Sulement 4 Permuttions, Legendre symbol nd qudrtic recirocity 1. Permuttions. If S is nite set contining n elements then ermuttion of S is one to one ming of S onto S. Usully S is the set f1; ; :::; ng nd ermuttion cn be reresented s follows: Thus if n 5 we could hve 1 ::: j ::: n (1) () ::: (j) ::: (n) 1 3 4 5 3 1 5 4 If f; bg S, 6 b then we cn de ne ermuttion of S s follows: < c if c f; bg (c) b if c : if c b We will cll such ermuttion trnsosition nd denote it (b). If nd re ermuttions of S then we write for the usul comosition of ms (() (())). Since ermuttion is one to one nd onto if is ermuttion then we cn de ne 1 by 1 (y) x if (x) y. If I is the identity m then 1 1 I. We lso note tht () () for ermuttions ; ;. Lemm 1 Every ermuttion,, cn be written s roduct of trnsositions. (Here roduct of no trnsositions is the identity m, I.) Proof. Let n be the number of elements in S. We rove the result by induction on j with n j the number of elements in S xed by. If j 0 then is the identity m nd thus is written s roduct of n 0 trnsositions. Assume tht we hve rove the theorem for ll t lest n j xed oints. We look t the cse when hs n j 1 xed oints. Let S be such 1
tht () b nd b 6. Then (b)(). Suose tht (c) c then c 6 nd c 6 b since if (c) b then () (c) which is contrry to our ssumtion. Thus if (c) c then (b)(c) c. This imlies tht the number of elements xed by (b) is t lest n j 1 + 1 n j. Thus (b)(c) cn be written s ( 1 b 1 ) ( r b r ). Now (b)(b) I the identity. Thus (b)(b) (b)( 1 b 1 ) ( r b r ): This comletes the inductive ste. Note tht the bove lemm gives method for clcultion. Consider given by 1 3 4 5 3 1 5 4 Then (13) is given by (3)(13) is given by (35)(3)(13) is given by 1 3 4 5 1 3 5 4 1 3 4 5 1 5 3 4 1 3 4 5 1 3 5 4 nd nlly (45)(35)(3)(13) is the identity. Since trnsosition is its own inverse we see tht (13)(3)(35)(45): Let x 1 ; :::; x n be n indetermintes. Set (x 1 ; :::; x n ) Q 1i<jn (x i x j ). Theorem Let be ermuttion of set S with n elements. Let f 1 ; ; :::; n g be listing of the elements of S: If de ne : f1; :::; ng! f1; :::; ng by ( i ) (i) then (1) Q 1i<jn (x (i) x (j) ) "()(x 1 ; :::; x n ) with "() f1; 1g. () If ; re ermuttions nd ( i ) (i) then ( i ) (i) nd "() "()"(). (3) Suose tht fb 1 ; :::; b n g is nother listing of the elements of S nd we de ne : f1; :::; ng! f1; :::; ng by (b i ) b (i). Then "() "(): This sys tht we cn de ne sgn() "() since the exression is indeendent of how we enumerte the elements of S.
