On zeros of cubic L-functions

Transcription

1 Journal of Number heory wwwelsevierom/loate/jnt On zeros of ubi L-funtions Honggang ia Department of Mathematis, Ohio State University, Columbus, OH 43210, USA Reeived 21 July 2005; revised 18 Otober 2005 Available online 22 November 2006 Communiated by J Brian Conrey Abstrat In this paper we obtain a zero density theroem for Heke L-funtions assoiated to ubi haraters by Heath-Brown s large sieve type inequality Using Patterson s result on ubi Gauss sums we also have the moments estimate for orresponding L-funtions Distribution of lass numbers of a family of degree 6 fields over Q is obtained as an appliation of the main results 2006 Elsevier In All rights reserved 1 Introdution Let k = Qω, where ω = LetO k be the integer ring of k It is known that k has lass number 1, and every ideal in O k oprime to 3 has a unique generator ongruent to 1 mod 3 For 1 O k whih is square-free and ongruent to 1 mod 9, letχ = be the ubi residue symbol So χ is trivial on units of O k by the law of ubi reiproity [IR, heorem 1b, p 114] χ an be viewed as a primitive harater on the ray lass group h = I /P, where I ={A I, A, = 1}, P ={a P, a 1mod} I and P are the group of frational ideals and the subgroup of prinipal ideals of O k respetively he Heke L-funtion assoiated with χ is defined by Ls, χ = A 0 χ ANA s address: xia@mathohio-stateedu /$ see front matter 2006 Elsevier In All rights reserved doi:101016/jjnt

2 416 H ia / Journal of Number heory for Res > 1 Here the sum is over nonzero ideals of O k, and NA is the norm of A Ls, χ admits analyti ontinuation to the whole s-plane as an entire funtion and has the funtional equation: where Wχ is the Gauss sum of χ : δ is the different of k, and Λs, χ = Wχ N 1/2 Λ1 s, χ, Wχ = where D k = 3 is the disriminant of k For 2, define a O k / χ ae ra/δ ; Λs, χ = D k N s/2 2π s ΓsLs,χ, Nσ,,χ = # { ρ Lρ, χ = 0inthestrip1> Rs σ and Is 0 } Heath-Brown already obtained the zero-density estimate by the large sieve inequality on real haraters whih is also established by himself [He2]: If SQ denotes the set of all real primitive haraters of ondutor at most Q, and Nσ,,χ is defined in the same way as above exept that χ is a quadrati harater, then Nσ,,χ ɛ Q ɛ Q 31 σ/2 σ 3 2σ/2 σ χ SQ for any ɛ>0 In this paper we establish the analogue of this result for ubi haraters: heorem 1 Let 1 >σ 1 2, 2χ = is the ubi residue symbol assoiated to, where O k, 1mod9, and is square-free hen we have Nσ,,χ Q 101 σ Q ɛ O k, 1 mod9, N Q One appliation of this zero density theorem is to study the moments distribution of speial values at the edge of the ritial strip of orresponding L-funtions For quadrati haraters, Jutila showed [Ju]: d [e,e] L r 1,χ d = br + O 1/2 log 2r, r 1, 3, where e = 1or 1, and br, 2 r are onstants depending on r only, and χ d is the quadrati symbol with ondutor d Let be the sum over 1mod9, is square-free in O k In this paper we will always use this notation to denote the restrition on unless speified

3 H ia / Journal of Number heory Write L s = Ls, χ Ls, χ 2, and note that L 1 = L1,χ 2 From heorem 1 we an derive: heorem 2 For m 1, we have L m 1 N C m, where C m is a positive onstant depending on m only and will be given liitly in Setion 3 Applying this moments result to a family of fields whih are ubi extensions of k, we an get a result related to the Brauer Siegel theorem [SL] It is an analog of the result of Siegel One onlusion of Brauer Siegel theorem shows that if F ranges over all number fields of fixed degree N over Q, then there is an asymptoti relation loghr log d 1/2 for d, where h, R, d are the lass number, regulator and disriminant of F respetively Siegel was the first to prove the Gauss onjeture for real quadrati fields He obtained the result 0<D<x hd log ε D = π 2 18ζ3 x3/2 + Ox log x, where hd is the narrow lass number of the order of disriminant D ontained in the quadrati field Q D, ε D is some fundamental unit of Q D, or in another word, the unit defined by ε D = t + u D/2, where t, u are the smallest positive integral solutions of t 2 Du 2 = 4 Note log ε D is essentially the regulator of Q D heorem 2 above implies results on moments of the produt of lass number and regulator for a family of degree 6 non-abelian extensions over Q Proposition 3 Let h, R be the lass number and regulator of k 1/3 respetively, m 1, then N 3 2m h R m C m m+1 2π Remark After I got the main theorems in this paper, I learned that Chinta, Friedberg and Hoffstein also obtained similar results in this respet [CFH] heir main result an be stated as follows For n 2, let F be a global field ontaining a full set of nth roots of unity, S is a finite set of plaes ontaining all arhimedean plaes, all plaes dividing n, and suh that the lass number of the ring of S-integers is 1 A is a lass on the ray lass group H C Fixs C with Rs > 1 1/r +1 and let π be an isobari automorphi representation of GL r A F, whih is unramified outside S hen if m>0 is a suffiiently large integer depending on F, n, r, for any ɛ>0 they showed: d A, μd 0, 1< d < L S s, π χ d 1 d / m S s = k O ɛ 1/2+ɛ

