Switchings of semifield multiplications
aa r X i v : . [ m a t h . C O ] J un SWITCHINGS OF SEMIFIELD MULTIPLICATIONS
XIANG-DONG HOU, FERRUH ¨OZBUDAK AND YUE ZHOU
Abstract.
Let B ( X, Y ) be a polynomial over F q n which defines an F q -bilinearform on the vector space F q n , and let ξ be a nonzero element in F q n . In thispaper, we consider for which B ( X, Y ), the binary operation xy + B ( x, y ) ξ defines a (pre)semifield multiplication on F q n . We prove that this questionis equivalent to finding q -linearized polynomials L ( X ) ∈ F q n [ X ] such thatTr q n /q ( L ( x ) /x ) = 0 for all x ∈ F ∗ q n . For n ≤
4, we present several families of L ( X ) and we investigate the derived (pre)semifields. When q equals a prime p , we show that if n > ( p − p − p + 4), L ( X ) must be a X for some a ∈ F p n satisfying Tr q n /q ( a ) = 0. Finally, we include a natural connectionwith certain cyclic codes over finite fields, and we apply the Hasse-Weil-Serrebound for algebraic curves to prove several necessary conditions for such kindof L ( X ). Introduction A semifield S is an algebraic structure satisfying all the axioms of a skewfieldexcept (possibly) the associativity. In other words, it satisfies the following axioms:(S1) ( S , +) is a group, with identity element 0;(S2) ( S \ { } , ∗ ) is a quasigroup;(S3) 0 ∗ a = a ∗ a ;(S4) The left and right distributive laws hold, namely for any a, b, c ∈ S ,( a + b ) ∗ c = a ∗ c + b ∗ c,a ∗ ( b + c ) = a ∗ b + a ∗ c ;(S5) There is an element e ∈ S such that e ∗ x = x ∗ e = x for all x ∈ S .A finite field is a trivial example of a semifield. Furthermore, if S does not necessar-ily have a multiplicative identity, then it is called a presemifield . For a presemifield S , ( S , +) is necessarily abelian [17]. A semifield is not necessarily commutative Date : October 22, 2018.2000
Mathematics Subject Classification.
Key words and phrases. cyclic code, finite field, linearized polynomial, semifield, the Hasse-Weil-Serre bound.Xiang-dong Hou is with the Department of Mathematics and Statistics, University of SouthFlorida, Tampa, FL 33620, USA; e-mail: [email protected]; research partially supported by NSAgrant H98230-12-1-0245.Ferruh ¨Ozbudak is with the Department of Mathematics and the Institute of Applied Mathe-matics, Middle East Technical University, Dumlupınar Bulvarı No. 1, 06800, Ankara, Turkey; e-mail: [email protected]; research partially supported by TUB˙ITAK under Grant no. TBAG-112T011.Yue Zhou is with the College of Science, National University of Defense Technology, YanwachiStreet No. 137, 410073, Changsha, China; e-mail: [email protected]. This work is par-tially supported by the National Natural Science Foundation of China (No. 61272484) and theNational Basic Research Program of China (No. 2013CB338002). or associative. However, by Wedderburn’s Theorem [27], in the finite case, asso-ciativity implies commutativity. Therefore, a non-associative finite commutativesemifield is the closest algebraic structure to a finite field. We refer to [18] for arecent and comprehensive survey.The first family of non-trivial semifields was constructed by Dickson [7] more thana century ago. In [17], Knuth showed that the additive group of a finite semifield S is an elementary abelian group, and the additive order of the nonzero elementsin S is called the characteristic of S . Hence, any finite semifield can be representedby ( F q , + , ∗ ), where q is a power of a prime p . Here ( F q , +) is the additive group ofthe finite field F q and x ∗ y can be written as x ∗ y = P i,j a ij x p i y p j , which forms amapping from F q × F q to F q .Geometrically speaking, there is a well-known correspondence, via coordinati-sation, between (pre)semifields and projective planes of Lenz-Barlotti type V.1,see [5, 13]. In [1], Albert showed that two (pre)semifields coordinatise isomorphicplanes if and only if they are isotopic. Definition 1.1.
Let S = ( F np , + , ∗ ) and S = ( F np , + , ⋆ ) be two presemifields. Ifthere exist three bijective linear mappings L, M, N : F np → F np such that M ( x ) ⋆ N ( y ) = L ( x ∗ y ) for any x, y ∈ F np , then S and S are called isotopic , and the triple ( M, N, L ) iscalled an isotopism between S and S . Let P = ( F p n , + , ∗ ) be a presemifield. We can obtain a semifield from it viaisotopisms in several ways, such as the well known Kaplansky’s trick (see [18, page2]). The following method was recently given by Bierbrauer [2]. Define a newmultiplication ⋆ by the rule(1.1) x ⋆ y := B − ( B ( x ) ∗ y ) , where B ( x ) := 1 ∗ x and B ( x ) ∗ ∗ x . We have x ⋆ B − ( B ( x ) ∗
1) = B − (1 ∗ x ) = x and 1 ⋆ x = B − ( B (1) ∗ x ) = B − (1 ∗ x ) = x , thus ( F p n , + , ⋆ ) is asemifield with identity 1. In particular, when P is commutative, B is the identitymapping.Let S = ( F p n , + , ∗ ) be a semifield. The subsets N l ( S ) = { a ∈ S : ( a ∗ x ) ∗ y = a ∗ ( x ∗ y ) for all x, y ∈ S } ,N m ( S ) = { a ∈ S : ( x ∗ a ) ∗ y = x ∗ ( a ∗ y ) for all x, y ∈ S } ,N r ( S ) = { a ∈ S : ( x ∗ y ) ∗ a = x ∗ ( y ∗ a ) for all x, y ∈ S } , are called the left, middle and right nucleus of S , respectively. It is easy to check thatthese sets are finite fields. The subset N ( S ) = N l ( S ) ∩ N m ( S ) ∩ N r ( S ) is called the nucleus of S . It is easy to see, if S is commutative, then N l ( S ) = N r ( S ) and N l ( S ) ⊆ N m ( S ), therefore N l ( S ) = N r ( S ) = N ( S ). In [13], a geometric interpretation ofthese nuclei is discussed. The subset { a ∈ S : a ∗ x = x ∗ a for all x ∈ S } is calledthe commutative center of S and its intersection with N ( S ) is called the center of S . Let G be a group and N a subgroup. A subset D of G is called a relativedifference set with parameters ( | G | / | N | , | N | , | D | , λ ), if the list of differences of D covers every element in G \ N exactly λ times, and no element in N \ { } . We call N the forbidden subgroup . WITCHINGS OF SEMIFIELD MULTIPLICATIONS 3
Jungnickel [15] showed that every semifield S of order q leads to a ( q, q, q, D in a group G which is not necessarily abelian. Assumethat S is commutative. If q = p n and p is odd, then G is isomorphic to theelementary abelian group C np ; if q = 2 n , then G ∼ = C n . ( C m is the cyclic group oforder m .)Let p be an odd prime. A function f : F p n → F p n is called planar , if the mapping x f ( x + a ) − f ( x )is a permutation of F p n for every a ∈ F ∗ p n . Planar functions were first defined byDembowski and Ostrom in [6]. It is not difficult to verify that planar functionsover F p n are equivalent to ( p n , p n , p n , C np . Planarfunctions over F n , introduced recently in [25, 29], has a slightly different definition:A function f : F n → F n is called planar , if the mapping x f ( x + a ) + f ( x ) + ax is a permutation of F n for every a ∈ F ∗ n . They are equivalent to (2 n , n , n , C n ; see [29, Theorem 2.1].Let f be a planar function over F q n , where q is a power of prime. A switching of f is a planar function of the form f + gξ where g is a mapping from F q n to F q and ξ ∈ F ∗ q n . Switchings of planar functions over F p n , where p is an odd prime,were investigated by Pott and the third author in [24]. In [29], it is proved thatswitchings of the planar function f ( x ) = 0 defined over F n can be written as affinepolynomials P a i x i + b , which are equivalent to f ( x ) itself.In the present paper, we will investigate the switchings of (pre)semifield multi-plications. To be precise, we will consider when the binary operation x ∗ y = x ⋆ y + B ( x, y ) ξ on F q n defines a (pre)semifield multiplication, where ⋆ is a given (pre)semifieldmultiplication, ξ ∈ F ∗ q n and B ( x, y ) is an F q -bilinear form from F q n × F q n to F q .