Commutative semifields from projection mappings

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1 Des. Codes Cryptogr. (2011) 61: DOI /s z Commutative semifields from projection mappings Jürgen Bierbrauer Received: 28 September 2009 / Revised: 9 September 2010 / Accepted: 20 October 2010 / Published online: 4 November 2010 Springer Science+Business Media, LLC 2010 Abstract We describe a general projection method to construct commutative semifields in odd characteristic. One application yields a family of commutative semifields of order q 2m with middle nucleus of order at least q 2 for every odd prime-power q and every odd integer m > 1. Another application of the method yields a generalization of the Budaghyan Helleseth family and also greatly simplifies the construction. Keywords Semifields Presemifields Symplectic semifields Planar functions Dembowski Ostrom polynomial Polarization Isotopy Nucleus Trace Mathematics Subject Classification (2000) 12K10 51E15 51A40 1 Introduction Definition 1 A semifield is a set F with two binary relations, addition and *, such that F is a commutative group with respect to addition. F is a loop under multiplication. 0 a = a 0 = 0foralla. The distributive law holds. There is e F such that e x = x e = x for all x F. If the last of those axioms is not necessarily satisfied one speaks of a presemifield. A presemifield is commutative if its multiplication is commutative. Given a commutative presemifield with multiplication x y it is easy to construct a commutative semifield of the same order by choosing e F and defining (x e) (y e) = x y (with unit element e e). It is Communicated by G. Lunardon. J. Bierbrauer (B) Department of Mathematical Sciences, Michigan Technological University, Houghton, MI 49931, USA jbierbra@mtu.edu

2 188 J. Bierbrauer easy to see that the order F of a finite semifield is a power of a prime p and that the additive group is elementary-abelian. For finite semifields the third axiom is a consequence of the others. We can choose F to be the finite field F p r such that the semifield addition equals the field addition. We can compare the field multiplication to the semifield multiplication x y. It can also be arranged that the unit element is e = 1. One approach to commutative semifields in odd characteristic p is by way of planar functions: Definition 2 Let F = GF(p r ) be a finite field in odd characteristic. A function f : F F is a planar function if for each 0 = a the mapping x x a = (1/2)( f (x + a) f (x) f (a)) is bijective. Definition 3 Let F be a finite field in characteristic p, f : F F. Write f as a polynomial f (x) = i a i x i. Then f is quadratic (or: a Dembowski Ostrom polynomial) if all its monomial terms x i for a i = 0havep-weight 2 (i issumoftwopowersofp). It has been observed by Coulter-Henderson [7] that commutative semifields in odd characteristic can be equivalently described by quadratic planar functions. The relation between the semifield multiplication and the quadratic planar function is formally identical to the classical relation (the polarization identity) between symmetric bilinear forms and quadratic forms, in odd characteristic. The term Dembowski-Ostrom polynomial stems from [9]where quadratic planar functions were used to describe the corresponding geometric objects, projective semifield planes. These are projective planes which are translation planes and also dual translation planes. Definition 4 Let F be a finite field of odd characteristic and f : F F a quadratic function. The polarization of f is defined by x y = ( f (x + y) f (x) f (y))/2. If f in Definition 4 is also planar, then the polarization is the corresponding presemifield multiplication. The paradigmatic planar function is f (x) = x 2. The corresponding semifield multiplication is x y = (1/2)((x + y) 2 x 2 y 2 ) = xy, the field multiplication. The notion of equivalence of semifields is motivated by the corresponding projective plane. Definition 5 Let F = F r p be the r-dimensional vector space over F p. Consider presemifields on F whose additions coincide with that of F. Two such presemifield multiplications and on F are isotopic if there exist α 1,α 2,β GL(r, p) such that β(x y) = α 1 (x) α 2 (y) always holds. They are strongly isotopic if we can choose α 2 = α 1. Two presemifields are isotopic if and only if the corresponding projective planes are isomorphic. Let F = F p r be the field of order p r. It is a commonly used method to replace a given commutative semifield of order p r by an isotopic copy which is defined on F and shares the additive structure and the unit element 1 with F. The question is then to which degree the semifield multiplication can be made to coincide with the field multiplication.

