Infinite Families of New Semifields

Infinite Families of New Semifields Gary L. Ebert∗, Giuseppe Marino†, Olga Polverino† and Rocco Trombetti† February 8, 2008

Abstract We construct six new infinite families of finite semifields, all of which are two-dimensional over their left nuclei. We give constructions for both even and odd characteristics when the left nucleus has odd dimension over the center. The characteristic is odd in the one family in which the left nucleus has even dimension over the center. Spread sets of linear maps are used in all the constructions.

1

Introduction

A finite semifield S is a finite algebraic structure satisfying all the axioms for a skewfield except (possibly) associativity. The subsets Nl = {a ∈ S | (ab)c = a(bc), ∀b, c ∈ S}, Nm = {b ∈ S | (ab)c = a(bc), ∀a, c ∈ S}, Nr = {c ∈ S | (ab)c = a(bc), ∀a, b ∈ S} and K = {a ∈ Nl ∩ Nm ∩ Nr | ab = ba, ∀b ∈ S} are fields and are known, respectively, as the left nucleus, middle nucleus, right nucleus and center of the semifield. A finite semifield is a vector space over its nuclei and its center (for more information on semifields see [6] and [3]). Throughout this paper we will assume the center of our semifield is the finite field Fq of order q, where q is some power of the prime p. If S satisfies all the axioms for a semifield, except that it does not have an identity element under multiplication, then S is called a pre-semifield. Two pre-semifields, say S = (S, +, ◦) and S0 = (S0 , +, ·), are said to be isotopic if there exist three Fp -linear maps g1 , g2 , g3 from S to S0 such that g1 (x) · g2 (y) = g3 (x ◦ y) ∗ This

author acknowledges the support of NSA grant H98230-06-1-0071 work was supported by the Research Project of MIUR (Italian Office for University and Research) “Strutture geometriche, combinatoria e loro applicazioni”and by the Research group GNSAGA of INDAM † This

1

for all x, y ∈ S. From any pre-semifield, one can naturally construct a semifield which is isotopic to it (see [12]). The survey [11] provides a concise description of the known finite semifields. There are surprisingly few infinite families. In this paper we use linearized polynomials and the notion of linear sets of maps (defined later) to construct six new infinite families of semifields, all of which are 2-dimensional over their left nuclei. In particular, we prove the following result. Theorem 1.1. The families FI , FII , FIII , FIV , FV and FV I constitute six new mutually non-isotopic families of semifields of order q 2n with left nucleus isomorphic to Fqn and center isomorphic to Fq . The semifields belonging to FI , FII and FIII are defined for all odd n > 1 and all odd prime powers q, while those belonging to FIV and FV are defined for all odd n > 1 and even q. The semifields belonging to these five families have middle and right nuclei both isomorphic to Fq , and the spread sets of linear maps defining them are given in (3) with conditions (I), (II), (III), (IV ) and (V ), respectively. The semifields belonging to FV I are defined for all odd prime powers q and even integers n > 2. They have middle and right nuclei both isomorphic to Fq2 , and the spread sets of linear maps defining them are given in (9) with condition (V I).

2

Preliminaries

A pre-semifield S, viewed as a vector space over some prime field Fp , can be used to coordinatize an affine (and hence a projective) plane of order |S| (see [6]). Albert [1] showed that the projective planes coordinatized by S and S0 are isomorphic if and only if the pre-semifields S and S0 are isotopic. Any projective plane π(S) coordinatized by a semifield (or pre-semifield) is called a semifield plane. Semifield planes are necessarily translation planes (see [6] for more information on translation planes). The kernel of a semifield plane, treated as a translation plane, is the left nucleus of the coordinatizing semifield. A semifield plane is desarguesian (classical) if and only if the coordinatizing semifield S is a field, in which case all nuclei as well as the center are equal to S. As discussed in [4], any translation plane can be obtained from a spread of an odd dimensional projective space. The translation planes are isomorphic if and only if the corresponding spreads are projectively equivalent. If the translation plane is at most 2-dimensional over its kernel, then the plane can be obtained from a spread of a finite projective 3-space; that is, a collection of lines which partition all the points of a projective 3-space. In particular, if a semifield S is 2-dimensional over its left nucleus, then the associated semifield plane π(S) can be obtained from a spread of a projective 3-space. Such spreads will be called semifield spreads. In this paper we only are concerned with finite semifields that are twodimensional over their left nuclei. We adopt the convention for all semifields under consideration that the left nucleus is a degree n extension of the center,

2

for some n > 1. Hence the left nucleus will be isomorphic to the finite field Fqn , and the associated semifield plane will arise from a spread S of PG(3, q n ). To each spread S of PG(3, q n ) there is an associated spread set of matrices; that is, a set CS of q 2n 2 × 2 matrices with entries from Fqn such that the difference of any two distinct matrices from the set is non-singular. One can always choose coordinates so that the zero matrix and the identify matrix are in the spread set. In particular, this implies that every nonzero matrix in the spread set is non-singular. Moreover, the spread S is a semifield spread if and only if the spread set CS is closed under addition. In this case CS defines a vector subspace, over some subfield of Fqn , of the vector space of all 2 × 2 matrices over Fqn . Denote by K the maximal subfield of Fqn with respect to which CS is a K-subspace of this vector space. The field K is isomorphic to the center K of the semifield S which coordinatizes the corresponding semifield plane (see [5], for instance), and hence K is isomorphic to Fq by our earlier assumption. We now consider the finite field Fq2n as a 2-dimensional vector space over its subfield Fqn , and fix a basis for Fq2n over Fqn . Then the spread set CS may be considered as a set C¯S of q 2n Fqn –linear maps of Fq2n containing the zero map and the identity map such that C¯S is closed under sums, any nonzero map in C¯S is non-singular, and Fq is the maximal subfield of Fqn with respect to which C¯S is a vector space. From now on we refer to C¯S as the spread set of linear maps for the semifield spread S, and simply denote it by S. The beauty of this approach is that we can quite easily, and naturally, describe sets S of linear maps which represent semifields (two-dimensional over their left nucleus with center Fq ), many of which form infinite families of new semifields. In this model any Fqn –linear map of Fq2n is of the form ϕη,ζ : Fq2n → Fq2n

n

via x → ηx + ζxq ,

for some η, ζ ∈ Fq2n . The use of linearized polynomials, in this case q n – polynomials, is an old idea going back at least to Ore [17], and has often been used in recent years to construct or characterize a wide variety of geometrical objects (see [2], for instance). As discussed above, a spread set S of linear maps representing one of our semifields will be a collection of q 2n such maps ϕη,ζ with the following properties: P1 S is closed under addition and Fq –scalar multiplication, with the usual point-wise operations on functions. P2 Fq is the largest subfield of Fqn with respect to which S is a vector subspace of the vector space of all Fqn –linear maps of Fq2n . P3 Every nonzero map in S is non-singular (that is, invertible). It should be noted that to show ϕη,ζ is non-singular, where we assume η and ζ are nonzero to avoid trivialities, it suffices to show that the kernel of this n linear operator is {0}. But this is equivalent to showing xq −1 6= −η ζ for all nonzero x ∈ Fq2n , which in turn is equivalent to showing N ( ηζ ) 6= 1, where N is 3

the norm from Fq2n to Fqn . This technique will be used often when our infinite families are introduced, and hence we display this fact for future reference: η ϕη,ζ is non-singular ⇔ N ( ) 6= 1, ζ

