ITALIAN JOURNAL OF PURE AND APPLIED MATHEMATICS N ( ) 451

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1 ITALIAN JOURNAL OF PURE AND APPLIED MATHEMATICS N ON THE k-normal ELEMENTS AND POLYNOMIALS OVER FINITE FIELDS Mahmood Alizadeh Department of Mathematics Ahvaz Branch Islamic Azad University Ahvaz Iran Mohammad Reza Darafsheh School of Mathematics Statistics and Computer Science College of Science University of Tehran Tehran Iran Saeid Mehrabi Department of Mathematics Farhangian University Tehran Iran saeid Abstract. An element α F q n is normal over F q if the set {α, α q,..., α qn 1 } is a basis of F q n over F q. The k-normal elements over finite fields are defined and characterized by Huczynska, Mullen, Panario and Thomson For 0 k n 1, the element α F q n is said to be a k-normal element if gcdx n 1, n 1 i=0 x n 1 i has degree αqi k. It is well known that a 0-normal element is a normal element. So, the k-normal elements are a generalization of normal elements. By analogy with the case of normal polynomials, a monic irreducible polynomial of degree n is called a k-normal polynomial if its roots are k-normal elements of F q n over F q. In this paper, a new characterization and construction of k-normal elements and polynomials over finite fields are given. Keywords: finite field, normal basis, k-normal element, k-normal polynomial. 1. Introduction Let F q be the Galois field of order q = p m, where p is a prime and m is a natural number, and F q be its multiplicative group. A normal basis of F q n over F q is a basis of the form N = {α, α q,..., α qn 1 }, i.e. a basis that consists of the algebraic conjugates of a fixed element α F q n. Such an element α F q n is. Corresponding author

2 452 MAHMOOD ALIZADEH, MOHAMMAD REZA DARAFSHEH and SAEID MEHRABI said to generate a normal basis for F q n over F q, and for convenience called a normal element. A monic irreducible polynomial F x F q [x] is called normal polynomial or N-polynomial if its roots are linearly independent over F q. Since the elements in a normal basis are exactly the roots of some N-polynomials, there is a canonical one-to-one correspondence between N-polynomials and normal bases. Normal bases have many applications, including coding theory, cryptography and computer algebra systems. For further details, see [9]. Recently, the k-normal elements over finite fields are defined and characterized by Huczynska et al [8]. For 0 k n 1, the element α F q n is called a k-normal element if deggcdx n 1, n 1 i=0 αqi x n 1 i = k. By analogy with the case of normal polynomials, a monic irreducible polynomial P x F q [x] of degree n is called a k-normal polynomial or N k - polynomial over F q if its roots are k-normal elements of F q n over F q. Here, P x has n distinct conjugate roots, of which n k are linearly independent. Recall that an element α F q n is called a proper element of F q n over F q if α / F q v for any proper divisor v of n. So, the element α F q n is a proper k-normal element of F q n over F q if α is a k-normal and proper element of F q n over F q. Using the above mention, a normal polynomial or element is a 0-normal polynomial or element. Since the proper k-normal elements of F q n over F q are the roots of a k-normal polynomial of degree n over F q, hence the k-normal polynomials of degree n over F q is just another way of describing the proper k-normal elements of F q n over F q. Some results on the constructions of special sequences of k-normal polynomials over F q, in the cases k = 0 and 1 can be found in [2, 4, 5, 10, 11] and [6], respectively. In this paper, in Sec. 2 some definitions, notes and results which are useful for our study have been stated. Section 3 is devoted to characterization and construction of k-normal elements. Finally, in Sec. 4 a recursive method for constructing k-normal polynomials of higher degree from a given k-normal polynomial is given. 2. Preliminary notes We use the definitions, notations and results given by Huczynska [8], Gao [7] and Kyuregyan [10, 11], where similar problems are considered. We need the following results for our further study. The trace of α in F q n over F q, is given by T r Fq n F q α = n 1 i=0 αqi. For convenience, T r Fq n F q is denoted by T r q n q. Let F be a field and fx = Σ n i=0 f ix i and gx = Σ m g jx j with all f i, g j F. The Sylvester matrix S f,g is the m + n m + n matrix given by:

3 ON THE k-normal ELEMENTS AND POLYNOMIALS OVER FINITE FIELDS S f,g = f n f n 1... f 1 f f n f f n f 0 g m g m 1... g 1 g g m g m g g m g 0 Proposition 2.1 [8]. Let F be a field. For two non-zero polynomials f, g F[x], we have ranks f,g = degf + degg deggcdf, g. Proposition 2.2 [8]. Let α F q n. Then the following properties are equivalent: i α is k-normal over F q ; ii α gives rise to a basis {α, α q,..., α qn k 1 } of a q-modules of degree n k over F q ; iii ranka α = n k, where A α = α α q... α qn 1 α q α q2... α.... α qn 1 α... α qn 2 Proposition 2.3 [6]. Let p divide n, then n = n 1 p e, for some e 1 and a, b F q. Theefore the element α is a proper k-normal element of F q n over F q if and only if a + bα is a proper k-normal element of F q n over F q. Let p denote the characteristic of F q and let n = n 1 p e = n 1 t, with gcdp, n 1 = 1 and suppose that x n 1 has the following factorization in F q [x] : 2 x n 1 = φ 1 xφ 2 x φ r x t, where φ i x F q [x] are the distinct irreducible factors of x n 1. For each s, 0 s < n, let there is a u s > 0 such that R s,1 x, R s,2 x,, R s,us x are all of the s degree polynomials dividing x n 1. So, from 2 we can write R s,i x = r j=1 φ j t ij x, for each 1 i u s, 0 t ij t. Let 3 ϕ s,i x = xn 1 R s,i x, for 1 i u s. Then, there is a useful characterization of the k-normal polynomials of degree n over F q as follows:.

4 454 MAHMOOD ALIZADEH, MOHAMMAD REZA DARAFSHEH and SAEID MEHRABI Proposition 2.4 [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 factor as 2 and let ϕ s,i x be as in 3. Then F x is a N k -polynomial over F q if and only if, there is j, 1 j u k, such that L ϕk,j α = 0, and also L ϕs,i α 0, for each s, k < s < n, and 1 i u s, where u s is the number of all s degree polynomials dividing x n 1 and L ϕs,i x is the linearized polynomial defined by n s n s L ϕs,i x = t iv x qv if ϕ s,i x = t iv x v. The following propositions and lemma are useful for constructing N k -polynomials over F q. Proposition 2.5 [3]. Let x p δ 2 x + δ 0 and x p δ 2 x + δ 1 be relatively prime polynomials in F q [x] and P x = n i=0 c ix i be an irreducible polynomial of degree n 2 over F q, and let δ 0, δ 1 F q, δ 2 F q, δ 0, δ 1 0, 0. Then x F x = x p δ 2 x + δ 1 n p δ 2 x + δ 0 P x p δ 2 x + δ 1 q 1 is an irreducible polynomial of degree np over F q if and only if δ 2 p 1 1 T r q p A p δ 1 δ 0 P 1 P 1 nδ 1 0, where A p 1 = δ 2, for some A F q. = 1 and Proposition 2.6 [1]. Let x p x + δ 0 and x p x + δ 1 be relatively prime polynomials in F q [x] and let P x be an irreducible polynomial of degree n 2 over F q, and 0 δ 1, δ 0 F p, such that δ 0 δ 1. Define F 0 x = P x F k x = x p x + δ 1 t k 1 x p x + δ 0 F k 1 x p, k 1 x + δ 1 where t k = np k denotes the degree of F k x. Suppose that δ1 δ 0 F 0 T r 1 nδ 1F 0 1 q p F 0 1 T r q p δ1 δ 0 F 0 δ 0 δ 1 + nδ 1 F 0 δ 0 F 0 δ 0 δ 1 δ 1 0. Then F k x k 0 is a sequence of irreducible polynomials over F q of degree t k = np k, for every k 0.

5 ON THE k-normal ELEMENTS AND POLYNOMIALS OVER FINITE FIELDS 455 Lemma 2.7. Let γ be a proper element of F q n and θ F p, where q = p m, m N. Then we have 4 p 1 1 γ + jθ = 1 γ p γ. Proof. By observing that p 1 it is enough to show that We note that 5 p 1 1 γ + jθ = 1 γ p γ p 1 γ p p 1 γ γ + jθ = where p 1 = j γ p γ, γ + jθ γ p γ γ + jθ = 1. p 1 γ + jθ p 1 1 = γ + jθ p 1 p 1 = θ j i j, j j=1 i=1 p 1! p 1 j!j!, j F p. On the other side, we know that { 6 i j 0 mod p, if p-1 j = 1 mod p, if p-1 j i=1 and also θ p 1 = 1. Thus by 5 and 6, the proof is completed. 3. Characterization and construction of k-normal elements In this section, we extend some existence results on the characterization and construction of normal elements into k-normal elements over finite fields. In the case k = 0, the following theorems had been obtained in [7] and [13]. Theorem 3.1. Suppose that α is a proper element of F q n over F q. Let α i = α qi and t i = T r q n qα 0 α i, 0 i n 1. Then α is a k-normal element of F q n over F q if and only if deggcdgx, x n 1 = k, where gx = Σ n 1 i=0 t ix i.

