1 Appendices We collect some results that might be covered in a first course in algebraic number theory. A. Quadratic Reciprocity Via Gauss Sums A1. Introduction In this appendix, p is an odd prime unless otherwise specified. A quadratic equation 2 modulo p looks like ax + bx + c =0inFp. Multiplying by 4a, we have 2 2ax + b ≡ b2 − 4ac mod p Thus in studying quadratic equations mod p, it suffices to consider equations of the form x2 ≡ a mod p. If p|a we have the uninteresting equation x2 ≡ 0, hence x ≡ 0, mod p. Thus assume that p does not divide a. A2. Definition The Legendre symbol a χ(a)= p is given by 1ifa(p−1)/2 ≡ 1modp χ(a)= −1ifa(p−1)/2 ≡−1modp. If b = a(p−1)/2 then b2 = ap−1 ≡ 1modp,sob ≡±1modp and χ is well-defined. Thus χ(a) ≡ a(p−1)/2 mod p. A3. Theorem a The Legendre symbol ( p ) is 1 if and only if a is a quadratic residue (from now on abbre- viated QR) mod p. Proof.Ifa ≡ x2 mod p then a(p−1)/2 ≡ xp−1 ≡ 1modp. (Note that if p divides x then p divides a, a contradiction.) Conversely, suppose a(p−1)/2 ≡ 1modp.Ifg is a primitive root mod p, then a ≡ gr mod p for some r. Therefore a(p−1)/2 ≡ gr(p−1)/2 ≡ 1modp, so p − 1 divides r(p − 1)/2, hence r/2 is an integer. But then (gr/2)2 = gr ≡ a mod p, and a isaQRmodp. ♣ 2 A4. Theorem The mapping a → χ(a) is a homomorphism from Fp to {±1}. Proof. We compute ab a b ≡ (ab)(p−1)/2 = a(p−1)/2b(p−1)/2 ≡ p p p so χ(ab)=χ(a)χ(b). ♣ A5. Theorem If g is a primitive root mod p, then χ(g)=−1, so g is a quadratic nonresidue (from now on abbreviated QNR) mod p. Consequently, exactly half of the integers 1,2,...,p − 1 are QR’s and half are QNR’s. Proof.Ifh2 ≡ g mod p then h is a primitive root with twice the period of g, which is impossible. Thus by (A4), χ(ag)=−χ(a), so a → ag gives a bijection between QR’s and QNR’s. ♣ A6. The First Supplementary Law From the definition (A2) and the fact that (p − 1)/2isevenifp ≡ 1 mod 4 and odd if p ≡ 3 mod 4, we have −1 1ifp ≡ 1mod4 =(−1)(p−1)/2 = p −1ifp ≡ 3mod4. A7. Definition Let K be a field of characteristic = p such that K contains the p-th roots of unity. Let ζ ∈ K be a primitive p-th root of unity. Define the Gauss sum by p−1 a τ = ζa. p p a=1 A8. Theorem 2 − (p−1)/2 τp =( 1) p. Proof. From the definition of Gauss sum and (A4) we have p−1 ab τ 2 = ζa+b. p p a,b=1 3 For each a, we can sum over all c such that b ≡ ac mod p. (As c ranges over 1,... ,p− 1, ac also takes all values 1,... ,p− 1.) Thus p−1 p−1 a2c τ 2 = ζa+ac. p p a=1 c=1 a2 Since p = 1, this simplifies to p−1 p−1 c τ 2 = ζa(1+c) . p p c=1 a=1 If 1 + c ≡ 0modp then 1,ζ1+c,ζ2(1+c),... ,ζ(p−1)(1+c) runs through all the roots (zeros) of Xp − 1 (note that ζp = 1). But the coefficient of Xp−1 is 0, so the sum of the roots is 0. Therefore the sum of ζ1+c,ζ2(1+c),... ,ζ(p−1)(1+c) is -1. If 1 + c ≡ 0modp, then we are summing p − 1 ones. Consequently, p−2 c −1 τ 2 = − +(p − 1) . p p p c=1 ≡ ↔ − − −1 (Note that 1 + c 0modp c = p 1.) We can sum from 1 to p 1ifweadd p , hence p−1 c −1 τ 2 = − + p . p p p c=1 The sum is 0 by (A5), and the result follows from (A6). ♣ A9. The Law of Quadratic Reciprocity Let p and q be odd primes, with p = q. Then p q =(−1)(p−1)(q−1)/4. q p Thus if either p or q is congruent to 1 mod 4, then p is a QR mod q if and only if q is a QR mod p; and if both p and q are congruent to 3 mod 4, then p isaQRmodq if and only if q is a QNR mod p. Proof. Let K have characteristic q and contain the p-th roots of unity. For