16 The Quadratic Reciprocity Law
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1 16 The Quadratic Recirocity Law Fix an odd rime If is another odd rime, a fundamental uestion, as we saw in the revious section, is to know the sign, ie, whether or not is a suare mod This is a very hard thing to know in general But Gauss noticed something remarkable, namely that knowing is euivalent to knowing ;they need not be eual however He found the recise law which governs this relationshi, called the Quadratic Recirocity Law Gauss was very roud of ths result and gave several roofs We will give one of his roofs, which incidentally introduces a very basic, ubiuitous sum in Mathematics called the Gauss sum We will also give an alternate roof, which is in some sendse more clever than the first, due to Eisenstein Theorem Gauss Quadratic recirocity Let, be distinct odd rimes Then Exlicitly, /// { 1, if or is 1 mod 4 1, if and are 3 mod 4 This theorem is very useful in comutations Examle It is not easy to comute mod 691 Better to use then: ;
2 Proof#1oftheorem:, odd rimes, Put ξ e πi C Then ξ 1, but ξ m 1ifm< ξ is called a rimitive th root of unity in C Allowersofξ will be on the unit circle In fact, we get a regular -gon by converting the oint 1,ξ,,ξ 1 cyclotong circledivision Put R {α a 0 + a 1 ξ + + a ξ 1 ξ 1 a 0,a 1,,a 1 Z} Clearly, R Z, hence R has 0 1 Let be in R Then 1 1 α a i ξ,β b i ξ i1 i1 1 α ± β a i + b i ξ i R i1 Since ξ 1,givenanyn Z we can write n l + r, 0 r 1by Euclidean algorithm in Z, and conclude that ξ n ξ r So R contains all the integral owers of ξ Then it also contains finite integral linear combinations of such owers Conseuently, αβ R if α, β R So R is very much like Z Itisa-dimensional analog of Z This allows us to define the divisibility in R To be recise, if α, β R, wesaythatβ divides α, β α iff γ R such that α βγ In articular, R, anditmakessensetoaskif divides some number in R Definition: Letα, β R Wesaythat α β mod iff a b inr
3 This allows us to do congruence arithmetic mod in R To study, Gauss introduced the following Gauss Sum : S a ξ a a mod Clearly, S R Aside Not art of roof of Quad Reci, but interesting S 1 a1 a ξ a + a ξ a 1 a ξ a So if and if ure read or im Lemma 1: Proof of Lemma 1: 1 1,S S 1 a1 a ξ a + ξ a R R 1 1, S R ir a mod a mod b mod S 1 1 a ξ a a ab c mod ξ c b ξ a+b a mod b mod ξ a ξ b b ξ b ac a 3
4 So S c mod c mod ξ c ξ c a mod a mod ac a a 1 a c, where a a 1mod But a 1 a c where fc? c 0mod: fc S 1 1 a mod f0 a 1 as a 0 mod c mod 1 a c a mod a 0 mod f0 1 c 0mod: Note that, in this case, the set ξ c fc, a 0mod 1 {1 a c a mod, a 0mod} 1 a c runs over elements of Z/ {1} exactly once Indeed, given any b Z/, b 1mod, we can solve a + b 1mod, and the solution is uniue Therefore, fc b mod b 1 mod b 4
5 We roved earlier that so when c 0mod Conseuently S 1 1 Claim: c mod ξc 0 Proof of claim: ξ c c mod 1 ξ 0 Proof of claim: c mod b mod fc c 1 mod c mod b 0 1 1, ξ c ξ c 0 c mod c mod c mod c 0 mod ξ c ξ c+1 ξ c mod ξ c ξ c 0 as claimed ξ c 1+ξ + + ξ 1 1 ξ 1 ξ 0sinceξ 1 By claim, S This roves Lemma 1 Lemma 1: S 1 1 5
6 Lemma : S 1 mod This haens in R mod Proof of Lemma : S a mod a mod a ξ a ξ a + w, w R a a a because a ±1 and is odd In other words, S a ξ a mod a mod Since, is invertible mod, andthemaa a is a ermutation of Z/, alsoa 0mod iffa 0mod so the sum over a mod can be relaced with the sume over a mod Write b for a mod Then a b mod, where 1mod S b mod But b Since 1mod, So b So gives S b b ξ b mod * 1 1 b ξ b b b mod S 6 mod
