# Non-extendable \(\mathbb{F}_{q}\)-Quadratic Perfect Nonlinear Maps

## Abstract

Let *q* be a power of an odd prime. We give examples of non-extendable \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps. We also show that many classes of \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps are extendable. We give a short survey of some related results and provide some open problems.

## 1 Introduction

*q*be a power of an odd prime. Let

*n*,

*m*be positive integers with

*m*dividing

*n*. We use \(\mathrm{Tr}_{\mathbb{F}_{q^{n}}/\mathbb{F}_{q^{m}}}\) and \(\mathrm{Norm}_{\mathbb{F}_{q^{n}}/\mathbb{F}_{q^{m}}}\) to denote the relative trace and relative norm functions from \(\mathbb{F}_{q^{n}}\) to \(\mathbb{F}_{q^{m}}\) given by

*m*= 1, we denote \(\mathrm{Tr}_{\mathbb{F}_{q^{n}}/\mathbb{F}_{q}}\) and \(\mathrm{Norm}_{\mathbb{F}_{q^{n}}/\mathbb{F}_{q}}\) as Tr and Norm in short.

*f*on \(\mathbb{F}_{q^{n}}\) is a map from \(\mathbb{F}_{q^{n}}\) to \(\mathbb{F}_{q}\) given by

*f*an \(\mathbb{F}_{q}\)-quadratic map from \(\mathbb{F}_{q^{n}}\) to \(\mathbb{F}_{q}\) as well.

*F*on \(\mathbb{F}_{q^{n}}\) is a map in Dembowski–Ostrom polynomial form given by

*f*

_{i}(

*x*) is an \(\mathbb{F}_{q}\)-quadratic form on \(\mathbb{F}_{q^{n}}\). Indeed if \(\{e_{1}^{{\ast}},e_{2}^{{\ast}},\ldots,e_{n}^{{\ast}}\}\) is the trace-orthogonal basis (see [18]), then

*i*≤

*n*.

*F*from \(\mathbb{F}_{q^{n}}\) to \(\mathbb{F}_{q^{n}}\) (or \(\mathbb{F}_{q}^{n}\)) as

*r*≤

*n*, we say that

*F*is an \(\mathbb{F}_{q}\)-quadratic map from \(\mathbb{F}_{q^{n}}\) to \(\mathbb{F}_{q}^{r}\) if

*F*can equivalently be considered as an \(\mathbb{F}_{q}\)-quadratic map from \(\mathbb{F}_{q^{n}}\) to \(\mathbb{F}_{q^{r}}\) given by

We give an important definition.

### Definition 1.1

*r*≤

*n*, let

*F*be an \(\mathbb{F}_{q}\)-quadratic map from \(\mathbb{F}_{q^{n}}\) to \(\mathbb{F}_{q}^{r}\). For \(a \in \mathbb{F}_{q^{n}}^{{\ast}}\), let

*D*

_{F, a}be the difference map from \(\mathbb{F}_{q^{n}}\) to \(\mathbb{F}_{q}^{r}\) given by

*F*is

*perfect nonlinear*or \((q^{n},q^{r})\)-

*bent*if the cardinality of the set

*q*

^{n−r}for all \(a \in \mathbb{F}_{q^{n}}^{{\ast}}\) and \(b \in \mathbb{F}_{q}^{r}\). We also call

*F*is (

*n*,

*r*)

*-bent*if

*q*is clear from the context. Moreover we say that

*F*is

*planar*if

*n*=

*r*and

*F*is

*bent*if

*r*= 1.

### Definition 1.2

*r*≤

*n*, let \(\left [\begin{array}{c} f_{1}\\ \vdots \\ f_{r}\end{array} \right ]\) and \(\left [\begin{array}{c} g_{1}\\ \vdots \\ g_{r}\end{array} \right ]\) be \(\mathbb{F}_{q}\)-quadratic maps from \(\mathbb{F}_{q^{n}}\) to \(\mathbb{F}_{q}^{r}\). We call that \(\left [\begin{array}{c} f_{1}\\ \vdots \\ f_{r}\end{array} \right ]\) and \(\left [\begin{array}{c} g_{1}\\ \vdots \\ g_{r}\end{array} \right ]\) are

*equivalent*if there exists an \(\mathbb{F}_{q}\)-linearized permutation polynomial \(L(x) \in \mathbb{F}_{q^{n}}[x]\) and an invertible

*r*×

*r*matrix [

*a*

_{ij}] with entries from \(\mathbb{F}_{q}\) such that

### Remark 1.3

There are more general notions of equivalence for arbitrary finite fields and more general maps between them: extended affine equivalence and Carlet–Charpin–Zinoviev equivalence [6]. However the equivalence in Definition 1.2 gives the same results if we use these two other equivalence notions for the \(\mathbb{F}_{q}\)-quadratic maps from \(\mathbb{F}_{q^{n}}\) to \(\mathbb{F}_{q}^{r}\) in this paper (see [21]).

