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

Chapter

## 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

Let 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
$$\displaystyle{\mathrm{Tr}_{\mathbb{F}_{q^{n}}/\mathbb{F}_{q^{m}}} = x + x^{q^{m} } + x^{2q^{m} } + \cdots + x^{q^{(n/m-1)m} }}$$
and
$$\displaystyle{\mathrm{Norm}_{\mathbb{F}_{q^{n}}/\mathbb{F}_{q^{m}}} = x \cdot x^{q^{m} } \cdot x^{2q^{m} }\cdots x^{q^{(n/m-1)m} }.}$$
When 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.
An arbitrary $$\mathbb{F}_{q}$$-quadratic form f on $$\mathbb{F}_{q^{n}}$$ is a map from $$\mathbb{F}_{q^{n}}$$ to $$\mathbb{F}_{q}$$ given by
$$\displaystyle{f(x) = \mathrm{Tr}(a_{0}x^{2} + a_{ 1}x^{q+1} + \cdots + a_{ \lceil \frac{n} {2} \rceil }x^{\lceil \frac{n} {2} \rceil +1})}$$
where $$a_{0},a_{1},\ldots,a_{\lceil \frac{n} {2} \rceil }\in \mathbb{F}_{q^{n}}$$. We call such f an $$\mathbb{F}_{q}$$-quadratic map from $$\mathbb{F}_{q^{n}}$$ to $$\mathbb{F}_{q}$$ as well.
An $$\mathbb{F}_{q}$$-quadratic map F on $$\mathbb{F}_{q^{n}}$$ is a map in Dembowski–Ostrom polynomial form given by
$$\displaystyle{F(x) =\sum _{ i,j=0}^{n-1}a_{ i,j}x^{q^{i}+q^{j} }}$$
where $$a_{i,j} \in \mathbb{F}_{q^{n}}$$.
Let $$\{e_{1},e_{2},\ldots,e_{n}\}$$ be an $$\mathbb{F}_{q}$$-basis of $$\mathbb{F}_{q^{n}}$$. Equivalently, an $$\mathbb{F}_{q}$$-quadratic map on $$\mathbb{F}_{q^{n}}$$ is a map given by
$$\displaystyle{ F(x) =\sum _{ i=1}^{n}f_{ i}(x)e_{i}. }$$
(1)
where fi(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
$$\displaystyle{f_{i}(x) = \mathrm{Tr}\left (e_{i}^{{\ast}}\sum _{ i_{1},i_{2}=0}^{n-1}a_{ i_{1},i_{2}}x^{q^{i_{1}}+q^{i_{2}} }\right )}$$
is an $$\mathbb{F}_{q}$$-quadratic form given in (1) for 1 ≤ i ≤ n.
We find it more convenient to represent the $$\mathbb{F}_{q}$$-quadratic map F from $$\mathbb{F}_{q^{n}}$$ to $$\mathbb{F}_{q^{n}}$$ (or $$\mathbb{F}_{q}^{n}$$) as
$$\displaystyle{F(x) = \left [\begin{array}{c} f_{1}(x)\\ \vdots \\ f_{n}(x) \end{array} \right ].}$$
In general, for 1 ≤ r ≤ n, we say that F is an $$\mathbb{F}_{q}$$-quadratic map from $$\mathbb{F}_{q^{n}}$$ to $$\mathbb{F}_{q}^{r}$$ if
$$\displaystyle{F(x) = \left [\begin{array}{c} f_{1}(x)\\ \vdots \\ f_{r}(x) \end{array} \right ].}$$
where $$f_{1}(x),\ldots,f_{r}(x)$$ are $$\mathbb{F}_{q}$$-quadratic forms on $$\mathbb{F}_{q^{n}}$$. Choosing a basis $$\{e_{1},\ldots,e_{n}\}$$ of $$\mathbb{F}_{q^{n}}$$ over $$\mathbb{F}_{q}$$, F can equivalently be considered as an $$\mathbb{F}_{q}$$-quadratic map from $$\mathbb{F}_{q^{n}}$$ to $$\mathbb{F}_{q^{r}}$$ given by
$$\displaystyle{F(x) = f_{1}(x)e_{1} + \cdots + f_{r}(x)e_{r}.}$$

We give an important definition.

### Definition 1.1

For 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 DF, a be the difference map from $$\mathbb{F}_{q^{n}}$$ to $$\mathbb{F}_{q}^{r}$$ given by
$$\displaystyle\begin{array}{rcl} D_{F,a}(x) = F(x + a) - F(x) - F(a)& & {}\\ \end{array}$$
We call F is perfect nonlinear or $$(q^{n},q^{r})$$-bent if the cardinality of the set
$$\displaystyle\begin{array}{rcl} \{x \in \mathbb{F}_{q^{n}}: D_{F,a}(x) = b\}& & {}\\ \end{array}$$
is equal to qnr 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

For 1 ≤ 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 [aij] with entries from $$\mathbb{F}_{q}$$ such that
$$\displaystyle{[a_{\mathit{ij}}]\left [\begin{array}{c} f_{1}(L(x))\\ \vdots \\ f_{r}(L(x)) \end{array} \right ] = \left [\begin{array}{c} g_{1}(x)\\ \vdots \\ g_{r}(x) \end{array} \right ]}$$
for all $$x \in \mathbb{F}_{q^{n}}$$.