6 456 MAHMOOD ALIZADEH, MOHAMMAD REZA DARAFSHEH and SAEID MEHRABI Proof. Let So, by setting = A α A T α = we get A α = α α q... α qn 1 α q α q2... α.... α qn 1 α... α qn 2. T r q n qα 0 α 0 T r q n qα 0 α 1... T r q n qα 0 α n 1 T r q n qα 1 α 0 T r q n qα 1 α 1... T r q n qα 1 α n T r q n qα n 1 α 0 T r q n qα 0 α 0... T r q n qα n 1 α n 1 = t 0 t 1... t n 1 t n 1 t 0... t n t 1 t 2... t 0, ranka α A α T = ranka α = rank. Now, it is enough to show that deggcdσ n 1 i=0 t ix i, x n 1 = k if and only if the matrix has rank n k. The Sylvester matrix S f,g see Equation 1 with fx = x n 1 can be converted, by a sequence of column operations, into the block matrix In 1 0 n 1 0 n 1 From this block decomposition, it follows that By Proposition 2.1,. ranks f,g = rank + ranki n 1 = rank + n 1. ranks f,g = n + n 1 deggcdfx, gx. Combining these two expressions yields The proof is complete. deggcdfx, gx = n rank. Theorem 3.2. Let α be a k-normal element of F q n over F q. Then the element γ = Σ n 1 i=0 a iα qi, where a i F q, is a k-normal element of F q n over F q if and only if the polynomial γx = Σ n 1 i=1 a ix i is relatively prime to x n 1.

7 ON THE k-normal ELEMENTS AND POLYNOMIALS OVER FINITE FIELDS 457 Proof. Since α is a k-normal element of F q n over F q, so by Proposition 2.2, ranka α = n k, where α α q... α qn 1 α q α q2... α A α =..... α qn 1 α... α qn 2 Let A γ = γ γ q... γ qn 1 γ q γ q2... γ.... γ qn 1 γ... γ qn 2 By Proposition 2.2, it is enough to show that ranka γ = n k. We note that A γ = A A α, where a 0 a 1... a n 1 a 1 a 2... a 0 A =..... a n 1 a 0... a n 2 Since γx = Σ n 1 i=0 a ix i is relatively prime to x n 1, thus A is non-singular and so ranka γ = ranka A α = ranka α = n k. The proof is complete. Theorem 3.3. Let t and v are two positive integers with 1 < t < v < 2t and α is a k-normal element of F q vt over F q for v t k t 1. If γ = T r q vt q t α is a proper element of F q t over F q, then γ is a proper k-normal element of F q t over F q. Proof. Since α is a k-normal element of F q vt over F q, so by Proposition 2.2 the elements α, α q, α q2,..., α qvt k 1 form a basis for a q-module of degree vt k over F q. By hypothesis and considering γ = T r q vt q t α, the elements γ, γq,..., γ qv k 1 are non-overlapping sums of the vt k conjugates of α, which are assumed to be linearly independent over F q. So the v k conjugates of γ are linearly independent over F q. On the other side, for each 0 s k 1, γ qv k+s = Σ v i=1α qvt k+v+s it. = Σ vt k i=1 c iα qvt k i, c i F q = Σ v k j=1 d jγ qv k j, d j F q t. So γ gives rise to a basis M = {γ, γ q,..., γ qv k 1 } of a q-modules of degree v k over F q. By Proposition 2.2, the proof is complete.