(One may identify F q n with F nq , although it is not necessary.) We call x ∗ y a switching neighbour of x ⋆ y . In particular, we will concentrate on the case in which ⋆ is the multiplication of a finite field.In Section 2, we show that finding B such that x ∗ y := xy + B ( x, y ) ξ de-fines a (pre)semifield multiplication is equivalent to finding q -linearized polynomials L ( X ) ∈ F q n [ X ] such that Tr q n /q ( L ( x ) /x ) = 0 for all x ∈ F ∗ q n . For n ≤
4, we give inSection 3 several q -linearized polynomials L ( X ) ∈ F q n [ X ] satisfying this conditionand we discuss the presemifields of the corresponding switchings. In Section 4, weprove that when q = p is a prime and n > ( p − p − p + 4) /
2, the only L ( X )satisfying the above condition are those of the form βX where Tr p n /p ( β ) = 0. InSection 5, we explore a connection of the q -linearized polynomials L ( X ) satisfyingthe above condition with certain cyclic codes over F q . Finally, in Section 6 wederive several necessary conditions for the existence of the q -linearized polynomials L ( X ) from the Hasse-Weil-Serre bound for algebraic curves over finite fields.2. Preliminary discussion
Let Tr q n /q be the trace function from F q n to F q . We define B ( x, y ) := Tr q n /q ( n − X i =0 b i xy q i ) , x, y ∈ F q n , XIANG-DONG HOU, FERRUH ¨OZBUDAK AND YUE ZHOU where b i ∈ F q n . It is easy to see that B ( x, y ) defines an F q -bilinear form from F q n × F q n to F q , and every such bilinear form can be written in this way.In the next theorem, we consider the switchings of a finite field multiplication. Theorem 2.1.
Let x ∗ y := xy + B ( x, y ) ξ , where B ( x, y ) := Tr q n /q ( P n − i =0 b i xy q i ) , b i ∈ F q n , and ξ ∈ F ∗ q n . Then ∗ defines a presemifield multiplication on F q n if andonly if for any a ∈ F ∗ q n , Tr q n /q ( M ( a ) /a ) = − , where M ( X ) := ξ P n − i =0 b i X q i ∈ F p n [ X ] .Proof. ( ⇒ ) Let x ∗ y be a presemifield multiplication. Assume to the contrary thatthere is a ∈ F ∗ q n such that Tr q n /q ( M ( a ) /a ) = − . We consider the equation x ∗ a = 0. It has a solution x if and only if there exists u ∈ F q such that xa = ξu and(2.1) B ( x, a ) = − u. (2.2)Plugging (2.1) into (2.2), we have B ( ξu/a, a ) = − u , which means that u Tr q n /q ξ n − X i =0 b i a q i − ! = − u, i.e. u Tr q n /q ( M ( a ) /a ) = − u, which holds for any u ∈ F q according to our assumption. Therefore, x ∗ a = 0has a nonzero solution. It contradicts our assumption that ∗ defines a presemifieldmultiplication.( ⇐ ) It is easy to see that the left and right distributivity of the multiplication ∗ hold. We only need to show that for any a = 0, x ∗ a = 0 if and only if x = 0. Thisis achieved by reversing the first part of the proof. (cid:3) Let x ∗ y be the multiplication defined in Theorem 2.1. Then it is straightforwardto verify that the presemifield ( F q n , + , ∗ ) is isotopic to ( F q n , + , ⋆ ), where x ⋆ y := xy + B ′ ( x, y )and B ′ ( x, y ) = Tr q n /q ( ξ P n − i =0 b i xy q i ). Therefore, we can restrict ourselves to theswitchings of finite field multiplications with ξ = 1.For the switchings x ⋆ y + B ( x, y ) ξ of a (pre)semifield multiplication ⋆ , it is difficulty to obtain explicit conditions on B ( x, y ). The reason is that generally we can not explicitly write down the solutionof x ⋆ a = ξu as we did for (2.1).Let α be an element in F q n such that Tr q n /q ( α ) = 1. To find M ( X ) satisfyingthe condition in Theorem 2.1, we only need to consider the q -linearized polynomial L ( X ) := M ( X ) + αX ∈ F q n [ X ] such that(2.3) Tr q n /q ( L ( x ) /x ) = 0 for all x ∈ F ∗ q n . Obviously, when L ( X ) = βX , where Tr q n /q ( β ) = 0, we have Tr q n /q ( L ( x ) /x ) = 0for every nonzero x . The question is whether there are other L ’s. We will giveseveral results concerning this question throughout Sections 3 – 6. WITCHINGS OF SEMIFIELD MULTIPLICATIONS 5
The proof of next proposition is also straightforward.
Proposition 2.2.
Let L ( X ) = P n − i =0 a i X q i ∈ F q n [ X ] . If Tr q n /q ( L ( x ) /x ) = 0 forall x ∈ F ∗ q n , then the mapping x L ( x ) is a permutation of F q n . We include several lemmas which will be used later to investigate the commuta-tivity of presemifield multiplications.
Lemma 2.3.
Let x ∗ y := xy + B ( x, y ) , where B ( x, y ) := Tr q n /q ( P n − i =0 b i xy q i ) , b i ∈ F q n . Then ∗ is commutative if and only if b i ∈ F q gcd( i,n ) for every i = 0 , , . . . , n − .Proof. Clearly, x ∗ y = y ∗ x if and only if B ( x, y ) = B ( y, x ), i.e.Tr q n /q n − X i =0 b i xy q i ! = Tr q n /q n − X i =0 b i yx q i ! , which means that Tr q n /q x n − X i =0 ( b i − b q i i ) y q i ! = 0for every x, y ∈ F q n . Therefore ∗ is commutative if and only if b i = b q i i for every i . (cid:3) It is possible that a non-commutative presemifield P is isotopic to a commu-tative presemifield. We can use the next criterion given by Bierbrauer [2], as ageneralization of Ganley’s criterion [8], to test whether this happens. Lemma 2.4.
A presemifield ( P , + , ∗ ) is isotopic to a commutative semifield if andonly if there is some nonzero v such that A ( v ∗ x ) ∗ y = A ( v ∗ y ) ∗ x , where A : F q n → F q n is defined by A ( x ) ∗ x . Given an arbitrary presemifield multiplication, it is not easy to get the explicitexpression for A ( x ). However, we can do it for the switchings of multiplications offinite fields. Lemma 2.5.
Let x ∗ y := xy + B ( x, y ) be a switching of F q n , where B ( x, y ) :=Tr q n /q ( P n − i =0 b i xy q i ) , b i ∈ F q n . Let A : F q n → F q n be such that A ( x ) ∗ x forevery x ∈ F q n . Then (2.4) A ( x ) = x + Tr q n /q (cid:18) − tx q n /q ( t ) (cid:19) , where t = P n − i =0 b i .Proof. First, we have u ∗ u + B ( u, u + Tr q n /q n − X i =0 b i u ! = u + Tr q n /q ( tu ) . XIANG-DONG HOU, FERRUH ¨OZBUDAK AND YUE ZHOU
It is worth noting that 1 ∗ q n /q ( t ) = 0. Let s := − t/ (1 + Tr q n /q ( t )).Replacing u by the expression in (2.4), we have A ( x ) ∗ x + Tr q n /q ( sx ) + Tr q n /q (cid:2) tx + t Tr q n /q ( sx ) (cid:3) = x + Tr q n /q (cid:2) s (1 + Tr q n /q ( t )) x + tx (cid:3) = x. (cid:3) Switchings of F q n for small n In this section, we investigate the switchings of finite fields ( F q n , + , · ) where n ≤ Lemma 3.1.
