Commutative semifields of rank 2 over their nucleus

This article is about finite commutative semifields that are of rank 2 over their nucleus, the largest subset of elements that is a finite field. These semifields have a direct correspondence to certain flocks of the quadratic cone in PG(3, q) and to certain ovoids of the parabolic space Q(4, q). We shall consider these links, the known examples and non-existence results. 1 Semifields A finite semifield S is a finite algebraic system that possesses two binary operations, addition and multiplication, which satisfy the following axioms. (S1) Addition is a group with identity 0. (S2) a(b+ c) = ab+ ac and (a+ b)c = ac+ bc for all a, b, c ∈ S. (S3) There exists an element 1 6= 0 such that 1a = a = a1 for all a ∈ S. (S4) If ab = 0 then either a = 0 or b = 0. Throughout this article the term semifield will refer to a finite semifield. The additive group of a semifield must be commutative. By (S2), (ac+ ad) + (bc+ bd) = (a+ b)(c+ d) = (ac+ bc) + (ad+ bd). Hence, ad + bc = bc+ ad and any elements that can be written as products commute under addition. By (S4) and finiteness, any element of S can be written as a product and so it follows that the additive group is abelian. Moreover it is not difficult to show that the group is elementary abelian. Let a 6= 0, and let p be the additive order of a. If p is not prime then we can write p = rs for r and s integers not equal to 1, and by observing that 0 = (pa)a = (rsa)a = (ra)(sa) we get a contradiction from (S4). The fact that every nonzero element has prime order suffices to show that the group is elementary abelian, and that all nonzero elements have the same prime order p. This number p is the characteristic of the semifield. An elementary abelian group can be viewed as a vector space over a finite field. In particular S has p elements where n is the dimension of S over the field GF (p). There are many examples of semifields known and some standard constructions can be found in Knuth [17]. If the order is p, the semifield must be GF (p). If the order is p, the semifield is GF (p). This is not difficult to see. Let {1, x} 2 S. Ball and M. Lavrauw be a basis for the semifield. Multiplication is determined by x = ax + b and the polynomial x − ax − b has no roots in GF (p) else we would have x − ax − b = (x − r)(x − s) = 0 contradicting (S4). Thus x − ax − b is irreducible and the multiplication is GF (p). This short argument comes again from [17] where it is also determined that the only semifield of order 8 is GF (8). And completing the question of existence Albert [1] and Knuth [17] construct semifields that are not finite fields for every other order q = p, that is h ≥ 3 if p is odd and h ≥ 4 if p = 2. The major motivation to study semifields in the 1960’s was their use in the construction of projective planes, see Hughes and Piper [14] or Hall [13]. Every semifield determines a projective plane and the projective plane is Desarguesian if and only if the semifield is a field. The incidence structure constructed from a semifield S with Points: (0, 0, 1) Lines: [0, 0, 1] (0, 1, a) [0, 1, a] a ∈ S (1, a, b) [1, a, b] a, b ∈ S such that the point (x1, x2, x3) is incident with the line [y1, y2, y3] if and only if y1x3 = x2y2 + x1y3 is a projective plane π(S) of order |S|. It is a simple matter to check that any two points of π(S) are incident with a unique line and dually that any two lines of π(S) are incident with a unique point and hence that π(S) is a projective plane. However it is harder to determine when two semifields S and S′ determine the same projective plane, i.e. π(S) ∼= π(S′). In [17] Knuth defines an isotopism from S to S′ and shows that an isotopism is equivalent to a set of three 1-1 maps (F,G,H) linear over GF (p) from S to S′, such that (ab)H = (aF )(bG) for all a, b, c ∈ S. Two semifields S and S′ are isotopic if there is an isotopism from S to S′. We have the following theorem due to Albert, a proof of which can be found in [17]. Theorem 1. Two semifields coordinatize the same projective plane if and only if they are isotopic. In his original work on semifields Dickson [10] considered constructing commutative semifields, that is semifields that satisfy (S5) ab = ba for all a and b in S. We define