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On Polynomial Reciprocity Law View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Elsevier - Publisher Connector Journal of Number Theory 101 (2003) 13–31 http://www.elsevier.com/locate/jnt On polynomial reciprocity law Chih-Nung Hsu Department of Mathematics, National Taiwan Normal University, 88 Sec. 4 Ting-Chou Road, Taipei, Taiwan Received 19 June 2001; revised 25 July 2002 Communicated by D. Goss Abstract The well-known law of quadratic reciprocity has over 150 proofs in print. We establish a relation between polynomial Jacobi symbols and resultants of polynomials over finite fields. Using this relation, we prove the polynomial reciprocity law and obtain a polynomial analogue of classical Burde’s quartic reciprocity law. Under the use of our polynomial Poisson summation formula and the evaluation of polynomial exponential map, we get a reciprocity for the generalized polynomial quadratic Gauss sums. r 2003 Elsevier Science (USA). All rights reserved. MSC: primary 11T55; 11T23; 11T24 Keywords: Polynomial rings; Finite fields; Reciprocity law 1. Introduction The law of quadratic reciprocity q p pÀ1 qÀ1 ¼ðÀ1Þ 2 2 p 2 q 2 was formulated by Euler and Legendre but Gauss was the first to provide a complete proof. Eisenstein used the period function f ðzÞ¼2i sin 2pz (cf. [7, Chapter 5]) to give an ingenius proof for the law of quadratic reciprocity. E-mail address: [email protected]. 0022-314X/03/$ - see front matter r 2003 Elsevier Science (USA). All rights reserved. doi:10.1016/S0022-314X(03)00020-9 14 C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31 In 1969, Burde [2] has proved the following interesting theorem about biquadratic reciprocity: if p ¼ a2 þ b2; q ¼ c2 þ d2; a c 1; b d 0 ðmod 2Þ; ab40; cd40; p and q are primes, and the quadratic residue symbol ðq=pÞ2 ¼ 1; then q p pÀ1 ad À bc ¼ðÀ1Þ 4 ; p 4 q 4 p 2 where ðq=pÞ4 ¼ 1orÀ1 according as the quadratic residue q is or is not a quartic residue modulo p: In [10], Siegel obtained the following reciprocity law for generalized quadratic Gauss sums: sffiffiffiffiffiffi jXcjÀ1 jXajÀ1 piðan2 þ bnÞ jcj piðjacjÀb2Þ piðcn2 þ bnÞ exp ¼ exp exp ; c a 4ac a n¼0 j j n¼0 where a; b and c are integers with aca0 and ac þ b even. In 1857, Dedekind [5] stated that quadratic reciprocity holds over function fields; this was proved later by Artin [1].In[9], Merrill and Walling use their inversion formula of the polynomial theta function to give another proof. In [3], Carlitz proved a more general reciprocity law for function fields which includes Dedekind’s quadratic law as a special case. In another paper [4], Carlitz use the Carlitz exponential map to reprove the polynomial reciprocity law. In Section 2, we establish a relation between polynomial Jacobi symbols and resultants of polynomials over finite fields. Using this relation, the polynomial reciprocity law is easy to prove. Let Fq be the finite field of q elements with q odd, let prime p be its characteristic and let wq be its quadratic character. Let A ¼ Fq½T be the polynomial ring in T over Fq: For any 0am; aAA; the leading coefficient of m is denoted by sgnðmÞ; the polynomial Jacobi symbols fa=mg and ða=mÞq;2 are defined to be no a ¼ the resultant of polynomials sgnðmÞÀ1m and a over F ; m q no a a ¼ wq : m q;2 m To use the properties of resultants of polynomials over finite fields, we obtain the law of polynomial reciprocity in Theorem 2.1 that if a; bAA satisfy aba0; then we have b a sgnðaÞdeg b ¼ sgnðbÞdeg aðÀ1Þdeg a deg b; a b b a qÀ1 deg b deg a 2 deg a deg b ¼ wqðsgnðaÞÞ wqðsgnðbÞÞ ðÀ1Þ : a q;2 b q;2 When P and Q are monic irreducibles in A of even degree, the polynomial quartic residue symbol ðQ=PÞq;4 is defined to be 1 or À1 according as the equation C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31 15 4 x Q ðmod PÞ is or is not solvable in xAFq½T: We obtain a polynomial analogue of Burde theorem in Theorem 2.3: if ðQ=PÞq;2 ¼ 1; then we have (a) If q 1 ðmod 4Þ; then P Q ¼ 1: Q q;4 P q;4 (b) If q 3 ðmod 4Þ; then A B À A B P Q A B À A B 1 2 2 1 ¼ ¼ 1 2 2 1 ; P q;2 Q q;4 P q;4 Q q;2 where A1; A2; B1; B2 are given in (2). Let KN ¼ Fqðð1=TÞÞ denote the completion