### 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 (qn, qr)-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

In this section we mainly consider $$\mathbb{F}_{q}$$-quadratic perfect nonlinear maps from $$\mathbb{F}_{q^{n}}$$ to $$\mathbb{F}_{q^{n}}$$. They are also called planar maps. The equivalence in Definition 1.2 of Sect. 1 becomes the following. Let $$f,g: \mathbb{F}_{q^{n}} \rightarrow \mathbb{F}_{q^{n}}$$ be $$\mathbb{F}_{q}$$-quadratic maps on $$\mathbb{F}_{q^{n}}$$. We say that 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
$$\displaystyle{f(x) = L_{1} \circ g \circ L_{2}\text{ for all }x \in \mathbb{F}_{q^{n}}.}$$
It is a difficult problem to decide whether a given map is planar or not in general. All $$\mathbb{F}_{q}$$-quadratic monic monomial maps on $$\mathbb{F}_{q^{3}}$$ are
$$\displaystyle{x^{2},x^{q+1},x^{2q},x^{q^{2}+1 },x^{q^{2}+q },x^{2q^{2} }.}$$
It is easy to prove that all these $$\mathbb{F}_{q}$$-quadratic monomial maps are planar. Recently the authors characterized all $$\mathbb{F}_{q}$$-quadratic planar binomial maps on $$\mathbb{F}_{q^{3}}$$ explicitly in [16]. Up to multiplication with a nonzero constant in $$\mathbb{F}_{q^{3}}$$, there are 15 distinct $$\mathbb{F}_{q}$$-quadratic binomial maps on $$\mathbb{F}_{q^{3}}$$:
1. (1)

$$x^{2} + \mathit{ux\,}^{q^{2}+q }$$

2. (2)

$$x^{q+1} + \mathit{ux\,}^{2q^{2} }$$

3. (3)

$$x^{2q} + \mathit{ux\,}^{q^{2}+1 }$$

4. (4)

$$x^{2} + \mathit{ux\,}^{q+1}$$

5. (5)

$$x^{2} + \mathit{ux\,}^{2q}$$

6. (6)

$$x^{2} + \mathit{ux\,}^{q^{2}+1 }$$

7. (7)

$$x^{2} + \mathit{ux\,}^{2q^{2} }$$

8. (8)

$$x^{q+1} + \mathit{ux\,}^{2q}$$

9. (9)

$$x^{q+1} + \mathit{ux\,}^{q^{2}+1 }$$

10. (10)

$$x^{q+1} + \mathit{ux\,}^{q^{2}+q }$$

11. (11)

$$x^{2q} + \mathit{ux\,}^{q^{2}+q }$$

12. (12)

$$x^{2q} + \mathit{ux\,}^{2q^{2} }$$

13. (13)

$$x^{q^{2}+1 } + \mathit{ux\,}^{q^{2}+q }$$

14. (14)

$$x^{q^{2}+1 } + \mathit{ux\,}^{2q^{2} }$$

15. (15)