### Definition 1.4

For 1 ≤ *r* ≤ *n* − 1, let *F* be an \(\mathbb{F}_{q}\)-quadratic (*q*^{n}, *q*^{r})-bent map. We call that *F* is *extendable* if there exists an \(\mathbb{F}_{q}\)-quadratic form *f* on \(\mathbb{F}_{q^{n}}\) such that the map \(\left [\begin{array}{c} F(x)\\ f(x) \end{array} \right ]\) is an \((q^{n},q^{r+1})\)-bent map. Otherwise we call that *F* is *non-extendable*.

Note that *F* is non-extendable if and only if *G* is non-extendable for any *G* equivalent to *F*. Moreover non-extendable \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps can be considered as a kind of “atomic” structures.

It is a difficult problem to characterize all \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps up to equivalence. In Sect. 2 we give a short survey on some of the results in this problem.

It seems also a difficult problem to characterize all \(\mathbb{F}_{q}\)-quadratic non-extendable \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps up to equivalence. In Sect. 3 we prove that many classes of \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps are extendable. In Sect. 4 we give some examples of non-extendable \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps.

There is a natural connection to finite semifields. We explain such a connection in Sects. 5 and 6. Throughout the chapter, we explain many open problems.

## 2 \(\mathbb{F}_{q}\)-Quadratic Perfect Nonlinear Maps

*f*and

*g*are

*equivalent*if there exist \(\mathbb{F}_{q}\)-linearized permutation polynomials \(L_{1},L_{2} \in \mathbb{F}_{q^{n}}[x]\) such that

- (1)
\(x^{2} + \mathit{ux\,}^{q^{2}+q }\)

- (2)
\(x^{q+1} + \mathit{ux\,}^{2q^{2} }\)

- (3)
\(x^{2q} + \mathit{ux\,}^{q^{2}+1 }\)

- (4)
\(x^{2} + \mathit{ux\,}^{q+1}\)

- (5)
\(x^{2} + \mathit{ux\,}^{2q}\)

- (6)
\(x^{2} + \mathit{ux\,}^{q^{2}+1 }\)

- (7)
\(x^{2} + \mathit{ux\,}^{2q^{2} }\)

- (8)
\(x^{q+1} + \mathit{ux\,}^{2q}\)

- (9)
\(x^{q+1} + \mathit{ux\,}^{q^{2}+1 }\)

- (10)
\(x^{q+1} + \mathit{ux\,}^{q^{2}+q }\)

- (11)
\(x^{2q} + \mathit{ux\,}^{q^{2}+q }\)

- (12)
\(x^{2q} + \mathit{ux\,}^{2q^{2} }\)

- (13)
\(x^{q^{2}+1 } + \mathit{ux\,}^{q^{2}+q }\)

- (14)
\(x^{q^{2}+1 } + \mathit{ux\,}^{2q^{2} }\)

- (15)
\(x^{q^{2}+q } + \mathit{ux\,}^{2q^{2} }\)

*u*is an arbitrary nonzero element in \(\mathbb{F}_{q^{3}}\). It is easier to decide whether the maps in the sublist 4, 5,

*…*, 15 are planar or not. First we recall that for \(u \in \mathbb{F}_{q^{3}}^{{\ast}}\), the polynomial maps

*x*

^{2}with a linearized polynomial map. For example, regarding the map in item 5, we have

*x*

^{2}if they are planar.

*x*

^{q+1}with a linearized polynomial map. For example, regarding the map in item 9, we have

*x*

^{q+1}if they are planar.