8 458 MAHMOOD ALIZADEH, MOHAMMAD REZA DARAFSHEH and SAEID MEHRABI Theorem 3.4. Let t and v are two positive integers with gcdv, t = 1 and α is a k-normal element of F q v over F q, for 0 k v 1. Then α is also a k-normal element of F q vt over F q t. Proof. Since α is a k-normal element of F q v over F q, so by Proposition 2.2, ranka α = v k, where A α = α α q... α qv 1 α q α q2... α.... α qv 1 α... α qv 2 The element α is also a k-normal element of F q vt over F q t if ranka α = v k, where α α qt... α qv 1t A α α = qt α q2t... α..... α qv 1t α... α qv 2t Since gcdv, t = 1, when j runs through 0,1,2,..., v 1 modulo v, tj also runs through 0,1,2,..., v 1 modulo v. Note that since α F q v, we have α qv = α and thus α qjt = α qk j whenever jt k j mod v and k j runs through 0,1,2,..., v 1. So ranka α = ranka α = v k and the proof is complete.. 4. Recursive construction N k -polynomials In this section we establish theorems which will show how propositions 2.4, 2.5 and 2.6 can be applied to produce infinite sequences of N k -polynomials over F q. Recall that, the polynomial P x = x n P 1 x is called the reciprocal polynomial of P x, where n is the degree of P x. In the case k = 0, some similar results of the following theorems have been obtained in [2], [4] and [5], Theorems and We use of an analogous technique to that used in the above results, where similar problems are considered. Theorem 4.1. Let P x = n i=0 c ix i be an N k -polynomial of degree n over F q, for each n = rp e, where e N and r equals 1 or is a prime different from p and q a primitive element modulo r. Suppose that δ F q and 7 F x = x p x + δ n x P p x x p. x + δ Then F x is an N k -polynomial of degree np over F q if k < p e and T r q p δ P 1 P 0. 1

9 ON THE k-normal ELEMENTS AND POLYNOMIALS OVER FINITE FIELDS 459 Proof. Since P x is an irreducible polynomial over F q, so Proposition 2.5 and theorem s hypothesis imply that F x is irreducible over F q. Let α F q n be a root of P x. Since P x is an N k -polynomial of degree n over F q by theorem s hypothesis, then α F q n is a proper k-normal element over F q. Since q is a primitive modulo r, so in the case r > 1 the polynomial x r x + 1 is irreducible over F q. Thus x n 1 has the following factorization in F q [x]: 8 x n 1 = φ 1 x φ 2 x t, where φ 1 x = x 1, φ 2 x = x r x + 1 and t = p e. Letting that for each 0 s < n and 1 i u s, R s,i x is the s degree polynomial dividing x n 1, where u s is the number of all s degree polynomials dividing x n 1. So, from 8, we can write R s,i x = x 1 s 1,i x r x + 1 s 2,i, where s = s 1,i + s 2,i r 1 for each 0 s 1,i, s 2,i t, except when s 1,i = s 2,i = t. So, we have 9 ϕ s,i x = xn 1 R s,i x = x n 1 x 1 s 1,i x r x + 1 s = n s t 2,i s,i,v x v. Since P x is an N k -polynomial of degree n over F q, so by Proposition 2.4, there is a j, 1 j u k, such that L ϕk,j α = 0, and also L ϕs,i α 0, for each k < s < n and 1 i u s. Further, we proceed by proving that F x is a k-normal polynomial. Let α 1 be a root of F x. Then β 1 = 1 α 1 is a root of its reciprocal polynomial F x. Note that by 8, the polynomial x np 1 has the following factorization in F q [x]: 10 x np 1 = φ 1 x φ 2 x pt, where φ 1 x = x 1, φ 2 x = x r x + 1 and t = p e. Letting that for each 0 s < np and 1 i u s, R s,i x is the s degree polynomial dividing x np 1, where u s is the number of all s degree polynomials dividing x np 1. So, from 10 we can write R s,i x = x 1 s 1,i x r x + 1 s 2,i, where s = s 1,i + s 2,i r 1 for each 0 s 1,i, s 2,i pt, except when s 1,i = s 2,i = pt. Therefore by considering 11 H s,i x = xnp 1 R s,i x,

