A note on primitive $1-$normal elements over finite fields

TL;DR

This paper extends the existence results of primitive 1-normal elements over finite fields to include the previously excluded case where q=2, broadening the understanding of such elements.

Contribution

It generalizes the existence theorem for primitive 1-normal elements to the case q=2, which was not covered in earlier work.

Findings

Existence of primitive 1-normal elements for q=2 established.

Extends previous results to include the case q=2.

Provides a more complete understanding of primitive 1-normal elements.

Abstract

Let $q$ be a prime power of a prime $p$, $n$ a positive integer and $\mathbb F_{q^n}$ the finite field with $q^n$ elements. The $k-$normal elements over finite fields were introduced and characterized by Huczynska et al (2013). Under the condition that $n$ is not divisible by $p$, they obtained an existence result on primitive $1-$normal elements of $\mathbb F_{q^n}$ over $\mathbb F_q$ for $q>2$. In this note, we extend their result to the excluded case $q=2$.

Equations

$$g_{\alpha }(x):=\sum _{i=0}^{n-1}{\alpha }^{q^i}{x}^{n-1-i}\text{and}{x}^n-1,$$

$${\Phi }_q(f)=\left|{\left(\frac{{\mathbb{F}}_q[x]}{⟨f⟩}\right)}^{\ast }\right|,$$

$$\sum _{\genfrac{}{}{0ex}{}{h|x^n-1}{\mathrm{deg}(h)=n-k}}{\Phi }_q(h),$$

$$W(T)\cdot W(q^n-1)<q^{n/2-1},$$

$$\frac{n-1-2s_2-3s_3-4s_4}{5}+s_2+s_3+s_4=\frac{n-1+3s_2+2s_3+s_4}{5}.$$

$$\phi (2^n-1)+{\Phi }_2\left(\frac{x^n-1}{x-1}\right)>2^n,$$

Full text

1 Introduction

Let $\mathbb{F}_{q^{n}}$ be the finite field with $q^{n}$ elements, where $q$ is a prime power and $n$ is a positive integer. Recall that an element $\alpha\in\mathbb{F}_{q^{n}}$ is said to be normal over $\mathbb{F}_{q}$ if $A=\{\alpha,\alpha^{q},\cdots,\alpha^{q^{n-1}}\}$ is a basis of $\mathbb{F}_{q^{n}}$ over $\mathbb{F}_{q}$; A is called a normal basis. Normal basis are frequently used in cryptography and computer algebra systems; sometimes it is useful to take normal basis composed by primitive elements, i.e., generators of the multiplicative group $\mathbb{F}_{q^{n}}^{*}$. The Primitive Normal Basis Theorem states that for any extension field $\mathbb{F}_{q^{n}}$ of $\mathbb{F}_{q}$, there exists a basis composed by primitive normal elements; this result was first proved by Lenstra and Schoof [6] and a proof without the use of a computer was latter given in [3].

A characterization of normal elements is given in ([7], Theorem 2.39): an element $\alpha\in\mathbb{F}_{q^{n}}$ is normal over $\mathbb{F}_{q}$ if and only if the polynomials

[TABLE]

are relatively prime over $\mathbb{F}_{q^{n}}$. Motivated by this characterization, in [5], the authors introduce $k-$normal elements:

Definition 1.1**.**

Let $\alpha\in\mathbb{F}_{q^{n}}^{*}$ and $g_{\alpha}(x)=\sum_{i=0}^{n-1}\alpha^{q^{i}}\cdot x^{n-1-i}$. We say that $\alpha$ is $k-$normal over $\mathbb{F}_{q}$ if the greatest common divisor of $x^{n}-1$ and $g_{\alpha}(x)$ over $\mathbb{F}_{q^{n}}$ has degree $k$.

From definition, $0-$normal elements correspond to normal elements in the usual sense. In the same paper, the authors give a characterization of $k-$normal elements and find a formula for their number. Also, they obtain an existence result on primtive $1-$normal elements:

Theorem 1.2** ([5], Theorem 5.10).**

Let $q=p^{e}$ be a prime power and $n$ a positive integer not divisible by $p$. Assume that $n\geq 6$ if $q\geq 11$ and that $n\geq 3$ if $3\leq q\leq 9$. Then there exists a primitive $1-$normal element of $\mathbb{F}_{q^{n}}$ over $\mathbb{F}_{q}$.

The authors propose an extension of the above theorem for all pairs $(q,n)$ with $n\geq 2$ as a problem ([5], Problem 6.2); they conjectured that such elements always exist. However, it was proved in [1] that for odd $q>3$ and $n=2$, there are no primitive $1-$normal elements of $\mathbb{F}_{q^{2}}$ over $\mathbb{F}_{q}$. The aim of this note is to extend Theorem 1.2 to the case when $q=2$ and $n$ is odd. Essentially, we show that the tools used in [5] to prove Theorem 1.2 can be adapted to that case.

2 Existence of primitive $1-$normal elements over $\mathbb{F}_{2}$

First, we present some definitions and results that will be useful in the rest of this paper.