*F*

_{u}be the binomial map given by

*F*

_{u}(

*x*) is planar for some \(u \in \mathbb{F}_{q^{3}}\) and equivalent to

*x*

^{2}. They also prove that

*F*

_{u}(

*x*) is planar for some \(u \in \mathbb{F}_{q^{3}}\) and equivalent to

*x*

^{q+1}. Therefore the binomial form of

*F*

_{u}(

*x*) is different from the sublist of the binomial forms in 4, 5,

*…*, 15 as they cannot be equivalent to both

*x*

^{2}and

*x*

^{q+1}for some different choices of \(u \in \mathbb{F}_{q^{3}}^{{\ast}}\). Moreover, it seems that proving planarity of

*F*

_{u}(

*x*) is quite difficult, and it has direct connections with arithmetic of some function fields [16]. Their main results is (see Theorem 4.5 in [16]):

### Theorem 2.1

*Let q be a power of an odd prime and*\(u \in \mathbb{F}_{q^{3}}\)

*. Let F*

_{u}

*(x) be the polynomial*\(x^{q^{2}+q } + ux^{2}\)

*in*\(\mathbb{F}_{q^{3}}[x]\)

*. If*\(q \equiv 1\mod 3\)

*, then let G and H be the subgroups of*\(\mathbb{F}_{q^{3}}^{{\ast}}\)

*with*\(\vert G\vert = q^{2} + q + 1\)

*and*\(\vert H\vert = (q^{2} + q + 1)/3\)

*. Then the polynomial F*

_{u}

*(x) is planar if and only if*\(q \equiv 1\mod 3\)

*and*

*Moreover we have the followings. Assume that*\(q \equiv 1\mod 3\).

*The polynomial F*_{u}*(x) is equivalent to x*^{2}*if and only if*\(u \in \frac{1} {2}(G\setminus H)\)*.**The polynomial F*_{u}*(x) is equivalent to x*^{q+1}*if and only if u ∈−(G∖H).*

We note that one of the important ingredients in the proof of Theorem 4.5 in [16] comes from the theory of (commutative) finite semifields. We will give more information about it in Sects. 5 and 6 below.

We also note that there are many related problems stated either explicitly or implicitly in [16]. We would like to refer to Remark 3.9 in [16] and Corollary 4.6 in [16].

*L*

_{1}(

*x*) and

*L*

_{2}(

*x*) are \(\mathbb{F}_{q}\)-linearized polynomials in \(\mathbb{F}_{q^{n}}[x]\). In [15] the planarity of such products of linearized polynomials is investigated. An important tool they use is the link between the set

Another important tool they use is Hasse–Weil–Serre Theorem (see, Theorem 4 in [15] and Theorem 5.3.7 in [24]). Using Hasse–Weil–Serre Theorem, they prove the following result.

### Theorem 2.2

*Let q be a power of an odd prime and n ≥ 2 be an integer. Let*\(\mathrm{Tr}_{n}(x): \mathbb{F}_{q^{n}} \rightarrow \mathbb{F}_{q}\)

*be the linearized polynomial*\(x\mapsto x + x^{q} + \cdots + x^{q^{n-1} }\)

*. For*\(a \in \mathbb{F}_{q^{n}}\)

*, if n ≥ 5, then the mapping*

*is not planar.*

There are planar maps of the form \(x(\mathrm{Tr}_{n}(x) + \mathit{ax})\) with \(a \in \mathbb{F}_{q^{n}}\) for *n* = 2 and *n* = 3. The case of *n* = 2 was completely characterized in Theorem 2 of [15].

*n*= 3, it was proved in [15] that

*F*(

*x*) in (3) is not planar on \(\mathbb{F}_{q^{3}}\). Moreover they conjectured in Conjecture 1 and Conjecture 2 that there is no planar map of the form (3) for the remaining values of \(a \in \mathbb{F}_{q^{3}}\). These conjectures were proved in [8] again using some facts from the theory of finite semifields.

In [8] it is commented that their method for resolving the conjecture for the case *n* = 3 would not work for the case *n* = 4. Indeed the conjecture for the case *n* = 4 was proved in [25] using a different approach. They also proved the subcase for *n* = 3 where \(a \in \{-1,-2\}\) using an elementary approach. It is still a problem whether an approach using exponential sums would work to prove the conjecture of the remaining situation of the subcase *n* = 3 that \(a \in \mathbb{F}_{q^{3}}\setminus \mathbb{F}_{q}\). We also refer to Sect. 2 in [15] for another open problem.

*n*,

*r*) = (2, 2) all \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps are equivalent to the map

*x*↦

*x*

^{2}. Moreover for (

*n*,

*r*) = (3, 3) all \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps are equivalent to one of the following maps:

*x*↦*x*^{2}(finite field),*x*↦*x*^{q+1}(twisted finite field).

