Finding normal bases over finite fields with prescribed trace self-orthogonal relations

TL;DR

This paper characterizes the existence of normal elements in finite fields with prescribed trace relations, providing conditions, algorithms, and existence results for special cases, especially over fields of characteristic two.

Contribution

It establishes necessary and sufficient conditions for constructing normal elements with prescribed trace vectors over finite fields, including algorithms and existence proofs for specific cases.

Findings

Normal elements exist with prescribed symmetric trace vectors under certain conditions.

An algorithm is provided for constructing such normal elements in specific cases.

Existence of normal elements with minimal Hamming weight (3) for fields with 4 dividing n.

Abstract

Normal bases and self-dual normal bases over finite fields have been found to be very useful in many fast arithmetic computations. It is well-known that there exists a self-dual normal basis of $\mathbb{F}_{2^n}$ over $\mathbb{F}_2$ if and only if $4\nmid n$. In this paper, we prove there exists a normal element $\alpha$ of $\mathbb{F}_{2^n}$ over $\mathbb{F}_{2}$ corresponding to a prescribed vector $a=(a_0,a_1,...,a_{n-1})\in \mathbb{F}_2^n$ such that $a_i={Tr}_{2^n|2}(\alpha^{1+2^i})$ for $0\leq i\leq n-1$, where $n$ is a 2-power or odd, if and only if the given vector $a$ is symmetric ($a_i=a_{n-i}$ for all $i, 1\leq i\leq n-1$), and one of the following is true. 1) $n=2^s\geq 4$, $a_0=1$, $a_{n/2}=0$, $\sum\limits_{1\leq i\leq n/2-1, (i,2)=1}a_i=1$; 2) $n$ is odd, $(\sum\limits_{0\leq i\leq n-1}a_ix^i,x^n-1)=1$. Furthermore we give an algorithm to obtain normal elements corresponding to prescribed vectors in the above two cases. For a general positive integer $n$ with $4|n$, some necessary conditions for a vector to be the corresponding vector of a normal element of $\mathbb{F}_{2^n}$ over $\mathbb{F}_{2}$ are given. And for all $n$ with $4|n$, we prove that there exists a normal element of $\mathbb{F}_{2^n}$ over $\mathbb{F}_2$ such that the Hamming weight of its corresponding vector is 3, which is the lowest possible Hamming weight.