(1)

where N is the norm from Fq2n to Fqn . From the above properties for the q 2n maps in S, we know that there is a unique element ϕ ∈ S such that ϕ(1) = y for each element y ∈ Fq2n . We call this uniquely determined map ϕy , and thus there is a natural one-to-one correspondence between the linear maps in S and the elements of the field Fq2n . If we now define an algebraic structure S = (Fq2n , +, ◦), where + is the sum operation in the field Fq2n and ◦ is defined as x ◦ y = ϕy (x), it turns out (for instance, see [8]) that S is a semifield with identity 1 and left nucleus Fqn that is isotopic to the semifield which coordinatizes the plane π(S) obtained from the semifield spread S. It is important to note that the spread set S of linear maps is not necessarily closed under composition of maps (our composition will always be read from right to left, and will be denoted by juxtaposition). However, if ϕz , ϕy ∈ S such that ϕz ϕy ∈ S, then necessarily ϕz ϕy = ϕϕz (y) . Namely, let ϕz ϕy = ϕ for some ϕ ∈ S. Then ϕ(1) = (ϕz ϕy )(1) = ϕz (ϕy (1)) = ϕz (y) and thus ϕ = ϕϕz (y) by definition. With this observation, one can easily compute the right and middle nuclei of S. For instance, the middle nucleus Nm may be computed as Nm

= = = = = =

{y {y {y {y {y {y

∈ Fq2n ∈ Fq2n ∈ Fq2n ∈ Fq2n ∈ Fq2n ∈ Fq2n

| (x ◦ y) ◦ z = x ◦ (y ◦ z) ∀x, z ∈ Fq2n } | ϕz (ϕy (x)) = ϕy◦z (x) ∀x, z ∈ Fq2n } | (ϕz ϕy )(x) = ϕϕz (y) (x) ∀x, z ∈ Fq2n } | ϕz ϕy = ϕϕz (y) ∀z ∈ Fq2n } | ϕz ϕy ∈ S ∀z ∈ Fq2n } | ϕϕy ∈ S ∀ϕ ∈ S}.

Using the identification of elements described above, let Nm = {ϕy ∈ S | y ∈ Nm } and Nr = {ϕy ∈ S | y ∈ Nr }. Our computation above, and a similar one for Nr , imply the following useful facts: Theorem 2.1. Let S be a semifield of order q 2n with center Fq that is 2dimensional over its left nucleus. Let S be the associated spread set of linear maps as above. Then (a) the right nucleus of S is isomorphic to the largest subfield Nr of S over which S is a left vector space, and (b) the middle nucleus of S is isomorphic to the largest subfield Nm of S over which S is a right vector space. 4

To construct such a semifield S, we must find a set S of q 2n Fqn -linear maps n ϕη,ζ : Fq2n → Fq2n via x → ηx + ζxq such that properties P1, P2, and P3, as defined above, hold for S. Letting V = V (2, q 2n ) be the left vector space of all Fqn –linear maps on Fq2n , these properties imply that S is a vector subspace of dimension 2n over the subfield Fq . Treating V as a 4-dimensional vector space over Fqn and letting Σ = PG(V, Fqn ) ∼ = PG(3, q n ) be the associated projective space, we define the linear set L(S) of S to be the set of projective points in Σ induced by the nonzero maps in S. Thus the linear set L(S) contains at most (q 2n − 1)/(q − 1) points of Σ, and we say the linear set is scattered if its cardinality meets this upper bound. Alternatively, treating V as a 4n-dimensional vector space over Fq , we see that S (being a 2n-dimensional vector space over Fq ) induces a projective subspace Γ of dimension 2n − 1 in Σ0 = PG(V, Fq ) ∼ = PG(4n − 1, q). Thus the points of the linear set L(S) in Σ ∼ = PG(3, q n ) can be “lifted” to subspaces of Γ ∼ = PG(2n − 1, q) inside Σ0 . If P = hϕi is a point of L(S), then P can be viewed as a 1-dimensional vector space over Fqn or an n-dimensional vector space WP = {uϕ : u ∈ Fqn } over Fq . We define the weight of P to be dimFq (WP ∩ S), where we are treating S as the underlying vector space for Γ. Thus L(S) is scattered, as previously defined, if and only if all of its points have weight 1. In general, we view a point of L(S) with weight i as an (i − 1)-dimensional subspace of Γ. The notion of attaching weights to the points of linear set L = L(S) will prove to be very useful. One particularly useful property, proved in [8, Property 3.1], is the following. Let r be any line of Σ ∼ = PG(3, q n ), and let L = L(S) be the linear set in Σ for some spread set S of linear maps. Then r ⊆ L ⇔ dimFq (R ∩ S) ≥ n + 1,

(2)

where R is the 2n-dimensional vector space over the subfield Fq underlying the line r. In what follows, for the sake of simplicity, we identify the elements of V with ordered pairs of elements from Fq2n ; namely, the map ϕη,ζ is identified with the pair (η, ζ) ∈ Fq2n × Fq2n . Then from Condition (1) we know that the nonzero singular maps in V induce the hyperbolic quadric Q = {h(η, ζ)i : η q

n

+1

= ζq

n

+1

}

in Σ. From property P3, every nonzero map in S is non-singular and thus the linear set L(S) is disjoint from Q. This geometric fact will also prove to be extremely useful, especially in sorting out isotopy issues. One can easily check that n n the two reguli lying on Q consist of the line sets {` : q +1 = 1} and {`0 : q +1 = n 1}, where ` = {h(d, d)i : d ∈ F∗q2n } and `0 = {h(a, aq )i : a ∈ F∗q2n }. We will come back to this hyperbolic quadric Q and its stabilizer subgroup when we discuss isotopy issues. We close this preliminary section by mentioning that linear sets have proved to be very useful in classifying semifields (hence semifield planes) which are 2dimensional over their left nuclei when the left nucleus itself has small dimension 5

over the center Fq . Denoting the left nucleus by Fqn as above, the complete classification for n = 2 is given in [5]. For n = 3, there are precisely six geometric configurations for a linear set of rank 6 in Σ ∼ = PG(3, q 3 ), where Fq is the maximal subfield of linearity (see [16]). These six different geometrical patterns for the linear set yield six different families of potential semifield planes, one of the families being further partitioned (again, geometrically) into three subfamilies, as discussed in [8]. Examples are known to exist for all families and subfamilies, and the classification of semifield planes in some of the families is now complete (see [16], [8], [7]). Nonetheless, work continues in the n = 3 case (see [7]).

3

Constructions

Rather than classifying semifields, in this paper we concentrate on constructing new infinite families of semifields that are 2-dimensional over the left nucleus. We first consider the case when the dimension of the left nucleus over the center is odd.

Case : n odd Let u be a fixed element of Fqn such that {1, u, u2 , . . . , un−1 } is an Fq –basis of Fqn . Since n is odd, {1, u, u2 , . . . , un−1 } is also an Fq2 –basis of Fq2n (n > 1). Let a, b, c be fixed elements of Fq2n with a, c 6∈ Fqn . Define S = S(u, a, b, c) to be the set consisting of the following Fqn –linear maps of Fq2n : (n−1)/2

(n−1)/2

{x 7→ (

X i=0

αi ui +a

X

(n−3)/2

(n−3)/2

X

βi ui )x+b(

X

γj uj +c

j=0

i=0

n

δj uj )xq : αi , βi , γj , δj ∈ Fq }

j=0

(3) To simplify the notation, we define α(u) β(u) γ(u) δ(u)

= α0 + α1 u + · · · + α n−1 u

n−1 2

2

= β0 + β1 u + · · · + β n−1 u

n−1 2

,

2

= γ0 + γ1 u + · · · + γ n−3 u

n−3 2

2

= δ0 + δ1 u + · · · + δ n−3 u 2

n−3 2

,

,

.