3 Commutative semifields from projection mappings 189 Definition 6 Let F = F p r and (F, ) a commutative semifield with unit 1 whose additive structure agrees with that of the field F.Define S = {c F c x = cx for all x F}. M = {c F (x c) y = x (c y) for all x, y F}. K = {c F c (x y) = (c x) y for all x, y F}. Here the dimensions of the middle nucleus M and of the kernel or left nucleus K of a commutative semifield are invariant under isotopy. The dimension of S depends on the embedding of the semifield in the field F. Asmentionedin[8] wehavek M. AsM is closed under semifield multiplication and is associative it is a field. The semifield multiplication on M can therefore be made to coincide with field multiplication. The same is true of the vector space structure of F over its subfield M. It follows that we can find a suitable isotope such that K M S. It seems that the known families of commutative semifields which are defined in arbitrary odd characteristic are the following: 1. The finite fields in odd characteristic. 2. The Dickson semifields [10]. 3. The Albert commutative twisted fields [1]. 4. The Zha Kyureghyan Wang semifields [12] (seealso[4]). 5. The family from [3]. 6. The Budaghyan Helleseth family, see [5,6]. 7. The Lunardon Marino Polverino Trombetti (LMPT)-family [11]. Here the LMPT-family is defined for each odd prime-power q, has order q 6, middle nucleus of order q 2 and left nucleus of order q. In Theorem 1 we present a generalization of the LMPT-construction. Our commutative semifields P(m, q) have order q 2m, middle nucleus of order at least q 2 and left nucleus of order at least q, for every odd prime-power q and every odd m > 1. The family of [11]isthe special case m = 3. Theorem 1 is proved in Sect. 2. The symplectic version of those semifields is explicitly described in Sect. 3. In Sect. 4 we describe a general projection method which contains the construction of the P(m, q) as a special case. Section 5 contains a simplification and generalization of the Budaghyan Helleseth family from [5]. It is shown that the generalized Budaghyan Helleseth presemifields BH( p, m, s, β) are another special case of the projection method from Sect. 4. The main result in this direction is Theorem 3. Theorem 4 explicitly describes the semifield product of BH( p, m, s, β) in a parametric special case. The final Sect. 6 contains an updated list of the currently known families of commutative semifields in arbitrary odd characteristic. It is proved that P(m, q) has F q in its left nucleus and F q 2 in its middle nucleus. In the generic case P(m, q) is not isotopic to a previously known semifield with the possible exception of BH(p, m, s,β). In the remainder of the present section we prepare the construction of P(m, q). Let q be a power of an odd prime p and m = 2k + 1 > 1. Consider the fields F = F q 2m L = F q m, K = F q 2 F q. Let μ F q be a non-square and ω K such that ω 2 = μ. Denote g i (x) = x qi for i = 0,...,2m 1, x = g m (x) and let the trace tr : F L be tr(x) = x + x.

4 190 J. Bierbrauer Lemma 1 Let s : L L be defined by s(x) = x + g 2 (x). Thens(x) is invertible and its inverse is k k 1 l(x) = (1/2) ( 1) i g 2i (x) + ( 1) k+ j+1 g 2 j+1 (x). i=0 j=0 Proof Let l(x) = i a i g i (x).thena i +a i+2 = 0fori = m 2, whereas a m 2 +a 0 = 1. Lemma 2 We have ker(tr) = Lω and F = L Lω. Proof This follows from tr(ω) = ω ω = 0 and the fact that {1,ω} form a basis of F over L. We will use x = a + bω for a, b L as standard representation for an element x F and refer to F = L Lω as the canonical decomposition of F. Proposition 1 Define h : F Fby h(x) = k k 1 ( 1) i g 2i (x) + ( 1) k+ j g 2 j+1 (x). i=0 Then h((b + g 2 (b))ω) = 2bω for b L. In particular Lω ker(h) = 0. Proof This follows from Lemma 1 and the fact that g r (ω) = ( 1) r ω for r = 0,...,m 1. In fact we have h(bω) = 2l(b)ω where l is from Lemma 1 and therefore j=0 h((b + g 2 (b))ω) = 2l(b + g 2 (b))ω = 2bω. Observe that h(x) involves only half the Galois group Gal(F F q ), the automorphisms g 0 = id, g 1,...,g m 1. Also all we need to know about h(x) is its restriction to Lω. Definition 7 Define a function f : F F by f (x) = (1/2)tr(x 2 ) + (1/2)G(x 1+q2 ), where G(x) = h(x x) and h is the function from Proposition 1. Theorem 1 f (x) is a planar function defined on F. The corresponding commutative semifield product is x y = (1/2)tr(xy) + (1/4)G(xg 2 (y) + g 2 (x)y). The corresponding semifield P(m, q) has F q in its left nucleus and K in its middle nucleus. It satisfies u x = ux for u K. 2 Proof of Theorem 1 The commutative product x y is the polarization of the function f (x) from Definition 7. Let us show that x 1 = x for all x F. Write x = a + bω in standard form. Then (1/2)tr(x) = a. By definition G vanishes on L. It follows G(x + g 2 (x)) = G((b + g 2 (b))ω) = h(2(b + g 2 (b))ω) = 4bω

