Advances in Mathematics of Communications doi:10.3934/amc.2016006 Volume 10, No. 2, 2016, 291–306

CONSTRUCTING COMMUTATIVE SEMIFIELDS OF SQUARE ORDER

Stephen M. Gagola III Department of Algebra Charles University Sokolovsk´a83, 186 75 Prague, Czech Republic

Joanne L. Hall∗ School of Mathematical Sciences Queensland University of Technology Brisbane, QLD, Australia

(Communicated by Tor Helleseth)

Abstract. The projection construction has been used to construct semifields of odd characteristic using a field and a twisted semifield [Commutative semi- fields from projection mappings, Designs, Codes and Cryptography, 61 (2011), 187–196]. We generalize this idea to a projection construction using two twist- ed semifields to construct semifields of odd characteristic. Planar functions and semifields have a strong connection so this also constructs new planar functions.

1. Introduction Semifields are algebraic structures satisfying most of the axioms of a field. The classification of finite fields has been concluded over a century ago, however the classification of finite semifields is far from complete. Semifields have connections with geometry [8, 16], and a connection with planar functions means that semifields have applications in classical cryptographic systems [10], quantum cryptographic systems [19], wireless communication [12], and coding theory [14]. Commutative semifields with odd characteristic are equivalent to those planar functions that are known as Dembowski-Ostrom Polynomials (DO polynomials) [8]. Planar functions belong to the larger class of highly nonlinear functions which are of use in the above mentioned applications as well as being of theoretical interest [7,9]. A look at a recent list of planar functions and semifields [5, 21] shows no obvious pattern. Computational searches can discover new planar functions and semifields [15], but algebraic work is required to significantly deepen our understanding. The motivation behind this work is to generalize the projection construction for planar functions of Bierbrauer [2]. We use the trace map with Dembowski-Ostrom polynomials which has also been used to explore Dembowski-Ostrom polynomials that are permutations [3].

2010 Mathematics Subject Classification: Primary: 94A60, 12K10; Secondary: 51E15, 05B25, 51A40. Key words and phrases: Planar function, Dembowski-Ostrom polynomial, trace, commutative semifield, projective plane. ∗ Corresponding author.

291 c 2016 AIMS 292 Stephen M. Gagola III and Joanne L. Hall

2r Let p be an odd prime. Suppose that Sg and Sh are semifields of order p associated with planar functions g and h respectively. Here Sg and Sh are isotopic to presemifields with multiplication operations

x ◦ y = g(x + y) − g(x) − g(y) and x4y = h(x + y) − h(x) − h(y) respectively.

Now we consider a new operation

(1) x ∗ y = x ◦ y + (x ◦ y) + x4y − (x4y)

r wherex ¯ = xp . Note that

x ∗ y = f(x + y) − f(x) − f(y) where

r r (2) f(x) = g(x) + g(x)p + h(x) − h(x)p .

Thus, if f(x) is also planar over Fp2r , then the multiplication operation in Equa- tion (1) forms another presemifield, called the projection construction. Equation (2) has been used to construct a planar function (and hence a semifield) by Bierbrauer [2]. The contrast with the current work is that in [2] the commutative semifield is associated with a planar function of the form (2) where g(x) = x2. Whereas, in this paper, we allow both g(x) and h(x) to be a wider variety of polynomials. We investigate the conditions on the polynomials g(x) and h(x) that result in the function f(x) being planar over Fp2r . Since we are primarily interested in constructing semifields, we have restricted our search for planar functions of the shape of Equation (2), to DO polynomials. We begin by showing that if g(x) and h(x) are planar Dembowski-Ostrom mono- 2 2 mials over Fp2r with either g(x) = x or h(x) = x , then f(x), as defined in Equation (2), is also planar over Fp2r . In Section3, we show that if one of the semifields Sg or Sh is a field and the other is a commutative semifield, then the semifield obtained using the projection construction (as in Equation (1)) is already known. The planar functions that these correspond to have either g(x) = x2 or h(x) = x2, where the other, namely g(x) or h(x), is another planar DO polynomial. Several known families of semifields are shown to also fit the projection construction. Section4 presents results on determining the middle and left nuclei of commu- tative semifields constructed using this projection construction. In Section5, some computationally derived semifields/planar functions are pre- sented. Computations of the sizes of the middle nuclei are used to show that these commutative semifields are new and not isotopic to previously known semifields.

2. Preliminaries Definition 2.1. A semifield S is an algebraic structure with two binary operations, addition, +, and multiplication, ∗, such that • (S, +) is an abelian group, • (S \{0}, ∗) is a loop, and • multiplication is distributive on both the left and right.

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A presemifield P is similar to a semifield, but with one of the axioms relaxed. Namely, (P \{0}, ∗) does not necessarily have a multiplicative identity and is there- fore a quasigroup. A presemifield is commutative if its multiplication is commuta- tive. Semifields can be non-commutative and non-associative but, by Wedderburn’s Theorem [20], if a semifield is finite, then associativity implies commutativity. Giv- en a commutative presemifield with multiplication x ∗ y it is easy to construct a commutative semifield. Lemma 2.2 ([1]). Every quasigroup (Q, ∗) is isotopic to a loop. For a, b ∈ Q, (Q, ∗) is isotopic to (Q, ◦) where

−1 −1 x ◦ y = (x)Rb ∗ (y)La ⇐⇒ (x ∗ b) ◦ (a ∗ y) = x ∗ y. Here (a ∗ b) ◦ x = x = x ◦ (a ∗ b) for any x ∈ Q.

