The Jacobi Symbol. q q 1 q 2 q n

Transcription

1 The Jacobi Symbol It s a little inconvenient that the Legendre symbol a is only defined when the bottom is an odd p prime You can extend the definition to allow an odd positive number on the bottom using the Jacobisymbol Most of the properties of Legendre symbols go through for Jacobi symbols, which makes Jacobi symbols very convenient for computation We ll see, however, that there is a price to pay for the greater generality: Euler s formula no longer works, and we lose part of the connection between the value of a symbol and the solvability of the corresponding uadratic congruence Definition Let p, Z, where (p,) = 1 and is a product of odd primes: = 1 n (The i need not be distinct) The Jacobi symbol p is defined by p p p p 1 n Note that the Jacobi symbol and the Legendre symbol coincide in the case where is a single odd prime That is why the same notation is used for both It s clear from the definition that p ±1 Lemma If is a product of odd primes and a is a uadratic residue mod, then a 1 Proof Write = 1 n, where each i is an odd prime Suppose a is a uadratic residue mod (a,) = 1 and x = a (mod ) has solutions Since i, it follows that (a, i ) = 1 and x = a (mod i ) for i = 1,n Hence, a 1 for i i = 1,n Therefore, a a a a = 1 1 n However, the converse is false: If p is a Jacobi symbol and p 1, it does not follow that p is a uadratic reside mod Example Show that 1, but is not a uadratic residue mod Since is not a suare mod 3 or mod 5, Therefore, ( 1)( 1) = However, here is a table of suares mod 15: x x (mod 15)

2 x x (mod 15) The table shows that is not a suare mod 15 The uadratic residues mod 15 are 1 and 4, as those are the suares that are relatively prime to 15 The results that follow amount to saying that the algebraic properties of Legendre symbols hold for Jacobi symbols and indeed, the proofs of these properties typically use those properties for Legendre symbols Theorem Let and be odd positive numbers, and suppose (pp, ) = 1 : (a) p p = p (b) p p pp (c) p p 1 (d) p p = p (e) If p = p (mod ), then p p Proof (a) Write and as products of odd primes: = 1 m and = 1 n p p ( p p p )( p p p ) 1 m 1 n = p (b) Write as a product of odd primes: = 1 m ( p p p p p )( p ) p p 1 m 1 m ( )( ) ( ) p p p p p p pp pp pp pp 1 1 m m 1 m (c) Write as a product of odd primes: = 1 m If k is an odd prime, then p 1 (as a Legendre symbol) Hence, k p p p p = 1 1 m Next, observe that if k is an odd prime, then p p (±1) = 1 k k

3 So p = p p p p 1 1 p m p m = = 1 (d) p p = p p (by (b)) = p (by (c)) = p p (by (a)) = p (by (c)) (e) Write as a product of odd primes: = 1 m Since p = p (mod ), I have p = p (mod k ) for k = 1,m Conseuently, p p (as k k Lengendre symbols) Therefore, p p p p = p p p p 1 m 1 m Example Show that if (a,) = 1 and is odd and positive, it does not follow that a a ( 1)/ (mod ) (Thus, the analog of Euler s lemma does not hold for Jacobi symbols) Note that (1)( 1) = But 7 (15 1)/ = 7 7 = 13 (mod 15) The next lemma will be used in the proofs of the formulas for 1 and, as well as in the proof that Quadratic Reciprocity holds for Jacobi symbols Lemma If m and n are odd, then m 1 + n 1 = mn 1 (mod ) Proof Since m and n are odd, I may write m = a+1 and n = b + 1 for a,b Z m 1 + n 1 = a b+1 1 = a+b