5 Commutative semifields from projection mappings 191 where the last equality follows from Proposition 1. It follows x 1 = x as claimed. Let now x = a + bω, y = c + dω such that x y = 0. It has to be shown that x = 0or y = 0. Observe that tr(xy) L and G(xg 2 (y) + g 2 (x)y) Lω. Because of the canonical decomposition we must have tr(xy) = G(xg 2 (y) + g 2 (x)y) = 0. The first condition is xy Lω. Letz = xg 2 (y) + g 2 (x)y. The second condition is G(z) = h(z z) = 0. This implies z z Lω ker(h) = 0, equivalently z L. As both conditions are homogeneous we can replace (x, y) by (λx,λy) for 0 = λ L. Without restriction a = 0ora = 1. Assume a = 0. Then d = 0 by the first condition. The second condition is bg 2 (c) + g 2 (b)c = 0. This is an Albert product in L. It follows that either b = 0orc = 0, hence x = 0ory = 0. Assume a = 1. The first condition is c = bdμ, the second is (g 2 (d)+d)(1 b q2 +1 μ) = 0. The second factor is = 0asμ is a non-square in L. The first factor shows d = 0which implies y = 0. We have shown that x y describes a commutative semifield. The general form of f (x) = x x shows that F q is in the kernel and K is in the middle nucleus, see [4]. Only one item of Theorem 1 remains to be proved: Proposition 2 Let u K.Then x u = xu for all x. Proof As F q is in the left nucleus and obviously satisfies our condition it suffices to prove it for u = ω.wehavex ω = (1/2)tr(ωx) + (1/4)G(z), where z = ω(x + g 2 (x)). It follows z z = (tr(x) + g 2 (tr(x))ω. Our defining condition shows G(z) = h(z z) = 2tr(x)ω and x ω = (1/2)ω(x x) + (1/4)2(x + x)ω = xω. 3 The symplectic version There is a canonical bijection between commutative semifields and symplectic semifields, see [2]. As the authors of [11] start from a symplectic semifield from which they derive its commutative version it may be of interest to give an explicit description of the symplectic semifield corresponding to the commutative semifield of Theorem 1. Definition 8 Let φ y (x) = 2m 1 r=0 b r g r (x) with b r F be the linearized polynomial such that φ y (x) = x y. Itsconjugate is φ y (x) = r g r (b 2m r )g r (x). Proposition 3 The coefficients b r of φ y are as follows: b 2i = (1/4)( 1) i (g 2i+2 (y) g 2i 2 (y)) for i = 0 and b 2 j+1 = (1/4)( 1) k+ j ((g 2 j+3 (y) g 2 j 1 (y)) for 2 j + 1 < m, b 2 j+1 = (1/4)( 1) k+ j+1 ((g 2 j+3 (y) g 2 j 1 (y)) for 2 j + 1 > m, whereas b 0 = y/2 + (1/4)(g 2 (y) + g 2m 2 (y)), b m = g m (y)/2 (1/4)(g m+2 (y) + g m 2 (y)). The coefficients of the conjugate are easy to compute: Lemma 3 Let φ y (x) = r b r g r (x). Then b 2i = (1/4)( 1) i+1 (g 2 (y) g 2 (y)), b 2 j+1 = (1/4)( 1) k+ j (g 2 (y) g 2 (y)) for 2 j + 1 > m, b 2 j+1 = (1/4)( 1) k+ j+1 (g 2 (y) g 2 (y)) for 2 j + 1 < m,