4 418 H ia / Journal of Number heory as, where the sum is over square-free integral ideals d in the ray lass A, and S s is some onstant his result gives the moments of twisted L-values at 1 under the ondition that S is suffiiently large to ensure S s is nonzero hus their result does not imply heorem 2 here he main ingredient to prove heorem 1 in Setion 2 is the large sieve inequality for ubi haraters by Heath-Brown in [He1] In Setion 3 we use the density theorem to study the moments of L 1, similar to the work of Luo [Lu1,Lu2] o get Proposition 3 in the last setion, whih is an appliation of heorem 2, we use the fat that h R is essentially L 1 by lass fields theory and the analyti lass number formula 2 Zero-density theorem for Ls, χ he goal of this setion is to prove heorem 1 Let M x s, χ = μaχ ANA s NA x hen Ls, χ M x s, χ = 1 + a A χ ANA s, NA>x where a A = B A,NB x μb and μ is the generalized Möbius funtion By Mellin transform we have e 1/y + a A χ ANA s e NA/y NA>x = 1 Ls + w,χ M x s + w,χ Γ wy w dw 1 l Define the integral in 1 to be Is,l, where l is the vertial line with real part equals l and s = σ + it, l + σ>1 Moving the line of the integral to Rew = 1/2 σ,1/2 <σ <1, we pass through the pole of the integrand at w = 0 and obtain by Cauhy s theorem that: Is,l= Ls, χ M x s, χ + Is,1/2 σ 2 If s = ρ = β + iγ is a zero of Ls, χ, with 1/2 β<1 and γ 0, from 1 and 2 we onlude that: e 1/y = a A χ ANA ρ e NA/y + Iρ,1/2 β 3 or l NA>x e 1/y = 1 1 Lρ + w,χ M x ρ + w,χ Γwy w dw + Iρ,1/2 β 4 Let 10 <y<q 1 and 10 <x<q 2, z = log 3, where 1 and 2 are onstants

5 H ia / Journal of Number heory From Stirling s formula: L1/2 + iγ + iν,χ M x 1/2 + iγ + iν,χ Γ 1/2 β + iνy 1/2 β+iν dν 1, 5 thus from the definition of Is,1/2 ρ, Iρ,1/2 β = 1 L1/2 + iγ + iν,χ M x Γ1/2 β + iνy 1/2 β+iν dν + O 1 6 On the other hand, moving the line of integration in 4 to Rew = l 0 = 1 β + ɛ and letting μ 0 = 1 + ɛ + iγ + iν, where ɛ is positive and small enough, we dedue that 1 Lμ0,χ M x μ 0,χ Γl 0 + iνy l0+iν dν 1 hus if is large enough, from 4 and 6 we have log y 1 β + y 1/2 β So by Cauhy s inequality we an infer Now let log 5 y 2 2β + y 1/2 β 1 Lμ 0,χ M x μ 0,χ dν L1/2 + iγ + iν,χ M x 1/2 + iγ + iν,χ dν 1 1 Lμ0,χ M x μ 0,χ 2 dν L1/2 + iγ + iν,χ M x 1/2 + iγ + iν,χ dν 1 7 N = # { N Q, 1mod9, O k for whih Ls, χ has a zero in the square σ Res σ + log Q 1,τ Ims τ + log Q 1} For τ, notie that β>σ, it follows that

