# An exponential bound on the number of non-isotopic commutative semifields

Faruk Göloğlu  Charles University in Prague<EMAIL_ADDRESS>and  Lukas
Kölsch  University of Rostock, Germany; University of South Florida
<EMAIL_ADDRESS>
###### Abstract.

We show that the number of non-isotopic commutative semifields of odd order
$p^{n}$ is exponential in $n$ when $n=4t$ and $t$ is not a power of $2$. We
introduce a new family of commutative semifields and a method for proving
isotopy results on commutative semifields that we use to deduce the
aforementioned bound. The previous best bound on the number of non-isotopic
commutative semifields of odd order was quadratic in $n$ and given by Zhou and
Pott [Adv. Math. 234 (2013)]. Similar bounds in the case of even order were
given in Kantor [J. Algebra 270 (2003)] and Kantor and Williams [Trans. Amer.
Math. Soc. 356 (2004)].

###### 2020 Mathematics Subject Classification:

Primary 12K10, 17A35; Secondary 51A35, 51A40

##  1\. Introduction

In this paper, we show that the number $N_{p^{n}}$ of non-isotopic commutative
semifields of odd order $p^{n}$ is exponential in $n$ when $n=4t$. To be
precise, we prove for every odd prime $p$,

$N_{p^{n}}\geq\frac{(\sigma(n)-1)(p^{n/4}-1)}{2n},$

when $\nu_{2}(n)\geq 2$, where we denote by $\sigma(n)$ the odd part of an
integer $n$ (i.e., $\sigma(n)=n/2^{\nu_{2}(n)}$), and by $\nu_{2}(n)$ the
$2$-adic valuation of $n$ (i.e., $2^{\nu_{2}(n)}|n$ and $2^{\nu_{2}(n)+1}\nmid
n$). For odd $p$, the previous best bound on $N_{p^{n}}$ was quadratic in $n$
and was proved in [35, Corollary 1]:

$N_{p^{n}}\geq\frac{n(\sigma(n)-1)}{8}+cn,$

when $\nu_{2}(n)\geq 1$ and $c$ a constant. When $p$ and $n$ are odd, the
known number for $N_{p^{n}}$ is linear in $n$. The problem of determining
whether the number $N_{p^{n}}$ can be bounded by a polynomial in $n$ has been
described [29, p. 180] as “the main problem in connection with commutative
semifields of [odd] order $p^{n}$.” Note that it is impossible to find
families with exponentially many non-isotopic commutative semifields of order
$p^{n}$ for arbitrary $p,n$. Indeed, by a result of Menichetti [27, Corollary
33], all commutative semifields of order $p^{n}$ with $n$ prime and $p$ large
enough are isotopic to the finite field or a twisted field (see Section 3.2).
It is thus impossible to give an exponential count for all $p,n$. The problem
in the characteristic $2$ case was solved almost two decades ago. Kantor [19,
Theorem 1.1] showed that the number of non-isotopic commutative semifields of
order $2^{km}$ is at least

$N_{2^{km}}\geq\frac{2^{km(\rho(m)-1)}}{k^{2}m^{4}},$

when $m>1$ is odd and $m$ is not a power of $3$ (where we denote by $\rho(m)$
the number of prime factors of $m$ counting multiplicities), using a
construction by Kantor and Williams [20, Theorem 1.7]. In these papers,
finding a large number of (commutative) semifields in odd characteristic and
finding a general approach to proving non-isotopy were posed as important open
problems [20, p. 936],[19, p. 112].

To prove the bound we introduce a new family of commutative semifields. These
semifields satisfy a property that we call biprojectivity, which also applies
to many known semifields of square order. The biprojective structure allows us
to develop a technique of determining isotopy between semifields. This
technique is key to proving the exponential bound on non-isotopic commutative
semifields.

In Section 2, we give the preliminaries. In Section 3, we define biprojective
semifields and give a quick survey on known commutative semifields and their
counts. Section 4 is devoted to proving the semifield property of our family
(Theorem 4.4). Section 5 introduces our technique of proving isotopy between
semifields (Theorem 5.10). In Section 6, we give the number of non-isotopic
semifields arising from our family (Theorem 6.2). Section 6 contains our main
result that the number of non-isotopic commutative semifields of odd order
$p^{n}$ is exponential in $n$ (Corollaries 6.4 and 6.5). In Section 7, we
compute the nuclei associated to our semifields (Theorem 7.2). Finally, in
Section 8, we show that our semifields are indeed new and not isotopic to most
known semifields (Theorem 8.1).

##  2\. Preliminaries

A finite semifield $\mathbb{S}=(S,+,\circ)$ is a set $S$ equipped with two
operations $(+,\circ)$ satisfying the following axioms.

  1. (S1)

$(S,+)$ is a group.

  2. (S2)

For all $x,y,z\in S$,

     * •

$x\circ(y+z)=x\circ y+x\circ z$,

     * •

$(x+y)\circ z=x\circ z+y\circ z$.

  3. (S3)

For all $x,y\in S$, $x\circ y=0$ implies $x=0$ or $y=0$.

  4. (S4)

There exists $\epsilon\in S$ such that $x\circ\epsilon=x=\epsilon\circ x$.

In this paper, we will be interested only in finite semifields. Henceforth,
when we say a semifield we will mean a finite semifield. An algebraic object
satisfying the first three of the above axioms is called a pre-semifield. If
$\mathbb{P}=(P,+,\circ)$ is a pre-semifield, then $(P,+)$ is an elementary
abelian $p$-group [22, p. 185], and $(P,+)$ can be viewed as an
$n$-dimensional $\mathbb{F}_{p}$-vector space $\mathbb{F}_{p}^{n}$. If $\circ$
is associative then $\mathbb{S}$ is the finite field $\mathbb{F}_{p^{n}}$ by
Wedderburn’s theorem which states that a finite division ring is a field. By a
result of Menichetti (known as Kaplansky’s conjecture [26]) when $n>2$, there
exist proper semifields of odd order $p^{n}$ where $\circ$ is non-associative.
There are no proper semifields of order $2^{3}$. For $n>3$, there exists
proper semifields of order $2^{n}$ [22]. A pre-semifield
$\mathbb{P}=(\mathbb{F}_{p}^{n},+,\circ)$ can be converted to a semifield
$\mathbb{S}=(\mathbb{F}_{p}^{n},+,\ast)$ using Kaplansky’s trick by defining
the new multiplication as

$(x\circ e)\ast(e\circ y)=(x\circ y),$

for any nonzero element $e\in\mathbb{F}_{p}^{n}$, making $(e\circ e)$ the
multiplicative identity of $\mathbb{S}$. A pre-semifield is an
$\mathbb{F}_{p}$-algebra, thus the multiplication is bilinear. Therefore we
have $\mathbb{F}_{p}$-bilinear
$B:\mathbb{F}_{p}^{n}\times\mathbb{F}_{p}^{n}\to\mathbb{F}_{p}^{n}$,
satisfying

$B(x,y)=x\circ y,$

and $\mathbb{F}_{p}$-linear left and right multiplications
$L_{x},R_{y}:\mathbb{F}_{p}^{n}\to\mathbb{F}_{p}^{n}$, with

$L_{x}(y):=B(x,y)=:R_{y}(x).$

The mapping $L_{x}$ (resp. $R_{y}$) is a bijection whenever $x\neq 0$ (resp.
$y\neq 0$) by (S3). Thus,

$R_{e}(x)\ast L_{e}(y)=x\circ y.$

Two pre-semifields $\mathbb{P}_{1}=(\mathbb{F}_{p}^{n},+,\circ_{1})$ and
$\mathbb{P}_{2}=(\mathbb{F}_{p}^{n},+,\circ_{2})$ are said to be isotopic if
there exist $\mathbb{F}_{p}$-linear bijections $L,M$ and $N$ of
$\mathbb{F}_{p}^{n}$ satisfying

$N(x\circ_{1}y)=L(x)\circ_{2}M(y).$

Such a triple $\gamma=(N,L,M)$ is called an isotopism between $\mathbb{P}_{1}$
and $\mathbb{P}_{2}$. If additionally $L=M$ holds, we call $\gamma$ a strong
isotopism and $\mathbb{P}_{1}$ and $\mathbb{P}_{2}$ strongly isotopic.
Isotopisms between a pre-semifield $\mathbb{P}$ and itself are called
autotopisms. Thus the pre-semifield $\mathbb{P}$ and the corresponding
semifield $\mathbb{S}$ constructed by Kaplansky’s trick are isotopic and even
strongly isotopic if $\mathbb{P}$ is commutative. Isotopy of pre-semifields is
an equivalence relation and the isotopism class of a pre-semifield
$\mathbb{P}$ is denoted by $[\mathbb{P}]$. Semifields coordinatize projective
planes and different semifields coordinatize isomorphic planes if and only if
they are isotopic ([1], see [22, Section 3] for a detailed treatment).
Semifields are further equivalent to maximum rank distance codes with certain
parameters (see e.g. [31]) and can be used to construct relative difference
sets (see [30]). Associative substructures of a semifield
$\mathbb{S}=(\mathbb{F}_{p}^{n},+,\ast)$, namely the left, middle and right
nuclei, are defined as follows:

$\displaystyle\mathbb{N}_{l}(\mathbb{S})$
$\displaystyle:=\\{x\in\mathbb{F}_{p}^{n}\ :\ (x\ast y)\ast z=x\ast(y\ast z),\
y,z\in\mathbb{F}_{p}^{n}\\},$ $\displaystyle\mathbb{N}_{m}(\mathbb{S})$
$\displaystyle:=\\{y\in\mathbb{F}_{p}^{n}\ :\ (x\ast y)\ast z=x\ast(y\ast z),\
x,z\in\mathbb{F}_{p}^{n}\\},$ $\displaystyle\mathbb{N}_{r}(\mathbb{S})$
$\displaystyle:=\\{z\in\mathbb{F}_{p}^{n}\ :\ (x\ast y)\ast z=x\ast(y\ast z),\
x,y\in\mathbb{F}_{p}^{n}\\}.$

It is easy to check that
$\mathbb{N}_{l}(\mathbb{S}),\mathbb{N}_{m}(\mathbb{S}),\mathbb{N}_{r}(\mathbb{S})\subseteq\mathbb{F}_{p^{n}}$
are finite fields and if $\mathbb{S}$ is commutative then
$\mathbb{N}_{l}(\mathbb{S})=\mathbb{N}_{r}(\mathbb{S})$. Nuclei are isotopy
invariants for semifields. Since every pre-semifield
$\mathbb{P}\in[\mathbb{S}]$ for some semifield $\mathbb{S}$, the nuclei can be
thought to extend to pre-semifields. Thus, when we speak of the nuclei of a
pre-semifield $\mathbb{P}$ we mean the nuclei of the isotopic semifield
$\mathbb{S}$.

Let $\operatorname{End}(\mathbb{F}_{p}^{n})$ denote the
$\mathbb{F}_{p}$-linear endomorphisms of the vector space
$\mathbb{F}_{p}^{n}$. Every $\mathbb{F}_{p}$-linear mapping
$L\in\operatorname{End}(\mathbb{F}_{p}^{n})$ can be written uniquely as a
linearized polynomial

$L(x)=\sum_{i=0}^{n-1}b_{i}x^{p^{i}},$

in the polynomial ring $\mathbb{F}_{p^{n}}[x]$. We will not make distinction
between mappings and the polynomials. Let $p$ be an odd prime and consider the
polynomials from $\mathbb{F}_{p^{n}}[x]$ of the form

$F(x)=\sum_{0\leq i,j<n}a_{ij}x^{p^{i}+p^{j}}.$

These polynomials are called Dembowski-Ostrom (DO) polynomials. The
polarization of a DO polynomial $F$ is defined as

$\Delta_{F}(x,y)=F(x+y)-F(x)-F(y).$

The mapping
$\Delta_{F}:\mathbb{F}_{p}^{n}\times\mathbb{F}_{p}^{n}\to\mathbb{F}_{p}^{n}$
is symmetric and $\mathbb{F}_{p}$-bilinear, thus if $\Delta_{F}(x,a)=0$
implies $x=0$ for all
$a\in\mathbb{F}_{p^{n}}^{\times}=\mathbb{F}_{p^{n}}\setminus\\{0\\}$, then
$\Delta_{F}(x,y)$ describes a commutative pre-semifield multiplication [10].
Conversely, by a counting argument, every commutative pre-semifield
multiplication can be written as $\Delta_{F}(x,y)$ for some DO polynomial $F$
[9]. In that case we will call $F$ a planar DO polynomial/mapping. Strong
isotopy between pre-semifields can be recognized also in the corresponding
planar DO polynomials:

######  Theorem 2.1.

[8, Theorem 3.5.] Let $F,G\in\mathbb{F}_{p^{n}}[x]$ be planar DO polynomials
and $\mathbb{P}_{1}$, $\mathbb{P}_{2}$ be the corresponding pre-semifields.
Then $\mathbb{P}_{1}$ and $\mathbb{P}_{2}$ are strongly isotopic via an
isotopism $\gamma=(N,L,L)$ if and only if $F=NGL^{-1}$.

Consequently, we say that two planar DO polynomials $F,G$ are equivalent if
bijective linear mappings $L_{1},L_{2}$ exist such that $F=L_{1}GL_{2}$. Note
that this type of equivalence is the most general equivalence known to
preserve the planarity of a DO polynomial, see [23].

##  3\. Biprojective planar mappings and commutative semifields

In this paper we are interested in planar DO polynomials of a specific form.
Let $\mathbb{F}=\mathbb{F}_{p^{n}}$ be a finite field of square odd order and
$\mathbb{M}=\mathbb{F}_{p^{m}}$ with $n=2m$. Let

$F(x,y)=\left[f(x,y),g(x,y)\right],$

where

$\displaystyle f(x,y)$
$\displaystyle=a_{0}x^{q+1}+b_{0}x^{q}y+c_{0}xy^{q}+d_{0}y^{q+1},$
$\displaystyle g(x,y)$
$\displaystyle=a_{1}x^{r+1}+b_{1}x^{r}y+c_{1}xy^{r}+d_{1}y^{r+1},$

with $q=p^{k},r=p^{l},1\leq k,l\leq m$. We will call $f(x,y)$ a
$q$-biprojective polynomial and $(x,y)\mapsto F(x,y)$ a $(q,r)$-biprojective
mapping (of $\mathbb{M}\times\mathbb{M}$). Note that $F(x,y)$ is a
$(q,r)$-biprojective polynomial pair. We will not make any distinction between
the polynomials and the mappings defined by them. We also let
${\overline{q}}=p^{m-k}$ and $\overline{r}=p^{m-l}$, so that
$q{\overline{q}}\equiv r\overline{r}\equiv 1\pmod{p^{m}-1}$. We are going to
use the shorthand notation

$\displaystyle f(x,y)$ $\displaystyle=(a_{0},b_{0},c_{0},d_{0})_{q},$
$\displaystyle g(x,y)$ $\displaystyle=(a_{1},b_{1},c_{1},d_{1})_{r}.$

We are going to refer to $f$ and $g$ as (left and right) components of $F$. We
refer the reader to [5] for projective polynomials over finite fields.

The polarization of a planar $(q,r)$-biprojective mapping defines a
$(q,r)$-biprojective (commutative) pre-semifield
$\mathbb{P}=(\mathbb{M}\times\mathbb{M},+,\ast)$. It is easy to see that both
components correspond to homogeneous operations due to biprojectivity:

(3.1) $\displaystyle(x,y)\ast(y,v)=(($ $\displaystyle
a_{0}u+b_{0}v)x^{q}+(a_{0}u^{q}+c_{0}v^{q})x+(c_{0}u+d_{0}v)y^{q}+(b_{0}u^{q}+d_{0}v^{q})y,$
$\displaystyle($ $\displaystyle
a_{1}u+b_{1}v)x^{r}+(a_{1}u^{r}+c_{1}v^{r})x+(c_{1}u+d_{1}v)y^{r}+(b_{1}u^{r}+d_{1}v^{r})y).$

Define

$\displaystyle{\mathsf{D}_{f}^{{0}}}(x,y)$
$\displaystyle=b_{0}x^{q}+c_{0}x+d_{0}y^{q}+d_{0}y,$
$\displaystyle{\mathsf{D}_{g}^{{0}}}(x,y)$
$\displaystyle=b_{1}x^{r}+c_{1}x+d_{1}y^{r}+d_{1}y,$

$\displaystyle{\mathsf{D}_{f}^{{\infty}}}(x,y)$
$\displaystyle=a_{0}x^{q}+a_{0}x+c_{0}y^{q}+b_{0}y,$
$\displaystyle{\mathsf{D}_{g}^{{\infty}}}(x,y)$
$\displaystyle=a_{1}x^{r}+a_{1}x+c_{1}y^{r}+b_{1}y,$

and for $u\in{\mathcal{P}^{1}}(\mathbb{M})\setminus\\{0,\infty\\}$,

$\displaystyle{\mathsf{D}_{f}^{{u}}}(x,y)$
$\displaystyle=(a_{0}u+b_{0})x^{q}+(a_{0}u^{q}+c_{0})x+(c_{0}u+d_{0})y^{q}+(b_{0}u^{q}+d_{0})y,$
$\displaystyle{\mathsf{D}_{g}^{{u}}}(x,y)$
$\displaystyle=(a_{1}u+b_{1})x^{r}+(a_{1}u^{r}+c_{1})x+(c_{1}u+d_{1})y^{r}+(b_{1}u^{r}+d_{1})y.$

The following lemma is straightforward.

######  Lemma 3.1.

Let $(x,y)\mapsto F(x,y)=[f(x,y),g(x,y)]$ be a $(q,r)$-biprojective mapping of
$\mathbb{M}\times\mathbb{M}$. Then $F$ is planar if and only if the pair of
equations

${\mathsf{D}_{f}^{{u}}}(x,y)=0={\mathsf{D}_{g}^{{u}}}(x,y)$

has exactly one solution for each $u\in{\mathcal{P}^{1}}(\mathbb{M})$.

###### Proof.

We need to show that the polarization
$\Delta_{F}((x,y),(u,v))=(x,y)\ast(u,v)=0$ has a unique zero for each
$(u,v)\in\mathbb{M}\times\mathbb{M}\setminus(0,0)$ if and only if
${\mathsf{D}_{f}^{{w}}}(x,y)=0={\mathsf{D}_{g}^{{w}}}(x,y)$ has a unique
solution for each $w\in{\mathcal{P}^{1}}(\mathbb{M})$. Inspecting Eq. (3.1),
one immediately sees that the case $v=0$ corresponds to
${\mathsf{D}_{f}^{{\infty}}}(x,y)=0={\mathsf{D}_{g}^{{\infty}}}(x,y)$ after
applying $x\mapsto xu$ and $y\mapsto yu$. For $v\in\mathbb{M}^{\times}$, apply
$x\mapsto xv$, $y\mapsto yv$ and $u\mapsto uv$ to get the remaining cases
${\mathsf{D}_{f}^{{w}}}(x,y)=0={\mathsf{D}_{g}^{{w}}}(x,y)$ for
$w\in\mathbb{M}$. ∎

In the following we will show that many known semifields fall into the
$(q,r)$-biprojective setting.