(4) Assume tht S S 1 [ S with S 1 \ S ;. Let i be ermuttion of S i for i 1;. De ne ermuttion of S by 1 (x) if x S (x) 1 : (x) if x S Then sgn() sgn( 1 )sgn( ). (5)If (b 1 c 1 ) (b r c r ) then "() ( 1) r. Proof. Let A f(i; jgj1 i < j n nd (i) < (j)g, nd B f(i; j)j1 i < j n nd (i) > (j)g. Since is one to one we see tht A [ B f(i; j)j1 i < j ng. Thus (x (i) x (j) ) (x (i) x (j) ) (x (i) x (j) ): 1i<jn (i;j)a (i;j)b Thus 1i<jn We now note tht (x (i) x (j) ) ( 1) jb j (i;j)a (x (i) x (j) ) (i;j)a (x (i) x (j) ) (i;j)b (x (j) x (i) ): (i;j)b (x (j) x (i) ) (x 1 ; :::; x n ): This roves 1 with "() ( 1) B. We will now rove : We hve (i) ( i ). Now (i) ( (i) ) (( i )): We hve (x (i) x (j) ) (x (i) x (j) ) (x (i) 1i<jn (i;j)a (i;j)b x (j) ) ( 1) jb j "() (x (i) (i;j)a x (j) ) (x (j) (i;j)b x (i) ) (x (i) x (j) ) "()"()(x 1 ; :::; x n ): 1i<jn This imlies tht "() "()"() roving. 3
We will now rove 3. For this we note tht b i (i) with ermuttion of f1; :::; ng. Thus (b i ) ( (i) ) ((i)) b 1 (((i))): Thus (i) 1 (i). Since "( 1 ) "(I) 1, we hve "() "( 1 )"()"() "( 1 )"()"() "(): To rove (4) we enumerte S by writing S 1 f 1 ; :::; r g nd S f r+1 ; :::; r+s g with r + s n. Then 1 ( i ) (i) for i 1; :::; r with ermuttion of f1; :::; sg nd ( r+j ) r+(j) with ermuttion of 1; :::; s: Then if is de ned by (x) if 1 x r (j) (x r) if r < x n : sgn() (x i x j ) (x (i) x (j) ) 1i<jr (x (i) x (j) ) 1i<jn The middle fctor is just The left most fctor is nd the right most is 1 i r 1 j s 1 i r 1 j s "() "() 1i<jn (x (i) x r+(j) ) 1i<jr 1i<js (x i x r+j ): (x i x j ) (x r+i x r+j ): 1i<js (x r+(i) x r+(j) ): Thus sgn() "()"() sgn( 1 )sgn( ). To rove (5) in light of. we need only show tht if ; b S nd 6 b then sgn((b)) 1. Let S 1 f; bg nd S S f; bg then (b) is given s in (4) with 1 (b) js1 nd the identity m on S. The result now follows since (x x 1 ) (x 1 x ). 4
De nition 3 We will cll sgn() the sign of the ermuttion. We will now look t the exmle bove with given by 1 3 4 5 3 1 5 4 we hve seen tht (13)(3)(35)(45) so sgn() 1 by (3) in the revious theorem. Note tht the method of roof (1),() of the bove theorem gives nother method. It indictes tht we should count the irs i < j such tht the corresonding elements re in the reverse order. This mens tht we cn red from left to right in the second row nd in this exmle note tht 3 > 1, 3 > ; 5 > nd 5 > 4. ll other irs re left in incresing order. Thus the method of the roof gives n lternte determintion tht sgn() 1. De nition 4 If 1 ::: j ::: n (1) () ::: (j) ::: (n) is ermuttion nd (i) > (j) with i < j then the ir (i); (j) is clled descent. If N is the number of descents then sgn() ( 1) N. In the next section we will need few clcultions of signs of ermuttions. Lemm 5 Let (i) i + 1 for 1 i n 1 nd (n) 1. Then sgn() ( 1) n 1. Proof. Here we re looking t 1 ::: j ::: n 1 n 3 ::: j + 1 ::: n 1 we note tht the descents re exctly (; 1); (3; 1); :::; (n; 1) nd thus there re exctly n 1 descents. Lemm 6 Assume tht n k consider given by 1 ::: k k + 1 k + ::: k 1 k 4 ::: k 1 3 ::: k 3 k 1 5
