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The Legendre Symbol (Z/Pz) to (Z/P Z) Quadratic Reciprocity the Second Supplement

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The Legendre Symbol (Z/Pz) to (Z/P Z) Quadratic Reciprocity the Second Supplement × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement The Legendre Symbol R. C. Daileda Trinity University Number Theory Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement Definitions Given an odd prime p and a 2 Z with p - a, we say a is a quadratic residue of p if a ≡ b2 (mod p) for some b. Otherwise a is a quadratic nonresidue. The Legendre symbol of a at p is ( a 1 if a is a quadratic residue of p; = p −1 otherwise. a is clearly p-periodic in a. Thus we can view p · :( =p )× ! {±1g: p Z Z Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement Euler's criterion immediately implies the next result. Theorem Let p be an odd prime, p - a. Then a ≡ a(p−1)=2 (mod p): p We can use this theorem to prove the following important fact. Theorem The Legendre symbol is completely multiplicative and induces a surjective homomorphism · :( =p )× ! {±1g: p Z Z Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement Proof. × We have already seen that exactly half of the elements of (Z=pZ) are squares a.k.a. quadratic residues. · Therefore p is surjective. Let a; b 2 Z be coprime to p. Then so is ab and ab a b ≡ (ab)(p−1)=2 ≡ a(p−1)=2b(p−1)=2 ≡ (mod p): p p p · Because the values of p belong to {±1g, ab a b − = 0; ±2: p p p But the left-hand side is divisible by the odd prime p, so ±2 are impossible. This proves the result. Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement Quadratic Reciprocity, First Supplement: a = −1 When a = −1, the first theorem tells us that −1 ≡ (−1)(p−1)=2 (mod p): (1) p Both sides of the congruence belong to {±1g. Because p is odd, 1 6≡ −1(mod p). Hence (1) must actually be an equality. Theorem (Quadratic Reciprocity, First Supplement) Let p be an odd prime. Then −1 = (−1)(p−1)=2: p Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement Reduction to Prime Argument Q ni Given a coprime to p, write a = i qi with qi 6= p distinct primes, 2 {±1g. According to the preceding theorem ni a Y qi Y qi = = : p p p p p i i ni odd The First Supplement evaluates p . The Second Supplement will 2 evaluate p . q We are therefore reduced to evaluating p where q 6= p is an odd prime. This is the subject of the Law of Quadratic Reciprocity. Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement Back to (Z=pmZ)× a Let a 2 Z be coprime to p. It turns out that p controls whether m or not a is a square (mod p ) for all m 2 N! Theorem a Suppose p = 1. Then there is a sequence of integers b1; b2; b3;::: so that: m bm+1 ≡ bm (mod p ) for all m ≥ 1; 2 m bm ≡ a (mod p ) for all m ≥ 1. m m × In particular, a + p Z 2 (Z=p Z) is a square for all m ≥ 1. Remark: These conditions imply that the sequence fb g1 p m m=1 converges in the ring Zp of p-adic integers to a. Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement Proof: We recursively construct the sequence of b's. 2 Since a is a quadratic residue of p, there is a b1 so that a ≡ b1 (mod p). Suppose we have found b1; b2;:::; bm as in the theorem. m m Consider bm + kp , which is ≡ bm (mod p ) for any choice of k. Moreover m 2 2 m 2 2m (bm + kp ) = bm + 2bmkp + k p 2 m m+1 ≡ bm + 2bmkp (mod p ) m m m+1 ≡ a + `p + 2bmkp (mod p ) m m+1 ≡ a + (` + 2bmk)p (mod p ): We need to choose k so that pj` + 2bmk, i.e. 2bmk ≡ −`(mod p). Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement 2bmk ≡ −`(mod p) is a linear congruence in the variable k. Since p - 2bm, we have (2bm; p) = 1, which means the congruence has a unique solution (mod p). Choose any element k of the solution set and define m bm+1 = bm + kp . Then bm+1 has the desired properties. Continuing this process indefinitely yields the sought after sequence. Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement Example Example Show that −1 is a square (mod 625) and find its two square roots. We have −1 = (−1)(5−1)=2 = (−1)2 = 1: 5 Hence −1 is a square (mod 5m) for all m ≥ 1. To find its square roots we implement the algorithm in the proof. 2 Since −1 ≡ 4(mod 5), clearly b1 = 2. And since 2 = −1 + 1 · 5, ` = 1. So b2 = b1 + 5k where 2b1k ≡ −` (mod 5) ) 4k ≡ −1 (mod 5) ) k ≡ 1 (mod 5); i.e. b2 = 7. Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement 2 2 Now b2 = −1 + 2 · 5 so that ` = 2. So we need to solve 2b2k ≡ −` (mod 5) ) 4k ≡ −2 (mod 5) ) k ≡ 2 (mod 5); 2 and b3 = b2 + 5 k = 57. 2 3 Finally, b3 = −1 + 26 · 5 yields ` = 26 and we solve for k: 2b3k ≡ −` (mod 5) ) 4k ≡ −1 (mod 5) ) k ≡ 1 (mod 5): 3 Thus b4 = b3 + 5 k = 182. Therefore the two square roots of −1(mod 625) are ±182 (mod 625) : Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement Gauss' Lemma Let's finally start heading toward the Law of Quadratic Reciprocity. Our first auxiliary result is the following. Lemma (Gauss' Lemma) Let p be an odd prime and suppose that p - a. For each p − 1 r 2 a; 2a; 3a;:::; a 2 n p−1 p−3 p−1 o choose the unique sr 2 − 2 ; − 2 ;:::; −1; 1; 2;:::; 2 so that r ≡ sr (mod p). Then a = (−1)ν; p where ν is the number of negative values of sr . Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement Proof of Gauss' Lemma n p−1 p−3 p−1 o Let Ip = − 2 ; − 2 ;:::; −1; 1; 2;:::; 2 . Note that if s 6= t 2 Ip then: (s; p) = (t; p) = 1; 0 < js − tj ≤ p − 1 < p ) p - s − t ) s 6≡ t (mod p); jIpj = p − 1 = '(p). Therefore the map × Ip ! (Z=pZ) s 7! s + pZ is a bijection and sr is well-defined. Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement 0 Suppose sr = −sr 0 . Then r ≡ −r (mod p). p−1 Thus ai ≡ −aj (mod p) for some 1 ≤ i; j ≤ 2 . Since (a; p) = 1, we can cancel it to obtain i ≡ −j (mod p) or i + j ≡ 0 (mod p) ) pji + j: But 0 < i + j < p − 1, so this is impossible. Therefore fsr g cannot contain both s and −s for any s 2 Ip. It follows that p − 1 fsr g = 1 · 1; 2 · 2; 3 · 3; : : : ; p−1 · 2 2 where each i 2 {±1g. Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement Now multiply together all of the r's and sr 's: (p−1)=2 p − 1 p − 1 a ! ≡ 12 ··· p−1 ! (mod p) 2 2 2 | {z } | {z } the r's the sr 's By an earlier theorem we therefore have a a ≡ (−1)ν (mod p) ) = (−1)ν p p since both sides belong to {±1g and p is odd. Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement Example p−1 Consider p = 11 and a = 7. We have 2 = 5 and 7 ≡ −4 (mod 11); 2 · 7 ≡ 3 (mod 11); 3 · 7 ≡ −1 (mod 11); 4 · 7 ≡ −5 (mod 11); 5 · 7 ≡ 2 (mod 11): Therefore, by Gauss' Lemma, 7 = (−1)3 = −1; 11 and 7 is a quadratic nonresidue (mod 11). Daileda The Legendre Symbol × m × The Legendre Symbol (Z=pZ) to (Z=p Z) Quadratic Reciprocity The Second Supplement Remark n p−1 o Let r 2 a; 2a; 3a;:::; 2 a . Use the Division Algorithm to write r = qp + r 0; 0 ≤ r 0 < p: 0 Since r ≡ r (mod p), the uniqueness of sr implies: 0 p 0 If r < 2 , then sr = r . 0 p p 0 0 If r > 2 , then − 2 < r − p < 0 and r ≡ r − p (mod p). 0 Hence sr = r − p. It follows that ν in Gauss' Lemma is the number of r 0 that exceed p=2. This alternate characterization of ν can sometimes be useful.
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