###  3.1. Dickson semifields $\mathcal{D}$

Dickson introduced the commutative semifields
$\mathbb{S}=(\mathbb{M}\times\mathbb{M},+,\circ)$ with

$(x,y)\circ(u,v)=(xu+ay^{q}v^{q},xv+yu)$

where $q=p^{k}$ with $0<k<l$ and
$a\in\mathbb{M}^{\times}\setminus{(\mathbb{M}^{\times})^{2}}$. Note that the
isotopic multiplication

$(x,y)\ast(u,v)=(xu+ayv,xv^{\overline{q}}+y^{\overline{q}}u)$

is $(1,{\overline{q}})$-biprojective and isotopic to the polarization of the
$(1,q)$-biprojective planar mapping

$F_{\mathcal{D}}=\left[(1,0,0,a)_{1},(0,1,0,0)_{q}\right].$

Different choices for
$a\in\mathbb{M}^{\times}\setminus{(\mathbb{M}^{\times})^{2}}$ produce isotopic
semifields and there are a total of $\lfloor\frac{n}{4}\rfloor$ non-isotopic
Dickson semifields [19, p.107].

###  3.2. Albert’s generalized twisted fields $\mathcal{A}$

Albert introduced [2] a family of commutative and noncommutative semifields.
The commutative ones may be given as $\mathbb{S}=(\mathbb{F},+,\circ)$ with

$X\circ U=X^{q}U+U^{q}X,$

where $q=p^{k}$ with $0<k<n$ satisfying $n/{\mathsf{gcd}({k},{n})}$ odd. When
$\mathbb{F}=\mathbb{M}(\xi)$ with $[\mathbb{F}:\mathbb{M}]=2$, one can write
$X=x\xi+y$ with $x,y\in\mathbb{M}$. One can choose
$\xi\in\mathbb{F}\setminus\mathbb{M}$ satisfying
$\xi^{2}=a\in\mathbb{M}^{\times}\setminus{(\mathbb{M}^{\times})^{2}}$.

$\displaystyle(x\xi+y)\circ(u\xi+v)$
$\displaystyle=(x\xi+y)^{q}(u\xi+v)+(u\xi+v)^{q}(x\xi+y)$
$\displaystyle=\xi^{q+1}(x^{q}u+u^{q}x)+\xi^{q}(x^{q}v+u^{q}y)+\xi(y^{q}u+v^{q}x)+(y^{q}v+v^{q}y)$
$\displaystyle=a^{(q+1)/2}(x^{q}u+u^{q}x)+a^{(q-1)/2}\xi(x^{q}v+u^{q}y)+\xi(y^{q}u+v^{q}x)+(y^{q}v+v^{q}y).$

Identifying $\xi\mathbb{M}+\mathbb{M}$ with $\mathbb{M}\times\mathbb{M}$, and

$\displaystyle(x,y)\circ(u,v)$
$\displaystyle=\left(a^{(q-1)/2}(x^{q}v+u^{q}y)+(y^{q}u+v^{q}x),\quad
a^{(q+1)/2}(x^{q}u+u^{q}x)+(y^{q}v+v^{q}y)\right),$

is $(q,q)$-biprojective and isotopic to the polarization of the
$(q,q)$-biprojective planar mapping

$F_{\mathcal{A}}=[(0,a^{(q-1)/2},1,0)_{q},(a^{(q+1)/2},0,0,1)_{q}].$

Different choices for
$a\in\mathbb{M}^{\times}\setminus{(\mathbb{M}^{\times})^{2}}$ produce isotopic
semifields and there are a total of $\frac{\sigma(n)-1}{2}$ non-isotopic
generalized twisted fields [19, 2].

###  3.3. Zhou-Pott semifields $\mathcal{ZP}$

Zhou and Pott [35] gave a family of pre-semifields
$S=(\mathbb{M}\times\mathbb{M},+,\circ)$ given by

$(x,y)\circ(u,v)=(x^{q}u+u^{q}x+a(y^{q}v+yv^{q})^{r},xv+yu),$

where $a\in\mathbb{M}\setminus{(\mathbb{M}^{\times})^{2}}$, $q=p^{k}$ and
$r=p^{j}$ with $0\leq j,k\leq m$ where $m/{\mathsf{gcd}({k},{m})}$ is odd. The
isotopic multiplication

$(x,y)\ast(u,v)=(x^{q}u+u^{q}x+a(y^{q}v+yv^{q}),x^{r}v+yu^{r}),$

is $(q,r)$-biprojective and isotopic to the polarization of the
$(q,r)$-biprojective planar mapping

$F_{\mathcal{ZP}}=[(1,0,0,a)_{q},(0,1,0,0)_{r}].$

Different choices for
$a\in\mathbb{M}^{\times}\setminus{(\mathbb{M}^{\times})^{2}}$ produce isotopic
semifields and there are a total of
$\left\lfloor\frac{\sigma(n)}{2}\right\rfloor\cdot\left\lceil\frac{n}{4}\right\rceil$
non-isotopic $\mathcal{ZP}$ semifields [35].

###  3.4. Budaghyan-Helleseth semifields
($\mathcal{BH},\mathcal{ZW},\mathcal{LMPTB}$)

These semifields were found in [6] and independently in [34]. The commutative
semifields given later in [24] and [4] were shown to be isotopic to the
previous ones [25]. We note that Bierbrauer’s construction in [4] gives also
non-commutative semifields. We will use the definition from [4]. Let
$S=(\mathbb{M}\times\mathbb{M},+,\circ)$ be the pre-semifield given by

$(x,y)\circ(u,v)=\left\\{\begin{array}[]{ll}(xv+yu,x^{q}u+xu^{q}+a(y^{q}v+yv^{q}))&\textrm{if
}m/{\mathsf{gcd}({k},{m})}\textrm{ is odd},\\\
(xu+ayv,x^{q}v+yu^{q}+a^{(q-1)/2}(xv^{q}+y^{q}u))&\textrm{if
}m/{\mathsf{gcd}({k},{m})}\textrm{ is even},\end{array}\right.$

where $a\in\mathbb{M}\setminus{(\mathbb{M}^{\times})^{2}}$ and $q=p^{k}$ with
$0<k<m$. The pre-semifield multiplication is $(1,q)$-biprojective. Similarly,
the corresponding $(1,q)$-biprojective planar mapping whose polarization is
isotopic to $\mathbb{S}$ is given by

$\displaystyle
F_{\mathcal{BH}}=\left\\{\begin{array}[]{ll}[(0,0,1,0)_{1},(1,0,0,a)_{q}]&\textrm{if
}m/{\mathsf{gcd}({k},{m})}\textrm{ is odd},\\\
(1,0,0,a)_{1},(0,1,a^{(q-1)/2},0)_{q}]&\textrm{if
}m/{\mathsf{gcd}({k},{m})}\textrm{ is even}.\end{array}\right.$

The number of non-isotopic semifields in this family is
$\lfloor\frac{n}{4}\rfloor$ which is proved in [14].

Family | Planar Mapping | $\\#\mathbb{S}$ | Notes | $(\\#\mathbb{N}_{l},\\#\mathbb{N}_{m})$ | Count | Proved in
---|---|---|---|---|---|---
$\mathbb{F}$ | $X^{2}$ | $p^{n}$ |  | $(p^{n},p^{n})$ | $1$ |
2-4[.8pt/0.5pt] | $[(0,1,0,0)_{1},(1,0,0,a)_{1}]$ |  | $a\in\mathbb{M}\setminus{(\mathbb{M}^{\times})^{2}},n=2m$ |  |  |
$\mathcal{A}$ | $X^{q+1}$ | $p^{n}$ |  |  $q=p^{k}$, $0<k<m$,
---
${\mathsf{gcd}({k},{n})}=d$, $n/d$ odd.
$(p^{d},p^{d})$ | $\left\lfloor\frac{\sigma(n)-1}{2}\right\rfloor$ | [2]
2-4[.8pt/0.5pt] | $[(0,a^{\frac{q-1}{2}},1,0)_{q},(a^{\frac{q+1}{2}},0,0,1)_{q}]$ |  | $a\in\mathbb{M}\setminus{(\mathbb{M}^{\times})^{2}},n=2m$ |  |  |
$\mathcal{D}$ | $[(1,0,0,a)_{1},(0,1,0,0)_{q}]$ | $p^{2m}$ |  |  $q=p^{k}$, $0<k<m$,
---
${\mathsf{gcd}({k},{m})}=d$,
$a\in\mathbb{M}\setminus{(\mathbb{M}^{\times})^{2}}$.
$(p^{d},p^{m})$ | $\left\lfloor\frac{n}{4}\right\rfloor$ | [12]
$\mathcal{ZP}$ | $[(1,0,0,a)_{q},(0,1,0,0)_{r}]$ | $p^{2m}$ |  |  $q=p^{k},r=p^{j}$, $0<j,k<m$,
---
${\mathsf{gcd}({k},{m})}=d,\quad{\mathsf{gcd}({j,k},{m})}=d^{\prime}$,
$m/d$ odd, $a\in\mathbb{M}\setminus{(\mathbb{M}^{\times})^{2}}$.
$(p^{d^{\prime}},p^{d})$ | $\left\lfloor\frac{\sigma(n)-1}{2}\right\rfloor\left\lfloor\frac{n}{4}\right\rfloor$ | [35]
$\mathcal{BH}$ | $[(0,1,0,0)_{1},(1,0,0,a)_{q}]$ | $p^{2m}$ |  |  $q=p^{k}$, $0<k<m$,
---
${\mathsf{gcd}({k},{m})}=d$, $m/d$ odd,
$a\in\mathbb{M}\setminus{(\mathbb{M}^{\times})^{2}}$.
$(p^{d},p^{2d})$ | $\left\lfloor\frac{n}{4}\right\rfloor$ | [6, 34, 14]
2-4[.4pt/1pt] | $[(1,0,0,a)_{1},(0,1,a^{\frac{q-1}{2}},0)_{q}]$ |  |  |  $q=p^{k}$, $0<k<m$,
---
${\mathsf{gcd}({k},{m})}=d$, $m/d$ even,
$a\in\mathbb{M}\setminus{(\mathbb{M}^{\times})^{2}}$.
|  |
$\mathcal{S}$ | $[(1,0,0,B)_{q},(0,1,\frac{a}{B},0)_{r}]$ | $p^{4l}$ |  |  $q=p^{k},\quad r=p^{k+l},\quad 0<k<l$,
---
$m=2l,\quad{\mathsf{gcd}({k},{m})}=e,\quad m/e\textrm{ odd}$,
$a\in\mathbb{L}^{\times},\quad
B\in\mathbb{M}\setminus{(\mathbb{M}^{\times})^{2}}$.
$(p^{e/2},p^{e})$ | $\geq\left\lfloor\frac{\sigma(n)-1}{2}\right\rfloor\left\lceil\frac{p^{l}-1}{n}\right\rceil$ |  Theorem 4.4
Table 1. Known infinite families of biprojective commutative semifields of
order $p^{n}$

The known infinite families of biprojective and other commutative semifields
and their planar representations are summarized in Tables 1 and 2. Families
$\mathcal{A},\mathcal{D},\mathcal{BH}$ reduce to $\mathbb{F}$ when
$k\in\\{0,m\\}$. Family $\mathcal{ZP}$ reduces to $\mathcal{D}$ when $k=0$, to
$\mathcal{BH}$ when $j=0$, and to $\mathbb{F}$ when $j=k=0$. Family
$\mathcal{S}$ reduces to $\mathcal{ZP}$ when $a=0$, and to $\mathcal{D}$ when
$k\in\\{0,l\\}$. We excluded those cases in the Notes and also in the Counts
columns of Table 1.

###  3.5. The Family $\mathcal{S}$

The main result of this paper is to prove that

  * •

Family $\mathcal{S}$ gives new commutative semifields, and

  * •

Family $\mathcal{S}$ contains an exponential number of non-isotopic
commutative semifields (in $n$).

An informal way to explain why Family $\mathcal{S}$ gives such a large number
of commutative semifields is that their polarizations admit only a few
$\mathbb{M}$-linear isotopisms (within the family) due to their complexity
—they contain two non-zero entries in either component of their planar
representations $(1,0,0,a)_{q}$ and $(0,1,b,0)_{r}$; and the underlying field
automorphisms $q$ and $r$ are nontrivial and are not simply related to each
other. Indeed, our method in Section 5 will show that $\mathbb{M}$-linear
isotopisms are essentially the only ones for biprojective semifields whose
autotopism groups satisfy a simply defined condition. The polarization of
$(0,1,0,0)_{q}$, which is a component polynomial of many other biprojective
semifields (except $\mathcal{A}$ and $\mathcal{BH}_{\textrm{even}}$), admits
more such isotopisms and that is the main reason why all $a\in\mathbb{M}$
allowed in these constructions lead to isotopic semifields. For
$\mathbb{F},\mathcal{D},\mathcal{BH}_{\textrm{even}}$ and $\mathcal{A}$ the
reasons for admitting only a small number of non-isotopic semifields include
the simplicity of the defining field automorphisms, e.g.,
$q\in\\{1,p^{m/2}\\}$; or having the same $(q,q)$ or conjugate
$(q,{\overline{q}})$ automorphisms. We will explain these in detail in Section
5 (see Theorem 5.10). We start by proving that Family $\mathcal{S}$ indeed
gives commutative semifields.

##  4\. The commutative semifield family $\mathcal{S}$

The following diagram and its annotations describe our setting.

$\mathbb{F}_{p}$$\mathbb{D}=\mathbb{F}_{p^{d}}$$\mathbb{E}=\mathbb{F}_{p^{e}}$$\mathbb{L}=\mathbb{F}_{p^{m/2}}$$\mathbb{M}=\mathbb{F}_{p^{m}}$$\mathbb{F}=\mathbb{F}_{p^{n}}$$d$$2$$\frac{m}{e}$$2$$2$

######  Notation 4.1.

  *   * •

$p$ is an odd prime.

  * •

$n=2m$, $m$ is even.

  * •

$Q=p^{m/2}$, $Q^{2}=p^{m}$.

  * •

$q=p^{k}$, $r=p^{k+m/2}=Qq$ with $1\leq k\leq m-1$.

  * •

$e={\mathsf{gcd}({k},{m})}$ with $m/e$ odd.

  * •

$d={\mathsf{gcd}({k+m/2},{m})}$.

  * •

$e=2d$ by Lemma 4.3.

  * •

$(\mathbb{M}^{\times})^{2}$ — the subgroup of non-zero squares in
$\mathbb{M}^{\times}$.

  * •

$\mathbb{L}^{\times}=(\mathbb{M}^{\times})^{Q+1}\leq{(\mathbb{M}^{\times})^{2}}\leq\mathbb{M}^{\times}$.

  * •

$(\mathbb{M}^{\times})^{Q-1}\leq{(\mathbb{M}^{\times})^{2}}\leq\mathbb{M}^{\times}$
— the subgroup of $(Q+1)^{\textrm{st}}$ roots of unity in
$\mathbb{M}^{\times}$.

  * •

$\mathbb{E}=\mathbb{F}_{q}\cap\mathbb{M}=\mathbb{F}_{q^{2}}\cap\mathbb{M}=\mathbb{F}_{r^{2}}\cap\mathbb{M}$
by Lemma 4.3.

  * •

$\mathbb{D}=\mathbb{F}_{r}\cap\mathbb{M}$.

We need the following well known result on the greatest common divisor of
$p^{i}\pm 1$ and $p^{m}-1$. A proof can be found in [28].

######  Lemma 4.2.

Let $i,m\in\mathbb{N}$ and $p$ be a prime. Then

  * •

${\mathsf{gcd}({p^{i}-1},{p^{m}-1})}=p^{{\mathsf{gcd}({i},{m})}}-1$.

  * •

${\mathsf{gcd}({p^{i}+1},{p^{m}-1})}=\begin{cases}1&\text{if
}m/{\mathsf{gcd}({i},{m})}\text{ odd, and }p=2,\\\ 2&\text{if
}m/{\mathsf{gcd}({i},{m})}\text{ odd, and }p>2,\\\
p^{{\mathsf{gcd}({i},{m})}}+1&\text{if }m/{\mathsf{gcd}({i},{m})}\text{
even}.\end{cases}$

First, we will prove a lemma.

######  Lemma 4.3.

We have,

  1. (i)

$(-1)\not\in(\mathbb{M}^{\times})^{q-1}$.

  2. (ii)

Any $x\in{(\mathbb{M}^{\times})^{2}}$ can be written (twice) as $x=cg$ where
$c\in\mathbb{L}^{\times}$ and $g\in(\mathbb{M}^{\times})^{Q-1}$.

  3. (iii)

${\mathsf{gcd}({k+m/2},{m})}={\mathsf{gcd}({k},{m})}/2$.

  4. (iv)

$\mathbb{E}=\mathbb{F}_{q}\cap\mathbb{M}=\mathbb{F}_{q^{2}}\cap\mathbb{M}=\mathbb{F}_{r^{2}}\cap\mathbb{M}$.

###### Proof.

  1. (i)

Recall that $\nu_{2}(x)$ denotes the index of $2$ in $x$, that is,
$\nu_{2}(x)=h$ if $2^{h}|x$ and $2^{h+1}\nmid x$. We have
$\nu_{2}(p^{e}-1)=\nu_{2}(Q^{2}-1)$ since
$Q^{2}-1=p^{m}-1=(p^{e}-1)\sum_{i=0}^{m/e-1}p^{i}$ and the fact that an odd
number of odd integers add up to an odd integer. Thus
$\nu_{2}((p^{m}-1)/2)=\nu_{2}(p^{e}-1)-1$. The result follows from Lemma 4.2
which shows ${\mathsf{gcd}({q-1},{Q^{2}-1})}=p^{e}-1$.

  2. (ii)

It is easy to see that
$(\mathbb{M}^{\times})^{Q-1}\cap\mathbb{L}^{\times}=\\{\pm 1\\}$ since
${\mathsf{gcd}({Q-1},{Q+1})}=2$ and $xg=yh$ if and only if $(x,g)=(y,h)$ or
$(x,g)=(-y,-h)$.

  3. (iii)

Let $k=2dk^{\prime}$ and $m=2dm^{\prime}$ with
${\mathsf{gcd}({k^{\prime}},{m^{\prime}})}=1$ and $m^{\prime}$ odd. Then
${\mathsf{gcd}({2dk^{\prime}+dm^{\prime}},{2dm^{\prime}})}=d\cdot{\mathsf{gcd}({2k^{\prime}+m^{\prime}},{2m^{\prime}})}=d\cdot{\mathsf{gcd}({2k^{\prime}+m^{\prime}},{m^{\prime}})}=d\cdot{\mathsf{gcd}({2k^{\prime}},{m^{\prime}})}=d$,
since $m^{\prime}+2k^{\prime}$ is odd.

