2 Toward the primitive completely normal basis theorem

We will split our exhibition of previous results depending on the means that workers that attacked Conjecture 1.4 have adopted in the past, rather than a strictly chronological order. The first and most obvious method is direct verification with the help of computers. This is in fact the method that Morgan and Mullen [Reference Morgan and Mullen27] used in order to support Conjecture 1.4.

Later still, with the help of more modern equipment, advanced computational methods, and theoretical arguments, Hachenberger and Hackenberg [Reference Hachenberger and Hackenberg21] extended the original Morgan–Mullen verifications as follows.

However, most of the advances toward establishing Conjecture 1.4 have been achieved with theoretical means. A first observation is that all completely normal elements are, by definition, normal, but the inverse is true only for certain finite field extensions. So, the following definition is essential.

For the case of completely basic extensions, Conjecture 1.4 is directly implied by Theorem 1.2, thus, if $\mathbb F_{q^n}/\mathbb F_q$ is completely basic, it possesses the MM property. Clearly, if r is prime, $\mathbb F_{q^r}/\mathbb F_q$ is completely basic, but this is not the only family of completely basic extensions. In fact, we have the following complete characterization of this family by Blessenohl and Johnsen [Reference Blessenohl and Johnsen7].

In particular, see [Reference Hachenberger, Mullen and Panario18, Example 5.4.19], this family includes cases, such as $n=r^2$ (where r is a prime number), $n\mid q-1$ or $n=p^k$ , where k is arbitrary and $p=\operatorname {\mathrm {char}}{\mathbb F_q}$ .

Another family that is of interest for us is the following.

Completely basic extensions are clearly regular extensions. But, see [Reference Hachenberger, Mullen and Panario18, Example 5.4.47], this family also include extensions $\mathbb F_{q^n}/\mathbb F_q$ , where the set $D = \{ r\mid n : r\text { prime} \}$ satisfies

1. (1) $|D|=1$ , or,

2. (2) for every $r\in D,$ we have that $r\mid q-1$ or $r\mid q$ , or,

3. (3) $D\subseteq \{7, 11, 13, 17, 19, 31, 41, 47, 49, 61, 73, 97, 101, 107, 109, 139, 151, 163, 167, 173, 179, 181, 193\}$ .

Also, $\mathbb F_{q^n}/\mathbb F_q$ is regular if n is a power of a Carmichael number. It has been established that regular extensions possess the MM property.

The above was initially partially established by Hachenberger [Reference Hachenberger16, Reference Hachenberger17] and, subsequently, fully established by the same author [Reference Hachenberger20]. Furthermore, we note that the establishment of Theorem 2.4 required deep understanding of the underlying module structure of finite fields and combines algebraic and character sum methods.

Last but not least, some efforts have been made with the adoption of combinatorial and character sum methods. Namely, in 2016, Hachenberger [Reference Hachenberger19] used elementary combinatorial arguments in order to obtain the following.

It is worth mentioning that the results of [Reference Hachenberger19] depend on an estimate of the number of $q^n/q$ -completely normal elements that derives from combinatorial arguments and does not depend on the prime decomposition of n and its relation with q.

Inspired by this work, the authors adjusted the character sum method accordingly for this setting. More precisely, in [Reference Garefalakis and Kapetanakis12, Reference Garefalakis and Kapetanakis13], the main term of the authors’ application of the method was manipulated in a way that it equaled the number of $q^n/q$ -completely normal elements. This lead to the following improvement of Theorem 2.5 in [Reference Garefalakis and Kapetanakis13].

Soon after, with more attention given to technical details, the above was further improved in [Reference Garefalakis and Kapetanakis12] by the same authors.

Here, we remark that, in this section, emphasis was given to the end results regarding the existence of primitive and $q^n/q$ -completely normal elements, which is also the object of this work. However, in addition to this question, there are many interesting related questions, such as the distribution of these elements. We refer the interested reader to the references of this work and in particular to related textbooks [Reference Hachenberger15, Reference Hachenberger, Mullen and Panario18, Reference Hachenberger and Jungnickel22] and the references therein.

3 Preliminaries

As we mentioned in Section 1, in this work, we will take advantage of the character sum method. In this method, characters and their sums play a crucial role in characterizing elements of finite fields with the desired properties and in estimating the number of elements that combine all the desired properties.

The finite field $\mathbb F_{q^n}$ is associated with its multiplicative and its additive group. From now on, we will call the characters of $\mathbb F_{q^n}^*$ multiplicative characters and the characters of $\mathbb F_{q^n}$ additive characters. Furthermore, we will denote by $\chi _0$ and $\psi _0$ the trivial multiplicative and additive character, respectively, and we will extend the multiplicative characters to zero with the rule

$$\begin{align*}\chi(0) := \begin{cases} 0, & \text{if }\chi\in\widehat{\mathbb F_{q^n}^*} \setminus \{ \chi_0 \} , \\ 1, & \text{if } \chi = \chi_0. \end{cases} \end{align*}$$

A character sum is a sum that involves characters. In this work, we will use the following incomplete character sum estimate.