10 460 MAHMOOD ALIZADEH, MOHAMMAD REZA DARAFSHEH and SAEID MEHRABI and Proposition 2.4, F x is an N k -polynomial of degree np over F q if and only if there is a j, 1 j u k, such that and also L H k,j β 1 = 0, L H s,i β 1 0, for each k < s < np and 1 i u s. Consider 12 H s,i x = xnp 1 R s,i x = xn 1 R s,i x x jn for each 0 s < n and 1 i u s. By 9 we obtain n s H s,i x = ϕ s,i x x jn = It follows that or n s L Hs,i β 1 = t s,i,v 1 n s 13 L Hs,i β 1 = L Hs,i = α 1 t s,i,v β 1 pjmn t s,i,v 1, pmv, α 1 x jn+v p jmn. pmv. From 7, if α 1 is a zero of F x, then α 1 p α 1 +δ α p 1 α 1 is a zero of P x, and therefore it may assume that α = α 1 p α 1 + δ α p, 1 α 1 or 14 α 1 δ = α 1 p α 1 1. Now, by 14 and observing that P x is an irreducible polynomial of degree n over F q, we obtain 15 α 1 δ It follows from 14 and 15 that = α 1 p mn p = α mn+1 p 1 α mn 1 1. δ 16 α 1 p mn+1 α 1 p mn 1 = α 1 p α 1 1.

11 ON THE k-normal ELEMENTS AND POLYNOMIALS OVER FINITE FIELDS 461 Also observing that F x is an irreducible polynomial of degree np over F q, we have α 1 p α 1 0 and α 1 p mn+1 α 1 p mn 0. Hence by α 1 p mn α 1 p = α 1 p mn α 1. It follows from 17 that α 1 p mn α 1 = θ F p. Hence α 1 p mn = α 1 + θ and α 1 p 2mn = α 1 + θ pmn = α 1 p mn + θ pmn = α 1 + θ + θ = α 1 + 2θ. It is easy to show that α 1 p jmn = α 1 + jθ, for 1 j p 1, or 18 1 α 1 p jmn = 1, for 1 j p 1. α 1 + jθ From 13 and 18, we immediately obtain n s 19 L Hs,i β 1 = t s,i,v Thus, by 14, 19 and Lemma 2.7 we have 20 n s L Hs,i β 1 = = 1 δ n s 1 α 1 + jθ t s,i,v 1 p mv α p 1 α 1 pmv. t s,i,v 1 α pmv = L ϕs,i 1 α δ Since α is a zero of P x, then α will be a k-normal element in F q n over F q. Thus according to Proposition 2.3, the element 1 α δ will also be a k-normal element. since 1 α δ is a root of P δx + 1, so by 20 and Proposition 2.4, there is a j, 1 j u k, such that L Hk,j β 1 = 0, and also L Hs,i β 1 0, for each s, k < s < n and 1 i u s. So, there is a j, 1 j u k, such that, L H β 1 = L k,j Hk,j β 1 = 0. On the other side, by 11 and 12, for each s, k < s < np and 1 i u s, there is s, k < s < n and 1 i u s such that H s,i x divide H s,i x. It follows that L H β 1 0, for each s, k < s < np s,i and 1 i u s. The proof is completed. In the following theorem, a computationally simple and explicit recurrent method for constructing higher degree N k -polynomials over F q starting from an N k -polynomial is described. Theorem 4.2. Let P x be an N k -polynomial of degree n over F q, for each n = rp e, where e N and r equals 1 or is a prime different from p and q a primitive element modulo r. Define F 0 x = P x.

12 462 MAHMOOD ALIZADEH, MOHAMMAD REZA DARAFSHEH and SAEID MEHRABI x 21 F u x = x p p x x + δ npu 1 F u 1 x p, x + δ where δ F p. Then Fux u 0 is a sequence of N k -polynomials of degree np u over F q if k < p e and P 0 P 1 T r q p P T r 0 q p P 0, 1 where P 0 and P 1 are the formal derivative of P x at the points x = 0 and x = 1, respectively. Proof. By Proposition 2.6 and hypotheses of theorem for each u 1, F u x is an irreducible polynomial over F q. Consequently, Fux u 0 is a sequence of irreducible polynomials over F q. The proof of k-normality of the irreducible polynomials Fux, for each u 1 is implemented by mathematical induction on u. In the case u = 1, by Theorem 4.1 F 1 x is a k-normal polynomial. For u = 2 we show that F 2 x is also a k-normal polynomial. To this end we need to show that the hypothesis of Theorem 4.1 are satisfied. By Theorem 4.1, F 2 x is a k-normal polynomial if T r q p F 1 1 F 1 1 since δ F p. We apply 21 to compute 0, 22 F u 0 = F u 1 = δ un P 0, u = 1, 2,.... We calculate the formal derivative of F 1 x at the points x = 0 and x = 1. According to 21 the first derivative of F 1 x is F 1 x = nx p x + δ n 1 x p x F 0 x p x + δ + x p x + δ n px p 1 1x p x + δ px p 1 1x p x F 0 x p x x p x + δ = δx p x + δ n 2 P x p x x p x + δ and at the points x = 0 and x = 1 x p x + δ 2, 23 F 1 0 = F 1 1 = δ n 1 P 0 which is not equal to zero by the condition T r q p P 0 P 0 0 in the hypothesis of theorem, since δ F p. From 23 and 22 F 24 T r 1 1 δ n 1 P 0 q p = T r F 1 1 q p δ n P = 1 0 δ T r P 0 q p P, 0