  4. (iv)

Obvious since $m/e$ is odd and $(qQ)^{2}=q^{2}\pmod{Q^{2}-1}$.

∎

Now we present the family of planar mappings.

######  Theorem 4.4.

Let $a\in\mathbb{L}^{\times}$ and
$B\in\mathbb{M}^{\times}\setminus{(\mathbb{M}^{\times})^{2}}$ and let

$F:\mathbb{M}\times\mathbb{M}\to\mathbb{M}\times\mathbb{M}$

be defined as

$F:(x,y)\mapsto F(x,y)=[(1,0,0,B)_{q},(0,1,a/B,0)_{r}].$

Then $F$ is planar.

###### Proof.

We are going to use Lemma 3.1. First,

$\displaystyle{\mathsf{D}_{f}^{{0}}}(x,y)$
$\displaystyle=B(y^{q}+y)=0,\quad\textrm{and},$
$\displaystyle{\mathsf{D}_{g}^{{0}}}(x,y)$
$\displaystyle=x^{r}+\frac{ax}{B}=0,$

imply $(x,y)=(0,0)$, since otherwise either of

$y^{q-1}=-1,\textrm{ and }x^{r-1}=-\frac{a}{B}$

lead to a contradiction, since $-1\not\in(\mathbb{M}^{\times})^{q-1}$ by Lemma
4.3, and $2|{\mathsf{gcd}({r-1},{Q^{2}-1})}$ and
$-a/B\not\in{(\mathbb{M}^{\times})^{2}}$. Similarly,

$\displaystyle{\mathsf{D}_{f}^{{\infty}}}(x,y)$
$\displaystyle=x^{q}+x=0,\quad\textrm{and},$
$\displaystyle{\mathsf{D}_{g}^{{\infty}}}(x,y)$
$\displaystyle=y+\frac{ay^{r}}{B}=0,$

has the unique common solution $(x,y)=(0,0)$ with the same argument after
changing variables. Now for $u\in\mathbb{M}^{\times}$,

$\displaystyle{\mathsf{D}_{f}^{{u}}}(x,y)$
$\displaystyle=ux^{q}+u^{q}x+B(y^{q}+y)=0,\quad\textrm{and},$
$\displaystyle{\mathsf{D}_{g}^{{u}}}(x,y)$
$\displaystyle=x^{r}+\frac{a}{B}x+\frac{a}{B}uy^{r}+u^{r}y=0.$

Or,

$\displaystyle{\mathsf{D}_{f}^{{u}}}(ux,y)$
$\displaystyle=u^{q+1}(x^{q}+x)+B(y^{q}+y)=0,\quad\textrm{and},$
$\displaystyle{\mathsf{D}_{g}^{{u}}}(ux,y)$
$\displaystyle=u^{r}(x^{r}+y)+\frac{a}{B}u(x+y^{r})=0.$

We will proceed to show that $(x,y)=(0,0)$ is the only common solution of
these equations for $x,y\in\mathbb{M}$. Now we can assume
$x,y\in\mathbb{M}^{\times}$, since $x=0$ implies $y=0$ and vice versa for
${\mathsf{D}_{f}^{{u}}}(ux,y)=0$. Furthermore, $x=-y^{r}$ implies
$y-y^{r^{2}}=0$ and
$x,y\in\mathbb{F}_{r^{2}}\cap\mathbb{M}=\mathbb{F}_{q^{2}}\cap\mathbb{M}=\mathbb{F}_{q}\cap\mathbb{M}=\mathbb{E}$
by Lemma 4.3. Thus $x^{q}+x=2x=-2y^{r}$ and $y^{q}+y=2y$, in turn

$2(-u^{q+1}y^{r}+By)=0,$

or

$y^{r-1}=\frac{B}{u^{q+1}}.$

This is impossible since $B\not\in{(\mathbb{M}^{\times})^{2}}$. The same
argument shows $x^{r}\neq-y$ and we can concentrate on

(4.1) $\displaystyle u^{q+1}$
$\displaystyle=-\frac{B(y^{q}+y)}{x^{q}+x},\textrm{ and},$ (4.2)
$\displaystyle u^{r-1}$ $\displaystyle=-\frac{a(x+y^{r})}{B(x^{r}+y)},$

for $x,y\in\mathbb{M}^{\times}$ with $x^{r}\neq-y$ and $x\neq-y^{r}$. Now
assume (4.1) and (4.2) hold for such $x,y\in\mathbb{M}^{\times}$, and let

(4.3) $\displaystyle\phi_{q}(x,y)$
$\displaystyle=\frac{y^{q}+y}{x^{q}+x}=\frac{\gamma}{cg},\textrm{ and},$ (4.4)
$\displaystyle\phi_{r}(x,y)$
$\displaystyle=\frac{x+y^{r}}{x^{r}+y}=\gamma^{Q}dh,$

for some $c,d\in\mathbb{L}^{\times}$, $g,h\in(\mathbb{M}^{\times})^{Q-1}$ and
a fixed $\gamma\in\mathbb{M}^{\times}\setminus{(\mathbb{M}^{\times})^{2}}$,
since $cg$ and $dh$ both run through ${(\mathbb{M}^{\times})^{2}}$
independently by Lemma 4.3. Note that (4.1) and (4.2) guarantees that
$\phi_{q}(x,y)$ and $\phi_{r}(x,y)$ are non-squares. Multiplying (4.3) and
(4.4) (and also (4.1) and (4.2)) we get

$\gamma^{Q+1}\frac{dh}{cg}=\phi_{q}(x,y)\phi_{r}(x,y)=\frac{u^{q+r}}{a}=\frac{u^{q(Q+1)}}{a}\in\mathbb{L}^{\times},$

and therefore $\epsilon g=h$ with $\epsilon\in\\{\pm 1\\}$ since
$\gamma^{Q+1}\in\mathbb{L}^{\times}$. Now let $z=y-x^{Q}$ and consider

$\displaystyle cg(z^{q}+z+(x^{q}+x)^{Q})$
$\displaystyle=\gamma(x^{q}+x),\textrm{ and},$ $\displaystyle(x^{q}+x)+z^{r}$
$\displaystyle=\epsilon\gamma^{Q}dg((x+x^{q})^{Q}+z),$

for $x\in\mathbb{M}^{\times}$, which is a rewriting of (4.3) and (4.4). Or,
equivalently,

(4.5) $\displaystyle cg(x^{q}+x)^{Q}-\gamma(x^{q}+x)$
$\displaystyle=-cg(z^{q}+z),\textrm{ and},$ (4.6)
$\displaystyle-\epsilon\gamma^{Q}dg(x^{q}+x)^{Q}+(x^{q}+x)$
$\displaystyle=-z^{r}+\epsilon\gamma^{Q}dgz.$

The two new equations generated by

  * •

Eq. (4.5) plus $\gamma$ times Eq. (4.6), and

  * •

$d\epsilon\gamma^{Q}$ times Eq. (4.5) plus $c$ times Eq. (4.6),

are as follows:

(4.7) $\displaystyle(c-\epsilon\gamma^{Q+1}d)(x^{q}+x)^{Q}g$
$\displaystyle=-\gamma z^{r}-cgz^{q}-(c-\epsilon\gamma^{Q+1}d)gz,\textrm{
and},$ (4.8) $\displaystyle(c-\epsilon\gamma^{Q+1}d)(x^{q}+x)$
$\displaystyle=-c(z^{r}-d\epsilon\gamma^{Q}gz^{q}).$

We will show that the common solutions of these equations for
$x\in\mathbb{M}^{\times}$ lead to a contradiction to our assumption that (4.1)
and (4.2) hold. Note that $x^{q}+x\neq 0$ since $x\in\mathbb{M}^{\times}$ and
$-1\not\in(\mathbb{M}^{\times})^{q-1}$. Now if $z=y-x^{Q}=0$, then (4.3)
becomes

$(x^{q}+x)^{Q-1}=\frac{\gamma}{cg},$

which is a contradiction since the left hand side is a square and the right
hand side is not. If $c=\epsilon\gamma^{Q+1}d$ then $z=0$ by Eq. (4.8), which
we have just handled, or $z^{r-q}=z^{q(Q-1)}=d\epsilon\gamma^{Q}g$ by Eq.
(4.8), which is again a contradiction since the right hand side is not a
square. Then $c\neq\epsilon\gamma^{Q+1}d$. Observe that
$c-\epsilon\gamma^{Q+1}d\in\mathbb{L}$. Comparing $g$ times Eq. (4.8) to the
power $Q$ with Eq. (4.7) yields

$\displaystyle(-c(z^{r}-d\epsilon\gamma^{Q}gz^{q}))^{Q}g$
$\displaystyle=-\gamma z^{r}-cgz^{q}-(c-\epsilon\gamma^{Q+1}d)gz$
$\displaystyle-cgz^{q}+cd\epsilon\gamma g^{Q+1}z^{r}$ $\displaystyle=-\gamma
z^{r}-cgz^{q}-(c-\epsilon\gamma^{Q+1}d)gz$ $\displaystyle cd\epsilon\gamma
z^{r}$ $\displaystyle=-\gamma z^{r}-(c-\epsilon\gamma^{Q+1}d)gz$
$\displaystyle\gamma(cd\epsilon+1)z^{r}$
$\displaystyle=-(c-\epsilon\gamma^{Q+1}d)gz.$

Now $cd\epsilon+1=0$ implies $z=0$ or $c=\epsilon\gamma^{Q+1}d$, which were
handled before. Thus, we have

$z^{r-1}=-\frac{g}{\gamma}\left(\frac{c-\epsilon\gamma^{Q+1}d}{cd\epsilon+1}\right),$

and noting again that $\gamma^{Q+1}\in\mathbb{L}^{\times}$, we reach another
contradiction since the right hand side is not a square. ∎

##  5\. A method to determine isotopy of biprojective pre-semifields

There are two usual ways to determine whether two semifields are isotopic. The
first one is to use isotopy invariants like the nuclei. Since there are less
than $n^{2}$ possible configurations for the left/right and central nucleus
for a commutative pre-semifield on $\mathbb{F}_{p}^{n}$, this method is not
enough to determine whether the number of non-isotopic pre-semifields grows
exponentially in $n$ or not. The second method works by directly considering
all possible isotopisms $(N,L,M)$. This is, however, in many cases not
feasible unless the pre-semifield has a very simple structure. Note that, for
commutative semifields, some general results were obtained in [8] that make
this approach slightly easier and allowed for instance to settle the isotopy
question inside the family of Zhou-Pott semifields [35]. However, the
semifields in our Family $\mathcal{S}$ are more delicate and such a direct
approach does not seem possible. In this section, we develop a new general
technique to determine whether two biprojective pre-semifields are isotopic or
not.

###  5.1. Group theoretic preliminaries

We denote the set of all autotopisms of a pre-semifield $\mathbb{P}$ by
$\operatorname{Aut}(\mathbb{P})$. It is easy to check that
$\operatorname{Aut}(\mathbb{P})$ is a group under component-wise composition,
i.e.,
$(N_{1},L_{1},M_{1})(N_{2},L_{2},M_{2})=(N_{1}N_{2},L_{1}L_{2},M_{1}M_{2})$.
We view $\operatorname{Aut}(\mathbb{P})$ as a subgroup of
$\operatorname{GL}(\mathbb{F})^{3}\cong\operatorname{GL}(\mathbb{M}\times\mathbb{M})^{3}\cong\operatorname{GL}(n,\mathbb{F}_{p})^{3}$.
Our approach is based on the following simple and well-known result.

######  Lemma 5.1.

Let $\mathbb{P}_{1}=(\mathbb{F}_{p}^{n},+,\ast_{1})$,
$\mathbb{P}_{2}=(\mathbb{F}_{p}^{n},+,\ast_{2})$ be isotopic pre-semifields
via the isotopism $\gamma\in\operatorname{GL}(\mathbb{F})^{3}$. Then
$\gamma^{-1}\operatorname{Aut}(\mathbb{P}_{2})\gamma=\operatorname{Aut}(\mathbb{P}_{1})$.

###### Proof.

Let $\gamma=(N_{1},L_{1},M_{1})\in\operatorname{GL}(\mathbb{F})^{3}$ be an
isotopism between $\mathbb{P}_{1}$ and $\mathbb{P}_{2}$ and
$\delta=(N_{2},L_{2},M_{2})\in\operatorname{Aut}(\mathbb{P}_{2})$. Then
$\gamma^{-1}\delta\gamma\in\operatorname{Aut}(\mathbb{P}_{1})$. Indeed

$\displaystyle(N_{1}^{-1}N_{2}N_{1})(x\ast_{1}y)$
$\displaystyle=(N_{1}^{-1}N_{2})(L_{1}(x)\ast_{2}M_{1}(y))$
$\displaystyle=N_{1}^{-1}((L_{2}L_{1}(x))\ast_{2}(M_{2}M_{1}(y)))$
$\displaystyle=(L_{1}^{-1}L_{2}L_{1}(x))\ast_{1}(M_{1}^{-1}M_{2}M_{1}(y)),$

so
$\gamma^{-1}\operatorname{Aut}(\mathbb{P}_{2})\gamma\subseteq\operatorname{Aut}(\mathbb{P}_{1})$.
The other inclusion follows by symmetry. ∎

The central idea of the technique we are going to develop is to identify large
abelian subgroups (in particular certain Sylow subgroups), in the autotopism
group of biprojective semifields. We then use tools from group theory to
obtain strong constraints on when the autotopism groups of two pre-semifields
are conjugate. This approach is inspired by a similar technique for
inequivalences of power functions on finite fields developed by Dempwolff [11]
and Yoshiara [32].

First we recall the well-known Sylow Theorems (see for instance [17, Chapter
4]).

######  Theorem 5.2 (Sylow Theorems).

Let $G$ be a group with order $p^{m}s$, with $p$ prime, $m>0$ and $p\nmid s$.
Then,

  1. (i)

$G$ has a subgroup of order $p^{m}$, called a Sylow $p$-subgroup of $G$.

  2. (ii)

Every $p$-subgroup of $G$ is contained in a Sylow $p$-subgroup of $G$.

  3. (iii)

The Sylow $p$-subgroups of $G$ are conjugate in $G$.

We will identify Sylow $p^{\prime}$-subgroups of
$\operatorname{Aut}(\mathbb{P})$ when $\mathbb{P}$ is a biprojective pre-
semifield. In order to do that, we need to find a suitable prime $p^{\prime}$,
for which we will employ Zsigmondy’s Theorem (see for instance [18, Chapter
IX., Theorem 8.3]).

######  Theorem 5.3 (Zsigmondy’s Theorem).

For every prime $p$ and $m>2$ except when $(p,m)=(2,6)$, there exists a
$p$-primitive divisor $p^{\prime}$ of $p^{m}-1$, that is, $p^{\prime}$ prime,
$p^{\prime}|p^{m}-1$ and $p^{\prime}\nmid p^{i}-1$ for all $1\leq i\leq m-1$.

We write $\mathbb{F}_{p}$-linear mappings $L\in\operatorname{End}(\mathbb{F})$
from $\mathbb{F}$ to itself as $2\times 2$ matrices of $\mathbb{F}_{p}$-linear
mappings in $\operatorname{End}(\mathbb{M})$. That is,

$L=\begin{pmatrix}L_{1}&L_{2}\\\ L_{3}&L_{4}\end{pmatrix},\textrm{ for
}L_{i}\in\operatorname{End}(\mathbb{M}).$

Set

$\gamma_{a}=(N_{a},L_{a},M_{a})\in\operatorname{GL}(\mathbb{F})^{3}\quad\textrm{
with }N_{a}=\begin{pmatrix}m_{a^{q+1}}&0\\\ 0&m_{a^{r+1}}\end{pmatrix},\quad
L_{a}=M_{a}=\begin{pmatrix}m_{a}&0\\\ 0&m_{a}\end{pmatrix},$

where $m_{a}$ denotes multiplication with the finite field element
$a\in\mathbb{M}^{\times}$. For simplicity, we write these diagonal matrices
also in the form $\operatorname{diag}(m_{a},m_{a})$, so

$\gamma_{a}=(\operatorname{diag}(m_{a^{q+1}},m_{a^{r+1}}),\operatorname{diag}(m_{a},m_{a}),\operatorname{diag}(m_{a},m_{a})).$

An important fact which follows immediately from biprojectivity is that
$\gamma_{a}\in\operatorname{Aut}(\mathbb{P})$ for all
$a\in\mathbb{M}^{\times}$ when $\mathbb{P}$ is a $(q,r)$-biprojective pre-
semifield, which can be readily verified using Eq. (3.1). We fix some further
notation that we will use from now on:

######  Notation 5.4.

  *   * •

Set $q=p^{k}$ and $r=p^{l}$.

  * •

Set $\overline{q}=p^{m-k}$ and $\overline{r}=p^{m-l}$, that is
$q\overline{q}\equiv r\overline{r}\equiv 1\pmod{p^{m}-1}$.

  * •

Define the cyclic group

$Z^{(q,r)}=\\{\gamma_{a}:a\in\mathbb{M}^{\times}\\},$

of order $p^{m}-1$.

  * •

Let $p^{\prime}$ be a $p$-primitive divisor of $p^{m}-1$. Such a prime
$p^{\prime}$ always exists if $m>2$ and $(p,m)\neq(2,6)$ by Zsigmondy’s
Theorem. In our case, we have $p>2$. We will also stipulate $m>2$. Note that
$p^{\prime}\neq 2$ since $p^{\prime}\nmid p-1$.

  * •

Let $R$ be the unique Sylow $p^{\prime}$-subgroup of $\mathbb{M}^{\times}$.

  * •

Define

$Z_{R}^{(q,r)}=\\{\gamma_{a}\colon a\in R\\},$

which is the unique Sylow $p^{\prime}$-subgroup of $Z^{(q,r)}$ with $|R|$
elements.

  * •

For a $(q,r)$-biprojective pre-semifield $\mathbb{P}$, denote by

$C_{\mathbb{P}}=C_{\operatorname{Aut}(\mathbb{P})}(Z_{R}^{(q,r)}),$

the centralizer of $Z_{R}^{(q,r)}$ in $\operatorname{Aut}(\mathbb{P})$.

  * •

Define

$S=\\{\operatorname{diag}(m_{a},m_{a})\colon a\in\mathbb{M}^{\times}\\},$

and

$S_{R}=\\{\operatorname{diag}(m_{a},m_{a})\colon a\in R\\}.$

We start by identifying a subgroup of the autotopism group of any
$(q,r)$-biprojective pre-semifield. The following lemma is straightforward,
but very important for our paper.

