Complete permutation polynomials of monomial type

Transcription

1 Complete permutation polynomials of monomial type Giovanni Zini (joint works with D. Bartoli, M. Giulietti and L. Quoos) (based on the work of thesis of E. Franzè) Università di Perugia Workshop BunnyTN 7 Trento, 16 novembre 2016

2 Outline 1 Permutation polynomials: an introduction

3 Outline 1 Permutation polynomials: an introduction 2 Monomial complete permutation polynomials: our results

4 Outline 1 Permutation polynomials: an introduction 2 Monomial complete permutation polynomials: our results 3 Particular cases: degree 8 and 9 in characteristic 2 and 3

5 Some definitions F l : finite field with l = p h elements Plane curve C : F (X, Y, T ) = 0 F l -rational point of C: P = (x, y, z) PG(2, l) such that F (x, y, z) = 0 Definition f (x) F l [x] is a permutation polynomial (shorlty, a PP) of F l if x f (x) is a bijection of F l (iff x f (x) is injective over F l ) Definition f (x) F l [x] is a complete permutation polynomial (shorlty, a CPP) of F l if both f (x) and f (x) + x are PPs of F l Definition f (x) F l [x] is an exceptional polynomial over F l if f (x) is a PP of an infinite number of extensions of F l

6 CPPs and Cryptography Definition f (x) F l [x] is a permutation polynomial (shorlty, a PP) of F l if x f (x) is a bijection of F l (iff x f (x) is injective over F l ). f (x) F l [x] is a complete permutation polynomial (shorlty, a CPP) of F l if both f (x) and f (x) + x are PPs of F l Definition f : F n 2 F 2 Boolean function is bent if x f (x + a) + f (x) is balanced a F n 2 ( f is PNF) bent-negabent if both x f (x + a) + f (x) and x f (x + a) + f (x) + Tr(ax) are balanced a F n 2 LINK: any PP of F 2 n gives rise to a bent function over F n 2 any CPP of F 2 n gives rise to a bent-negabent function over F n 2

7 Link with curves f (x) F l [x] C f : f (x) f (y) x y = 0 f (x) is a PP of F l = C f has no affine F l -rational points (a, b) with a b

8 Link with curves f (x) F l [x] C f : f (x) f (y) x y = 0 f (x) is a PP of F l = C f has no affine F l -rational points (a, b) with a b Theorem C absolutely irreducible curve of degree d defined over F l The number N l of F l -rational points satisfies N l l + 1 (d 1)(d 2) l for l large enough: f (x) is a PP of F l C f has no F l -rat. abs. irr. components distinct from X = Y

9 Conversely: Theorem (Cohen 1970) C f contains no F l -rational abs. irr. component distinct from X = Y f (x) is an exceptional polynomial over F l

10 Conversely: Theorem (Cohen 1970) C f contains no F l -rational abs. irr. component distinct from X = Y f (x) is an exceptional polynomial over F l It is not difficult to construct PP without any prescribed structure Remark f (x) is a PP of F l αf (γx + δ) + β is a PP of F l (α, β, γ, δ F l, α, γ 0) PP-equivalence : f (x) αf (γx + δ) + β, α, β, γ, δ F l, α, γ 0

11 The monomial case b 1 x d is a PP of F l (d, l 1) = 1 b 1 x d is a CPP of F l (d, l 1) = 1 and x d + bx is a PP of F l

12 The monomial case b 1 x d is a PP of F l (d, l 1) = 1 b 1 x d is a CPP of F l (d, l 1) = 1 and x d + bx is a PP of F l f b (x) = b 1 x qn 1 q 1 +1 has been studied as CPP of F q n for n = 2, 3, 4 and partially for n = 6

13 The monomial case b 1 x d is a PP of F l (d, l 1) = 1 b 1 x d is a CPP of F l (d, l 1) = 1 and x d + bx is a PP of F l f b (x) = b 1 x qn 1 q 1 +1 has been studied as CPP of F q n for n = 2, 3, 4 and partially for n = 6 EXPLICIT LIST of all b F q n such that f b is a CPP of F q n, in the cases: n = 7, for arbitrary q (E. Franzè, Master Thesis) n = 6, for arbitrary q (Bartoli-Giulietti-Z., FFA 2016)

14 The monomial case b 1 x d is a PP of F l (d, l 1) = 1 b 1 x d is a CPP of F l (d, l 1) = 1 and x d + bx is a PP of F l f b (x) = b 1 x qn 1 q 1 +1 has been studied as CPP of F q n for n = 2, 3, 4 and partially for n = 6 EXPLICIT LIST of all b F q n such that f b is a CPP of F q n, in the cases: n = 7, for arbitrary q (E. Franzè, Master Thesis) n = 6, for arbitrary q (Bartoli-Giulietti-Z., FFA 2016) Conjecture (Wu-Li-Helleseth-Zhang 2015) If n + 1 is prime, n + 1 p, gcd(n + 1, q 2 1) = 1, then: there exist CPPs of F q n of type b 1 x qn 1 q 1 +1