The conditions on the parameters immediately imply that |S| = q 2n . Moreover, from its definition the set S = S(u, a, b, c) is a vector subspace over the subfield Fq , and hence Property P1 holds for S. From Condition (1) we see that Property P3 will hold if and only if ! α(u) + aβ(u) N 6= N (b) (4) γ(u) + cδ(u) 6

for every choice of αi , βi , γj , δj in Fq with γ(u) + cδ(u) 6= 0. Before addressing this concern, we first show that Property P2 always holds for the set S. Proposition 3.1. The largest subfield of Fqn with respect to which S(u, a, b, c) is a vector subspace of the vector space of all Fqn –linear maps of Fq2n is the subfield Fq . Proof. We have already observed that S = S(u, a, b, c) is a subspace over the subfield Fq . Suppose that there exist a subfield Fqt of Fqn containing Fq with respect to which S is a linear set, and let λ ∈ F∗qt . Since neither a nor c is in Fqn we see that λ[1, u, . . . , u and λ[1, u, . . . , u

n−1 2

n−3 2

]Fq = [1, u, . . . , u ]Fq = [1, u, . . . , u

n−1 2

n−3 2

]Fq , ]Fq .

Straightforward computations show that this is possible if and only if λ ∈ Fq , and the result follows. Thus S = S(u, a, b, c) is a linear set of maps that determines a semifield S = S(u, a, b, c) of order q 2n which is 2-dimensional over its left nucleus and has center Fq if and only if Equation (4) is satisfied. In the next section we will exhibit several choices for the parameters u, a, b, and c that guarantee this condition is satisfied. For now, we assume that Equation (4) holds and determine further characteristics of the resulting semifield S. Let L = L(S) be the associated linear set in Σ, namely L = L(u, a, b, c) = {h(α(u) + aβ(u), b(γ(u) + cδ(u)))i : αi , βi , γj , δj ∈ Fq }, and let r be the line of Σ defined by the subspace {(η, 0) : η ∈ Fq2n }. Then r⊥ = {h(0, ζ)i : ζ ∈ F∗q2n } is its polar line with respect to the polarity induced by the quadric Q. Proposition 3.2. The Fq –linear set L associated with S satisfies the following properties: (i) the points of L ∩ r have weight at most points of L ∩ r with weight n+1 2 ; (ii) the points of L ∩ r⊥ have weight at most points of L ∩ r⊥ with weight n−1 2 ;

n+1 2 ,

n−1 2 ,

and there are at least q + 1 and there are at least q + 1

(iii) the line r is the unique line of Σ contained in L. Proof. We use the shorthand notation previously established, and work with the underlying subspaces over the subfield Fq . In particular, we identify r and L with their associated subspaces over Fq . Thus L ∩ r = {(α(u) + aβ(u), 0) : αi , βi ∈ Fq } 7

and L ∩ r⊥ = {(0, b(γ(u) + cδ(u))) : γj , δj ∈ Fq }. So dimFq (L ∩ r) = n + 1, and this implies that r ⊂ L by Equation (2). Similarly, dimFq (L ∩ r⊥ ) = n − 1. Moreover, if P = h(1, 0)i and Pa = h(a, 0)i, then P ∩ L = {(α(u), 0) : αi ∈ Fq } and Pa ∩ L = {(aβ(u), 0) : βi ∈ Fq }. Hence the points P and Pa both have weight n+1 2 . Since dimFq (L ∩ r) = n + 1, this implies that any other point of L ∩ r has weight at most n+1 2 . Also any point Pt+a = h(t + a, 0)i, with t ∈ Fq , has weight exactly n+1 2 since Pt+a ∩ L = {(α(u)(t + a), 0) : αi ∈ Fq }. Therefore we conclude that there are at least q +1 points of weight n+1 2 , showing (i) is true. By using similar arguments, (ii) also follows. Now suppose that there exists a line ` contained in L such that ` 6= r. Since ` ⊂ L, we know that dimFq (l ∩ L) ≥ n + 1 by Equation (2) and hence ` ∩ r 6= ∅. Let π = h`, ri and T = ` ∩ r. Since T ∈ r, the weight of T is at most n+1 2 by (i), and hence dimFq (π ∩ L) ≥ dimFq hr ∩ L, ` ∩ Li ≥ n + 1 + n + 1 −

3n + 3 n+1 = . 2 2

Since L = hπ ∩ L, r⊥ ∩ LiFq and points of r⊥ ∩ L have weight at most (ii), we have 2n = dimFq hπ ∩ L, r⊥ ∩ Li ≥

n−1 2

by

3n + 3 n−1 +n−1− = 2n + 1, 2 2

a contradiction. We now discuss the nuclei of such a semifield S. Proposition 3.3. If the set S(u, a, b, c) defines a semifield S = S(u, a, b, c), then Nm (S) = Nr (S). Moreover, either Nm (S) = Nr (S) ∼ = Fq or Nm (S) = Nr (S) ∼ = Fq2 ; the latter case occurs if and only if a and c both belong to Fq2 . Proof. We work with the corresponding nuclei Nr and Nm , defined as subsets of S = S(u, a, b, c). Let ϕ ∈ Nr , so that ϕ = ϕA,B for some A, B ∈ Fq2n . By Theorem 2.1 we know that Nr is the largest subfield over which S is a left vector space. So, for any map ϕ ∈ S we have ϕϕ ∈ S. That is, n

n

A(α(u) + aβ(u)) + Bbq (γ(u) + cq δ(u)) ∈ [1, . . . , u

n−1 2

, a, . . . , au

n−1 2

]Fq

(5)

and n

Ab(γ(u)+cδ(u))+B(α(u)+aq β(u)) ∈ b[1, u, . . . , u

n−3 2

, c, cu, . . . , cu

n−3 2

]Fq , (6)

n−3 for all αi , βi ∈ Fq , i = 0, 1, . . . n−1 2 , and all γj , δj ∈ Fq , j = 0, 1, . . . , 2 .

8

If γj = δj = 0 for j = 0, 1, . . . , n−3 and βi = 0 for i = 0, 1, . . . , n−1 2 2 , from Equation (6) we get n−3 n−3 n−1 B [1, u, . . . , u 2 ]Fq ⊆ [1, u, . . . , u 2 , c, cu, . . . , cu 2 ]Fq . b

Since c ∈ Fq2n \ Fqn , we may write Equation (7) we get S[1, . . . , u and T [1, . . . , u

n−1 2

n−1 2

B b

(7)

= S + T c with S, T ∈ Fqn , and from

]Fq ⊆ [1, u, . . . , u ]Fq ⊆ [1, u, . . . , u

n−3 2

n−3 2

]Fq , ]Fq .

n−1 2

Since 1, u, . . . , u are linearly independent over Fq , the last two conditions imply that S = T = 0, and hence B = 0. n−3 Now, if βi = 0 for i = 0, . . . , n−1 2 and γj = δj = 0 for j = 0, . . . , 2 , from Equation (5) we get

A[1, u, . . . , u

n−1 2

]Fq ⊆ [1, u, . . . , u

n−1 2

, a, au, . . . , au

n−1 2

]Fq .

(8)

Since a ∈ Fq2n \ Fqn , we may write A = L + M a with L, M ∈ Fqn . From Equation (8) we have L[1, u, . . . , u and M [1, u, . . . , u

n−1 2

n−1 2

]Fq ⊆ [1, u, . . . u

n−1 2

]Fq ⊆ [1, u, . . . , u

]Fq ,

n−1 2

]Fq .

n−1 2

]Fq for i = 0, 1, . . . n−1 Using the fact that Lui and M ui belong to [1, u, . . . , u 2 , and that {1, u, . . . , un−1 } is an Fq –basis of Fqn , we obtain L, M ∈ Fq . In the same way, from Equation (6) we obtain A = L0 + M 0 c for some L0 , M 0 ∈ Fq . Now, express a2 = F + Ga for some F, G ∈ Fqn . From Equation (5) with αi = 0 for i = 0, 1, . . . n−1 2 , we obtain Aa[1, u, . . . , u

n−1 2

]Fq ⊆ [1, u, . . . , u

n−1 2

, a, au, . . . , au

n−1 2

]Fq .