######  Lemma 5.5.

Let $\mathbb{P}$ be any $(q,r)$-biprojective pre-semifield. Then

$Z_{R}^{(q,r)}\leq Z^{(q,r)}\leq\operatorname{Aut}(\mathbb{P}).$

###### Proof.

Follows directly from Eq. (3.1). ∎

We continue with a simple observation on $R$.

######  Lemma 5.6.

Let $a\in R$, $a\neq 1$. Then $a$ is not contained in a proper subfield of
$\mathbb{M}$.

###### Proof.

Clearly, $a$ is contained in the subfield $\mathbb{F}_{p^{i}}$ if and only if
$a^{p^{i}-1}=1$, i.e. the multiplicative order of $a$ is a divisor of
$p^{i}-1$. The order of $a$ is a power of $p^{\prime}$ and $p^{\prime}$ does
not divide $p^{i}-1$ for any $i<m$, so $a$ is not contained in a proper
subfield of $\mathbb{F}_{p^{m}}$. ∎

Now we observe that the normalizer and the centralizer of certain subgroups of
$\operatorname{GL}(\mathbb{F})$ must have certain shape.

######  Lemma 5.7.

Let $N_{\operatorname{GL}(\mathbb{F})}(S_{R})$,
$N_{\operatorname{GL}(\mathbb{F})}(S)$ and
$C_{\operatorname{GL}(\mathbb{F})}(S_{R})$,
$C_{\operatorname{GL}(\mathbb{F})}(S)$ be the normalizers and the centralizers
of $S_{R}$ and $S$ in $\operatorname{GL}(\mathbb{F})$. Then

  1. (a)

$N_{\operatorname{GL}(\mathbb{F})}(S_{R})=N_{\operatorname{GL}(\mathbb{F})}(S)=\left\\{\begin{pmatrix}m_{c_{1}}\tau&m_{c_{2}}\tau\\\
m_{c_{3}}\tau&m_{c_{4}}\tau\end{pmatrix}\colon
c_{1},c_{2},c_{3},c_{4}\in\mathbb{M},\tau\in\operatorname{Gal}(\mathbb{M}/\mathbb{F}_{p})\right\\}\cap\operatorname{GL}(\mathbb{F})$,

  2. (b)

$C_{\operatorname{GL}(\mathbb{F})}(S_{R})=C_{\operatorname{GL}(\mathbb{F})}(S)=\left\\{\begin{pmatrix}m_{c_{1}}&m_{c_{2}}\\\
m_{c_{3}}&m_{c_{4}}\end{pmatrix}\colon
c_{1},c_{2},c_{3},c_{4}\in\mathbb{M}\right\\}\cap\operatorname{GL}(\mathbb{F})$.

###### Proof.

We present the proof only for the more delicate case $S_{R}$. The proof for
$S$ is identical with $\mathbb{M}^{\times}$ substituting $R$ throughout.

Let $\begin{pmatrix}A_{1}&A_{2}\\\ A_{3}&A_{4}\end{pmatrix}\in
N_{\operatorname{GL}(\mathbb{F})}(S_{R})$, where
$A_{1},A_{2},A_{3},A_{4}\in\operatorname{End}(\mathbb{M})$. Then

$\begin{pmatrix}A_{1}&A_{2}\\\
A_{3}&A_{4}\end{pmatrix}\operatorname{diag}(m_{a},m_{a})=\operatorname{diag}(m_{b},m_{b})\begin{pmatrix}A_{1}&A_{2}\\\
A_{3}&A_{4}\end{pmatrix},$

for all $a\in R$ and some $b=\pi(a)$ where $\pi\colon R\rightarrow R$ is a
bijection. Simple matrix multiplication implies $A_{i}(ax)=bA_{i}(x)$ for
$i\in\\{1,2,3,4\\}$.

We now write the mappings as linearized polynomials, i.e.
$A_{i}=\sum_{j=0}^{m-1}c_{j,i}x^{p^{j}}$ for $i\in\\{1,2,3,4\\}$. The
equations above then immediately yield

$\sum_{j=0}^{m-1}c_{j,i}a^{p^{j}}x^{p^{j}}=b\sum_{j=0}^{m-1}c_{j,i}x^{p^{j}}$

for $i\in\\{1,2,3,4\\}$ and all $a\in R$. We now compare the coefficients of
these polynomials. If $a\neq 1$, it is not contained in any proper subfields
of $\mathbb{M}$ by Lemma 5.6, so we have $b=a^{p^{j}}$ only for at most one
$j\in\\{0,1,\dots,m-1\\}$. So $A_{1},A_{2},A_{3},A_{4}$ are zero or monomials
of the same degree $p^{j}$, which proves the statement for the normalizer. In
the case of the centralizer, we have $b=a$, so we get the same possible
mappings except with $j=0$ forced. ∎

We have shown in Lemma 5.5 that $Z_{R}^{(q,r)}$ is a subgroup of
$\operatorname{Aut}(\mathbb{P})$. Now we show that, under a certain condition
which is key to our proofs, it is not just a Sylow $p^{\prime}$-subgroup of
$Z^{(q,r)}$, but even a Sylow $p^{\prime}$-subgroup of
$\operatorname{Aut}(\mathbb{P})$.

######  Lemma 5.8.

Let $\mathbb{P}$ be a $(q,r)$-biprojective pre-semifield. Assume that
$C_{\mathbb{P}}$ contains $Z^{(q,r)}$ as an index $I$ subgroup such that
$p^{\prime}$ does not divide $I$. Then $Z_{R}^{(q,r)}$ is a Sylow
$p^{\prime}$-subgroup of $\operatorname{Aut}(\mathbb{P})$.

###### Proof.

First define

$U=\\{(\operatorname{diag}(m_{a},m_{b}),\operatorname{diag}(m_{c},m_{d}),\operatorname{diag}(m_{e},m_{f}))\colon
a,b,c,d,e,f\in R\\}.$

Clearly, $|U|=|R|^{6}$. We will now show that $U$ is a Sylow
$p^{\prime}$-subgroup of $\operatorname{GL}(\mathbb{F})^{3}$. We have

$|\operatorname{GL}(\mathbb{F})|=|\operatorname{GL}(2m,\mathbb{F}_{p})|=p^{m(2m-1)}\prod_{i=1}^{2m}(p^{i}-1).$

Clearly, $p^{m+i}-1\equiv p^{i}-1\pmod{p^{m}-1}$. As $p^{\prime}$ is a
$p$-primitive divisor of $p^{m}-1$, all integers $p^{j}-1$ with $j$ in
$[1,2m]$ are coprime to $p^{\prime}$, except for $j\in\\{m,2m\\}$.
Furthermore, the $p^{\prime}$-part of $p^{2m}-1=(p^{m}-1)(p^{m}+1)$ is $|R|$
since $p^{\prime}\neq 2$. Thus the $p^{\prime}$-part of
$|\operatorname{GL}(\mathbb{F})|$ is $|R|^{2}$ and $U$ is a Sylow
$p^{\prime}$-subgroup of $\operatorname{GL}(\mathbb{F})^{3}$ as claimed. All
Sylow $p^{\prime}$-subgroups of $\operatorname{GL}(\mathbb{F})^{3}$ are
abelian since $U$ is abelian, by Sylow Theorem (iii). Note that any
$p^{\prime}$-subgroup of a group $G$ is contained in a Sylow
$p^{\prime}$-subgroup of $G$ by Sylow Theorem (ii). Let $T$ be a Sylow
$p^{\prime}$-subgroup of $\operatorname{Aut}(\mathbb{P})$ that contains the
$p^{\prime}$-group $Z_{R}^{(q,r)}$. Then $T$ itself is (again by Sylow Theorem
(ii)) contained in a Sylow $p^{\prime}$-subgroup of
$\operatorname{GL}(\mathbb{F})^{3}$, say $U^{\prime}$. In particular, $T$ is
abelian. This implies that $T$ is a subgroup of the centralizer
$C_{\mathbb{P}}$ of $Z_{R}^{(q,r)}$ in $\operatorname{Aut}(\mathbb{P})$. By
assumption, $Z^{(q,r)}$ is an index $I$ subgroup of $C_{\mathbb{P}}$ and
$p^{\prime}$ does not divide $I$. Moreover $Z_{R}^{(q,r)}$ is a Sylow
$p^{\prime}$-subgroup of $Z^{(q,r)}$ and therefore
$p^{\prime}\nmid[Z^{(q,r)}:Z_{R}^{(q,r)}]=I_{1}$. Let
$[T:Z_{R}^{(q,r)}]=I_{2}=p^{\prime h}$ for $h\geq 0$, since both are
$p^{\prime}$-groups. Since $I_{2}|I_{1}I$, and $p^{\prime}\nmid I_{1}I$, we
must have $p^{\prime}\nmid I_{2}$ and $I_{2}=1$. Thus, $Z_{R}^{(q,r)}=T$ and
$Z_{R}^{(q,r)}$ is a Sylow $p^{\prime}$-subgroup of
$\operatorname{Aut}(\mathbb{P})$. ∎

$Z_{R}^{(q,r)}$$Z^{(q,r)}$$C_{\mathbb{P}}$$\operatorname{Aut}(\mathbb{P})$$T$$U^{\prime}$$\operatorname{GL}(\mathbb{F})^{3}$$I_{1}$$I$$I_{2}$

###  5.2. A theorem on isotopisms between biprojective semifields

Now we are going to show that if two biprojective pre-semifields are isotopic,
an isotopism $(N,L,M)$ that satisfies strong requirements on the shape of the
linearized polynomials $N,L$ and $M$ has to exist, whenever the condition
appearing in the assumption of Lemma 5.8 is satisfied. We will name it
Condition (C). First we need a simple lemma.

######  Lemma 5.9.

Let $(x,y)\mapsto
F(x,y)=[f(x,y),g(x,y)]=[(a_{0},b_{0},c_{0},d_{0})_{q},(a_{1},b_{1},c_{1},d_{1})_{r}]$
be a $(q,r)$-biprojective mapping for arbitrary values of $q,r$. If $F$ is
planar, then $(a_{0},a_{1})\neq(0,0)$ and $(d_{0},d_{1})\neq(0,0)$. That is to
say, all biprojective pre-semifield polarizations $\Delta_{F}((x,y),(u,v))$
have a component that contains both monomials $x^{\sigma}u$ and $xu^{\sigma}$
and a component (not necessarily different) that contains both $y^{\tau}v$ and
$yv^{\tau}$ for $\sigma,\tau\in\\{q,r\\}$ depending on the component.

###### Proof.

If $a_{0}=a_{1}=0$, then
${\mathsf{D}_{f}^{{\infty}}}(x,y)=0={\mathsf{D}_{g}^{{\infty}}}(x,y)$ for
$y=0$ and arbitrary $x$, contradicting Lemma 3.1 for $u=\infty$. The
contradiction for $d_{0},d_{1}$ is obtained similarly by considering Lemma 3.1
for $u=0$. The statement on the corresponding pre-semifield follows
immediately by Eq. (3.1) and the fact that $(a_{0},a_{1})\neq(0,0)$. ∎

We are now ready to prove the main result of this section: If two biprojective
semifields are isotopic, then there exists an isotopism between them of a very
specific form.

######  Theorem 5.10.

Let $\mathbb{P}_{1}=(\mathbb{M}\times\mathbb{M},+,\ast_{1})$ and
$\mathbb{P}_{2}=(\mathbb{M}\times\mathbb{M},+,\ast_{2})$ be $(q_{1},r_{1})$\-
and $(q_{2},r_{2})$-biprojective pre-semifields, respectively, such that
$q_{1}\not\in\\{r_{1},\overline{r_{1}}\\}$, $1\notin\\{q_{1},r_{1}\\}$ and
$Q\not\in\\{q_{1},r_{1}\\}$, where $q_{i}=p^{k_{i}}$ and $r_{i}=p^{l_{i}}$ for
$i\in\\{1,2\\}$. Assume that

(C) $C_{\mathbb{P}_{1}}\text{ contains }Z^{(q_{1},r_{1})}\text{ as an index
}I\text{ subgroup such that }p^{\prime}\text{ does not divide }I.$

If $\mathbb{P}_{1},\mathbb{P}_{2}$ are isotopic, then there exists an
isotopism $\gamma=(N,L,M)\in\operatorname{GL}(\mathbb{F})^{3}$, with the
following properties:

  * •

All non-zero subfunctions of $N,L$ and $M$ are monomials.

  * •

All non-zero subfunctions of $L$ and $M$ have the same degree $p^{t}$.

  * •

We have,

    * –

either $k_{1}\equiv\pm k_{2}\pmod{m}$ and $l_{1}\equiv\pm l_{2}\pmod{m}$,

    * –

or $k_{1}\equiv\pm l_{2}\pmod{m}$ and $l_{1}\equiv\pm k_{2}\pmod{m}$.

More precisely, we have either,

  * •

$N_{2}=N_{3}=0$ and $N_{1},N_{4}\neq 0$,

  * •

$k_{1}\equiv\pm k_{2}\pmod{m}$ and $l_{1}\equiv\pm l_{2}\pmod{m}$,

  * •

if $k_{1}\equiv k_{2}\pmod{m}$ (resp. $l_{1}\equiv l_{2}\pmod{m}$) then
$N_{1}$ (resp. $N_{4}$) is a monomial of degree $p^{t}$,

  * •

if $k_{1}\equiv-k_{2}\pmod{m}$ (resp. $l_{1}\equiv-l_{2}\pmod{m}$) then
$N_{1}$ (resp. $N_{4}$) is a monomial of degree $p^{t+k_{2}}$ (resp.
$p^{t+l_{2}}$),

or,

  * •

$N_{1}=N_{4}=0$ and $N_{2},N_{3}\neq 0$,

  * •

$k_{1}\equiv\pm l_{2}\pmod{m}$ and $l_{1}\equiv\pm k_{2}\pmod{m}$,

  * •

if $k_{1}\equiv l_{2}\pmod{m}$ (resp. $l_{1}\equiv k_{2}\pmod{m}$) then
$N_{3}$ (resp. $N_{2}$) is a monomial of degree $p^{t}$,

  * •

if $k_{1}\equiv-l_{2}\pmod{m}$ (resp. $l_{1}\equiv-k_{2}\pmod{m}$) then
$N_{3}$ (resp. $N_{2}$) is a monomial of degree $p^{t+l_{2}}$ (resp.
$p^{t+k_{2}}$).

###### Proof.

Set

$\displaystyle(x,y)\ast_{1}(u,v)$
$\displaystyle=(f_{1}(x,y,u,v),g_{1}(x,y,u,v)),\textrm{ and}$
$\displaystyle(x,y)\ast_{2}(u,v)$
$\displaystyle=(f_{2}(x,y,u,v),g_{2}(x,y,u,v)).$

By Lemma 5.5, we have
$Z_{R}^{(q_{1},r_{1})}\leq\operatorname{Aut}(\mathbb{P}_{1})$ and
$Z_{R}^{(q_{2},r_{2})}\leq\operatorname{Aut}(\mathbb{P}_{2})$. Assume
$\mathbb{P}_{1}$ and $\mathbb{P}_{2}$ are isotopic via the isotopism
$\delta\in\operatorname{GL}(\mathbb{F})^{3}$ that maps $\mathbb{P}_{1}$ to
$\mathbb{P}_{2}$. Then
$\delta^{-1}\operatorname{Aut}(\mathbb{P}_{2})\delta=\operatorname{Aut}(\mathbb{P}_{1})$
by Lemma 5.1. Observe that
$|\delta^{-1}Z_{R}^{(q_{2},r_{2})}\delta|=|R|=|Z_{R}^{(q_{1},r_{1})}|$, so
$Z_{R}^{(q_{1},r_{1})}$ and $\delta^{-1}Z_{R}^{(q_{2},r_{2})}\delta$ are Sylow
$p^{\prime}$-subgroups of $\operatorname{Aut}(\mathbb{P}_{1})$ by Lemma 5.8 as
long as Condition (C) holds. In particular, these two subgroups are conjugate
in $\operatorname{Aut}(\mathbb{P}_{1})$ by Sylow Theorem (iii), i.e., there
exists a $\lambda\in\operatorname{Aut}(\mathbb{P}_{1})$ such that

(5.1)
$\lambda^{-1}\delta^{-1}Z_{R}^{(q_{2},r_{2})}\delta\lambda=(\delta\lambda)^{-1}Z_{R}^{(q_{2},r_{2})}(\delta\lambda)=Z_{R}^{(q_{1},r_{1})}.$

Set $\gamma=(N,L,M)=\delta\lambda$. Note that $\gamma$ is an isotopism between
$\mathbb{P}_{1}$ and $\mathbb{P}_{2}$ since
$\lambda\in\operatorname{Aut}(\mathbb{P}_{1})$. Eq. (5.1) then immediately
implies that

$\displaystyle\operatorname{diag}(m_{a^{q_{2}+1}},m_{a^{r_{2}+1}})N$
$\displaystyle=N\operatorname{diag}(m_{b^{q_{1}+1}},m_{b^{r_{1}+1}})$
$\displaystyle\operatorname{diag}(m_{a},m_{a})L$
$\displaystyle=L\operatorname{diag}(m_{b},m_{b})$
$\displaystyle\operatorname{diag}(m_{a},m_{a})M$
$\displaystyle=M\operatorname{diag}(m_{b},m_{b}),$

for all $a\in R$ and $b=\pi(a)$ where $\pi\colon R\rightarrow R$ is a
permutation. In particular, $L$ and $M$ are in the normalizer of
$S_{R}=\\{\operatorname{diag}(m_{a},m_{a})\colon a\in R\\}$. By Lemma 5.7, all
of the four subfunctions of $L$ and $M$ are zero or monomials of the same
degree, say $p^{t_{2}}$ and $p^{t_{3}}$, respectively. Then, for all
$(x,y),(u,v)\in\mathbb{M}^{2}$,

$\displaystyle L(x,y)\ast_{2}M(u,v)$
$\displaystyle=(a_{2}x^{p^{t_{2}}}+b_{2}y^{p^{t_{2}}},c_{2}x^{p^{t_{2}}}+d_{2}y^{p^{t_{2}}})\ast_{2}(a_{3}u^{p^{t_{3}}}+b_{3}v^{p^{t_{3}}},c_{3}u^{p^{t_{3}}}+d_{3}v^{p^{t_{3}}}),$
$\displaystyle=(h_{1}(x,y,u,v),h_{2}(x,y,u,v))$

for some $a_{2},b_{2},c_{2},d_{2},a_{3},b_{3},c_{3},d_{3}\in\mathbb{M}$. We
also have

$\displaystyle N((x,y)\ast_{1}(u,v))=($ $\displaystyle
N_{1}(f_{1}(x,y,u,v))+N_{2}(g_{1}(x,y,u,v)),$ $\displaystyle
N_{3}(f_{1}(x,y,u,v))+N_{4}(g_{1}(x,y,u,v))).$