13 ON THE k-normal ELEMENTS AND POLYNOMIALS OVER FINITE FIELDS 463 which is not equal to zero by hypothesis of theorem. Hence the polynomial F2 x is a k-normal polynomial. If induction holds for u 1, then it must hold also for u, that is by assuming that Fu 1 x is a k-normal polynomial, we show that Fux is also a k-normal polynomial. Let u 3. By Theorem 4.1, Fux is a k-normal polynomial if F T r u 1 1 q p 0, F u 1 1 since δ F p. We calculate the formal derivative of F u 1 x at the points 1 and 0. By 21 the first derivative of F u 1 x is px F u 1x = x p p 1 1x p x + δ px p 1 1x p x x + δ npu 2 x p x + δ 2 x F u 2 p x x p x + δ = δx p x + δ npu 2 2 x F u 2 p x x p, x + δ and at the point x = 0 and x = 1 F u 10 = F u 11 = δ npu 2 1 F u 20 = δ n 1 F u 20. So we have F u 10 = F u 11 = 1 u 2 δ n 1u 2 F 10, which is not equal to zero by 23 and the condition T r q p hypothesis of theorem, since δ F p. Also F T r u 1 1 q p = 1 u 2 1 F u 1 1 δ u 2 T r q p P 0 P 0 F 1 1, F in the which is not equal to zero by 24 and hypothesis of theorem. The theorem is proved. References [1] S. Abrahamyan, M. K. Kyureghyan, A recurrent method for constructing irreducible polynomials over finite fields, Proceeding of the 13th international conference on computer algebra in scientific computing, 2011, 1-9. [2] S. Abrahamyan, M. K. Kyureghyan, New recursive construction of normal polynomials over finite fields, Proceeding of the 11th international conference on finite fields and their applications, 2013, 1-10.

14 464 MAHMOOD ALIZADEH, MOHAMMAD REZA DARAFSHEH and SAEID MEHRABI [3] S. Abrahamyan, M. Alizadeh, M. K. Kyureghyan, Recursive constructions of irreducible polynomials over finite fields, Finite Fields Appl., , [4] M. Alizadeh, S. Abrahamyan, S. Mehrabi, M. K. Kyuregyan, Constructing of N-polynomials over finite fields, International Journal of Algebra, 5 29, 2011, [5] M. Alizadeh, Construction of Irreducible and Normal Polynomials over Finite Fields, Ph.D. Thesis, National Academy of Sciences of Armenia, Yerevan, [6] M. Alizadeh, Some notes on the k-normal elements and k-normal polynomials over finite fields, Journal of Algebra and Its Applications, Vol. 16, No. 1, , 11 pages, [7] S. Gao, Normal bases over finite fields, Ph.D. Thesis, Waterloo, [8] S. Huczynska, G. L. Mullen, D. Panario and D. Thomson, Existence and properties of k-normal elements over finite fields, Finite Fields Appl., , [9] D. Jungnickel, Finite Fields: Structure and Arithmetics, Wissenschaftsverlag, Mannheim, [10] M. K. Kyuregyan, Iterated construction of irreducible polynomials over finite fields with linearly independent roots, Finite Fields Appl., , [11] M. K. Kyuregyan, Recursive construction of N-polynomials over GF 2 s, Discrete Applied Mathematics, , [12] H. W. Lenstra and R. Schoof, Primitive normal bases for finite fields, Mathematics of Computation, , [13] S. Perlis, Normal bases of cyclic fields of prime-power degree, Duke Math. J., , [14] S. Schwartz, Irreducible polynomials over finite fields with linearly independent roots, Math. Slovaca, , Accepted:

Constructing N-Polynomials over Finite Fields

International Journal of Algebra, Vol. 5, 2011, no. 29, 1437-1442 Constructing N-Polynomials over Finite Fields Mahmood Alizadeh Islamic Azad University- Ahvaz Branch Ahvaz, Iran Alizadeh@iauahvaz.ac.ir