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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 Fq½TŠ 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 xAFq½TŠ: 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 - Â 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 Fq½TŠ be the polynomial ring in T over % % % Fq: For any polynomials m; aAFq½TŠ; 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 AFq½TŠ: 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 AFq½TŠ; j¼0 Xn j % aðaÞ¼ cja AFq: j¼0

Let P be a monic irreducible polynomial in Fq½TŠ of degree d and let aAFq½TŠ: 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 Â ¼ > 1if Fq ; P q;2 :> P À1 otherwise:

If 0amAFq½TŠ and m ¼ sgnðmÞP1P2?Pr with each Pi monic irreducible in Fq½TŠ; 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; aAFq½TŠ and let P be a monic irreducible polynomial in Fq½TŠ: 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 xAFq½TŠ 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; bAFq½TŠ satisfies aba0; then we have  no b a sgnðaÞdeg b ¼ sgnðbÞdeg aðÀ1Þdeg a deg b: a b

(e) If a; bAFq½TŠ 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

Fq½TŠ=ðPÞ-Fq

dÀ1 ½xŠ/x1þqþ?þq ðmod PÞ is the norm map from the finite field Fq½TŠ=ðPÞ onto the finite field Fq; there exists a polynomial xAFq½TŠ 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. Hsu / Journal of Number Theory 101 (2003) 13–31 19

By (a), we obtain that

xr  a ðmod PÞ:

The proof of (c) follows from the definition of resultant. To prove (d), we have  b sgnðaÞdeg b ¼ RESða=sgnðaÞ; bÞ sgnðaÞdeg b a ¼ RESðb; a=sgnðaÞÞ sgnðaÞdeg bðÀ1Þdeg a deg b

¼ RESðb=sgnðbÞ; aÞ sgnðbÞdeg aðÀ1Þdeg a deg b no a ¼ sgnðbÞdeg aðÀ1Þdeg a deg b: b

ðqÀ1Þ=2 The proof of (e) follows from (d) and wqðÀ1Þ¼ðÀ1Þ : &

Remark. The polynomial quadratic reciprocity (e) is a special case of (d) and (e) can also be derived from product formula of the quadratic Hilbert symbols for rational function field FqðTÞ (cf. [6, Theorem 0.4]). Let P be a monic irreducible polynomial in Fq½TŠ of degree 2d: Let Fq2 be the % 2 unique subfield of Fq with q elements and let P ¼ P1P2 where the Pi are monic irreducible polynomials in Fq2 ½TŠ: Basic facts for Pi are

ð1Þ deg P1 ¼ deg P2 ¼ d and P2 ¼ P1 :

Proposition 2.2. Let notation be as above. Then we have

2 (1) The resultant RESðP1; P2Þ satisfies 0aRESðP1; P2ÞAFq2 ; RESðP1; P2Þ AFq and  RESðP1; P2ÞAFq if and only if 4j deg P: (2) The polynomial P1 þ P2AFq½TŠ and the polynomial quadratic residue symbol ðP þ P Þ=2 1 2 ¼ 1 P q;2

if and only if 2jdor2[d and q  3 ðmod 4Þ: À1 (3) The leading coefficient c ¼ sgnðP1 À P2Þ and monic polynomial B ¼ c ðP1 À A e  e  2A  P2Þ satisfy B Fq½TŠ; c Fq2 ; c Fq ; but c Fq : (4) Let B be as in (3). Then the polynomial quadratic residue symbol B ¼ 1: P q;2 20 C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31

 Â2 (5) There exist monic polynomials A; BAFq½TŠ and gAFq such that geFq ;

P ¼ A2 À gB2;

deg A ¼ d and deg Bod: The solutions A; B and g above are unique.