Theorem 3.1 Let $\mathbb F_{q^n}/\mathbb F_q$ be a finite field extension and take $\chi $ a multiplicative character of $\mathbb F_{q^n}$ and $\psi $ an additive character of $\mathbb F_q$ , such that not both of them are trivial. Further, let $a\in \mathbb F_{q^n}$ be such that $\mathbb F_q(a)=\mathbb F_{q^n}$ . Then

$$\begin{align*}\left| \sum_{c\in\mathbb F_q} \chi(a +c)\psi(c) \right| \leq n q^{1/2}. \end{align*}$$

The above follows from the main result of [Reference Fu and Wan11]. However, an alternative simplified proof was established recently (see [Reference Mazumder, Kapetanakis, Kala and Basnet26]).

The additive and the multiplicative groups of $\mathbb F_{q^n}$ can also be seen as modules. In particular, $\mathbb F_{q^n}^*$ (the multiplicative group) can be seen as a $\mathbb Z$ -module under the rule $r\circ a = a^r$ , where $r\in \mathbb Z$ and $a\in \mathbb F_{q^n}$ . The fact that $\mathbb F_{q^n}$ contains primitive elements implies that this module is cyclic.

By using Vinogradov’s formula for generators of cyclic modules over Euclidean domains, it follows that the characteristic function for $a\in \mathbb F_{q^n}^*$ being primitive is

$$\begin{align*}\omega(a) := \theta(q') \sum_{d\mid q'} \frac{\mu(d)}{\phi(d)} \sum_{\chi\in\widehat{\mathbb F_{q^n}^*},\ \operatorname{\mathrm{ord}}(\chi)=d} \chi(a), \end{align*}$$

where $q'=q^n-1$ , $\theta (r)=\phi (r)/r$ , $\mu $ is the Möbius function, $\phi $ is the Euler function, and the order of a multiplicative character is defined as its multiplicative order in $\widehat {\mathbb F_{q^n}^*}$ . Also, we stress that $\mathbb F_{q^n}^*$ contains $\phi (q')$ primitive elements.

Regarding the additive group, for any $d\mid n$ , $\mathbb F_{q^{d}}$ can be seen as an $\mathbb F_{q}[x]$ -module under the rule $f\circ a = \mathfrak f(a)$ , where $\mathfrak f$ stands for the q-associate of $f\in \mathbb F_{q}[x]$ , that is, for $f=\sum _{i=0}^k f_ix^i$ , then $\mathfrak f = \sum _{i=0}^k f_ix^{q^{i}}$ .

Theorem 1.1 implies that for all $d\mid n$ , the corresponding module is cyclic. Similarly, the characteristic function for $a\in \mathbb F_{q^{d}}$ being ${q^{d}}/{q}$ -normal is

$$\begin{align*}\varOmega_{d}(a) := \theta_q(F_d') \sum_{f\mid F_{d}'} \frac{\mu_q(f)}{\phi_q(f)} \sum_{\psi\in\widehat{\mathbb F_{q^d}}, \ \operatorname{\mathrm{ord}}(\psi)=f} \psi(a) , \end{align*}$$

where $F_{d}'=x^{d}-1\in \mathbb F_{q}[x]$ , $\theta _{q}(f):= \phi _{q}(f)/q^{\deg (f)}$ , and $\mu _{q}$ and $\phi _{q}$ are the Möbius and Euler functions in $\mathbb F_{q}[x]$ , respectively, the first sum extends over the monic divisors of $F_d'$ in $\mathbb F_{q}[x]$ and the second sum runs through the additive characters of $\mathbb F_{q}$ of q-order f. The q-order of some $\psi \in \widehat {\mathbb F_{q^{d}}}$ , denoted by $\operatorname {\mathrm {ord}}(\psi )$ , is defined as the lowest degree monic polynomial $g\in \mathbb F_{q}[x]$ such that $\psi ( g\circ a ) = 1$ for all $a\in \mathbb F_{q^{d}}$ . We note that the q-order of an additive character of $\mathbb F_{q^{d}}$ divides $F_d'$ and that $\mathbb F_{q^d}$ contains $\phi _q(F_d')$ elements that are $q^d/q$ -normal.

Last but not least, it follows immediately from the orthogonality relations that the characteristic function for some $a\in \mathbb F_q$ to be nonzero is

$$\begin{align*}z(a) = \frac 1q \sum_{\eta\in\widehat{\mathbb F_q}}(1- \eta(a)). \end{align*}$$

4 A construction of completely normal elements

Before we proceed, we start with the following definition.

Definition 4.1 Take some finite field extension $\mathbb F_{q^n}/\mathbb F_q$ and some $a\in \mathbb F_{q^n}$ . The set of translates of a is defined as

$$\begin{align*}\mathcal T_a = \{ a + c : c\in\mathbb F_q\}. \end{align*}$$

In this section, we will modify the following proposition, which plays an important role in the original proof of Theorem 1.3.

