QUADRATIC RECIPROCITY

Transcription

1 QUADRATIC RECIPROCITY JORDAN SCHETTLER Abstract. The goals of this roject are to have the reader(s) gain an areciation for the usefulness of Legendre symbols and ultimately recreate Eisenstein s slick roof of Gauss s Theorema Aureum of quadratic recirocity. 1. Quadratic Residues and Legendre Symbols Definition 0.1. Let m, n Z with (m, n) 1 (recall: the gcd (m, n) is the nonnegative generator of the ideal mz + nz). Then m is called a quadratic residue mod n if m x 2 (mod n) for some x Z, and m is called a quadratic nonresidue mod n otherwise. Prove the following remark by considering the kernel and image of the ma x x 2 on the grou of units (Z/nZ) {m + nz : (m, n) 1}. Remark 1. For 2 < n N the set {m+nz : m is a quadratic residue mod n} is a subgrou of the grou of units of order ϕ(n)/2 where ϕ(n) #(Z/nZ) is the Euler totient function. If n is an odd rime, then the order of this grou is equal to ϕ()/2 ( 1)/2, so the equivalence classes of all quadratic nonresidues form a coset of this grou. Definition 1.1. Let be an odd rime and let n Z. The Legendre symbol (n/) is defined as 1 if n is a quadratic residue mod 1 if n is a quadratic nonresidue mod 0 if n. The law of quadratic recirocity (the main theorem in this roject) gives a recise relationshi between the recirocal Legendre symbols (/q) and (q/) where, q are distinct odd rimes. We ll rove quadratic recirocity in section 2, and we ll see alications thereof to Diohantine equations and comutations of Legendre symbols in section 3. In the meantime, use remark 1 to establish the following roosition. Proosition 2. Let be an odd rime and let m, n Z. Then ( ) ( ) ( ) mn m n (1.1) and (1.2) m n (mod ) ( ) m. Show the following lemma by recalling that an element of a finite grou has order dividing the size of the grou. Lemma 3 (Fermat s Little Theorem). Let be an odd rime and let n Z with (n, ) 1. Then (1.3) n 1 1 (mod ). 1

2 2 JORDAN SCHETTLER Now aly lemma 3 to get the roceeding fundamental result about Legendre symbols (Hint: factor n 1 1 (n ( 1)/2 1)(n ( 1)/2 + 1) and use the fact that the grou of units of a finite field is cyclic). Theorem 4 (Euler s Criterion). Let be an odd rime and let n Z with (n, ) 1. Then (1.4) n ( 1)/2 (mod ). Next, lug in n 1 into theorem 4 to get the following immediate consequence. Corollary 5. Let be an odd rime. Then ( ) 1 (1.5) ( 1) ( 1)/2. 2. Quadratic Recirocity Flesh out the sketch of the roof for the crucial lemma which follows where for x R we take x : max{n Z : n x}. Lemma 6 (Eisenstein). Let be an odd rime and let n Z with (n, ) 1. Then (2.1) ( 1) s where s ( 1)/2 2kn Sketch. For each k {1,..., ( 1)/2} define 2kn 2kn r k : 2kn Then n ( 1)/2 ( 1)/2 (2k) ( 1)/2 ( 1) k r k. r k ( 1) k r k ( 1)/2 (2k) so by theorem 4 we get (n ) n ( 1)/2 ( 1) k r k ( 1) s ( 1)/2 (mod ), (mod ). [( 1) r k r k ] Deduce the following corollary which is traditionally established with a result called Gauss s lemma. Corollary 7. Let be an odd rime. Then ( ) 2 (2.2) ( 1) (2 1)/8.

3 Proof. Note that ( 1)/2 QUADRATIC RECIPROCITY 3 #{k N : /4 < k /2}, 2 4 so if 8m ± 1, then ( 1)/2 4m ± 1/2 2m ± 1/4 2m and if 8m ± 3, then ( 1)/2 4m ± 3/2 2m ± 3/4 2m ± (mod 2), Finally, fill in the details of the following roof (originally suggested by Eisenstein) of the quadratic recirocity theorem (originally roved by Gauss). Theorem 8 (Quadratic Recirocity). Let, q be distinct odd rimes. Then ( ) ( ) q (2.3) ( 1) [( 1)/2][(q 1)/2]. q Sketch. Consider the following diagram in the (x, y)-lane. Let µ be the number of lattice oints (i.e., oints in the lane with integer coefficients) in the interior of the triangle EF G. Note that number of lattice oints with odd x-coordinates