$$x^{q^{2}+q } + \mathit{ux\,}^{2q^{2} }$$

Here 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
$$\displaystyle{ x^{q+1} + \mathit{ux}^{2} \in \mathbb{F}_{ q^{3}}[x]\text{ and }x^{q^{2}+1 } + \mathit{ux}^{2} \in \mathbb{F}_{ q^{3}}[x] }$$
(2)
are not planar (see Lemma 5.1 and page 643 in [12]). It is clear that the items 4, 6, 8, 11, 14, and 15 are all equivalent to one of the maps in (2). For example, regarding the map in item 8, we have
$$\displaystyle{x^{q+1} + \mathit{ux}^{2q} = (x^{q^{2}+1 } + \mathit{ux}^{2}) \circ x^{q}\mod (x^{q^{3} } - x).}$$
Therefore, the maps in items 4, 6, 8, 11, 14, and 15 are not planar for any $$u \in \mathbb{F}_{q^{3}}^{{\ast}}$$. Next we consider the remaining maps in the sublist. The maps in items 5, 7, and 12 are compositions of x2 with a linearized polynomial map. For example, regarding the map in item 5, we have
$$\displaystyle{x^{2} + \mathit{ux}^{2q} = (x + \mathit{ux}^{q}) \circ x^{2}}$$
Therefore the map in item 5 is planar if and only if the map $$x\mapsto x + \mathit{ux}^{q^{2} }$$ is a permutation map. Similarly it is easy to decide whether the maps in items 7 and 12 are planar or not depending on the corresponding linearized polynomials in terms of $$u \in \mathbb{F}_{q^{3}}$$. Moreover they are equivalent to x2 if they are planar.
The maps in items 9, 10, and 13 are compositions of xq+1 with a linearized polynomial map. For example, regarding the map in item 9, we have
$$\displaystyle{x^{q+1} + \mathit{ux}^{q^{2}+1 } \equiv (x + \mathit{ux}^{q^{2} }) \circ x^{q+1}\mod (x^{q^{3} } - x)}$$
Therefore the map in item 9 is planar if and only if the map $$x\mapsto x + ux^{q^{2} }$$ is a permutation map. Similarly the maps in items 10 and 13 are planar if and only if the corresponding linearized polynomials give permutation maps, which depend on $$u \in \mathbb{F}_{q^{3}}^{{\ast}}$$. Moreover they are equivalent to xq+1 if they are planar.
It remains to consider the maps in items 1, 2, and 3. These maps are quite different from the rest of the maps in the list. First the maps in items 2 and 3 are equivalent to the map in item 1 as
$$\displaystyle{x^{q+1} + \mathit{ux}^{2q^{2} } \equiv (x^{q^{2}+q } + \mathit{ux}^{2}) \circ x^{q^{2} }\mod (x^{q^{3} } - x)}$$
and
$$\displaystyle{x^{2q} + \mathit{ux}^{q^{2}+1 } \equiv (x^{2} + \mathit{ux}^{q^{2}+q }) \circ x^{q}\mod (x^{q^{3} } - x).}$$
Hence for $$u \in \mathbb{F}_{q^{3}}^{{\ast}}$$ let Fu be the binomial map given by
$$\displaystyle{F_{u}(x) = x^{q^{2}+q } + \mathit{ux}^{2} \in \mathbb{F}_{ q^{3}}[x].}$$
In [16] the authors prove that Fu(x) is planar for some $$u \in \mathbb{F}_{q^{3}}$$ and equivalent to x2. They also prove that Fu(x) is planar for some $$u \in \mathbb{F}_{q^{3}}$$ and equivalent to xq+1. Therefore the binomial form of Fu(x) is different from the sublist of the binomial forms in 4, 5, , 15 as they cannot be equivalent to both x2 and xq+1 for some different choices of $$u \in \mathbb{F}_{q^{3}}^{{\ast}}$$. Moreover, it seems that proving planarity of Fu(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 Fu(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 Fu(x) is planar if and only if$$q \equiv 1\mod 3$$and
$$\displaystyle{u \in \left (-(G\setminus H) \cup \frac{1} {2}(G\setminus H)\right ).}$$
Moreover we have the followings. Assume that$$q \equiv 1\mod 3$$.
• The polynomial Fu(x) is equivalent to x2if and only if$$u \in \frac{1} {2}(G\setminus H)$$.

• The polynomial Fu(x) is equivalent to xq+1if 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].

A special type of Dembowski–Ostrom is of the form
$$\displaystyle{F(x) = L_{1}(x)L_{2}(x),}$$
where both L1(x) and L2(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
$$\displaystyle{\mathcal{M}(L):\{\alpha \in \mathbb{F}_{q}: L(x) +\alpha x\text{ is bijective on }\mathbb{F}_{q^{n}}\},}$$
which is studied in finite geometry [1, 4, 23].

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
$$\displaystyle{F(x) = x(\mathrm{Tr}_{n}(x) + \mathit{ax})}$$
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].

For the case n = 3, it was proved in [15] that
$$\displaystyle{ F(x) = x(\mathrm{Tr}_{3}(x) + \mathit{ax}) }$$
(3)
is planar for $$a \in \{-1,-2\}$$. Moreover for certain values of $$a \in \mathbb{F}_{q}$$, it was proved 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.
The authors in [15] also conjectured that there is no planar map on $$a \in \mathbb{F}_{q^{4}}$$ of the form
$$\displaystyle{F(x) = x(\mathrm{Tr}_{4}(x) + \mathit{ax})\text{ with }a \in \mathbb{F}_{q^{4}}.}$$

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.

Recall that we define the equivalence of $$\mathbb{F}_{q}$$-quadratic perfect nonlinear maps from $$\mathbb{F}_{q^{n}}$$ to $$\mathbb{F}_{q}^{r}$$ in Definition 1.2 of Sect. 1. Up to recently, such maps were classified completely only in the following cases:
• n ≥ 1, r = 1 (nondegenerate $$\mathbb{F}_{q}$$-quadratic form),

• $$n = 2,r = 2$$ ([10], finite fields),

• $$n = 3,r = 3$$ ([20], finite fields or twisted finite field).

Namely, for (n, r) = (2, 2) all $$\mathbb{F}_{q}$$-quadratic perfect nonlinear maps are equivalent to the map xx2. Moreover for (n, r) = (3, 3) all $$\mathbb{F}_{q}$$-quadratic perfect nonlinear maps are equivalent to one of the following maps:
• xx2 (finite field),

• xxq+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.