Let L ( X ) = a X q + a X ∈ F q [ X ] . Then the polynomial f ( X ) = Tr q /q ( L ( X ) /X ) has no root in F ∗ q if and only if the equation x q − = y has no solution x ∈ F ∗ q forevery y ∈ F q satisfying (3.1) a y + Tr q /q ( a ) y + a q = 0 . Proof.
Let y := x q − , where x ∈ F ∗ q . ThenTr q /q ( L ( x ) /x ) = Tr q /q ( a x q − + a )= Tr q /q ( a y + a )= a q y q + a y + Tr q /q ( a )= y q ( a y + Tr q /q ( a ) y + a q )since y q +1 = 1. Therefore, f has a nonzero root if and only if there exists a ( q − F ∗ q satisfying (3.1). (cid:3) Theorem 3.2.
Let L ( X ) = a X q + a X ∈ F q [ X ] . Then (3.2) f ( X ) = Tr q /q ( L ( X ) /X ) has no root in F ∗ q if and only if g ( X ) = X + Tr q /q ( a ) X + a q +11 ∈ F q [ X ] has twodistinct roots in F q .Proof. If a = 0, then f ( X ) = Tr q /q ( a ) and g ( X ) = X + Tr q /q ( a ) X . It is clearthat f has no nonzero roots if and only if g has two distinct roots.In the rest of the proof, we assume that a = 0.( ⇐ ) Let a y ∈ F q ( y ∈ F q ) be a root of g . By Lemma 3.1, it suffices to showthat y q +1 = 1. Case 1.
Assume that q is even. Since g has two distinct roots, we haveTr q /q ( a ) = 0. Since( a y ) q +1 = ( a y ) = Tr q /q ( a ) a y + a q +11 , we have y q +1 = 1 + Tr q /q ( a ) ya q = 1 . Case 2.
Assume that q is odd. We have y = a ( − Tr q /q ( a ) + d ), where d ∈ F ∗ q and d = Tr q /q ( a ) − a q +11 . Suppose to the contrary that y q +1 = 1. It followsthat ( − Tr q /q ( a ) + d ) q +1 = 4 a q +11 , WITCHINGS OF SEMIFIELD MULTIPLICATIONS 7 which means Tr q /q ( a ) + d − d Tr q /q ( a ) = 4 a q +11 . Hence 2 d − d Tr q /q ( a ) = 0 . Therefore d = Tr q /q ( a ). But then d = Tr q /q ( a ) = Tr q /q ( a ) − a q +11 , whichis a contradiction.( ⇒ ) We first show that g is reducible in F q [ x ]. Otherwise, let a y ∈ F q \ F q bea root of g . Then ( a y ) q +1 = a q +11 , thus y q +1 = 1. By Lemma 3.1, f has nonzeroroots.It remains to show that Tr q /q ( a ) − a q +11 = 0. Assume to the contrary thatTr q /q ( a ) − a q +11 = 0. Case 1.
Assume that q is even. It follows that Tr q /q ( a ) = 0. Write a = x ,where x ∈ F q , and let y = x q − . Then a y is a root of g , which leads to acontradiction. Case 2.
Assume that q is odd. Then a y = − Tr q /q ( a ) / g , and y q +1 = Tr q /q ( a ) a q +11 = 1 , which is impossible by Lemma 3.1. (cid:3) Remark.
When n = 2, if there is some L ( X ) such that (3.2) has no root in F ∗ q ,then we can define a presemifield multiplication ∗ over F q via Theorem 2.1. Let S = ( F q , + , ⋆ ) be a semifield which is isotopic to ( F q , + , ∗ ). We may assumethat ⋆ is defined by (1.1) and hence S has identity 1. There are a ij ∈ F q suchthat x ∗ y = P i,j a ij x q i y q j for all x, y ∈ F q . Thus there are b ij ∈ F q such that x ⋆ y = P i,j b ij x q i y q j for all x, y ∈ F q . It follows that the center of S contains F q .(For x ∈ F q and y ∈ F q , we have x ⋆ y = x (1 ⋆ y ) = xy and y ⋆ x = x ( y ⋆
1) = xy .This implies that F q is contained in both the commutative center and the nucleusof S .) Due to the classification of two-dimensional finite semifields by Dickson [7], S is isotopic to a finite field. Theorem 3.3.
Let q be a power of odd prime and let L ( X ) = a X q + a X ∈ F q [ X ] with a = 0 . Then Tr q /q ( L ( X ) /X ) has no root in F ∗ q if and only if a q +11 is asquare in F ∗ q and Tr q /q ( a ) = 0 .Proof. Let b = Tr q /q ( a ). Let x ∈ F ∗ q and set y := x q − and z := a y + a q /y .Then Tr q /q ( L ( x ) /x ) = Tr q /q ( a x q − + a )= a y + a q y q + a q /y + a q /y q + Tr q /q ( a )= z + z q + b. = (cid:18) z + b (cid:19) q + (cid:18) z + b (cid:19) . (3.3)Thus Tr q /q ( L ( x ) /x ) = 0 if and only if ( z + b ) q − = − z = t − b for some t ∈ T := { t ∈ F q : t q = − t } ⊂ F q . Since z = a y + a q /y , we see that z = t − b if XIANG-DONG HOU, FERRUH ¨OZBUDAK AND YUE ZHOU and only if(3.4) a y + (cid:18) b − t (cid:19) y + a q = 0 . By the proof of Theorem 3.2, we see that { x ∈ F ∗ q : y = x q − satisfies (3.4) } 6 = ∅ if and only if g ( X ) := X + (cid:18) b − t (cid:19) X + a q +11 has two distinct roots in F q . Therefore, to sum up, Tr q /q ( L ( x ) /x ) has no rootin F ∗ q if and only if g ( X ) has two distinct roots in F q for every t ∈ T . We nowproceed to prove the “if” and the “only if” portions of the theorem separately.( ⇐ ) Assume b = 0 and a q +11 is a square in F ∗ q . Then a q +11 = t for all t ∈ T .Hence ∆ := (cid:18) b − t (cid:19) − a q +11 = t − a q +11 ∈ F ∗ q . It follows that g has two distinct roots in F q .( ⇒ ) Assume that Tr q /q ( L ( X ) /X ) has no root in F ∗ q . We want to show R1. b = 0, and R2. a q +11 is a square in F ∗ q . Equivalently, a q +11 is in F q and there is no t ∈ T such that t = 4 a q +11 .Now we assume that ∆ = (cid:0) b − t (cid:1) − a q +11 = 0 always has a square root in F q ,for every t ∈ T . Choose an element ξ of F q \ F q , such that ξ q − = −
1. Then everyelement of F q can be written as z + wξ , where z , w ∈ F q , and T = { xξ : x ∈ F q } .We write a q +11 = A + A ξ . As ∆ is always a square in F ∗ q , the equation(3.5) ( z + wξ ) = ( xξ − b/ − ( A + A ξ )in ( z, w ) has solutions for every x ∈ F q . Expanding (3.5), we have z + w α = x α + b / − A , (3.6) 2 wz = − xb − A , (3.7)where α = ξ ∈ F q .If we can show that b = 0 and A = 0, then the proof is complete ( R2 can beeasily derived from the condition that ∆ = 0). Suppose to the contrary that atleast one of b and A is not 0. Then there exists at most one x = x ∈ F q such that w = 0 by (3.7). Now assume that w = 0. From (3.7) we have z = − xb + A w . Plugging it into (3.6), we get( xb + A ) w + w α = x α + b − A , i.e., α ( w ) − ( x α + b − A ) w + ( xb + A ) . WITCHINGS OF SEMIFIELD MULTIPLICATIONS 9
For every given x ∈ F q \{ x } , this equation always has a solution w in F q . It followsthat f ( x ) = ( x α + b − A ) − α ( xb + A ) is always a square in F q . Let ψ be the multiplicative character of F q of order 2, andfor convenience we set ψ (0) = 0. Then we have(3.8) X c ∈ F q ψ ( f ( c )) ≥ q − . On the other hand, by Theorem 5.41 in [19] (it is routine to verify all the conditionsfor f ( x ), because ( b, A ) = (0 ,
0) and ( A , A ) = (0 , X c ∈ F q ψ ( f ( c )) ≤ √ q. Therefore q − ≤ √ q , which means that q = 3 , , , , , , ,
19. We can useMAGMA [3] to show that f ( x ) is not always a square for x ∈ F q \{ x } when q ≤ b = A = 0, which completes the proof. (cid:3) Theorem 3.4.