6 420 H ia / Journal of Number heory log 5 y 2 2σ + y 1/2 σ N Q N Q 1 Lμ0,χ M x μ 0,χ 2 du 1 L 2 + iu,χ where is the notation mentioned in the introdution part For the seond integral in 8, by Cauhy s inequality we have: y 1/2 σ N Q y 1/2 σ y 1/2 σ 1 L 2 + iu,χ N Q N Q By the estimation from [Lu1]: 1 mod9 M x iu,χ du 1 2 1/2 L 2 + iu,χ 1 L 2 + iu,χ 1 2 L 2,χ M x iu,χ du N, 8 N Q 2 1/2 du N y 1 2 1/2 M x 2 + iu,χ du N Q y 1+ɛ, 1/2 M x 2 dν 9 it follows that: N Q 1 2 L 2,χ Q 1+ɛ hough Luo s result is for L1/2,χ, his proof an be applied to L1/2 + it,χ for any t So the seond integral in 8 is y 1/2 σ Q 1+ɛ 2 log 3/2 y 1/2 σ Q 1+ɛ 2 log 3/2 M2 x N Q 1 1/2 2 + iu,χ du N Q 1<NA x μaχ ANA 1/2 iu 2 du 1/2 Now we appeal the large sieve inequality established by Heath-Brown [He1] for the ubi haraters: Nm M Nn N n 2 n ɛ M + N + MN 2/3 NM ɛ n 2 10 m Nn N

7 H ia / Journal of Number heory for any ɛ>0, where denotes summation over square-free elements of O k whih are ongruent to 1 mod 3 We onlude that the seond integral in 8: y 1/2 σ Q 1+ɛ 2 log 3/2 Q + x + Qx 2/3 11 For the first integral in 8 by Cauhy s inequality again we have: log 5 y 2 2σ log 8 y 2 2σ z N Q NA>x max a A χ ANA μ 2 0 du x B e log3 N Q B<NA 2B a A χ ANA μ 2 0 By the fat that a A NA ɛ for any ɛ>0 and 10, we dedue the above sum: log 8 y 2 2σ z So finally by 11 and 12 we have max x B e log3 B + Q + BQ 2/3 B 1 Q ɛ ɛ y 2 2σ x + Q + xq 2/3 x 1 Q ɛ 12 N ɛ y 2 2σ x + Q + xq 2/3 x 1 Q ɛ + y 1/2 σ Q 1+ɛ 2 log 3 Q + x + Qx 2/3 Now let x = Q 2,y = Q 5, we get N Q 101 σ Q ɛ heorem 1 then follows from this bound immediately 3 Distribution of L m s at 1 We have L m s = Ls, χ L s,χ 2 m = a = a = A 0 τ m aχ a Na s ab=a b τ m bχ 2 b Nb s τ m aτ m bχ aχ 2 b NA s, where τ m is the generalized divisor funtion for O k χ a m Na s b χ 2b m Nb s

8 422 H ia / Journal of Number heory Consider the integral with 1 < 2 1 L m s + 1Γ ss dx 2 Moving the line of the integral to Res = γ, 1 <γ <0, we pass through the pole of integrand at s = 0 with residue L m 1 and obtain: L m 1 = NA 1 τ m aτ m bχ ab 2 NA A ab=a 1 L m s + 1Γ ss dx γ Denote I = 1 L m s + 1Γ ss dx γ Summing over 1mod9, where is square-free in O k, we an get: L m 1 N = A NA/ NA χ aχ b 2 N ab=a τ m aτ m b For the first term of the right side, hange the order of summation then we dedue: = A = a NA/ NA b τ m aτ m b Nab N I τm aτ m bχ aχ b 2 N ab=a Nab χ ab 2 N Now write a = 1 ω p a 1, b = 1 ω q b 1, where a, 1 ω = b, 1 ω = 1 We see that the above sum is = p=0 q=0 a 1,b 1 m+q 1 q m+p 1 p χ a1 b 2 1 N τm a 1 τ m b 1 3 q+p Na 1 b 1 3 p+q Na 1 b 1 Write a 1 = r 3 1 a 2, b 1 = r 3 2 b 2, where a 2, b 2 are ubi free, the above sum is