7 This is justified because S 1 S 0mod, mod which follows from lemma 1 ProofofTheorem: Comute S 1 in different ways On the one hand, by lemma 1, ie, 1 S 1 S Euler mod 1 1 S 1 mod, S On the other hand, by lemma, S 1 So, utting them together we get mod mod Last time, gave a roof of Quadratic Recirocity law More recisely we roved: Theorem Gauss Let, be distinct, odd rimes Then
8 Examle: Check if 9 is a suare mod 43: 9 and 43 are distinct odd rimes, so by definition 9 mod 43 iff byqrl, as9 5mod QRL So 9 mod 43 Remark: QRL tells you a way to know whether is a suare mod or not But when it is a suare, it gives no rocedure to find the suare root One can use QRL to check whether a number is a rime, similar to the way one uses Fermat s little theorem For examle, one can show that m 179 is not a rime by looking at y def mod 179 Note: So,ifmis a rime, y 11 modm 179 Since 179 1mod4,byQRL, as 11 3 mod 8 on the other hand, one can check using PARI, or by successively suaring mod m 179, that modm 8
9 This is art of a homework roblem Get a contradiction! So the only ossibility is that 179 is not a rime which is easy to verify directly as But this method is helful, when it works, for larger numbers A histoical remark: GHHardy went to see Ramanujan, when the latter was dying of TB in England Then Ramanujan asked Hardy if the number of the taxicab Hardy came in was an interesting number Hardy said No, not interesting, just 179 Ramanujan relied immediately, saying, On the contrary, the number is interesting because it is the first number which can be written as a sum of cubes in two different ways Indeed we have A second roof of uadratic reci Eisenstein Eisenstein s trignometric lemma Lemma: Letn be a ositive, odd integer Then sin nx sin x n 1 4 n 1 j1 sin x sin πj n Proof: U to us Hint: treat as a olynomial in sin x: Examle: n 3 sin 3x sinx + x LHS sin x sin x sin x cos x +cosxsin x sin x sinxcos x +1 sin xsinx sin x 1 sin x+1 sin x3 4sin x RHS 4sin x sin π }{{ 3 } 3/ 3 4sin x 4 sin x 3 4 9
10 Sketch of roof of lemma: Use induction on n to show that sin nx sin x f nsin x, where f n is a olynomial in sin x of degree n 1 f 0 t 1,f 3 t 3 4t, On the other hand, the RHS of lemma is also of the form g n sin x, where g n is the exlicitly given olynomial in sin x of degree n 1 So it suffices to show that f n and g n have the same roots and that the leading coefficient of f n is 4 n 1 So when we use induction on n, check that the leading coefficient is 4 n 1 and that its roots are { sin πj n 1 j n 1 } Alternatively, check the constant coefficient by checking at x 0 Recall Gauss lemma: e s s S where S {1,,, 1 } and e x {±1} defined by s e s s, with s S Alying sin π, we get πs πes s sin sin πs e s sin since sin is an odd function So e s sin sin πs πs 10
11 By Gauss lemma, sin πs πs s S sin s S sin πs s S sin πs Note the ma S S is a ermutation of S So, πs sin πs sin s S s S 1 i1 sin πi sin πi 1 Alying Eisenstein s trig lemma with n and sub in 3, we get i 1 1 i 1 πi πj sin sin Can get everything we need from this without comuting the sines: Reversing the roles of and, weget i 1 i 1 Comaring 3 and 4, we see that πj πi sin sin
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