Since Aa = M F + (M G + L)a, using the same arguments as above, we obtain M F ∈ Fq and M G + L ∈ Fq . Thus, either M = 0 and hence A = L ∈ Fq , or M 6= 0 and hence F, G ∈ Fq . In the latter case we have a ∈ Fq2 . A similar argument, using Equation (6), shows that either A ∈ Fq or c ∈ Fq2 . If a, c ∈ Fq2 , then Equation (5) and Equation (6) are satisfied for each element A = L + M a = L0 + M 0 c, with L, M, L0 , M 0 ∈ Fq . In this case Nr = {x 7→ ηx : η ∈ Fq2 }. Therefore, taking all cases into consideration, we see that either Nr = {x 7→ Ax : A ∈ Fq } ' Fq or Nr ' Fq2 , and this latter case occurs precisely when a, c ∈ Fq2 . In a similar way it can be proven that either Nm = {x 7→ Ax : A ∈ Fq } ' Fq or Nm = {x 7→ ηx : η ∈ Fq2 } ' Fq2 , and again the latter case occurs precisely when a, c ∈ Fq2 . This completes the proof. 9

Remark 3.4. If a, c ∈ Fq2 , the linear maps of Type (3) can be written in this way (n−1)/2 (n−3)/2 X X n i {ϕ : x 7→ ( Ai u )x + b( Bj uj )xq : Ai , Bj ∈ Fq2 }, i=0

j=0

and hence we obtain the family Su,b already studied in [9, Section 4]. In that paper such examples are called generalized cyclic semifields.

Case : n even We now assume that n, the dimension of the left nucleus Fqn over the center Fq , is even. Again let u be a fixed element of Fqn such that {1, u, u2 , . . . , un−1 } is an Fq –basis of Fqn , and let a, b, c be fixed elements of Fq2n with a, c 6∈ Fqn . Define S = S(u, a, b, c) to be the set consisting of the following Fqn –linear maps of Fq2n : (n−2)/2

{x 7→ (

X i=0

(n−2)/2

αi ui +a

X i=0

(n−2)/2

βi ui )x+b(

X

(n−2)/2

X

γj uj +c

j=0

n

δj uj )xq : αi , βi , γj , δj ∈ Fq }

j=0

(9) To simplify the notation, we again define α(u)

= α0 + α1 u + · · · + α n−2 u

β(u)

= β0 + β1 u + · · · + β n−2 u

γ(u) δ(u)

n−2 2

2

n−2 2

,

2

= γ0 + γ1 u + · · · + γ n−2 u

n−2 2

2

= δ0 + δ1 u + · · · + δ n−2 u 2

n−2 2

,

,

.

In a similar fashion to the case when n is odd, we see that S = S(u, a, b, c) is a linear set of maps that determines a semifield S = S(u, a, b, c) of order q 2n which is 2-dimensional over its left nucleus and has center Fq if and only if Equation (4) holds. Just as before, assuming such a semifield S is determined, let L = L(u, a, b, c) be the associated Fq –linear set in Σ ∼ = P G(3, q n ), let r be ⊥ the line of Σ defined by {(η, 0) : η ∈ Fq2n }, and let r = {h(0, ζ)i : ζ ∈ F∗q2n } be its polar line with respect to the polarity induced by the hyperbolic quadric Q. Proposition 3.5. The Fq –linear set L associated with S satisfies the following property: • the points of L ∩ r (or L ∩ r⊥ ) have weight at most least q + 1 such points.

n 2,

and there are at

Proof. The proof is completely analogous to that given for Proposition 3.2, using the facts that dimFq (L ∩ r) = dimFq (L ∩ r⊥ ) = n and the points P = h(1, 0)i and Pa = h(a, 0)i both have weight n2 .

10

4

The infinite families

We first describe several families with n odd. Let S(u, a, b, c) be a set of linear maps of Type (3). Let f (x) = xn − Bn−1 xn−1 − · · · − B1 x − B0 be the minimal polynomial for u over Fq . Thus un = B0 +B1 u+· · ·+Bn−1 un−1 , where Bi ∈ Fq . Family FI Let q be an odd  (i)      (ii) (I)      (iii)

prime power, and assume u is a non-square in Fqn , a = c and a2 = u, N (b) = Cu, where −C is a non-square in Fq .

Theorem 4.1. For any odd prime power q and any odd integer n ≥ 3, the set S(u, a, b, c) of Type (3) satisfying (i), (ii) and (iii) above defines a semifield S. Such a semifield has Nl ∼ = Fqn and Nr = Nm = K ∼ = Fq . n

Proof. The conditions on a and u imply that aq = −a. To show that S(u, a, b, c) determines a semifield, it suffices to show that Equation (4) is satisfied. To the contrary, suppose that for some values of αi , βi , γi , δi ∈ Fq , not all zero, we have N (α(u) + aβ(u)) = N (b)N (γ(u) + aδ(u)). Taking into account the conditions on the parameters, the above equality becomes α2 (u) − uβ 2 (u) = Cu(γ 2 (u) − uδ 2 (u)). (10) Recall that we can write un = B0 + B1 u + · · · + Bn−1 un−1 , where Bi ∈ Fq . q n −1

Moreover, since u is a non-square in Fqn , we know B0 = u q−1 is a non-square in Fq . Expanding Equation (10) and substituting for un , we conclude that α02 − β 2n−1 B0 = 0. Since B0 is a non-square in Fq , this implies that α0 = β n−1 = 0. 2 2 Looking at the coefficient of u in the above equation, this further implies that −β02 = Cγ02 . Since −C is a non-square in Fq , we necessarily have β0 = γ0 = 0. In a similar fashion we see that αi = βi = 0 for i = 0, 1, . . . n−1 2 , and γj = δj = 0 for j = 0, 1, . . . n−3 . This contradiction shows that Equation (4) holds, and hence 2 S determines a semifield S of order q 2n with left nucleus isomorphic to Fqn and center isomorphic to Fq . The result now follows from Proposition 3.3. Family FII Let q be an odd prime power, and assume

11

 (i) u is a non-square in Fqn ,      (ii) a2 = u and c ∈ Fq2 \ Fq , (II)      (iii) N (b) = Ru + T u2 , where R, T ∈ Fq such that −R/T is a non-square. Theorem 4.2. For any odd prime power q and any odd integer n ≥ 3, the set S(u, a, b, c) of Type (3) satisfying (i), (ii) and (iii) above defines a semifield S. Moreover, such a semifield has Nl ∼ = Fq . = Fqn and Nr = Nm = K ∼ Proof. Since c ∈ Fq2 \ Fq , we have [1, u, u2 , . . . , u

n−3 2

, c, cu, cu2 , . . . , cu

n−3 2

]Fq = [1, u, u2 , . . . , u

n−3 2

]Fq2 ,

and hence the elements of S(u, a, b, c) are of the form n

ϕ : x 7→ (α(u) + aβ(u))x + bA(u)xq ,

(11)

n−3

where A(u) = A0 + A1 u + · · · + A n−3 u 2 with Aj ∈ Fq2 for j = 0, 1, 2, . . . , n−3 2 . 2 Suppose that for some values of αi , βi ∈ Fq and Aj ∈ Fq2 , not all zero, we have N (α(u) + aβ(u)) = N (b)N (A(u)). Taking into account the conditions on the parameters, the above equality becomes n α2 (u) − uβ 2 (u) = (Ru + T u2 )A(u)Aq (u). (12) As in the proof of the previous theorem, the above equation implies that α0 = β n−1 = 0. Now, since the maximum power of u that appears in (12) is n − 1 and 2 {1, u, u2 , . . . , un−1 } is an Fq2 –basis of Fq2n , we obtain the following polynomial identity over Fq2 : ˆ α2 (x) − xβ 2 (x) = (Rx + T x2 )A(x)A(x), where α(x) = α1 x + · · · + α n−1 x 2