9 = p,q=0 r 1,r 2 a 2,b 2 H ia / Journal of Number heory m+q 1 q χ a2 b 2 2 N m+p 1 p τm r1 3a 2τr2 3b 2 3 p+q Nr1 3 3 p+q Nr1 3r3 2 a r3 2 a 2b 2 2b 2 Write a 2 = ra 3, b 2 = rb 3, where r = a 2,b 2 he sum above an be rewritten: = p,q=0 a 3,b 3 =1 r 1,r 2,r m+q 1 q χ a3 b 2 3 N For the most inner sum in the ase a 3 b3 2 = 1, we have: N m+p 1 p τm r1 3ra 3τr2 3rb 3 3 p+q Nr p+q Nr1 3 r3 2 r2 a 3 b 3 r3 2 r2 a 3 b 3 = #h9 13 Γs s 1 N s ds 1 Γs s χa μa Na s ds, a 0 χ mod 9 where χ runs over all ray lass haraters mod 9, μ is the Möbius funtion By moving the line of the integration to Res = 1/2 + ɛ, we find that the above sum is asymptotially C + O ɛ 1/2+ɛ and C = res s=1ζ k s #h 9 ζ k Np 1 1, p 9 ζ k s being the Dedekind zeta funtion of the field k If a 3 b3 2 1, then the most inner sum in 13 is: N χ a3 b3 2 = N χa3 b3 2 = N χ a3 b3 2 μd 19 d 2 d 13 = d 13 Nd<B μdχ a3 b 2 3 d 2 d 2 9 χ a3 b 2 3 Nd 2

10 424 H ia / Journal of Number heory b 13 = R + S χ a3 b 2 3 b 2 d b,nd>b d 13 μd b 2 9 Nb 2 χ a3 b3 2 Here B> 1/2 1/2, and it will be hosen optimally Using the ray lass haraters mod 9 to detet the ongruene ondition d 2 mod 9: R = d 13 Nd<B = d 13 Nd<B μdχ a3 b 2 3 μdχ a3 b 2 3 d 2 13 d 2 1 #h9 Nd 2 χ a3 b3 2 1 χ 9 χ 9 d 2 #h9 χ 9 χ 9 13 χ a3 b3 2χ 9χ 9 d 2 Nd 2 By an analogue of the Polya Vinogradov inequality see [HP, Lemma 2], for any ɛ>0 We have 1 mod3 R N χ a ɛ Na 1/2+ɛ y d 1 mod3 Nd<B BN a 3 b 2 3 1/2+ɛ Let a 3 = a 3 s2, where a 3 is the square-free part of a 3 S = b 13 χ a 3 s 2 b 2 3 he ontribution of S to 13 is: p,q=0 r 1,r 2,r,b a 3,s,b 3 χ a 3 sb 2 3 b 2 m+q 1 q d b,nd>b d 13 b 2 d b,nd>b d 13 μdχa3 b3 2 d 2 N a 3 b3 2 1/2+ɛ μd b 2 9 Nb 2 χ a 3 s 2 b3 2 m+p 1 p τm r1 3ra 3 sτr3 2 rb 3 3 p+q Nr1 3 3 p+q Nr1 3r3 2 r2 a 3 sb r3 2 r2 a 3 sb 3 3 μd b 2 9 Nb 2 N χ a 3 χ s 2 b3 2

11 H ia / Journal of Number heory Let C a 3 m + q 1 m + p 1 = τ m r 3 q p 1 ra 3 s τ r2 3 rb 3 By Cauhy s inequality the above sum is: p,q,r 1,r 2 r,b 3,b,s a 3 Ca 3 3p+q Nr1 3r3 2 r2 a 3 s2 b 3 / 3 p+q Nr1 3r3 2 r2 a 3 s2 b 3 a 3 Nb 2 χ a 3 b9 By Heath-Brown s large sieve inequality 10 again: p,q,r 1,r 2 r,b 3,b,s 2 3 p+q Nr1 3r3 2 r2 a 3 s2 b 3 / a 3 Ns 2 b p+q Nr1 6r6 2 a 2 3 s 4 b3 2 Nb 2 + b,b 3 1/2 Nb 3 Nb 2 3 b 1/2+ɛ + b 3 Nb 3 Nb 3 1/2+ɛ + 1/3 5/6 B 2/3 Ns 2 b 2 b Nb 3 Nb 2 b 3 + 1/3 5/6 B 2/3 χ s 2 b /2 1/2 2/3 1/2 1/2 Nb 2 2 1/2 + 1/3 5/6 Nb 3 Nbb 3 5/3 For the R part, we an hoose Na 3 b 2 3 <3/2, and the ontribution of R to 13 is 3/4 B Altogether the nontrivial harater part in 13 has ontribution 3/4 B + 1/2 1/2+ɛ + 5/6+ɛ 1/3 B 2/3 Choose B = 1/2 1/4 by letting 3/4 B = 1/3 5/6, and we get 1/2 1/2+ɛ for the ontribution from the part where a 3 b3 2 1 For the part where a 3b3 2 = 1wehave B 2/3 m+q 1 m+p 1 q p τm r1 3ra 3τ m r2 3rb 3 p,q=0 r 1,r 2,r 3 a 3,b 3 =1 p+q Nr1 3r3 2 a 3b 3 3 p+q Nr1 3 r3 2 a 3b 3 C + Oɛ 1/2+ɛ = C m + O ɛ 1/2+ɛ