Two presemifields S1 = (S, +, ∗) and S2 = (S, +, ◦) are called isotopic if there exist three linearized permutation polynomials L1,L2,L3 over S such that L1(x) ◦ L2(y) = L3(x ∗ y) for any x, y ∈ S. Let S be a semifield. The subsets

Nl(S) = {a ∈ S | a(xy) = (ax)y for all x, y ∈ S} , Nm(S) = {a ∈ S | x(ay) = (xa)y for all x, y ∈ S} , Nr(S) = {a ∈ S | x(ya) = (xy)a for all x, y ∈ S} , are called the left, middle and right nucleus of S, respectively. These sets are finite fields [16]. The intersection N(S) = Nl(S) ∩ Nm(S) ∩ Nr(S) is called the nucleus of S. In the case where S is commutative it can easily be shown that N(S) = Nl(S) = Nr(S) ⊆ Nm(S). The size of the nuclei is invariant under isotopism, a fact which is used in Section5 to show non isotopism of some semifields. We will be using polynomials and on finite fields to explore semifields. Let

Df (x, a) = f(x + a) − f(x) − f(a).

Definition 2.3. Let Fpr be a field of characteristic p. A function f : Fpr → Fpr is ∗ called a planar function if for every a ∈ Fpr the map x 7→ Df (x, a) is a bijection. Note that the definition of planar function does not require p to be odd, however planar functions cannot exist on fields of even characteristic [11], thus the focus of this work is on polynomials on fields of odd characteristic. A polynomial f(x) ∈ Fpr [x] is a Dembowski-Ostrom polynomial if its reduced form has the shape k X pi+pj f(x) = aijx . i,j=0 pr Any polynomial f(x) ∈ Fpr [x] may be reduced modulo x − x, which yields a r polynomial function of degree less than p that induces the same function on Fpr . A polynomial which is planar over Fpr and also planar when restricted to the subfield Fp is planar over every subfield.

Lemma 2.4 ([8]). Let f(x) be a DO polynomial on Fpr . Then f(x) is planar if ∗ and only if Df (x, y) 6= 0 for all x, y ∈ Fpr . Theorem 2.5 ([8]). Let p be an odd prime. Each equivalence class of DO planar polynomials on Fpr constructs a unique isotopy class of commutative semifields of

Advances in Mathematics of Communications Volume 10, No. 2 (2016), 291–306 294 Stephen M. Gagola III and Joanne L. Hall order pr. Furthermore all commutative semifields of order pr can be constructed from a planar DO polynomial on Fpr . Theorem 2.5 shows that commutative semifields can be investigated by investigat- ing planar DO polynomials. Notions of equivalence of polynomials are of importance in applications [6]. Planar polynomials Π and Π0 with

0 Π (x) = (L1 ◦ Π ◦ L2)(x) + L3(x) are said to be EA-equivalent if L1(x), L2(x), L3(x) are affine polynomials and L1(x), L2(x) are bijections [17]; and linear equivalent if L1(x), L2(x), L3(x) are linearized polynomials and L1(x),L2(x) are bijections. CCZ equivalence [6] is of interest in cryptographic applications. For planar Dembowski-Ostrom polynomials CCZ, EA and linear equivalence are equivalent [4]. It has been shown that linear equivalent planar functions construct isotopic semi- fields [8]. The proof of Theorem 2.5 is constructive: let x, y ∈ Fq and ∗ be multi- plication on a presemifield of order q, then there exists a planar DO polynomial f such that 0 x ∗ y = Df (x, y) forms a presemifield isotopic to (Fq, +, ∗). So in showing the non-equivalence of planar DO polynomials it is sufficient to show that their corresponding semifields are non-isotopic. There are several known planar functions that fit the shape of Equation (2) for which g(x) and h(x) are different (see Section3). The following theorem establishes necessary and sufficient conditions on g(x) and h(x) to form a planar function using Equation (2).

Theorem 2.6. Suppose p is an odd prime. Let g(x) and h(x) be functions on Fp2r and let r r f(x) = g(x) + (g(x))p + h(x) − (h(x))p , ∗ then the polynomial f(x) is planar if and only if, for any a, b ∈ Fp2r either pr Dg(a, b) + (Dg(a, b)) 6= 0 or Dh(a, b) 6∈ Fpr .

Proof. Since Fpr is a subfield of Fp2r , the trace function pr T r(x) = T r r (x) = x + x Fp2r /Fp can be used in rewriting f(x):

r f(x) = T r (g(x)) + h(x) − (h(x))p .

By Lemma 2.4, the polynomial f(x) is planar if and only if Df (a, b) 6= 0 for all ∗ a, b ∈ Fp2r . Now assume that Df (a, b) = 0. Then

pr T r (Dg(a, b)) + Dh(a, b) − (Dh(a, b)) = 0(3) pr  pr  =⇒ T r (Dg(a, b)) + Dh(a, b) − (Dh(a, b)) = 0(4)

pr (5) =⇒ T r (Dg(a, b)) + (Dh(a, b)) − Dh(a, b) = 0.

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By adding Equations (3) and (5) it follows that Df (a, b) = 0 if and only if

T r (Dg(a, b)) = 0 and(6) pr (7) Dh(a, b) − (Dh(a, b)) = 0. ∗ Thus, f(x) is planar if and only if, for any a, b ∈ Fp2r , either Equation (6) or Equation (7) does not hold.

3. Known planar functions that fit the projection construction The LMPTB functions [2] fit the shape of the projection construction; they are easily recognizable as being of the form of Equation (2).

1 2 Theorem 3.1 ([2, Theorem 1]). Let g(x) = 2 x and k k−1 1 X 2 2i 1 X 2 2j+1 (8) h(x) = (−1)ix(1+p )p + (−1)k+jx(1+p )p , 2 2 i=0 j=0 then the function f(x) as defined in Equation (2) is a planar function on Fp2(2k+1) for k > 0. The LMPTB semifields are an example of the projection construction as de- scribed in [2]. The projection construction can be used to construct a new semifield from two or more semifields. The polynomials that are used to construct a planar polynomial in Equation (2) are not always planar polynomials (See Example1). Hence, the semifields that are derived from the planar polynomials of the shape of Equation (2) may not be constructible from other semifields. The LMPTB semifields are isotopic to the Budaghyan-Helleseth semifields [18]. Under certain conditions Equation (2) contains generalized Budaghyan-Helleseth polynomials with g(x) = x2, and under other conditions, contains planar functions with h(x) = x2 which have corresponding semifields that are isotopic to commuta- tive semifields constructed by Zhou & Pott [21]. Theorem 3.2. Let g(x) and h(x) be planar Dembowski-Ostrom monomials over 2 2 Fp2r . If either g(x) = x or h(x) = x , then r r f(x) = g(x) + g(x)p + h(x) − h(x)p is planar over Fp2r . The following lemma will be used to prove Theorem 3.2.