A(x) = A0 + A1 x + · · · + A n−3 x

(13)

n−1 2

n−3 2

2

n−3

, β(x) = β0 + β1 (x) + · · · + β n−3 x 2 , 2 n−3 ˆ and A(x) = Aq + Aq x + · · · + Aqn−3 x 2 . 0

1

2

ˆ Since −R/T is an element of Fq and A(x) is the conjugate of the polynomial A(x) over Fq2 , the algebraic multiplicity of − R T as a root of the polynomial 2 ˆ (Rx + T x )A(x)A(x) is an odd integer. Let s and t (s, t ≥ 0) be the algebraic multiplicities of − R T as a root of α(x) 2l and β(x), respectively, and let l = min{s, t}. Then (x + R ) |(α2 (x) − xβ 2 (x)). T α(x) β(x) ¯ Defining α ¯ (x) = (x+ , we have R l and β(x) = ) (x+ R )l T

α ¯ 2 (x) − xβ¯2 (x) =

T

ˆ (Rx + T x2 )A(x)A(x) . R 2l (x + T ) 12

2 ˆ Since the multiplicity of − R T as a root of (Rx + T x )A(x)A(x) is odd, we obtain

α ¯ 2 (−

R R R ) + β¯2 (− ) = 0. T T T

¯ R Since − R ¯ (− R T is a non–square element in Fq , it follows α T ) = β(− T ) = 0, a contradiction. Thus Equation (4) is satisfied, and the result follows from our work in the previous section. Family FIII Let q be an odd prime power, and assume  (i) u is a non-square in Fqn ,      (ii) c2 = u and a ∈ Fq2 \ Fq , (III)      (iii) N (b) = R + T u, where R, T ∈ Fq such that −R/T is a non-square. Theorem 4.3. For any odd prime power q and any odd integer n ≥ 3, the set S(u, a, b, c) of Type (3) satisfying (i), (ii) and (iii) above defines a semifield S. Such a semifield has Nl ∼ = Fqn and Nr = Nm = K ∼ = Fq . Proof. The proof is completely analogous to that of Theorem 4.2, using the polynomial identity n

A(x)Aq (x) = (R + T x)(γ 2 (x) − xδ 2 (x)) over Fq2 . Family FIV Let q be a power  (i)      (ii) (IV )      (iii)

of 2, and assume T rqn /2 (u) = 1, where T rqn /2 denotes the trace function of Fqn over F2 , a ∈ Fq2 \ Fq and c is a root of the polynomial x2 + x + u, N (b) = R + T u, where R, T ∈ Fq such that T rq/2 (R/T ) = 1.

Theorem 4.4. For any even prime power q and any odd integer n ≥ 3, the set S(u, a, b, c) of Type (3) satisfying (i), (ii) and (iii) above defines a semifield S. Such a semifield has Nl ∼ = Fqn and Nr = Nm = K ∼ = Fq . Proof. Since a ∈ Fq2 \ Fq , the elements of S(u, a, b, c) are of the form n

ϕ : x 7→ A(u)x + b(γ(u) + cδ(u))xq ,

13

n−1

where A(u) = A0 + A1 u + A2 u2 + · · · + A n−1 u 2 with Ai ∈ Fq2 for i = 2 0, 1, 2, . . . , n−1 2 . Suppose that for some values of γj , δj ∈ Fq and Ai ∈ Fq2 , not all zero, we have N (A(u)) = N (b)N (γ(u) + cδ(u)). The condition on u implies that x2 + x + u is an irreducible polynomial over Fqn , n n and hence we know that c + cq = 1 and cq +1 = u. Thus the above equality becomes n A(u)Aq (u) = (R + T u)(γ 2 (u) + γ(u)δ(u) + uδ 2 (u)). As in the proof of Theorem 4.2, we obtain the following polynomial identity over Fq2 : ˆ A(x)A(x) = (R + T x)(γ 2 (x) + γ(x)δ(x) + xδ 2 (x)). ˆ Since R/T is an element of Fq and A(x) is the conjugate of the polynomial 2 A(x) over Fq2 , the algebraic multiplicity of R T as a root of the polynomial γ (x)+ 2 γ(x)δ(x) + xδ (x) is an odd integer. If s and t (s, t ≥ 0) are the algebraic multiplicities of R T as a root of γ(x) and δ(x), respectively, and l = min{s, t}, 2l 2 ) |(γ (x) + γ(x)δ(x) + xδ 2 (x)). then (x + R T δ(x) γ(x) ¯ , we have Letting γ¯ (x) = (x+ R l and δ(x) = ) (x+ R )l T

T

¯ γ¯ 2 (x) + γ¯ (x)δ(x) + xδ¯2 (x) = Since

R T

ˆ A(x)A(x) . R 2l (x + T ) (R + T x)

ˆ has even multiplicity as a root of A(x)A(x) we obtain γ¯ 2 (

R R ¯ R R R ) + γ¯ ( )δ( ) + δ¯2 ( ) = 0. T T T T T

¯ R But T rq/2 ( R ¯( R T ) = 1 then implies that γ T ) = δ( T ) = 0, a contradiction. The result now follows as in the proof of Theorem 4.2. Family FV Let q be a power of 2, and assume  (i) T rqn /2 (u) = 1,      (ii) c ∈ Fq2 \ Fq and a is a root of the polynomial x2 + x + u, (V )      (iii) N (b) = R + T u, where R, T ∈ Fq with T rq/2 (R/T ) = 1. Theorem 4.5. For any even prime power q and any odd integer n ≥ 3, the set S(u, a, b, c) of Type (3) satisfying (i), (ii) and (iii) above defines a semifield S. Such a semifield has Nl ∼ = Fqn and Nr = Nm = K ∼ = Fq .

14

Proof. As in the proof of Theorem 4.4, we obtain the equation n

(α2 (u) + α(u)β(u) + uβ 2 (u)) = (R + T u)A(u)Aq (u).

(14)

Expanding this equality and substituting un = B0 + B1 u + · · · + Bn−1 un−1 ,

B i ∈ Fq ,

the linear independence of {1, u, u2 , . . . , un−1 } over Fq2 implies that α2n−1 + α n−1 β n−1 + β 2n−1 Bn−1 = 0. 2

2

2

2

Noting that Bn−1 = T rqn /q (u) and T rq/2 (Bn−1 ) = T rq/2 (T rqn /q (u)) = T rqn /2 (u) = 1, it follows that α n−1 = β n−1 = 0. From this it follows that the maximum power 2 2 of u appearing in (14) is n − 1, and hence we obtain the following polynomial identity over Fq2 : ˆ (α2 (x) + α(x)β(x) + xβ 2 (x) = (R + T x)(A(x)A(x)). The result now follows as in the previous proof. In our final family, we assume that n is even, and we let S(u, a, b, c) be a set of linear maps of Type (9). Family FV I Let q be an odd prime power, and assume  (i) u is a non-square in Fqn ,      (ii) a = c and a2 = u, (V I)      (iii) N (b) = C, where C is a non-square in Fq . Theorem 4.6. For any odd prime power q and any even integer n > 2, the set S(u, a, b, c) of Type (9) satisfying (i), (ii) and (iii) above defines a semifield S of order q 2n . Such a semifield has Nl ∼ = Fq n , N m = N r ∼ = Fq2 and K ∼ = Fq . Proof. The same proof as that given for Theorem 4.1 allows us to prove the first statement in this theorem, as well as the facts that the left nucleus of S is isomorphic to Fqn and the center is isomorphic to Fq . Now, in order to determine the middle and the right nuclei of S, we work with the corresponding nuclei Nr and Nm , defined as subsets of S = S(u, a, b, c). Let ϕ ∈ Nr , so that ϕ = ϕA,B for some A, B ∈ Fq2n . By Theorem 2.1 we know that Nr is the largest 15

subfield over which S is a left vector space. So, for any map ϕ ∈ S we have ϕϕ ∈ S. That is, n