6 192 J. Bierbrauer for i = 0, 2 j + 1 = m, whereas b 0 = y/2 + (1/4)(g 2 (y) + g 2 (y)), b m = y/2 (1/4)(g 2 (y) + g 2 (y)), The corresponding symplectic semifield product is defined by x y = φ x (y) = r b r (x)g r (y). After reordering x y = y + g m(y) x + (1/4) { y g m (y) + α y 2 } +β y + γ y g2 (x) + (1/4) { } y g m (y) α y β y γ y g 2 (x) where α y = m 1 i=1 ( 1)i+1 g 2i (y), β y = k 1 j=0 ( 1)k+ j+1 g 2 j+1 (y), γ y = m 1 g 2 j+1 (y). Observe that the only conjugates of x that occur are x, g 2 (x), g 2 (x). j=k+1 ( 1)k+ j 4 The general structure The proof of Theorem 1 relied on the canonical decomposition F = L Lω and on the fact that tr(xy) = G(xg 2 (y) + g 2 (x)y) = 0 implies xy = 0. This can be generalized: Definition 9 Let F = F q r, f 1, f 2 : F F quadratic functions and x 1 y, x 2 y the corresponding polarizations. Let S i, A i be F q -subspaces such that A i S i F and S i contains all x i y. Then ( 1, S 1, A 1 ) and ( 2, S 2, A 2 ) are compatible if x 1 y A 1 and x 2 y A 2 imply xy = 0. Observe that x i y in Definition 9 satisfies the axioms of a commutative presemifield multiplication (see Definition 1) with the possible exception of the central axiom (x i y = 0 xy = 0). Proposition 4 Let the polarizations x 1 y and x 2 y of quadratic functions be defined on F = F q r such that ( 1, S 1, A 1 ) and ( 2, S 2, A 2 ) are compatible. Assume dim GF(q) (S 1 /A 1 )+ dim GF(q) (S 2 /A 2 ) r. Choose α i : S i F such that α i is F q -linear, K er(α i ) A i and α 1 (S 1 ) α 2 (S 2 ) = 0. Then defines a commutative presemifield on F. x y = α 1 (x 1 y) + α 2 (x 2 y) The proofof Proposition 4 is trivial. Observe that the assumption on the dimensions guarantees that α 1,α 2 can be chosen as required. The choice of α 1,α 2 is irrelevant as the resulting semifields are all isotopic. Observe that Definition 9 and Proposition 4 can be generalized in an obvious way to non-commutative presemifields. Again the products x i y will have to satisfy the axioms of a presemifield multiplication with the possible exception of the non-existence of zero divisors. In Theorem 1 we have S 1 = S 2 = F, r = 2m, x 1 y = xy (the field multiplication) and x 2 y = xg 2 (y) + g 2 (x)y, the polarization of f (x) = x q2 +1, corresponding to an Albert

7 Commutative semifields from projection mappings 193 presemifield. The core of the proof is the observation that ( 1, F, Lω) and ( 2, F, L) are compatible when r = 2m and m > 1 odd. In the next section we are going to see that the constructions of [5] are special cases of Proposition 4 as well. 5 The generalized Budaghyan Helleseth family In this section let L = F p m F=F p 2m for an odd prime p and tr, N : F L the trace and norm. Denote g i (x) = x pi. Also let x i y = xg i (y) + g i (x)y, the polarization of 2x pi +1. In the special case i = m we have x m y=tr(xy). In[5] the following types of functions are considered: Definition 10 Let tr(ω) = 0 and consider the canonical decomposition F=L Lω. Let h(x) = m 1 i=0 h i g i (x) be an invertible linear mapping : F F where h i L and k(x) = m 1 i=0 k i g i (x) be a linear mapping : F F where k i L.Lets be a positive integer and 0 = β 1,β 2,β 3 F such that β 3 / L. Consider the following functions f 1, f 2 : F F : and the corresponding polarizations f 1 (x) = (β 1 x) ps +1 (β 1 x) ps +1 + h(n(x)) (1) f 2 (x) = tr(β 2 x ps +1 ) + β 3 N(x) + k(n(x)) (2) 2(x 1 y) = β ps +1 p 1 (xg s (y) + g s (x)y) β s +1 1 xgs (y) + g s (x)y + h(x y + xy) (3) 2(x 2 y) = tr(β 2 (xg s (y) + g s (x)y)) + β 3 (x y + xy) + k(x y + xy)) (4) It is proved in [5] that f 1 (x), f 2 (x) are PN-functions under suitable conditions on the parameters. We simplify using strong isotopy, see Definition 5. Theorem 2 Each of the quadratic functions f 1 (x), f 2 (x) is strongly isotopic to a function f (x) = N(x) + tr(βx ps +1 )ω where ω = 0 is fixed such that tr(ω) = 0 and 0 = β F. Proof Start with the second family, f 2 (x) and x 2 y.letβ 3 = u + vω where v = 0. Let α be the invertible linear function defined by α(a + bω) = a (u/v)b + (b/v)ω where a, b L. Thenα(x 2 y) shows that we can assume β 3 = ω. Concatenation with another obvious linear mapping shows that without restriction k(x) is the 0 mapping. This leads to the form tr(β 2 (x s y)) + tr(x y)ω. Finally concatenation with a + bω b + aω leads to the standard form tr(x y) + tr(β 2 (x s y))ω. Consider the first family, f 1 (x) and x 1 y.usingα(x) = x/β 1 in (1) shows that we can assume β 1 = 1. The difference of the first two terms in (1) isinlω, the last term is in L. Concatenation of x 1 y with a linear mapping, which is the identity on Lω and acts as the inverse of h on L shows that we can assume h(x) to be the identity. Up to isotopy the first product has the form x 1 y = tr(x y) + x s y x s y As tr(x s y x s y) = 0, clearly this can be expressed as claimed choosing β = 1/ω.

