Commutative Presemifields and Semifields

Introduction

A presemifields is a ring with left and right distributivity and with no zero divisor. A presemifield with a multiplicative identity is called a semifield. Any finite presemifield can be represented by $\mathbb {S} =(\mathbb {F} _{p^{n}},+,\star )$, for $p$ a prime, $n$ a positive integer, $\mathbb {S} =(\mathbb {F} _{p^{n}},+)$ additive group and $x\star y$ multiplication linear in each variable.

Two presemifields $\mathbb {S} _{1}=(\mathbb {F} _{p^{n}},+,\star )$ and $\mathbb {S} _{2}=(\mathbb {F} _{p^{n}},+,\circ )$ are called isotopic if there exist three linear permutation $T,M,N$ of $\mathbb {F} _{p^{n}}$ such that $T(x\star y)=M(x)\circ N(y)$, for any $x,y\in \mathbb {F} _{p^{n}}$. If $M=N$ then they are called strongly isotopic. Each commutative presemifields of odd order defines a planar DO polynomial and viceversa:

• given $\mathbb {S} =(\mathbb {F} _{p^{n}},+,\star )$ let $F_{\mathcal {S}}(x)={\frac {1}{2}}(x\star x)$;
• given $F$ let $\mathbb {S} _{F}=(\mathbb {F} _{p^{n}},+,\star )$ defined by $x\star y=F(x+y)-F(x)-F(y)$.

Hence two quadratic planar functions $F,F'$ are isotopic equivalent if their corresponding presemifields are isotopic. Moreover, we have:

• $F,F'$ are CCZ-equivalent if and only if $\mathbb {S} _{F},\mathbb {S} _{F'}$ are strongly isotopic;
• for $n$ odd, isotopic coincides with strongly isotopic;
• if $F,F'$ are isotopic equivalent, then there exists a linear map $L$ such that $F'$ is EA-equivalent to $F(x+L(x))-F(x)-F(L(x))$.