15 The monomial case b 1 x d is a PP of F l (d, l 1) = 1 b 1 x d is a CPP of F l (d, l 1) = 1 and x d + bx is a PP of F l f b (x) = b 1 x qn 1 q 1 +1 has been studied as CPP of F q n for n = 2, 3, 4 and partially for n = 6 GOAL : to characterize for any n the b F q n is a CPP of F q n such that f b = b 1 x qn 1 q 1 +1

16 The monomial case b 1 x d is a PP of F l (d, l 1) = 1 b 1 x d is a CPP of F l (d, l 1) = 1 and x d + bx is a PP of F l f b (x) = b 1 x qn 1 q 1 +1 has been studied as CPP of F q n for n = 2, 3, 4 and partially for n = 6 GOAL : to characterize for any n the b F q n is a CPP of F q n such that f b = b 1 x qn 1 q 1 +1 WE OBTAIN : complete classification for n 4 < q = p m with the exception of the cases n + 1 = p r, with r > 1 n + 1 = p r (p r 1)/2, with p {2, 3}, r > 1, gcd(r, 2m) = 1

17 b F q n = A i (b) := 0 j 1 <j 2 <...<j i n 1 b qj 1 +q j q j i F q i-th elementary symmetrical polynomial in b, b q,..., b qn 1

18 b F q n = A i (b) := 0 j 1 <j 2 <...<j i n 1 b qj 1 +q j q j i F q i-th elementary symmetrical polynomial in b, b q,..., b qn 1 Proposition (Wu-Li-Helleseth-Zhang 2013) If n 4 < q, then: b 1 x qn 1 q 1 +1 is a CPP of F q n gcd(n + 1, q 1) = 1, x n+1 + A 1 (b)x n + + A n (b)x is an exceptional polynomial over F q

19 b F q n = A i (b) := 0 j 1 <j 2 <...<j i n 1 b qj 1 +q j q j i F q i-th elementary symmetrical polynomial in b, b q,..., b qn 1 Proposition (Wu-Li-Helleseth-Zhang 2013) If n 4 < q, then: b 1 x qn 1 q 1 +1 is a CPP of F q n gcd(n + 1, q 1) = 1, x n+1 + A 1 (b)x n + + A n (b)x is an exceptional polynomial over F q Remark b 1 x qn 1 q 1 +1 is a CPP of F q n b qi x qn 1 q 1 +1 is a CPP of F q n

20 Proposition (Wu-Li-Helleseth-Zhang 2013) If n 4 < q, then: b 1 x qn 1 q 1 +1 is a CPP of F q n gcd(n + 1, q 1) = 1, x n+1 + A 1 (b)x n + + A n (b)x is an exceptional polynomial over F q Definition Let g(x) = x n+1 + λ 1 x n + λ n 1 x 2 + λ n x F q [x], λ n 0, be a PP of F q. g(x) is good if the roots of v g (x) := g( x) x = x n λ 1 x n ( 1) n 1 λ n 1 x + ( 1) n λ n form a unique orbit under the Frobenius map z z q.

21 Proposition If n 4 < q, then: b F q n \ F q is such that b 1 x qn 1 q 1 +1 is a CPP of F q n b is a root of v g (x) = g( x) x for some g good exceptional pol. of degree n + 1 over F q with g(0) = 0 and g (0) 0

22 Proposition If n 4 < q, then: b F q n \ F q is such that b 1 x qn 1 q 1 +1 is a CPP of F q n b is a root of v g (x) = g( x) x for some g good exceptional pol. of degree n + 1 over F q with g(0) = 0 and g (0) 0 Definition An exceptional polynomial g is decomposable if g(x) = g 1 (g 2 (x)) with g 1, g 2 exceptional pol., deg(g 1 ), deg(g 2 ) > 1 Proposition g good exceptional polynomial = g indecomposable

23 Idea In order to classify all CPPs of type f (x) = b 1 x qn 1 q 1 +1 take all the good indecomposable exceptional polynomials and determine the roots of v g (x)

24 Idea In order to classify all CPPs of type f (x) = b 1 x qn 1 q 1 +1 take all the good indecomposable exceptional polynomials and determine the roots of v g (x) Unfortunately: the complete classification of indecomposable exceptional polynomials is not known!