Let q be a power of an odd prime. Let a ∈ F ∗ q such that a q +11 isa square in F ∗ q and let ˜ a be an element in F q such that Tr q /q (˜ a ) = − . Define x ∗ y = xy + Tr q /q ( a xy q + ˜ a xy ) . According to Theorem 2.1 and Theorem 3.3, ( F q , + , ∗ ) forms a presemifield. Fur-thermore, it is isotopic to a commutative semifield.Proof. According to Lemma 2.4, we only have to show that there exists some v such that A ( v ∗ x ) ∗ y = A ( v ∗ y ) ∗ x for every x , y ∈ F q , where A is given by (2.4).Using the same notation as in Lemma 2.5, we set t = a + ˜ a and s = − t/ (1 +Tr q /q ( t )). Now, A ( v ∗ x ) = A ( vx + Tr q /q ( a vx q + ˜ a vx ))= vx + Tr q /q ( a vx q + ˜ a vx ) + Tr q /q (cid:2) s ( vx + Tr q /q ( a vx q + ˜ a vx )) (cid:3) = vx + (1 + Tr q /q ( s ))Tr q /q ( a vx q + ˜ a vx ) + Tr q /q ( svx )= vx + Tr q /q ( a vx q + ˜ a vx )1 + Tr q /q ( a + ˜ a ) − Tr q /q (( a + ˜ a ) vx )1 + Tr q /q ( a + ˜ a )= vx + Tr q /q ( a vx q − a vx )1 + Tr q /q ( a + ˜ a ) . For convenience, let r ( x ) denote A ( v ∗ x ) − vx . Then A ( v ∗ x ) ∗ y = vxy + r ( x ) y + Tr q /q ( a vxy q + ˜ a vxy ) + r ( x )Tr q /q ( a y q + ˜ a y )= vxy + Tr q /q ( a vx q − a vx )1 + Tr q /q ( a + ˜ a ) ( y + Tr q /q ( a y q + ˜ a y ))+ Tr q /q ( a vxy q + ˜ a vxy ) . It is not difficult to see that if v is an element in F q such that a v ∈ F q , then A ( v ∗ x ) ∗ y = A ( v ∗ y ) ∗ x , from which it follows that ( F q , + , ∗ ) is isotopic to acommutative semifield. (cid:3) Theorem 3.5.
Let q be a power of an odd prime. Let a ∈ F ∗ q such that a q +11 isa square in F ∗ q and let ˜ a be an element in F q such that Tr q /q (˜ a ) = − . Let x ∗ y be defined as in Theorem 3.4, i.e., x ∗ y = xy + Tr q /q ( a xy q + ˜ a xy ) . Then the presemifield ( F q , + , ∗ ) is isotopic to Dickson’s semifield.Proof. We have already shown in Theorem 3.4 that ( F q , + , ∗ ) is isotopic to a com-mutative semifield, which is denoted by S . Next we are going to prove that itsmiddle nucleus N m ( S ) is of size q and its left nucleus N l ( S ) is of size q . Further-more, as S is commutative, we have N r ( S ) = N l ( S ) . Due to the classification ofsemifields planes of order q with kernel F q and center F q by Cardinali, Polverinoand Trombetti in [4], ( F q , + , ∗ ) is isotopic to Dickson’s semifield.To determine the middle and left nuclei of S , we need to introduce anotherpresemifield multiplication x ◦ y , which corresponds to the dual spread of the spreaddefined by x ∗ y . (For more details on the dual spread, see [16].) Actually, x ◦ y isdefined as(3.9) x ◦ y := xy + ( a y q + ˜ a y )Tr q /q ( x ) . It is straightforward to verify that Tr q /q ( x ( z ◦ y ) − z ( x ∗ y )) = 0. Let S ′ denotea semifield which is isotopic to the presemifield defined by x ◦ y . According tothe interchanging of nuclei of semifields in the so called Knuth orbit ([16] and [18,Section 1.4]), we have N l ( S ′ ) ∼ = N m ( S ) and N m ( S ′ ) ∼ = N l ( S ).To determine N l ( S ′ ) and N m ( S ′ ), we use the connection between certain homol-ogy groups as described in [13, Theorem 8.2] and [14, Result 12.4]. To be precise,we want to find every q -linearized polynomial A ( X ) over F q such that for every y ∈ F q , there is a y ′ ∈ F q satisfying A ( x ) ◦ y = x ◦ y ′ for every x ∈ F q . The set M ( S ′ ) of all such A ( X ) is equivalent to the middle nucleus N m ( S ′ ).First, it is routine to verify that A ( X ) = uX with u ∈ F q is in M ( S ′ ). Next weshow that there are no other A ( X ) in M ( S ′ ).Assume that(3.10) A ( x ) y + Tr q /q ( A ( x ))( a y q + ˜ a y ) = xy ′ + Tr q /q ( x )( a y ′ q + ˜ a y ′ )holds for every x ∈ F q .Let x ∈ F ∗ q be such that Tr q /q ( x ) = Tr q /q ( A ( x )) = 0. Then A ( x ) y = x y ′ . It means that y ′ = uy holds for each y ∈ F q , where u = A ( x ) /x . Plugging itinto (3.10), we have A ( x ) y + Tr q /q ( A ( x ))( a y q + ˜ a y ) = uxy + Tr q /q ( x )( a ( uy ) q + ˜ a uy ) . From this equation we can deduce that A ( x ) − ux + (Tr q /q ( A ( x )) − Tr q /q ( x ) u )˜ a = 0 , (3.11) (Tr q /q ( A ( x )) − Tr q /q ( x ) u q ) a = 0 . (3.12) WITCHINGS OF SEMIFIELD MULTIPLICATIONS 11
Since a = 0, from (3.12) we see that(3.13) Tr q /q ( A ( x )) = u q Tr q /q ( x )for every x ∈ F q . From (3.13) it follows that u ∈ F q . Therefore, by (3.11), we have A ( x ) = ux where u ∈ F q . Hence | N l ( S ) | = | N m ( S ′ ) | = q .Next we determine every q -linearized polynomial A ( X ) over F q such that forevery y ∈ F q , there is a y ′ ∈ F q satisfying A ( x ◦ y ) = x ◦ y ′ for every x ∈ F q . Theset of all such A ( X ) is equivalent to the left nucleus N l ( S ′ ).Assume that(3.14) A ( xy + Tr q /q ( x )( a y q + ˜ a y )) = xy ′ + Tr q /q ( x )( a y ′ q + ˜ a y ′ ) . It is readily verified that when A ( X ) = cX for some c ∈ F q , (3.14) holds forall x and y in F q with y ′ = cy . Hence F q is a subfield contained in N l ( S ′ ).On the other hand, N l ( S ′ ) has to be a proper subfield of F q , for otherwise S ′ would be a finite field, which would lead to a contradiction. Therefore, we have | N m ( S ) | = | N l ( S ′ ) | = q , which completes the proof. (cid:3) Theorem 3.6.