8 194 J. Bierbrauer The first family is the special case of the second corresponding to the choice β = ω/μ, where μ = ω 2 = N(ω), inotherwordsβ = 1/ω. The families of PN-functions constructed in [5] are a special case of the following: Theorem 3 The following are equivalent: 1. The function f from Theorem 2 is planar. 2. (x m y, L, 0) is compatible with (β(x s y), F, Lω). 3. The following hold β pm 1 is not contained in the subgroup of order (p m + 1)/ gcd(p m + 1, p s + 1). There is no 0 = a F such that tr(a) = 0 and g s (a) = a. Proof The equivalence of the first two items is obvious. The rest of the proof follows the lines of the proofs of Theorems 1 and 2 in [5]. Assume there are nonzero elements x, y F such that tr(x y) = 0andβ(x s y) Lω. The second condition can be rewritten as xg s (y)(β β y (pm 1)(p s +1) ) = g s (x)y(β β y (pm 1)(p s +1) ). This is satisfied if the common factor vanishes. Assume this is not the case. Cancel this factor. Let a = x/y. The first condition is tr(a) = 0, the second says now g s (a) = a. Denote the generalized Budaghyan Helleseth presemifields obtained from Theorem 3 and Proposition 4 by BH(p, m, s,β). The first condition in item (3) is satisfied in particular when β F is a non-square. The first family of [5] corresponds to β = 1/ω and hence tr(β) = 0. Lemma 4 Let tr(β) = 0 in Theorem 3. We can choose without restriction β = 1/ω. Necessarily p m 1(mod 4). Proof Recall μ = ω 2 = N(ω) L. We can choose β = (1/μ)ω = 1/ω. The first condition of Theorem 3 is that (p m +1)/ gcd(p m +1, p s +1) must be odd. Assume p m 3(mod 4), Then p 3(mod 4) and the condition implies s odd. We can choose μ F p. This implies ω p = ωand g s (ω) = ( 1) s ω.ass is odd a contradiction to the second condition in 3 of Theorem 3 is obtained. In case tr(β) = 0 we give a more explicit description of the semifield multiplication. Theorem 4 Let L = F p m F = F p 2m for an odd prime p such that p m 1( mod 4). Let the integer s such that there is no 0 = a F such that tr(a) = 0 and g s (a) = a. Let 0 = ω such that tr(ω) = 0. Letμ = ω 2 L. Then the semifield multiplication corresponding to BH(p, m, s, 1/ω) is x y = (1/2)tr(x y) + (1/2)k(tr((x s y)/ω))ω where k(b) : L L is the inverse of b b + g s (b)μ (ps 1)/2. Proof Observe that the linear mapping b b + g s (b)μ (ps 1)/2 is indeed invertible on L as a zero b implies g s (bω) = bω contradicting the second condition if b = 0. It only needs to be checked that x 1 = x.letx = a + bω.then(1/2)tr(x) = a and x s 1 = x + g s (x) and tr((x s 1)/ω)) = (1/μ)tr((bω + g s (bω))ω) = 2b + 2g s (b)μ (ps 1)/2.