Let us now assume that $N((x,y)\ast_{1}(u,v))=L(x,y)\ast_{2}M(u,v)$. We
consider the first component, i.e.
$h_{1}(x,y,u,v)=N_{1}(f_{1}(x,y,u,v))+N_{2}(g_{1}(x,y,u,v))$. Lemma 5.9
implies both monomials $x^{q}u$ and $xu^{q}$ occur in one of the two
components of $\mathbb{P}_{1}$. Since switching the components clearly
preserves isotopy, we can assume without loss of generality that they occur in
the first component $f_{1}$. Let us for now assume $N_{1}\neq 0$. We know that
$N_{1}(f_{1}(x,y,u,v))$ has then terms of the form

(5.2) $x^{p^{k_{1}+t}}u^{p^{t}}\text{ and }x^{p^{t}}u^{p^{k_{1}+t}}$

for at least one $0\leq t\leq m-1$. Observe that the differences of $p$-adic
valuations of exponents in the $x$\- and $u$-degrees of the monomials are
$k_{1}+t-t=k_{1}$ and $t-k_{1}-t=-k_{1}$, respectively. In particular, if
$q_{1}\neq r_{1}$, $q_{1}\neq\overline{r_{1}}$, then these terms cannot be
canceled out by $N_{2}(g_{1}(x,y,u,v))$. Since $\mathbb{P}_{2}$ is a
$(q_{2},r_{2})$-biprojective pre-semifield, all possible terms in $h_{1}$ are
of the form

(5.3) $w^{p^{k_{2}+t_{2}}}z^{p^{t_{3}}}\text{ or
}w^{p^{t_{2}}}z^{p^{k_{2}+t_{3}}},$

where $w\in\\{x,y\\}$, $z\in\\{u,v\\}$ and $0\leq t_{2},t_{3}\leq m-1$.
Comparing Eqs. (5.2) and (5.3) gives either

$\displaystyle k_{1}+t$ $\displaystyle\equiv k_{2}+t_{2}\pmod{m},$
$\displaystyle t$ $\displaystyle\equiv t_{3}\pmod{m},$ $\displaystyle t$
$\displaystyle\equiv t_{2}\pmod{m},$ $\displaystyle k_{1}+t$
$\displaystyle\equiv k_{2}+t_{3}\pmod{m},\textrm{ or,}$

$\displaystyle k_{1}+t$ $\displaystyle\equiv t_{2}\pmod{m},$ $\displaystyle t$
$\displaystyle\equiv k_{2}+t_{3}\pmod{m},$ $\displaystyle t$
$\displaystyle\equiv k_{2}+t_{2}\pmod{m},$ $\displaystyle k_{1}+t$
$\displaystyle\equiv t_{3}\pmod{m}.$

The first possibility is equivalent to $t=t_{2}=t_{3}$, $k_{1}\equiv
k_{2}\pmod{m}$ and the second implies $t_{2}=t_{3}$, $t=t_{2}+k_{2}$ and
$k_{1}\equiv-k_{2}\pmod{m}$. Note in particular that in any case
$t_{2}=t_{3}$. Moreover, both cases cannot occur simultaneously since, by
assumption, $k_{1}\not\equiv m/2\pmod{m}$. We conclude that $N_{1}$ is a
monomial with the same degree $p^{t_{2}}$ as the $L_{i},M_{i}$ if $k_{1}\equiv
k_{2}\pmod{m}$, and a monomial with the degree $p^{t_{2}+k_{2}}$ if
$k_{1}\equiv-k_{2}\pmod{m}$.

Now assume $N_{2}\neq 0$ and observe that the terms of $N_{2}(g_{1}(x,y,u,v))$
are of the form

(5.4) $w^{p^{l_{1}+t}}z^{p^{t}}\text{ or }w^{p^{t}}z^{p^{l_{1}+t}}$

, where $w\in\\{x,y\\}$, $z\in\\{u,v\\}$ and some $0\leq t\leq m-1$. In
particular, the difference of $p$-adic valuations of exponents of the two
monomials is $l_{1}$ or $-l_{1}$. This however yields a contradiction since
the difference of $p$-adic valuations of exponents in Eq. (5.3) is $k_{2}$ or
$-k_{2}$, that is by the considerations above, $-k_{1}$ or $k_{1}$, which
leads to $k_{1}\equiv\pm l_{1}\pmod{m}$ which is not possible since $q_{1}\neq
r_{1}$, $q_{1}\neq\overline{r_{1}}$. We conclude that $N_{2}=0$.

Let us now consider the second component. Since $N_{2}=0$, we must have
$N_{4}\neq 0$ since $N$ is bijective. The terms of $N_{4}(g_{1}(x,y,u,v))$ are
then again of the same form as in Eq. (5.4). Similar to Eq. (5.3), the terms
in $h_{2}$ are of the form (using $t_{2}=t_{3}$)

(5.5) $w^{p^{l_{2}+t_{2}}}z^{p^{t_{2}}}\text{ or
}w^{p^{t_{2}}}z^{p^{l_{2}+t_{2}}},$

where $w\in\\{x,y\\}$, $z\in\\{u,v\\}$. A comparison between the Eqs. (5.4)
and (5.5) yields either $t=t_{2}$ and $l_{1}\equiv l_{2}\pmod{m}$ or
$t=l_{2}+t_{2}$ and $l_{1}\equiv-l_{2}\pmod{m}$. Again, this means that
$N_{4}$ is a monomial of degree $p^{t_{2}}$ or $p^{l_{2}+t_{2}}$ as both cases
cannot occur simultaneously since $l_{1}\not\equiv m/2\pmod{m}$.

Since $q_{1}\neq r_{1}$, $q_{1}\neq\overline{r_{1}}$, we can again deduce
$N_{3}=0$ with the same argument we used to prove $N_{2}=0$ before. This
concludes the case $N_{1}\neq 0$.

Now assume $N_{1}=0$. Since $N$ is bijective, this implies $N_{3}\neq 0$. We
can then employ the entire argument, just starting with the second component
and exchanging $k_{2}$ and $l_{2}$, $N_{1}$ and $N_{3}$, and $N_{2}$ and
$N_{4}$ throughout. We conclude that in this case $N_{3},N_{2}\neq 0$ and
$N_{1}=N_{4}=0$.

∎

######  Remark 5.11.

  1. (i)

We exclude the cases $q_{1}=r_{1}$, $q_{1}=\overline{r_{1}}$,
$1\in\\{q_{1},r_{1}\\}$, and $Q\in\\{q_{1},r_{1}\\}$. It is possible to give
(slightly weaker) versions of Theorem 5.10 also in the excluded cases. We
avoided these cases to simplify the exposition. For instance, when we allow
$Q\in\\{q_{1},r_{1}\\}$, then the subfunctions of $N$ may be binomials of the
form $N_{i}=ax^{p^{t}}+bx^{p^{t+m/2}}$. We will showcase an isotopy of this
kind in Remark 8.3 in Section 8. Observe that in the version we have given,
all non-zero subfunctions of $N,L,M$ are monomials. We chose this presentation
of the theorem to avoid listing unnecessary special cases that we do not need
in this paper.

  2. (ii)

We will mainly use Theorem 5.10 to determine the number of isotopy classes in
Family $\mathcal{S}$. Of course, it can also be used to give alternative (in
most cases simpler) proofs of the number of isotopy classes of the known
commutative biprojective pre-semifields.

Theorem 5.10 enables us to settle the isotopy question of biprojective pre-
semifields with relative ease as long as Condition (C) is satisfied. In the
next section we will first show that Condition (C) is satisfied for our family
and then use Theorem 5.10 to determine the number of non-isotopic semifields
in the family. Note that the condition $m>2$ we stipulate in this section is
not restrictive when considering $\mathcal{S}$ since it does not yield
semifields when $m$ is a power of $2$.

##  6\. Isotopisms within the Family $\mathcal{S}$

In this section we will show that the number of non-isotopic semifields within
the Family $\mathcal{S}$ is exponential in $n$ (when $n=4t$, where $t$ is not
a power of $2$). We first need to check Condition (C) in Theorem 5.10 for the
pre-semifields in the Family $\mathcal{S}$. The following lemma does that in a
straightforward manner.

######  Lemma 6.1.

Let $n=2m$ and $\mathbb{P}=(\mathbb{M}\times\mathbb{M},+,\ast)$ be a
$(q,r)$-biprojective pre-semifield in the Family $\mathcal{S}$. Then

$\displaystyle|C_{\mathbb{P}}|$
$\displaystyle=(p^{m}-1)(p^{{\mathsf{gcd}({k},{m})}}-1),\textrm{ or}$
$\displaystyle|C_{\mathbb{P}}|$
$\displaystyle=2(p^{m}-1)(p^{{\mathsf{gcd}({k},{m})}}-1).$

In particular, Condition (C) is always satisfied.

###### Proof.

If $(N,L,M)\in C_{\mathbb{P}}$ then, by Lemma 5.7, the subfunctions $L_{i}$
and $M_{i}$ for $i\in\\{1,2,3,4\\}$ are zero or monomials of degree $1$, we
write

$\displaystyle L_{1}(x)=a_{2}x,\quad L_{2}(x)=b_{2}x,\quad
L_{3}(x)=c_{2}x,\quad L_{4}(x)=d_{2}x,$ $\displaystyle M_{1}(x)=a_{3}x,\quad
M_{2}(x)=b_{3}x,\quad M_{3}(x)=c_{3}x,\quad M_{4}(x)=d_{3}x.$

We then have

$\displaystyle L(x,y)\ast M(u,v)=$
$\displaystyle(a_{2}x+b_{2}y,c_{2}x+d_{2}y)\ast(a_{3}u+b_{3}v,c_{3}u+d_{3}v)$
$\displaystyle=$
$\displaystyle((a_{2}x+b_{2}y)^{q}(a_{3}u+b_{3}v)+(a_{2}x+b_{2}y)(a_{3}u+b_{3}v)^{q}$
$\displaystyle+B((c_{2}x+d_{2}y)^{q}(c_{3}u+d_{3}v)+(c_{2}x+d_{2}y)(c_{3}u+d_{3}v)^{q}),$
$\displaystyle(a_{2}x+b_{2}y)^{qQ}(c_{3}u+d_{3}v)+(c_{2}x+d_{2}y)(a_{3}u+b_{3}v)^{qQ}$
(6.1)
$\displaystyle+\frac{a}{B}((a_{2}x+b_{2}y)(c_{3}u+d_{3}v)^{qQ}+(c_{2}x+d_{2}y)^{qQ}(a_{3}u+b_{3}v))).$

Similarly, we have

$\displaystyle N((x,y)\ast(u,v))=($ $\displaystyle
N_{1}(x^{q}u+xu^{q}+B(y^{q}v+yv^{q}))+N_{2}(x^{qQ}v+yu^{qQ}+\frac{a}{B}(xv^{qQ}+y^{qQ}u)),$
$\displaystyle
N_{3}(x^{q}u+xu^{q}+B(y^{q}v+yv^{q}))+N_{4}(x^{qQ}v+yu^{qQ}+\frac{a}{B}(xv^{qQ}+y^{qQ}u))).$

By comparing the degrees, it is then easy to see that
$N((x,y)\ast(u,v))=L(x,y)\ast M(u,v)$ implies $N_{2}=N_{3}=0$ and
$N_{1}=a_{1}x$, $N_{4}=d_{1}x$ for some $a_{1},d_{1}\in\mathbb{M}^{\times}$.
Then

(6.2)
$N((x,y)\ast(u,v))=(a_{1}(x^{q}u+xu^{q}+B(y^{q}v+yv^{q})),d_{1}(x^{qQ}v+yu^{qQ}+\frac{a}{B}(xv^{qQ}+y^{qQ}u))).$

We compare the coefficients of $x^{q}u$, $xu^{q}$, $x^{q}v$, $xv^{q}$,
$y^{q}u$, $yu^{q}$, $y^{q}v$, $yv^{q}$ in the first component of Eqs. (6.1)
and (6.2) to get the following $8$ equations:

(6.3) $\displaystyle a_{1}$ $\displaystyle=a_{2}^{q}a_{3}+Bc_{2}^{q}c_{3}$
(6.4) $\displaystyle a_{1}$ $\displaystyle=a_{2}a_{3}^{q}+Bc_{2}c_{3}^{q}$
(6.5) $\displaystyle 0$ $\displaystyle=a_{2}^{q}b_{3}+Bc_{2}^{q}d_{3}$ (6.6)
$\displaystyle 0$ $\displaystyle=a_{2}b_{3}^{q}+Bc_{2}d_{3}^{q}$

(6.7) $\displaystyle 0$ $\displaystyle=b_{2}^{q}a_{3}+Bd_{2}^{q}c_{3}$ (6.8)
$\displaystyle 0$ $\displaystyle=b_{2}a_{3}^{q}+Bd_{2}c_{3}^{q}$ (6.9)
$\displaystyle Ba_{1}$ $\displaystyle=b_{2}^{q}b_{3}+Bd_{2}^{q}d_{3}$ (6.10)
$\displaystyle Ba_{1}$ $\displaystyle=b_{2}b_{3}^{q}+Bd_{2}d_{3}^{q}.$

And similarly the $8$ equations that come from comparing the coefficients in
the second component:

(6.11) $\displaystyle 0$ $\displaystyle=a_{2}^{qQ}c_{3}+(a/B)c_{2}^{qQ}a_{3}$
(6.12) $\displaystyle 0$ $\displaystyle=(a/B)a_{2}c_{3}^{qQ}+c_{2}a_{3}^{qQ}$
(6.13) $\displaystyle d_{1}$
$\displaystyle=a_{2}^{qQ}d_{3}+(a/B)c_{2}^{qQ}b_{3}$ (6.14) $\displaystyle
d_{1}a/B$ $\displaystyle=(a/B)a_{2}d_{3}^{qQ}+c_{2}b_{3}^{qQ}$

(6.15) $\displaystyle d_{1}a/B$
$\displaystyle=b_{2}^{qQ}c_{3}+(a/B)d_{2}^{qQ}a_{3}$ (6.16) $\displaystyle
d_{1}$ $\displaystyle=(a/B)b_{2}c_{3}^{qQ}+d_{2}a_{3}^{qQ}$ (6.17)
$\displaystyle 0$ $\displaystyle=b_{2}^{qQ}d_{3}+(a/B)d_{2}^{qQ}b_{3}$ (6.18)
$\displaystyle 0$ $\displaystyle=(a/B)b_{2}d_{3}^{qQ}+d_{2}b_{3}^{qQ}.$

Let us first assume that $a_{2},c_{2},a_{3},c_{3}\neq 0$. Then by Eqs. (6.11)
and (6.12), we have

$-\frac{a}{B}=\frac{a_{2}^{qQ}c_{3}}{c_{2}^{qQ}a_{3}}=\frac{c_{2}a_{3}^{qQ}}{a_{2}c_{3}^{qQ}}.$

Setting $a_{3}=\omega_{1}a_{2}$, $c_{3}=\omega_{2}c_{2}$ gives

$-\frac{a}{B}=\left(\frac{a_{2}}{c_{2}}\right)^{qQ-1}\frac{\omega_{2}}{\omega_{1}}=\left(\frac{a_{2}}{c_{2}}\right)^{qQ-1}\left(\frac{\omega_{1}}{\omega_{2}}\right)^{qQ}.$

This implies $\omega_{1}^{qQ+1}=\omega_{2}^{qQ+1}$, that is
$\omega_{2}=\zeta\omega_{1}$ where $\zeta$ is a $(qQ+1)^{\textrm{st}}$ root of
unity. Substituting this into the previous equation yields

(6.19) $-\frac{a}{B}=\left(\frac{a_{2}}{c_{2}}\right)^{qQ-1}\zeta.$

By Lemmas 4.2 and 4.3 (iii), we have
${\mathsf{gcd}({qQ+1},{p^{m}-1})}=p^{{\mathsf{gcd}({k},{m})}/2}+1$, so $\zeta$
is a $(p^{{\mathsf{gcd}({k},{m})}/2}+1)^{\textrm{st}}$ root of unity. In
particular, $\zeta\in\mathbb{E}$ since $p^{{\mathsf{gcd}({k},{m})}/2}+1$
divides $p^{{\mathsf{gcd}({k},{m})}}-1$. The
$(p^{{\mathsf{gcd}({k},{m})}/2}+1)^{\textrm{st}}$ roots of unity in
$\mathbb{E}$ are precisely the
$(p^{{\mathsf{gcd}({k},{m})}/2}-1)^{\textrm{st}}$ powers in $\mathbb{E}$. In
particular, $\zeta$ is a square. This is however a contradiction to Eq. (6.19)
since the left hand side is a non-square (recall that $-1$ and $a$ are
squares), and the right hand side is a square. We conclude that
$a_{2}c_{2}a_{3}c_{3}=0$. We can proceed identically with
$b_{2},d_{2},b_{3},d_{3}$ and Eqs. (6.17) and (6.18) which yields
$b_{2}d_{2}b_{3}d_{3}=0$. The conditions in Eqs. (6.11), (6.12), (6.17),
(6.18) and the bijectivity of $L,M$ then only leave two possibilities: Either
$a_{2}=a_{3}=d_{2}=d_{3}=0$ and $b_{2},b_{3},c_{2},c_{3}\neq 0$ or, the other
way round, $a_{2},a_{3},d_{2},d_{3}\neq 0$ and $b_{2}=b_{3}=c_{2}=c_{3}=0$. We
will deal with these two cases separately. Note that in both cases, Eqs.
(6.5), (6.6), (6.7), (6.8), (6.11), (6.12), (6.17), (6.18) are always
satisfied.