pi+pk Lemma 3.3 ([9, Theorem 3.3]). Let h(x) = x ∈ Fpn [x] with 0 ≤ i < k, then the following are equivalent: (i) h(x) is planar; ∗ (ii) Dh(x, a) 6= 0 for all x, a ∈ Fpn ; k i p −p ∗ (iii) hb i is odd for any b ∈ Fpn ; (iv) n/gcd(n, i − k) is odd. Proof of Theorem 3.2. Assume that f(x) is not planar. Then there exist elements ∗ ∗ x, a ∈ Fp2r such that Df (x, a) = 0. Thus, from Theorem 2.6, there exist x, a ∈ Fp2r such that T r (Dg(x, a)) = 0 and Dh(x, a) ∈ Fpr . Thus,  pr −1 pr (9) Dg(x, a) 1 + Dg(x, a) = Dg(x, a) + Dg(x, a) = 0.

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pr −1 By Lemma 2.4, Dg(x, a) 6= 0, hence Equation (9) implies that Dg(x, a) = −1. ∗ r Let G be the unique subgroup of Fp2r such that |G| = 2 Fpr = 2 (p − 1). Note that n r o p −1 G = x ∈ Fp2r x = ±1 . pr −1 ∗ Since Dg(x, a) = −1, Dg(x, a) ∈ G \ Fpr . Now let c = xa and note that i k k i i i  k i k i  xp ap + xp ap = xp ap ap −p + xp −p ,

i  k i i k k i  (10) = cp cp −p xp −p + xp −p .

pi+pk 2 pi pk pk pi Case 1. Suppose g(x) = x and h(x) = x . Since Dg(x, a) = x a + x a ∈ ∗ pk−pi pi−pk pk−pi G \ Fpr and Dh(x, a) = 2ax = 2c ∈ Fpr , by Equation (10), c x + x is ∗ contained in G \ Fpr . 2 pi+pk ∗ Case 2. Suppose g(x) = x and h(x) = x . Since Dg(x, a) = 2ax = 2c ∈ G\Fpr pi pk pk pi pk−pi pi−pk pk−pi and Dh(x, a) = x a + x a ∈ Fpr , by Equation (10), c x + x ∈ ∗ G \ Fpr . pk−pi pi−pk pk−pi ∗ Therefore, in either case the sum (c x + x ) is contained in G \ Fpr . pk−pi pk−pi ∗ By Lemma 3.3, both c and x are of odd order. Since Fpr ≤ G with |G| = ∗ pk−pi pk−pi ∗ pk−pi pi−pk pk−pi ∗ 2|Fpr |, we know that c , x ∈ Fpr . Therefore, (c x +x ) ∈ Fpr ∗ forming a contradiction. Hence, there do not exist x, a ∈ Fp2r such that Df (x, a) = 0 and therefore, f(x) is a planar function. We now show that all of the polynomials described in Theorem 3.2 are known.

2 pk+1 Theorem 3.4. Let h(x) = x ∈ Fp2r [x], g(x) = x be a planar function on Fp2r , and f(x) be of the form (2), then the semifield associated with f(x) is isotopic to the Zhou-Pott commutative semifield, Sk,id [21].

pk+1 (pk+1)pr 2 2pr Proof. For f(x) = x + x + x − x ∈ Fp2r [x], the binary operation on the corresponding presemifield is r k k  k k p r r x ∗ y = xp y + xyp + xp y + xyp + 2xy − 2xp yp .

∗ pr Let ω ∈ Fp2r such that ω = −ω and denote any x ∈ Fp2r by x = x1 + x2ω for pr pr x1, x2 ∈ Fpr . Then, since x = (x1 + x2ω) = x1 − x2ω,

x ∗ y = (x1 + x2ω) ∗ (y1 + y2ω), pk pk pk pk = 2x1 y1 + 2x1y1 + 2(x2ω) (y2ω) + 2(x2ω)(y2ω)

+ 4x1(y2ω) + 4(x2ω)y1.

By [21, Theorem 1], the Zhou-Pott presemifield Pk,σ = (Fp2r , +, ◦) has multipli- cation

x ◦ y = (x1 + x2ω) ◦ (y1 + y2ω), σ pk pk  pk pk  (11) = x1 y1 + x1y1 + α x2 y2 + x2y2 + (x1y2 + x2y1)ω, where α is a non-square element in Fpr and σ is a field automorphism of Fpr . Let k ωp +1 2r k α = 4 . Since gcd(2r,k) is odd, k is even, and therefore, p + 1 is not divisible by

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k r pk+1 p +1 k p 2 p +1 four. Since ω = −ω and 2 is odd, ω ∈/ Fpr . Hence, ω is a non-square element of Fpr , hence α is non-square. Let σ be the identity map, then Equation (11) becomes

pk pk 1 pk 1 pk (x1 + x2ω) ◦ (y1 + y2ω) = x1 y1 + x1y1 + 4 (x2ω) (y2ω) + 4 (x2ω)(y2ω) + x1(y2ω) + (x2ω)y1.

Let L1 and L2 be linearized permutation polynomials over Fp2r where L1(x) = pr 3 1 pr 3x−x and L2(x) = 2 x− 2 x . To show that L1 is a permutation polynomial it is r r enough to show that the only solution to 3x − xp = 0, is x = 0. If 3x = xp , then r pr −1 pr −1
p pr −1 pr −1 pr +1 2 x = x , and thus, x ∈ Fpr . Since x , x ∈ Fpr , then x is pr 2 2 also contained in Fpr . By squaring both sides of 3x = x , it follows that 9x = x . Hence, 8x2 = 0 and x = 0. Since

L1(x1 + x2ω) ◦ L2(y1 + y2ω) = (2x1 + 4x2ω) ◦ (y1 + 2y2ω), pk pk 8 pk = 2x1 y1 + 2x1y1 + 4 (x2ω) (y2ω) 8 pk + 4 (x2ω)(y2ω) + 4x1(y2ω) + 4(x2ω)y1, = (x1 + x2ω) ∗ (y1 + y2ω),

pk+1 (pk+1)pr 2 2pr the presemifield associated with f(x)=x +x +x −x is isotopic to Sk,id.

