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arXiv 1303.2283math.NT
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求具有给定迹self-orthogonal关系的有限域上的正规基

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

Xiyong Zhang, Rongquan Feng, Qunying Liao, Xuhong Gao

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摘要：

有限域上的正规基和自对偶正规基在许多快速算术计算中是非常有用的。众所周知，存在$\mathbb{F}_2$上的$\mathbb{F}_{2^n}$的自对偶正规基当且仅当$4\nmid n$ .本文证明了$\mathbb{F}_{2^n}$在$\mathbb{F}_{2}$上存在一个正规元$α$，它对应于一个给定的向量$a=(a_0,a_1,...,a_{n-1})\in \mathbb{F}_2^n$，使得$a_i={Tr}_{2^n|2}(α^{1+2^i})$ for $0\leq i\leq n-1$，其中$n$是2次幂或奇数，当且仅当给定的向量$a$是对称的（$a_i=a_{n-i}$ for所有$i, 1\leq i\leq n-1$），并且下列之一成立. 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）$n$是奇数，$(\sum\limits_{0\leq i\leq n-1}a_ix^i,x^n-1)=1$。此外，我们还给出了在上述两种情况下获得与指定向量对应的法向元素的算法。对于一般正整数$n$，给出了$4|n$为$\mathbb{F}_{2}$的正整数$n$的正规元的对应向量为$\mathbb{F}_{2^n}$的正规元的必要条件.对于所有具有$4|n$的$n$，证明了存在$\mathbb{F}_2$上的$\mathbb{F}_{2^n}$的正规元，使得其对应向量的汉明重量为3，这是可能的最低汉明重量.

英文摘要：

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 $α$ 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 $a_i={Tr}_{2^n|2}(α^{1+2^i})$ 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.

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