Then sgn() ( 1) k(k+1). Proof. The descents re (; 1); (4; 1); (4; 3); (6; 1); (6; 3); (6; 5); :::; (k; 1); (k; 3); :::; (k; k 1): If we count them u we get 1 + + ::: + k k(k+1) : Thus sgn() ( 1) k(k+1) s sserted.. The Legendre symbol nd ermuttions. In this section will denote n odd rime. Let F be the eld ZZ f1; ; :::; 1g, s usul. If F f0g then we consider the elements f 1; ; :::; ( 1)g then since 6 0 this de nes ermuttion of F f0g. We note tht b b for ; b F f0g. Our key observtion is Lemm 7 If F f0g then sgn( ). Proof. Let F f0g be rimitive element. Then F f0g f1; ; ; :::; g. We write i i 1 for i 1; :::; 1. Then with this enumertion corresonds to the ermuttion 1 ::: j ::: 1 3 ::: j + 1 ::: 1 1 which, by Lemm 5, hs sign ( 1) ( 1) 1 ( 1) 1 since is odd. We now note tht if F f0g then j with 0 j. Thus sgn( ) sgn( j) sgn( ) j ( 1) j. We hve seen in sulement 3 tht ( 1) j. Corollry ( 1) 1 : Proof. We note tht i i for i 1. Also ( 1 + j) 1 + j + (j 1): 6
Thus ( 1 1 1 +j) j 1 mod for j 1; :::;. This imlies tht if k then is given by 1 ::: k k + 1 k + ::: k 1 k 4 ::: k 1 3 ::: k 3 k 1 nd so Lemm 6 imlies tht sgn( ) ( k + 1 +1 follows from Lemm 7. hence k(k+1) ( 1) (+1) 1) k(k+1). Now k 1 nd which is 1. The corollry now Our next result is usully clled Guss Lemm. He roved it s rt of his third roof of the lw of qudrtic recirocity. Lemm 9 Let Z be such tht -. Let N be the number of elements j (1; ; :::; 1 1 g such tht j r mod with < r 1. Then ( 1) N. Proof. Let b be the element of F corresonding to. We write out the miniml ositive residues mod s 1; ; :::; 1 ; 1; ; :::; 1 if we reduce modulo then we cn write out F f0g s 1; ; :::; 1 ; 1; ; :::; 1 : Now if 1 i 1 nd bi u i mod which we tke to be in the rnge 1 u i 1. We note tht b( i) u i mod. We de ne v i s follows. If 1 u i 1 then v i u i. If 1 < u i < then we de ne v i by u i v i mod with 1 v i 1 : Then if b(i) v i then b ( i) v i nd if b (i) v i then b ( i) v i. Thus if T is the set of 1 i 1 nd 1 < u i < then if i T then b (i) v i. Assume tht T fj 1 ; :::; j N g. We multily b by the roduct of N trnsositions tht interchnge v i nd v i for i T (v j1 ( v j1 )) (v jn ( v jn )) b 7
nd get the ermuttion 1 1 1 ::: 1 ::: v 1 v ::: v 1 v 1 v ::: v 1 this ermuttion hs sign equl to 1 by 4. in Theorem since if we write S 1 f1; ; :::; 1 g nd S 1 f 1; :::; g then if we lbel S by 1 1; ; :::; 1 1 then we see tht is of the form of 4. in Theorem with the sme ermuttion on both rts. Thus by rt 3 of Theorem. sgn( b ) sgn((v j1 ( v j1 )) (v jn ( v jn ))) ( 1) N Here is n exmle of the method of roof of the bove lemm. We ssume tht 7 we tke b 3: Then the multiles re 3; 6; 9 3; 6; 9 if we reduce modulo 7 we hve 3; 6; ; 3; 6; now if we relce the elements lrger thn 1 3 by negtive residue we relce 6 by 1 thus we get 1 3 1 3 3 1 3 1 If we multily this by the trnsosition of 1 nd 1 we hve 1 3 1 3 3 1 3 1 We now hve 3 de nitions of the Legendre symbol we will now give fourth. For this we need new nottion: if is rel number then we set [] mxfj Zjj g. Thus [ 5 4 ] 6. Lemm 10 If Z nd - set Then ( 1) U(;) : U(; ) X 1 i1 i : ( 1) U(;) ( 1) 1 ( 1). In rticulr, if is odd then