Case $b_{2},b_{3},c_{2},c_{3}=0$: We set $a_{3}=\omega_{1}a_{2}$,
$d_{3}=\omega_{2}d_{2}$. Then Eqs. (6.3), (6.4), (6.9), (6.10) become

$a_{1}=a_{2}^{q+1}\omega_{1}=a_{2}^{q+1}\omega_{1}^{q}=d_{2}^{q+1}\omega_{2}=d_{2}^{q+1}\omega_{2}^{q},$

which is satisfied if and only if $\omega_{1},\omega_{2}\in\mathbb{E}$ and

(6.20) $(a_{2}/d_{2})^{q+1}=\omega_{2}/\omega_{1}.$

Similarly, from Eq. (6.13), (6.14), (6.15), (6.16), we get immediately (using
$\omega_{1},\omega_{2}\in\mathbb{E}$),

$d_{1}=a_{2}^{qQ}d_{2}\omega_{2}=a_{2}d_{2}^{qQ}\omega_{2}^{Q}=a_{2}d_{2}^{qQ}\omega_{1}=a_{2}^{qQ}d_{2}\omega_{1}^{Q}.$

This is equivalent to $\omega_{1}=\omega_{2}^{Q}$,
$(a_{2}/d_{2})^{qQ-1}=\omega_{2}^{Q-1}$. Multiplying this with Eq. (6.20)
gives

$(a_{2}/d_{2})^{q(Q+1)}=1,$

i.e., $a_{2}/d_{2}\in(\mathbb{M}^{\times})^{Q-1}$, say
$\zeta^{Q-1}=a_{2}/d_{2}$. Rewriting Eq. (6.20) gives

$(a_{2}/d_{2})^{q+1}=(\zeta^{q+1})^{Q-1}=\omega_{1}^{Q-1}.$

The equation cannot be satisfied if $\omega_{1}$ is a non-square. If
$\omega_{1}$ is a square, then $d_{2}$ is uniquely determined up to the sign
from $\omega_{1}$ and $a_{2}$. Thus we have in total $p^{m}-1$ choices for
$a_{2}$, $(p^{{\mathsf{gcd}({k},{m})}}-1)/2$ choices for $\omega_{1}$ and $2$
choices for $d_{2}$, making in total
$(p^{m}-1)(p^{{\mathsf{gcd}({k},{m})}}-1)$ choices in this case.
Case $a_{2},a_{3},d_{2},d_{3}=0$: Similarly to the previous case, we set
$b_{3}=\omega_{1}b_{2}$, $c_{3}=\omega_{2}c_{2}$. We get from the first set of
equations:

$a_{1}=Bc_{2}^{q+1}\omega_{2}=Bc_{2}^{q+1}\omega_{2}^{q}=(1/B)b_{2}^{q+1}\omega_{1}=(1/B)b_{2}^{q+1}\omega_{1}^{q},$

which is equivalent to $\omega_{1},\omega_{2}\in\mathbb{E}$ and

(6.21) $(b_{2}/c_{2})^{q+1}=B^{2}\omega_{2}/\omega_{1}.$

The second set of equations gives

$d_{1}=(a/B)c_{2}^{qQ}b_{2}\omega_{1}=(B/a)c_{2}b_{2}^{qQ}\omega_{1}^{Q}=(B/a)c_{2}b_{2}^{qQ}\omega_{2}=(a/B)c_{2}^{qQ}b_{2}\omega_{2}^{Q}.$

This again implies $\omega_{1}^{Q}=\omega_{2}$ and
$(b_{2}/c_{2})^{qQ-1}=(1/\omega_{1}^{Q-1})(a/B)^{2}$. Multiplying this with
Eq. (6.21) gives

$(b_{2}/c_{2})^{q(Q+1)}=\frac{\omega_{2}}{\omega_{1}^{Q}}a^{2}=a^{2}=a^{Q+1}.$

Thus, $b_{2}/c_{2}$ is determined up to multiplication with a
$(Q+1)^{\textrm{st}}$ root of unity, that is, a $(Q-1)^{\textrm{st}}$ power,
say $(b_{2}/c_{2})^{q}=\zeta^{Q-1}a$ . Eq. (6.21) can be rewritten as

$(b_{2}/c_{2})^{q+1}=a^{\overline{q}+1}(\zeta^{\overline{q}+1})^{Q-1}=B^{2}\omega_{1}^{Q-1}.$

This equation has a solution if and only if $(a^{\overline{q}+1})/B^{2}$ is a
$(Q-1)^{\textrm{st}}$ power, say $(a^{\overline{q}+1})/B^{2}=\rho^{Q-1}$. In
this case, there are $(p^{{\mathsf{gcd}({k},{m})}}-1)/2$ possible choices for
$\omega_{1}$, either all squares or all non-squares, depending on if $\rho$ is
a square or not. Then $\zeta$ is determined up to the sign, so there are
$p^{m}-1$ possible choices for $b_{2}$, $2$ possible choices for $c_{2}$ and
$(p^{{\mathsf{gcd}({k},{m})}}-1)/2$ possible choices for $\omega_{1}$. This
case thus contributes either $0$ or $(p^{m}-1)(p^{{\mathsf{gcd}({k},{m})}}-1)$
elements. Both cases together show that $|C_{\mathbb{P}}|$ is either
$(p^{m}-1)(p^{{\mathsf{gcd}({k},{m})}}-1)$ or
$2(p^{m}-1)(p^{{\mathsf{gcd}({k},{m})}}-1)$.

Now it is clear that
$p^{\prime}\nmid[C_{\mathbb{P}}:Z^{(q,r)}]\in\\{p^{{\mathsf{gcd}({k},{m})}}-1,2(p^{{\mathsf{gcd}({k},{m})}}-1)\\}$
by our assumption that $p^{\prime}$ is $p$-primitive (recall $p^{\prime}\neq
2$). ∎

We can now apply Thereom 5.10 to the pre-semifields in the Family
$\mathcal{S}$.

######  Theorem 6.2.

Let $\mathbb{P}_{q,B,a}=(\mathbb{M}\times\mathbb{M},+,\ast_{1})$ and
$\mathbb{P}_{q^{\prime},B^{\prime},a^{\prime}}=(\mathbb{M}\times\mathbb{M},+,\ast_{2})$
be pre-semifields from the Family $\mathcal{S}$. Then

  1. (i)

$\mathbb{P}_{q,B,a}$ and $\mathbb{P}_{q^{\prime},B^{\prime},a^{\prime}}$ are
isotopic if and only if they are strongly isotopic.

  2. (ii)

$\mathbb{P}_{q,B,a}$ is isotopic to $\mathbb{P}_{\overline{q},B,a^{\prime}}$
for $a^{\prime}=B^{Q+1}/a$ and arbitrary $q$.

  3. (iii)

$\mathbb{P}_{q,B,a}$ is isotopic to $\mathbb{P}_{q,B^{\prime},a^{\prime}}$ for
arbitrary $q,B,B^{\prime},a$ and a suitable choice for $a^{\prime}$.

  4. (iv)

If $\mathbb{P}_{q,B,a}$ is isotopic to $\mathbb{P}_{q,B,a^{\prime}}$, then it
is also isotopic to $\mathbb{P}_{q,B,-a^{\prime}}$.

  5. (v)

There are at most $2m=n$ different $a^{\prime}$ such that $\mathbb{P}_{q,B,a}$
is isotopic to $\mathbb{P}_{q,B,a^{\prime}}$.

  6. (vi)

No other isotopisms exist.

###### Proof.

Let $(N,L,M)$ be an isotopism between $\mathbb{P}_{q,B,a}$ and
$\mathbb{P}_{q^{\prime},B^{\prime},a^{\prime}}=(\mathbb{M}\times\mathbb{M},+,\ast_{2})$.
All subfunctions of $N,L,M$ are zero or monomials by Theorem 5.10. Moreover,
$\mathbb{P}_{q,B,a}$ and $\mathbb{P}_{q^{\prime},B^{\prime},a^{\prime}}$ can
only be isotopic if $q^{\prime}=q$, $q^{\prime}=\overline{q}$,
$q^{\prime}=qQ$, or $q^{\prime}=\overline{q}Q$. Note that if
$m/{\mathsf{gcd}({k},{m})}$ is odd, then $m/{\mathsf{gcd}({k+m/2},{m})}$ is
even by Lemma 4.3 (iii), so the cases $q^{\prime}=qQ$,
$q^{\prime}=\overline{q}Q$ do not satisfy the conditions of Theorem 4.4 and
need not be considered.

We first show the isotopy in the case $q^{\prime}=\overline{q}$. We have

$(x,y)\ast_{1}(u,v)=(x^{q}u+xu^{q}+B(y^{q}v+yv^{q}),x^{qQ}v+yu^{qQ}+(a/B)(xv^{qQ}+y^{qQ}u)).$

A transformation with

$N_{1}=x,N_{4}=(B^{Q}/a)x^{Q},N_{2}=N_{3}=0$

and raising $x,y,u,v$ to the $\overline{q}$-th power yields

$N((x^{\overline{q}},y^{\overline{q}})\ast_{1}(u^{\overline{q}},v^{\overline{q}}))=(xu^{\overline{q}}+x^{\overline{q}}u+B(yv^{\overline{q}}+y^{\overline{q}}v),(B^{Q}/a)(xv^{\overline{q}Q}+y^{\overline{q}Q}u)+x^{\overline{q}Q}v+yu^{\overline{q}Q}).$

Observe that one can write $B^{Q}/a=a^{\prime}/B$ for some
$a^{\prime}\in\mathbb{L}$ (indeed this is equivalent to $B^{Q+1}=aa^{\prime}$
which has always a solution since $B^{Q+1}\in\mathbb{L}$). We conclude that
there is always a strong isotopism between $\mathbb{P}_{q,B,a}$ and
$\mathbb{P}_{\overline{q},B,B^{Q+1}/a}$. Thus we have proved Part (ii) of the
theorem.

It thus only remains to deal with the case $q^{\prime}=q$. By Theorem 5.10, we
only need to consider isotopisms $(N,L,M)$ with subfunctions

$\displaystyle N_{1}=a_{1}x^{p^{t}},\quad N_{4}=d_{1}x^{p^{t}},$
$\displaystyle L_{1}=a_{2}x^{p^{t}},\quad L_{2}=b_{2}x^{p^{t}},\quad
L_{3}=c_{2}x^{p^{t}},\quad L_{4}=d_{2}x^{p^{t}},$ $\displaystyle
M_{1}=a_{3}x^{p^{t}},\quad M_{2}=b_{3}x^{p^{t}},\quad
M_{3}=c_{3}x^{p^{t}},\quad M_{4}=d_{3}x^{p^{t}},$

for some $t\in\\{0,\dots,m-1\\}$. Then

$\displaystyle L(x,y)\ast_{2}M(u,v)=$
$\displaystyle(((a_{2}^{\prime}x+b_{2}^{\prime}y)^{q}(a_{3}^{\prime}u+b_{3}^{\prime}v)+(a_{2}^{\prime}x+b_{2}^{\prime}y)(a_{3}^{\prime}u+b_{3}^{\prime}v)^{q})^{p^{t}}$
$\displaystyle+B^{\prime}((c_{2}^{\prime}x+d_{2}^{\prime}y)^{q}(c_{3}^{\prime}u+d_{3}^{\prime}v)+(c_{2}^{\prime}x+d_{2}^{\prime}y)(c_{3}^{\prime}u+d_{3}^{\prime}v)^{q})^{p^{t}},$
$\displaystyle((a_{2}^{\prime}x+b_{2}^{\prime}y)^{qQ}(c_{3}^{\prime}u+d_{3}^{\prime}v)+(c_{2}^{\prime}x+d_{2}^{\prime}y)(a_{3}^{\prime}u+b_{3}^{\prime}v)^{qQ})^{p^{t}}$
$\displaystyle+(a^{\prime}/B^{\prime})((a_{2}^{\prime}x+b_{2}^{\prime}y)(c_{3}^{\prime}u+d_{3}^{\prime}v)^{qQ}+(c_{2}^{\prime}x+d_{2}^{\prime}y)^{qQ}(a_{3}^{\prime}u+b_{3}^{\prime}v))^{p^{t}}),$

where $a_{i}^{\prime}=a_{i}^{p^{m-t}}$ and similarly for the other
coefficients $b_{i},c_{i},d_{i}$. We also obtain

$N((x,y)\ast_{1}(u,v))=(a_{1}(x^{q}u+xu^{q}+B(y^{q}v+yv^{q}))^{p^{t}},d_{1}(x^{qQ}v+yu^{qQ}+(a/B)(xv^{qQ}+y^{qQ}u))^{p^{t}}).$

We compare the coefficients $(x^{q}u)^{p^{t}}$, $(xu^{q})^{p^{t}}$,
$(x^{q}v)^{p^{t}}$, $(xv^{q})^{p^{t}}$, $(y^{q}u)^{p^{t}}$,
$(yu^{q})^{p^{t}}$, $(y^{q}v)^{p^{t}}$, $(yv^{q})^{p^{t}}$ in the first
component to get the following $8$ equations.

(6.22) $\displaystyle a_{1}$
$\displaystyle=a_{2}^{q}a_{3}+B^{\prime}c_{2}^{q}c_{3}$ (6.23) $\displaystyle
a_{1}$ $\displaystyle=a_{2}a_{3}^{q}+B^{\prime}c_{2}c_{3}^{q}$ (6.24)
$\displaystyle 0$ $\displaystyle=a_{2}^{q}b_{3}+B^{\prime}c_{2}^{q}d_{3}$
(6.25) $\displaystyle 0$
$\displaystyle=a_{2}b_{3}^{q}+B^{\prime}c_{2}d_{3}^{q}$

(6.26) $\displaystyle 0$
$\displaystyle=b_{2}^{q}a_{3}+B^{\prime}d_{2}^{q}c_{3}$ (6.27) $\displaystyle
0$ $\displaystyle=b_{2}a_{3}^{q}+B^{\prime}d_{2}c_{3}^{q}$ (6.28)
$\displaystyle B^{p^{t}}a_{1}$
$\displaystyle=b_{2}^{q}b_{3}+B^{\prime}d_{2}^{q}d_{3}$ (6.29) $\displaystyle
B^{p^{t}}a_{1}$ $\displaystyle=b_{2}b_{3}^{q}+B^{\prime}d_{2}d_{3}^{q}.$

And similarly the $8$ equations that come from comparing the coefficients in
the second component:

(6.30) $\displaystyle 0$
$\displaystyle=a_{2}^{qQ}c_{3}+(a^{\prime}/B^{\prime})c_{2}^{qQ}a_{3}$ (6.31)
$\displaystyle 0$
$\displaystyle=(a^{\prime}/B^{\prime})a_{2}c_{3}^{qQ}+c_{2}a_{3}^{qQ}$ (6.32)
$\displaystyle d_{1}$
$\displaystyle=a_{2}^{qQ}d_{3}+(a^{\prime}/B^{\prime})c_{2}^{qQ}b_{3}$ (6.33)
$\displaystyle d_{1}(a/B)^{p^{t}}$
$\displaystyle=(a^{\prime}/B^{\prime})a_{2}d_{3}^{qQ}+c_{2}b_{3}^{qQ}$

(6.34) $\displaystyle d_{1}(a/B)^{p^{t}}$
$\displaystyle=b_{2}^{qQ}c_{3}+(a^{\prime}/B^{\prime})d_{2}^{qQ}a_{3}$ (6.35)
$\displaystyle d_{1}$
$\displaystyle=(a^{\prime}/B^{\prime})b_{2}c_{3}^{qQ}+d_{2}a_{3}^{qQ}$ (6.36)
$\displaystyle 0$
$\displaystyle=b_{2}^{qQ}d_{3}+(a^{\prime}/B^{\prime})d_{2}^{qQ}b_{3}$ (6.37)
$\displaystyle 0$
$\displaystyle=(a^{\prime}/B^{\prime})b_{2}d_{3}^{qQ}+d_{2}b_{3}^{qQ}.$

Note that Eqs. (6.30), (6.31), (6.36), (6.37) are identical to Eqs. (6.11),
(6.12), (6.17), (6.18) in the proof of Lemma 6.1, just with $a/B$ substituted
by $(a^{\prime}/B^{\prime})$. We can thus conclude with the same reasoning as
in the proof of Lemma 6.1 that either $b_{2}=b_{3}=c_{2}=c_{3}=0$ or
$a_{2}=a_{3}=d_{2}=d_{3}=0$.
Case $b_{2}=b_{3}=c_{2}=c_{3}=0$: Here, we also proceed similarly to the proof
of Lemma 6.1. We set $a_{3}=\omega_{1}a_{2}$, $d_{3}=\omega_{2}d_{2}$ and get
from Eqs. (6.22), (6.23), (6.28), (6.29)

$a_{1}=a_{2}^{q+1}\omega_{1}=a_{2}^{q+1}\omega_{1}^{q}=(B^{\prime}/B^{p^{t}})d_{2}^{q+1}\omega_{2}=(B^{\prime}/B^{p^{t}})d_{2}^{q+1}\omega_{2}^{q},$

which is satisfied if and only if $\omega_{1},\omega_{2}\in\mathbb{E}$ and

(6.38)
$\left(\frac{a_{2}}{d_{2}}\right)^{q+1}=\frac{\omega_{2}B^{\prime}}{\omega_{1}B^{p^{t}}}.$

Similarly, from Eq. (6.32), (6.33), (6.34), (6.35), we get immediately (using
$\omega_{1},\omega_{2}\in\mathbb{E}$)

$d_{1}=a_{2}^{qQ}d_{2}\omega_{2}=\frac{a^{\prime}B^{p^{t}}}{a^{p^{t}}B^{\prime}}a_{2}d_{2}^{qQ}\omega_{2}^{Q}=\frac{a^{\prime}B^{p^{t}}}{a^{p^{t}}B^{\prime}}a_{2}d_{2}^{qQ}\omega_{1}=a_{2}^{qQ}d_{2}\omega_{1}^{Q}.$

This is equivalent to $\omega_{1}^{Q}=\omega_{2}$ and
$(a_{2}/d_{2})^{qQ-1}=\omega_{2}^{Q-1}(a^{\prime}B^{p^{t}})/(a^{p^{t}}B^{\prime})$.
Multiplying the second condition with Eq. (6.38) gives

(6.39)
$\left(\frac{a_{2}}{d_{2}}\right)^{q(Q+1)}=\frac{\omega_{2}^{Q}a^{\prime}}{\omega_{1}a^{p^{t}}}=\frac{a^{\prime}}{a^{p^{t}}}.$

Using $\omega_{1}^{Q}=\omega_{2}$, we rewrite Eq. (6.38):

(6.40)
$\left(\frac{a_{2}}{d_{2}}\right)^{q+1}=\frac{\omega_{1}^{Q-1}B^{\prime}}{B^{p^{t}}}.$

Observe that $B,B^{\prime},t,\omega_{1}$ uniquely determine $(a_{2}/d_{2})$ up
to the sign from Eq. (6.40). Since $(a_{2}/d_{2})^{q(Q+1)}\in\mathbb{L}$,
there is thus for each $B,B^{\prime},t,\omega_{1},a$ precisely one
$a^{\prime}$ that satisfies all conditions. For all $\omega_{1}$ that are
squares (i.e., all $(q+1)^{\textrm{st}}$ powers), this $a^{\prime}$ is the
same since $\omega_{1}^{(Q-1)(Q+1)}=1$. Similarly, for all $\omega_{1}$ that
are non-squares, we have $\omega_{1}^{(Q-1)(Q+1)/2}=-1$, so they also all
yield the same $a^{\prime}$, and in fact precisely the same $a^{\prime}$ as
when $\omega_{1}$ is a square, just with different sign. In particular, we
conclude that a pre-semifield $\mathbb{P}_{q,B,a}$ is always isotopic to
$\mathbb{P}_{q,B^{\prime},a^{\prime}}$ for arbitrary $B^{\prime}$ and a
suitable choice of $a^{\prime}$. Since we can choose
$\omega_{1}=\omega_{2}=1$, we can even choose $a^{\prime}$ such that the pre-
semifields are strongly isotopic. Thus, we have proved Part (iii) of our
theorem.

