THE QUADRATIC RECIPROCITY LAW OF DUKE-HOPKINS PETE L CLARK Circ 1870, Zolotrev observed tht the Legendre symbol ( p ) cn be interpreted s the sign of multipliction by viewed s permuttion of the set Z/pZ He used this observtion to give strikingly originl proof of qudrtic reciprocity [2] We shll not discuss Zolotrev s proof per se, but rther 2005 pper of W Duke nd K Hopkins which explores the connection between permuttions nd qudrtic symbols in more mbitious wy En route, we explore qudrtic reciprocity s expressed in terms of the Kronecker symbol 1 The Kronecker Symbol The Jcobi symbol ( n ) is n extension of the Legendre symbol ( p ) which is defined for ny positive odd integer n by ( 1 ) = 1 for ll Z; if n = r i=1 p i, then ( r ( ) n) i=1 For n integer, define ( ) 0 0 (mod 2) = 1 1, 7 (mod 8) 2, 1 3, 5 (mod 8) ( ) 0 = 0 = 1 > 0 1, 1 < 0 ( ) { } 0 1 = 0 1 = 1 With these dditionl rules there is unique extension of the Jcobi symbol to symbol ( n ) defined for ny n, Z such tht for ll integers n,, b, we hve ( n b ) = ( n )( n b b ) One lso hs ( n ) = ( n )( b n ), ie, the symbol is bi-multiplictive This extension of the Jcobi symbol is known s the Kronecker symbol When n is not odd nd positive, some uthors (eg [1]) define ( n ) only when 0, 1 (mod 4) It is not worth our time to discuss these two conventions, but we note tht ll of our results involve only this restricted Kronecker symbol For odd n Z +, define n = ( 1) n 1 2 n Full qudrtic reciprocity ie, the usul QR lw together with its First nd Second Supplements is equivlent to one elegnt identity: for Z nd n odd positive n Z, (1) p i n n) 1
2 PETE L CLARK 2 The Duke-Hopkins Reciprocity Lw Let be finite commuttive group (written multiplictively) of order n define n ction of (Z/nZ) on, by ( mod n) g := g By Lgrnge s Theorem, g n = 1, so tht g = g if (mod n) nd is well defined It is immedite tht ech gives homomorphism from to ; moreover, since (Z/nZ), there exists b (Z/nZ) such tht b 1 (mod n), nd then b = b = Id, so tht ech is n utomorphism of As for ny group ction on set, this determines homomorphism from (Z/nZ) to the group Sym( of permuttions of, the ltter group being isomorphic to S n, the symmetric group on n elements Recll tht there is unique homomorphism from S n to the cyclic group Z 2 given by the sign of the permuttion Therefore we hve composite homomorphism which we will denote by (Z/nZ) Sym( Z 2 (mod n) ( Exmple 21 (Zolotrev): Let p be n odd prime nd ( = is the cyclic group of order p The mpping (Z/pZ) Z 2 given by is nothing else thn the usul Legendre symbol ( p ) Indeed, the group (Z/pZ) is cyclic of even order, so dmits unique surjective homomorphism to the group Z 2 = {±1}: if g is primitive root mod p, we send g to 1 nd hence every odd power of g to 1 nd every even power of g to +1 This precisely describes the Legendre ( ) symbol ( p ) Thus it suffices to see tht for some (Z/pZ) we hve = 1, ie, the sign of the permuttion n n is 1 To see this, switch to dditive nottion, viewing s the isomorphic group (Z/pZ, +); the ction in question is now just multipliction by nonzero element If g is primitive root modulo p, multipliction by g fixes 0 nd cycliclly permutes ll p 1 nonzero elements, so is cycle of even order nd hence n odd permuttion: thus ( g ) = 1 ) The next result shows tht the symbol ( ) is lso bi-multiplictive Proposition 1 For i = 1, 2 let i be finite commuttive group of order n i nd (Z/n 1 n 2 Z) Then ( ) ( ) ( ) (mod n1 ) (mod n2 ) = 1 2 1 2 Proof: If (Z/n 1 n 2 Z), then (mod n2) (g 1, g 2 ) = (g1, g2) (mod n1) = (g1, g2 ) After identifying 1 (resp 2 ) with the subset 1 e 2 (resp e 1 2 ) of 1 2, the permuttion tht induces on 1 2 is the product of the permuttion tht (mod n) 1 induces on 1 with the permuttion tht (mod n) 2 induces on 2 Let us now consider the ction of 1 on Sym( Let r 1 be the number of fixed We
THE QUADRATIC RECIPROCITY LAW OF DUKE-HOPKINS 3 points of 1 More concretely, 1 g = g 1 = g iff g hs order 1 or 2 Note tht r 1 1 becuse of the identity element The n r 1 other elements of re ll distinct from their multiplictive inverses, so there exists positive integer r 2 such tht n r 1 = 2r 2 Definition: We put = ( 1) r2 r1 = ( 1) r2 n r1 Lemm 2 For ny finite commuttive group, we hve 0 or 1 (mod 4) Proof: Let n = If n is odd, then by Lgrnge the only g with g 1 = g is the identity, so tht r 1 = 1 nd r 2 = n 1 2 In this cse = = ( 1) n 1 2 n 1 (mod 4) If n is even, then n r 1 = 2r 2 0 (mod 2), so r 1 is even nd hence is t lest 2, so = ( 1) r2 n r1 0 (mod 4) ( ) So the Kronecker symbol is lwys