Commutative Presemifields and Semifields: Difference between revisions
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=Introduction= | =Introduction= | ||
A <span class="definition">semifields</span> | A <span class="definition">semifields</span> is a ring with left and right distributivity and with no zero divisor. | ||
A presemifield with a multiplicative identity is called a <span class="definition">semifield</span>. | A presemifield with a multiplicative identity is called a <span class="definition">semifield</span>. | ||
Any finite presemifield can be represented by <math>\mathbb{S}=(\mathbb{F}_{p^n},+,\star)</math>, | Any finite presemifield can be represented by <math>\mathbb{S}=(\mathbb{F}_{p^n},+,\star)</math>, | ||
Revision as of 10:26, 29 August 2019
Introduction
A semifields 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 <math>\mathbb{S}=(\mathbb{F}_{p^n},+,\star)</math>, for <math>p</math> a prime, <math>n</math> a positive integer, <math>\mathbb{S}=(\mathbb{F}_{p^n},+)</math> additive group and <math>x\star y</math> multiplication linear in each variable.
Two presemifields <math>\mathbb{S}_1=(\mathbb{F}_{p^n},+,\star)</math> and <math>\mathbb{S}_2=(\mathbb{F}_{p^n},+,\circ)</math> are called isotopic if there exist three linear permutation <math>T,M,N</math> of <math>\mathbb{F}_{p^n}</math> such that <math>T(x\star y)=M(x)\circ N(y)</math>, for any <math>x,y\in\mathbb{F}_{p^n}</math>. If <math>M=N</math> then they are called strongly isotopic. Each commutative presemifields of odd order defines a planar DO polynomial and viceversa:
- given <math>\mathbb{S}=(\mathbb{F}_{p^n},+,\star)</math> let <math>F_\mathcal{S}(x)=\frac{1}{2}(x\star x)</math>;
- given <math>F</math> let <math>\mathbb{S}_F=(\mathbb{F}_{p^n},+,\star)</math> defined by <math>x\star y=F(x+y)-F(x)-F(y)</math>.
Hence two quadratic planar functions <math>F,F'</math> are isotopic equivalent if their corresponding presemifields are isotopic. Moreover, we have:
- <math>F,F'</math> are CCZ-equivalent if and only if <math>\mathbb{S}_F,\mathbb{S}_{F'}</math> are strongly isotopic;
- for <math>n</math> odd, isotopic coincides with strongly isotopic;
- if <math>F,F'</math> are isotopic equivalent, then there exists a linear map <math>L</math> such that <math>F'</math> is EA-equivalent to <math>F(x+L(x))-F(x)-F(L(x))</math>.