Let q be a power of prime and let u, v be elements in F ∗ q such that N q /q ( − v/u ) = 1 . For every β ∈ B , where B := { x ∈ F q : Tr q /q ( u q v q x ) = u q + q +1 + v q + q +1 } , the equation (3.15) ux q − + vx q − + β = 0 has no solution in F ∗ q . Let L ( X ) := u q v q ( ua q − X q + va q − X q + θX ) , where θ ∈ B and a ∈ F ∗ q . Then the polynomial Tr q /q ( L ( X ) /X ) has no root in F ∗ q .Proof. When β = 0, (3.15) becomes x q − ( ux q ( q − + v ) = 0. If there exists x ∈ F ∗ q such that ux q ( q − + v = 0, then N q /q ( − v/u ) = N q /q ( x q ( q − ) = 1, which leads toa contradiction.Now suppose β = 0. Assume to the contrary that (3.15) has a solution x ∈ F ∗ q .Let y := x q − . Then we have uy q +1 + vy + β = 0. It follows that(3.16) y q = − vy − βuy , and y q = v q ( vy + β ) − β q uy − u q ( vy + β ) . Hence y q y q y = v q ( vy + β ) − β q uyu q +1 , which is equal to 1 since y = x q − . Therefore,(3.17) ( v q +1 − β q u ) y + v q β = u q +1 . Suppose that uβ q = v q +1 . Then u q v q β = v q +1 v q , and Tr q /q ( u q v q β ) = 3 v q + q +1 .On the other hand, we also have u q +1 = v q β from (3.17). It follows that Tr q /q ( u q v q β ) =3 u q + q +1 . All together with β ∈ B , we have that u q + q +1 + v q + q +1 = 3 v q + q +1 = 3 u q + q +1 , which can not holds for 3 ∤ q . Moreover, if 3 | q , then u q + q +1 = − v q + q +1 whichcontradicts the assumption that N q /q ( − v/u ) = 1. Hence uβ q = v q +1 .Since uβ q = v q +1 , from (3.17) we obtain(3.18) y = u q +1 − v q βv q +1 − β q u Plugging (3.18) into (3.16), we have u q + q − v q β q v q + q − β q u q = vu q − β q +1 v q β − u q +1 . Hence u q + q v q β − u q +2 q +1 + u q +1 v q β q − v q + q β q +1 = v q + q +1 u q − β q vu q − v q + q β q +1 + β q + q +1 u q . Dividing it by u q , we have β q + q +1 − ( u q vβ q + uv q β q + u q v q β ) + u q + q +1 + v q + q +1 = 0 . It follows from Tr q /q ( u q v q β ) = u q + q +1 + v q + q +1 that β q + q +1 = 0 . Hence β = 0, which is a contradiction. Therefore, (3.15) has no solution in F ∗ q .Furthermore, if Tr q /q ( L ( X ) /X ) has a root x ∈ F ∗ q , then u q v q ( u ( ax ) q − + v ( ax ) q − + θ ) = γ for some γ ∈ F q satisfying Tr q /q ( γ ) = 0. We write γ as γ = u q v q τ for some τ ∈ F q . Then θ − τ ∈ B and u ( ax ) q − + v ( ax ) q − + θ − τ = 0 , which contradicts the fact that (3.15) has no solution in F ∗ q . (cid:3) For given u and v , it is not difficult to see that for different a , we obtain isotopicsemifields via Theorem 3.6: Let the multiplication corresponding to a = 1 be xy + B ( x, y ). Then for other a ∈ F ∗ q , the semifield multiplication is axy + B ( x,ay ) a .Furthermore, when u , v ∈ F q and a = 1, it follows from Lemma 2.3 that thepresemifield P derived from L ( x ) in Theorem 3.6 is commutative. It is worth notingthat, up to isotopism, we can obtain non-commutative semifields via Theorem 3.6.For instance, let q = 4 and let ξ be a primitive element of F q which is a rootof X + X + X + X + 1. Setting u = ξ , v = ξ and β = ξ , we can usecomputer to show that the presemifield P derived from Theorem 3.6 is not isotopicto a commutative one.According to the classification of semifields of order q with center containing F q in [21], the presemifield obtained via Theorem 3.6 is either finite field or generalizedtwisted field.Besides all the L ’s described in this section, we did not find any other examples.Thus we propose the following question: Question 3.7.
For n > , is there a q -linearized polynomial L ( X ) = P n − i =0 a i X q i ∈ F q n [ X ] with ( a , . . . , a n − ) = (0 , . . . , satisfying (2.3) ? WITCHINGS OF SEMIFIELD MULTIPLICATIONS 13 Switchings of F p n for large n The main result of this section is a negative answer to Question 3.7 when q = p (prime) and n is large. Theorem 4.1.
Let q = p , where p is a prime, and assume n ≥ ( p − p − p + 4) .If L ( X ) = P n − i =0 a i X p i ∈ F p n [ X ] satisfies (2.3) , i.e., Tr p n /p (cid:0) L ( x ) /x (cid:1) = 0 for all x ∈ F ∗ p n , then a = · · · = a n − = 0 . In 1971, Payne [22] considered a similar problem which calls for the determinationof all 2-linearized polynomials L = P n − i =0 a i X i ∈ F n [ X ] such that both L ( X ) and L ( X ) /X are permutation polynomials of F n . Such linearized polynomials give riseto translation ovoids in the projective plane PG(2 , F n ) [23]. Payne later solvedthe problem by showing that such linearized polynomials can have only one term[23]. For a different proof of Payne’s theorem, see [11, § q -ary versionof Payne’s theorem, see [12].4.1. Preliminaries.
Let L ( X ) = P n − i =0 a i X q i ∈ F q n [ X ]. For x ∈ F ∗ q n , we haveTr q n /q (cid:16) L ( x ) x (cid:17) = Tr q n /q (cid:16) n − X i =0 a i x q i − (cid:17) = X ≤ i,j ≤ n − a q j i x q j ( q i − . Therefore (2.3) is equivalent to (cid:20) X ≤ i,j ≤ n − a q j i X q j ( q i − (cid:21) q − ≡ Tr q n /q ( a ) q − + h − Tr q n /q ( a ) q − i X q n − (mod X q n − X ) . (4.1)Let Ω = { , , . . . , q n − } and Ω = { , , . . . , q n − q − } . For α, β ∈ Ω , define α ⊕ β ∈ Ω such that α ⊕ β ≡ α + β (mod q n − q − ) and α ⊕ β = ( α = β = 0 , q n − q − if α + β ≡ q n − q − ) and ( α, β ) = (0 , . For d , . . . , d n − ∈ Z , we write( d , . . . , d n − ) q = n − X i =0 d i q i . When q is clear from the context, we write ( d , . . . , d n − ) q = ( d , . . . , d n − ). For j, i ∈ Z , i ≥
0, let s ( j, i ) = ( · · · j · · · | {z } i · · · n − q , where the positions of the digits are labeled modulo n and the string of 1’s maywrap around. For example, with n = 4, s (1 ,
3) = (0 1 1 1) , s (3 ,
2) = (1 0 0 1) . Note that s ( j, i ) ≡ q j q i − q − q n − . For each α ∈ Ω , let C ( α ) denote the coefficient of X α ( q − in the left side of (4.1)after reduction modulo X q n − X . Then we have(4.2) C ( α ) = X ≤ j ,i ,...,j q − ,i q − ≤ n − s ( j ,i ) ⊕···⊕ s (( j q − ,i q − )= α a q j i · · · a q jq − i q − . Let S = { s ( j, i ) : 0 ≤ j ≤ n − , ≤ i ≤ n − } . If C ( α ) = 0, we can derive from (4.2) useful information about a i ’s if we know thepossible ways to express α as an ⊕ sum of q − S ∪ { } .Let α = ( d , . . . , d n − ) q ∈ Ω, where 0 ≤ d i ≤ q −
1. If d i > d i − ( d i < d i − ),where the subscripts are taken modulo n , we say that i is an ascending ( descending )position of α with multiplicity | d i − d i − | . The multiset of ascending (descending)positions of α is denoted by Asc( α ) (Des( α )). The multiset cardinality | Asc( α ) | (= | Des( α ) | ) is denoted by asc( α ). For example, if α = (2 0 1 1 3 0), thenAsc( α ) = { , , , , } , Des( α ) = { , , , , } , asc( α ) = 5 . Assume that α ∈ Ω has asc( α ) = q −
1. Then α cannot be a sum of less than q − S . Moreover, if α = s ( j , i ) + · · · + s ( j q − , i q − ) , where 0 ≤ j , . . . , j q − ≤ n − ≤ i , . . . , i q − ≤ n −
1, we must have { j , . . . , j q − } = Asc( α ) and { j + i , . . . , j q − + i q − } = Des( α ), where j k + i k istaken modulo n .4.2. Proof of Theorem 4.1.Lemma 4.2.