Recently we proved in [21] that all \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps for (*n*, *r*) = (3, 2) are equivalent. We used some results from algebraic geometry, in particular Bezout’s Theorem in our proof. We also presented an explicit algorithm for finding the corresponding equivalence using a geometric method.

### Proposition 2.3

*Let q be a power of an odd prime. For*\(\theta,w \in \mathbb{F}_{q^{3}}\)

*consider the*\(\mathbb{F}_{q}\)

*-quadratic map*\(F: \mathbb{F}_{q^{3}} \rightarrow \mathbb{F}_{q}^{2}\)

*given by*

*Then F is perfect nonlinear if and only if the polynomial*

*where*

*is irreducible over*\(\mathbb{F}_{q}\)

*.*

For some special \(\mathbb{F}_{q}\)-quadratic maps, it is easy to decide whether they are perfect nonlinear or not (also equivalence among them is much easier to find). The following was proved in Sect. 2 of [21].

### Theorem 2.4

*Let*\(\{w_{1},w_{2}\},\{w_{3},w_{4}\} \subseteq \mathbb{F}_{q^{3}}\)

*be*\(\mathbb{F}_{q}\)

*-linearly independent sets. Let F*

_{1}

*and F*

_{2}

*be the*\(\mathbb{F}_{q}\)

*-quadratic maps given by*

*For*\(w \in \mathbb{F}_{q^{3}}\setminus \mathbb{F}_{q}\)

*let G and H be the*\(\mathbb{F}_{q}\)

*-quadratic maps*

*Then all of the maps F*

_{1}

*,F*

_{2}

*,G, and H are perfect nonlinear and equivalent to each other.*

The main result of the [21] is the following theorem.

### Theorem 2.5

*Let*\(F: \mathbb{F}_{q^{3}} \rightarrow \mathbb{F}_{q}^{2}\)

*be an*\(\mathbb{F}_{q}\)

*-quadratic perfect nonlinear map. We can assume that there exist*\(\theta,w \in \mathbb{F}_{q^{3}}\)

*such that*

*without loss of generality (up to equivalence). Let*\(\mathit{PG}(2, \mathbb{F}_{q^{3}})\)

*denote the projective plane over*\(\mathbb{F}_{q^{3}}\)

*. Let*\(\mathcal{T}_{1}\)

*be the set of consisting*\((x_{0}: x_{1}: x_{2}) \in \mathit{PG}(2, \mathbb{F}_{q^{3}})\)

*such that*

*Let*\(\mathcal{T}_{2}\)

*be the set of consisting of*\((x_{0}: x_{1}: x_{2}) \in \mathit{PG}(2, \mathbb{F}_{q^{3}})\)

*such that*

*Let*\(E: \mathit{PG}(2, \mathbb{F}_{q^{3}}) \rightarrow PG(2, \mathbb{F}_{q^{3}})\)

*be the action defined as*

*Then we have the following:*

- (1)
*There exist*\(Q,P_{1} \in \mathit{PG}(2, \mathbb{F}_{q^{3}})\)*such that*$$\displaystyle{\mathcal{T}_{1} \cap \mathcal{T}_{2} =\{ Q,P_{1},E(P_{1}),(E \circ E)(P_{1})\},}$$*E(Q) = Q and*\(\vert \mathcal{T}_{1} \cap \mathcal{T}_{2}\vert = 4\)*.* - (2)
*Put*\(P_{1} = (a_{0}: a_{1}: a_{2}) \in \mathit{PG}(2, \mathbb{F}_{q^{3}})\)*with*\(a_{0},a_{1},a_{2} \in \mathbb{F}_{q^{3}}\)*. Let L(x) be the*\(\mathbb{F}_{q}\)*-linearized polynomial in*\(\mathbb{F}_{q^{3}}[x]\)*given by*$$\displaystyle{L(x) = a_{0}x + a_{1}^{q}x^{q} + a_{ 2}^{q^{2} }x^{q^{2} }.}$$*Then there exist*\(\delta _{1},\delta _{2} \in \mathbb{F}_{q^{3}}\)*such that*$$\displaystyle{\left [\begin{array}{l} \mathrm{Tr}(\delta _{1}x^{q+1}) \\ \mathrm{Tr}(\delta _{2}x^{q+1})\end{array} \right ]\text{ is perfect nonlinear}}$$*and*$$\displaystyle{\left [\begin{array}{l} \mathrm{Tr}(\delta _{1}x^{q+1}) \\ \mathrm{Tr}(\delta _{2}x^{q+1})\end{array} \right ] = F(L(x))\text{ for all }x \in \mathbb{F}_{q^{3}}.}$$

*F*(

*x*) to an \(\mathbb{F}_{q}\)-quadratic perfect nonlinear map in the form

*F*

_{1},

*F*

_{2},

*G*, and

*H*in Theorem 2.4, we easily obtain a geometric algorithm for finding an equivalence from

*F*(

*x*) to, for example,

*G*(

*x*). We refer to Sect. 3 in [21] for such an explicit algorithm.

