# Linearized trinomials with maximum kernel

Paolo Santonastaso and Ferdinando Zullo

###### Abstract

Linearized polynomials have attracted a lot of attention because of their
applications in both geometric and algebraic areas. Let $q$ be a prime power,
$n$ be a positive integer and $\sigma$ be a generator of
$\mathrm{Gal}({\mathbb{F}}_{q^{n}}\colon{\mathbb{F}}_{q})$. In this paper we
provide closed formulas for the coefficients of a $\sigma$-trinomial $f$ over
${\mathbb{F}}_{q^{n}}$ which ensure that the dimension of the kernel of $f$
equals its $\sigma$-degree, that is _linearized polynomials with maximum
kernel_. As a consequence, we present explicit examples of linearized
trinomials with maximum kernel and characterize those having $\sigma$-degree
$3$ and $4$. Our techniques rely on the tools developed in [24]. Finally, we
apply these results to investigate a class of rank metric codes introduced in
[8], to construct quasi-subfield polynomials and cyclic subspace codes,
obtaining new explicit constructions to the conjecture posed in [37].

AMS subject classification: 11T06, 15A04.

Keywords: linearized polynomial, recursive formula, closed formula, subspace
polynomial, rank metric code, cyclic subspace code, quasi-subfield polynomial.

##  1 Introduction

Let $q=p^{h}$, where $h$ is a positive integer and $p$ is a prime. Linearized
polynomials over ${\mathbb{F}}_{q^{n}}$ correspond to
${\mathbb{F}}_{q}$-linear transformations of the $n$-dimensional
${\mathbb{F}}_{q}$-vector space ${\mathbb{F}}_{q^{n}}$. For this reason they
have been intensively investigated and used to describe geometric and
algebraic objects such as rank metric codes, ${\mathbb{F}}_{q}$-linear sets,
see e.g. [1, 3, 7, 9, 25, 26, 30, 36, 40].

A $\sigma$-_linearized polynomial_ (for short $\sigma$-_polynomial_) over
${\mathbb{F}}_{q^{n}}$ is a polynomial of shape

$f(x)=\sum_{i=0}^{t}a_{i}x^{\sigma^{i}},$

where $a_{i}\in{\mathbb{F}}_{q^{n}}$, $t$ is a positive integer and $\sigma$
is a generator of the Galois group
$\mathrm{Gal}({\mathbb{F}}_{q^{n}}\colon{\mathbb{F}}_{q})$. Furthermore, if
$a_{t}\neq 0$ we say that $t$ is the $\sigma$-_degree_ of $f(x)$. We will
denote by $\mathcal{L}_{n,q,\sigma}$ the set of all $\sigma$-polynomials over
${\mathbb{F}}_{q^{n}}$ (or simply by $\mathcal{L}_{n,q}$ if $\sigma\colon
x\in{\mathbb{F}}_{q^{n}}\mapsto x^{q}\in{\mathbb{F}}_{q^{n}}$). In the
remainder of this paper we shall always silently identify the elements of
${\mathcal{L}}_{n,q,\sigma}$ with the endomorphisms of ${\mathbb{F}}_{q^{n}}$
they represent and, as such, speak also of _kernel_ and _rank_ of a
polynomial. Clearly, the kernel of $f(x)\in{\mathcal{L}}_{n,q,\sigma}$
coincides with the set of the roots of $f(x)$ over ${\mathbb{F}}_{q^{n}}$.

As a consequence of [16, Theorem 5] and [17, Theorem 10], for
$\sigma$-linearized polynomials we have the following bound on the number of
roots.

######  Theorem 1.1.

Consider

$f(x)=a_{0}x+a_{1}x^{\sigma}+\cdots+a_{k-1}x^{\sigma^{k-1}}+a_{k}x^{\sigma^{k}}\in\mathcal{L}_{n,q,\sigma},$

with $k\leq n-1$ and let $a_{0},a_{1},\ldots,a_{k}$ be elements of
${\mathbb{F}}_{q^{n}}$ not all of them zero. Then

$\dim_{{\mathbb{F}}_{q}}(\ker(f(x)))\leq k.$

Furthermore, if $\dim_{{\mathbb{F}}_{q}}(\ker(f(x)))=k$ then
$\mathrm{N}_{q^{n}/q}(a_{0})=(-1)^{nk}\mathrm{N}_{q^{n}/q}(a_{k})$.