The generalized Budaghyan-Helleseth functions require some rearranging to show they can be constructed using Equation (2). From [2, Theorem 3], r s r s r (12) B(x) = xp +1 + ωβxp +1 + ωβp x(p +1)p ps ∗ pr −1 where T r(ω) = 0, x 6= −x for all x ∈ Fp2r and β is not in the subgroup of order (pr + 1)/gcd(pr + 1, ps + 1). Their corresponding presemifields are usually denoted by BH(p, m, s, β). Remark 1. Note that the generalized Budaghyan-Helleseth functions can be writ- ten as r r s  s p B(x) = xp +1 + ωβxp +1 − ωβx(p +1)

r r since T r(ω) = 0. Thus, B(x) + B(x)p = 2xp +1.

2 pk+pi Theorem 3.5. Let g(x) = x ∈ Fp2r [x], h(x) = x be a planar function on Fp2r and f(x) be of the form (2). If either r is even or p ≡ 1 (mod 4), then f(x) is equivalent to a generalized Budaghyan-Helleseth function. Proof. Note that since either r is even or p ≡ 1 (mod 4), there exists an element 2 pk+pi j ∈ Fpr such that j = −1. Also note that since k − i is even, j = −1 and r pk pi p pr ∗ j = j . Let L1(x) = c0x + jc0 x with c0 ∈ Fp2r . Then pr pr +1 pr +1 pk+pi pk+pi  pk+pi pk+pi  f(L1(x)) = 4jc0 x + 2c0 x − 2c0 x . 1 Case 1. Suppose i = 0 and L2(x) = pr +1 x. Then 4jc0 r k r k r p r p −p k  p −p k  p +1 c0 p +1 c0 p +1 L2(f(L1(x))) = x + 2j x − 2j x .

Advances in Mathematics of Communications Volume 10, No. 2 (2016), 291–306 298 Stephen M. Gagola III and Joanne L. Hall

1 pr Case 2. Suppose i = r and L2(x) = pr +1 x . Then 4jc0 r k+r r k+r r p r p −p k+r  p −p k+r  p +1 c0 p +1 c0 p +1 L2(f(L1(x))) = x + 2j x − 2j x .

1 1 pr Case 3. Now suppose that i∈ / {0, r} and L2(x) = pr +1 x + pr +1 x 8jc0 8jc0 p2r−i pr−i +d2r−ix − d2r−ix where d2r−i ∈ Fpr . Note that if r is odd and p ≡ 1 (mod 4), then the Budaghyan-Helleseth semifields belong to the class of semifields Sk as defined in Theorem 3.4([21]). 2 pk+1 Theorem 3.6. Let g(x) = x ∈ Fp2r [x], h(x) = x a planar function on Fp2r and f(x) be of the form (2). If p ≡ 3 (mod 4) and r is odd, then the semifield associated with f(x) is isotopic to a Zhou-Pott semifield, Sk,id. ∗ 4 k Proof. Let β ∈ Fp2r such that β = −1. Since k is even and p ≡ 3 (mod 4), p +1 ≡ k pr 2 (mod 8). Thus, βp +1 = β2. Furthermore, since β2 has order four, β2
= −β2. Therefore, r k k r (βx)2 + (βx)2p + (βx)p +1 − (βx)(p +1)p

r r k r k r = β2x2 + (β2)p xp + β2xp +1 − (β2)p x(p +1)p

r k k r = β2x2 − β2xp + β2xp +1 + β2x(p +1)p

 k k r r  = β2 xp +1 + x(p +1)p + x2 − xp

k k r r and f(x) is equivalent to xp +1 + x(p +1)p + x2 − x2p . Hence, by Theorem 3.4, the semifield corresponding to f(x) is isotopic to a Zhou-Pott semifield, Sk,id. Theorems 3.4, 3.5, and 3.6, show that all of the planar functions described by Theorem 3.2 have semifields that are isotopic to previously known semifields.

4. Nuclei of semifields In this section we explore some results on the nuclei of semifields which can be constructed using the projection construction of Equation (1). 4.1. Middle nuclei. The following lemma is explicit but important. Lemma 4.1. Let f(x) be a planar function over a finite field F. If the middle nucle- us of the commutative semifield associated with f is F, then the semifield associated with f is F, and hence, f(x) is EA-equivalent to x2 ∈ F[x]. In order to determine the middle nuclei of semifields from the projection con- struction we will need the following proposition.

n n r m m r Proposition 1. Suppose f(x) = xp +1 + x(p +1)p + xp +1 − x(p +1)p is planar pm+1 2 over Fp2r . If x is planar and f(x) is EA-equivalent to x over Fp2r , then m xp +1 = x2 and n ≡ 0 (mod r).

p p2 p2r−1 p Proof. Let L1(x) = c0x+c1x +c2x +···+c2r−1x and let L2(x) = d0x+d1x + p2 p2r−1 2 d2x + ··· + d2r−1x be permutation polynomials such that L1(x) = L2(f(x)). Furthermore, let S1 be a commutative semifield associated with f(x) and S2 be a 2 commutative semifield associated with f(x)|Fpr . Since f(x) is EA-equivalent to x 2r over Fp , the nucleus of S1 is S1. Thus, the nucleus of S2 is S2 and f(x)|Fpr is

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2 pn+1 r EA-equivalent to x over Fp . Since f(x)|Fpr = 2x , n ≡ 0 (mod r).