Proof. We note tht the division lgorithm cn be rehrsed s j j + r j with 0 j <. If 1 j 1 then r j 6 0. since doesn t divide j or. Let T fjj 1 < r j < g. Then if j T we cn write r j s j with 1 s j 1. We therefore see tht X j 1 j1 X 1 i1 i + X r j + jt j j T X s j. Now the set of elements fr j jj T g [ fs j jj T g f1; ; :::; 1 g. Thus jt X j T r j X s j jt X j 1 j1 X jt s j : If we observe tht P 1 j1 j 1 +1 then we hve 1 1 X i ( 1) + jt j X s j : i1 jt If we consider this eqution modulo tking ccount of the fct tht 1 mod we hve 1 ( 1) P 1 h i i i1 + jt j mod. This imlies tht jt j + 1 ( 1) U(; ) mod. Thus by Guss Lemm (Lemm 9) we hve ( 1) 1 ( 1) ( 1) jt j ( 1) 1 ( 1) ( 1) U(;) : Qudrtic Recirocity. We re nlly redy to stte nd rove the Lw of Qudrtic Recirocity. This theorem is usully ttributed to Guss who gve the rst roof of the theorem nd in the course of his life gve 7 roofs. However, both Legendre nd Euler hd thought tht the lw ws very robble nd ech gve t lest one incorrect roof. 9
Theorem 11 Let nd q be distinct odd rimes then q ( 1) 1 q 1 : q Proof. We consider the set S of ll irs (i; j) with 1 i 1 nd 1 j q 1. If (i; j) S we ssert tht qi 6 j. Since if qi j then since 6 q the equlity would imly tht qjj but j is not 0 nd is too smll. Thus if (i; j) S then qi 6 j. Let A f(i; j) Sjqi > jg nd B f(i; j) Sjqi < jg then S is the disjoint union of A nd B. We will now count the number of elements of A. We count the number with rst coordinte i. If (i; j) A then qi > j. This imlies tht qi > j. Thus h i h i qi j 1 nd since qi > qi we see tht every such j ers. Hence the h i qi number of elements of A with rst coordinte i is. Hence the number of elements in A is 1 X qi U(q; ) i1 (see Lemm 10). The sme rgument imlies tht the number of elements in B is U(; q): But S hs 1 q 1 elements. Hence ( 1) U(;q) ( 1) U(q;) ( 1) 1 q 1 : The result now follows since ( 1) U(;q) q nd ( 1) U(q;) We will now show how one cn use this result s owerful tool for the comuttion of Legendre symbols. Consider, 11 131 10 ( 1) 565 ( 5 11 1 ) ( 1) ( 5 131 11 11 11 11 11 ) by Corollry. But 10 15 so we hve 11 5 11 ( 1) 5 131 11 5 Thus 11 is squre in F 131. 1 1: 5 q. 10
Exercises. 1. Write the ermuttion s roduct of trnsositions. 1 3 ::: n n 1 n 3 4 ::: n 1 n 1. Prove tht if S hs 1 n < 1 elements then the number of ermuttions of S is n!. Use this to rove tht if n then n n > n!. (Hint. How mny functions re there from S to S?). 3. Comute the sign of the following ermuttion by counting the descents nd by writing it s roduct of trnsositions 1 3 4 5 6 7 3 7 6 4 1 5 : 4. Wht is the sign of the ermuttion 1 ::: k k + 1 k + ::: k 1 3 ::: k 1 4 ::: k 5. Let q be n odd rime. Show tht is qudrtic residue modulo q if nd only if q l 1. 6. Let ; q be odd rimes. Show tht of qj( 1) then j(q 1) nd q l1 for some integer l. (Hint: Is qudrtic residue mod q?) 7. Clculte the following Legendre symbols: 31 ) 641. b) 7 79 : c) 105 1009 :. Let be n odd rime. Let ; b; c F with 6 0 nd let f(t) t +bt+c then f hs root in F if nd only of 4bc 6 1. How mny roots does f(t) hve in F if 1? 4bc 9. Let be n odd rime show tht the smllest ositive integer tht is not squre modulo is rime. 11
