A New Approach to Permutation Polynomials over Finite Fields

Transcription

1 A New Approach to Permutation Polynomials over Finite Fields Joint work with Dr. Xiang-dong Hou and Stephen Lappano Department of Mathematics and Statistics University of South Florida Discrete Seminar November 19, 2012

2 Permutation polynomials Let F q be the finite field with q elements. Definition A polynomial f (x) F q [x] is called a permutation polynomial (PP) over finite field F q if the mapping x f (x) is a permutation of F q. Facts Every linear polynomial over F q is a permutation polynomial of F q. The monomial x n is a permutation polynomial of F q if and only if gcd(n, q 1) = 1.

3 Dickson polynomial Let n 0 be an integer. Elementary symmetric polynomials x 1 + x 2 and x 1 x 2 form a Z basis of the ring of symmetric polynomials in Z[x 1, x 2 ]. There exists D n (x, y) Z[x, y] such that x n 1 + x n 2 = D n (x 1 + x 2, x 1 x 2 ) D n (x, y) = n 2 i=0 n n i ( n i i ) ( y) i x n 2i

4 Dickson polynomial D 0 (x, y) = 2, D 1 (x, y) = x, D n (x, y) = xd n 1 (x, y) yd n 2 (x, y), n 2. [Dickson 1897] For fixed a F q, D n (x, a) F q [x] is the Dickson polynomial of degree n and parameter a. When a = 0, D n (x, a) = x n, PP if and only if (n, q 1) = 1. When 0 a F q, PP if and only if (n, q 2 1) = 1.

5 Reversed Dickson polynomial [Hou, Mullen, Sellers, Yucas 2009] Fix a F q. Interchanged the roles of x and a. D n (a, x) F q [x] - reversed Dickson polynomial. When a = 0, D n (0, x) is a PP if and only if n = 2k with (k, q 1) = 1. When a 0, D n (a, x) = a n D n (1, x a 2 ) D n (a, x) is a PP on F q if and only if D n (1, x) is a PP on F q. The nth Reversed Dickson Polynomial D n (1, x) Z[x] is defined by D n (1, x(1 x)) = x n + (1 x) n

6 Polynomial g n,q q = p k, n 0. There exists a unique polynomial g n,q F p [x] such that (x + a) n = g n,q (x q x) a F q Question : When is g n,q a permutation polynomial(pp) of F q e? If g n,q is a PP, we call triple (n, e; q) desirable.

7 Outline Basic properties of the polynomial g n,q The Case e = 1 The Case n = q a q b 1, 0 < b < a < pe. Results with even q

8 Polynomial g n,q g n,q (x) = n q l n q 1 ( n l l n l(q 1) ) x n l(q 1)

9 The Polynomial g n,q Recurrence g 0,q =... = g q 2,q = 0, g q 1,q = 1, g n,q = xg n q,q + g n q+1,q, n q

10 When n < 0 Recurrence relation for n 0 can be used to define g n,q for n < 0 : g n,q = 1 x (g n+q,q g n+1,q ). For n < 0, there exists a g n,q F p [x, x 1 ] such that (x + a) n = g n,q (x q x) a F q Recurrence relation holds for all n Z.

11 The Polynomial g n,q g n,q and Reversed Dickson polynomial The nth Reversed Dickson Polynomial D n (1, x) Z[x] is defined by D n (1, x(1 x)) = x n + (1 x) n When q = 2, g n,2 (x(1 x)) = x n + (1 x) n F 2 [x] g n,2 = D n (1, x) F 2 [x].

12 Desirable Triples Equivalence (1) g pn,q = g p n,q. (2) If n 1, n 2 > 0 are integers such that n 1 n 2 (mod q pe - 1 ), then g n1,q g n2,q (mod x qe x). (3) If m, n > 0 belong to the same p-cyclotomic coset modulo q pe - 1, we say that two triples (m, e; q) and (n, e; q) are equivalent and write (m, e; q) (n, e; q). If (m, e; q) (n, e; q), g m,q is a PP if and only if g n,q is a PP.

13 Desirable Triples Some Necessary Conditions [Hou 2011] If (n, e; 2) is desirable, gcd(n, 2 2e 1) = 3. If (n, e; q) is desirable, gcd(n, q 1) = 1.

14 Generating function A Quick Reminder : g 0,q =... = g q 2,q = 0, g q 1,q = 1, g n,q = xg n q,q + g n q+1,q, n q g n,q t n t q 1 = 1 t q 1 xt q n 0

15 Theorem g n,qt n (xt) q 1 1 (xt) q 1 (xt) + (1 q xq 1 ) tq 1 (mod x q x). 1 t q 1 n 0 Namely, modulo x q x, g n,q(x) a nx n + { x q 1 1 if n > 0, n 0 (mod q 1), 0 otherwise, where a nt n t q 1 = 1 t q 1 t q n 0