Toward this end, we will need the following auxiliary lemmas and the notion of the trace function. The $q^n/q$ -trace is the function

$$\begin{align*}\operatorname{\mathrm{Tr}}_{q^n/q} : \mathbb F_{q^n}\to\mathbb F_q , \ a \mapsto \sum_{i=0}^{n-1} a^{q^i} = \sum_{\sigma\in\operatorname{\mathrm{Gal}}(\mathbb F_{q^n}/\mathbb F_q)} \sigma(a). \end{align*}$$

Proof Since a is $q^n/q$ -normal, the map

$$\begin{align*}\pi : \mathbb F_q^n \to \mathbb F_{q^n} ,\, (\lambda_0,\ldots ,\lambda_{n-1}) \mapsto \sum_{i=0}^{n-1} \lambda_i a^{q^i} \end{align*}$$

is a bijection. The statement of the lemma amounts to

$$\begin{align*}\pi^{-1}(\mathbb F_q) = \{(\lambda, \ldots, \lambda) : \lambda \in \mathbb F_q\}. \end{align*}$$

Indeed, if $\lambda _i=\lambda \in \mathbb F_q$ for all $i=0,\ldots ,n-1$ , then

$$\begin{align*}\sum_{i=0}^{n-1} \lambda_i a^{q^i} = \lambda \operatorname{\mathrm{Tr}}_{q^n/q}(a)\in\mathbb F_q. \end{align*}$$

Therefore, $\{(\lambda ,\ldots , \lambda ) : \lambda \in \mathbb F_q\} \subseteq \pi ^{-1}(\mathbb F_q)$ . Since $|\pi ^{-1}(\mathbb F_q)| = q$ , the claim follows.

We prove the following lemma.

Proof Take

$$\begin{align*}\sum_{i=0}^{n-1} \lambda_i (a+c)^{q^i} = 0 , \end{align*}$$

for some $\lambda _i\in \mathbb F_q$ , for $i=0,\ldots ,n-1$ . Recall that $a+c$ is $q^n/q$ -normal if and only if the above holds only for $\lambda _i=0$ for all $i=0,\ldots ,n-1$ . Now, the latter is equivalent to

$$\begin{align*}\sum_{i=0}^{n-1} \lambda_i a^{q^i} =- c \sum_{i=0}^{n-1} \lambda_i. \end{align*}$$

Further, Lemma 4.5 entails that $\lambda _i=\lambda $ for every $i=0,\ldots ,n-1$ , for some $\lambda \in \mathbb F_q$ . Finally, the last equation becomes

$$\begin{align*}\lambda \left( \operatorname{\mathrm{Tr}}_{q^n/q}(a) +cn\right) = 0 , \end{align*}$$

which implies the desired result, with Lemma 4.4 in mind.

Proof Recall that $a_1a_2$ is $q^n/q$ -completely normal by Proposition 4.1. Next, fix some $c\in \mathbb F_q$ . Clearly, if $c=-n\operatorname {\mathrm {Tr}}_{q^n/q}(a_1a_2)$ (in the case $p\nmid n$ ), then, by Lemma 4.6, $a_1a_2+c$ cannot be $q^n/q$ -completely normal. So, from now on, we assume that either $p\mid n$ , or $\operatorname {\mathrm {Tr}}_{q^n/q}(a_1a_2+c)\neq 0$ if $p\nmid n$ .

Now, take some $d\mid n$ . If $p\mid (n/d)$ , Lemma 4.6 yields that $a_1a_2+c$ is $q^n/q^d$ -normal. So, we may focus on the case $p\nmid (n/d)$ .

In this case, Lemma 4.6 implies that $a_1a_2+c$ is $q^n/q^d$ -normal if $\operatorname {\mathrm {Tr}}_{q^n/q^d}(a_1a_2+c)\neq 0$ . However, if this is not the case, then

$$\begin{align*}\operatorname{\mathrm{Tr}}_{q^n/q}(a_1a_2+c) = \operatorname{\mathrm{Tr}}_{q^d/q}\left( \operatorname{\mathrm{Tr}}_{q^n/q^d}(a_1a_2+c) \right) = \operatorname{\mathrm{Tr}}_{q^d/q}(0) = 0, \end{align*}$$

a contradiction.

As a corollary of Proposition 4.7, we obtain the main result of this section.

5 Main condition

Before we proceed further, we introduce the following family of finite field extensions that will be of interest for us.

We will also need the following technical definition.

We are now in position to prove the main results of this section.

Theorem 5.2 Let $\mathbb F_{q^n}/\mathbb F_q$ be an $(n_1,n_2)$ -partially completely basic extension, such that $p\nmid n$ and

$$\begin{align*}\frac{q-2}{q-1}\cdot q^{n_2/2} \geq 2 n_1 W(q')W(F_{n_2}'). \end{align*}$$

Then $\mathbb F_{q^n}/\mathbb F_q$ possesses the MM property.

Proof Theorem 1.3 ensures the existence of some $a\in \mathbb F_{q^{n_1}}$ that is $q^{n_1}/q$ -completely normal. The fact that $\mathbb F_{q^{n_2}}/\mathbb F_q$ is completely basic and Proposition 4.7 implies that if we identify some $b\in \mathbb F_{q^{n_2}}$ and $c\in \mathbb F_q^*$ such that

1. (1) b is $q^{n_2}/q$ -normal,

2. (2) $ab+c$ is primitive, and

3. (3) $\operatorname {\mathrm {Tr}}_{q^n/q}(ab+c)\neq 0$ ,

then $ab+c$ is primitive and $q^n/q$ -completely normal. In particular, $\mathbb F_{q^n}/\mathbb F_q$ possesses the MM property.

