Constructing N-Polynomials over Finite Fields

Transcription

1 International Journal of Algebra, Vol. 5, 2011, no. 29, Constructing N-Polynomials over Finite Fields Mahmood Alizadeh Islamic Azad University- Ahvaz Branch Ahvaz, Iran Sergey Abrahamyan Institute for Informatics and Automation Problems Yerevan, Armenia Saeid Mehrabi Esfahan Payame Noor University Esfahan, Iran saeid Melsik K. Kuyregyan Institute for Informatics and Automation Problems National Academy of Sciences of Armenia Yerevan, Armenia Abstract This paper is devoted to the composition of constructing families of Normal polynomials over the finite field of Characteristic three. Mathematics Subject Classification: 12A20 Keywords: Galois fields, N-polynomials, Irreducible polynomials 1 Introduction The problem of normality of polynomials over Galois fields is a case of spacial interest and plays an important role in modern engineering.

2 1438 M. Alizadeh, S. Abrahamyan, S. Mehrabi and M. K. Kuyregyan Let F q, be a Galois field of order q = p s, where p is a prime and s is a natural number and F q, be its multiplicative group. A normal basis N for F q n over F q is a basis of the form N = {α, α q,..., α qn 1 } for some element α F q n. A monic irreducible polynomial F (x) F q (x) is called normal if its roots form a normal basis or, equivalently, if they are linearly independent over F q. This paper presents a result on the theory of the synthesis of Normal polynomials (N-Polynomials) over F 3 s. Some results regarding computationally simple costructions of N-polynomials over F q can be found in [3, 6, 7]. Also kyuregyan in [4, 5] cosidered constructions which yield sequences of normal irreducible polynomials. 2 Preliminary Notes we ll begin with recalling some definitions and basic results on the irreducibility and normality of polynomials that will be helpful to derive our main result. Definition 2.1 Let F q n be a finite extension field of the finite field F q.for α F q n, the trace Tr q n q(α) over F q is defined by Tr q n q(α) = Let n = n 1 p e = n 1 t, with gcd(p, n 1 ) = 1 and suppose that x n 1 has the following factorization in F q Set n 1 α qi. x n 1=(x n 1 1) t =(ϕ 1 (x)ϕ 2 (x)...ϕ r (x)) t. (1) φ i (x) = xn mi 1 ϕ i (x) = t iv x v We will need the following theorem which allows us to check if an irreducible polynomial is an N-polynomial. Proposition 2.1(Theorem 4.8, [6]). Let F (x) be an irreducible polynomial of degree n over F q and α be a root of it. Let x n 1 factors as above and let φ i (x) be as in above. Then F (x) is an N-polynomial over F q if and only if L φi (α) 0for i =1, 2,..., r where L φi (x) is the linearized polynomial defined by L φi (x) = t iv x qv if φ i (x) = t iv x v.

3 Constructing N-polynomials over finite fields 1439 Proposition 2.2 ([1], Theorem 1). Let P (x) = n c i x i be irreducible over F q of degree n and let δ 0,δ 1 F q, δ 0 δ 1. then F (x) =(x p x + δ 1 ) n P ( xp x + δ 0 x p x + δ 1 ) is irreducible polynomial of degree pn over F q if and only if Tr q p ((δ 1 δ 0 ) P (1) P (1) nδ 1) 0 Lemma 2.2 (D. Jungnickel, [2]) Let f(x) = n c i x i be a N-polynomial of degree n over F q (q = p s ). Then the polynomial g(x) =f( x a ) is a N- b polynomial if and only if na bc n 1 0. Proof. Let n = n 1 p e = n 1 t, then by (1), x n 1 has the following factorization in F q x n 1=(x n 1 1) t =(ϕ 1 (x)ϕ 2 (x)...ϕ r (x)) t, where put ϕ 1 (x) =x 1. Set for i =2, 3,..., r where Hence φ i (x) = xn 1 ϕ i (x) =(x 1)t s i (x) =(x 1)s i(x), s i(x) =(x 1) t 1 s i (x) = t iv x v. φ i (x) = t iv x v+1 t iv x v. Because f(x) is N-polynomial, we have L φi (α) 0 for i =1, 2,..., r, where α is a root of f(x). We show that we can derive also L φi (a + bα) 0, for i =2, 3,..., r, where a + bα is a root of g(x). Since L φi (a + bα) = t iv (a + bα) qv+1 t iv (a + bα) qv = a t iv + b t iv α qv+1 a t iv b t iv α qv = b( t iv α qv+1 t iv α qv )=bl φi (α) 0