% ð1Þ ð2Þ ð2dÞ Proof. To prove (1), if aAFq is a root of P ¼ 0; then a ; a ; y; a ¼ a are distinct and are all the roots of P ¼ 0: We may assume that að1Þ; að3Þ; y; að2dÀ1Þ (resp. ð2Þ ð4Þ ð2dÞ a ; a ; y; a ) are the roots of P1 ¼ 0 (resp. P2 ¼ 0). Thus, we have

Y Y  ðiÞ ðjÞ RESðP1; P2Þ¼ a À a a0: i odd j even 1pip2dÀ1 2pjp2d

This implies that

q ð1Þ d2 RESðP1; P2Þ ¼ RESðP1; P2Þ ¼ðÀ1Þ RESðP1; P2Þ and the proof of (1) is complete. To prove (2), the first assertion follows from

ð1Þ ð1Þ ð1Þ ð1Þ ð2Þ ð1Þ ðP1 þ P2Þ ¼ðP1 þ P1 Þ ¼ P1 þ P1 ¼ P1 þ P1:

We have

 ðP þ P Þ=2 2À1 P þ P 1 2 ¼ 1 2 P P P P 1 2 2À1 P P ¼ 2 1 P P1 P2 À2d d2 2 ¼ 2 ðÀ1Þ RESðP1; P2Þ :

ðqÀ1Þ=2 The second assertion follows from (1) and wqðÀ1Þ¼ðÀ1Þ : ð2Þ ð1Þ ð1Þ ð2Þ ð1Þ To prove (3), since P1 ¼ P1; ðP1 À P2Þ ¼ P1 À P1 ¼ P1 À P1 ¼ÀðP1 À P2Þ ð2Þ ð1Þ ð2Þ and ðP1 À P2Þ ¼ P1 À P2: Thus c ¼Àc and c ¼ c: Combining these, we have A  e  2A  A c Fq2 ; c Fq ; c Fq and B Fq½TŠ: C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31 21

To prove (4), we have   B aÀ1 P À P ¼ 1 2 P P P P 1 2  À1 P P ¼ðaÀ2Þd 2 1 P1 P1 P2 À2 d d2þd 2 ¼ða Þ ðÀ1Þ RESðP1; P2Þ

À2 d 2 ¼ða Þ RESðP1; P2Þ :

The proof of (4) follows from (1) and (3). To prove (5), let A ¼ðP1 þ P2Þ=2: Then we have

a2 P ¼ A2 À B2: 4

Since the decomposition of P ¼ P1P2 in Fq2 ½TŠ is unique, the remainder is easy to prove. &

Suppose that P and Q are monic irreducible polynomials in Fq½TŠ of even degrees. By polynomial reciprocity law, we know Q P qÀ1 ¼ðÀ1Þ 2 deg P deg Q ¼ 1: P q;2 Q q;2

Thus if ðQ=PÞq;2 ¼ 1; then ðP=QÞq;2 ¼ 1: Under this situation, the polynomial quartic residue symbol ðQ=PÞq;4 is defined to be 1 or À1 according as the equation 4 x  Q ðmod PÞ is or is not solvable in xAFq½TŠ: If q  3 ðmod 4Þ; then since wqðÀ1Þ¼À1; by Proposition 2.2(5), we may write

2 2 2 2 P ¼ A1 þ B1; Q ¼ A2 þ B2; ð2Þ where A1; A2 are monic polynomials in Fq½TŠ with deg A1 ¼ deg P=2; deg A2 ¼ deg Q=2 and B1; B2 are polynomials in Fq½TŠ with deg B1odeg P=2; deg B2o Â2 Â2 deg Q=2; sgnðB1ÞAFq and sgnðB2ÞAFq :

Theorem 2.3 (Polynomial Burde theorem). Let P; Q; A1; A2; B1 and B2 be as above. Suppose Q ¼ 1: P q;2 22 C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31

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

Proof. Let P ¼ P1P2; Q ¼ Q1Q2 where P1; P2; Q1 and Q2 are monic irreducible polynomials in Fq2 ½TŠ: Since the finite field Fq½TŠ=ðPÞ is isomorphic to the finite field Fq2 ½TŠ=ðPiÞ via the nature mapping

Fq½TŠ=ðPÞ-Fq2 ½TŠ=ðPiÞ

a ðmod PÞ/a ðmod PiÞ;