From the characteristic functions of the corresponding properties, as presented in Section 3, it follows that the number of $b\in \mathbb F_{q^{n_2}}$ that combine the desired properties as above is

$$\begin{align*}\mathcal N = \sum_{b\in\mathbb F_{q^{n_2}}}\sum_{c\in\mathbb F_q^*} \omega (ab+c) \varOmega_{n_2}(b) z\left( \operatorname{\mathrm{Tr}}_{q^n/q}(ab+c) \right) , \end{align*}$$

where we note that, given the restrictions on a, b, and c, we have that $ab+c\neq 0$ . So, for our purposes, it suffices to show that $\mathcal N\neq 0$ . We replace $\omega $ and $\varOmega _{n_2}$ with their expressions from Section 3 and obtain

(5.1) $$ \begin{align} \frac{\mathcal N}{\theta(q')} & = \sum_{d\mid q'} \frac{\mu(d)}{\phi(d)} \sum_{\substack{\chi\in\widehat{\mathbb F_{q^n}^*} \\ \operatorname{\mathrm{ord}}(\chi) = d}} \sum_{b\in\mathbb F_{q^{n_2}}} \sum_{c\in\mathbb F_q^*} \chi(ab+c) \varOmega_{n_2}(b) z\left( \operatorname{\mathrm{Tr}}_{q^n/q}(ab+c) \right) \nonumber \\ & = S_1 + S_2 , \end{align} $$

where $S_1$ is the part that corresponds to $d=1$ and $S_2$ the one that corresponds to $d\neq 1$ .

First, we focus on $S_1$ . We have that

(5.2) $$ \begin{align} S_1 = \sum_{b\in\mathbb F_{q^{n_2}}} \varOmega_{n_2}(b) \sum_{c\in\mathbb F_q^*} z\left( \operatorname{\mathrm{Tr}}_{q^n/q}(ab+c) \right) = (q-2) \phi_q(F_{n_2}'). \end{align} $$

Then, we turn our attention to $S_2$ . We have that

$$\begin{align*}S_2 = \frac{\theta_q(F_{n_2}')}{q} \sum_{\substack{d\mid q' \\ d\neq 1}} \sum_{f\mid F_{n_2}'} \frac{\mu(d)\mu_q(f)}{\phi(d)\phi_q(f)} \sum_{\substack{\chi\in\widehat{\mathbb F_{q^n}^*} \\ \operatorname{\mathrm{ord}}(\chi) = d}} \sum_{\substack{\psi\in\widehat{\mathbb F_{q^{n_2}}} \\ \operatorname{\mathrm{ord}}(\psi)=f}} \sum_{\eta\in\widehat{\mathbb F_q}} \sum_{c\in\mathbb F_q^*} \mathcal C(\chi,\psi,\eta,c) , \end{align*}$$

where

$$\begin{align*}\mathcal{C}(\chi,\psi,\eta,c) := \sum_{b\in\mathbb F_{q^{n_2}}} \chi(ab+c)\psi(b)\left( 1-\eta(\operatorname{\mathrm{Tr}}_{q^n/q}(ab+c)) \right). \end{align*}$$

It follows that

$$ \begin{align*} &|\mathcal{C}(\chi,\psi,\eta,c)| \\ &\leq|\chi(a)|\cdot \left| \sum_{b\in\mathbb F_{q^{n_2}}} \chi(a^{-1}c+b)\psi(b) \right| + \left| \chi(a)\eta\left( \operatorname{\mathrm{Tr}}_{q^n/q}(c)\right)\right| \cdot \left| \sum_{b\in\mathbb F_{q^{n_2}}} \chi(a^{-1}c+b)\rho(b) \right| , \end{align*} $$

where $\rho (b) = \psi (b)\eta (\operatorname {\mathrm {Tr}}_{q^n/n}(ab))$ is an additive character of $\mathbb F_{q^{n_2}}$ . Further, observe that, since a is $q^{n_1}/q$ -normal, Lemma 4.2 entails that it is also $q^n/q^{n_2}$ -normal, thus $\mathbb F_{q^{n_2}}(a) = \mathbb F_{q^n}$ , that is, $\mathbb F_{q^{n_2}}(a^{-1}c) = \mathbb F_{q^n}$ . It follows from Theorem 3.1 (applied on the extension $\mathbb F_{q^n}/\mathbb F_{q^{n_2}}$ ) that, since $\chi $ is not trivial, $|\mathcal {C}(\chi ,\psi ,\eta ,c)|\leq 2 n_1 q^{n_2/2}$ . Furthermore, recall that there are exactly $\phi (d)$ (resp. $\phi _q(f)$ ) multiplicative (resp. additive) characters of order d (resp. q-order f). It follows that

(5.3) $$ \begin{align} |S_2| & \leq \frac{\theta_q(F_{n_2}')}{q} (W(q')-1)W(F_{n_2}') q (q-1) 2n_1 q^{n_2/2} \nonumber \\ & < \frac{\phi_q(F_{n_2}')}{q^{n_2/2}} W(q')W(F_{n_2}') (q-1) 2n_1. \end{align} $$

We combine Equations (5.1)–(5.3) and obtain that $\mathcal N\neq 0$ if

$$\begin{align*}(q-2) \phi_q(F_{n_2}') \geq \frac{\phi_q(F_{n_2}')}{q^{n_2/2}} W(q')W(F_{n_2}') (q-1) 2 n_1 , \end{align*}$$

from where the result follows.