Case 1. Suppose n ≡ 0 (mod 2r). Note that 2r−1 2 X 2 2pj X pj +p` (13) (L1(x)) = cj x + 2cjc`x j=0 0≤j<`<2r and

L2(f(x)) r−1 X  2pj 2pj+r   (pm+1)pj (pm+1)pj+r  = (dj + dj+r) x + x + (dj − dj+r) x − x . j=0 pm+1 Since x is planar over Fp2r , m 6≡ r (mod 2r). Therefore, 2cjcj+r = 0 for all pm+1 2 0 ≤ j < r. So for every 0 ≤ j < r either cj = 0 or cj+r = 0. If x 6= x , then 2 2 cj = (dj + dj+r) = cj+r for all 0 ≤ j < r. Hence, cj = cj+r = 0 for any 0 ≤ j < r pm+1 2 and L1(x) = 0. By contradiction, x = x .

Case 2. Suppose n ≡ r (mod 2r). Then r−1 X (pr +1)pj  (pm+1)pj (pm+1)pj+r  L2(f(x)) = 2(dj + dj+r)x + (dj − dj+r) x − x . j=0 pm+1 2 2 If x 6= x , then cj = 0 for all 0 ≤ j < 2r and L1(x) = 0. By contradiction, m xp +1 = x2. The following lemma will be also needed. Lemma 4.2. Suppose r and ` are integers such that `|r. If x is an element of a 2` r pr p` field, such that x = x and 2 - ` , then x = x . p` r pr−` pr  pr−`  p` Proof. Since 2 - ` , (2`) | (r − `) and x = x. Hence, x = x = x .

n n r m m r Proposition 2. Suppose f(x) = xp +1 + x(p +1)p + xp +1 − x(p +1)p is planar k over Fp2r . Let p be the order of the middle nucleus of S, a commutative semifield pm+1 associated with f. If x is planar over Fp2r , then gcd(2n, m, 2r) | k. Proof. The binary operation on the corresponding presemifield is

n n r n+r n+r r x ∗ y = xyp + xp y + xp yp + xp yp

m m r m+r m+r r + xyp + xp y − xp yp − xp yp .

n m n+r m+r With L(x) = 1 ∗ x = 2x + xp + xp + xp − xp the multiplication ◦ of the corresponding commutative semifield is defined by (x ∗ 1) ◦ (1 ∗ y) = x ∗ y ⇐⇒ L(x) ◦ L(y) = x ∗ y (14) ⇐⇒ x ◦ y = L−1(x) ∗ L−1(y) .

Let t =gcd(2n, m, 2r) and a ∈ Fpt . By definition, L(a) is contained in the middle nucleus if and only if (x ◦ L(a)) ◦ y = x ◦ (L(a) ◦ y) for all x, y ∈ S. Since L is a permutation and ◦ is commutative, L(a) is contained in the middle nucleus if and

Advances in Mathematics of Communications Volume 10, No. 2 (2016), 291–306 300 Stephen M. Gagola III and Joanne L. Hall only if (L(x) ◦ L(a)) ◦ L(y) = (L(y) ◦ L(a)) ◦ L(x) for all x, y ∈ S. From Equation (14), this is equivalent to (15) L−1(x ∗ a) ∗ y = L−1(y ∗ a) ∗ x . Note that r n r n+r L(x) + L(x)p = 2x + 2xp + 2xp + 2xp

r m r m+r and L(x) − L(x)p = 2x + 2xp − 2xp − 2xp . Thus, since L(x) is additive,

r  n r n+r  (16) x + xp = 2L−1 x + xp + xp + xp

r  m r m+r  (17) and x − xp = 2L−1 x + xp − xp − xp .

2n m Hence, from Equations (16) and (17) along with the facts that ap = a and ap = a it follows that  n n r n+r n+r r L−1(x ∗ a) = L−1 xap + xp a + xp ap + xp ap

m m r m+r m+r r  +xap + xp a − xp ap − xp ap

n n r r (xap ) + (xap )p (xa) − (xa)p (18) = + . 2 2 pm+1 2r 2r Since x is planar over the field Fp , 2 - gcd(m,2r) . Likewise, since f(x)|Fpr = pn+1 r 2x is planar over Fpr (but not necessarily planar over Fp2r ), 2 - gcd(n,r) . In other words, the largest power of 2 dividing 2r also divides m and the largest power of 2 dividing r also divides n. Thus, either gcd(n, m, 2r) = gcd(2n, m, 2r) or the greatest power of 2 dividing n equals the greatest power of 2 dividing r. If gcd(n, m, 2r) = gcd(n,m,2r) n gcd(2n, m, 2r), then ap = a, and therefore, ap = a. Otherwise, if the great- r est power of 2 dividing n equals the greatest power of 2 dividing r, then 2 m - gcd(n, 2 ,r) n gcd(n,m/2,r) r n n r and, by Lemma 4.2, ap = ap = ap . So either ap = a or ap = ap .

n Case 1. Suppose ap = a. Then from Equation (18) L−1(x ∗ a) = xa and L−1(x ∗ a) ∗ y = (xa) ∗ y = (ya) ∗ x = L−1(y ∗ a) ∗ x. n r Case 2. Suppose ap = ap . Then from Equation (18) r r r xa + ap
+ xp a − ap
L−1(x ∗ a) = 2 and pr pr pr −1 x(a+a )+x (a−a ) L (x ∗ a) ∗ y = 2 ∗ y r r r y(a+ap )+yp (a−ap ) (19) = 2 ∗ x = L−1(y ∗ a) ∗ x.

Since Equation (15) is satisfied for any x, y ∈ Fp2r and any a ∈ Fpt , the field Fpt is contained in the middle nucleus of the commutative semifield associated with f.

Advances in Mathematics of Communications Volume 10, No. 2 (2016), 291–306 Constructing commutative semifields 301

n n r m m r Theorem 4.3. Suppose f(x) = xp +1 + x(p +1)p + xp +1 − x(p +1)p is planar pm+1 over Fp2r . If x is planar over Fp2r , then the middle nucleus of a commutative semifield associated with f has order equal to pgcd(2n,m,2r).

