A new proof to complexity of dual basis of a type I optimal normal basis
aa r X i v : . [ c s . D M ] J a n A new proof to complexity of dual basis of atype I optimal normal basis ∗ Baofeng Wu † , Kai Zhou ‡ , Zhuojun Liu § Abstract
The complexity of dual basis of a type I optimal normal basis of F q n over F q was determined to be 3 n − n − q is even orodd, respectively, by Z.-X. Wan and K. Zhou in 2007. We give a newproof to this result by clearly deriving the dual of a type I optimalnormal basis with the aid of a lemma on the dual of a polynomialbasis. Keywords.
Normal basis; Complexity; Type I optimal normal ba-sis; Dual basis; Polynomial basis.
Let F q n be an extension of the finite field F q . A basis of the form { α, α q , . . . , α q n − } is called a normal basis, and α is called a normal basis generator in this case.Normal bases have great advantages against the common polynomial basesin arithmetic of finite fields. A simple property about a normal basis isthat it has normal dual basis. By dual basis { β ∗ , . . . , β ∗ n } of a given basis { β , . . . , β n } , we mean Tr F qn / F q ( β i β ∗ j ) = δ ij , 1 ≤ i, j ≤ n . ∗ Partially supported by National Basic Research Program of China (2011CB302400). † Key Laboratory of Mathematics Mechanization, AMSS, Chinese Academy of Sciences,Beijing 100190, China. Email: [email protected] ‡ Key Laboratory of Mathematics Mechanization, AMSS, Chinese Academy of Sciences,Beijing 100190, China. Email: [email protected] § Key Laboratory of Mathematics Mechanization, AMSS, Chinese Academy of Sciences,Beijing 100190, China. Email: [email protected] α , denote α i = α q i , i = 0 , . . . , n −
1, and N = { α , . . . , α n − } . Let α · α i = n − X j =0 t ij α j , t ij ∈ F q , ≤ i ≤ n − . (1)The number of nonzero elements in the n × n matrix T = ( t ij ) is called thecomplexity of the normal basis N . We denote it by C N . It can be shownthat C N ≥ n −
1, and N is called optimal when the lower bound is attained.A type I optimal normal basis of F q n over F q is the normal basis formedby the n nonunit ( n + 1)-th roots of unity, where n + 1 is a prime and q isprimitive modulo n + 1 [1, 3]. In [4] the dual of a type I optimal normalbasis was studied, and its complexity was determined to be 3 n − n − q is even or odd, respectively.Two bases { β , . . . , β n } and { β ′ , . . . , β ′ n } of F q n over F q are called equiv-alent, or weakly equivalent, if β i = cβ ′ i , i = 1 , . . . , n for some c ∈ F q , or F q n ,respectively. It is easy to see that equivalent normal bases share the samecomplexity.In this paper, we clearly derive the dual M of a type I optimal normalbasis N through a new approach and thus rediscover the complexity of it.The main auxiliary lemma we use will be presented in section 2, and in section3, we will propose our method to get M and its complexity, which is totallydifferent from that in [4] and seems more simple and easier to understand. The following lemma, proposed as an exercise in [2, Chapter 2], is the mainresult 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 2.1.
Let F q n = F q ( α ) be an extension of the finite field F q , and f ( x ) ∈ F q [ x ] be the minimal polynomial of α . Assume f ( x ) x − α = β + β x + · · · + β n − x n − ∈ F q n [ x ] . Then the dual basis of the polynomial basis { , α, . . . , α n − } is f ′ ( α ) { β , β , . . . , β n − } := { β f ′ ( α ) , β f ′ ( α ) , . . . , β n − f ′ ( α ) } , where f ′ is the formal derivative of f . roof. Let σ i be the i -th Frobenius automorphism of F q n / F q and α i = σ i ( α ) = α q i , 0 ≤ i ≤ n −
1. Then f ( x ) = Q n − i =0 ( x − α i ). For l = 0 , , . . . , n − F l ( x ) = n − X i =0 f ( x ) x − α i α li f ′ ( α i ) − x l . It is easy to check F l ( α i ) = 0 for i = 0 , , . . . , n −
1. Thus F l ( x ) ≡ F l ( x ) ≤ n −
1. On the other hand, x l = x l + F l ( x ) = n − X i =0 σ i (cid:16) f ( x ) x − α α l f ′ ( α ) (cid:17) = n − X i =0 σ i (cid:16) n − X j =0 β j x j α l f ′ ( α ) (cid:17) = n − X j =0 n − X i =0 σ i (cid:16) β j α l f ′ ( α ) (cid:17)! x j = n − X j =0 Tr F qn / F q (cid:16) β j f ′ ( α ) α l (cid:17) x j . So we get Tr F qn / F q (cid:0) β j f ′ ( α ) α l (cid:1) = δ jl for any 0 ≤ j, l ≤ n −
1, which is what wewant to prove. (cid:3)
Remark 2.2.
In fact, the lemma and the proof hold for any separable alge-braic extension K ( α ) of a field K with generator α . In the remaining part of the paper, we suppose n + 1 is a prime and q isprimitive modulo n + 1. Let α be a nonunit ( n + 1)-th root of unity. Thenthe set of all nonunit ( n + 1)-th roots of unity is N = { α, α , . . . , α n } . Lemma 3.1. [3] The minimal polynomial of α is f ( x ) = P ni =0 x i . Further-more, N = { α, α q , . . . , α q n − } and it forms an optimal normal basis of F q n over F q . N in Lemma 3.1 is often called a type I optimal normalbasis. Since N = { α, α , . . . , α n } = α { , α, . . . , α n − } , we find it weaklyequivalent to the polynomial basis { , α, . . . , α n − } . Thus if we can computethe dual of the polynomial basis, say f ′ ( α ) { β , β , . . . , β n − } by Lemma 2.1,then we can get the dual of N of the form M = αf ′ ( α ) { β , β , . . . , β n − } . Theremaining task is just to compute αf ′ ( α ) and β , . . . , β n − . Lemma 3.2. αf ′ ( α ) = n + 1 α − . Proof.