For the case $p\mid n$ , we have the following improved version of Theorem 5.2.

Theorem 5.3 Let $\mathbb F_{q^n}/\mathbb F_q$ be an $(n_1,n_2)$ -partially completely basic extension, such that $p\mid n$ . Then, if

$$\begin{align*}q^{n_2/2} \geq n_1 W(q')W(F_{n_2}') , \end{align*}$$

$\mathbb F_{q^n}/\mathbb F_q$ possesses the MM property.

6 An asymptotic result

In this section, we will use Theorems 5.2 and 5.3, in order to obtain an asymptotic condition for an $(n_1,n_2)$ -partially completely basic extension to possess the MM property. We will need the following classic result.

We will also need the following lemmas.

For small values of q, we will use the following.

Lemma 6.3 [Reference Aguirre and Neumann1, Lemma 3.7]

Let q be a prime power and n be a positive integer. Then, $W(F_n') \leq 2^{\frac {n+a}{b}}$ for some $a,b\in \mathbb Z$ . In particular, for $q\geq 29$ , we have $(a,b)=(0,1)$ ; for $7\leq q\leq 27$ , we have $(a,b)=(q-1,2)$ and for $q\leq 5,$ we have the following values for $a,b$ :

$$\begin{align*}\begin {array}{lll|lll} q & a & b & q & a & b\\ \hline 2 & 14 & 5 & 3 & 20 & 4 \\ 4 & 12 & 3 & 5 & 18 & 3 \end {array} \end{align*}$$

We are now in position to prove the main result of this section.

Proof First, we assume $q>2$ . Fix some $n_1\geq 2$ . Theorem 5.2, along with Lemmas 6.1 and 6.2, implies that an $(n_1,n_2)$ -partially completely basic extension possesses the MM property, given that

$$\begin{align*}\left( \frac{\sqrt{q}}{2} \right)^{n_2}> 4 n_1 \cdot o(q^{n_2/4}). \end{align*}$$

This implies the result for $q>16$ .

For $4 \leq q \leq 16$ , Lemma 6.3 yields $W(F_{n_2}') \leq 2^{\frac {n_2+q-1}{2}}$ and a sufficient condition implied by Theorem 5.2 is

$$\begin{align*}q^{\frac{n_2}{2}} \geq 2^{\frac{q+1}{2}} n_1 o(q^{\delta}) 2^{\frac{n_2}{2}},\end{align*}$$

which is satisfied for $n_2$ large enough.

For $q=3$ , Lemma 6.3 yields $W(F_{n_2}') \leq 2^{\frac {n_2+20}{3}}$ and the above condition is replaced by

$$\begin{align*}\left( \frac{\sqrt{q}}{\sqrt[3]{2}} \right)^{n_2}> 2^{26/3} n_1 \cdot o(q^{n_2/4}) , \end{align*}$$

which settles the case $q=3$ .

Finally, for $q=2$ , we are confined to n even, thus we may use Theorem 5.3 instead. Now, Lemma 6.3 entails $W(F_{n_2}') \leq 2^{\frac {n_2+14}{5}}$ and the above condition is replaced by

$$\begin{align*}2^{3n_2/10}> 2^{14/5} n_1 \cdot o(2^{n_2/5}) , \end{align*}$$

which completes our proof.

7 Computational aspects

In this section, we explore the computational aspects of Theorem 5.2. First, with the condition of Theorem 5.2 in mind, we see that a concrete expression of Lemma 6.1 is needed.

Let $p_1,\ldots ,p_s$ be all the primes less or equal to $2^\nu $ . We note that, in this section, in addition to the number $\mathcal C_\nu ^{(\alpha )}$ defined above, we will use the following two numbers:

1. (1) $\mathcal C_\nu := \prod _{i=1}^{s} 2/(p_i^{1/\nu })$ . Clearly, $\mathcal C_\nu ^{(\alpha )} \leq \mathcal C_\nu $ for all $\alpha \in \mathbb Z$ .

2. (2) For any $1\leq j\leq s$ , $\mathcal C_\nu ^{[p_j]} := \mathcal C_{\nu } p_j^{1/\nu }/2$ . Clearly, $\mathcal C_\nu ^{(\alpha )} \leq \mathcal C_\nu ^{[p_j]}$ for all $\alpha \in \mathbb Z$ with $p_j\nmid \alpha $ .

Furthermore, the number $\mathcal C_\nu ^{(\alpha )}$ in the statement of Lemma 7.1 can be freely replaced by either $\mathcal C_\nu $ , or, if the prime $r\leq 2^\nu $ does not divide $\alpha $ , by $\mathcal C_\nu ^{[r]}$ .

7.2 Concrete results

We successfully applied the algorithm of Section 7.1 for $n_1=2$ and $n_1=3$ , while we obtained partial results for $n_1=4$ . Namely, we establish the following theorem.