A(α(u)+aβ(u))+Bbq (γ(u)−aδ(u)) ∈ [1, u, . . . , u

n−2 2

, a, au, . . . , au

n−2 2

]Fq (15)

and Ab(γ(u)+aδ(u))+B(α(u)−aβ(u)) ∈ b[1, u, . . . , u

n−2 2

, a, au, . . . , au

n−2 2

]Fq , (16) n

q for all αi , βi , γi , δi ∈ Fq , i = 0, 1, . . . n−2 = −a, 2 . Note that above we used a as in the proof of Theorem 4.1 . Using the same type of arguments as in the proof of Proposition 3.3, one shows that Equations (15) and (16) are equivalent to

A = L1 + M1 a B = L2 + M2 a b n Bbq a = L3 + M3 a Aa = L4 + M4 a B a = L5 + M5 a b n Bbq = L6 + M6 a,

(17) (18) (19) (20) (21) (22)

where Li , Mi ∈ Fq for i = 1, . . . , 6. Comparing Equations (17) and (20), we obtain M1 a2 + (L1 − M4 )a − L4 = 0 and hence, since a ∈ / Fq2 , necessarily M1 = 0, L4 = 0, and L1 = M4 . In particular, A ∈ Fq . Comparing Equations (18) and (19), we obtain N (b)M2 a2 + (N (b)L2 − M3 )a − L3 = 0 and hence, since N (b) ∈ Fq and a ∈ / Fq2 , necessarily M2 = 0, L3 = 0, and N (b)L2 = M3 . In particular, B = bL2 . Thus we have A ∈ Fq and B = bA0 , with A0 ∈ Fq . n If ϕ¯ = ϕA,bA0 with A, A0 ∈ Fq , since bq +1 ∈ Fq , we have Equations (17)– n (22) satisfied. This implies that Nr = {x 7→ Ax + bA0 xq : A, A0 ∈ Fq } ∼ = Fq 2 . Similar arguments show that Nm = Nr , completing the proof. Remark 4.7. When n = 2, Conditions (i), (ii) and (iii) still produce a semifield, but using results found in [5] enables one to prove this semifield is isotopic to a Knuth semifield.

5

Isotopism Issues

In what follows we prove that the six families of semifields presented above are pairwise non–isotopic. Since only Family FV I contains semifields in which the 16

left nucleus has even dimension over the center, it suffices to concentrate on Families FI − FV . We will make strong use of the hyperbolic quadric Q disjoint from each linear set L = L(S) in Σ, as defined in Section 2. In particular, we let G denote the index two subgroup of Aut(Q) which leaves the reguli of Q invariant. First we explicitly describe this group G. Let ϕ be any nonzero map in V, n and let hϕi be the corresponding projective point in Σ. If ϕ : x 7→ ηx + ζxq , then for any σ ∈ Aut(Fq2n ), we let ϕσ denote the Fqn -linear map of Fq2n defined n by the rule ϕσ : x 7→ η σ x + ζ σ xq . Then for any non-singular Fqn -linear maps ψ and φ of Fq2n , we define Iψσφ to be the collineation of Σ induced by the semilinear map on V whose rule is ϕ 7→ ψϕσ φ. Since ϕ is singular if and only if ψϕσ φ is singular, we see that Iψσφ maps Q to Q. Straightforward computations show that Iψσφ leaves invariant the reguli on Q, and thus Iψσφ ∈ G. A counting argument then implies that G = {Iψσφ : σ ∈ Aut(Fq2n ); ψ, φ non-singular Fqn -linear maps of Fq2n }. We now use the fact that for each of the Families FI –FV , by Proposition 3.2 the line r is the unique line of Σ contained in the linear set L(u, a, b, c) of S(u, a, b, c). The following result then follows immediately. Proposition 5.1. If I ∈ G maps a linear set L(u, a, b, c) to a linear set L(u0 , a0 , b0 , c0 ), then I leaves invariant the line r of Σ. We now characterize the elements of G that leave the line r invariant. Proposition 5.2. If Iψσφ leaves the line r invariant, then either ψ : x 7→ η1 x or ψ : x 7→ ζ1 xq

n

and

φ : x 7→ η2 x

and

φ : x 7→ ζ2 xq ,

n

where η1 , η2 , ζ1 , and ζ2 are nonzero elements of Fq2n . n

n

Proof. Let φ : x 7→ η1 x + ζ1 xq , ψ : x 7→ η2 x + ζ2 xq , and let ϕ : x 7→ ξx induce an arbitrary point of the line r, where ξ is an arbitrary nonzero element of Fq2n . Since Iψσφ leaves the line r invariant, the mapping n

n

n

n

ψϕσ φ : x 7→ (η2 ξ σ η1 + ζ2 ξ σq ζ1q )x + (η2 ξ σ ζ1 + ζ2 ξ σq η1q )xq

n

n

n

induces another point of the line r. This implies that η2 ξ σ ζ1 + ζ2 ξ σq η1q = 0 n for any nonzero ξ ∈ Fq2n . This in turn implies that η2 ζ1 = 0 = ζ2 η1q . The result now follows from the fact that φ and ψ are non-singular maps. A version of the following result may be found in [5]. 17

Theorem 5.3. [5, Thm. 2.1] Two semifield spreads S1 and S2 of PG(3, q n ) are projectively equivalent (hence the semifield planes π(S1 ) and π(S2 ) are isomorphic) if and only if the associated Fq –linear sets are equivalent with respect to the subgroup G of P ΓO+ (4, q n ) fixing the reguli of the hyperbolic quadric Q of Σ. It now follows from Proposition 5.2 and Theorem 5.3 that any semifield of type (3) isotopic to a semifield S(u, a, b, c) of Type (3) is defined by a spread set of Fqn –linear maps with one of the following two forms: n

n

(∗) {x 7→ η1 η2 (α(uσ )+aσ β(uσ ))x+η1 η2q bσ (γ(uσ )+cσ δ(uσ ))xq : αi , βi , γj , δj ∈ Fq }, for some nonzero η1 , η2 ∈ Fq2n and some σ ∈ Aut(Fq2n ), n

n

n

n

n

(∗∗) {x 7→ ζ1 ζ2q (α(uσ )+aσq β(uσ ))x+ζ1 ζ2 bσq (γ(uσ )+cσq δ(uσ ))xq : αi , βi , γj , δj ∈ Fq }, for some nonzero ζ1 , ζ2 ∈ Fq2n and some σ ∈ Aut(Fq2n ). Of course, the same shorthand notation as that introduced after Equation (3) is being used above. In fact, one can obtain form (∗) from form (∗∗) by replacing ζ1 by η1 , ζ2 by qn η2 , and σq n by σ. Thus one may concentrate solely on form (∗). Thus we have the following result. Proposition 5.4. Any semifield of Type (3) isotopic to the semifield S(u, a, b, c) of Type (3) is defined by a spread set of Fqn –linear maps of the form n {x 7→ λ(α(uσ ) + aσ β(uσ ))x + ¯bλ(γ(uσ ) + cσ δ(uσ ))xq : αi , βi , γj , δj ∈ Fq },

for some λ ∈ F∗q2n , ¯b ∈ Fq2n and σ ∈ Aut(Fq2n ) such that N (¯b) = N (b)σ . Proof. Let λ = η1 η2 and ¯b = bσ η2q

n

−1

in form (∗).