Consequently, it is enough to consider isotopisms in the case $B=B^{\prime}$
for an arbitrary non-square $B$. When $B=B^{\prime}$, every possible choice of
$t$ yields an $a^{\prime}$ such that a pre-semifield $\mathbb{P}_{q,B,a}$ is
strongly isotopic to $\mathbb{P}_{q,B,a^{\prime}}$ and isotopic to
$\mathbb{P}_{q,B,-a^{\prime}}$. Assume the choice of $t$ in the previously
described procedure leads to a strong isotopy between $\mathbb{P}_{q,B,a}$ and
$\mathbb{P}_{q,B,a^{\prime}}$. We now show that choosing $t^{\prime}$ defined
by $t^{\prime}-t\equiv m/2\pmod{m}$ in the same procedure gives a strong
isotopy to $\mathbb{P}_{q,B,-a^{\prime}}$, i.e. $\mathbb{P}_{q,B,a}$ and
$\mathbb{P}_{q,B,-a^{\prime}}$ are not just isotopic but also strongly
isotopic.

Let $(a_{2}/d_{2})^{q+1}$ be determined by $\omega_{1}=1$ and fixed
$B=B^{\prime}$, $t$ via Eq. (6.40), i.e.

$\left(\frac{a_{2}}{d_{2}}\right)^{q+1}=\frac{1}{B^{p^{t}-1}}.$

Similarly, let $(a_{2}^{\prime}/d_{2}^{\prime})^{q+1}$ be determined by
$\omega_{1}=1$, the same $B=B^{\prime}$ and $t^{\prime}$:

$\left(\frac{a_{2}^{\prime}}{d_{2}^{\prime}}\right)^{q+1}=\frac{1}{B^{p^{t^{\prime}}-1}}.$

We then have

$\left(\frac{a_{2}^{\prime}}{d_{2}^{\prime}}\right)^{q+1}=\left(\frac{a_{2}}{d_{2}}\right)^{q+1}\cdot\frac{1}{B^{p^{t^{\prime}}-p^{t}}}.$

Since
$B^{p^{t^{\prime}}-p^{t}}=(B^{Q-1})^{p^{t}}\in(\mathbb{M}^{\times})^{Q-1}$, we
have $(a_{2}^{\prime}/d_{2}^{\prime})^{q+1}=(\zeta a_{2}/d_{2})^{q+1}$ where
$\zeta^{q+1}=1/(B^{p^{t^{\prime}}-p^{t}})\in(\mathbb{M}^{\times})^{Q-1}$. Note
that $\zeta\notin(\mathbb{M}^{\times})^{Q-1}$ since $B$ is a non-square, so
$B^{(Q-1)(Q+1)/2}\neq 1$. In particular, we have $\zeta^{Q+1}=-1$. Then by Eq.
(6.39),
$(a_{2}^{\prime}/d_{2}^{\prime})^{q(Q+1)}=\zeta^{q(Q+1)}(a_{2}/d_{2})^{q(Q+1)}=-(a_{2}/d_{2})^{q(Q+1)}=-a^{\prime}/a^{p^{t}}$.
We conclude that $\mathbb{P}_{q,B,a}$ and $\mathbb{P}_{q,B,a^{\prime}}$ are
strongly isotopic if and only if $\mathbb{P}_{q,B,a}$ and
$\mathbb{P}_{q,B,-a^{\prime}}$ are strongly isotopic.
Case $a_{2},a_{3},d_{2},d_{3}=0$: Similarly to the previous case, we set
$b_{3}=\omega_{1}b_{2}$, $c_{3}=\omega_{2}c_{2}$. Since we know from the
previous case that different $B,B^{\prime}$ always lead to strongly isotopic
pre-semifields (for suitable choices of $a,a^{\prime}$), we only consider the
case $B=B^{\prime}$ without loss of generality. We get from the first set of
equations:

$a_{1}=Bc_{2}^{q+1}\omega_{2}=Bc_{2}^{q+1}\omega_{2}^{q}=(1/B^{p^{t}})b_{2}^{q+1}\omega_{1}=(1/B^{p^{t}})b_{2}^{q+1}\omega_{1}^{q},$

which is equivalent to $\omega_{1},\omega_{2}\in\mathbb{E}$ and

(6.41)
$\left(\frac{b_{2}}{c_{2}}\right)^{q+1}=B^{p^{t}+1}\cdot\frac{\omega_{2}}{\omega_{1}}.$

The second set of equations gives

$d_{1}=(a^{\prime}/B)c_{2}^{qQ}b_{2}\omega_{1}=(B/a)^{p^{t}}c_{2}b_{2}^{qQ}\omega_{1}^{Q}=(B/a)^{p^{t}}c_{2}b_{2}^{qQ}\omega_{2}=(a^{\prime}/B)c_{2}^{qQ}b_{2}\omega_{2}^{Q}.$

This again implies $\omega_{1}^{Q}=\omega_{2}$ and
$(b_{2}/c_{2})^{qQ-1}=(1/\omega_{1}^{Q-1})(a/B)^{p^{t}}(a^{\prime}/B)$.
Multiplying this with Eq. (6.41) gives

$\left(\frac{b_{2}}{c_{2}}\right)^{q(Q+1)}=\frac{\omega_{2}}{\omega_{1}^{Q}}a^{p^{t}}a^{\prime}=a^{p^{t}}a^{\prime}.$

Eq. (6.41) can be rewritten as

$\left(\frac{b_{2}}{c_{2}}\right)^{q+1}=\omega_{1}^{Q-1}B^{p^{t}+1}.$

These two equations are structurally identical to Eqs. (6.39) and (6.40) from
the previous case. With the same argumentation, we conclude that every
possible choice of $t$ yields an $a^{\prime}$ such that a pre-semifield
$\mathbb{P}_{q,B,a}$ is strongly isotopic to $\mathbb{P}_{q,B,a^{\prime}}$ and
isotopic to $\mathbb{P}_{q,B,-a^{\prime}}$. Again, choosing $t^{\prime}$ such
that $t^{\prime}-t\equiv m/2\pmod{m}$ gives also strong isotopy between
$\mathbb{P}_{q,B,a}$ and $\mathbb{P}_{q,B,-a^{\prime}}$. This proves Parts
(iv) and (i) of our theorem. Now we can simply prove Parts (v) and (vi).
Considering both cases together, there are thus at most $2m=n$ different
$a^{\prime}$ such that $\mathbb{P}_{q,B,a}$ is strongly isotopic to
$\mathbb{P}_{q,B,a^{\prime}}$. We have considered all cases thus there are no
more isotopisms. ∎

######  Remark 6.3 (Planar equivalence and strong isotopy).

Instead of the exposition we chose based on isotopy, we could have developed
an approach based on (in)equivalences of planar $(q,r)$-biprojective mappings.
Recall that Theorem 2.1 states that strong isotopy of pre-semifields
corresponds to equivalence of the corresponding planar mappings. We give a
very brief sketch of such an approach: One can define the automorphism group
$\operatorname{Aut}(F)$ of a planar mapping $F$ of
$\mathbb{F}\cong\mathbb{M}\times\mathbb{M}$ from a DO polynomial as the set of
all $(N,L)\in\operatorname{GL}(\mathbb{F})^{2}$ such that $NFL^{-1}=F$. Note
that, by Theorem 2.1,
$\operatorname{Aut}(F)\cong\operatorname{Aut}_{S}(\mathbb{P})$, where
$\operatorname{Aut}_{S}(\mathbb{P})$ is the group of all strong autotopisms of
the pre-semifield $\mathbb{P}$ belonging to $F$. It is then clear (identically
to Lemma 5.5) that the set

$Z_{F}^{(q,r)}=\\{(\operatorname{diag}(m_{a^{q+1}},m_{a^{r+1}}),\operatorname{diag}(m_{a},m_{a}))\colon
a\in\mathbb{M}^{\times}\\}\cong Z^{(q,r)}$

is a subgroup of $\operatorname{Aut}(F)$. The same group theoretic machinery
can then be applied, with $Z_{F}^{(q,r)}$ and $\operatorname{Aut}(F)$ taking
the role of $Z^{(q,r)}$ and $\operatorname{Aut}(\mathbb{P})$ in the approach
we presented, proving an analogue of Theorem 5.10, with the conclusion that
all subfunctions of $N$ and $L$ are zeros or monomials.

Then, the planar mappings from Family $\mathcal{S}$ can be tested for
equivalence similar to Theorem 6.2 by comparing the coefficients of the
polynomial equation $NF=FL$. One obtains the same set of equations as Eqs.
(6.22)-(6.37), just with the simplification that $M=L$. Then the same
argumentation of the proof of Theorem 6.2 can be followed, with the result
that the only possible equivalences that need to be considered are
equivalences via $N,L$ where the subfunctions are

$N_{1}=a_{1}x^{p^{t}},\quad N_{2}=N_{3}=0,\quad N_{4}=d_{1}x^{p^{t}}$

and either

$L_{1}=a_{2}x^{p^{t}},\quad L_{2}=L_{3}=0,\quad L_{4}=d_{2}x^{p^{t}}$

or

$L_{1}=0,\quad L_{2}=b_{2}x^{p^{t}},\quad L_{3}=c_{2}x^{p^{t}},\quad L_{4}=0.$

The conditions on the coefficients are then identical to the ones in the proof
of Theorem 6.2 (e.g. Eq. (6.39) and (6.40)), just with
$\omega_{1}=\omega_{2}=1$. This way, one obtains the same result as Theorem
6.2, except that one only gets information on _strong_ isotopy and not regular
isotopy.

Since an isotopy class of a commutative semifields contains at most $2$
strong-isotopy classes [8, Theorem 2.6.], this approach would suffice to prove
the exponential count. With some more effort, the planar mapping approach can
also be used to find all isotopisms between commutative semifields (not just
strong isotopisms). Indeed, by [8, Theorem 2.6.], if two commutative
semifields with corresponding planar mapping $F:x\mapsto x\ast_{1}x$ and
$G:x\mapsto x\ast_{2}x$ are isotopic, then either $F$ and $G$ are equivalent
(the semifields are then strongly isotopic) or $F$ is equivalent to
$G^{\prime}:x\mapsto x\ast_{2}(a\ast_{2}x)$ where $a$ is an arbitrary non-
square element in the middle nucleus of the semifield. So in order to settle
the isotopy question one could check equivalence between $F$ and both $G$ and
$G^{\prime}$, yielding an alternative proof of Theorem 5.10. The two
approaches are essentially equivalent and require similar effort.

The isotopy approach we chose has the advantage that it can be extended
naturally to non-commutative semifields where the connection to planar
mappings does not exist.

The number of distinct isotopy classes can now be counted.

######  Corollary 6.4.

Let $N_{\mathcal{S}}(p,n)$ be the number of non-isotopic pre-semifields in
Family $\mathcal{S}$ on $\mathbb{F}_{p}^{n}$. Then

$\frac{\sigma(n)-1}{2}\cdot\frac{p^{n/4}-1}{n}\leq
N_{\mathcal{S}}(p,n)\leq\frac{\sigma(n)-1}{2}\left(p^{n/4}-1\right).$

###### Proof.

This follows directly from Theorem 6.2 (i),(ii),(iv) and (v): There are
$\sigma(n)-1$ admissible values for $q$, and only $q,{\overline{q}}$ yield
isotopic pre-semifields. Then there are $p^{n/4}-1$ admissible values for $a$,
with at most $n$ of them yielding isotopic pre-semifields. ∎

In particular, $\mathcal{S}$ is the first known family of commutative
(pre-)semifields that yields exponentially many non-isotopic (pre-)semifields.
Since non-isotopic pre-semifields lead to inequivalent planar mappings (see
Theorem 2.1), this also shows that the number of inequivalent planar mappings
grows exponentially in $n$.

######  Corollary 6.5.

The number of non-isotopic commutative semifields of order $p^{n}$ and the
number of inequivalent planar DO mappings of $\mathbb{F}_{p^{n}}$ are
exponential in $n$ for a fixed odd prime $p$ and $n$ divisible by $4$.

##  7\. The nuclei

In this section we will compute the nuclear parameters of Family
$\mathcal{S}$. As explained in Section 2, the nuclei are defined for
semifields and not for pre-semifields. However, the nuclei of the isotopic
semifield can be computed using the following theorem of Marino and Polverino
[25, Theorem 2.2] (we give the commutative version of their general theorem)
that allows computing the nuclei directly from the pre-semifield.

Let $\mathbb{P}=(\mathbb{F}_{p}^{n},+,\ast)$ be a commutative pre-semifield
with right multiplication defined as

$R_{U}:X\mapsto X\ast U,\quad\textrm{for }U\in\mathbb{F}_{p}^{n}.$

Then the spread set associated to $\mathbb{P}$ is defined as

${\mathcal{L}}=\\{R_{U}:U\in\mathbb{F}_{p}^{n}\\}.$

In the following $\mathbb{N}_{j}(\mathbb{P})$ denotes the corresponding
nucleus of the semifield isotopic to $\mathbb{P}$, for $j\in\\{l,m,r\\}$.

######  Theorem 7.1.

[25, Theorem 2.2] Let
${\mathcal{N}}_{0},{\mathcal{N}}_{1}\subset\operatorname{End}(\mathbb{F}_{p}^{n})$
be the largest sets (and then necessarily fields) such that

${\mathcal{L}}{\mathcal{N}}_{0}\subseteq{\mathcal{L}}\textrm{ and
}{\mathcal{N}}_{1}{\mathcal{L}}\subseteq{\mathcal{L}}.$

Then

$\mathbb{N}_{m}(\mathbb{P})\cong{\mathcal{N}}_{0}\textrm{ and
}\mathbb{N}_{l}(\mathbb{P})=\mathbb{N}_{r}(\mathbb{P})\cong{\mathcal{N}}_{1}.$

Now, let $\mathbb{P}=(\mathbb{M}\times\mathbb{M},+,\ast)$ be a pre-semifield
in Family $\mathcal{S}$. Then
${\mathcal{L}}=\\{R_{u,v}:(u,v)\in\mathbb{M}\times\mathbb{M}\\}$, where

$R_{u,v}:(x,y)\mapsto(R^{(1)}_{u,v}(x,y),R^{(2)}_{u,v}(x,y)),$

with

$R^{(1)}_{u,v}(x,y)=x^{q}u+xu^{q}+B(y^{q}v+yv^{q})\textrm{ and
}R^{(2)}_{u,v}(x,y)=x^{r}v+Axv^{r}+Ay^{r}u+yu^{r}.$

We write again $L\in\operatorname{End}(\mathbb{M}\times\mathbb{M})$ as
$L:(x,y)\mapsto(\alpha(x)+\beta(y),\gamma(x)+\delta(y))$, where
$\alpha,\beta,\gamma,\delta\in\operatorname{End}(\mathbb{M})$.

######  Theorem 7.2.

The left, middle and right nuclei
$\mathbb{N}_{l}(\mathbb{P}),\mathbb{N}_{m}(\mathbb{P}),\mathbb{N}_{r}(\mathbb{P})$
satisfy $\mathbb{N}_{l}(\mathbb{P})=\mathbb{N}_{r}(\mathbb{P})\cong\mathbb{D}$
and $\mathbb{N}_{m}(\mathbb{P})\cong\mathbb{E}$.

###### Proof.

We have nonzero $L\in{\mathcal{N}}_{1}$, if and only if, for every
$(u,v)\in\mathbb{M}\times\mathbb{M}$ there exists
$(w,t)\in\mathbb{M}\times\mathbb{M}$ such that

$\displaystyle\alpha R^{(1)}_{u,v}+\beta R^{(2)}_{u,v}$
$\displaystyle=R^{(1)}_{w,t},\textrm{ and},$ $\displaystyle\gamma
R^{(1)}_{u,v}+\delta R^{(2)}_{u,v}$ $\displaystyle=R^{(2)}_{w,t},$

that is

$\displaystyle\alpha(x^{q}u+xu^{q}+B(y^{q}v+yv^{q}))+\beta(x^{r}v+Axv^{r}+Ay^{r}u+yu^{r})$
$\displaystyle=x^{q}w+xw^{q}+B(y^{q}t+yt^{q}),\textrm{ and},$
$\displaystyle\gamma(x^{q}u+xu^{q}+B(y^{q}v+yv^{q}))+\delta(x^{r}v+Axv^{r}+Ay^{r}u+yu^{r})$
$\displaystyle=x^{r}t+Axt^{r}+Ay^{r}w+yw^{r}.$

This implies (after a routine comparison of degrees of $x,y,u,v$ as in
previous sections) that $\beta=\gamma=0$ and $\alpha(x)=z_{1}x$ and
$\delta(x)=z_{4}x$ for some $z_{1},z_{4}\in\mathbb{M}^{\times}$. Now, the
above equations become

$\displaystyle z_{1}(x^{q}u+xu^{q}+B(y^{q}v+yv^{q}))$
$\displaystyle=x^{q}w+xw^{q}+B(y^{q}t+yt^{q}),\textrm{ and},$ $\displaystyle
z_{4}(x^{r}v+Axv^{r}+Ay^{r}u+yu^{r})$
$\displaystyle=x^{r}t+Axt^{r}+Ay^{r}w+yw^{r},$

or (the case $uv=0$ is easy to see)

$\displaystyle z_{1}$ $\displaystyle=w/u=(w/u)^{q}=t/v=(t/v)^{q},\textrm{
and},$ $\displaystyle z_{4}$ $\displaystyle=w/u=(w/u)^{r}=t/v=(t/v)^{r}.$

Thus for every $(u,v)\in\mathbb{M}\times\mathbb{M}$ there exists
$(w,t)\in\mathbb{M}\times\mathbb{M}$ if and only if
$z_{4}=z_{1}=z_{1}^{q}=z_{1}^{r}$ if and only if
$z_{1}\in\mathbb{F}_{q}\cap\mathbb{F}_{r}\cap\mathbb{M}=\mathbb{D}$. That is
to say $L\in{\mathcal{N}}_{1}$ if and only if $L(x,y)=(zx,zy)$ for
$z\in\mathbb{D}$. Now Theorem 7.1 implies
$\mathbb{N}_{l}(\mathbb{P})=\mathbb{N}_{r}(\mathbb{P})\cong\mathbb{D}$.