16 Proof of the Theorem a n t n t q 1 = 1 t q 1 t q n 0 t q 1 1 t q 1 xt q and t q 1 1 t q 1 xt q (xt) q 1 1 (xt) q 1 (xt) + (1 x q 1 ) tq 1 q 1 t (mod x q 1 1) q 1 (xt) q 1 1 (xt) q 1 (xt) + (1 x q 1 ) tq 1 q 1 t (mod x) q 1

17 The case e = 1 (n, 1; q) is desirable if and only if gcd(n, q 1) = 1 and a n 0. Proof. By the Theorem, we have Namely, modulo x q x, { g n,q (x) a n x n x q 1 1 if n > 0, n 0 (mod q 1), + 0 otherwise, g n,q (x) = a n x n for all x F q. ( ) Since g n,q is PP, by a previous fact, gcd(n, q 1) = 1. So g n,q (x) a n x n (mod x q x) which implies a n 0. ( ) gcd(n, q 1) = 1 and a n 0 g n,q (x) a n x n (mod x q x) g n,q is PP. Open question : Determine n s.t. a n 0.

18 The case e = 1 contd Assume q = 2. (n, 1; 2) is desirable if and only if a n = 0(in F 2 ). Proof. By the Theorem, we have g n,2 a n x + x 1(mod x 2 x).

19 The case n = q a q b 1, 0 < b < a < pe.

20 g qa q b 1,q Define S a = x + x q + + x qa 1 for every integer a 0. For 0 < b < a < pe, we have Assume e 2. Write g qa q b 1,q = 1 x q 1 (Sb 1)S qb x qb +1 a b a b = a 0 + a 1 e, b = b 0 + b 1 e, where a 0, a 1, b 0, b 1 Z and 0 a 0, b 0 < e. Then we have Namely modulo x qe x, g q a q b 1,q x qe 2 x qe q b 0 2 (a 1 S e + S qb 0 a 0 ) ( (b 1 S e + S b0 ) q 1 1 )..

21 The case b = 0 If b = 0 and a > 0, we have n q a 2 (mod q pe 1). g q a 2,q = x q 2 + x q x qa 1 2. Conjecture 1 Let e 2 and 2 a < pe. Then (q a 2, e; q) is desirable if and only if (i) a = 3 and q = 2, or (ii) a = 2 and gcd(q 2, q e 1) = 1.

22 The case e 3 Conjecture 2 Let e 3 and n = q a q b 1, 0 < b < a < pe. Then (n, e; q) is desirable if and only if (i) a = 2, b = 1, and gcd(q 2, q e 1) = 1, or (ii) a b 0 (mod e).

23 The Case b = p Theorem Let p be an odd prime and q a power of p. (i) F q 2 \ F q consists of the roots of (x x q ) q (ii) Let 0 < i 1 2 (p 1) and n = qp+2i q p 1. Then (2i 1)x q 2 if x F q, g n,q (x) = 2i 1 + 2i x x q if x F q 2 \ F q. (iii) For the n in (ii), (n, 2; q) is desirable if and only if 4i 1 (mod p).

24 The Case b = p Theorem Let p be an odd prime and q a power of p. (i) Let 0 < i 1 2 (p 1) and n = qp+2i 1 q p 1. Then 2(i 1)x q 2 if x F q, g n,q (x) = 2i 1 + 2i 2 x x q if x F q 2 \ F q. (ii) For the n in (i), (n, 2; q) is desirable if and only if i > 1 and 4i 3 (mod p).

25 Results Result 1 q = 5 2x 3 + 2x 19 is a PP of F q 2. Result 2 q = 13 6x x 155 is a PP of F q 2. Result 3 q = 121 6x x is a PP of F q 2. Result 4 q = 11 8x 9 + 2x 109 is a PP of F q 2. Result 5 q = 29 7x x 811 is a PP of F q 2.

26 Conjecture 3 Let f = x q 2 + tx q2 q 1, t F q. Then f is a PP of F q 2 one of the following occurs: (i) t = 1, q 1 (mod 4); (ii) t = 3, q ±1 (mod 12); (iii) t = 3, q 1 (mod 6). if and only if

27 Results Let p be an odd prime and q a power of p. Result 1 q 1 (mod 4) (q p+5 q 9 1, 2; q) is desirable. Result 2 Result 3 Result 4 (q p+3 q 5 1, 2; q) is desirable. (q p+4 q 7 1, 2; q) is desirable. (q p+6 q 11 1, 2; q) is desirable.

28 A most recent result Theorem Let p be an odd prime, q = p k, n = q p+i+1 q 2i+1 1. If ( ) { 2i : if i is odd, = q ( 1) q 1 2 : if i is even. ( ) a where is the Jacobian symbol, then (q p+i+1 q 2i+1 1, 2; q) is b desirable.