25 Remark f (x) is a good PP of F l αf (γx) + β is a good PP of F l (α, β, γ F l, α, γ 0) CPP-equivalence : f (x) αf (γx) + β, α, β, γ F l, α, γ 0

26 Remark f (x) is a good PP of F l αf (γx) + β is a good PP of F l (α, β, γ F l, α, γ 0) CPP-equivalence : f (x) αf (γx) + β, α, β, γ F l, α, γ 0 We use the known partial classification of indecomposable exceptional polynomial, up to CPP-equivalence

27 Classification of indecomposable exceptional polynomials, up to CPP-equivalence A) n + 1 q 1 is a prime different from p and A1) g(t) = (t + e) n+1 e n+1, e F q A2) g(t) = D n+1 (t + e, a) D n+1 (e, a), a, e F q, a 0, n + 1 q 2 1 D n+1 (t, a) Dickson polynomial of degree n + 1 ) B) n + 1 = p and g(t) = (t + e) ((t + e) p 1 r ) r a e (e p 1 r r a r p 1, a, e F q, a r(q 1)/(p 1) 1. C) n + 1 = s(s 1)/2 p {2, 3}, q = p m, r > 1, s = p r > 3 and (r, 2m) = 1. D) n + 1 = p r with r > 1.

28 Case A1 n + 1 is prime, n + 1 p, n + 1 does not divide q 1 ζ n+1 := (n + 1)-th primitive root of unity Proposition Let e F q. Then g(t) = (t + e) n+1 e n+1 is good exceptional over F q ord n+1 (q) = n If ord n+1 (q) = n, then for each e F q and i {1,..., n} ( e(ζ i n+1 1) ) 1 x q n 1 q 1 +1 is a CPP of F q n

29 Case A2 n + 1 is prime, n + 1 p, n + 1 does not divide q 2 1 (Dickson polynomials) Proposition D n+1 (t, a) = n/2 k=0 n + 1 n + 1 k ( n + 1 k k ) ( a) k t n+1 2k g(x) = D n+1 (x + e, a) D n+1 (e, a), e, a F q, a 0, D n+1 (e, a) 0, is good exceptional over F q if and only if one of the following cases occurs: i) 4 n and ord n+1 (q) = n { e 2 4a / q, ord n+1 (q) = n/2 ii) 4 n and e 2 4a q, ord n+1 (q) = n

30 Case B n + 1 = p N Fq/F p : the norm map F q F p, x x 1+p+p2 + +q/p. Theorem Let n 4 < q. Then b 1 x qn 1 q 1 +1 is a CPP of F q n for some r n, one of the following cases occurs: i) b {ζq 1 i gcd(r, i) = 1} ii) b {(v 0 λu 0 ) r e λ F p, e, u p 1 0 F q, u q 1 r 0 1, ( )) v0 r u = e, ord (N p 1 0 Fq/Fp e (p 1)/r = p 1}

31 n + 1 = 8, p = 2 F (x) F q [x] monic of degree 8 Proposition F (x) is good exceptional over F q if and only if F (x) = x 8 + ax 4 + bx 2 + cx is additive and x 7 + ax 3 + bx + c is irreducible over F q.

32 n + 1 = 9, p = 3 No classification is known! When is F (x) = x 9 + A 1 x 8 + A 2 x 7 + A 3 x 6 + A 4 x 5 + A 5 x 4 + A 6 x 3 + A 7 x 2 + A 8 x Theorem (Cohen 1970) good exceptional? C F contains no F l -rational component distinct from X = Y F (x) is an exceptional polynomial over F l Determine when C F := F (x) F (y) x y = 0 has only non-rational components (other than x y) Study when the roots of v F (x) are in a unique orbit under Frobenius

33 n + 1 = 9, p = 3 Proposition F (x) is good exceptional over F q if and only if i) F (x) = x 9 + A 6 x 3 + A 8 x and x 8 + A 6 x 2 + A 8 irreducible over F q ; ii) F (x) = x 9 + A 3 x 6 + A 4 x 5 + A 5 x 4 + ( ) ( + 2A 3 A A3 5 x 2 + A A 4 0, ( A A 3 A3 5 A 3 4 2A 3 A 5 + A A4 5 A 3 4 ) + A2 5 A 4 x 3 ) x, 2 the polynomial x 8 + 2A 3 x 2 + 2A 4 F q [x] has no roots in F q 4; ( iii) F (x) = x 9 + A 2 x 7 + A 3 x 6 + A 5 x 4 + A A 3A 5 ( 2A 2 A A3 3 A 2 ) x 2 + A 2 ) x 3 + ( A A 3A 5 + A2 5 A 2 + A4 3 A 2 2 ) x, 1 2A 2 is not a square in F q, 2 the polynomial v F (x) = F ( x)/( x) is irreducible over F q.

34 Thank you for your attention!