12 426 H ia / Journal of Number heory as goes to infinity, where C m = C p,q=0 r 1,r 2,r a 3,b 3 =1 m+q 1 q m+p 1 p τm r1 3ra 3τ m r2 3rb 3 3 p+q Nr1 3r3 2 a 3b 3 Sine τn < n ɛ for any positive ɛ if n is large enough, and m+q 1 q q m, we an see that Cm is onvergent to a finite number So finally from we have: L m 1 N = C m + O 1/2 + 1 N I Aording to whether L m s is zero-free in the domain 1 >σ 1 η, t log3,theset {; N, 1mod9, square-free} is divided into two parts J 1 and J 2 If J 1, then we take γ = η/2 ini Lemma 1 L m s ɛ for σ 1 η/2, t log 2 Proof It follows from the proof of Lemma 2 in Luo s paper [Lu2] While for σ 1 η/2, t log 2, it follows that L m s t 2 for some 2 > 0 from onvexity bound Hene by Stirling s formula: I η/20 η/2 + A for J 1 and any A>0 On the other hand, by taking γ = ɛ in I we get J 2 I J 2 1/20 where J 2 =#{: N, 1mod9, is square-free and L m s has a zero in the region σ 1 η, t log 3 } By heorem 1 we an get J 2 1/5 if η is small enough hus: So we onlude that L m 1 N J 2 I 1/4 = C m + O 1/2 1/ η/20 η/2 Let = 22/23, we obtain L m 1 N his ompletes the proof of heorem 2 = C m + O 45/46

13 H ia / Journal of Number heory Class numbers of degree 6 extensions Let K = k 1/3, where O k, 1mod9 and is square-free Reall the analyti lass number formula: lim s 1ζ Ks = 2r12π r2h K R K s 1 ω K DK Sine K /k is an abelian extension atually, ubi yli, by lass field theory f [SL, Chapter VI] we have: ζ K s = ζ k sls, χ Ls, χ = ζ k sl s Computing ζ K s/ζ k s by lass number formula given above we have: ζ K s lim s 1 ζ k s = 2π2 3h R D with h, R, D the lass number, regulator, disriminant of K It is easy to show that K is the splitting field of x 3 N = 0 over Q hus K is the omposite field of Q 3 N and k SoD K = D 3 k D2 2, where D 2 is the disriminant of Q 3 N D k = 3 Sine 1mod9, N 1mod9 and is square-free, by [Na, heorem 28, p 62], D 2 = 3N Consequently D K = 3 5 N 2 hus L 1 = 2π2 h R 3 2 N so we have: h R m N N 3 2m = C m + o 2π By the well-known auberian theorem see [Ha, heorem 108], we have: On the other hand, N N h R m 3 2m = C m + o N 2π h R m = t m d h R m N 1 N t = m N h R m + 1 t m 1 = m h R m + o N N t h R m dt + O1

14 428 H ia / Journal of Number heory so we have: N his ompletes the proof of Proposition 3 Referenes 3 2m h R m = C m m+1 + o m+1 2π [CFH] Gautam Chinta, Solomon Friedberg, Jeffrey Hoffstein, Asymptotis for sums of twisted L-funtions and appliation: Progress and prospets, in: Automorphi Representation, L-Funtions and Appliation, in: OSU MRI Publ, vol 11, de Gruyter, Berlin, 2005, pp [Ha] GH Hardy, Divergent Series, Clarendon Press, Oxford, 1949 [He2] DR Heath-Brown, A mean value estimate for real harater sums, Ata Arith LII [He1] DR Heath-Brown, Kummer s onjeture for ubi Gauss sums, Israel J Math [HP] DR Heath-Brown, SJ Patterson, he distribution of Kummer sums at prime arguments, J Reine Angew Math [IR] Kenneth F Ireland, M Rosen, A Classi Introdution to Modern Number heory, Grad exts in Math, vol 84, Springer, New York, 1990 [Ju] Matti Jutila, On harater sums and lass numbers, J Number heory [SL] Serge Lang, Algebrai Number heory, Grad exts in Math, vol 110, Springer, New York, 1996 [Lu2] W Luo, Values of symmetri square L-funtions at 1, J Reine Angew Math [Lu1] W Luo, On Heke L-series assoiated with ubi haraters, Compos Math [Na] W Narkiewiz, Elementary and Analyti heory of Algebrai Numbers, seond ed, PWN/Springer, 1990