First we note that, up to equivalence, an $$\mathbb{F}_{q}$$-quadratic perfect nonlinear map from $$\mathbb{F}_{q^{3}}$$ to $$\mathbb{F}_{q}^{2}$$ has to be of the form
$$\displaystyle{F(x) = \left [\begin{array}{l} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\theta x^{2} + \mathit{wx}^{q+1}) \end{array} \right ]}$$
for some $$\theta,w \in \mathbb{F}_{q^{3}}$$ (see Sect. 3 in [21]). However it is perfect nonlinear only for some $$\theta,w \in \mathbb{F}_{q^{3}}$$. We also provided a characterization result as follows (see Proposition 3.1 in [21]).

### 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
$$\displaystyle{F(x) = \left [\begin{array}{l} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\theta x^{2} + \mathit{wx}^{q+1}) \end{array} \right ].}$$
Then F is perfect nonlinear if and only if the polynomial
$$\displaystyle{T^{3} + A_{ 2}T^{2} + A_{ 1}T + A_{0} \in \mathbb{F}_{q}[T],}$$
where
$$\displaystyle{\begin{array}{l} A_{2} = \mathrm{Tr}(2\theta ), \\ A_{1} = \mathrm{Tr}((2\theta )^{q+1} - w^{2}), \\ A_{0} = \mathrm{Norm}(2\theta ) -\mathrm{Tr}(2\theta w^{2q}) + 2\mathrm{Norm}(w) \end{array} }$$
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 F1and F2be the$$\mathbb{F}_{q}$$-quadratic maps given by
$$\displaystyle{F_{1}(x) = \left [\begin{array}{l} \mathrm{Tr}(w_{1}x^{2}) \\ \mathrm{Tr}(w_{2}x^{2}) \end{array} \right ]\text{ and }F_{2}(x) = \left [\begin{array}{l} \mathrm{Tr}(w_{3}x^{q+1}) \\ \mathrm{Tr}(w_{4}x^{q+1}) \end{array} \right ]}$$
For$$w \in \mathbb{F}_{q^{3}}\setminus \mathbb{F}_{q}$$let G and H be the$$\mathbb{F}_{q}$$-quadratic maps
$$\displaystyle{G(x) = \left [\begin{array}{l} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{2}) \end{array} \right ]\text{ and }H(x) = \left [\begin{array}{l} \mathrm{Tr}(x^{q+1}) \\ \mathrm{Tr}(\mathit{wx}^{q+1}) \end{array} \right ].}$$
Then all of the maps F1,F2,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
$$\displaystyle{F(x) = \left [\begin{array}{l} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\theta x^{2} + \mathit{wx}^{q+1}) \end{array} \right ]}$$
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
$$\displaystyle{\mathcal{T}_{1}: x_{0}^{2} + x_{ 1}^{2} + x_{ 2}^{2} = 0.}$$
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
$$\displaystyle{\mathcal{T}_{2}: (x_{0}^{2}\theta + x_{ 1}^{2}\theta ^{q^{2} } + x_{2}^{2}\theta ^{q}) + (x_{ 0}x_{1}w^{q^{2} } + x_{0}x_{1}w + x_{1}x_{2}w^{q}) = 0.}$$
Let$$E: \mathit{PG}(2, \mathbb{F}_{q^{3}}) \rightarrow PG(2, \mathbb{F}_{q^{3}})$$be the action defined as
$$\displaystyle{(x_{0}: x_{1}: x_{2})\mapsto (x_{0}^{q}: x_{ 1}^{q}: x_{ 2}^{q}).}$$
Then we have the following:
1. (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. (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
$$\displaystyle{\left [\begin{array}{l} \mathrm{Tr}(\delta _{1}x^{q+1}) \\ \mathrm{Tr}(\delta _{2}x^{q+1})\end{array} \right ] = F(L(x))\text{ for all }x \in \mathbb{F}_{q^{3}}.}$$

Note that Theorem 2.5 gives an explicit geometric method for finding an equivalence from F(x) to an $$\mathbb{F}_{q}$$-quadratic perfect nonlinear map in the form
$$\displaystyle{\left [\begin{array}{l} \mathrm{Tr}(\delta _{1}x^{q+1}) \\ \mathrm{Tr}(\delta _{2}x^{q+1}) \end{array} \right ].}$$
As it is easy to find equivalence for the $$\mathbb{F}_{q}$$-quadratic perfect nonlinear maps in between the maps F1, F2, 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