field of the rational function field FqðTÞ at the infinite place 1=T; in other words, every aAKN; if aa0; then a can be expressed as XÀN i a ¼ ciT ; i¼d where ciAFq and cd a0: The sign, degree, and absolute value of a are defined by d sgn a ¼ cd ; deg a ¼ d; and jaj¼q : The residue of a at infinite place is denoted by resN a ¼ cÀ1: Let Trq : Fq-Fp be the trace map of the finite field Fq onto the finite -  field Fp (identifying with Z=pZ). Let cq : Fq C be the standard additive character of Fq defined by 2pi Tr ðaÞ c ðaÞ¼exp q : q p  The polynomial exponential map E :KN-C is defined by EðaÞ¼cqðresNaÞ: Let P; Q; aAA satisfy PQa0 and set X Qx2 þ ax S ðQ; aÞ¼ E : P P xAA deg xodeg P In Section 4, we explore the generalized polynomial quadratic Gauss sums SPðQ; aÞ: In Section 3, we establish a polynomial Poisson summation formula in Theorem 3.2. Using our polynomial Poisson summation formula and the evaluation of the polynomial exponential map E; the sums SPðQ; aÞ enjoy a reciprocity law in 16 C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31 Theorem 4.1: 8 sffiffiffiffiffiffiffi > > a2 jQj > E À S ðQ; aÞ if 2j degðPQÞ; < 4PQ jPj P SQðP; aÞ¼ sffiffiffiffiffiffiffi > 2 > a jQj :> w ðsgnðPQÞÞrqE À SPðQ; aÞ if 2[degðPQÞ; q 4PQ jPj where the polynomial quadratic sign function rq is given in (7). 2. Polynomial Jacobi symbols and resultants of polynomials over finite fields Let Fq be the finite field of q elements, let p be its characteristic satisfying pa2 and % % let Fq denote the algebraic closure of Fq: Let Fq½T be the polynomial ring in T over % % % Fq: For any polynomials m; aAFq½T; let RESðm; aÞAFq denote the resultant of % polynomials m and a in T over Fq (cf. [8, Chapter IV]); i.e., Y ÀÁ deg a deg m % RESðm; aÞ¼sgnðmÞ sgnðaÞ ai À bj AFq; 1pipdeg m 1pjpdeg a where ai (resp. bj) are roots (counting multiplicities) of m ¼ 0 (resp. a ¼ 0) in % % variable T over Fq: If ma0; then we define the polynomial Jacobi symbol fa=mgAFq to be no a ¼ RESðsgnðmÞÀ1m; aÞ: m % Let aAFq and Xn j % a ¼ cjT AFq½T: j¼0 For any integer iX0; we define aðiÞ; aðiÞ and aðaÞ to be ðiÞ q i % a ¼ a AFq; Xn ðiÞ q i j % a ¼ cj T AFq½T; j¼0 Xn j % aðaÞ¼ cja AFq: j¼0 Let P be a monic irreducible polynomial in Fq½T of degree d and let aAFq½T: Let % ð0Þ ð1Þ ðdÀ1Þ aAFq be a root of P ¼ 0inT: Then a ¼ a ; a ; y; a are the roots of P ¼ 0in C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31 17 T and we have no a ¼ RESðP; aÞ P ¼ aðað0ÞÞaðað1ÞÞ?aðaðdÀ1ÞÞ 1þq1þ?þqdÀ1 a ðmod PÞAFq: ð1Þ The polynomial quadratic residue symbol ða=PÞq;2 is defined to be 8 > 0ifPja; <> no a a 2 A  ¼ > 1if Fq ; P q;2 :> P À1 otherwise: If 0amAFq½T and m ¼ sgnðmÞP1P2?Pr with each Pi monic irreducible in Fq½T; then, by definition, we have no a a a a ¼ ? m P1 P2 Pr and we define ða=mÞq;2 to be a a a a ¼ ? : m q;2 P1 q;2 P2 q;2 Pr q;2 - We denote the quadratic character of the finite field Fq by wq; i.e., wq : Fq C satisfies 8 > <> 0ifa ¼ 0; A Â2 wqðaÞ¼> 1ifa Fq ; :> À1 otherwise: The relation between functions fg; ðÞq;2 and wq is given by no a a wq ¼ : P P q;2 Theorem 2.1 (Polynomial reciprocity law). Let r be a positive integer satisfy- ing rjðq À 1Þ; let 0am; aAFq½T and let P be a monic irreducible polynomial in Fq½T: Then (a) Pja if and only if fa=Pg¼0: [ Âr (b) If P a; then fa=PgAFq if and only if there exists a polynomial xAFq½T such that xr a ðmod PÞ: 18 C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31 (c) We have no a AF : m q (d) If a; bAFq½T satisfies aba0; then we have no b a sgnðaÞdeg b ¼ sgnðbÞdeg aðÀ1Þdeg a deg b: a b (e) If a; bAFq½T satisfies aba0; then we have b a qÀ1 deg b deg a 2 deg a deg b ¼ wq ðsgnðaÞÞ wq ðsgnðbÞÞ ðÀ1Þ : a q;2 b q;2 Proof. The proof of (a) follows from (1). To prove (b), if P[a and xr a ðmod PÞ; then by (1), we have no a ? dÀ1 ¼ðxrÞ1þqþ þq ðmod PÞ P 1þqþ?þqdÀ1 r Âr ¼ðx ðmod PÞÞ AFq : r  Conversely, let fa=Pg¼a for some aAFq : Since Fq½T=ðPÞ-Fq dÀ1 ½x/x1þqþ?þq ðmod PÞ is the norm map from the finite field Fq½T=ðPÞ onto the finite field Fq; there exists a polynomial xAFq½T such that dÀ1 x1þqþ?þq a ðmod PÞ: This implies that no dÀ1 a dÀ1 xrð1þqþ?þq Þ a1þqþ?þq ðmod PÞ: P Thus xr À a ¼ 0: P C.-N.
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