Since f ( x ) = Q ni =1 ( x − α i ), we know that f ′ ( α ) = n Y i =2 ( α − α i ) = α n − n − Y i =1 (1 − α i ) = α n − f (1)1 − α n = α n − − α n ( n + 1) . Then αf ′ ( α ) = α n − α n ( n + 1) = α n +1 α − α n +1 ( n + 1) = n + 1 α − . (cid:3) Lemma 3.3. β i = α n − i − α − , i = 0 , , . . . , n − . Proof.
Since f ( x ) x − α = Q ni =2 ( x − α i ) = P n − i =0 β i x i , we get β n − = 1 and β n − i =( − i − s i − ( α , α , . . . , α n ) for i = 2 , . . . , n , where s k stands for the k -thelementary symmetric polynomial.As f ( x ) = Q ni =1 ( x − α i ) = P ni =0 x i , we know that1 = ( − i s i ( α, α , α , . . . , α n )= ( − i [ s i ( α , α , . . . , α n ) + αs i − ( α , α , . . . , α n )]= ( − i s i ( α , α , . . . , α n ) − α ( − i − s i − ( α , α , . . . , α n )= β n − i − − αβ n − i . Thus β n − i − = 1 + αβ n − i = 1 + α (1 + αβ n − i +1 )4 1 + α + α β n − i +1 = · · · = 1 + α + α + · · · + α i β n − = 1 + α + · · · + α i = α i +1 − α − i = 1 , . . . , n −
1. So β i = α n − i − α − for i = 0 , , . . . , n − (cid:3) From Lemma 3.2 and Lemma 3.3, we can clearly get that the dual of N is M = 1 αf ′ ( α ) { β , β , . . . , β n − } = 1 n + 1 { α n − , α n − − , . . . , α − } . The normal basis M can be generated by ˜ α = α − n +1 , the minimal polynomialof which is ˜ f ( x ) = n Y i =1 ( x − α i − n + 1 )= 1( n + 1) n n Y i =1 (( n + 1) x + 1 − α i )= 1( n + 1) n f (( n + 1) x + 1)= 1( n + 1) n n X i =0 (( n + 1) x + 1) i = 1( n + 1) n n X i =0 i X j =0 (cid:18) ij (cid:19) ( n + 1) j x j = n X j =0 (cid:18) n X i = j (cid:0) ij (cid:1) ( n + 1) n − j (cid:19) x j . Theorem 3.4.
The complexity of the dual of a type I optimal normal basisis n − when q is even, and n − when q is odd.Proof. We use the notations above. We only need to compute the com-plexity of ( n + 1) M = { α n − , α n − − , . . . , α − } as it is equivalent to M . 5ince( α − α i −
1) = α i +1 − α − α i + 1= ( α i +1 − − ( α i − − ( α − ( α − − α −
1) if i = 1( α i +1 − − ( α i − − ( α −
1) if 2 ≤ i ≤ n − − ( α n − − ( α −
1) if i = n, we can get that the complexity of ( n + 1) M is 2 + 3( n −
2) + 1 = 3 n − q is even, and is 2 + 3( n −
2) + 2 = 3 n − q is odd. (cid:3) The matrix T determined by (1) for a normal basis is often called its mul-tiplication table. For the type I optimal normal basis N = { α, α q , . . . , α q n − } ,its multiplication table is of the form T α = P − − · · · − P − , where P is an n × n permutation matrix with q q ... q n − ≡ P n mod ( n + 1) . (More properties on this matrix P can be found in [1].) From the proof ofTheorem 3.4, we actually obtain that the multiplication table of the dualbasis M = { ˜ α, ˜ α q , . . . , ˜ α q n − } is of the form T ˜ α = 1 n + 1 P − − − − − − − P − . emark 3.5. Note that the main observation we make in deriving the dual ofa type I optimal normal basis generated by α is the weak equivalence between itand the polynomial basis generated by α , so it is a natural question that whenwill a normal basis generator β generate a normal basis weakly equivalentto the polynomial basis generated by it in a general finite field F q n . Weremark that this happens if and only if β is a type I optimal normal basisgenerator. The reason is simple: if there exists some γ ∈ F q n such that N = { β, β q , . . . , β q n − } = γ { , β, . . . , β n − } , we know that β = γβ k for some k ∈ Z , ≤ k ≤ n − . Thus γ = β − k . If k = 0 , γβ k − = β = 1 , whichis impossible since it is an element of N . So γ = β . A normal basis of theform { β, β , . . . , β n } must be a type I optimal normal basis, as its complexityis no more than n − . References [1] Q.Y. Liao and L. You, On multiplication tables of optimal normal basesover finite fields,
Acta Mathematica Sinica, Chinese Series , 2005, 48(5):947–954.[2] R. Lidl and H. Niederreiter,
Finite Fields, second ed. (EncyclopediaMath. Appl., vol. 20), Cambridge University Press, Cambridge, 1997.[3] A.J. Menezes, I.F. Blake, X.H. Gao, R.C. Mullin, S.A. Vanstone and T.Yaghoobian,
Applications of Finite Fields , Kluwer Academic, Boston,MA, 1993.[4] Z.-X. Wan and K. Zhou, On the complexity of the dual basis of a typeI optimal normal basis,