defined (even in the restricted sense) Theorem 3 (Duke-Hopkins Reciprocity Lw) For finite commuttive group nd n integer, we hve The proof will be given in the next section Corollry 4 ) Suppose hs odd order n Then for ny (Z/nZ), we hve n b) Tking = Z n we recover (1) c) We hve ( ) = 1 for ll (Z/nZ) iff n is squre Proof of Corollry 4: In the proof of Lemm 2 we sw tht = n ; prt ) then follows immeditely from the reciprocity lw By prt ), the symbol ( ) cn be computed using ny group of order n, so fctor n into product p 1 p r of not necessrily distinct primes nd pply Exmple 21: we get ( ) = r i=1 ( p i ) = ( n ) This gives prt b) Finlly, using the Chinese Reminder Theorem it is esy to see tht there is some such tht ( n ) = 1 iff n is not squre 3 The Proof Enumerte the elements of s g 1,, g n nd the chrcters of s χ 1,, χ n Let M be the n n mtrix whose (i, j) entry is χ i (g j ) Since ny chrcter χ X( hs vlues on the unit circle in C, we hve χ 1 = χ Therefore the number r 1 of fixed points of 1 on is the sme s the number of chrcters χ such tht χ = χ, ie, rel-vlued chrcters Thus the effect of complex conjugtion on the chrcter mtrix M is to fix ech row corresponding to rel-vlued chrcter nd to otherwise swp the ith row with the jth row where χ j = χ i In ll r 2 pirs of rows get swpped, so Moreover, with M = (M) t, we hve det(m) = det(m) ( 1) r2 MM = ni n,
4 PETE L CLARK so tht so det(m) det(m) = n n, (2) det(m) 2 = ( 1) r2 n n = ( 1) r2 n r1 n 2r2 = l 2, where l = n r2 (In prticulr det(m) 2 is positive integer Note tht det(m) itself lies in Q( ), nd is not rtionl if n is odd) So for ny Z, we hve ( ) ( ) det(m) 2 (3) The chrcter mtrix M hs vlues in the cyclotomic field Q(ζ n ), which is lois extension of Q, with lois group isomorphic to (wht concidence!) (Z/nZ), n explicit isomorphism being given by mking (Z/nZ) correspond to the unique utomorphism σ of Q(ζ n ) stisfying σ (ζ n ) = ζ n (All of this is elementry lois theory except for the more number-theoretic fct tht the cyclotomic polynomil Φ n is irreducible over Q) In prticulr the group (Z/nZ) lso cts by permuttions on the chrcter group X(, nd indeed in exctly the sme wy it cts on : g, ( χ)(g) = χ(g ) = (χ(g)) = χ (g), so χ = χ This hs the following beutiful consequence: For (Z/nZ), pplying the lois utomorphism σ to the chrcter mtrix M induces permuttion of the rows which is the sme s the permuttion of In prticulr the signs re the sme, so ( (4) det(σ M) = det(m) Combining (2) nd (4), we get tht for ll (Z/nZ), σ ( ( ) ) = Now, by the multiplictivity on both sides it is enough to prove Theorem 3 when = p is prime not dividing n nd when = 1 Proposition 5 Let p be prime not dividing n TFAE: ) σ p ( ) = b) p splits in Q( ) c) ( p ) = 1 The proof of this stndrd result in lgebric number theory is omitted for now We deduce tht ( ) ( p p Finlly, when = 1, σ 1 is simply complex conjugtion, so ( ) 1 = σ { } ( ) 1( ) = > 0 =, < 0 1 so ( ) ( ) 1 1 This completes the proof of Theorem 3
THE QUADRATIC RECIPROCITY LAW OF DUKE-HOPKINS 5 4 In fct the rel Duke-Hopkins reciprocity lw is n ssertion bout group of order n which is not necessrily commuttive In this cse, the mp g g need not be n utomorphism of, so more sophisticted pproch is needed Rther, one considers the ction of (Z/nZ) on the conjugcy clsses {C 1,, C m } of : if g = xhx 1 then g = xh x 1, so this mkes sense We further define r 1 to be the number of rel conjugcy clsses C = C 1 nd ssume tht in our lbelling C 1,, C r1 re ll rel nd define r 2 by the eqution m = r 1 + 2r 2 Then in plce of our (nottion which is not used in [1]), one hs the discriminnt d( = ( 1) r2 n r1 r 1 j=1 C j 1 The Duke-Hopkins reciprocity lw sserts tht for (Z/nZ), d( The proof is very similr, except the group X( of one-dimensionl chrcters gets replced by the set {χ 1,, χ m } of chrcters (ie, trce functions) of the irreducible complex representtions of Perhps surprisingly, the only prt of the proof which looks truly deeper is the clim tht d( 0, 1 (mod 4) which is required, ccording to the conventions of [1], for the Kronecker symbol ( d( ) cn be defined Duke nd Hopkins suggest this s n nlogue of Stickelberger s theorem in lgebric number theory which sserts tht the discriminnt of ny number field is n integer which is 0 or 1 modulo 4; moreover they dpt 1928 proof of tht theorem due to Issi Schur References [1] W Duke nd K Hopkins, Qudrtic reciprocity in finite group Amer Mth Monthly 112 (2005), no 3, 251 256 [2] Zolotrev, Nouvelle démonstrtion de l loi de réciprocité de Legendre Nouvelles Ann Mth (2) 11 (1872) 354-362