbols of the type ½u; vÞ ðpx Þ; ½u; vÞ ðdx Þ; ½u; px Þ ðdx Þ; with u; vAA : By Lemma 2.7 (below), the extension of a0 to L is unramified at S: Let nS denote the discrete valuation of S; @S denote the residue homomorphism at S and mS denote the maximal ideal of S: By (2.5), ½u; vÞ ðpx Þ; ½u; vÞ ðdx Þ are unramified at S: Suppose that the residue field of S contains a quadratic extension of kðxÞ: Then @S ð½u; px Þ ðdx ÞÞ contains the quadratic extension given by X 2 þ X þ u: % Since the
ARTICLE IN PRESS 88
V. Suresh / Journal of Number Theory 107 (2004) 80–94
unique quadratic extension of kðxÞ is contained in the residue field of S; X 2 þ X þ u% is reducible over the residue field of S: Therefore @S ðaL Þ ¼ 0: Suppose that px ¼ wy2 for some wAS such that w% ¼ ll21 with lAkðxÞ ; and yAS: Then ð½u; px Þ ðdx ÞÞL ¼ ð½u; wÞ ðdx ÞÞL : We have @S ð½u; px Þ ðdx ÞÞ ¼ nS ðdx Þ½u; % wÞ % ¼ nS ðdx Þ½u; % lÞ: Since u; % lAkðxÞ ; as before, it follows that ½u; % lÞ ¼ 0: Similarly, one can prove that if dx ¼ wy2 ; with w; y as above, then @S ð½u; px Þ ðdx ÞÞ ¼ 0: Suppose that px dx ¼ wy2 ; with w; y as above. Since ½u; px Þ ðdx Þ ¼ ½u; px dx Þ ðdx Þ; we have ð½u; px Þ ðdx ÞÞL ¼ ð½u; wÞ ðdx ÞÞL and @S ðð½u; px Þ ðdx ÞÞL Þ ¼ nS ðpx Þ½u; % wÞ % ¼ 0: Therefore a is unramified at S: Suppose that the valuation of px and dx are even. Then by (2.3), it is easy to see that a is unramified at S: & Lemma 2.7. Let k and K be as in (2.6). Let A be a regular local ring of dimension 2 with K as its quotient field. Let L be a finite extension of K and S a discrete valuation ring with quotient field L: Suppose S dominates A: Then the map H 3 ðK; Z=2ð2ÞÞ-H 3 ðL; Z=2ð2ÞÞ restricts to a map 3 3 Hnr ðK=SpecðAÞ; Z=2ð2ÞÞ-Hnr ðL=SpecðSÞ; Z=2ð2ÞÞ:
Proof (Parimala). Let aAH 3 ðKÞ: Since K is a function field in one variable over a padic field, we have I 4 ðKÞ ¼ 0 and e3 : I 3 ðKÞ-H 3 ðKÞ is an isomorphism [AEJ, Corollary 4 and Theorem 2]. Let q0 AI 3 ðKÞ be such that e3 ðq0 Þ ¼ a: Let q be an anisotropic quadratic form over K which represents q0 in I 3 ðKÞ: Let P be any height one prime ideal of A and R ¼ AP : Since a is unramified at R; by [K, Lemma 1.4 (3)], ˆ ¼ 0: Since e3 is an isomorphism, q0 #Kˆ ¼ 0: Hence q#Kˆ is hyperbolic. e3 ðq0 #KÞ ˆ the two forms q over K ¼ ðAP Þ (p a uniformizing Since q is hyperbolic over K; p P xi yi over A#P patch over the Kˆ and give a quadratic form QP parameter in AP ) and over AP such that QP #KCq [CO]. Now, by the general non-sense of the torsors over two-dimensional regular rings [CS], q comes from a quadratic form Q over A: Therefore aL ¼ e3 ðQ#S#L) and hence unramified at S: & Lemma 2.8. With the notation as above, let L be a quadratic extension of K: Let S be a discrete valuation ring with L as its quotient field. Assume that S-K ¼ BðpÞ ; where p is a prime element in B giving a local equation for a component C1 of C: If C1 -Ea|; let C1 -E ¼ fx1 ; y; xr g and dxi a local equation of E at xi ; 1pipr: Suppose that either pffiffiffi C1 -E ¼ | or L ¼ Kð f Þ with f AB satisfying one of the following conditions: (i) f is a parameter in BðpÞ ; (ii) f is a unit in BðpÞ and f% is not a square in B=ðpÞ; (iii) Let Bi be the local ring at ðp; dxi Þ and B* i is the integral closure of Bi in L: Then B* i =ðp; dxi Þ contains the quadratic extension of Bi =ðp; dxi Þ: Then aL is unramified at S:
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
89
Proof. Let A ¼ BðpÞ : Then the residue field kðpÞ of A is the quotient field of B=ðpÞ: Since ramX aCC þ E and C1 is regular curve on X; it follows from the complex [K, 1.7] @
H 3 ðK; Z=2ð2ÞÞ !