Theorem 7.2 Let $\mathbb F_{q^n}/\mathbb F_q$ be an $(n_1,n_2)$ -partially completely basic extension. Then $\mathbb F_{q^n}/\mathbb F_q$ possesses the MM property in the following cases:

1. (1) $n_1=2$ or $3$ .

2. (2) $n_1=4$ and either

   • • $q\geq 199211272511189639\,(\approx 2\cdot 10^{17})$ ,

   • • $n_2\geq n_1^{n_1} = 256$ and $q\geq 22271$ , or,

   • • $n_2\geq n_1^6 = 4096$ .

For our calculations, we used the SageMath system, running on a modern laptop and the source files are available online.Footnote ² In the following sections, we present the most important details of our calculations.

7.2.1 The cases $n_1=2$ and $n_1=3$

The most essential information of our implementation for the cases $n_1=2$ and $n_1=3$ is summarized in Table 1.

Table 1

Summary of the implementation of the algorithm of Section 7.1 for $n_1=2$ and $n_1=3$ .

Table 1 Long description

The table consists of 11 columns. The headers are n sub 1, n sub min, C sub 4 n sub 1, q sub max, followed by ‘No. of pairs after step’ which spans columns for steps 4, 5, 6, 8.4, 8.5, and 8.6, and a final column for Time.

* Row 1 (n sub 1 equals 2): n sub min is 102; C sub 4 n sub 1 is 4,514.6265; q sub max is 11. The number of pairs after step 4 is 15. For steps 5, 6, 8.4, 8.5, and 8.6, the count is 0. Time is approximately 1 second.

* Row 2 (n sub 1 equals 3): n sub min is 67; C sub 4 n sub 1 is 1.057 dot 10 super 24; q sub max is 487. The number of pairs after step 4 is 3,011. For steps 5 and 6, the count is 0. For step 8.4, the count is 111. For steps 8.5 and 8.6, the count is 0. Time is approximately 30 seconds.

We note that the zeros appearing in the columns under Steps 6 and 8.6 show that the application was successful. The zeros in the columns under Steps 5 and 8.5 show that we never had to resort to computing $W(q')$ , which is an expensive computation, as the computer should first factor $q^{n_1n_2}-1$ into primes. Furthermore, the zero in the entry that corresponds to $n_1=2$ , and Step 8.4 is the result of the fact that, in this case, we expect $n_2$ to be both odd (as it has to be relatively prime to $n_1=2$ ) and even (as it has to be divided by $p=2$ ).

Next, in Table 2, we present the numbers appearing in Step 3 for $n_1=2$ and, in Tables 3–5, we present the numbers appearing in Step 3 for $n_1=3$ . We note that in Table 2, the case $q=2$ is missing, as, we already explained, this case is irrelevant and, in Table 3, the entries that correspond to $q=2$ derive from Step 8.3.

Table 2

Prime powers $3 \leq q<q_{\max }$ , the corresponding $n_{\max }^{(q)}$ , and the corresponding value of $\mathcal C_{4n_1}^{[p]}$ for $n_1=2$ .

Table 2 Long description

The table is organized into three repeating sets of three columns: q, n sub max to the q power, and C sub 4 n sub 1 to the p power.

Row 1:

- First set: q is 3, n sub max to the q power is 133, C is 2,589.5959.

- Second set: q is 4, n sub max to the q power is 108, C is 2,461.6176.

- Third set: q is 5, n sub max to the q power is 81, C is 2,760.3433.

Row 2:

- First set: q is 7, n sub max to the q power is 84, C is 2,878.9167.

- Second set: q is 8, n sub max to the q power is 68, C is 2,461.6176.

- Third set: q is 9, n sub max to the q power is 60, C is 2,589.5959.

Table 3

Prime powers $2 \leq q\leq 128$ , the corresponding $n_{\max }^{(q)}$ , and the corresponding value of $\mathcal C_{4n_1}^{[p]}$ for $n_1=3$ .

Table 3 Long description

The table is organized into two identical sets of three columns. The columns are q, n sub max super q, and C sub 4 n sub 1 super p.

Selected data points from the first set:

* q equals 2, n sub max super q equals 1,666, C sub 4 n sub 1 super p equals 5.6008 dot 10 super 23.

* q equals 4, n sub max super q equals 517, C sub 4 n sub 1 super p equals 5.6008 dot 10 super 23.

* q equals 7, n sub max super q equals 421, C sub 4 n sub 1 super p equals 6.2172 dot 10 super 23.

* q equals 13, n sub max super q equals 207, C sub 4 n sub 1 super p equals 6.5463 dot 10 super 23.

* q equals 31, n sub max super q equals 344, C sub 4 n sub 1 super p equals 7.0380 dot 10 super 23.

* q equals 64, n sub max super q equals 163, C sub 4 n sub 1 super p equals 5.6008 dot 10 super 23.

* q equals 127, n sub max super q equals 110, C sub 4 n sub 1 super p equals 7.9156 dot 10 super 23.

Selected data points from the second set:

* q equals 3, n sub max super q equals 599, C sub 4 n sub 1 super p equals 5.7933 dot 10 super 23.

* q equals 8, n sub max super q equals 341, C sub 4 n sub 1 super p equals 5.6008 dot 10 super 23.

* q equals 19, n sub max super q equals 162, C sub 4 n sub 1 super p equals 6.7566 dot 10 super 23.