We now have sufficient tools to sort out any possible isotopisms between our constructed families. Theorem 5.5. The Families FI , FII , FIII , FIV , FV and FV I are pairwise non-isotopic. Proof. Suppose first that a semifield S = S(u, a, b, c) of Family FI is isotopic to a semifield S0 = S(u0 , a0 , b0 , c0 ) of Family FII . Then by Proposition 5.4 there exist λ ∈ F∗q2n , ¯b ∈ F∗q2n and σ ∈ Aut(Fq2n ) with N (¯b) = N (b)σ such that n {x 7→ λ(α(uσ ) + aσ β(uσ ))x + ¯bλ(γ(uσ ) + cσ δ(uσ ))xq : αi , βi , γj , δj ∈ Fq } n

= {x 7→ (α(u0 ) + a0 β(u0 ))x + b0 (γ(u0 ) + c0 δ(u0 ))xq : αi , βi , γj , δj ∈ Fq }, where we are using the same notation as that discussed when defining Families FI and FII . As in the proof of Theorem 4.2, the last spread set can be written as n

{x 7→ (α(u0 ) + a0 β(u0 ))x + b0 A(u0 )xq : αi , βi ∈ Fq , Aj ∈ Fq2 }, 18

where Aj = γj + c0 δj for j = 1, . . . , n−3 2 . This then implies that λ[1, uσ , . . . , u

n−1 2 σ

, aσ , aσ uσ , . . . , aσ u

n−1 2 σ

]Fq = [1, u0 , . . . , u0

n−1 2

, a0 , a0 u0 , . . . , a0 u0

n−1 2

and n−3 n−3 n−3 λ¯b[1, uσ , . . . , u 2 σ , aσ , aσ uσ , . . . , aσ u 2 σ ]Fq = b0 [1, u0 , . . . , u0 2 ]Fq2 . n−3

From the second equality we see that [1, uσ , . . . , u 2 σ , aσ , aσ uσ , . . . , aσ u is also an Fq2 –vector subspace of dimension n−1 2 . In fact, we have [1, uσ , . . . , u

n−3 2 σ

, aσ , aσ uσ , . . . , aσ u

n−3 2 σ

]Fq = [1, uσ , . . . , u

n−3 2 σ

n−3 2 σ

]Fq

]Fq2 .

Hence, for i = 0, 1, . . . , n−3 2 we have (i)

(i)

(i)

aσ uσi = A0 + A1 uσ + . . . + A n−3 u

n−3 2 σ

,

(23)

2

(i)

where Aj ∈ Fq2 for j = 0, 1, . . . , n−3 2 . From Equation (23), as in the proof of Proposition 3.3, it follows that aσ ∈ Fq2 and hence a ∈ Fq2 . But this is a contradiction since a2 = u and u ∈ Fqn \ Fq . Thus we must have that Family FI and Family FII are not isotopic. Next suppose that a semifield S = S(u, a, b, c) of Family FII is isotopic to a semifield S0 = S(u0 , a0 , b0 , c0 ) of Family FIII . Again from Proposition 5.4, we obtain n

{x 7→ λ(α(uσ ) + aσ β(uσ ))x + ¯bλA(uσ )xq : αi , βi ∈ Fq , Aj ∈ Fq2 } n

= {x 7→ (α(u0 ) + a0 β(u0 ))x + b0 (γ(u0 ) + c0 δ(u0 ))xq : αi , βi , γj , δj ∈ Fq } n

= {x 7→ A(u0 )x + b0 (γ(u0 ) + c0 δ(u0 ))xq : γj , δj ∈ Fq , Ai ∈ Fq2 }, for some λ, ¯b ∈ Fq2n and some σ ∈ Aut(Fq2n ) such that N (¯b) = N (b)σ . Then, as in the previous case, this implies that [1, uσ , . . . , u

n−1 2 σ

, aσ , aσ uσ , . . . , aσ u

n−1 2 σ

]Fq = [1, uσ , . . . , u

n−1 2 σ

]Fq2

and hence a ∈ Fq2 , a contradiction. That is, Family FII is not isotopic to Family FIII . In a similar way one shows that Families FI and FIII are non-isotopic, and Families FIV and FV (where q is even) are non-isotopic. Since Family FV I is the only constructed family of semifields in which the left nucleus has even dimension over the center, the six families FI − FV I are pairwise non-isotopic.

6

Comparison with Known Examples

In the previous sections we constructed six infinite families of finite semifields, which were shown to be pairwise non-isotopic. In all cases the semifields have 19

]Fq

order q 2n , left nucleus isomorphic to Fqn , and center isomorphic to Fq . In the Families FI − FV , where the left nucleus has odd dimension n over the center, we showed that the middle and right nuclei are isomorphic to the center, namely Fq . In Family FV I , where this dimension n is even, the middle and right nuclei are isomorphic to Fq2 . None of our examples is symplectic since the associated Fq –linear sets are not contained in a plane of Σ ∼ = PG(3, q n ) (see [14]), nor commutative since in no family are the left nucleus, right nucleus and center all equal. Moreover, none of the associated linear sets (if n > 2) is scattered by Propositions 3.2 and 3.5. Looking at the known finite semifields, say as categorized in [11], we now show that none of Families FI − FV I , if n > 2, is isotopic to a known family, nor isotopic to a Knuth “derivative” of a known family. Recall that Knuth’s symmetric group S3 , obtained from his cubical array (see [12]), creates six semifields from a given semifield, one of which is the original and some of which may be isotopic to one another. For lack of better terminology, we will call these six semifields derivatives of the original semifield. The group S3 permutes the nuclei (up to isomorphism) of the six derivatives [15]. Any Albert generalized twisted field (see Section 3.3 of [11]) which is twodimensional over its left nucleus is shown in [16] to have an associated linear set which is scattered. Thus none of our examples is a generalized twisted field. Moreover, none of our examples is a Knuth derivative of a generalized twisted field since Proposition 5.2 in [10] shows that any Knuth derivative of a generalized twisted field is again a generalized twisted field. In [12] Knuth gives four multiplications which, under certain conditions on various parameters, yield finite semifields of order q 2n . The multiplications are given in (2) − (5) of Section 3.4 in [11]. The semifields arising from (3) − (5) all are two–dimensional over two of their nuclei (see [12, Thm. 7.4.1]), and, since the group S3 permutes the size of the nuclei, the same occurs for their Knuth derivatives. So these examples are not isotopic to any of the families FI − FV I . Regarding semifields whose multiplication is defined by (2), direct computations show that their three nuclei are equal to the center and the same occurs for their Knuth derivatives. So none of our families FI − FV I is isotopic to one of them. In [13] Knuth defines a class of “binary” semifields (see Section 3.5 in [11]). These were later generalized in [10], where it was shown in Theorem 4.7 of that paper that all such proper semifields have left nucleus F2 . Thus these semifields, and their Knuth derivatives, are not isotopic to any of our families. The so-called cyclic semifields of Jha and Johnson, studied in [9], which have order q 2n (n odd) and are two-dimensional over their left nuclei, all have right and middle nuclei equal to Fq2 . Thus no Knuth derivative of these semifields will be isotopic to one in our families FI − FV I . Hence we are only left with the generalized Dickson semifields, whose multiplication is given in (1) of Section 3.4 in [11] (see also [6, p. 241], Multiplication (15) with g = 0). Equation 5.11 in [10] shows that Knuth’s transpose operation (corresponding to taking the dual spread of the given semifield spread) maps a generalized Dickson semifield to another generalized Dickson semifield. 20