4 A ðQ=PÞq;4 ¼ 1 if and only if the equation x  Q ðmod PiÞ is solvable in x Fq2 ½TŠ: By Theorem 2.1 ðbÞ and 4jðq2 À 1Þ; we have  Q Q 4 A  ¼ 1 if and only if Fq2 : P q;4 Pi

Similarly, we have  P P Â4 ¼ 1 if and only if AF 2 ; Q Q q q;4 i Q Q Â2 ¼ 1 if and only if AF 2 ; P P q q;2 i P P 2 A Â ¼ 1 if and only if Fq2 : Q q;2 Qi

Since ðQ=PÞ ¼ 1; we have ðP=QÞ ¼ 1 (follows from polynomial reciprocity law), q;2 q;2  P Q 2 A Â ; Fq2 ; Q2 P1 and  À1 P Q P Q 4 A Â ¼ 1 if and only if Fq2 : ð3Þ Q q;4 P q;4 Q2 P1 C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31 23

By Theorem 2.1(d), we have

  P Q À1 P P Q À1 Q À1 ¼ 1 2 1 2 Q2 P1 Q2 Q2 P1 P1 () Pð1Þ Q À1 1 deg P1 deg Q2 1 1 ¼ðÀ Þ ð1Þ Q P1 1  P ð1Þ Q À1 ¼ðÀ1Þdeg P1 deg Q2 1 1 Q1 P1  Q qÀ1 ¼ðÀ1Þdeg P1 deg Q2 ðÀ1Þdeg P1 deg Q1 1 P1  Q qÀ1 ¼ 1 : ð4Þ P1

Combining (3) and (4), we have 8 <> 1ifq  1 ðmod 4Þ; P Q ¼ > Q1 ð5Þ Q q;4 P q;4 : if q  3 ðmod 4Þ: P1 q2;2

If q  3 ðmod 4Þ; then we have

2 2 P ¼ P1P2 ¼ A1 þ B1; P1 ¼ A1 þ aB1; P2 ¼ A1 À aB1;

2 2 Q ¼ Q1Q2 ¼ A2 þ B2; Q1 ¼ A2 þ aB2; Q2 ¼ A2 À aB2;

A  2 where a Fq2 satisfies a ¼À1: Thus we have

P1B2  A1B2 þ aB1B2 ðmod Q1Þ

 A1B2 À A2B1 ðmod Q1Þ:

Since Fq½TŠ=ðQÞ is isomorphic to Fq2 ½TŠ=ðQ1Þ via the nature mapping

Fq½TŠ=ðQÞ-Fq2 ½TŠ=ðQ1Þ

a ðmod QÞ/a ðmod Q1Þ; 24 C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31 we have A B À A B A B À A B 1 2 2 1 ¼ 1 2 2 1 Q Q 2 q;2 1 q ;2 P B ¼ 1 2 Q 2 1 q;2 P B ¼ 1 2 Q 2 Q 2 1 q ;21 q ;2 P B ¼ 1 2 Q1 q2;2 Q q;2 by Proposition 2.2(4) P ¼ 1 Q1 q2;2 by Theorem 2.1(e) q2À1 Q1 ¼ðÀ1Þ 2 deg P1 deg Q1 : P1 q2;2

Since q  3 ðmod 4Þ; we have A B À A B Q 1 2 2 1 ¼ 1 : Q q;2 P1 q2;2

By symmetry, we have A B À A B Q P A B À A B 1 2 2 1 ¼ 1 ¼ 1 ¼ 1 2 2 1 : Q q;2 P1 q2;2 Q1 q2;2 P q;2

Combining this with (5), the proof is complete. &

3. Integral of polynomial exponential mapping and polynomial Poisson summation formula

In this and the next sections, let Fq be a finite field with q elements and q is odd, let p be its characteristic. Let A ¼ Fq½TŠ (resp. K ¼ FqðTÞ) be the polynomial ring (resp. rational function field) with coefficients in Fq: Let KN ¼ Fqðð1=TÞÞ denote the completion field of K at infinite place, in other words, every aAKN; if aa0; then a can be expressed as