Let q = p , where p is a prime, and assume L = P n − i =0 a i X p i ∈ F p n [ X ] satisfies (2.3) . Then for all ≤ i < · · · < i p − and ≤ t p − ≤ · · · ≤ t with i p − + t ≤ n − , we have (4.3) X τ a i p − + τ ( p − a p ip − − ip − i p − + τ ( p − · · · a p ip − − i i + τ (1) = 0 , where ( τ (1) , . . . , τ ( p − runs through all permutations of ( t , . . . , t p − , .Proof. Let α = ( i p − − i p − z }| { · · · · · · i − i z }| { p − · · · p − i z }| { p − · · · p − p − · · · p − | {z } t p − p − · · · p − | {z } t p − − t p − · · · · · · | {z } t − t · · · | {z } n − i p − − t ≥ ) ∈ Ω . For 1 ≤ k ≤ p −
2, we have α + ( k · · · k )= ( i p − z }| { k + 1 · · · k + 1 · · · p − · · · p − · · · · · · t z }| { · · · d n − i p − − t z }| { e k · · · k | {z } ≥ ) , where e = k + 1 or k , depending on whether it receives a carry from the precedingdigit. If e = k + 1, then asc( α + ( k · · · k )) ≥ p − − k + k + 1 = p . If e = k , then WITCHINGS OF SEMIFIELD MULTIPLICATIONS 15 t > d ≥ k + 1, which also implies that asc( α + ( k · · · k )) ≥ p . Therefore α + ( k · · · k ) is not a sum of ≤ p − S , i.e.,not a sum of p − S ∪ { } .On the other hand, we have asc( α ) = p − α ) = { , i p − − i p − , . . . , i p − − i } , Des( α ) = { i p − , i p − + t p − , . . . , i p − + t } . Therefore, the only possible ways to express α as a sum of p − S ∪ { } are α = s (0 , i p − + τ ( p − s ( i p − − i p − , i p − + τ ( p − · · · + s ( i p − − i , i + τ (1)) , where ( τ (1) , . . . , τ ( p − t , . . . , t p − , ≤ k ≤ p − α + ( k · · · k ) is not a sum of p − S ∪ { } , we have proved that α = α ⊕ · · · ⊕ α p − , α i ∈ S ∪ { } , if and only if { α , . . . , α p − } = (cid:8) s (0 , i p − + τ ( p − , s ( i p − − i p − , i p − + τ ( p − , . . . , s ( i p − − i , i + τ (1)) (cid:9) , where ( τ (1) , . . . , τ ( p − t , . . . , t p − , C ( α ) (by (4.1))= ( p − X τ a i p − + τ ( p − a p ip − − ip − i p − + τ ( p − · · · a p ip − − i i + τ (1) (by (4.2)) , (4.4)which gives (4.3). (cid:3) Proof of Theorem 4.1. ◦ We first show that for all 1 ≤ k ≤ p − k − X j =0 j ≤ i k < · · · < i p − ≤ n − k − , we have a i k · · · a i p − = 0 . We use induction on k . When k = 1, the conclusion follows from Lemma 4.2with t p − = · · · = t = 0. Assume 2 ≤ k ≤ p −
1. In Lemma 4.2, let t = k − t = k − , . . . , t k − = 1, t k = · · · = t p − = 0, i k − = i k − i k − = i k − , . . . , i = i k − ( k − i p − + t = i p − + k − ≤ n −
2. We have(4.5) X τ a ∗ i p − + τ ( p − · · · a ∗ i + τ (1) = 0 , where ( τ (1) , . . . , τ ( p − k − , k − , . . . , , , . . . , ∗ ’s are suitable powers of p . (In general, we use a ∗ to denote a positive inte-ger exponent whose exact value is not important.) Multiplying (4.5) by a i k · · · a i p − gives(4.6) a ∗ i k · · · a ∗ i p − + X τ ( τ (1) ,...,τ ( k − =( k − ,..., a i k · · · a i p − a ∗ i p − + τ ( p − · · · a ∗ i + τ (1) = 0 . When ( τ (1) , . . . , τ ( k − = ( k − , . . . , i + τ (1) , . . . , i p − + τ ( p − i ′ k − , is less than i k . Also note that i ′ k − ≥ i = i k − ( k − ≥ · · · +( k − a i ′ k − a i k · · · a i p − = 0. Thus the P in (4.6) equals 0, which gives a i k · · · a i p − = 0.2 ◦ Let k = p − ◦ . We have a i = 0 for all 1 + 12 ( p − p − ≤ i ≤ n − p. ◦ We claim that a i = 0 for all 1 ≤ i ≤
12 ( p − p − . Assume to the contrary that this is not true. Let 1 ≤ l ≤ ( p − p −
1) be thelargest integer such that a l = 0. Let α = (1 · · · | {z } l · · · | {z } p +1 · · · | {z } l · · · | {z } p +1 · · · · · · | {z } l · · · | {z } p +1 | {z } p − · · · ∈ Ω . (Here we used the assumption that n ≥ ( p − (cid:2) ( p − p −
1) + p + 1 (cid:3) .) For0 ≤ k ≤ p −
2, we have asc( α + ( k · · · k )) = p − (cid:0) α + ( k · · · k ) (cid:1) = (cid:8) , l + p + 1 , l + p + 1) , . . . , ( p − l + p + 1) (cid:9) , Des (cid:0) α + ( k · · · k ) (cid:1) = (cid:8) l, l + p + 1 + l, l + p + 1) + l, . . . , ( p − l + p + 1) + l (cid:9) . If α + ( k · · · k ) is expressed as a sum of p − S , the expression must be of the form(4.7) α + ( k · · · k ) = s (0 , i ) + s ( l + p + 1 , i ) + · · · + s (( p − l + p + 1) , i p − ) , where i , . . . , i p − ∈ { , . . . , n − } , and in modulus n (cid:8) i , l + p + 1 + i , . . . , ( p − l + p + 1) + i p − (cid:9) = (cid:8) l, l + p + 1 + l, l + p + 1) + l, . . . , ( p − l + p + 1) + l (cid:9) . (4.8)We further require a i · · · a i p − = 0, which implies that i , . . . , i p − ∈ { , . . . , l } ∪{ n − p + 1 , . . . , n − } . It follows from (4.8) that i = · · · = i p − = l . Thus we have0 = C ( α ) (by (4.1))= ( p − a p l a p l + p +1 l · · · a p ( p − l + p +1) l (by (4.2) and (4.7)) , (4.9)which is a contradiction.4 ◦ Finally, we show that a i = 0 for all n − p + 1 ≤ i ≤ n − . Assume to the contrary that this is not true. Let n − l ∈ { n − p + 1 , . . . , n − } bethe smallest integer such that a n − l = 0. Let α = (1 · · · · · · | {z } l · · · · · · | {z } l · · · | {z } l | {z } p − ) ∈ Ω . WITCHINGS OF SEMIFIELD MULTIPLICATIONS 17
For 0 ≤ k ≤ p −
2, we have asc( α + ( k · · · k )) = p − (cid:0) α + ( k · · · k ) (cid:1) = (cid:8) n − , n − − ( l + 1) , . . . , n − − ( p − l + 1) (cid:9) , Des (cid:0) α + ( k · · · k ) (cid:1) = (cid:8) n − − l, n − − l − ( l + 1) , . . . , n − − l − ( p − l + 1) (cid:9) . If α + ( k · · · k ) is expressed as a sum of p − S , the expression must be of the form α + ( k · · · k )= s ( n − , i ) + s ( n − − ( l + 1) , i ) + · · · + s ( n − − ( p − l + 1) , i p − ) , (4.10)where i , . . . , i p − ∈ { , . . . , n − } , and in modulus n (cid:8) n − i , n − − ( l + 1) + i , . . . , n − − ( p − l + 1) + i p − (cid:9) = (cid:8) n − − l, n − − l − ( l + 1) , . . . , n − − l − ( p − l + 1) (cid:9) . (4.11)We further require a i · · · a i p − = 0, which implies that i , . . . , i p − ∈ { n − l, . . . , n − } . Under this restriction, it is easy to see that (4.11) forces i = · · · i p − = n − l .Thus we have0 = C ( α ) (by (4.1))= ( p − a p n − n − l a p n − − ( l +1) n − l · · · a p n − − ( p − l +1) n − l (by (4.2) and (4.10)) , (4.12)which is a contradiction. (cid:3) It appears that the assumption that n ≥ ( p − p − p + 4) in Theorem 4.1 maybe weakened. On the other hand, when q is not a prime, the proofs of Lemma 4.2and Theorem 4.1 fail for the following reason: In (4.4), (4.9) and (4.12), ( p − q − F q . When q = p e , (4.1) becomes (cid:20) e − Y k =0 X ≤ i,j ≤ n − a p k q j X p k q j ( q i − (cid:21) p − ≡ Tr q n /q ( a ) q − + h − Tr q n /q ( a ) q − i X q n − (mod X q n − X ) . The question is how to decipher this equation.5.