Proof. First note that f(x) is a planar function over any subfield of Fp2r . Let t = gcd(2n, m, 2r) and pk be the order of the middle nucleus. By Proposition2, pm+1 t | k. Assume that Fpt  Fpk ≤ Fp2r . Note that since x is planar over Fp2r , m 2 | m. Thus, 2 | t, and therefore, k = 2` for some integer `, namely, ` = gcd(n, 2 , r). pn+1 r r r Since f(x)|Fpr = 2x is planar over Fp , 2 - gcd(n,r) , and therefore, 2 - ` . By r ` Lemma 4.2, if b ∈ Fp2` , then b = b . Thus, n n r m m r f(x)| = xp +1 + x(p +1)p + xp +1 − x(p +1)p Fp2` n n ` m m ` = xp +1 + x(p +1)p + xp +1 − x(p +1)p .

Since f(x) is planar over Fp2` , the middle nucleus of the commutative semifield associated with f(x)| is k = 2` . By Lemma 4.1, f(x) is EA-equivalent to Fp2` Fp Fp 2 pm+1 2 x over Fpk . But, by Proposition1, this is not possible unless x = x over Fp2` and n ≡ 0 (mod `). Hence, (2`) | m and ` | n. In other words, k | m and k | (2n) contradicting the assumption that Fpt  Fpk . 4.2. Left nuclei. The following proposition can be used to calculate the left nuclei of some semifields constructed using the projection construction

n n 3n 2n 2n 3n Lemma 4.4. Suppose f(x) = xp +1 + x(p +1)p + xp +1 − x(p +1)p is planar over Fp6n . If ∗ is the binary operation of the corresponding presemifield and L(x) = n 2n 4n 5n x ∗ 1 = 2x + xp + xp + xp − xp , then −1 1 1 pn 1 p5n L (x) = 4 x − 4 x + 4 x . 1 1 pn 1 p5n Proof. Let F (x) = 4 x − 4 x + 4 x ∈ Fp6n [x]. Then 1 1 pn 1 p2n 1 p4n 1 p5n F (L(x)) = 2 x + 4 x + 4 x + 4 x − 4 x 1 1 pn 1 p2n 1 p3n 1 p5n + 4 x − 2 x − 4 x − 4 x − 4 x 1 1 pn 1 p3n 1 p4n 1 p5n + 4 x + 4 x + 4 x − 4 x + 2 x , = x.

n n 3n 2n 2n 3n Theorem 4.5. Suppose f(x) = xp +1 + x(p +1)p + xp +1 − x(p +1)p is planar over Fp6n . If S is its corresponding semifield, then the left nucleus of S is equal to the middle nucleus and has order of p2n.

Proof. By Theorem 4.3, the middle nucleus of S is Fp2n . Since the left nucleus of S is a subfield of the middle nucleus, it is enough to show that any element of Fp2n is pn p3n contained in the left nucleus. Let a ∈ Fp2n . Since a = a , we are in case 2 of the proof of Proposition2. Thus, by using Equation (19) with m = 2n, r = 3n and 3n n ap = ap , n 3n 4n 3n   3n 3n 3n xp a+ap +xp ap −a x a+ap +xp a−ap −1 pn L (x ∗ a) ∗ y = 2 y + 2 y 4n 3n n 3n 3n 3n  3n  xp a+ap +xp a−ap xp a+ap +x ap −a p3n p4n + 2 y + 2 y 2n 3n 5n 3n  3n 3n 3n xp a+ap +xp a−ap x a+ap +xp a−ap p2n + 2 y + 2 y

Advances in Mathematics of Communications Volume 10, No. 2 (2016), 291–306 302 Stephen M. Gagola III and Joanne L. Hall

5n 3n 2n 3n  3n 3n  3n  xp a+ap +xp ap −a xp a+ap +x ap −a p3n p5n − 2 y − 2 y n n 4n n 2n n 5n n xp (a+ap )+xp (ap −a) xp (a+ap )+xp (a−ap ) = 2 y + 2 y pn p3n pn pn p3n pn x(a+a )+x (a−a ) pn x(a+a )+x (a−a ) p2n + 2 y + 2 y p4n pn pn pn p5n pn p2n pn x (a+a )+x (a−a ) p3n x (−a−a )+x (a−a ) p3n + 2 y + 2 y p3n pn pn p3n pn pn x (a+a )+x(a −a) p4n x (−a−a )+x(a−a ) p5n (20) + 2 y + 2 y

for any x, y ∈ Fp6n . −1 1 1 pn 1 p5n By Lemma 4.4, L (x) = 4 x − 4 x + 4 x . Therefore, since n n 3n 4n 4n 3n 2n 2n 3n 5n 5n 3n x ∗ y = xyp +xp y+xp yp +xp yp +xyp +xp y−xp yp −xp yp ,

n −1  pn p2n p4n p5n y  p2n p3n p5n yp L (x ∗ y) = x +x +x +x 4 + x−x −x +x 4 2n 3n  pn p3n p4n yp  pn p2n p4n p5n yp + x−x +x −x 4 + −x +x +x −x 4 4n 5n  p2n p3n p5n yp  pn p3n p4n yp (21) + x−x +x −x 4 + x+x −x −x 4 .

Note that for any z ∈ Fp6n , n n 3n 4n 4n 3n 2n 2n 3n 5n 5n 3n z ∗ a = zap +zp a+zzp ap +zp ap +zap +zp a−zp ap −zp ap

 n 2n 3n  3n 4n 5n n (22) = z+zp +zp +zp a + z−zp +zp −zp ap .