4 1440 M. Alizadeh, S. Abrahamyan, S. Mehrabi and M. K. Kuyregyan So for being g(x), N-polynomial we need to find a condition that L φ1 (a+bα) 0. But we have So φ 1 (x) = xn 1 x 1 = xn 1 + x n 2 n x +1= x i. This completes the proof. n 1 L φ1 (a + bα) = (a + bα) qi = n 1 n 1 a + b α qi = na + bt r q n q(α) =na bc n 1. 3 constructing N-polynomials over finite fields The following theorem shows if P (x) be a N-polynomial of degree n over F 3 s, how proposition 2.2 can be used to produce a new N-polynomial of degree 3n over F 3 s. Theorem 3.1 Let P (x) = n c i x i be an irreducible polynomial of degree n over F 3 s and P (x) be a N-polynomial over F 3 s. Also let F (x) =(x 3 x +1) n x 3 x P ( ). (2) x 3 x +1 Then F (x) is a N-polynomial of degree 3n over F 3 s (n + c 1 ) Tr q p ( P (1) n) 0 c 0 P (1) if and only if Proof. proposition 2.2 and theorem s hypothesis implies that F (x) is irreducible over F 3 s. On the other side by (1) We have x 3n 1=(ϕ 1 (x) ϕ 2 (x)... ϕ r (x)) 3t. where ϕ i (x) F 3 s[x] are distinct irreducible factors of x n 1. Let H i (x) = x3n 1 ϕ i (x) = (xn 1)(x 2n + x n +1) ϕ i (x) = t iv x v (x 2n + x n + 1)) = t iv (x 2n+v + x n+v + x v )) (3)

5 Constructing N-polynomials over finite fields 1441 where (x n 1) ϕ i (x) = t iv x v. Suppose that α 1 be a root of F (x). Then β 1 = 1 α 1 is a root of F (x).we must show that L Hi (β 1 ) 0 But L Hi (β 1 )= t iv (β (3s ) 2n+v 1 + β (3s ) n+v 1 + β (3s ) v ) 1 ) = t iv (( 1 ) 32sn +( 1 ) 3sn +( 1 )) 3sv. (4) α 1 α 1 α 1 We note that by (2), α 1 3 α 1 α 1 3 α 1 +1 α = α 1 3 α 1 α 13 α 1 +1, where α is a root of P (x). Hence we have is a root of P (x). So we may assume that α 1= α 1 3 α 1 α 13 α = 1 α 13 α 1 +1 or equivalently α 1= (α 3 1 α 1 +1) 1 (5) so by (5) we have α 3 1 α 1 = α (6) 1 α also by (5) (α 1) 3sn 3 = α 1= (α sn α sn 1 +1) 1. (7) Then (7) and (5) implies that (α 1 3 sn+1 α 1 3 sn +1) 1 =(α 1 3 α 1 +1) 1. But (α 1 3 α 1 +1) 0, so we have (α 1 3 sn+1 α 1 3 sn +1)=(α 1 3 α 1 +1) or 3 (α sn 1 α 1 ) 3 3 =(α sn 1 α 1 ). 3 Then α sn 1 α 1 = θ F 3, and it is easy to show that α 1 3 ksn = α 1 + kθ. (8)

6 1442 M. Alizadeh, S. Abrahamyan, S. Mehrabi and M. K. Kuyregyan So from (4) and (8) we have 1 L Hi (β 1 )= t iv (( α 1 +2θ )+( 1 α 1 + θ )+( 1 )) 3sv α 1 = 1 = t iv ( ) 3sv. α 13 α 1 Thus by (6) we have L Hi (β 1 )= t iv ( α 1 3sv α ) t iv ( α 1(α 1 + θ)+α 1 (α 1 +2θ)+(α 1 +2θ)(α 1 + θ) ) 3sv α 1 (α 1 + θ)(α 1 +2θ) =( t iv (1 1 3 sv. α )) Denote P (x) byg(x), that is N-polynomial. Also it is clear that by theorem s hypothesis and lemma 2.2, G( x + 1) is a N-polynomial. But 1 1 is a root α of G( x + 1). Then we have and proof is completed. References t iv (1 1 α )3sv 0 [1] M.Alizadeh, Contructing Methods for irreducible Polynomials, Mathematical Problems of Computer Scinces, vol. XXXV.2011, PP ,. [2] D. Jungnickel, Trace-Orthogonal normal bases, Discrete applied mathematics, , 47(1993). [3] S.Gao. Normal bases over finite fields, Ph.D. Thesis, Waterloo, (1993). [4] Melsik, k. Kyuregyan. Iterated constructions of irreducible polynomials over finite fields with linearly independent roots. Finite fields and their applications. 10, (2004). [5] Melsik, k. Kyuregyan, Recursive constructions of N-polynomials over GF (2 s ), Discrete Applied Mathematics, 156(2008) [6] A.J. Menezes, I.F.Blake, X.Gao, R.C.Mullin, S.A.Vanstone, T.Yaghoobian, Applications of finite fields, Kluwer Academic publishers, Boston, Dordrecht, Lancaster, [7] H.Meyn, Explicit N-polynomial of 2-power degree over finite fields, Designs, Codes and Cryptography, 6, (1995), Received: July, 2011