M ZAX
1
@
H 2 ðkðZÞ; Z=2ð1ÞÞ !
M yAX
H 1 ðkðyÞ; Z=2Þ
2
that @ðpÞ ðaÞ is possibly ramified only at the discrete valuations of kðpÞ corresponding to C1 -E: Suppose that C1 -E ¼ |: Then it follows that @ðpÞ ðaÞ is unramified at every discrete valuation ring of kðpÞ: Since kðpÞ is either a global field of positive characteristic (so that there are no archimedean primes) or a local field, by class field theory, we have @ðpÞ ðaÞ ¼ 0 and hence aL is unramified at S: Suppose that C1 -Ea|: Suppose that f is a parameter in A: By (1.3), a ½u; vÞ ðf Þ is unramified at A for some units u; vAA: Since f is a square in L; ½u; vÞ ðf Þ is trivial over L and hence aL is unramified at S: Assume that f is as in (ii) or (iii). Suppose kðpÞ is a p-adic field. Then the residue field k of S is a quadratic extension of kðpÞ and @S ðaL Þ ¼ @p ðaÞ#k: Since the unique quaternion division algebra over kðpÞ is split over any quadratic extension of kðpÞ (cf. [Sc, p. 353]), we have @S ðaS Þ ¼ @p ðaÞ#k ¼ 0: If kðpÞ is a function field of a curve over a finite field, then by (2.1), it follows that a is unramified at S: & Lemma 2.9. Let A be regular local ring of dimension 2 with quotient field K and with a finite residue field k: Let m be the maximal ideal of A and a; bAA units. Then there exist p; dAm; positive integer i and u0 ; v0 ; w0 AA=ðp2iþ1 ; dÞ such that m ¼ ðp; dÞ and pffiffiffi % 2 abw2 modulo ðp2iþ1 ; dÞÞ; if L ¼ Kð f Þ and A˜ for any f AA with f% ¼ au % 2 þ bv ˜ the integral closure of A in L; then A=m contains the unique quadratic extension of A=m ¼ k: Proof. Since only finitely many prime elements of A divide 2 and there are infinitely many prime elements in m\m2 ; we can write m ¼ ðp; dÞ; where d does not divide 2: pffiffiffi Then A1 ¼ A=ðdÞ is a dyadic ring with p% a parameter. Let lAA1 be such that A1 ½ l is the unramified extension of A1 : The square class of l is uniquely determined by its image modulo 4p: % Since the discriminant l of the quadratic form /a; b; ab; lS is not a square in A1 ; /a; b; ab; lS is isotropic (cf. [Sc, p. 187]). Thus /a; b; abS represents l over A1 : In particular there exists u0 ; v0 ; w0 AA1 =ð4pÞ such that l ¼ au20 þ bv20 abw20 in A1 =ð4pÞ: Let i be the valuation of 2 in A1 : Then ð4p; dÞ ¼ ðp2iþ1 ; dÞ: Let f AA be any unit such that f ¼ au20 þ bv20 abw20 modulo ðp2iþ1 ; dÞ: pffiffiffi Then it is easy to see that the quadratic extension Kð f Þ has the required property mentioned in the lemma. & Theorem 2.10. Let k be a dyadic field, X a smooth, projective, irreducible curve over k: Let K ¼ kðX Þ and ai AH 3 ðK; Z=2ð2ÞÞ; 1pipn: Let a; bAK : Then there exists f AK pffiffiffi which is a value of the quadratic form /a; b; abS such that ai #Kð f Þ ¼ 0 for 1pipn:
ARTICLE IN PRESS 90
V. Suresh / Journal of Number Theory 107 (2004) 80–94
Proof. Let X be a regular, projective model of X over Ok such that [ SuppðramX ðai ÞÞCSuppðC þ EÞ; Suppð2Þ,SuppðaÞ,SuppðbÞ, 1pipn
where C and E are regular curves on X with only normal crossings. Let T ¼ C-E: Let B be the semi-local ring at T: For xAT; let px ; dx AB be local equations for C and E at x; respectively. Since B is a unique factorization domain with quotient field K; without loss of generality, we assume that a; b are square free in B and SuppSpecðBÞ ðabÞCSuppðC þ EÞ: Let cAB be the greatest common divisor of a and b; so that a ¼ ca0 ; b ¼ cb0 ; with a0 ; b0 AB: Since a and b are square free, c; a0 ; b0 are pairwise coprime. For xAT; choose ux ; vx ; wx as follows: (i) Suppose cðxÞ ¼ 0: Let mx denote the maximal ideal of B at x: Since c; a0 ; b0 are pairwise coprime and the only prime elements of Bmx which divide ca0 b0 are px ; dx ; at least one of a0 and b0 is coprime with px and dx ; and hence is a unit at x: Thus a0 ðxÞa0 or b0 ðxÞa0: Let ux ; vx AkðxÞ be such that a0 ðxÞu2x þ b0 ðxÞv2x a0: Let wx ¼ 0: (ii) Suppose that cðxÞa0 and a0 b0 ðxÞ ¼ 0: Let ux ¼ vx ¼ 1 and wx ¼ 0: (iii) Suppose that cðxÞa0 ðxÞb0 ðxÞa0: Then a; b are units in Bmx : Let ix ¼ i; p0x ¼ 2
2
p; d0x ¼ d and ux ¼ v0 =a0 ; vx ¼ u0 =b0 ; wx ¼ w0 ABmx =ðp2iþ1 ; dÞ be as in (2.9). Let u; v; wAB be such that for any xAT; if x as in (i) or (ii) uðxÞ ¼ ux ; vðxÞ ¼ vx and wðxÞ ¼ wx ; and if x is as in (iii), u ¼ ux ; v ¼ vx ; w ¼ wx modulo ðpx2ix þ1 ; d0x Þ: Let f ¼ ca0 b0 ða0 u2 þ b0 v2 cw2 Þ: Clearly f is a value of ca0 b0 /a0 ; b0 ; cS ¼ /a; b; abS: pffiffiffi pffiffiffi We now show that a#Kð f Þ ¼ 0: Let L ¼ Kð f Þ and k0 be the field of constants of L: Let X 0 be a smooth, projective, irreducible curve over k0 with k0 ðX 0 Þ ¼ L: Let X0 be a regular proper model of X 0 over Ok0 and yAX0 be a point of codimension one. Let S ¼ OX0 ;y be the discrete valuation ring at y: Since X is projective over Ok ; there exists a point zAX of codimension 1 or 2, such that S dominates the local ring A ¼ OX;z : Fix i; 1pipn and let a ¼ ai : Suppose dimðAÞ ¼ 1: Then A is a discrete valuation ring. Suppose that z does not correspond to a component of C or E: Then a is unramified at A and hence aL is unramified at S: Let z correspond to a component C1 of C: The case where z corresponds to a component of E is similar. Suppose that C1 -E ¼ |: Then by (2.8), aL is unramified at S: Suppose that C1 -Ea|: Let p be a prime element of B corresponding to the component C1 : Then A ¼ BðpÞ : Since c; a0 ; b0 are pairwise coprime in B; it follows that at most one of c; a0 ; b0 