Let f1(x) = F(x). Note that f1(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
$$\displaystyle{\left [\begin{array}{c} f_{1}(x) \\ f_{2}(x)\\ \vdots \\ f_{n}(x) \end{array} \right ]}$$
is an $$\mathbb{F}_{q}$$-quadratic perfect nonlinear map (or planar map) from $$\mathbb{F}_{q^{n}}$$ to $$\mathbb{F}_{q}^{n}$$. This shows that F is extendable.
Assume first that 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 g1(x) and h1(x) are nondegenerate $$\mathbb{F}_{q}$$-quadratic forms on $$\mathbb{F}_{q^{n}}$$. Moreover the discriminants of g1(x) and h1(x) are not both square or both nonsquare in $$\mathbb{F}_{q}^{{\ast}}$$. Here we use the fact that n is odd and λn 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
$$\displaystyle{f_{1}(L(x)) = g_{1}(x)\text{ or }h_{1}(x)\text{ for all }x \in \mathbb{F}_{q^{n}}.}$$
As xx2 and xλ x2 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
$$\displaystyle{\left [\begin{array}{c} g_{1}(x) \\ g_{2}(x)\\ \vdots \\ g_{n}(x) \end{array} \right ]\text{ and }\left [\begin{array}{c} h_{1}(x) \\ h_{2}(x)\\ \vdots \\ h_{n}(x) \end{array} \right ]}$$
are $$\mathbb{F}_{q}$$-quadratic nonlinear maps from $$\mathbb{F}_{q^{n}}$$ to $$\mathbb{F}_{q}^{n}$$. Therefore
$$\displaystyle{\left [\begin{array}{c} f_{1}(x) \\ g_{2}(L^{-1}(x))\\ \vdots \\ g_{n}(L^{-1}(x)) \end{array} \right ]\text{ or }\left [\begin{array}{c} f_{1}(x) \\ h_{2}(L^{-1}(x))\\ \vdots \\ h_{n}(L^{-1}(x)) \end{array} \right ]}$$
is an $$\mathbb{F}_{q}$$-quadratic perfect nonlinear map from $$\mathbb{F}_{q^{n}}$$ to $$\mathbb{F}_{q}^{n}$$ if f1(L(x)) = g1(x) or $$f_{1}(L(x)) = h_{1}(x)$$, respectively. This completes the proof if n is odd.
Assume next then 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 g1(x) and h1(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
$$\displaystyle{\eta (u) = \left \{\begin{array}{ll} 1 &\text{if }u\text{ is a square in }\mathbb{F}_{q}^{{\ast}}, \\ - 1&\text{if }u\text{ is not a square in }\mathbb{F}_{q}^{{\ast}}. \end{array} \right.}$$
Let $$Q: \mathbb{F}_{q^{n}} \rightarrow \mathbb{F}_{q}$$ be an arbitrary nondegenerate $$\mathbb{F}_{q}$$-quadratic form on $$\mathbb{F}_{q^{n}}$$ with discriminant $$\Delta \in \mathbb{F}_{q}^{{\ast}}$$. Let $$b \in \mathbb{F}_{q^{{\ast}}}$$ be an arbitrary nonzero element. It is well known that (see, e.g., Theorem 6.26 in [18])
$$\displaystyle{ \vert \{x \in \mathbb{F}_{q^{n}}: Q(x) = b\}\vert = q^{n-1} - q^{\frac{n-2} {2} }\eta (-1)^{n/2}\Delta. }$$
(4)
Note that we also have
$$\displaystyle{ \begin{array}{lll} 2q^{n-1} & =&2\vert \{y \in \mathbb{F}_{q^{n}}: \mathrm{Tr}(y) = b\}\vert \\ \quad & =&\vert \{x \in \mathbb{F}_{q^{n}}: \mathrm{Tr}(x^{2}) = b\}\vert + \vert \{x \in \mathbb{F}_{q^{n}}: \mathrm{Tr}(\lambda x^{2}) = b\}\vert \end{array} }$$
(5)
Here we use the fact that if $$y \in \mathbb{F}_{q^{n}}^{{\ast}}$$ is a square, then
$$\displaystyle{\vert \{x \in \mathbb{F}_{q^{n}}: y = x^{2}\}\vert = 2\text{ and }\vert \{x \in \mathbb{F}_{ q^{n}}: y =\lambda x^{2}\}\vert = 0;}$$
otherwise if $$y \in \mathbb{F}_{q^{n}}^{{\ast}}$$ is not a square, then
$$\displaystyle{\vert \{x \in \mathbb{F}_{q^{n}}: y = x^{2}\}\vert = 0\text{ and }\vert \{x \in \mathbb{F}_{ q^{n}}: y =\lambda x^{2}\}\vert = 2.}$$
Assume the contrary that $$\eta (\Delta (g_{1})) =\eta (\Delta (h_{1}))$$. Using (4) and (5), we obtain that
$$\displaystyle{2q^{n-1} = 2(q^{n-1} \pm q^{\frac{n-2} {2} })}$$
which is a contradiction. Therefore $$\eta (\Delta (g_{1}))\neq \eta (\Delta (h_{1}))$$, which completes the proof. □

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

By Theorems 2.4 and 2.5 of Sect. 2, we know that F is equivalent to
$$\displaystyle{\left [\begin{array}{l} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{2}) \end{array} \right ]\text{ and }\left [\begin{array}{l} \mathrm{Tr}(x^{q+1}) \\ \mathrm{Tr}(\mathit{wx}^{q+1}) \end{array} \right ]}$$
for $$w \in \mathbb{F}_{q^{3}}\setminus \mathbb{F}_{q}$$. Hence F is extendable to both of the maps xx2 and xxq+1 on $$\mathbb{F}_{q^{3}}$$. □

### Example 3.3

Let 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
$$\displaystyle{x\mapsto x^{4} + x^{10} - x^{36}\text{ and }x\mapsto x^{2}.}$$
Using computer, we have verified that all $$\mathbb{F}_{3}$$-quadratic (q4, q2)-bent maps are extendable.

