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A new proof to the complexity of the dual basis of a type I optimal normal basis ∗ Baofeng Wu , Kai Zhou, Zhuojun Liu 1 1 Academy of Mathematics and Systems Science, Chinese Academy of Sciences, 0 Beijing 100190, China 2 c e D 2 2 Abstract F M] The complexity of the dual basis of a type I optimal normal basis of qn over F was determined to be 3n − 3 or 3n − 2 according as q is even or odd, q D respectively, by Z.-X. Wan and K. Zhou in 2007. We give a new proof to this . s result by clearly deriving the dual of a type I optimal basis with the aid of a c lemma on the dual of a polynomial basis. [ 2 Keywords: Finite fields; Normal basis; Complexity; Type I optimal normal v basis; Dual basis; Polynomial basis. 3 5 1 3 1. Introduction . 2 1 Let F be an extension of the finite field F . A basis of the form qn q 1 {α,αq,...,αqn−1} is called a normal basis, and α is called a normal basis 1 : generator in this case. Normal bases have great advantages against the com- v i mon polynomial bases in arithmetic of finite fields. A simple property about X ∗ ∗ a normal basis is it has normal dual basis. By dual basis {β ,...,β } of a r 1 n a given basis {β1,...,βn}, we mean TrFqn/Fq(βiβj∗) = δij, 1 ≤ i,j ≤ n. For a normal basis generator α, denote α = αqi, i = 0,...,n − 1 and i N = {α0,...,αn−1}. Let n−1 α·α = t α , t ∈ F , 0 ≤ i ≤ n−1. i ij j ij q j=0 X ∗ Corresponding author Email address: [email protected](Baofeng Wu) Preprint submitted to Elsevier December 23, 2011 The number of nonzero elements in the n×n matrix T = (t ) is called the ij complexity of the normal basis N. We denote it by C . It can be shown N that C ≥ 2n−1, and N is called optimal when the lower bound is attained. N F F A type I optimal normal basis of over is the normal basis formed qn q by the n nonunit (n + 1)-th roots of unity, where n + 1 is a prime and q is primitive modulo n + 1 [2]. In [3] the dual of a type I optimal normal basis was studied, and its complexity was determined to be 3n−3 or 3n−2 according as q is even or odd, respectively. Two bases {β ,...,β } and {β′,...,β′} of F over F are called equiv- 1 n 1 n qn q alent, or weakly equivalent, if β = cβ′, i = 1,...,n for some c ∈ F , or i i q F , respectively. It is easy to see equivalent normal bases share the same qn complexity. In this paper, we clearly derive the dual M of a type I optimal normal basis N through a new approach and thus rediscover the complexity of it. Themainauxiliarylemmaweusewillbepresentedinsection2, andinsection 3, we will propose our method to get M and its complexity, which is totally different from that in [3] and seems more simple and easier to understand. 2. Auxiliary lemma on the dual of a polynomial basis The following lemma, proposed as an exercise in [1, Chapter 2], is the main result we will need in computing the dual of a type I optimal normal basis. We will include a short proof for it here for completeness. Lemma 1. Let F = F (α) be an extension of the finite field F , and f(x) ∈ qn q q F [x] be the minimal polynomial of α. Assume q f(x) = β0 +β1x+···+βn−1xn−1 ∈ Fqn[x]. x−α Thenthe dual basisofthe polynomialbasis{1,α,...,αn−1} is 1 {β ,β ,..., f′(α) 0 1 βn−1} := {f′β(0α), f′β(1α),..., fβ′n(−α1)}, where f′ is the formal derivative of f. Proof. Let σ be the i-th Frobenius automorphism of F /F and α = i qn q i σ (α) = αqi, 0 ≤ i ≤ n−1. Then f(x) = n−1(x−α ). Forl = 0,1,...,n−1, i i=0 i we construct the auxiliary polynomial Q n−1 f(x) αl F (x) = i −xl. l x−α f′(α ) i i i=0 X 2 It’s easy to check F (α ) = 0 for i = 0,1,...,n − 1. Thus F (x) ≡ 0 since l i l deg F (x) ≤ n−1. On the other hand, l n−1 f(x) αl F (x) = σ l i x−αf′(α) Xi=0 (cid:16) (cid:17) n−1 n−1 αl = σ β xj i j f′(α) Xi=0 (cid:16)Xj=0 (cid:17) n−1 n−1 αl = σ β xj i jf′(α) ! Xj=0 Xi=0 (cid:16) (cid:17) n−1 β = TrFqn/Fq f′(jα)αl xj. Xj=0 (cid:16) (cid:17) So we get TrFqn/Fq f′β(jα)αl = δjl for any 0 ≤ j,l ≤ n−1, which is what we (cid:3) want to prove. (cid:0) (cid:1) Remark 1. In fact, the lemma and the proof hold for any separable alge- braic field extension K(α)/K. 3. Dual basis of a type I optimal normal basis and its complexity F F Recall that a type I optimal normal basis of over is of the form qn q N = {α,α2,...,αn} = α{1,α,...,αn−1} where α is a root of the primitive polynomial f(x) = n xi [2], i.e. it is weakly equivalent to the polynomial i=0 basis {1,α,...,αn−1}. Thus if we can compute the dual of the polynomial P basis, say f′(1α){β0,β1,...,βn−1} by Lemma 1, then we can get the dual of N of the form M = αf′1(α){β0,β1,...,βn−1}. The remaining task is just to ′ compute αf (α) and β0,...,βn−1. Lemma 2. n+1 ′ αf (α) = . α−1 Proof. Since f(x) = n (x−αi), we know that i=1 n Q n−1 f(1) αn−1 f′(α) = (α−αi) = αn−1 (1−αi) = αn−1 = (n+1). 1−αn 1−αn i=2 i=1 Y Y 3 Then αn αn+1 n+1 ′ αf (α) = (n+1) = (n+1) = . 1−αn α−αn+1 α−1 (cid:3) Lemma 3. αn−i −1 β = , i = 0,1,...,n−1. i α−1 Proof. Since xf−(xα) = ni=2(x − αi) = in=−01βixi, we get βn−1 = 1 and βn−i = (−1)i−1σi−1(α2,α3,...,αn) for i = 2,...,n, where σk stands for the Q P k-th elementary symmetric polynomial. As f(x) = n (x−αi) = n xi, we know that i=1 i=0 1 = Q(−1)iσ (α,α2,α3P,...,αn) i = (−1)i[σi(α2,α3,...,αn)+ασi−1(α2,α3,...,αn)] = (−1)iσi(α2,α3,...,αn)−α(−1)i−1σi−1(α2,α3,...,αn) = βn−i−1 −αβn−i. Thus βn−i−1 = 1+αβn−i = 1+α(1+αβn−i+1) = 1+α+α2βn−i+1 = ··· = 1+α+α2 +···+αiβn−1 = 1+α+···+αi αi+1 −1 = α−1 for i = 1,...,n−1. So β = αn−i−1 for i = 0,1,...,n−1. (cid:3) i α−1 From Lemma 2 and Lemma 3, we can clearly get that the dual of N is 1 1 M = αf′(α){β0,β1,...,βn−1} = n+1{αn −1,αn−1−1,...,α−1}. 4 Theorem 4. The complexity of the dual of a type I optimal normal basis is 3n−3 when q is even, and 3n−2 when q is odd. Proof. We use the notations above. We only need to compute the complexity of (n + 1)M = {αn − 1,αn−1 − 1,...,α − 1} as it is equivalent to M. Since (αn −1)(αi −1) = αn+i −αn −αi +1 = (αi−1 −1)−(αn −1)−(αi −1) −(αn −1)−(α−1) if i = 1 = (αi−1 −1)−(αn −1)−(αi −1) if 2 ≤ i ≤ n−1  (αn−1 −1)−2(αn −1) if i = n,  we can get the complexity of (n+1)M is 2+3(n−2)+1 = 3n−3 when q  is even, and is 2+3(n−2)+2 = 3n−2 when q is odd. (cid:3) Remark 2. Note that the main observation we make in deriving the dual of a type I optimal normal basis generated by α is the weak equivalence betweenitandthepolynomialbasisgeneratedbyα,soitisanaturalquestion that when will a normal basis generator β generate a normal basis weakly equivalent to the polynomial basis generated by it in a general finite field F . We remark that this happens if and only if β is a type I optimal normal qn basis generator. The reason is simple: if there exists some γ ∈ F such that qn N = {β,βq,...,βqn−1} = γ{1,β,...,βn−1}, we know that β = γβk for some k ∈ Z, 0 ≤ k ≤ n−1. Thus γ = β1−k. If k 6= 0, γβk−1 = β0 = 1, which is impossible since it is an element of N. So γ = β. A normal basis of the form {β,β2,...,βn} must be a type I optimal normal basis, as its complexity is no more than 2n−1. References [1] R. Lidl, H. Niederreiter, Finite Fields, second ed., Encyclopedia Math. Appl., vol. 20, Cambridge University Press, Cambridge, 1997. [2] A.J. Menezes, I.F. Blake, X.H. Gao, R.C. Mullin, S.A. Vanstone, T. Yaghoobian, Applications of Finite Fields, Kluwer Academic, Boston, MA, 1993. [3] Z.-X. Wan, K. Zhou, On the complexity of the dual basis of a type I optimal normal basis, Finite Fields Appl. 13(2007) 411-417. 5