is divisible by p: Suppose p divides c: Let xAC1 -E: Then by (i), a0 u2 þ b0 v2 cw2 is a unit in OX;x : Since A is a localization of OX;x ; a0 u2 þ b0 v2 cw2 is a unit in A: Further, since p divides c; both a0 and b0 are units in A: Therefore f is a parameter in A and hence by (2.8), aL is unramified at S: Suppose p does not divide c and divides a0 or b0 : Let xAC1 -E: If cðxÞ ¼ 0; then by (i), a0 u2 þ b0 v2 cw2 is a unit at x and hence it is a unit in A: Assume that cðxÞa0:
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
91
Suppose p divides a0 (the other case is similar). If b0 ðxÞa0; then by (ii), ux ¼ 1 and b0 ðxÞv2x cwx ¼ b0 ðxÞa and hence a0 u2 þ b0 v2 cw2 is a unit in A: If b0 ðxÞ ¼ 0; since a0 ; b0 ; c0 are pairwise coprime, we have a0 ¼ a00 p and b0 ¼ b00 dx for some units a00 ; b00 in Bmx : Since wx ¼ 0; write w ¼ w1 p þ w2 dx for some w1 ; w2 in Bmx : We have b0 v2 cw2 ¼ dx ððb00 v2 Þ cw22 dx Þ pðw21 p þ 2w1 w2 dx Þ: Since vx ¼ 1; it follows that b00 v2 cw22 dx is a unit in Bmx and hence a0 u2 þ b0 v2 cw2 is a unit in BðpÞ ¼ A: Therefore, as above, f is a parameter in A and aL is unramified at S: Suppose that p does not divide ca0 b0 : Let xAC1 -E: Suppose that dx divides ab: If dx divides a and b; then dx divides c and does not divide a0 b0 : If dx divides only one of a and b; then dx divides only one of a0 ; b0 and does not divide c: By (i), a0 u2 þ b0 v2 cw2 is a unit. In particular f is a unit at px and not a square modulo px : Therefore by (2.8), a is unramified at S: Suppose that dx does not divide ab: Then a and b are units at x: Let C1 -E ¼ fx ¼ x1 ; y; xr g: Suppose that dxi divides ab for some xi ; then as above, it follows that f is a unit at px and not a square modulo px : Thus, by (2.8), a is unramified at S: Suppose that dxi does not divide ab for all i: Then, by the choice of u; v; w; it follows that f is as in (2.8, (iii)) and hence by (2.8), a is unramified at S: Suppose dimðAÞ ¼ 2: Then z is a closed point of X: If zeC,E; then a is unramified on A and hence unramified at S (2.7). Assume that zAC,E: If zeC-E; then by (2.3 and 2.5), aL is unramified at S: Suppose that zAC-E: Then A ¼ Bmz ; where mz is the maximal ideal of B at z: Suppose that cðzÞ ¼ 0: Then, by the choice of u; v; w; a0 u2 þ b0 v2 cw2 is a unit at z: Since the only prime elements of A which divide ca0 b0 are pz ; dz and c; a0 ; b0 are pairwise coprime, f ¼ ca0 b0 ða0 u2 þ b0 v2 cw2 Þ is of the form w0 pz or w0 dz or w0 pz dz ; with w0 AA : Since f AL2 ; by (2.6), aL is unramified at S: Suppose that cðzÞa0 and a0 ðzÞb0 ðzÞ ¼ 0: If a0 ðzÞ or b0 ðzÞ is not zero, then, as above, one shows that either pz or dz or pz dz is as in (2.6, (ii)) and