A connection to some cyclic codes for general F q In this section we prove certain necessary conditions for a q -linearized polynomi-als L ( X ) ∈ F q n [ X ] to satisfy Tr q n /q ( L ( x ) /x ) = 0 for all x ∈ F ∗ q n , where q is a primepower. In particular, we give a natural connection to some cyclic codes. There isalso a connection of such cyclic codes to some algebraic curves. In the next section,we will use this connection to algebraic curves to get some necessary conditions forsuch q -linearized polynomials L ( X ) ∈ F q n [ X ].If L ( X ) = a X ∈ F q n [ X ], then Tr q n /q ( L ( x ) /x ) = 0 for all x ∈ F ∗ q n if and only ifTr q n /q ( a ) = 0. Hence we assume that L ( X ) = a X + a X q + · · · + a n − X q n − ∈ F q n [ X ] with ( a , a , . . . , a n − ) = (0 , , . . . , N = q n −
1. A code of length N over F q is just a nonempty subsetof F Nq . It is called a linear code if it is a vector space over F q . The set C ⊥ of all N -tuples in F Nq orthogonal to all codewords of a linear code C with respect to the usual inner product on F Nq is called the dual code of C . The Hamming weight ofan arbitrary N -tuple u = ( u , u , . . . , u N − ) ∈ F Nq is || u || = |{ ≤ i ≤ N − u i = 0 }| . A cyclic code of length N over F q is an ideal C of the quotient ring R = F q [ X ] / h X N − i . Here a codeword ( c , c , . . . , c N − ) ∈ F Nq of C corresponds to an element c + c X + · · · + c N − X N − + h X N − i ∈ C . All ideals of R are principal. Themonic polynomial g ( X ) of the least degree such that C = h g ( X ) i / h X N − i is calledthe generator polynomial of C . The dual C ⊥ is cyclic with generator polynomial X deg h h ( X − ) /h (0), where h ( X ) = ( X N − /g ( X ).If θ ∈ F q n is a root of g ( X ), then so is θ q . A set B ⊂ F q n is called a basic zeroset of C if both of the following conditions are satisfied: • { θ q i : θ ∈ B, ≤ i ≤ n − } is the set of the roots of g ( X ). • If θ , θ ∈ B with θ q i = θ for some integer i , then θ = θ .The following proposition gives a natural connection to some cyclic codes. Somearguments in its proof will also be used in the next section. Proposition 5.1.
Let γ be a primitive element of F ∗ q n . Let C be the cyclic code oflength N = q n − over F q whose dual code C ⊥ has { , γ q − , γ q − , . . . , γ q n − − } as a basic zero set. We have the following: There exists a q -linearized polyno-mial L ( X ) = a X + a X q + · · · + a n − X q n − ∈ F q n [ X ] with ( a , a , . . . , a n − ) =(0 , , . . . , such that Tr q n /q ( L ( x ) /x ) = 0 for all x ∈ F ∗ q n if and only if thecyclic code C has a codeword ( c , c , . . . , c N − ) of Hamming weight N such that ( c , c , . . . , c N − ) = u (1 , , . . . , for any u ∈ F ∗ q . Moreover the dimension of C over F q is n − n + 1 .Proof. We first show that { , γ q − , γ q − , . . . , γ q n − − } is a basic zero set. Thismeans that the exponents 0 , q − , q − , . . . , q n − − q -cyclotomiccosets modulo q n −
1. For 0 ≤ d < q n −
1, let ψ ( d ) be the base q digits of d , i.e., ψ ( d ) = ( d , d , . . . , d n − ), where 0 ≤ d i ≤ q − d = P n − i =0 d i q i . Let 0 , q − , q − , . . . , q n − − q -cyclotomic cosets of0 , q − , q − , . . . , q n − − q n −
1. Their images under ψ are ψ (0) = { (0 , , . . . , } ,ψ ( q −
1) = { ( q − , , , . . . , , (0 , q − , , . . . , , . . . , (0 , , , . . . , , q − } ,ψ ( q −
1) = { ( q − , q − , , . . . , , (0 , q − , q − , . . . , , . . . , ( q − , , . . . , , q − } , ... ψ ( q n − −
1) = { ( q − , . . . , q − , , (0 , q − , . . . , q − , . . . , ( q − , . . . , q − . } . Note that the elements in each row are obtained via cyclic shifts of the first elementof the row. This proves that 0 , q − , q − , . . . , q n − − q -cyclotomiccosets modulo q n −
1. Moreover the cardinality of the union of their q -cyclotomiccosets modulo q n − n − n = n − n + 1 . WITCHINGS OF SEMIFIELD MULTIPLICATIONS 19
Therefore the dimensions of C is n − n + 1. Finally using Delsarte’s Theorem [26,Theorem 9.1.2] we obtain that the codewords of C in F Nq are C = (cid:26)(cid:16) Tr q n /q ( a + a x q − + · · · + a n − x q n − − ) (cid:17) x ∈ F ∗ qn : a , a , . . . , a n − ∈ F q n (cid:27) . Note that Tr q n /q ( L ( x ) /x ) = u for all x ∈ F ∗ q n if and only if Tr q n /q ( L ( X ) /X ) ≡ u (mod X q n − X ), from which it follows that ( a , a , . . . , a n − ) = (0 , , . . . , (cid:3) Some conditions via the Hasse-Weil-Serre bound for general F q In this section we obtain some necessary conditions for the q -linearized polyno-mials L ( X ) ∈ F q n [ X ] such that Tr q n /q ( L ( x ) /x ) = 0 for all x ∈ F ∗ q n .The Hasse-Weil-Serre bound for algebraic curves over finite fields implies upperand lower bounds on the Hamming weights of codewords of cyclic codes (see [10,28]). Using this method we obtain Theorem 6.1.First we introduce further notations. Let Res : Z → { , , . . . , q n − } be themap such that Res ( j ) ≡ j (mod q n − q = p m with m ≥
1, where p isthe characteristic of F q . Let Lead : { , , . . . , p mn − } → { , , . . . , p mn − } bethe map sending j to the smallest integer k in { , , . . . , p mn − } such that k ≡ jp u (mod p mn −
1) for some integer u ≥
0. In other words, Lead( j ) is the smallestnonnegative integer in the p -cyclotomic coset of j modulo p mn −
1. It is importantto note that if 0 < j < p mn −
1, then Lead( j ) is a nonnegative integer which iscoprime to p . Theorem 6.1.