By using Equality (21) and letting z = L−1(x ∗ y),

n 2n 3n z+zp +zp +zp n  pn p2n p4n p5n y  p3n yp = 2x +2x −2x +2x 4 + 2x+2x 4 2n 3n  p3n yp  pn p2n p4n p5n yp + 2x+x 4 + 2x +2x +2x −2x 4 4n 5n  p3n yp  p3n yp (23) + −2x+2x 4 + 2x−2x 4 and n 2n 3n z−zp +zp −zp n  pn p2n p4n p5n y  p3n yp = 2x +2x +2x −2x 4 + 2x−2x 4 2n 3n  p3n yp  pn p2n p4n p5n yp + 2x−x 4 + −2x −2x +2x −2x 4 4n 5n  p3n yp  p3n yp (24) + 2x+2x 4 + −2x−2x 4 . Hence, from Equalities (22), (23) and (24), it follows that

n 2n 4n 5n n n 2n n 4n n 5n n −1 xp a+xp a−xp a+xp a xp ap +xp ap +xp ap −xp ap L (x ∗ y) ∗ a = 2 y + 2 y 3n n 3n n 3n n 3n n xa+xp a+xap −xp ap pn xa+xp a+xap −xp ap p2n + 2 y + 2 y n 2n 4n 5n n n 2n n 4n n 5n n xp a+xp a+xp a−xp a p3n −xp ap −xp ap +xp ap −xp ap p3n + 2 y + 2 y 3n n 3n n 3n n 3n n −xa+xp a+xap +xp ap p4n xa−xp a−xap −xp ap p5n (25) + 2 y + 2 y .

Advances in Mathematics of Communications Volume 10, No. 2 (2016), 291–306 Constructing commutative semifields 303

−1 −1 From Equations (20) and (25), L (x ∗ a) ∗ y = L (x ∗ y) ∗ a for any x, y ∈ Fp6n . Hence, a is contained in the left nucleus of S.

5. Additional semifields We now present several planar polynomials obtained from Equation (2). By examining their corresponding semifields, we show that they are new. With the aid of an algebra package such as GAP [13] it is possible to check if a function is planar. Additionally, the following lemma reduces the amount of computations required.

pi+pj ps+pt Lemma 5.1. Let g(x) = x and h(x) = x be DO monomials over Fp2r ∗ and let f(x) be defined as in Equation (2). Then for any c ∈ Fpr , Df (cx, ca) = 0 if and only if Df (x, a) = 0. ∗ Proof. For any c ∈ Fpr ,

Df (x, a) = 0 pr pr ⇐⇒ Df (x, a) + Df (x, a) = 0 and Df (x, a) − Df (x, a) = 0 i j j i i+r j+r j+r i+r ⇐⇒ 2xp ap + 2xp ap + 2xp ap + 2xp ap = 0 s t t s s+r t+r t+r s+r and 2xp ap + 2xp ap − 2xp ap − 2xp ap = 0

1  i j j i i+r j+r j+r i+r  ⇐⇒ 2(cx)p (ca)p + 2(cx)p (ca)p + 2(cx)p (ca)p + 2(cx)p (ca)p = 0 cpi+pj 1  s t t s s+r t+r t+r s+r  and 2(cx)p (ca)p + 2(cx)p (ca)p − 2(cx)p (ca)p − 2(cx)p (ca)p = 0 cps+pt pr pr ⇐⇒ Df (cx, ca) + Df (cx, ca) = 0 and Df (cx, ca) − Df (cx, ca) = 0

⇐⇒ Df (cx, ca) = 0.

From Lemma 5.1, one only has to check one value of x for each of the pr + 1 ∗ ∗ cosets of Fpr in Fp2r . Let α be a primitive element of Fp2r , then by using Lemma 5.1 it can be determined if

i j i j r s t s t r f(x) = xp +p + x(p +p )p + xp +p − x(p +p )p k r is planar over Fp2r just by checking if α is a root of Df (x, a) ∈ Fp2r [x] for 0 ≤ k ≤ p r ∗ (or 1 ≤ k ≤ p + 1) and a ∈ Fp2r . Thus, using Lemma 5.1 along with GAP [13], one can easily verify that the following are examples of planar functions of the form of Equation (2). Example 1. The following are examples of planar functions of the form

pi+1 (pi+1)pr ps+1 (ps+1)pr x + x + x − x ∈ Fp2r [x]. 6 (6)53 26 (26)53 1. x + x + x − x ∈ F56 [x] 8 (8)73 50 (50)73 2. x + x + x − x ∈ F76 [x] 10 (10)36 82 (82)36 3. x + x + x − x ∈ F312 [x] 10 (10)36 28 (28)36 4. x + x + x − x ∈ F312 [x] Note that in the first three functions in Example1, the function is of the form of Equation (2) where h(x) is a planar monomial. Whereas, h(x) is not planar in the fourth function in Example1. Further note that the first three functions are of pn+1 (pn+1)p3n p2n+1 (p2n+1)p3n the form f(x) = x + x + x − x ∈ Fp2r [x] where r = 3n in which case Theorem 4.3 may be used to determine the sizes of the middle nuclei of the corresponding semifields.

Advances in Mathematics of Communications Volume 10, No. 2 (2016), 291–306 304 Stephen M. Gagola III and Joanne L. Hall

Corollary 1. The orders of the middle nuclei of the semifields corresponding to the first three functions in Example1 are 25, 49 and 81 respectively. The nuclei of semifield corresponding to the fourth function of Example1 may also be calculated.

Theorem 5.2. If S is a corresponding semifield of the planar function f(x) = 10 (10)36 28 (28)36 x + x + x − x ∈ F312 [x], then the middle nucleus of S is F9 and the left nucleus of S is F3.