Similarly for the middle nucleus, nonzero $L\in{\mathcal{N}}_{0}$, if and only
if,

$\displaystyle R^{(1)}_{u,v}(\alpha(x)+\beta(y),\gamma(x)+\delta(y))$
$\displaystyle=R^{(1)}_{w,t}(x,y),\textrm{ and},$ $\displaystyle
R^{(2)}_{u,v}(\alpha(x)+\beta(y),\gamma(x)+\delta(y))$
$\displaystyle=R^{(2)}_{w,t}(x,y),$

that is

$\displaystyle(\alpha(x)+\beta(y))^{q}u+(\alpha(x)+\beta(y))u^{q}+B(\gamma(x)+\delta(y))^{q}v+B(\gamma(x)+\delta(y))v^{q}$
$\displaystyle=x^{q}w+xw^{q}+B(y^{q}t+yt^{q}),\textrm{ and},$
$\displaystyle(\alpha(x)+\beta(y))^{r}v+A(\alpha(x)+\beta(y))v^{r}+A(\gamma(x)+\delta(y))^{r}u+(\gamma(x)+\delta(y))u^{r}$
$\displaystyle=x^{r}t+Axt^{r}+Ay^{r}w+yw^{r}.$

This implies (after a routine comparison of degrees of $x,y,u,v$) that
$\alpha(x)=z_{1}x,\beta(y)=z_{2}y,\gamma(x)=z_{3}x$ and $\delta(y)=z_{4}y$ for
$z_{1},z_{2},z_{3},z_{4}\in\mathbb{M}$. Now, the $x$-part of the first of the
above equation implies

$z_{1}^{q}x^{q}u+z_{1}xu^{q}+B(z_{3}^{q}x^{q}v+z_{3}xv^{q})=x^{q}w+xw^{q},$

in other words,

$z_{1}^{q}u+Bz_{3}^{q}v=w\textrm{ and }z_{1}u^{q}+Bz_{3}v^{q}=w^{q},$

which implies

$(z_{1}^{q^{2}}-z_{1})u^{q}+(B^{q}z_{3}^{q^{2}}-Bz_{3})v^{q}=0,$

for all $u,v\in\mathbb{M}$. That is to say
$z_{1}\in\mathbb{F}_{q^{2}}\cap\mathbb{M}=\mathbb{E}$. The $x$-part of the
second equation yields (after simple calculations)

$A^{r}z_{3}^{r^{2}}-\frac{z_{3}}{A}=0.$

That is to say, if $z_{3}\neq 0$ then,

$z_{3}^{q^{2}-1}=\frac{1}{B^{q-1}}\textrm{ and
}z_{3}^{r^{2}-1}=\frac{1}{A^{r+1}}.$

By definition of $\mathcal{S}$, $B$ is a non-square in
$\mathbb{M}=\mathbb{F}_{Q^{2}}$ and $A=a/B$ where
$a\in\mathbb{F}_{Q}^{\times}$. Recalling that $q^{2}\equiv r^{2}\pmod{Q^{2}}$,
we reach

$B^{q(Q+1)}=B^{q+r}=a^{r+1}.$

Note that since $B$ is a non-square in $\mathbb{F}_{Q^{2}}$ we have
$B^{(Q+1)(Q-1)/2}=-1$ and $B^{Q+1}$ is a non-square in $\mathbb{F}_{Q}$. But
$a^{r+1}$ is a square in $\mathbb{F}_{Q}$ and we get $z_{3}=0$. By the
$y$-parts of the equations we similarly reach $z_{2}=0$ and
$z_{4}\in\mathbb{E}$. Thus,

$\displaystyle z_{1}^{q}x^{q}u+z_{1}xu^{q}+B(z_{4}^{q}y^{q}v+z_{4}yv^{q}))$
$\displaystyle=x^{q}w+xw^{q}+B(y^{q}t+yt^{q}),\textrm{ and},$ $\displaystyle
z_{1}^{r}x^{r}v+Az_{1}xv^{r}+Az_{4}^{r}y^{r}u+z_{4}yu^{r})$
$\displaystyle=x^{r}t+Axt^{r}+Ay^{r}w+yw^{r},$

implying (the case $uv=0$ is easy to see)

$\displaystyle z_{1}^{q}=w/u$ $\displaystyle\textrm{ and }z_{1}=(w/u)^{q},$
$\displaystyle z_{4}^{q}=t/v$ $\displaystyle\textrm{ and }z_{4}=(t/v)^{q},$
$\displaystyle z_{1}^{r}=t/v$ $\displaystyle\textrm{ and }z_{1}=(t/v)^{r},$
$\displaystyle z_{4}^{r}=w/u$ $\displaystyle\textrm{ and }z_{4}=(w/u)^{r}.$

Thus for every $(u,v)\in\mathbb{M}\times\mathbb{M}$ there exists
$(w,t)\in\mathbb{M}\times\mathbb{M}$ if and only if
$z_{4}^{q}=z_{1}=z_{1}^{q^{2}}=z_{1}^{r^{2}}$ if and only if
$z_{1}\in\mathbb{F}_{q^{2}}\cap\mathbb{M}=\mathbb{E}$. That is to say
$L\in{\mathcal{N}}_{0}$ if and only if $L(x,y)=(z^{q}x,zy)$ for
$z\in\mathbb{E}$. Now Theorem 7.1 implies
$\mathbb{N}_{m}(\mathbb{P})\cong\mathbb{E}$. ∎

##  8\. Comparison to other commutative semifields and concluding remarks

Table 2 lists known commutative semifields that are not biprojective. We
should say here that these commutative semifields are not obviously
represented as biprojective semifields. When the order is square, there might
be isotopic semifields that can be biprojective, but we are not aware of such
isotopisms.

Family | Planar Mapping | $\\#\mathbb{S}$ | Notes | $(\\#\mathbb{N}_{l},\\#\mathbb{N}_{m})$ | Count | Proved in
---|---|---|---|---|---|---
$\mathcal{ZKW}$ | $X^{q+1}-a^{Q-1}X^{qQ+Q^{2}}$ | $p^{3s}$ |  | $Q=p^{s},\quad q=p^{t},$
---
$d={\mathsf{gcd}({s},{t})},\quad s^{\prime}=s/d,\quad t^{\prime}=t/d,$
$s^{\prime}$ odd, $\quad s^{\prime}+t^{\prime}\equiv 0\pmod{3}$,
$\langle a\rangle=\mathbb{F}_{p^{3s}}^{\times}$.
$(p^{d},p^{d})$ [25] | $\geq 1$ | [33]
$\mathcal{B}_{3}$ | $X^{q+1}-a^{Q-1}X^{qQ+Q^{2}}$ | $p^{3s}$ |  | $Q=p^{s},q=p^{t},$
---
$d={\mathsf{gcd}({s},{t})},\quad s/d$ odd,
$q\equiv Q\equiv 1\pmod{3}$,
$\langle a\rangle=\mathbb{F}_{p^{3s}}^{\times}$.
$(p^{d},p^{d})$ [25] | $\leq 9\sigma(s)$ | [3]
$\mathcal{B}_{4}$ | $X^{q+1}-a^{Q-1}X^{qQ+Q^{3}}$ | $p^{4s}$ |  | $Q=p^{s},q=p^{t},$
---
$d={\mathsf{gcd}({2s},{t})},\quad 2s/d$ odd,
$q\equiv Q\equiv 1\pmod{4}$,
$\langle a\rangle=\mathbb{F}_{p^{4s}}^{\times}$.
$(p^{d/2},p^{d})$ [25] | $\leq 8\sigma(s)$ | [3]
$\mathcal{CG}$ | $(x^{2}+y^{10},xy-y^{6})$ | $3^{2m}$ |  $m\geq 3$ odd | $(3,3^{m})$ | $1$ | [7]
$\mathcal{G}$ | $(x^{2}+ay^{2}+a^{3}y^{18},xy-ay^{6})$ | $3^{2m}$ |  $m\geq 3$, $a\in\mathbb{F}_{3^{m}}^{\times}\setminus(\mathbb{F}_{3^{m}}^{\times})^{2}$ | $(3,3)$ | $1$ | [15]
$\mathcal{CM}/\mathcal{DY}$ | $X^{10}\pm X^{6}-X$ | $3^{m}$ |  $m\geq 5$ odd | $(3,3)$ | $2$ | [9, 13]
Table 2. Known infinite families of (non-biprojective) commutative semifields
of order $p^{n}$

We now consider isotopisms between the new Family $\mathcal{S}$ and other
commutative pre-semifields.

######  Theorem 8.1.

Let $\mathbb{P}_{q,B,a}=(\mathbb{M}\times\mathbb{M},+,\ast)$ be a pre-
semifield in the Family $\mathcal{S}$. $\mathbb{P}_{q,B,a}$ is not isotopic to
any other known commutative semifield, except possibly semifields from Family
$\mathcal{B}_{4}$. Family $\mathcal{S}$ yields new examples of commutative
semifields.

###### Proof.

The non-isotopy with the biprojective pre-semifields follows directly from
Theorem 5.10, except for possible isotopisms between the families
$\mathcal{S}$ and $\mathcal{ZP}$ when the coefficients $q,r$ coincide. We
exclude this case by again applying Theorem 5.10: Consider the Zhou-Pott pre-
semifield $\mathbb{P}_{\alpha}=(\mathbb{M}\times\mathbb{M},+,\star)$ with
multiplication

$(x,y)\star(u,v)=(x^{q}u+u^{q}x+\alpha(y^{q}v+yv^{q}),x^{qQ}v+yu^{qQ})$

for some (arbitrary) non-square $\alpha$. Note that it is not possible to use
the parameter $qQ$ in the first component and $q$ in the second component
since $\gcd(k+m/2,m)=\gcd(k,m)/2$ by Lemma 4.3 (iii), contradicting the
necessary conditions of a Zhou-Pott pre-semifield. If $\mathbb{P}_{\alpha}$ is
isotopic to $\mathbb{P}_{q,B,a}$, then (using Theorem 5.10), there is an
isotopism $(N,L,M)$, where

$\displaystyle N_{1}=a_{1}x^{p^{t}},\quad N_{4}=d_{1}x^{p^{t}},\quad
N_{2}=N_{3}=0,$ $\displaystyle L_{1}=a_{2}^{p^{t}}x^{p^{t}},\quad
L_{2}=b_{2}^{p^{t}}x^{p^{t}},\quad L_{3}=c_{2}^{p^{t}}x^{p^{t}},\quad
L_{4}=d_{2}^{p^{t}}x^{p^{t}},$ $\displaystyle
M_{1}=a_{3}^{p^{t}}x^{p^{t}},\quad M_{2}=b_{3}^{p^{t}}x^{p^{t}},\quad
M_{3}=c_{3}^{p^{t}}x^{p^{t}},\quad M_{4}=d_{3}^{p^{t}}x^{p^{t}},$

where $a_{1},d_{1}\neq 0$. Then (only considering the second components), we
have

$L((x,y))\star M((u,v))=(\rule{10.00002pt}{0.5pt},\
(a_{2}x+b_{2}y)^{qQ+p^{t}}(c_{3}u+d_{3}v)^{p^{t}}+(c_{2}x+d_{2}y)^{p^{t}}(a_{3}u+b_{3}v)^{qQ+p^{t}})$

and

$N((x,y)\ast(u,v))=(\rule{10.00002pt}{0.5pt},\
d_{1}(x^{qQ}v+yu^{qQ}+(a/B)(xv^{qQ}+y^{qQ}u))^{p^{t}}).$

Comparing the coefficients of $(x^{qQ}v)^{p^{t}}$, $(xv^{qQ})^{p^{t}}$,
$(x^{qQ}u)^{p^{t}}$ and $(xu^{qQ})^{p^{t}}$ yields the following four
equations:

$\displaystyle\left(a_{2}^{qQ}d_{3}\right)^{p^{t}}$ $\displaystyle=d_{1}$
$\displaystyle\left(c_{2}b_{3}^{qQ}\right)^{p^{t}}$
$\displaystyle=d_{1}(a/B)^{p^{t}}$ $\displaystyle a_{2}^{qQ}c_{3}$
$\displaystyle=0$ $\displaystyle c_{2}a_{3}^{qQ}$ $\displaystyle=0.$

The bijectivity of $L$ and $M$ induces the conditions $(a_{2},c_{2})\neq(0,0)$
and $(a_{3},c_{3})\neq(0,0)$. Thus, the last two equations only allow
$a_{2}=a_{3}=0$ or $c_{2}=c_{3}=0$. Both cases contradict the first two
equations. We conclude that $\mathbb{P}_{q,B,a}$ is not isotopic to a Zhou-
Pott pre-semifield.

The pre-semifields from $\mathcal{S}$ are also not isotopic to the ones from
$\mathcal{CG}$, $\mathcal{G}$, $\mathcal{CM}/\mathcal{DY}$, $\mathcal{ZKW}$,
$\mathcal{B}_{3}$ by considering the order of the semifields and their nuclei
(see Table 2). Furthermore, the Family $\mathcal{S}$ is not contained in
$\mathcal{B}_{4}$ since we can choose $p,m,q$ in a way that the conditions for
$\mathcal{B}_{4}$ in Table 2 are violated. ∎

Although the parameters $p,m,q$ for the pre-semifields from Family
$\mathcal{S}$ are more general than that of Family $\mathcal{B}_{4}$, for
suitable choices of $p,m,q$ the parameters may coincide. The next proposition
shows that even in that case Family $\mathcal{S}$ contains new semifields
thanks to its exponential count. More precisely, we show that the number of
non-isotopic semifields from Families $\mathcal{B}_{3}$ and $\mathcal{B}_{4}$
of order $p^{3s}$ and $p^{4s}$, respectively, is linear in $s$.

######  Proposition 8.2.

The number of non-isotopic pre-semifields in Family $\mathcal{B}_{3}$ (and
$\mathcal{B}_{4}$ resp.) of order $p^{3s}$ (and $p^{4s}$ resp.) is at most
$9\sigma(s)$ (and $8\sigma(s)$ resp.).

###### Proof.

The $\mathcal{B}_{4}$ planar mappings are of the form

$f(X)=X^{q+1}-a^{Q-1}X^{qQ+Q^{3}},$

where $a$ generates $\mathbb{F}_{p^{4s}}^{\times}$. We count the number of
different $a$’s which give inequivalent planar mappings. Consider the change
of variable $X\mapsto BX$, and rescaling of $f$ to get

$X^{q+1}-B^{Q^{3}+qQ-q-1}a^{Q-1}X^{qQ+Q^{3}}.$

Note that $Q^{3}+qQ-q-1=(Q-1)(Q^{2}+Q+q+1)$. We have by [3, Lemma 6],

${\mathsf{gcd}({Q^{2}+Q+q+1},{Q^{3}+Q^{2}+Q+1})}={\mathsf{gcd}({Q^{3}-q},{Q^{3}+Q^{2}+Q+1})}=4,$

when the semifield conditions on $q,Q$ appearing on Table 2 is satisfied. Thus
the number of inequivalent planar mappings in the Family $\mathcal{B}_{4}$ for
a given $q$ is at most $4$. This means that (using Theorem 2.1) that for a
given $q$, the number of pre-semifields, that are not strongly isotopic, is
also at most $4$. Any isotopy class of a commutative semifield contains at
most $2$ strong-isotopy classes ([8, Theorem 2.6.]), so for a given $q$ there
are at most $8$ non-isotopic pre-semifields. Thus the total number of non
isotopic pre-semifields in the Family $\mathcal{B}_{4}$ of order $p^{4s}$ is
bounded by $8\sigma(s)$. The $\mathcal{B}_{3}$ case is essentially the same
using [3, Lemma 5]. In this case (again with [8, Theorem 2.6.]) strong isotopy
and isotopy coincide. ∎

For the Family $\mathcal{ZKW}$ we are not aware of any result on the exact
value or a bound on the number of non-isotopic pre-semifields.

######  Remark 8.3.

We remark that we could also allow $q=1$ in $\mathcal{S}$. However, in that
case the resulting pre-semifields are strongly isotopic to Dickson semifields.
Indeed, consider the planar mapping $F=[(1,0,0,B)_{1},(0,1,A,0)_{Q}]$ with
$A=a/B$ where $B$ is a non-square and $a\in\mathbb{L}^{\times}$. Note that
$A\notin(\mathbb{M}^{\times})^{Q-1}$ since it is a non-square. Define $N$ via
its subfunctions $N_{1}=x$, $N_{2}=N_{3}=0$,
$N_{4}=d_{1}x+d_{1}^{\prime}x^{Q}$ with $d_{1}=1/(1-A^{Q+1})$ and
$d_{1}^{\prime}=-A/(1-A^{Q+1})$. Note that $x\mapsto\alpha x-\beta x^{Q}$ is
bijective if and only if $\alpha/\beta\not\in(\mathbb{M}^{\times})^{Q-1}$.
Therefore, $N_{4}$ is bijective since
$((A^{Q+1}-1)/(A(1-A^{Q+1}))^{Q+1}=(-1/A)^{Q+1}\neq 1$ since
$A\notin(\mathbb{M}^{\times})^{Q-1}$. We conclude that $N$ is bijective. The
subfubction $N_{4}$ is chosen such that $d_{1}+A^{Q}d_{1}^{\prime}=1$ and
$Ad_{1}+d_{1}^{\prime}=0$. Then

$\displaystyle NF$
$\displaystyle=[(1,0,0,B)_{1},d_{1}(0,1,A,0)_{Q}+d_{1}^{\prime}(0,A^{Q},1,0)_{Q}]$
$\displaystyle=[(1,0,0,B)_{1},(0,d_{1}+A^{Q}d_{1}^{\prime},d_{1}A+d_{1}^{\prime},0)_{Q}]=[(1,0,0,B)_{1},(0,1,0,0)_{Q}],$

so $F$ is equivalent to a planar mapping belonging to a Dickson pre-semifield
and the corresponding semifields are strongly isotopic by Theorem 2.1. It
makes thus sense to exclude the case $q=1$ so that the different families do
not intersect (as proven in Theorem 8.1). Note that the same choice of $N$
also yields equivalence between the Budaghyan-Helleseth planar mapping and the
planar mappings associated with Dickson semifields for the parameter $q=Q$.

######  Remark 8.4.

Recall that Kantor [19] gave a family that contains an exponential number of
non-isotopic commutative semifields in characteristic two using a construction
of Kantor and Williams [20]. We remark that Family $\mathcal{S}$ (and in
general, a planar mapping) does not exist in characteristic two. However, a
conceptual analogue of planar functions in characteristic two is possible.
These are the so-called almost perfect nonlinear (APN) functions (whose
polarizations are $2$-to-$1$) that parallel planar mappings (whose
polarizations are $1$-to-$1$) without the connection to semifields. In a
follow-up work to this one, we give an analogous method for determining
equivalence of biprojective APN functions and an analogous family that
contains an exponential number of inequivalent APN functions in [16]. The
first result to show that an APN family contains an exponential number of
inequivalent functions was given recently by Kaspers and Zhou [21] using a
different method.

##  9\. Acknowledgments

The authors would like to thank an anonymous reviewer for informing them of
Theorem 7.1 which leads to a proof of the nuclei that is simpler and more
coherent with the paper than our original proof and for several other comments
that improved the presentation.

The first author was supported by the GAČR Grant 18-19087S - 301-13/201843.
The second author was supported by the National Science Foundation under grant
No. 2127742.

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