4 On the other hand, mn 1 = (a+1)(b+1) 1 mn 1 = 4ab+a+b+1 1 mn 1 = 4ab+a+b mn 1 = ab+a+b mn 1 = ab+ m 1 + n 1 mn 1 = m 1 + n 1 (mod ) Corollary If m 1,m,m n are odd, then m m 1 Proof Use the previous lemma and induction + + m n 1 = m 1m m n 1 (mod ) The way this corollarywill be used in the following proofis the simple observation that if s = t (mod ), then ( 1) = ( 1) t Theorem Let be an odd positive number 1 ( 1) ( 1)/ Proof Write as a product of odd primes: = 1 m n The terms on the right are Legendre symbols, for which I know 1 ( 1) ( k 1)/ for k = 1,n k Thus, 1 ( 1) (1 1)/ ( 1) ( 1)/ ( 1) (n 1)/ = ( 1) S, where S = Using the preceding corollary, k 1 S = k 1 = n k 1 = 1 (mod ) Therefore, 1 ( 1) ( 1)/ Theorem (Quadratic Reciprocity) Suppose p and are odd positive integers and (p,) = 1 p ( 1) [(p 1)/][( 1)/] p 4

5 Proof I ll prove the euivalent statement p ( 1) [(p 1)/][( 1)/] p (To get from either this statement to the original one or vice versa, multiply both sides by and p note that = 1) p Write p and as products of odd primes: p = p 1 p p m and = 1 n p m p m n i p i i=1 Here s what the last double product looks like, multiplied out: p 1 p p m p 1 p p m p 1 n p n p m n By Quadratic Reciprocity for Legendre symbols, p i j ( 1) [(p i 1)/][( j 1)/] p i j i=1j=1 j Taking the product over i and j on both sides, I get p m n p m n i j ( 1) [(p i 1)/][( j 1)/] = m n ( 1) [(pi 1)/][(j 1)/] p i p Taking the product of powers of 1 causes the powers to add So n m ( 1) [(pi 1)/][(j 1)/] = ( 1) S, where S = j=1 By the preceding corollary, j p i 1 j 1 That is, j=1 p i 1 j 1 = m i=1 p i 1 j=1 = p 1p p m 1 = p 1 1 j 1 1 n 1 (mod ) ( 1) S = ( 1) [(p 1)/][( 1)/] 5

6 Hence, p ( 1) [(p 1)/][( 1)/] p Remark In computational terms, this version of reciprocity is like the one for Legendre symbols Thus, suppose p and are odd and relatively prime If either p or euals 1 mod 4, then p p If both p and eual 3 mod 4, then p p Next, I ll derive a formula for, where is an odd prime The proof is similar to the proof of the formula for 1, except that I have slightly different preliminary lemmas Lemma If m and n are odd, then m 1 Proof Since m and n are odd, I may write + n 1 = m n 1 (mod ) So m 1 m = a+1 and n = b + 1 for a,b Z m = 4a +4a+1 and n = 4b +4b+1 m 1 = 4a +4a and n 1 = 4b +4b + n 1 = 4a +4a + 4b +4b = a +a + b +b (Note that a +a is even because it s the sum of two odd numbers, so a +a is an integer Likewise, b +b is an integer) Now m n = (4a +4a+1)(4b +4b+1) m n = 16a b +16a b+16ab +16ab+4a +4b +4a+4b+1 m n 1 = 16a b +16a b+16ab +16ab+4a +4b +4a+4b m n 1 m n 1 m n 1 m n 1 = 16a b +16a b+16ab +16ab+4a +4b +4a+4b = a b +a b+ab +ab+ a +a = a +a = m 1 Corollary If m 1,m,m n are odd, then + b +b + n 1 (mod ) (mod ) + b +b m m m n 1 = m 1 m m n 1 (mod ) 6

7 Proof Use the previous lemma and induction Theorem Let be an odd positive number ( 1) ( 1)/ Proof Write as a product of odd primes: = 1 m 1 n The terms on the right are Legendre symbols, for which I know ( 1) ( k 1)/ for k = 1,n k Thus, ( 1) ( 1 1)/ ( 1) 1)/ ( 1) ( ( n 1)/ = ( 1) S, where S = Using the preceding corollary, k 1 S = k 1 n = k 1 = 1 (mod ) Therefore, ( 1) ( 1)/ Example Compute the Jacobi symbol = ( 1)(1) = Example Compute the Legendre symbol Jacobi symbols can often be used to simplify the computation of Legendre symbols ( 1) = c 015 by Bruce Ikenaga 7