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Lecture 19: Quadratic Reciprocity (Contd.) Overview 1 Proof Of CS681 Computational Number Theory Lecture 19: Quadratic Reciprocity (contd.) Instructor: Piyush P Kurur Scribe: Ramprasad Saptharishi Overview Last class we proved one part of the Quadratic Reciprocity Theorem. We shall first finish the proof of the other part and then get to a generalization of the Legendre symbol - the Jacobi symbol. 1 Proof of Reciprocity Theorem (contd.) Recall the statement of the theorem. If p 6= q are odd primes, then p q p−1 q−1 = (−1) 2 2 q p The idea of the proof is just the same. We choses τ = 1 + i last time and p−1 p we got a 2 2 term while computing τ . It would be the same here as well. Let ζ be a principle q-th root of unity. We shall work with the field Fp(ζ). Let X a τ = ζa ? q a∈Fq Just as before, we compute τ 2. 2 X a a X b b τ = ζ ζ ? q ? q a∈Fq b∈Fq X a b = ζa+b ? q q a,b∈Fq X a b−1 = ζa+b ? q q a,b∈Fq X ab−1 = ζa+b ? q a,b∈Fq 1 Let us do a change of variable, by putting c = ab−1 X c τ 2 = ζbc+b ? q c,b∈Fq X c X X −1 = (ζc+1)b + ? q ? ? q −16=c∈Fq b∈Fq b∈Fq ? c+1 Since both p and q are primes and −1 6= c ∈ Fq, ζ is also a principle q-th P c+1 b root of unity. And therefore, b(ζ ) = −1. Therefore, X c −1 τ 2 = + (q − 1) ? q q −16=c∈Fq ? Look at the first term. If one were to sum over all elements of Fq, then half c of the q would be 1 and the other would be −1 thus fully cancelling off. −1 Since we are just excluding c = −1, the first term is just q . −1 −1 −1 τ 2 = + (q − 1) = q q q q Now to evaluate τ p. X a τ p = ζap ? q a∈Fq X cp−1 = ζc (c = ap) ? q c∈Fq p−1 X c = ζc q ? q c∈Fq p = τ q p 2 p−1 τ = τ(τ ) 2 p−1 p p−1 −1 2 =⇒ τ = τq 2 q q q p−1 q−1 = τ (−1) 2 2 p p q p−1 q−1 =⇒ = (−1) 2 2 q p 2 2 Jacobi Symbol m The legendre symbol n can be naturally generalized to the case when m and n are odd and coprime numbers. α1 α2 αk Definition 1. Suppose n = p1 p2 ··· pk . Then the Jacobi symbol, also repre- m sented as n , is defined as follows ( m 0 if (m, n) 6= 1 = αi n Qk m otherwise i=1 pi The Jacobi symbol also satisfies some nice multiplicative properties m1m2 m1 m2 • n = n n • m = m m n1n2 n1 n2 Using the above properties, we can get a generalize the theorem on the legendre symbol as well. Theorem 1. If m, n are odd numbers such that (m, n) = 1, then 2 n2−1 = (−1) 8 n m n m−1 n−1 = (−1) 2 2 n m We shall prove this theorem in the next class. The proof will just be us- ing induction on the factors of m and n. m WARNING: Please note that the Jacobi symbol n doesn’t say any- thing about whether or not m is a square modulo n. For example, if n = p1p2 and m was chosen such that m is not a square modulo p1 or p2. Then m = (−1)(−1) = 1 m p p . the Jacobi symbol p1p2 but is not a square in 1 2 3.
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