example, take p K to be the splitting field of X − 1 over Fq. Then q 2 (q−1)/2 τp =(τp ) τp which by (A8) is (q−1)/2 (p−1)/2 (−1) p τp. 4 Thus q − (p−1)(q−1)/4 (q−1)/2 τp =( 1) p τp. But by the binomial expansion applied to the definition of τp in (A7), p−1 a τ q = ζaq p p a=1 (Recall that K has characteristic q,soaq = a in K.) Let c ≡ aq mod p and note that q−1 q = p p 1 because the product of the two terms is ( p ) = 1. Thus p−1 c q q τ q = ζc = τ . p p p p p c=1 q We now have two expressions (not involving summations) for τp ,so q (−1)(p−1)(q−1)/4 p(q−1)/2 = . p (q−1)/2 p F Since p =(q ) by (A2), the above equation holds not only in K but in q, hence p can be written as a congruence mod q. Finally, multiply both sides by ( q ) to complete the proof. ♣ A10. The Second Supplementary Law 2 2 =(−1)(p −1)/8 p so 2 1ifp ≡±1mod8 = p −1ifp ≡±3mod8. Thus if p ≡ 1 or 7 mod 8, then 2 is a QR mod p, and if p ≡ 3or5mod8,then2isa QNR mod p. Proof. Let K be a field of characteristic p containing the 8th roots of unity, and let ζ be a primitive 8th root of unity. Define τ = ζ + ζ−1. Then τ 2 = ζ2 + ζ−2 +2. Now ζ2 and ζ−2 are distinct 4th roots of unity, not ±1, so they must be negatives of each other (analogous to i and −i in C). Therefore τ 2 = 2. Modulo p we have 2 2 τ p =(τ 2)(p−1)/2 τ =2(p−1)/2 τ = τ = (ζ + ζ−1). p p 5 But by definition of τ, τ p =(ζ + ζ−1)p = ζp + ζ−p. Now (again as in C) ζp + ζ−p and ζ + ζ−1 will coincide if p ≡±1 mod 8, and will be negatives of each other if p ≡±3mod 8. In other words, 2 ζp + ζ−p =(−1)(p −1)/8(ζ + ζ−1). Equating the two expressions for τ p, we get the desired result. We can justify the appeal to the complex plane by requiring that K satisfy the constraints ζ2 + ζ−2 = 0 and 2 ζp + ζ−p =(−1)(p −1)/8(ζ + ζ−1). ♣ A11. Example We determine whether 113 is a QR mod 127: 113 127 ≡ ( 127 )=(113 ) because 113 1mod4; 127 14 a ( 113 )=(113 ) because ( p ) depends only on the residue class of a mod p; 14 2 7 7 ≡ ( 113 )=(113 )( 113 )=(113 ) by (A4) and the fact that 113 1mod8; 7 113 ≡ ( 113 )=( 7 ) because 113 1mod4; 113 1 ≡ ( 7 )=(7 ) because 113 1mod7 1 and since ( 7 ) = 1 by inspection, 113 is a QR mod 127. Explicitly, 113 +13(127) = 1764 = 42 2. A12. The Jacobi Symbol Let Q be an odd positive integer with prime factorization Q = q1 ···qs. The Jacobi symbol is defined by s a a = . Q q i=1 i It follows directly from the definition that a a a a a aa = , = . Q Q QQ Q Q Q Also, a a a ≡ a mod Q ⇒ = Q Q because if a ≡ a mod Q then a ≡ a mod qi for all i =1,... ,s. Quadratic reciprocity and the two supplementary laws can be extended to the Jacobi symbol, if we are careful. 6 A13. Theorem If Q is an odd positive integer then −1 =(−1)(Q−1)/2. Q Proof. We compute − s − s 1 1 (q −1)/2 s (q −1)/2 = = (−1) j =(−1) j=1 j . Q q j=1 j j=1 Now if a and b are odd then ab − 1 a − 1 b − 1 (a − 1)(b − 1) − + = ≡ 0mod2, 2 2 2 2 hence a − 1 b − 1 ab − 1 + ≡ mod 2. 2 2 2 We can apply this result repeatedly (starting with a = q1,b = q2) to get the desired formula. ♣ A14. Theorem If Q is an odd positive integer, then 2 2 =(−1)(Q −1)/8. Q Proof. As in (A13), s s 2 2 (q2−1)/8 = = (−1) j . Q q j=1 j j=1 But if a and b are odd then a2b2 − 1 a2 − 1 b2 − 1 (a2 − 1)(b2 − 1) − + = ≡ 0mod8. 8 8 8 8 a2−1 (a−1)(a+1) 2 − ≡ In fact 8 = 8 is an integer and b 1 0 mod 8. (Just plug in b =1, 3, 5, 7.) Thus a2 − 1 b2 − 1 a2b2 − 1 + ≡ mod 8 8 8 8 and we can apply this repeatedly to get the desired result.