Proof. It is easy to check, by direct computation using GAP [13], that neither F27 nor F81 lies in the middle nucleus of S. Furthermore, by using GAP and letting a be a generator of F9, it is easy to show that a, and therefore the field F9, lies in the middle nucleus of S. Since the left nucleus of S is a subfield of the middle nucleus, the left nucleus of S is either F3 or F9. By direct computation, F9 does not lie in the left nucleus of S. Hence, the left nucleus of S is F3. By Corollary1 and Lemma 5.2 along with [18, §3] and [21], the semifields associ- ated with the functions of Example1 are either new or otherwise isotopic to either the Budaghyan-Helleseth BH(p, r, s, β) or the Zhou-Pott Sk,σ semifields. We now proceed to show that the semifields corresponding to the functions in Example1 are not isotopic to either the Budaghyan-Helleseth or the Zhou-Pott semifields. Theorem 5.3. The semifields corresponding to the first three functions in Ex- ample1 are not isotopic to either the Budaghyan-Helleseth BH(p, r, s, β) or the Zhou-Pott Sk,σ semifields and are therefore not isotopic to any previously known semifields. Proof. It is known from [18, Theorem 4.1] that the Budaghyan-Helleseth presemi- fields BH(p, r, s, β) have middle nuclei of order p2gcd(r,s) and left nuclei of order pgcd(r,s). Since their left nuclei are proper subfields of their middle nuclei, by Theo- rem 4.5, the semifields of the first three functions in Example1 are not isotopic to the Budaghyan-Helleseth semifields BH(p, r, s, β). By Theorem 2 of [21], the order of the middle nucleus of a Zhou-Pott semifield 2gcd(r,k) gcd(r,k) Sk,σ is p when σ = 1 and p when σ 6= 1. But by Theorem 4.3, the semifields of the first three functions in Example1 have middle nuclei of order gcd(2n,m,2r) gcd(2n,2n,6n) 2n r p = p = p . Since 2 - gcd(n,r) , the largest power of 2 dividing 2r divides 2n but does not divide gcd(r, k). Therefore, the semifields of the first three functions in Example1 are not isotopic to a Zhou-Pott semifield Sk,σ unless σ = 1. But if σ = 1, then, by Theorem 3 of [21], the order of the left nucleus gcd(r,k,0) gcd(r,k) of the Zhou-Pott semifield Sk,σ is p = p . Hence, if σ = 1, then the left nucleus of Sk,σ is a proper subfield of the middle nucleus. Therefore, the semifields of the first three functions in Example1 are not isotopic to the Zhou-Pott semifields Sk,σ.

10 (10)36 28 (28)36 Theorem 5.4. The planar function f(x) = x + x + x − x ∈ F312 [x] is not EA-equivalent to any previously known planar functions and its corresponding semifield is not isotopic to any previously known semifields. Proof. Suppose f(x) is EA-equivalent to a planar function, ϕ(x), associated with ∗ a Zhou-Pott presemifield Pk,σ = (Fp2r , +, ∗) where p = 3 and r = 6. Let ω ∈ Fp2r

Advances in Mathematics of Communications Volume 10, No. 2 (2016), 291–306 Constructing commutative semifields 305

pr such that ω = −ω and denote any x ∈ Fp2r by x = x1 + x2ω for x1, x2 ∈ Fpr . By [21, Theorem 1],

ϕ(x) = x ∗ x = (x1 + x2ω) ∗ (x1 + x2ω), σ pk+1  pk+1 = 2x1 + α 2x2 + 2x1x2ω,

k k σ p +1  p +1 1  2 2pr  = 2x1 + 2α x2 + 2 (x1 + x2ω) − (x1 + x2ω) ,

k k σ p +1  p +1 1 2 1 2pr = 2x1 + 2α x2 + 2 x − 2 x ,

r r and therefore, ϕ(x) − ϕ(x)p = x2 − x2p . p p2 p2r−1 Now let L1(x) = c0x + c1x + c2x + ··· + c2r−1x and L2(x) be linear permutation polynomials such that ϕ(L1(x)) = L2(f(x)). Note that pr 2 2pr ϕ(L1(x)) − ϕ(L1(x)) = L1(x) − L1(x)

X 2 2pr 2pj 2 2pr 2pj+r = (cj − cj+r)x + (cj+r − cj )x 0≤j

r pr pr 2 2p Since ϕ(L1(x)) − ϕ(L1(x)) = L2(f(x)) − L2(f(x)) , cj − cj+r = 0 and 2cjcj+1 − pr pr pr pr pr 2cj+rcj+1+r = 0 for any 0 ≤ j < r. Thus, cj+r = ±cj and cj+rcj+1+r = cjcj+1 for pr pr pr all 0 ≤ j < r. So if cj+r = cj, then cj+1+r = cj+1 and, likewise, if cj+r = −cj, pr pr then cj+1+r = −cj+1. Hence, there exists an a ∈ {±1} such that cj+r = acj for all 2 2pr 2 0 ≤ j < r. Therefore, L1(x) −L1(x) = 0 meaning L1(x) ∈ Fpr for any x ∈ Fp2r . This contradicts the fact that L1(x) ∈ Fp2r [x] is a permutation polynomial. Hence, f(x) is not EA-equivalent to a planar function, ϕ(x), corresponding to a Zhou-Pott presemifield. Now assume that f(x) is EA-equivalent to the planar function, r s r s r B(x) = xp +1 + ωβxp +1 + ωβp x(p +1)p , p p2 p2r−1 where p = 3 and r = 6. Let L1(x) = c0x + c1x + c2x + ··· + c2r−1x and L2(x) be linear permutation polynomials such that B(L1(x)) = L2(f(x)). Since r r B(x) + B(x)p = 2xp +1, pr pr +1 B(L1(x)) + B(L1(x)) = 2L1(x)

X  2pj pr  2pj+r = 2 cj cj+rx + cj+rcj x 0≤j

pr pr 2pj Since B(L1(x))+B(L1(x)) = L2(f(x))+L2(f(x)) , the coefficient of x , name- r p pj +pj+r ly cjcj+r, is zero for any 0 ≤ j < r. Likewise, the coefficient of x , namely

Advances in Mathematics of Communications Volume 10, No. 2 (2016), 291–306 306 Stephen M. Gagola III and Joanne L. Hall

pr +1 pr +1 cj + cj+r , is zero for all 0 ≤ j < r. Thus, cj = 0 = cj+r for all 0 ≤ j < r, meaning L1(x) = 0. This contradicts the fact that L1(x) is a permutation poly- nomial. Hence, f(x) is not EA-equivalent to the generalized Budaghyan-Helleseth planar function.

Acknowledgments Thanks to the anonymous referees who’s suggestions have been valuable in im- proving this paper.

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