There is a consequence of Theorems 2.4 and 2.5 given in Sect. 3 below.

## 3 Extendable \(\mathbb{F}_{q}\)-Quadratic Perfect Nonlinear Maps

In this section we show that many classes of \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps are extendable.

### Proposition 3.1

*Let n ≥ 2 and*\(F: \mathbb{F}_{q^{n}} \rightarrow \mathbb{F}_{q}\)*be an*\(\mathbb{F}_{q}\)*-quadratic perfect nonlinear (or bent) map. Then F is extendable.*

### Proof

*f*

_{1}(

*x*) =

*F*(

*x*). Note that

*f*

_{1}(

*x*) is a nondegenerate \(\mathbb{F}_{q}\)-quadratic form on \(\mathbb{F}_{q^{n}}\). We will show that there exist \(\mathbb{F}_{q}\)-quadratic forms \(f_{2}(x),\ldots,f_{n}(x)\) on \(\mathbb{F}_{q^{n}}\) such that

*F*is extendable.

*n*is odd. Let \(\lambda \in \mathbb{F}_{q}^{{\ast}}\) be a nonsquare. Let \(g_{1}(x) = \mathrm{Tr}(x^{2})\) and \(h_{1}(x) =\lambda \mathrm{Tr}(x^{2})\). Note that both

*g*

_{1}(

*x*) and

*h*

_{1}(

*x*) are nondegenerate \(\mathbb{F}_{q}\)-quadratic forms on \(\mathbb{F}_{q^{n}}\). Moreover the discriminants of

*g*

_{1}(

*x*) and

*h*

_{1}(

*x*) are not both square or both nonsquare in \(\mathbb{F}_{q}^{{\ast}}\). Here we use the fact that

*n*is odd and

*λ*

^{n}is a nonsquare in \(\mathbb{F}_{q}^{{\ast}}\). It is well known that there are exactly two nondegenerate \(\mathbb{F}_{q}\)-quadratic forms on \(\mathbb{F}_{q^{n}}\) (also for

*n*is even) up to a choice of \(\mathbb{F}_{q}\)-basis of \(\mathbb{F}_{q}^{n}\); they are determined by whether the discriminant is square or not in \(\mathbb{F}_{q}^{{\ast}}\) (see, e.g., Theorem 4.9 in [11]). Hence, there exists an \(\mathbb{F}_{q}\)-linearized polynomial \(L(x) \in \mathbb{F}_{q^{n}}[x]\) such that

*x*↦

*x*

^{2}and

*x*↦

*λ x*

^{2}are \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps from \(\mathbb{F}_{q^{n}}\) to \(\mathbb{F}_{q^{n}}\), there exist \(\mathbb{F}_{q}\)-quadratic forms \(g_{2}(x),h_{2}(x),\ldots,g_{n}(x),h_{n}(x)\) on \(\mathbb{F}_{q^{n}}\) such that

*f*

_{1}(

*L*(

*x*)) =

*g*

_{1}(

*x*) or \(f_{1}(L(x)) = h_{1}(x)\), respectively. This completes the proof if

*n*is odd.

*n*is even. Let \(\lambda \in \mathbb{F}_{q^{n}}^{{\ast}}\) be a nonsquare. Let \(g_{1}(x) = \mathrm{Tr}(x^{2})\) and \(h_{1}(x) = \mathrm{Tr}(\lambda x^{2})\). Note that

*g*

_{1}(

*x*) and

*h*

_{1}(

*x*) are again nondegenerate \(\mathbb{F}_{q}\)-quadratic forms on \(\mathbb{F}_{q^{n}}\). Comparing with the proof of the case

*n*is odd above, it is enough to show that the discriminants \(\Delta (g_{1})\) and \(\Delta (h_{1})\) are distinct modulo squares in \(\mathbb{F}_{q}^{{\ast}}\). Let \(\eta: \mathbb{F}_{q}^{{\ast}}\rightarrow \{ 1,-1\}\) be the quadratic character

The following is a corollary of Theorems 2.4 and 2.5 of Sect. 2.