### Example 3.4

Let 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
• xx2.

• xxq+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 (q5, q2)-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 x2 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 E5 below. Only E5 extends to both x2 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

$$\displaystyle{E_{1}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{10}) \\ \mathrm{Tr}(w^{2}x^{10} + w^{5}x^{4}) \end{array} \right ],\;E_{2}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{10}) \\ \mathrm{Tr}(w^{2}x^{2}) \end{array} \right ],\;E_{3}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{10}) \\ \mathrm{Tr}(w^{8}x^{2}) \end{array} \right ],}$$
$$\displaystyle{E_{4}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(w^{13}x^{10}) \\ \mathrm{Tr}(w^{5}x^{4}) \end{array} \right ],\;E_{5}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{4}) \\ \mathrm{Tr}(w^{2}x^{4}) \end{array} \right ].}$$

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

$$\displaystyle\begin{array}{rcl} & & \qquad \mathit{NE}_{1}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{10}) \\ \mathrm{Tr}(wx^{4}) \end{array} \right ],\:\mathit{NE}_{2}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{10}) \\ \mathrm{Tr}(w^{5}x^{4}) \end{array} \right ],\:\mathit{NE}_{3}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{10}) \\ \mathrm{Tr}(w^{7}x^{4}) \end{array} \right ], {}\\ & & \mathit{NE}_{4}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{10}) \\ \mathrm{Tr}(w^{3}x^{10} + w^{8}x^{4} + \mathit{wx}^{2}) \end{array} \right ],\:\mathit{NE}_{5}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{10}) \\ \mathrm{Tr}(w^{3}x^{10} + w^{25}x^{4} + \mathit{wx}^{2}) \end{array} \right ], {}\\ & & \quad \mathit{NE}_{6}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{10}) \\ \mathrm{Tr}(w^{3}x^{10} + \mathit{wx}^{4} + w^{2}x^{2}) \end{array} \right ],\:\mathit{NE}_{7}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{10}) \\ \mathrm{Tr}(w^{30}x^{4} + w^{6}x^{2}) \end{array} \right ], {}\\ & & \qquad \qquad \quad \mathit{NE}_{8}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{10}) \\ \mathrm{Tr}(w^{15}x^{2}) \end{array} \right ],\:\mathit{NE}_{9}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{10}) \\ \mathrm{Tr}(w^{4}x^{4} + w^{15}x^{2}) \end{array} \right ], {}\\ & & \qquad \qquad \quad \mathit{NE}_{10}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(w^{13}x^{10}) \\ \mathrm{Tr}(\mathit{wx}^{4}) \end{array} \right ],\:\mathit{NE}_{11}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{4}) \\ \mathrm{Tr}(w^{2}x^{10} + \mathit{wx}^{2}) \end{array} \right ], {}\\ & & \mathit{NE}_{12}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{4}) \\ \mathrm{Tr}(w^{14}x^{10} + w^{8}x^{4} + \mathit{wx}^{2}) \end{array} \right ],\:\mathit{NE}_{13}(x) = \left [\begin{array}{ll} \mathrm{Tr}(x^{2}) \\ \mathrm{Tr}(\mathit{wx}^{4}) \\ \mathrm{Tr}(w^{4}x^{10} + w^{8}x^{4} + w^{2}x^{2}) \end{array} \right ]. {}\\ \end{array}$$

These are the non-extendable maps.

We summarize these results in Fig. 1 in the form of a lattice.

## 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.

Recall that a field is a nonempty set $$\mathbb{F}$$ with two binary operations + and × satisfying the following axioms:
• $$(\mathbb{F},+)$$ 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

Let $$\mathbb{R}$$ be the field of real numbers. Let $$\hat{i},\hat{j},\hat{k}$$ be the symbols satisfying
$$\displaystyle{ \hat{i}^{2} =\hat{ j}^{2} =\hat{ k}^{2} = -1,\:\hat{i}\hat{j} = -\hat{j}\hat{i} =\hat{ k},\:\hat{j}\hat{k} = -\hat{k}\hat{j} =\hat{ i},\:\hat{k}\hat{i} = -\hat{i}\hat{k} =\hat{ j}. }$$
(6)
Let $$\mathbb{H}$$ be the set consisting of
$$\displaystyle{a + \mathit{b\hat{i}} + \mathit{c\hat{j}} + \mathit{d\hat{k}}\text{ with }a,b,c,d \in \mathbb{R}.}$$
We define + on $$\mathbb{H}$$ componentwise. We define × on $$\mathbb{H}$$ using (6), distributive law and multiplication on $$\mathbb{R}$$. Then $$\mathbb{H}$$ is called the Hamilton quaternions, and it is a skew field, which is not a field.