* q equals 32, n sub max super q equals 327, C sub 4 n sub 1 super p equals 5.6008 dot 10 super 23.

* q equals 81, n sub max super q equals 140, C sub 4 n sub 1 super p equals 5.7933 dot 10 super 23.

* q equals 128, n sub max super q equals 109, C sub 4 n sub 1 super p equals 5.6008 dot 10 super 23.

Table 4

Prime powers $131 \leq q\leq 289$ , the corresponding $n_{\max }^{(q)}$ , and the corresponding value of $\mathcal C_{4n_1}^{[p]}$ for $n_1=3$ .

Table 4 Long description

The table contains 15 rows of data across two parallel sections. Each section has three columns: q, n sub max super open parenthesis q close parenthesis, and C sub 4 n sub 1 super open parenthesis p close parenthesis.

* Row 1: q = 131, n sub max = 109, C = 7.9361 dot 10 super 23 | q = 137, n sub max = 106, C = 7.9658 dot 10 super 23.

* Row 2: q = 139, n sub max = 106, C = 7.9754 dot 10 super 23 | q = 149, n sub max = 102, C = 8.0217 dot 10 super 23.

* Row 3: q = 151, n sub max = 102, C = 8.0306 dot 10 super 23 | q = 157, n sub max = 100, C = 8.0568 dot 10 super 23.

* Row 4: q = 163, n sub max = 98, C = 8.0820 dot 10 super 23 | q = 167, n sub max = 97, C = 8.0983 dot 10 super 23.

* Row 5: q = 169, n sub max = 97, C = 6.5463 dot 10 super 23 | q = 173, n sub max = 96, C = 8.1222 dot 10 super 23.

* Row 6: q = 179, n sub max = 95, C = 8.1453 dot 10 super 23 | q = 181, n sub max = 94, C = 8.1528 dot 10 super 23.

* Row 7: q = 191, n sub max = 92, C = 8.1894 dot 10 super 23 | q = 193, n sub max = 92, C = 8.1966 dot 10 super 23.

* Row 8: q = 197, n sub max = 91, C = 8.2106 dot 10 super 23 | q = 199, n sub max = 91, C = 8.2175 dot 10 super 23.

* Row 9: q = 211, n sub max = 89, C = 8.2577 dot 10 super 23 | q = 223, n sub max = 87, C = 8.2958 dot 10 super 23.

* Row 10: q = 227, n sub max = 86, C = 8.3081 dot 10 super 23 | q = 229, n sub max = 86, C = 8.3142 dot 10 super 23.

* Row 11: q = 233, n sub max = 85, C = 8.3262 dot 10 super 23 | q = 239, n sub max = 85, C = 8.3439 dot 10 super 23.

* Row 12: q = 241, n sub max = 84, C = 8.3497 dot 10 super 23 | q = 243, n sub max = 84, C = 5.7933 dot 10 super 23.

* Row 13: q = 251, n sub max = 83, C = 8.3780 dot 10 super 23 | q = 256, n sub max = 82, C = 5.6008 dot 10 super 23.

* Row 14: q = 257, n sub max = 82, C = 8.3945 dot 10 super 23 | q = 263, n sub max = 82, C = 8.4107 dot 10 super 23.

* Row 15: q = 269, n sub max = 81, C = 8.4265 dot 10 super 23 | q = 271, n sub max = 81, C = 8.4317 dot 10 super 23.

* Row 16: q = 277, n sub max = 80, C = 8.4471 dot 10 super 23 | q = 281, n sub max = 80, C = 8.4572 dot 10 super 23.

* Row 17: q = 283, n sub max = 80, C = 8.4622 dot 10 super 23 | q = 289, n sub max = 79, C = 6.6943 dot 10 super 23.

Table 5

Prime powers $293 \leq q<q_{\max }$ , the corresponding $n_{\max }^{(q)}$ , and the corresponding value of $\mathcal C_{4n_1}^{[p]}$ for $n_1=3$ .

Table 5 Long description

The table contains data for n sub 1 equals 3 across two parallel sections of three columns each.

* Row 1: q equals 293, n sub max equals 79, mathcal C equals 8.4867 dot 10 super 23 | q equals 307, n sub max equals 78, mathcal C equals 8.5198 dot 10 super 23.

* Row 2: q equals 311, n sub max equals 77, mathcal C equals 8.5290 dot 10 super 23 | q equals 313, n sub max equals 77, mathcal C equals 8.5336 dot 10 super 23.

* Row 3: q equals 317, n sub max equals 77, mathcal C equals 8.5426 dot 10 super 23 | q equals 331, n sub max equals 76, mathcal C equals 8.5734 dot 10 super 23.

* Row 4: q equals 337, n sub max equals 75, mathcal C equals 8.5863 dot 10 super 23 | q equals 343, n sub max equals 74, mathcal C equals 6.2172 dot 10 super 23.

* Row 5: q equals 347, n sub max equals 74, mathcal C equals 8.6072 dot 10 super 23 | q equals 349, n sub max equals 74, mathcal C equals 8.6113 dot 10 super 23.

* Row 6: q equals 353, n sub max equals 74, mathcal C equals 8.6195 dot 10 super 23 | q equals 359, n sub max equals 74, mathcal C equals 8.6316 dot 10 super 23.