### Corollary 3.2

*Let*\(F: \mathbb{F}_{q^{3}} \rightarrow \mathbb{F}_{q^{2}}\)*be an*\(\mathbb{F}_{q}\)*-quadratic perfect nonlinear map. Then F is extendable.*

### Proof

*F*is equivalent to

*F*is extendable to both of the maps

*x*↦

*x*

^{2}and

*x*↦

*x*

^{q+1}on \(\mathbb{F}_{q^{3}}\). □

### Example 3.3

*q*= 3 and

*n*= 4. There are exactly 2 distinct \(\mathbb{F}_{q}\)-quadratic planar maps from \(F: \mathbb{F}_{3^{4}}\) to \(\mathbb{F}_{3^{4}}\) up to equivalence. They are the maps

*q*

^{4},

*q*

^{2})-bent maps are extendable.

### Example 3.4

*q*= 3 and

*n*= 5. There are exactly 7 distinct \(\mathbb{F}_{q}\)-quadratic planar maps from \(F: \mathbb{F}_{3^{5}}\) to \(\mathbb{F}_{3^{5}}\) up to equivalence. They are the maps

*x*↦*x*^{2}.*x*↦*x*^{q+1}.\(x\mapsto x^{q^{2}+1 }.\)

\(x\mapsto x^{10} + x^{6} - x^{2}.\)

\(x\mapsto x^{10} - x^{6} - x^{2}.\)

\(x\mapsto x^{90} + x^{2}.\)

\(x\mapsto - (x^{3} - x) + D(x^{3} - x) + \frac{1} {2}x^{2}\) with \(D(x) = -x^{36} + x^{28} + x^{12} + x^{4}\).

Using computer, we have also verified that all \(\mathbb{F}_{3}\)-quadratic (*q*^{5}, *q*^{2})-bent maps are extendable.

In the next section we give the first examples of non-extendable \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps in the literature as far as we know.

## 4 Non-extendable \(\mathbb{F}_{q}\)-Quadratic Perfect Nonlinear Maps

In this section we give some non-extendable \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps. We note that finding non-extendable \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps seems to be more difficult than finding \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps. The reason is that non-extendable maps require that they are impossible to extend, which is an extra condition.

### Example 4.1

Let *q* = 3 and *n* = 4. Recall that, up to equivalence, there are exactly two \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps from \(\mathbb{F}_{q^{4}}\) to \(\mathbb{F}_{q^{4}}\), which are the maps given by the polynomials *x*^{2} and \(x^{4} + x^{10} - x^{36}\). We also recall that all \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps \(F: \mathbb{F}_{q^{4}} \rightarrow \mathbb{F}_{q^{2}}\) are extendable (see Sect. 3). In fact, up to equivalence, there are exactly \(7\ \mathbb{F}_{q}\)-quadratic perfect nonlinear maps from \(\mathbb{F}_{q^{4}}\) to \(\mathbb{F}_{q^{2}}\).

However there are non-extendable \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps from \(\mathbb{F}_{q^{4}}\) to \(\mathbb{F}_{q^{3}}\). There are exactly \(18\ \mathbb{F}_{q}\)-quadratic perfect nonlinear maps from \(\mathbb{F}_{q^{4}}\) to \(\mathbb{F}_{q^{3}}\) up to equivalence. Only 5 of them are extendable. We define them as \(E_{1},E_{2},E_{3},E_{4}\), and *E*_{5} below. Only *E*_{5} extends to both *x*^{2} and \(x^{4} + x^{10} - x^{36}\). The others extend only to \(x^{4} + x^{10} - x^{36}\). The remaining 13 are \(\mathbb{F}_{q}\)-quadratic perfect nonlinear maps from \(\mathbb{F}_{q^{4}}\) to \(\mathbb{F}_{q^{3}}\) are non-extendable. We define them as \(\mathit{NE}_{1},\mathit{NE}_{2},\ldots,\mathit{NE}_{13}\) below.

Now we give these maps explicitly. Let *w* be a primitive element of \(\mathbb{F}_{q^{4}}\) such that \(w^{4} + 2w^{3} + 2 = 0\). Let \(E_{1},E_{2},E_{3},E_{4},E_{5}: \mathbb{F}_{q^{4}} \rightarrow \mathbb{F}_{q}^{3}\) be the maps given by

These are the extendable maps.