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

A finite 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 ab = 0, then a = 0 or b = 0.

If $$\mathbb{S}$$ is a presemifield and there exists $$e \in \mathbb{S}\setminus \{0\}$$ such that
$$\displaystyle{e {\ast} a = a {\ast} e = a\text{ for all }a \in \mathbb{S},}$$
then $$\mathbb{S}$$ is called semifield.
If $$\mathbb{S}$$ is presemifield (or semifield) and
$$\displaystyle{a {\ast} b = b {\ast} a\text{ for all }a,b \in \mathbb{S},}$$
then $$\mathbb{S}$$ is called commutative presemifield (or commutative semifield).

The additive group of a finite semifield is elementary abelian. Therefore the order of $$\mathbb{S}$$ is pn 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

Two finite presemifields $$(\mathbb{S},+,{\ast})$$ and $$(\mathbb{S},+,\star )$$ of order pn are isotopic if there exist linearized permutation polynomials L, M, N over $$\mathbb{F}_{p^{n}}$$ such that
$$\displaystyle{M(x) \star N(y) = L(x {\ast} y)\text{ for all }x,y \in \mathbb{S}.}$$

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.

There is not much restriction in considering semifields instead of presemifields. Indeed any presemifield is isotopic to a semifield. Let $$(\mathbb{S},+,{\ast})$$ be a presemifield. Choose any nonzero element $$a \in \mathbb{S}$$. Let $$(\mathbb{S},+,\star )$$ be the semifield with the binary operation $$\star: \mathbb{S} \times \mathbb{S} \rightarrow \mathbb{S}$$ defined using
$$\displaystyle{x {\ast} y = (x {\ast} a) \star (a {\ast} y).}$$
Namely, if $$\alpha,\beta \in \mathbb{S}$$, then there exists uniquely determined $$x,y \in \mathbb{S}$$ such that xa = α and ay = β. Then αβ is defined as
$$\displaystyle{\alpha \star \beta = x {\ast} y.}$$
It is now clear that e = aa 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.
Let $$(\mathbb{S},+,{\ast})$$ be a semifield. The subsets
$$\displaystyle\begin{array}{rcl} & & \mathcal{N}_{l}(\mathbb{S}) =\{ a \in \mathbb{S}: (a {\ast} x) {\ast} y = a {\ast} (x {\ast} y)\text{ for all }x,y \in \mathbb{S}\} {}\\ & & \mathcal{N}_{m}(\mathbb{S}) =\{ a \in \mathbb{S}: (x {\ast} a) {\ast} y = x {\ast} (a {\ast} y)\text{ for all }x,y \in \mathbb{S}\} {}\\ & & \mathcal{N}_{r}(\mathbb{S}) =\{ a \in \mathbb{S}: (x {\ast} y) {\ast} a = x {\ast} (y {\ast} a)\text{ for all }x,y \in \mathbb{S}\} {}\\ \end{array}$$
are called the left, middle, and right nucleus of $$\mathbb{S}$$, respectively. They are in fact finite fields. The subset
$$\displaystyle{\mathcal{N}(\mathbb{S}) = \mathcal{N}_{l}(\mathbb{S}) \cap \mathcal{N}_{m}(\mathbb{S}) \cap \mathcal{N}_{r}(\mathbb{S})}$$
is called the nucleus (or associative center) of $$\mathbb{S}$$. The center$$C(\mathbb{S})$$ of $$\mathbb{S}$$ is defined by
$$\displaystyle{C(\mathbb{S}) = \mathcal{N}(\mathbb{S}) \cap \{ c \in \mathbb{S}: c {\ast} a = a {\ast} c\text{ for any }a \in \mathbb{S}\}.}$$
It is known that $$\mathbb{S}$$ is a division algebra over its center. Moreover the nuclei are invariant under isotopism.
Let $$f: \mathbb{F}_{p^{n}} \rightarrow \mathbb{F}_{p^{n}}$$ be an $$\mathbb{F}_{p}$$-quadratic polynomial map. If f is planar, then we get a commutative presemifield $$(\mathbb{S}_{f},+,{\ast})$$ defined as
$$\displaystyle{x {\ast} y = \frac{1} {2}(f(x + y) - f(x) - f(y)).}$$
Conversely if $$(\mathbb{S},+,\star )$$ is a finite commutative semifield of odd characteristic with order pn, 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 pn starting from a given finite semifield of order pn.