Straightforward computations similarly show that the Knuth operation corresponding to reversing the order of multiplication in the given semifield again maps a generalized Dickson semifield to a generalized Dickson semifield. As these two operations generate Knuth’s S3 , we see that any Knuth derivative of a generalized Dickson semifield is a generalized Dickson semifield. Thus it suffices to show that any generalized Dickson semifield which is two-dimensional over its left nucleus has an associated linear set which is not G-equivalent to any of the linear sets associated with our families, where G is the group defined in Theorem 5.3. Theorem 6.1. Let D = (D, +, ◦) be a generalized Dickson semifield which is two–dimensional over its left nucleus. Then D is not isotopic to any of the semifields from Families FI , FII , FIII , FIV , FV or FV I . Proof. We construct the linear set L of D by using the spread set of matrices defining D (see [5, Section 4]), and show that it is not equivalent to any of the linear sets from Section 4 under the group G of Theorem 5.3. We begin by computing all possible points of L that have weight greater than 1. The semifield D = (Fq0 × Fq0 , +, ◦) has the following general form for its multiplication: (x, y) ◦ (u, v) = (xu + y α v β k, xv + yuσ ), (24) α−1 β+1 where α, β, σ ∈ Aut(Fq0 ) and k ∈ / Fσ+1 if that is possible (see [6], p. q 0 F q 0 Fq 0 241). If βσ = 1 and σα = β, Multiplication (24) becomes the Knuth operation (18) with g = 0 [6, p. 241]. So we can assume either βσ 6= 1 or σα 6= 1. In this case D is two–dimensional over its left nucleus if and only if α = 1. When this occurs we have Nl = {(x, 0) : x ∈ Fq0 } and K = {(x, 0) : x ∈ F ix(σ) ∩ F ix(β)}. Let K = Fq and Fq0 = Fqn . The linear map ϕ(u,v) defined via (x, y) 7→ (x, y) ◦ (u,  v) is represented in the Nl –basis {(1, 0), (0, 1)} of D by the matrix  u vβ k , and hence the Fq –linear set of Σ ' PG(3, q n ) associated with D is v uσ

LD = {h(u, v β k, v, uσ )i : u, v ∈ Fqn }. Let P(u,v) be a point of LD and suppose that the weight of P(u,v) is greater than 1. Then there exists λ ∈ Fqn \ Fq such that λu = u0 , λv β k = v 0β k, λv = v 0 and λuσ = u0σ , for some u0 , v 0 ∈ Fqn . These conditions imply that λuσ = λσ uσ and λv β = λβ v β .

(25)

Since λ ∈ / Fq = F ix(σ) ∩ F ix(β), Equation (25) implies that either u = 0 or v = 0. So the points of LD having weight greater than 1 belong to the two secant lines of Q with equations l : x1 = x2 = 0 and l0 : x0 = x3 = 0. Now F ix(σ) = Fqs and F ix(β) = Fqt , with s and t two coprime integers dividing n. Thus all points of LD ∩ l have weight s (over Fq ), while all points of LD ∩ l0 have weight t (over Fq ). Note that if s = t = 1, then LD is of scattered type. This means that a line of Σ which is external to the quadric Q contains at most two points of LD with weight greater than 1. Since for any of our examples 21

FI − FV I , there exists an external line to the quadric Q containing at least q + 1 points of the associated linear set with weight greater than 1 (see Propositions 3.2 and 3.5), we see that LD cannot be G–equivalent to any of the linear sets associated with Families FI , FII , FIII , FIV , FV or FV I . Thus we have indeed shown that Families FI −FV I are not related by Knuth operations or isotopism to any of the semifield families currently in print.

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Concluding Remarks

The use of spread sets of linear maps seems to be a very robust method for producing semifields that are two–dimensional over their left nuclei. The classification of such semifields of order q 6 with center Fq is not yet completed, and work continues on this project in [7]. As discussed in [11], there are many more types of construction for finite semifields in odd characteristic than for finite semifields in even characteristic, although “almost all” (in the probabilistic sense) known semifield planes have even order. The six new families constructed here only amplify that situation. The delicate isotopism calculations within each given family are still ongoing. Finally, the general processes of algebraic lifting (as discussed in Section 4.6 of [11]) is not very well understood, in the sense that it seems to be a very difficult problem to determine if applying a sequence of one or more liftings of a known family results in a family isotopic to one of those constructed here. While this does not seem to be likely, we currently have no proof of this fact.

References [1] A.A. Albert: Finite division algebras and finite planes, Proc. Symp. Appl. Math., 10 (1960), 53–70. [2] R.B. Baker, J.M. Dover, G.L. Ebert, K.L. Wantz: Perfect Baer subplane partitions and three-dimensional flag-transitive planes, Designs, Codes, Cryptogr., 21 (2000), 19–39. [3] M. Biliotti, V. Jha, N.L. Johnson: Foundations of Translation Planes, Pure and Applied Mathematics, Vol. 243, Marcel Dekker, New York, Basel, 2001, 1–552. [4] R.H. Bruck, R.C. Bose: The construction of translation planes from projective spaces, J. Algebra, 1 (1964), 85–102. [5] I. Cardinali, O. Polverino, R. Trombetti: Semifield planes of order q 4 with kernel Fq2 and center Fq , European J. Combin., 27 (2006), 940–961. [6] P. Dembowski: Finite Geometries, Springer Verlag, Berlin, 1968.

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[7] G.L. Ebert, G. Marino, O. Polverino, R. Trombetti: Some new semifields of order q 6 with center Fq , preprint. [8] N.L. Johnson, G. Marino, O. Polverino, R. Trombetti: Semifields of order q 6 with left nucleus Fq3 and center Fq , Finite Fields and Their Applications, to appear, (available online 8 May 2007). [9] N.L. Johnson, G. Marino, O. Polverino, R. Trombetti: On a generalization of cyclic semifields, submitted. [10] W.M. Kantor: Commutative semifields and symplectic spreads, J. Algebra, 270 (2003), 96–114. [11] W.M. Kantor: Finite semifields, pp. 103–114, in: Finite Geometries, Groups, and Computation (Proc. of Conf. at Pingree Park, CO, Sept. 2005), de Gruyter, Berlin-New York, 2006. [12] D.E. Knuth: Finite semifields and projective planes, J. Algebra, 2 (1965), 182–217. [13] D.E. Knuth: A class of projective planes, Trans. AMS, 115 (1965), 541– 549. [14] G. Lunardon: Translation ovoids, J. Geom., 76 (2003), 200–215. [15] G. Lunardon: Symplectic spreads and finite semifields, Designs, Codes, Cryptogr., to appear. [16] G. Marino, O. Polverino, R. Trombetti: On Fq –linear sets of PG(3, q 3 ) and semifields, J. Combin. Theory (Ser. A), to appear. [17] O. Ore: On a special class of polynomials, Trans. Amer. Math. Soc., 35 (1933), 559-584.

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G. L. Ebert Dept. of Mathematical Sciences University of Delaware Newark, DE 19716, USA [email protected] G. Marino, R. Trombetti Dip. di Matematica e Applicazioni Universit` a degli Studi di Napoli “Federico II” I– 80126 Napoli, Italy [email protected], [email protected] O. Polverino Dip. di Matematica Seconda Universit` a degli Studi di Napoli I– 81100 Caserta, Italy [email protected], [email protected]

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