XÀN i a ¼ ciT ; i¼d C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31 25 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 M be the set of aAKN with deg ao0: Let Trq : Fq-Fp be the trace map of the finite field Fq onto the finite field Fp - Â (identifying with Z=pZ). Let cq : Fq C be the standard additive character of the finite field Fq defined by 2pi Tr ðaÞ c ðaÞ¼exp q : q p

 The exponential map E :KN-C is defined by

EðaÞ¼cqðresNaÞ:

 The exponential map E is a non-trivial additive character from KN to C and the d Pontryagin (self) duality KN ¼ KN is deduced by the bilinear map

 KN  KN-C

ða; bÞ/Eða Á bÞ:

The Haar measure for KN is defined to be Z 1 da ¼ 1: aAM

Let aAA: We define a þ M to be the set fa þ xjxAMg:

 Theorem 3.1. Let cAFq and let d be a positive integer with ðp; dÞ¼1: Then

(a) Let r be an integer,0aaAA and aAKN with deg a ¼ r: Then we have Z ( ÀÁ 0 if ðd À 1Þ deg a þ rX0; E a xd dx Á ¼ d aþM cqðresNða Á a ÞÞ if ðd À 1Þ deg a þ rp À 1:

(b) If n is an integer, then we have 8 Z <> qn if no0; d X Eðc Á x Þ dx ¼ > 0 if n 0 and d ¼ 1; deg x n : o 1 if nX0 and da1: 26 C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31

(c) If integer r satisfies 0orod; then we have 8 > n > q if np À 1; Z <> ÀÁ1 P r d d EcÁ T Á x dx ¼ cqðcc1 Þ if nX0 and r ¼ d À 1; > q A deg xon > c1 Fq :> 1 if nX0 and rad À 1:

(d) If aAKN satisfies deg a ¼ dn þ r ð0prodÞ; then we have 8 > 1 if n 0; > o > Z <> 0 if nX0; d ¼ 1; Eða Á xd Þ dx ¼ 1 P > c ðsgnðaÞcd Þ if nX0; dX2; r ¼ d À 1; M > qnþ1 q 1 > c1AFq :> qÀn if nX0; dX2; rad À 1:

Proof. To prove (a), we may assume sgnðaÞ¼sgnðaÞ¼1 and set deg a ¼ nX0: Since p[d; we may write a ¼ T rbd for some bA1 þ M: Thus, in this assumption, we have Z Z Eða Á xd Þ dx ¼ EðT rðbxÞd Þ dx aþM ZaþM ¼ EðT rxd Þ dx: aþM

We write

X0 n i a ¼ T þ aiT ; i¼nÀ1 where aiAFq: If

À1 À2 x ¼ a þ cÀ1T þ cÀ2T þ ?Aa þ M; then we have

r d resNðT Á x Þ ( r d resNðT Á a ÞþdcÀðdnÀnþrþ1Þ þ C if ðd À 1Þn þ rX0; ¼ r d resNðT Á a Þ if ðd À 1Þn þ rp À 1; C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31 27

[ where C is a polynomial in cÀ1; cÀ2; y; cÀðdnÀnþrÞ; ai over Fq: Since p d; it is clear that Z ( 0ifðd À 1Þn þ rX0; E T r xd dx ð Á Þ ¼ r d aþM cqðresNðT Á a ÞÞ if ðd À 1Þn þ rp À 1:

To prove (b), it follows from (a) and Z ( 0ifd ¼ 1; Eðxd Þ dx ¼ M 1ifdX1:

To prove (c), the first assertion follows from the fact that Eðc Á T r Á xd Þ¼1 for all deg xo À 1: The second and final assertions follow from (a). To prove (d), since p[d; we may write

a ¼ sgnðaÞT rðT nbÞd for some bA1 þ M: Thus we have Z Z Eða Á xd Þ dx ¼ EðsgnðaÞT rðT nbxÞd Þ dx M ZM ¼ EðsgnðaÞT rðT nxÞd Þ dx M Z ¼ qÀn EðsgnðaÞTrxd Þ dx: deg xon

The remainder follows from (b) and (c). &

For any aAA; we define Ea :KN-C to be

EaðbÞ¼Eða Á bÞ:

The polynomial Poisson summation formula states in

Theorem 3.2. Let F :KN-C be a locally constant map of complex values on KN: Then for any integer dX0; we have X X Z FðaÞ¼ FðaÞEaðaÞ da: aAA aAA deg aod deg aod

Proof. This formula easily deduced from the basic theorem concerning Fourier series for the non-archimedean complete field KN: Let f ðaÞ coincide with FðaÞ for 28 C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31 aAM: Since the mapping

M Â A-C

ða; aÞ/EaðaÞ # deduces the Pontryagin dual M ¼ A; the Fourier series f ðaÞ is given by X ˆ f ðaÞ¼ f ðaÞEaðaÞ: aAA

Thus

Fð0Þ¼f ð0Þ X ¼ fˆðaÞ aAA X Z ¼ f ðaÞEaðaÞ da aAA M X Z ¼ FðaÞEaðaÞ da: aAA M

The remainder is easy to deduce. &

4. A generalized polynomial quadratic Gauss sums

The value of quadratic Gauss sum was conjectured and proved by Gauss in the form ( pffiffiffi XpÀ1 2pia2 p if p  1 ðmod 4Þ; exp ¼ pffiffiffi p a¼0 i p if p  3 ðmod 4Þ:

Using the well-known theorem of Davenport and Hasse, we obtain that the Gauss sum for the finite field Fq is given by X X ffiffiffi 2 p cqða Þ¼ wqðaÞcqðaÞ¼rq q; ð6Þ aAFq aAFq where the quadratic sign function rq is given by ( ÀðÀ1ÞdimFp Fq if p  1 ðmod 4Þ; rq ¼ ð7Þ ÀðÀiÞdimFp Fq if p  3 ðmod 4Þ: C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31 29

 If bAFq ; then, by (6), we have X ffiffiffi 2 p cqðba Þ¼wqðbÞrq q: ð8Þ aAFq

Let P; Q; aAA satisfy PQa0: We define the generalized polynomial quadratic Gauss sum SPðQ; aÞ to be X Qx2 þ ax S ðQ; aÞ¼ E : P P xAA deg xodeg P

Like (6), if P is a monic irreducible polynomial, Q ¼ 1 and a ¼ 0inA; then we get the polynomial quadratic Gauss sum   X x2 X x x SPð1; 0Þ¼ E ¼ E : P P q;2 P x ðmod PÞ x ðmod PÞ

In this section, we discuss the generalized polynomial quadratic Gauss sums SPðQ; aÞ: The sums SPðQ; aÞ enjoy a reciprocity law in Theorem 4.1. The key point of the proof of Theorem 4.1 is our polynomial Poisson summation formula.

Theorem 4.1. Let P; Q; aAA satisfy PQa0 and deg aodeg ðPQÞ: Then we have 8 sffiffiffiffiffiffiffi > > a2 jQj > E À S ðQ; aÞ if 2j deg ðPQÞ; < 4PQ jPj P SQðP; aÞ¼ sffiffiffiffiffiffiffi > 2 > a jQj :> w ðsgnðPQÞÞrqE À SPðQ; aÞ if 2[deg ðPQÞ; q 4PQ jPj where the quadratic sign function rq is given in (7).

Proof. First, we evaluate ! X 2 ða þ bQÞ E À 4PQ deg bodeg P ! X 2 a2 Qðb=2Þ þ aðb=2Þ ¼ E À E À 4PQ P deg bodeg P a2 ¼ E À S ðQ; aÞ: ð9Þ 4PQ P 30 C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31