10. Prove tht if is n odd rime with > 3 then the congruence x 3 mod hs solution of nd only if 1 mod 3. Show tht if is rime such tht 1 mod 3 nd 1 mod 4 then two solutions to re 3 +1 4. 11. Use Lemm 10 to clculte x 3 mod for nd odd rime. 1. Derive the rst rt of roblem 10 bove from Lemm 10. 13. A Fermt rime is rime q r + 1 with r ositive integer. Use the lw of qudrtic recirocity to show tht if q r + 1 is n odd rime then r must be even nd 1. 3 q 1
The Jcobi symbol As it turns out the rguments in the revious two sections on the Legendre symbol work eqully well for the (more generl) Jcobi symbol. We will describe the modi ctions in this section. Let b Z be odd number with b > 1. We will set R b ZbZ. If Z nd gcd(; b) 1 we note tht multiliction by de nes ermuttion of R b f0g which we denote by ;b. Then we de ne the Jcobi symbol b sgn(;b ). If b 1 we set 1 1 nd if gcd(; b) 6 1 then set b 0. Lemm 7 imlies tht if b is rime then the Jcobi symbol grees with the Legendre symbol. Lemm 1 The if 0 mod b then Jcobi symbol b 0 b : If b is ositive nd odd then 1 1 : b b b Proof. The rst ssertion is direct consequence of the de nition. If b 1 or if ny one of the terms in the sserted equlity is 0 then the equlity is obviously true. iin ll the other cses b > 1 nd gcd( 1 ; b) gcd( ; b) 1 nd the result follows from 1 ;b ;b 1 ;b. Theorem 13 If b Z >0 is odd then b ( 1) b 1 : Proof. We my ssume tht b > 1. The roof is exctly the sme s tht of Corollry with relced by ;b, relced by b, k b 1 nd Lemm 7 relced by the de nition of the Jcobi symbol. Lemm 14 Let b be odd nd b > 1 nd let Z be such tht gcd(; b) 1. Let N be the number of elements j (1; ; :::; b 1 g such tht j r mod b with b 1 < r b 1. Then b ( 1) N. Proof. The roof is the sme s tht of Lemm 9 however since the nottion in tht roof overls tht in the bove sttement we will give the detils. We write out the miniml ositive residues mod b s 1; ; :::; b 1 ; b 1; b ; :::; b b 1 if we reduce modulo b then we cn write out R b f0g s 1; ; :::; b 1 ; 1; ; :::; b 1 : 13
Now if 1 i b 1 nd i u i mod b which we tke to be in the rnge 1 u i b 1. We note tht ( i) u i mod b. We de ne v i s follows. If 1 u i b 1 then v i u i. If b 1 < u i < b then we de ne v i by u i v i mod b with 1 v i b 1: Then if ;b(i) v i then ;b ( i) v i nd if ;b (i) v i then ;b ( i) v i. Thus if T is the set of 1 i b 1 nd b 1 < u i < b then if i T then ;b (i) v i. Assume tht T fj 1 ; :::; j N g. We multily ;b by the roduct of N trnsositions, (v i ( v i )) tht interchnge v i nd v i for i T (v j1 ( v j1 )) (v jn ( v jn )) b nd get the ermuttion given by b 1 b 1 1 ::: 1 ::: v 1 v ::: v b 1 v 1 v ::: v b 1 this ermuttion hs sign equl to 1 by (4) in Theorem since if we write S 1 f1; ; :::; b 1g nd S b 1 f 1; :::; g then if we lbel S by 1 1; ; :::; b 1 b 1 then we see tht is of the form of (4) in Theorem with the sme ermuttion on both rts. Thus sgn( ;b ) sgn((v j1 ( v j1 )) (v