29 Outline of the proof e = 2, a = p + i + 1, b = 2i + 1. Case 1 - i is odd. a 0 = 0, a 1 = p i 2, b 0 = 1, b 1 = i g x q2 2 + i 2 x q2 q 2 (x + x q )[((i + 1)x + ix q ) q 1 1]. Clearly, g(0) = 0. When x F q, 2 g(x) = i((i + 1)w q iw) 2 + i((i + 1)w q iw) 2q + 2(i + 1)((i + 1)w q iw) q+1 (4i + 2) 2. w

30 Outline of the proof contd i((i + 1)w q iw) 2 + i((i + 1)w q iw) 2q + 2(i + 1)((i + 1)w q iw) q+1 Assume c 0. Let t = wc t F q. w = c t = where u = (i + 1)c q ic 1. t is unique w is unique. 1 i(u) 2q + 2(i + 1)(u) q+1 + i(u) 2,

31 Outline of the proof contd i((i + 1)w q iw) 2 + i((i + 1)w q iw) 2q + 2(i + 1)((i + 1)w q iw) q+1 Now assume c = 0. w = c i((i + 1)w q iw) 2q 2 + 2(i + 1)((i + 1)w q iw) q 1 + i = 0. Let z = ((i + 1)w q iw) q 1 F q 2. Then iz 2 + 2(i + 1)z + i = 0. (1) Since i is odd 2i + 1 is a square in F q. So (1) implies that z F q. Then we have z 2 = z q+1 = ((i + 1)w q iw) q2 1 = 1. So z = ±1 and it contradicts (1).

32 Outline of the proof contd Case 2 - i is even a 0 = 1, a 1 = p i 1 2, b 0 = 1, b 1 = i

33 Results with even q.

34 A Desirable Family Theorem Let q = 2 s, s > 1, e > 0, and let n = (q 1)q 0 + q 2 qe 1 + q 2 qe. We have g n,q = x + Tr q e /q(x) + x 1 2 qe Tr q e /q(x) 1 2 q. When e is odd, g n,q is a PP of F q e.

35 Example Let q = 4, e > 1 and Let n = 3q 0 + 2q e 1 + 2q e. Then g 3q 0 +2q e 1 +2q e = x + Tr q e /q(x) + x 2qe 1 Tr 2 q e /q(x) We claim that when e is odd, g n,q is a PP of F q e. Assume that there exist x, a F q e such that g(x) = g(x + a). Then we have a + Tr qe /q(a) + Tr 2 q /q(x)a e 2qe 1 + Tr 2 q /q(a)x e 2qe 1 + Tr 2 q /q(a)a e 2qe 1 = 0 Tr q e /q(a) = 0. a + Tr 2 q /q(x)a e 2qe 1 = 0 a = 0 in all three following cases Tr qe /q(x) = 0 Tr qe /q(x) = 1 Tr qe /q(x) 0, 1

36 Conjecture 4 Let q = 4, e = 3k, k 1, and n = 3q 0 + 3q 2k + q 4k. Then g n,q x + S 2k + S 4k + S 4k S2k 3 x + S q2k 2k + S qk +3 2k (mod x qe x). g n,q is a PP of F q e.

37 Theorem Let e = 3k, k 1, q = 2 s, s 2, and n = (q 3)q 0 + 2q 1 + q 2k + q 4k. Then g n,q x 2 + S 2k S 4k (mod x qe x), and g n,q is a PP of F q e.

38 A most recent result Theorem Let q = p 2, e > 0, and n = (p 2 p 1)q 0 + (p 1)q e + pq a + q b, a, b 0. Then g n,q = Sa p S b Se p 1. Assume that a + b 0 (mod p) and gcd ( x 2a+1 + 2x a+1 + x ɛ(x b + 1) 2, (x + 1)(x e + 1) ) = (x + 1) 2, for ɛ = 0, 1. Then g n,q is a PP of F q e.

39 Example Let q = 4, e > 3, n = q 0 + 2q 1 + q 2 + q e. Then, g n,q x 2 + xtr qe /q(x) + x q Tr qe /q(x) (mod x qe x). We claim that when gcd(1 + x + x 2, x e + 1) = 1, g n,q is a PP of F q e. Assume that g n,q (x) = g n,q (y), x, y F q e. Tr q e /q(g n,q (x)) = Tr q e /q(g n,q (y)) Tr q e /q(x) = Tr q e /q(y). Let Tr qe /q(x) = Tr qe /q(y) = a F q. If a = 0, then x = y. If a 0, then g n,q (x) = g n,q (y) becomes z 2 = a(z + z q ) where z = x + y. Substituting the above equation to itself we get (z 2 ) q = (z 2 ) + (z 2 ) q2. Since gcd(1 + x + x 2, x e + 1) = 1, we have z = 0.

40 References X. Hou, G. L. Mullen, J. A. Sellers, J. L. Yucas, Reversed Dickson polynomials over finite fields, Finite Fields Appl. 15 (2009), X. Hou, Two classes of permutation polynomials over finite fields, J. Combin. Theory A, 118 (2011), X. Hou, A new approach to permutation polynomials over finite fields, Finite Fields Appl. 18 (2012) N. Fernando, X. Hou, S. D. Lappano, A new approach to permutation polynomials over finite fields, II, submitted (2012), Preprint available at R. Lidl and H. Niederreiter, Finite Fields, 2nd ed., Cambridge Univ. Press, Cambridge (1997).

41 Thank You!