the nucleus of a commutative semifield to be N := {x | (ax)b = a(xb), ∀a, b ∈ S}. Commutative semifields 3 It is clear that N contains the field GF (p) where p is the characteristic and that N is itself a finite field. Moreover, S can be viewed as a vector space over its nucleus. Dickson [11] gave a construction of a commutative semifield of rank 2 over its nucleus. It is as follows. Let S := {(x, y) | x, y ∈ GF (q)} and let σ be an automorphism of GF (q) where q is odd. Addition is defined component-wise and multiplication by (x, y)(u, v) = (xv + yu, yv +mxu) where m is a non-square in GF (q). The only axiom that requires much thought is (S4) and we shall check this in a more general setting shortly. In this article we shall only be concerned with commutative semifields that are of rank 2 over their nucleus which are in direct correspondence with certain useful geometric objects. Cohen and Ganley [8] made significant progress in the investigation of commutative semifields of rank 2 over their nucleus. They put Dickson’s construction in the following more general setting. Let S be a commutative semifield of order q with nucleus GF (q). Then there is an α ∈ S\GF (q) such that {1, α} is a basis for S. Addition in S is component-wise and multiplication is defined as (xα+ y)(uα+ v) = xuα + (xv + yu)α+ yv (1) where xα = g(x)α+ f(x), f and g are functions from GF (q) → GF (q). The distributive laws are satisfied if and only if both f and g are linear maps, in other words, f(x + y) = f(x) + f(y) and g(x + y) = g(x) + g(y) for all x, y in GF (q). Thus we must check (S4). Suppose that (xα + y)(uα+ v) = 0 and that x, y, u and v are non-zero. It follows that g(xu) + xv + yu = 0 and f(xu) + yv = 0 and eliminating y that xv + vg(xu)− uf(xu) = 0. Writing xu = z and v/u = w zw + g(z)w − f(z) = 0. If one or more of x, y, u or v is zero it follows immediately that at least one of (x, y) or (u, v) is (0, 0). Hence we have proved the following theorem which comes from [8]. 4 S. Ball and M. Lavrauw Theorem 2. Let S be a commutative semifield of rank 2 over its nucleus GF (q). Then there exist linear functions f and g such that multiplication in S is defined as in (1) and zw + g(z)w− f(z) = 0 has no solutions for all w, z ∈ GF (q) and z 6= 0. If q is odd then this quadratic in w will have no solutions in GF (q) if and only if g(z) + 4zf(z) is a non-square for all z ∈ GF (q)∗. Cohen and Ganley [8] prove the following theorem for q even. Theorem 3. For q even the only commutative semifield of rank 2 over its nucleus GF (q) is the finite field GF (q). In light of this theorem we restrict ourselves to the case q is odd. Let us consider again the example of Dickson. We have g = 0 and f(z) = mz where m is a non-square. We had only to check that (S4) is satisfied and this is clear since g(z)+4zf(z) = 4mz is a non-square for all z ∈ GF (q)∗. 2 Flocks of the Quadratic Cone Let q be an odd prime power and let K be a quadratic cone of PG(3, q) with vertex v and base a conic C. The quadratic cones of PG(3, q) are equivalent under the action of PGL(4, q) so we can assume that v is the point 〈0, 0, 0, 1〉 and the conic C in the plane π with equation X3 = 0, is the set of zeros of X0X1 = X 2 2 . A flock F of K is a partition of K \ {v} into q conics. We call the planes that contain conics of the flock the planes of the flock. A flock F is equivalent to a flock F ′ if there is an element in the stabiliser group of the quadratic cone that maps the planes of the flock F to the planes of the flock F ′. If all the planes of the flock share a line then the flock is called linear. Let a0X0 + a1X1 + a2X2 + a3X3 = 0 be a plane of the flock. Since 〈0, 0, 0, 1〉 is disjoint from any plane of the flock a3 6= 0 and hence we may assume that a3 = 1. The point 〈1, 0, 0,−a0〉 is incident with the quadratic cone and this plane and hence the coefficients of X0 in the planes of the flock are distinct. Hence we can parameterise by the elements of GF (q) so that the planes of the flock are πt : tX0 − f(t)X1 + g(t)X2 +X3 = 0 where t ∈ GF (q) and f and g are functions from GF (q) → GF (q). The points that are incident with the line that is the intersection of two planes of the flock πt and πs are incident with the plane (t− s)X0 − (f(t)− f(s))X1 + (g(t)− g(s))X2 = 0. Commutative semifields 5 The points that are incident with the coneK satisfy the equationX0X1 = X 2 . If the equation (t− s)X 2 − (f(t)− f(s))X 1 + (g(t)− g(s))X1X2 = 0 has a solution then we can find a line on the cone, by choosing the X0 coordinate appropriately, that would be contained in the plane above, and hence a point on the cone and incident with both the planes πt and πs. The flock property implies that no such point exists and hence that this equation has no solutions. There is no solution with X1 = 0 as this would imply that X2 = 0 and that t = s. Hence we can put w = X2/X1 and we have the forward implication of the following theorem which is due to Thas [24]. Theorem 4. Let F be a flock of the quadratic cone with vertex 〈0, 0, 0, 1〉 and base X0X1 = X 2 2 . Then there exists functions f and g from GF (q) → GF (q) such that the planes of the flock are tX0 − f(t)X1 + g(t)X2 +X3 = 0 where t ∈ GF (q) and F is a flock if and only if (t− s)w + (g(t)− g(s))w − (f(t)− f(s)) = 0 has no solution for all s and t ∈ GF (q), s 6= t. If f and g are linear then the condition of the theorem says that F is a flock if and only if zw + g(z)w − f(z) = 0 has no solutions for w ∈ GF (q) and z ∈ GF (q)∗. A flock with this property is called a semifield flock as such a flock is in one-to-one correspondence with a commutative semifield of rank 2 over its nucleus. This is clear from Theorem 2. The commutative semifield S = {(x, y) | x, y ∈ GF (q)} where addition is defined component-wise and multiplication is defined by (x, y)(u, v) = (xv + yu+ g(xu), yv + f(xu)) is the semifield associated to the flock F . name g(x) f(x) q = p linear 0 mx all Dickson [10] Kantor [16] Knuth [17] 0 mx Cohen-Ganley [8] Thas-Payne [26] x mx+mx 3 Penttila-Williams [22], [3] x x 3 Table 1. The known examples of semifield flocks up to equivalence 6 S. Ball and M. Lavrauw The known examples of semifield flocks up to equivalence are listed in Table 2. In all relevant cases m is taken to be a non-square in GF (q) and σ is a nontrivial automorphism of GF (q). Some of the links between the commutative semifields, certain ovoids of Q(4, q), semifield flocks of the quadratic cone and semifield translation planes were not known until recently and hence in most cases more than one person or persons is accredited with the discovery of the functions f and g. In fact in the second case Dickson [10] discovered the semifield, Kantor [16] the ovoid and Knuth [17] the semifield plane. In the third example Cohen and Ganley [8] discovered the semifield while Thas and Payne found the ovoid [26]. And in the fourth example Penttila and Williams discovered the ovoid [22] and details concerning the corresponding flock were investigated by Bader, Lunardon and Pinneri [3]. We shall discuss these equivalent objects in the following sections and explain the links between them and how this can be of use. Firstly however we shall check that the last two examples in Table 2 do indeed satisfy the condition of Theorem 2 and Theorem 4. In the Cohen-Ganley Thas-Payne example g(x) + 4xf(x) = g(x) + xf(x) = x +m−1x2 +mx = m(x −m−1x)2 which is a non-square for all x ∈ GF (3h)∗. The Penttila-Williams example is somewhat more difficult to prove. The following comes from [2]. We have that g(x) + 4xf(x) = g(x) + xf(x) = x + x = x(1 + x) and since 3 − 1 = 242 = 2.11 we need to show that 1 + ǫ is a non-square for all ǫ such that ǫ = 1. Now (q − 1)/2 = 121 = 1 + 3 + 3 + 3 + 3 and in GF (3) (1 + ǫ) = (1 + ǫ)(1 + ǫ)(1 + ǫ)(1 + ǫ)(1 + ǫ). The set {1, 3, 9, 27, 81} are the squares modulo 11 and each non-zero integer modulo 11 can be written exactly 3 times as the sum of elements of {1, 3, 9, 27, 81} modulo 11. Hence in GF (3) (1 + ǫ) = 2 = −1 and 1 + ǫ is a non-square for all ǫ such that ǫ = 1. The following theorem comes from [12]. Theorem 5. The projective planes obtained from the flocks F and F ′ are isomorphic if and only if the flocks F and F ′ are equivalent. The projective planes in Theorem 5 are constructed, via the Bruck Bose André method, from the spread {〈(y, x, 1, 0), (f(x), y + g(x), 0, 1)〉 | x, y ∈ GF (q)} ∪ {〈(1, 0, 0, 0), (0, 1, 0, 0)〉}. Commutative semifields 7 This plane is a semifield plane. Following [9, (5.1.2)] the spread comes from the spread set D = {(