hence, by (2.6), aL is unramified at S: Suppose that a0 ðzÞ ¼ b0 ðzÞ ¼ 0: Since the only prime elements of A which divide a0 ; b0 are pz ; dz and a0 ; b0 are coprime and non-units at z; we have a0 ¼ w0 pz and b0 ¼ w00 dz or a0 ¼ w0 dz and b0 ¼ w00 pz for some w0 ; w00 AA : Consider the case where a0 ¼ w0 pz and b0 ¼ w00 dz ; with w0 ; w00 AA (the other case being similar). Let nS denote the valuation at S: Since S dominates A; we have nS ða0 ÞX1 and nS ðb0 ÞX1: We assume without loss of generality that nS ða0 ÞpnS ðb0 Þ: Then, b0 =a0 AS: Since wAm and a0 em2 ; w2 =a0 is in the maximal ideal of S: We have
0
2 b w f ¼ cb0 ðu2 Þ þ 0 v2 c 0 ða0 Þ2 : a a 0
2
Suppose that nS ða0 ÞonS ðb0 Þ: Then u2 þ ba0 v2 c wa0 AS : Since b0 ¼ w00 dz ; w00 AA ; cAA and f AL2 ; it follows that dz is as in (2.6, (ii)) and aL is unramified at S: 0 2 Suppose that nS ða0 Þ ¼ nS ðb0 Þ: If u2 þ ba0 v2 c wa0 AS ; then nS ðf Þ ¼ nS ðb0 Þ þ 2nS ða0 Þ ¼ 3nS ðb0 Þ: Since nS ðf Þ is even, it follows that nS ða0 Þ ¼ nS ðb0 Þ is even. In particular nS ðpz Þ ¼ nS ðdz Þ is even. Therefore by (2.6, (iii)), a is unramified at S: Assume that 0 2 u2 þ ba0 v2 c wa0 is not a unit in S: Let n ¼ nS ða0 Þ ¼ nS ðb0 Þ: Let y be a parameter in S and write a0 ¼ w1 yn ; b0 ¼ w2 yn ; with w1 ; w2 AS : By (ii), u; vAA and wAmx : Since
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
92 0
2
2
2
u2 þ ba0 v2 c wa0 ¼ u2 þ ww21 v2 c wa0 is not a unit in S and c wa0 is also not a unit in S; we have u% 2 w2 ¼ : v%2 w1 In particular, the image of w1 w2 in the residue field of S has its square class coming from kðzÞ: We have ww0 pz dz ¼ a0 b0 ¼ w1 w2 y2n : Therefore pz dz as in (2.6. (ii)). Therefore by (2.6), a is unramified at S: Suppose that cðzÞa0 ðzÞb0 ðzÞa0: Then by the choice of u; v; w it follows that f is as in (2.8. (iii)). Therefore, by (2.8), a is unramified at S: This implies 3 3 that a#LAHnr ðL=X0 ; Z=2ð2ÞÞ: Since Hnr ðL=X0 ; Z=2ð2ÞÞ ¼ 0 by a result of Kato [K, Proposition 5.2], we have a#L ¼ 0: This completes the proof of the theorem. & 3. Number fields Theorem 3.1. Let k be a number field and K a function field of a curve over k: Let a1 ; y; an AH 3 ðK; Z=2ð2ÞÞ and a; b; dAK : Then there exists f AK such that (1) f is a value of /a; b; abdS; (2) for every finite non-dyadic place v of k; ðai Þv ¼ ðf Þ,bi for some bi AH 2 ðKv ; Z=2ð1ÞÞ; (3) for every dyadic place v of k; where d is a square in Kv ; ðai Þv ¼ ðf Þ,bi for some bi AH 2 ðKv ; Z=2ð1ÞÞ: Proof. Let Ok be the ring of integers of k: Let X be a smooth, integral, projective curve over k and K ¼ kðX Þ: Let X be a regular, projective model of X over Ok with SuppX ð2Þ,SuppX ðaÞ,SuppX ðbÞ,SuppX ðdÞ,
n [
SuppðramX ai ÞCSuppðC þ EÞ;