Definition 2.1**.**

Let $f(x)$ be a monic polynomial with coefficients in $\mathbb{F}_{q}$. The Euler Phi Function for polynomials over $\mathbb{F}_{q}$ is given by

[TABLE]

where $\langle f\rangle$ is the ideal generated by $f(x)$ in $\mathbb{F}_{q}[x]$. 2. 2.

If $t$ is a positive integer (or a monic polynomial over $\mathbb{F}_{q}$), $W(t)$ denotes the number of square-free (monic) divisors of $t$.

We have an interesting formula for the number of $k-$normal elements over finite fields:

Lemma 2.2** ([5], Theorem 3.5).**

The number $N_{k}$ of $k-$normal elements of $\mathbb{F}_{q^{n}}$ over $\mathbb{F}_{q}$ is given by

[TABLE]

where the divisors are monic and polynomial division is over $\mathbb{F}_{q}$.

In particular, if $n\geq 2$, the number of $1-$normal elements of $\mathbb{F}_{q^{n}}$ over $\mathbb{F}_{q}$ is at least equal to $\Phi_{q}(T)$, where $T=\frac{x^{n}-1}{x-1}$.

2.1 A sieve inequality

The proof of Theorem 1.2 is based in an application of the Lenstra-Schoof method, introduced in [6]; this method has been used frequently in the characterization of elements in finite fields with particular properties like being primitive, normal and of zero-trace. For more details, see [3] and [5]. In particular, from Corollary 5.8 of [5], we can easily deduce the following:

Lemma 2.3**.**

Suppose that $q$ is a power of a prime $p$, $n\geq 2$ is a positive integer not divisible by $p$ and $T(x)=\frac{x^{n}-1}{x-1}$. If

[TABLE]

then there exist $1-$normal elements of $\mathbb{F}_{q^{n}}$ over $\mathbb{F}_{q}$.

Inequality (2) is an essential step in the proof of Theorem 1.2 and it was first studied in [2]; under the condition that $n\geq 6$ for $q\geq 11$ and $n\geq 3$ for $3\leq q\leq 9$, this inequality is not true only for a finite number of pairs $(q,n)$ (see Theorem 4.5 of [2]). Here we extend the study of inequality (2) to the case when $q=2$ and $n$ is odd. First, we have the following:

Proposition 2.4**.**

Suppose that $n\geq 3$ is odd and $T(x):=\frac{x^{n}-1}{x-1}\in\mathbb{F}_{2}[x]$. Then $W(T)\leq 2^{\frac{n+9}{5}}.$

Proof.

For each $2\leq i\leq 4$, let $s_{i}$ be the number of irreducible factors of degree $i$ dividing $T(x)$. Since $n$ is odd, $T(x)$ has no linear factor. By a direct verification we see that the number of irreducible polynomials over $\mathbb{F}_{2}$ of degrees $2,3$ and $4$ is $1,2$ and $3$, respectively. Hence $s_{2}\leq 1,s_{3}\leq 2$ and $s_{4}\leq 3$. In particular, the number of irreducible factors of $T(x)$ over $\mathbb{F}_{2}$ is at most

[TABLE]

Since $\frac{n-1+3s_{2}+2s_{3}+s_{4}}{5}\leq\frac{n-1+3+4+3}{5}=\frac{n+9}{5}$, we conclude the proof. ∎

According to Lemma 7.5 in [4], $W(2^{n}-1)<2^{\frac{n}{7}+2}$ if $n$ is odd. In particular, we obtain the following:

Corollary 2.5**.**

Suppose that $n\neq 15$ is odd, $q=2$ and $T(x)=\frac{x^{n}-1}{x-1}\in\mathbb{F}_{2}[x]$. For $n>9$, inequality (2) holds.

Proof.

Notice that $\frac{n+9}{5}+\frac{n}{7}+2<\frac{n}{2}-1$ for $n\geq 31$. From Proposition 2.4 and Lemma 7.5 of [4], it follows that inequality (2) holds for odd $n\geq 31$. The remaining cases can be verified directly. ∎

We are ready to state and prove our result:

Theorem 2.6**.**

Suppose that $n\geq 3$ is odd. Then there exist a primitive $1-$normal element of $\mathbb{F}_{2^{n}}$ over $\mathbb{F}_{2}$.

Proof.

According to Lemma 2.3 and Corollary 2.5, this statement is true for $n>9$ if $n\neq 15$. For the remaining cases $n=3,5,7,9$ and $15$ we use the following argument. Let $P$ be the number of primitive elements of $\mathbb{F}_{2^{n}}$ and $N_{1}$ the number of $1-$normal elements of $\mathbb{F}_{2^{n}}$ over $\mathbb{F}_{2}$; if $P+N_{1}>2^{n}$, there exists a primitive $1-$normal element of $\mathbb{F}_{2^{n}}$ over $\mathbb{F}_{2}$. Notice that $P=\varphi(2^{n}-1)$ and, according to Lemma 2.2, $N_{1}\geq\Phi_{2}\left(\frac{x^{n}-1}{x-1}\right)$. By a direct calculation we see that

[TABLE]

for $n=3,5,7,9$ and $15$. This completes the proof. ∎