Let $$F: \mathbb{F}_{q^{n}} \rightarrow \mathbb{F}_{q^{n}}$$ be a perfect nonlinear and $$\mathbb{F}_{q}$$-quadratic map. For $$a \in \mathbb{F}_{q^{n}}$$, the corresponding difference map is
$$\displaystyle{ \begin{array}{lrl} D_{a}:&\mathbb{F}_{q^{n}} & \rightarrow \mathbb{F}_{q^{n}} \\ & x&\mapsto F(x + a) - F(x) - F(a). \end{array} }$$
Note that D0 is the zero map and Da 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
$$\displaystyle{ x \circ y = D_{y}(x). }$$
Let $$e_{1},e_{2},\ldots,e_{n}$$ be a basis of $$\mathbb{F}_{q^{n}}$$ over $$\mathbb{F}_{q^{n}}.$$ The presemifield is determined by the operations
$$\displaystyle{ D_{e_{j}}(e_{i}) = e_{i} \circ e_{j} =\sum _{ k=1}^{n}a_{\mathit{ ijk}}e_{k}\text{ for }1 \leq i,j \leq n. }$$
Here $$(a_{\mathit{ijk}}) =\mathop{ (a_{\mathit{ijk}})}\limits_{1 \leq i,j,k \leq n}$$ is Knuth’s cubical array with entries from $$\mathbb{F}_{q}.$$ Namely, up to a choice of basis, $$(\mathbb{S},+,\circ )$$ is determined by Knuth’s cubical array (aijk).

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

For 1 ≤ i, j, k ≤ n, the slice $$M(\mathbb{S},j)$$ of the cubical array (aijk) is an n × n matrix with entries from $$\mathbb{F}_{q}$$ defined as
$$\displaystyle{ M(\mathbb{S},j) =\mathop{ (a_{\mathit{ijk}})}\limits_{1 \leq i,k \leq n}. }$$
The matrix $$M(\mathbb{S},j)$$ is the matrix of the linear transformation
$$\displaystyle{ D_{e_{j}}(x) = y }$$
via left multiplication
$$\displaystyle{ \underline{x} \in \mathbb{F}_{q}^{n}\mapsto \underline{y} =\underline{ x} \cdot M(\mathbb{S},j). }$$
Here we have $$\underline{x} = (x_{1},x_{2},\ldots,x_{n}),\ \underline{y} = (y_{1},y_{2},\ldots,y_{n}),\ x =\sum \limits _{ k=1}^{n}x_{ i}e_{i},\ y =\sum \limits _{ k=1}^{n}y_{ i}e_{i}$$, and $$D_{e_{j}}(x) = y.$$ The matrices $$M(\mathbb{S},1),M(\mathbb{S},2),\ldots,M(\mathbb{S},n)$$ are linearly independent over $$\mathbb{F}_{q}.$$ In fact their linear span is a spread set, which is defined, for example, in [2]. A spread set also determines $$\mathbb{S}.$$
The matrices $$M(\mathbb{S},1),M(\mathbb{S},2),\ldots,M(\mathbb{S},n)$$ are not necessarily symmetric. Recall that $$M(\mathbb{S},j)$$ is obtained from the cubical array (aijk) by fixing j. Let us fix k instead. For 1 ≤ k ≤ n,  the slice $$\hat{M}(\mathbb{S},k)$$ of the cubical array (aijk) is an n × n matrix with entries from $$\mathbb{F}_{q}$$ defined as
$$\displaystyle{ \hat{M}(\mathbb{S},k) =\mathop{ (a_{\mathit{ijk}})}\limits_{1 \leq i,j \leq n}. }$$
(7)
As $$\mathbb{S}$$ is commutative, $$\hat{M}(\mathbb{S},k)$$ is symmetric for all 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 (aijk) 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.

The permuted cubical array (ajik) corresponds to the permutation (12) and gives the opposite semifield $$\mathbb{S}^{{\ast}}$$. The permuted cubical array (akji) corresponds to the permutation (13) and gives the dual semifield $$\mathbb{S}^{d}$$. In fact we have the lattices in Fig. 2 in general.
Moreover if $$\mathbb{S}$$ is commutative, then we only have the lattice in Figure 3.
As $$\mathbb{S}$$ is commutative, it is well known that $$\mathbb{S}^{d{\ast}}$$ is symplectic (see [2]). This means that the matrices (cf [13, 17] )$$M(\mathbb{S}^{d{\ast}},1),M(\mathbb{S}^{d{\ast}},2),\ldots,M(\mathbb{S}^{d{\ast}},n)$$ are all symmetric (see [13]). It is not difficult to observe that for the matrices $$\hat{M}(\mathbb{S},1),\hat{M}(\mathbb{S},2),\ldots,\hat{M}(\mathbb{S},n)$$ obtained in (7), we have
$$\displaystyle{ \hat{M}(\mathbb{S},1) = M(\mathbb{S}^{d{\ast}},1),\hat{M}(\mathbb{S},2) = M(\mathbb{S}^{d{\ast}},2),\ldots,\hat{M}(\mathbb{S},n) = M(\mathbb{S}^{d{\ast}},n). }$$

In other words slicing Knuth’s cubical array (aijk) 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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