i¼1
where C and E are regular curves on X with only normal crossings. Let T be the finite set of closed points of X containing C-E and one closed points from each component of C and E: Let B be the semi-local ring at T: For xAT; let px ; dx AB be local equations for C and E at x; respectively. Since B is a unique factorization domain with quotient field K; without loss of generality, we assume that a; b; d are square free in B and SuppSpecðBÞ ðabdÞCSuppðC þ EÞ: Let cAB be the greatest common divisor of a and b; so that a ¼ ca0 ; b ¼ cb0 ; with a0 ; b0 AB: Since a and b are square free, c; a0 ; b0 are pairwise coprime. For xAT; choose ux ; vx ; wx as follows: (i) Suppose cðxÞ ¼ 0: Let mx denote the maximal ideal of B at x: Since c; a0 ; b0 are pairwise coprime and the only prime elements of Bmx which divide ca0 b0 are px ; dx ; at least one of a0 and b0 is coprime with px and dx ; and hence is a unit at x: Thus a0 ðxÞa0 or b0 ðxÞa0: Let ux ; vx AkðxÞ be such that a0 ðxÞu2x þ b0 ðxÞv2x a0: Let wx ¼ 0:
ARTICLE IN PRESS V. Suresh / Journal of Number Theory 107 (2004) 80–94
93
(ii) Suppose that cðxÞa0 and a0 b0 ðxÞ ¼ 0: Let ux ¼ vx ¼ 1 and wx ¼ 0: (iii) Suppose that cðxÞa0 ðxÞb0 ðxÞa0 and charðkðxÞÞa2: Then every element of kðxÞ is represented by the quadratic form /a0 ðxÞ; b0 ðxÞS: Let ux ; vx AkðxÞ be such that cðxÞa0 ðxÞb0 ðxÞða0 ðxÞu2x þ b0 ðxÞv2x ÞekðxÞ2 : Let wx ¼ 0: (iv) Suppose that cðxÞa0 ðxÞb0 ðxÞa0 and charðkðxÞÞ ¼ 2: Let A ¼ Bmx : As in (2.9), let p0x ; d0x Amx be such that p0x does not divide 2. Let A1 ¼ A=ðp0x Þ: Then A1 is a ring of integers in a number field. For each dyadic place n of A1 with d a square in pffiffiffiffiffi A1n ; let ln AA1n be such that A1n ½ ln is unramified. For each such dyadic place % 2xn abdw2xn (cf. proof of 2.9). Let n; let uxn ; vxn ; wxn be such that ln ¼ au % 2xn þ bv ux ; vx ; wx AA1 which are very close to uxn ; vxn ; wxn : Let ix be the maximum positive integer among the positive integers given in (2.9) corresponding to the completions at dyadic places n: Let u; v; wAB be such that for any xAT; if x as in (i) or (ii) or (iii), uðxÞ ¼ ux ; vðxÞ ¼ vx and wðxÞ ¼ wx and if x is as in (iv), u ¼ ux =a% 2 ; v ¼ vx =b%2 ; w ¼ wx modulo ðpx02ix þ1 ; d0x Þ: Let f ¼ ca0 b0 ða0 u2 þ b0 v2 cdw2 Þ: Clearly f is a value of ca0 b0 /a0 ; b0 ; cdS ¼ /a; b; abdS: Let n be any finite place of k: Xn ¼ X kn ; Xn ¼ X On ; Cn ¼ C On and En ¼ E On ; where kn is the completion of k at n and On is the ring of integers of kn : If kn is non-dyadic, as in the proof of Proposition 4.4 in [PS] pffiffiffi and (2.9), it follows that ðai Þn #kn ðXn Þð f Þ ¼ 0: Therefore for every finite nondyadic place v of k; we have ðai