Let \(\mathit{NE}_{1},\mathit{NE}_{2},\ldots,\mathit{NE}_{13}: \mathbb{F}_{q^{4}} \rightarrow \mathbb{F}_{q^{3}}\) be the maps given by

These are the non-extendable maps.

## 5 Semifields

In this section we give a short introduction to finite semifields. We refer to [3, 5, 7, 9, 13, 14, 17, 19, 20, 22, 26] for further information.

\((\mathbb{F},+)\) is an abelian group with 0.

\((\mathbb{F}\setminus \{0\},\times )\) is an abelian group with 1.

- If \(a,b,c \in \mathbb{F}\), then$$\displaystyle{a \times (b + c) = a \times b + a \times c\text{ and }(a + b) \times c = a \times c + b \times c.}$$

If \(\mathbb{F}\) is infinite, it is possible to weaken the commutativeness condition above. \(\mathbb{F}\) is called a *skew field (or a division ring)* if \((\mathbb{F}\setminus \{0\},\times )\) is a group with 1 (not necessarily abelian), and all the other conditions are satisfied.

### Example 5.1

Note that \(\mathbb{H}\) in Example 5.1 is an infinite set. The following important results of Wedderburn (see, e.g., [18]) says that we have no such an example for finite sets.

### Theorem 5.2

*If*\(\mathbb{F}\)*is a finite set and*\(\mathbb{F}\)*is a skew field, then*\(\mathbb{F}\)*must be a finite field.*

However, if \(\mathbb{F}\) is finite, then it is possible to weaken associativity condition of \((\mathbb{F}\setminus \{0\},\times )\) instead of the commutativity condition.

### Definition 5.3

*presemifield*\(\mathbb{S}\) is a finite set with two binary operations + and ∗ satisfying the following axioms:

\((\mathbb{S},+)\) is an abelian group with 0.

- If \(a,b,c \in \mathbb{S}\), then$$\displaystyle{a {\ast} (b + c) = a {\ast} b + a {\ast} c\text{ and }(a + b) {\ast} c = a {\ast} c + b {\ast} c.}$$
If \(a,b \in \mathbb{S}\) and

*a*∗*b*= 0, then*a*= 0 or*b*= 0.

*semifield*.

*commutative presemifield*(or

*commutative semifield*).

The additive group of a finite semifield is elementary abelian. Therefore the order of \(\mathbb{S}\) is *p*^{n} for a prime *p* and integer *n* ≥ 1. Here *p* is called the characteristic of \(\mathbb{S}\). In fact there is no finite semifield \(\mathbb{S}\), which is not a field, with \(\vert \mathbb{S}\vert = p^{2}\) or \(\vert \mathbb{S}\vert = 8\), where *p* is a prime. However, for each prime *p* and integer *n* ≥ 3 with (*p*, *n*) ≠ (2, 3), there exists a semifield \(\mathbb{S}\) which is not a finite field with \(\vert \mathbb{S}\vert = p^{n}\).

Any finite presemifield can be represented by \((\mathbb{S},+,{\ast})\), where \(\mathbb{S}\) is the underlying set of a finite field \(\mathbb{F}_{p^{n}}\) and the addition of \(\mathbb{S}\) coincides with the addition of \(\mathbb{F}_{p^{n}}\). The notion to classify finite presemifield is isotopism:

### Definition 5.4

*p*

^{n}are

*isotopic*if there exist linearized permutation polynomials

*L*,

*M*,

*N*over \(\mathbb{F}_{p^{n}}\) such that

Note that two presemifields can be isotopic only if their orders are the same. Finite fields are classified up to isomorphism by their orders. However there are many non-isotopic semifields of the same order. It is a big open problem to classify all finite semifields up to isotopism.

*x*∗

*a*=

*α*and

*a*∗

*y*=

*β*. Then

*α*⋆

*β*is defined as

*e*=

*a*∗

*a*is the identity of \(\mathbb{S}\setminus \{0\}\) under ⋆, and hence \((\mathbb{S},+,\star )\) is a semifield. This is known as Kaplansky’s trick. Here if \((\mathbb{S},+,{\ast})\) is a commutative presemifield, then \((\mathbb{S},+,\star )\) is a commutative semifield. In general isotopism does not preserve commutativity.