jn ( v jn ))) ( 1) N by rt (3) of Theorem. We lso hve Lemm 15 If b is odd nd b > 1 nd if Z >0 nd gcd(; b) 1 set U(; b) X b 1 i1 i b then b ( 1) U(;b) ( 1) ( 1) b 1. In rticulr, if is odd then b ( 1) U(;b). Proof. As bove the roof of this result is essentilly the sme s tht of Lemm 10 if we relce by b: We will go through the rgument since the stes re the sme but some of the justi ctions re di erent. We note tht the division lgorithm cn be rehrsed s j j b + r j b 14
with 0 j < b. If 1 j b 1 then r j 6 0. since gcd(; b) 1 nd j is too smll for b to divide it. Let T fjj b 1 < r j < bg. Then if j T we cn write r j b s j with 1 s j b 1. We note tht X j b 1 j1 b 1 ( b 1 + 1) b 1 : X j b b 1 j1 X b 1 i1 i + X r j + bjt j b j T X s j. Now the set of elements fr j jj T g [ fs j jj T g f1; ; :::; b 1 g. Thus jt X j T r j X s j jt X j b 1 j1 X jt s j b 1 X jt s j : Also P b 1 j1 j P b 1 j1 j b 1 thus. b b 1 1 X i ( 1) b + bjt j X s j : b i1 jt If we consider this eqution modulo tking ccount of the fct tht b 1 mod we hve b 1 ( 1) P b 1 i i1 b + jt j mod. This imlies tht jt j + b 1( 1) U(; b) mod. Thus by Lemm 14 we hve ( 1) b 1 ( 1) ( 1) jt j ( 1) b 1 ( 1) ( 1) U(;b) : b Finlly we hve the recirocity theorem. Theorem 16 Let nd b be odd numbers such tht gcd(; b) 1 with ; b > 1 then b ( 1) 1 b 1 : b 15
Proof. The roof is essentilly the sme s tht of Theorem 11 we will give the detils since the chnge in nottion could be confusing. We consider the set S of ll irs (i; j) with 1 i 1 nd 1 j b 1. If (i; j) S we ssert tht bi 6 j. Since if i bj then since gcd(; b) 1 the equlity would imly tht jj but j is not 0 nd is too smll. Thus if (i; j) S then bi 6 j. Let A f(i; j) Sjbi > jg nd B f(i; j) Sjbi < jg then S is the disjoint union of A nd B. We will now count the number of elements of A by counting the number with rst coordinte i. If (i; j) A then bi > j. This imlies tht bi > j. Thus bi j 1 nd since bi > bi (since - (bi). we see tht every such j ers. Hence the number of elements of A with rst coordinte i is bi. Hence the number of elements in A is P 1 bi i1 U(b; ). The sme rgument imlies tht the number of elements in B is U(; b): But S hs 1 b 1 elements. Hence ( 1) U(;b) ( 1) U(b;) ( 1) 1 b 1 : The result now follows since ( 1) U(;b) b nd ( 1) U(b;) b. This theorem imlies tht our de nition of the Jcobi symbol is equivlent to the stndrd one (see Rose,.71-7 nd do Exercise 17). The bove recirocity lw llows us to comute Legendre symbols without rime fctoriztion (excet for the highest ower of which is obvious if we write out numbers bse ). Here is n exmle. ou cn check tht 7919 nd 37 re rimes. Thus we hve 37 46 7919 7919 117 7919 7919 0 117 117 117 5 117 117 7919 7919 7919 4 117 1: 5 Exercises. 14. Write out the detils of the roof of Theorem 1. 5 5 117 117 15. Clculte the following Legendre symbols using the Jcobi symbol. ) 357 579. 16
b) 311 6133. 16. Why is it good ide to write numbers in bse in the clcultion of Jcobi symbols? 17. Show (using the theorems bove) tht our de nition of the Jcobi symbol is equivlent with tht in Rose. 17