* Row 7: q equals 361, n sub max equals 73, mathcal C equals 6.7566 dot 10 super 23 | q equals 367, n sub max equals 73, mathcal C equals 8.6475 dot 10 super 23.

* Row 8: q equals 373, n sub max equals 73, mathcal C equals 8.6592 dot 10 super 23 | q equals 379, n sub max equals 72, mathcal C equals 8.6707 dot 10 super 23.

* Row 9: q equals 383, n sub max equals 72, mathcal C equals 8.6783 dot 10 super 23 | q equals 389, n sub max equals 72, mathcal C equals 8.6896 dot 10 super 23.

* Row 10: q equals 397, n sub max equals 71, mathcal C equals 8.7043 dot 10 super 23 | q equals 401, n sub max equals 71, mathcal C equals 8.7116 dot 10 super 23.

* Row 11: q equals 409, n sub max equals 71, mathcal C equals 8.7259 dot 10 super 23 | q equals 419, n sub max equals 70, mathcal C equals 8.7435 dot 10 super 23.

* Row 12: q equals 421, n sub max equals 70, mathcal C equals 8.7470 dot 10 super 23 | q equals 431, n sub max equals 70, mathcal C equals 8.7641 dot 10 super 23.

* Row 13: q equals 433, n sub max equals 70, mathcal C equals 8.7675 dot 10 super 23 | q equals 439, n sub max equals 69, mathcal C equals 8.7776 dot 10 super 23.

* Row 14: q equals 443, n sub max equals 69, mathcal C equals 8.7842 dot 10 super 23 | q equals 449, n sub max equals 69, mathcal C equals 8.7941 dot 10 super 23.

* Row 15: q equals 457, n sub max equals 68, mathcal C equals 8.8070 dot 10 super 23 | q equals 461, n sub max equals 68, mathcal C equals 8.8134 dot 10 super 23.

* Row 16: q equals 463, n sub max equals 68, mathcal C equals 8.8166 dot 10 super 23 | q equals 467, n sub max equals 68, mathcal C equals 8.8229 dot 10 super 23.

* Row 17: q equals 479, n sub max equals 67, mathcal C equals 8.8416 dot 10 super 23.

7.2.2 The case $n_1=4$

Finally, we comment on our attempt in implementing the algorithm of Section 7.1 on the $n_1=4$ case. The main information of our attempt on the $n_1=4$ case is summarized in Table 6.

Table 6

Summary of the implementation of the algorithm of Section 7.1 for $n_1=4$ , with various choices of $n_{\min }$ .

Table 6 Long description

The table consists of 9 columns. The first two columns are n sub min and q sub max. The next six columns are grouped under the heading No. of pairs after step, with sub-headings for steps 4, 5, 6, 8.4, 8.5, and 8.6. The final column is Time.

* Row 1: n sub min is 50. q sub max is approximately 2 times 10 super 17. All other columns contain dashes.

* Row 2: n sub min is 256. q sub max is 22,271. Step 4 is 82,148. All other columns contain dashes.

* Row 3: n sub min is 1,024. q sub max is 27. Step 4 is 5,936. All other columns contain dashes.

* Row 4: n sub min is 4,096. q sub max is 4. Step 4 is 188. Step 5 is 0. Step 6 is 0. Step 8.4 is 3,317. Step 8.5 is 0. Step 8.6 is 0. Time is approximately 8 minutes.

First, we attempted to run the algorithm of Section 7.1 as it is. So, we computed $n_{\min } = 50$ , $\mathcal C_{4n_1} = 1.9356\cdot 10^{200}$ , and $q_{\max } = 199211272511189639 \approx 2\cdot 10^{17}$ . However, the vast size of $q_{\max }$ makes it clear that even storing the numbers $n_{\max }^{(q)}$ for each $3\leq q\leq q_{\max }$ is not straightforward and, in fact, our computer crashed during the computation of these numbers.

Then, we confined ourselves (arbitrarily) to $n_{\min } = n_1^{n_1} = 256$ . In this case, we computed $q_{\max } = 22,271$ . This made the completion of Step 3 feasible with a total number of 82,148 possible exceptional pairs. However, due to the large size of these numbers, our computer failed to complete Step 4 after having examined about 2,200 pairs, with no exceptional pair found until this point.

Our attempt on the $n_{\min } = n_1^5 = 1,024$ case faced similar difficulties, although, in this case, we got a significantly smaller $q_{\max } = 27$ and after Step 3 we had 5,936 possible exceptional pairs. This time, only around 1,650 of them were in fact checked during Step 4, before our computer’s crash, without any exceptions up to this point.

Finally, we successfully run the algorithm with the twist that we chose $n_{\min }=n_1^6 = 4,096$ , where we obtained $q_{\max } = 4$ . Also, we note that in this case, we got $n_{\max }^{(2)} = 13,385$ , $\mathcal C_{4n_1}^{(2)} = 1.0106\cdot 10^{200}$ , $n_{\max }^{(3)} = 4,605$ , and $\mathcal C_{4n_1}^{(3)} = 1.0366\cdot 10^{200}$ , respectively. This completes the proof of Theorem 7.2.