*left, middle*, and

*right nucleus*of \(\mathbb{S}\), respectively. They are in fact finite fields. The subset

*nucleus*(or

*associative center*) of \(\mathbb{S}\). The

*center*\(C(\mathbb{S})\) of \(\mathbb{S}\) is defined by

*f*is planar, then we get a commutative presemifield \((\mathbb{S}_{f},+,{\ast})\) defined as

*p*

^{n}, then there exists an \(\mathbb{F}_{p}\)-quadratic polynomial \(f \in \mathbb{F}_{p^{n}}[x]\) such that \((\mathbb{S}_{f},+,{\ast})\) is isotopic to \((\mathbb{S},+,\star )\). Therefore the classification of finite commutative semifields and classification of \(\mathbb{F}_{p^{n}}\)-quadratic planar maps on \(\mathbb{F}_{p^{n}}\) are the same problem.

In the sections above, we mainly considered nondegenerate symmetric quadratic forms over \(\mathbb{F}_{q}\). They are directly linked to symmetric bilinear forms (or symmetric matrices) over \(\mathbb{F}_{q}\) as the characteristic is odd. Moreover planarity is related to commutative semifields as explained in this section. However, symmetric matrices are known to be connected symplectic semifields. In the next section, we explain a geometric connection among these concepts.

## 6 Knuth’s Cubical Array

In this section we explain Knuth’s cubical array. This also gives an important and well-known method to construct up to six non-isotopic semifields of order *p*^{n} starting from a given finite semifield of order *p*^{n}.

*D*

_{0}is the zero map and

*D*

_{a}is an \(\mathbb{F}_{q^{n}}\)-linear map for all \(a \in \mathbb{F}_{q^{n}}\). The corresponding presemifield \((\mathbb{S},+,\circ )\) has the operation

*a*

_{ijk}).

Note that \((a_{\mathit{ijk}}) = (a_{\mathit{jik}})\) as \(\mathbb{S}\) is commutative in this chapter. In general \(\mathbb{S}\) can be noncommutative.

*i*,

*j*,

*k*≤

*n*, the slice \(M(\mathbb{S},j)\) of the cubical array (

*a*

_{ijk}) is an

*n*×

*n*matrix with entries from \(\mathbb{F}_{q}\) defined as

*a*

_{ijk}) by fixing

*j*. Let us fix

*k*instead. For 1 ≤

*k*≤

*n*, the slice \(\hat{M}(\mathbb{S},k)\) of the cubical array (

*a*

_{ijk}) is an

*n*×

*n*matrix with entries from \(\mathbb{F}_{q}\) defined as

*k*. In fact \(\{\hat{M}(\mathbb{S},k): 1 \leq k \leq n\}\) corresponds to a basis of the spread set \(\mathrm{span}\{\hat{M}(\mathbb{S}^{d{\ast}},k): 1 \leq k \leq n\}\) of another semifield \(\mathbb{S}^{d{\ast}}\) obtained via Knuth operations \(\mathbb{S}\mapsto \mathbb{S}^{d}\) and \(\mathbb{S}^{d}\mapsto \mathbb{S}^{d{\ast}}\) that we explain below.

It is well known that any permutation of the indices of the cubical array (*a*_{ijk}) gives a cubical array corresponding to a (pre)semifield. These permutations also respect semifield isotopism. Since a presemifield is isotopic to a semifield, we assume that \(\mathbb{S}\) is a semifield without loss of generality here.

*a*

_{jik}) corresponds to the permutation (12) and gives the opposite semifield \(\mathbb{S}^{{\ast}}\). The permuted cubical array (

*a*

_{kji}) corresponds to the permutation (13) and gives the dual semifield \(\mathbb{S}^{d}\). In fact we have the lattices in Fig. 2 in general.

In other words slicing Knuth’s cubical array (*a*_{ijk}) by fixing *k* and forming the symmetric matrices \(\hat{M}(\mathbb{S},1),\hat{M}(\mathbb{S},2),\ldots,\hat{M}(\mathbb{S},n)\) of the commutative semifield \(\mathbb{S}\) corresponds to forming a basis \(M(\mathbb{S}^{d{\ast}},1),M(\mathbb{S}^{d{\ast}},2),\ldots,M(\mathbb{S}^{d{\ast}},n)\) of a spread set for the symplectic semifield \(\mathbb{S}^{d{\ast}}.\)

## Notes

### Acknowledgements

Ferruh Özbudak is partially supported by TUBİTAK under Grant no. TBAG-112T011.

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