4 4 JORDAN SCHETTLER in EF G is equal to the number of lattice oints with even x-coordinates in BCE, which has the same arity as the number of lattice oints with even x-coordinates in CEGH. Hence µ #{P : P is a lattice oint in CF H and has even x-coordinate} (mod 2), but for each ositive integer m <, the number of lattice oints in CF H with x-coordinate m is mq/, so ( 1)/2 2kq µ Therefore lemma 6 now imlies A symmetric argument shows that ( ) q ( 1) µ. ( ) ( 1) ν q where ν is the number of lattice oints in DEF. Thus the statement follows from the observation µ + ν #{P : P is a lattice oint in DEGF } 1 q Examles and Alications Remark 9. For a fixed odd rime, we can reasonably obtain the quadratic residues mod by simly squaring the integers 1,..., ( 1)/2 and reducing modulo. On the other hand, the inverse roblem of determining those rimes for which a fixed integer n is a quadratic residue mod requires recirocity as the next examle illustrates. Examle 10. To determine all odd rimes for which 5 is a quadratic residue mod we note that ( ) ( ) 5 ( 1) 5 [( 1)/2][(5 1)/2] 1, so (5/) 1 (/5) 1 ±1 (mod 5). Exercise 11. Determine all odd rimes for which n 3 is a quadratic residue mod. Do the same for n 7 and n 6. The recirocity law can also be utilized to comute Legendre symbols with large arguments as seen in the following examle. Examle 12. Suose we wanted to comute (21/53). We could start squaring the integers 1,..., 26 and reducing modulo 53, but this would be laborious. It s easier to accomlish this by reeated alications of theorem 8 and roosition 2; e.g., ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 1) 1 26 ( 1) ( ) 2 2 ( 1) ( 1)( 1) (72 1)/4 ( 1)( 1)

5 QUADRATIC RECIPROCITY 5 Exercise 13. Use the techniques of the above examle to comute (143/409). Another use of quadratic recirocity includes (as one would exect) finding integer solutions to degree two olynomial equations. Prove the following theorem by filling the details of the outline. Theorem 14. Let be a rime. Then the equation x 2 3y 2 has a solution in integers x, y if and only if 1 (mod 12). Furthermore, the equation 3y 2 x 2 has a solution in integers x, y if and only if either 1 (mod 12) or {2, 3}. Proof. First handle the cases 2 and 3 using reduction modulo 3 and 4, resectively. Now suose > 3. If is of the form ±(x 2 3y 2 ) for some integers x, y, then y is nonzero modulo, so 3 is a square modulo. Conversely, suose 3 is a square modulo. Then there are integers m, n such that m n 2 3 (n 3)(n + 3). Thus divides the roduct (n 3)(n+ 3) in the ring Z[ 3], but does not divide n± 3 in Z[ 3], so is not a rime element in Z[ 3]. Hence is not an irreducible element in Z[ 3] since Z[ 3] is a UFD (justify this), so there are x, y, s, t Z such that neither x+y 3 nor s + t 3 is a unit in Z[ 3] and (x + y 3)(s + t 3). Moreover, the norm ma N : Z[ 3] Z given by a + b 3 a 2 3b 2 is multilicative, so 2 N() (x 2 3y 2 )(s 2 3t 2 ), but this imlies x 2 3y 2 ± since only units have norm ±1. Now by exercise 11 (which used quadratic recirocity) we know that ( ) 3 1 if and only if ±1 (mod 12). Finally, distinguish cases by reducing the exression ±(x 2 3y 2 ) modulo 3. References [1] Pete L. Clark, Instructor s Name. Quadratic Reciroctity I. Course notes. Math 4400/6400 Number Theory. Course home age. Deartment of Mathematics, University of Georgia. htt://math.uga.edu/ ete/4400qrlaw.df. [2] Kenneth Ireland and Michael Rosen, A Classical Introduction to Modern Number Theory, 2th ed., Sringer, [3] Reinhard C. Laubenbacher and David J. Pengelley, Eisenstein s Misunderstood Geometric Proof of the Quadratic Recirocity Theorem, The College Mathematics Journal, Vol. 25, No. 1, (Jan., 1994),