9 Commutative semifields from projection mappings 195 6Conclusion Here is an updated list of the families of known commutative semifields which exist in arbitrary odd characteristic: 1. The finite fields in odd characteristic. 2. The Dickson semifields [10]. 3. The Albert commutative twisted fields [1]. 4. The Zha Kyureghyan Wang semifields [12] (seealso[4]). 5. The family from [3]. 6. The generalized Budaghyan Helleseth semifields BH( p, m, s, β) of Theorem The semifields P(m, q) of Theorem 1 where q is an arbitrary odd prime-power and m > 1 odd. Proposition 5 The left nucleus of P(m, q) has odd dimension over F q. Proof Assume the left nucleus K has even dimension over F q. As it is contained in the middle nucleus it follows K K. It suffices to show that ω/ K. Recall that ω x = ωx for all x. We are assuming ω(x y) = (ωx) y. With z = xg 2 (y) + g 2 (x)y this means concretely (1/2)tr(xy)ω + (1/4)G(z)ω = (1/2)tr(xyω) + (1/4)G(zω). Collecting the terms in the direct summand Lω this implies tr(xy)ω = (1/2)G(zω). By definition G(zω) = h((z + z)ω) = h(tr(z)ω). Our assumption reads h(tr(z)ω) = 2tr(xy)ω. We know from Proposition 1 that the restriction of h to Lω is a bijective mapping : Lω Lω and its inverse maps 2bω (b+g 2 (b))ω for all b L. In our situation, with b = tr(xy), this shows tr(z) = tr(xy) + g 2 (tr(xy)). For x = y we have z = 2x 1+q2 and the above equation yields a polynomial identity which clearly is not satisfied. Corollary 1 Let m be prime. Then P(m, q) has left nucleus of order q and middle nucleus of order q 2. Corollary 2 The commutative semifields P(m, q) are not isotopic to commutative Albert semifields. They are not quadratic over the middle nucleus. Proof The Albert commutative semifields have all nuclei identical which by Proposition 5 is not the case for our semifields. Our semifields have F q 2 in their middle nucleus. If they were quadratic over the middle nucleus, the middle nucleus would be all of F. As the dimensions of the semifields from [12]and[3] are divisible by 3 and 4, respectively, it follows that in the generic case P(m, q) is not isotopic to any of those. The isotopy relations involving P(m, q) and BH(p, m, s,β)remain to be studied. We expect those families to be non-isotopic in general.

10 196 J. Bierbrauer References 1. Albert A.A.: On nonassociative division algebras. Trans. Am. Math. Soc. 72, (1952). 2. Ball S., Brown M.R.: The six semifield planes associated with a semifield flock. Adv. Math. 189, (2004). 3. Bierbrauer J.: New semifields, PN and APN functions. Des. Codes Cryptogr. 54, (2010). 4. Bierbrauer J.: New commutative semifields and their nuclei. In: Bras-Amorós M., Høholdt T. (eds.) Proceedings of AAECC-18, Lecture Notes in Computer Science, vol. 5527, pp Tarragona, Spain (2009). 5. Budaghyan L., Helleseth T.: New commutative semifields defined by new PN multinomials, Cryptography and Communications (to appear). 6. Budaghyan L., Helleseth T.: New perfect nonlinear multinomials over F p 2k for any odd prime p. In: Sequences and their applications-seta LNCS, vol. 5203, pp Springer, Heidelberg (2008). 7. Coulter R.S., Henderson M.: Commutative presemifields and semifields. Adv. Math. 217, (2008). 8. Coulter R.S., Henderson M., Kosick P.: Planar polynomials for commutative semifields with specified nuclei. Des. Codes Cryptogr. 44, (2007). 9. Dembowski P., Ostrom T.G.: Planes of order n with collineation groups of order n 2.Math.Z.103, (1968). 10. Dickson L.E.: On commutative linear algebras in which division is always uniquely possible. Trans. Am. Mat. Soc 7, (1906). 11. Lunardon G., Marino G., Polverino O., Trombetti R.: Symplectic spreads and quadric veroneseans (manuscript) 12. Zha Z., Kyureghyan G.M., Wang X.: Perfect nonlinear binomials and their semifields. Finite Fields Appl. 15, (2009).

COMMUTATIVE SEMIFIELDS OF ORDER 243 AND 3125

COMMUTATIVE SEMIFIELDS OF ORDER 243 AND 3125 ROBERT S. COULTER AND PAMELA KOSICK Abstract. This note summarises a recent search for commutative semifields of order 243 and 3125. For each of these two orders,