Þn ¼ ðf Þ,ðbi Þ for some bi AH 2 ðKn ; Z=2ð1ÞÞ: Let n be any dyadic place with d a square in Kv : Then we have /a; b; abdS ¼ /a; b; abS: pffiffiffi As in the proof of (2.10), it follows that ðai Þn #kn ðXn Þð f Þ ¼ 0: This completes the proof. & Theorem 3.2. Let k be a number field and K a function field of a curve over k: Let a1 ; y; an AH 3 ðK; Z=2ð2ÞÞ: Then there exist f AK and bi AH 2 ðK; Z=2ð1ÞÞ such that ai ¼ ðf Þ,bi : Proof. Let a ¼ b ¼ d ¼ 1: Let f be a value of / 1; 1; 1S as in (3.1). Let n pffiffiffi be a real place of k: Since f is a sum of squares, by [Se], cd2 ðKn ð f ÞÞ ¼ 1: pffiffiffi pffiffiffi Therefore H 3 ðKn ð f Þ; Z=2ð2ÞÞ ¼ 0: In particular, ðai Þn #Kn ð f Þ ¼ 0: Since d ¼ 1 is pffiffiffi a square, by (3.1), if n is a finite place of k; then ai #Kv ð f Þ ¼ 0: Thus, by [K, pffiffiffi Theorem 0.8], ai #Kð f Þ ¼ 0 and hence there exist bi AH 2 ðK; Z=2ð1ÞÞ such that ai ¼ ðf Þ,bi : &
Acknowledgments I thank R. Parimala, A. Wadsworth and J.-L. Colliot-The´le`ne for various helpful discussions during the preparation of this paper. I also thank the referee for many suggestions.
ARTICLE IN PRESS 94
V. Suresh / Journal of Number Theory 107 (2004) 80–94
References [AEJ] J.K. Arason, R. Elman, B. Jacob, Fields of cohomological 2-dimension three, Math. Ann. 274 (1986) 649–657. [CF] J.W.S. Cassels, A. FrO¨hlich, Algebraic Number Theory, Academic Press, London, 1967. [CO] J.-L. Colliot-The´le`ne, M. Ojanguren, Espaces principaux homoge`nes localement triviaux, Inst. Hautes Etudes Sci. Publ. Math. 75 (1992) 97–122. [CS] J.-L. Colliot-The´le`ne, J.-J. Sansuc, Fibre´s quadratiques et composantes connexes re´elles, Math. Ann. 244 (1979) 105–134. [G] M.J. Greenburg, Lectures on Forms in Many Variables, W.A. Benjamin, Inc., New York, Amsterdam, 1969. [K] K. Kato, A Hasse principle for two-dimensional global fields, J. Reine Angew. Math. 366 (1986) 142–181. [PP] R. Parimala, R. Preeti, Hasse principle for classical groups over function field of curves over number fields, J. Number Theory 101 (2003) 151–184. [PS] R. Parimala, V. Suresh, Isotropy of quadratic forms over function fields of p-adic curves, Inst. Hautes Etudes Sci. Publ. Math. 88 (1998) 129–150. [S1] D.J. Saltman, Division algebras over p-adic curves, J. Ramanujan Math. Soc. 12 (1) (1997) 25–47. [S2] D.J. Saltman, Correction to: division algebras over p-adic curves, J. Ramanujan Math. Soc. 13 (2) (1998) 125–129. [Sc] W. Scharlau, Quadratic and Hermitian Forms, Grundlehren Mathematischen Wissenschaften, Vol. 270, Springer-Verlag, Berlin–Heidelberg–New York, 1985. [Se] J.-P. Serre, Cohomologie Galoisienne, in: Lecture Notes in Mathematics, Vol. 5, Springer, Berlin, 1964, 1994.