Let L ( X ) = a X + a X q + · · · + a n − X q n − ∈ F q n [ X ] be a q -linearized polynomial with ( a , . . . , a n − ) = (0 , . . . , . For each ≤ j ≤ q n − with gcd( j, q n −
1) = 1 , let ℓ ( j ) = max { Lead(Res ( j ( q i − ≤ i ≤ n − and a i = 0 } . Moreover, let (6.1) ℓ = min j ℓ ( j ) , where the minimum is over all integers ≤ j ≤ q n − with gcd( j, q n −
1) = 1 .Then we have the following: • Case Tr q n /q ( a ) = 0 : If (6.2) q n + 1 − ( q − ℓ − ⌊ q n/ ⌋ > , then it is impossible that Tr q n /q ( L ( x ) /x ) = 0 for all x ∈ F ∗ q n . • Case Tr q n /q ( a ) = 0 : If (6.3) q n + 1 − ( q − ℓ − ⌊ q n/ ⌋ > q + 1 , then it is impossible that Tr q n /q ( L ( x ) /x ) = 0 for all x ∈ F ∗ q n .Proof. If γ is a primite element of F ∗ q n , then γ j is also a primitive element of F ∗ q n for all 1 ≤ j ≤ q n − j, q n −
1) = 1. Note thatTr q n /q ( L ( x ) /x ) = Tr q n /q ( a + a x q − + · · · + a n − x q n − − ) = 0 for all x ∈ F ∗ q n , if and only ifTr q n /q ( L ( x j ) /x j ) = Tr q n /q ( a + a x j ( q − + · · · + a n − x j ( q n − − ) = 0 for all x ∈ F ∗ q n . Moreover, x j ( q i − = x Res ( j ( q i − for x ∈ F ∗ q n , 1 ≤ i ≤ n − ≤ j ≤ q n − ℓ is defined in (6.1). We choose and fix an integer 1 ≤ j ≤ q n − j, q n −
1) = 1 such that ℓ = ℓ ( j ).Let a t , . . . , a t s be the nonzero coefficients among a , . . . , a n − . (Note that s ≥ a , . . . , a n − ) = (0 , . . . , , q t − , . . . , q t s − p -cyclotomic cosets modulo q n − j, q n −
1) = 1, we have that 0 , j ( q t − , . . . , j ( q t s −
1) belong to different p -cyclotomic cosets modulo q n −
1. ThusRes ( j ( q t i − j i p u i , where u i ≥ p ∤ j i , 1 ≤ i ≤ s , and j , . . . , j s are distinct.We may assume 0 < j < j < · · · < j s = ℓ . We have a + a X Res ( j ( q − + · · · + a n − X Res ( j ( q n − − = a + b X j p u + · · · + b s X j s p us , where b i = a t i , 1 ≤ i ≤ s .Let χ be the Artin-Shreier type algebraic curve over F q n given by χ : Y q − Y = a + b X j p u + · · · + b s X j s p us . Let S ⊂ F ∗ p mn be a complete set of coset representatives of F ∗ p in F ∗ p mn . For µ ∈ S ,let χ µ be the Artin-Shreier type algebraic curve over F q n given by χ µ : Y p − Y = µ ( a + b X j p u + · · · + b s X j s p us ) . Note that χ µ is a degree p covering of the projective line. Using [9, Theorem 2.1]the genus g ( χ ) of χ is computed in terms of the genera of χ µ as g ( χ ) = X µ ∈ S g ( χ µ ) . (6.4)Now we determine the genus g ( χ µ ) of χ µ . We choose and fix µ ∈ S . Let c , c , . . . , c s ∈ F ∗ p mn be such that c p u = µb , c p u = µb , . . . , c p us s = µb s . Let χ ′ µ be the Artin-Schreier type algebraic curve over F q n given by χ ′ µ : Y p − Y = µa + c X j + · · · + c s X j s . We observe that χ µ and χ ′ µ are birationally isomorphic and hence the genera g ( χ µ )and g ( χ ′ µ ) are the same. Indeed, if u ≥
1, then Y p − Y = µa + c p u X j p u + c p u X j p u + · · · + c p us s X j s p us = µa + (cid:16) c p u − X j p u − (cid:17) p + c p u X j p u + · · · + c p us s X j s p us and hence h Y − (cid:16) c p u − X j p u − (cid:17)i p − h Y − (cid:16) c p u − X j p u − (cid:17)i = µa + c p u − X j p u − + c p u X j p u + · · · + c p us s X j s p us . This gives a birational isomorphism between χ µ and the curve given by Y p − Y = µa + c p u − X j p u − + c p u X j p u + · · · + c p us s X j s p us . WITCHINGS OF SEMIFIELD MULTIPLICATIONS 21
By induction on u we obtain a birational isomorphism between χ µ and the curvegiven by Y p − Y = µa + c X j + c p u X j p u + · · · + c p us s X j s p us . Applying the same method to the monomials c p u X j p u , . . . , c p us s X j s p us we con-clude that the curves χ µ and χ ′ µ are birationally isomorphic.Recall that the integers 0 , j , . . . , j s are in distinct p -cyclotomic cosets modulo q n −
1. As c s = 0 and gcd( j s , p ) = 1 we obtain that χ ′ µ is absolutely irreducibleover F q n . Moreover s ≥ j s = ℓ . Hence by [26, Proposition 3.7.8] we have g ( χ µ ) = g ( χ ′ µ ) = ( p − ℓ − / , which is independent from the choice of µ ∈ S . Using (6.4) for the genus g ( χ ) of χ we obtain that g ( χ ) = X µ ∈ S g ( χ µ ) = | S | ( p − ℓ − / q − ℓ − / . Assume that Tr q n /q ( L ( x ) /x ) = 0 for all x ∈ F ∗ q n . The number N ( χ ) of F q n -rational points of χ is(6.5) N ( χ ) = 1 + q |{ x ∈ F q n : Tr( L ( x ) /x ) = 0 }| = (cid:26) q n /q ( a ) = 0 ,q + 1 if Tr q n /q ( a ) = 0 . The Hasse-Weil-Serre lower bound on N ( χ ) (see, for example, [26, Theorem 5.3.1])implies that(6.6) N ( χ ) ≥ q n + 1 − ( q − ℓ − ⌊ q n/ ⌋ . Combining (6.2), (6.3), (6.5) and (6.6), we complete the proof. (cid:3)
The following corollary, which is a restatement of Theorem 6.1, shows that thedistribution of the nonzero coefficients of a q -linearized polynomial L satisfyingTr q n /q ( L ( x ) /x ) = 0 for all x ∈ F ∗ q n is subject to certain restrictions. Corollary 6.2.
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