Next, by polynomial Poisson summation formula, we have X Z Px2 þ ax S ðP; aÞ¼ E E ðxÞ dx Q Q b bAA deg xodeg Q X Z Px2 þ ax ¼ Ebxþ dx Q bAA deg xodeg Q X Z ða þ bQÞx ¼ jQj EPQx2 þ dx PQ bAA M !Z ! X 2 2 ða þ bQÞ a þ bQ ¼ jQjE À EPQxþ dx 4PQ 2PQ bAA M !Z ! X 2 2 ða þ bQÞ a þ bQ ¼ jQjE À EPQxþ dx 4PQ 2PQ deg bXdeg P M !Z ! X 2 2 ða þ bQÞ a þ bQ þ jQjE À EPQxþ dx 4PQ 2PQ deg bodeg P M by Theorem 3.1(a) and deg aodeg ðPQÞ !Z ! X 2 2 ða þ bQÞ a þ bQ ¼ jQjE À EPQxþ dx 4PQ 2PQ deg bodeg P M !Z X 2 ða þ bQÞ ¼ jQjE À EðPQx2Þ dx 4PQ deg bodeg P M by (9) Z 2 a 2 ¼jQjE À SPðQ; aÞ EðPQx Þ dx: 4PQ M In order to evaluate the integral above, we consider the following two cases: (1) If 2j deg ðPQÞ; then, by Theorem 3.1(d) with d ¼ 2 and n ¼ deg ðPQÞ=2; we obtain a2 deg ðPQÞ S ðP; aÞ¼jQjE À S ðQ; aÞqÀ 2 Q 4PQ P

sffiffiffiffiffiffiffi a2 jQj ¼ E À S ðQ; aÞ: 4PQ jPj P C.-N. Hsu / Journal of Number Theory 101 (2003) 13–31 31

(2) If 2[deg ðPQÞ; then, again by Theorem 3.1(d) with d ¼ 2; n ¼ðdeg ðPQÞÀ 1Þ=2; r ¼ 1 and by (8), we obtain sffiffiffiffiffiffiffi X jQj ÀÁa2 S ðP; aÞ¼qÀ1=2 c sgnðPQÞc2 E À S ðQ; aÞ Q jPj q 4PQ P cAFq sffiffiffiffiffiffiffi a2 jQj ¼ w ðsgnðPQÞÞrqE À SPðQ; aÞ: q 4PQ jPj

Thus the proof is complete. &

References

[1] E. Artin, Quadratische Ko¨ rper im Gebiete der ho¨ heren Kongruenzen, Math. Zeit. 19 (1924) 153–246. [2] K. Burde, Ein rationales biquadratisches Reziprozitatsgesetz, J. Reine Angew. Math. 235 (1969) 175–184. [3] L. Carlitz, The arithmetic of polynomials in a Galois field, Amer. J. Math. 54 (1932) 39–50. [4] L. Carlitz, On certain functions connected with polynomials in a Galois field, Duke Math. J. 1 (1935) 137–168. [5] R. Dedekind, Abriss einer Theorie der ho¨ heren Congruenzen in Bezug auf einer reellen Primzahl- Modulus, J. Reine Angew. Math. 54 (1857) 1–26. [6] J. Hoffstein, M. Rosen, Average values of L-series in function fields, J. Reine Angew. Math. 426 (1992) 117–150. [7] K. Ireland, M. Rosen, A Classical Introduction to Modern Number Theory, Springer, 1992. [8] S. Lang, Algebra, Addison–Wesley, Reading, MA, 1993. [9] K.D. Merrill, L.H. Walling, On quadratic reciprocity over function fields, Pacific J. Math. 173 (1996) 147–150. [10] C.L. Siegel, U¨ ber das quadratische Reziprozita¨ tsgestz algebraischen Zahlko¨ rpern, Nachr. Azad. Wiss. Go¨ ttingen Math.-Phys. Kl. II (1960) 1–16.

Further reading B.C. Berndt, R.J. Evans, The determination of Gauss sums, Bull. Amer. Math. Soc. 5 (2) (1981) 107–129. M. Gerstenhaber, The 152nd proof of the law of quadratic reciprocity, Amer. Math. Monthly 70 (1963) 397–398. E. Lehmer, On the quadratic character of some quadratic surds, J. Reine Angew. Math. 250 (1971) 42–48. E. Lehmer, On the quartic character of quadratic units, J. Reine Angew. Math. 268/269 (1974) 294–301. K.D. Merrill, L.H. Walling, Sums of squares over function fields, Duke Math. J. 71 (3) (1993) 665–684.