Particular attention was paid for those polynomials attaining the bound of
Theorem 1.1, which are called _linearized polynomials with maximum kernel_ or
_subspace polynomials_ when $\sigma\colon x\in{\mathbb{F}}_{q^{n}}\mapsto
x^{q}\in{\mathbb{F}}_{q^{n}}$. The coefficients of a $\sigma$-linearized
polynomials have been characterized as follows.

######  Theorem 1.2.

([9, Theorem 1.2 and Corollary 3.2],[25, Theorem 7]) Consider

$f(x)=a_{0}x+a_{1}x^{\sigma}+\cdots+a_{k-1}x^{\sigma^{k-1}}-x^{\sigma^{k}}\in\mathcal{L}_{n,q,\sigma},$

with $k\leq n-1$. Then $f(x)$ has maximum kernel if and only if

$C_{f}C_{f}^{\sigma}\cdot\ldots\cdot C_{f}^{\sigma^{n-1}}=I_{k},$ (1)

where

$C_{f}=\left(\begin{matrix}0&0&\cdots&0&a_{0}\\\ 1&0&\cdots&0&a_{1}\\\
0&1&\cdots&0&a_{2}\\\ \vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&\cdots&1&a_{k-1}\end{matrix}\right)$

and is called the $\sigma$-companion matrix of $f$, $C_{f}^{\sigma^{i}}$ is
the matrix obtained from $C_{f}$ by applying to each of its entries the
automorphism $x\mapsto x^{\sigma^{i}}$ and $I_{k}$ is the identity matrix of
order $k$. Furthermore, Equation (1) holds if and only if

$C_{f}C_{f}^{\sigma}\cdot\ldots\cdot
C_{f}^{\sigma^{n-1}}\mathbf{e}_{0}=\mathbf{e}_{0},$ (2)

where $\mathbf{e}_{0}=(1,0,\ldots,0)^{T}\in\mathbb{F}_{q^{n}}^{k\times 1}$.

Linearized polynomials attracted a lot of attention because of their
applications to construct MRD codes [36], to study list decodability of rank
metric codes [32, 38, 39], cyclic subspace codes [4, 29] and affine dispersers
[5] and Elliptic curve discrete logarithm problem [19]. Motivated by these
applications, we further investigated the results on linearized trinomials
with maximum kernel of [24]. More precisely, we first refine their results for
$\sigma$-polynomials proving the following.

######  Theorem 1.3.

Let $d,n$ and $s$ be positive integers with $\gcd(s,n)=1$ such that $d\geq 3$.
Let denote by $\sigma$ the automorphism $x\in\mathbb{F}_{q^{n}}\mapsto
x^{q^{s}}\in\mathbb{F}_{q^{n}}$.

  1. (a)

If $n\leq d(d-1)$ and $d\nmid n$, then
$\dim_{\mathbb{F}_{q}}(\ker(ax+bx^{\sigma}-x^{\sigma^{d}}))<d$, for any
$a,b\in\mathbb{F}_{q^{n}}$.

  2. (b)

If $n\leq d(d-1)$ and $d\mid n$, then
$\dim_{\mathbb{F}_{q}}(\ker(ax+bx^{\sigma}-x^{\sigma^{d}}))=d$ with
$a,b\in\mathbb{F}_{q^{n}}$ if and only if $b=0$ and
$\mathrm{N}_{q^{n}/q^{d}}(a)=1$.

  3. (c)

If $n=d(d-1)+1$, then
$\dim_{\mathbb{F}_{q}}(\ker(ax+bx^{\sigma}-x^{\sigma^{d}}))=d$ with
$a,b\in\mathbb{F}_{q^{n}}$ if and only if

$\left\\{\begin{array}[]{lll}\mathrm{N}_{q^{n}/q}(a)=(-1)^{d-1},\\\
b=-a^{\sigma f_{1}}\,\,\text{where}\,\,f_{1}=\sum_{i=0}^{d-1}\sigma^{id},\\\
d-1\,\,\text{is a power of}\,\,p.\end{array}\right.$

Then we characterize $\sigma$-trinomials with maximum kernel giving closed
formulas involving their coefficients.

######  Theorem 1.4.

Let $d,n$ and $s$ be positive integers with $\gcd(s,n)=1$ such that $d\geq 3$
and $n=d(d-1)+g$, for some $g\in\\{1,\ldots,d-1\\}$. Let denote by $\sigma$
the automorphism $x\in\mathbb{F}_{q^{n}}\mapsto
x^{q^{s}}\in\mathbb{F}_{q^{n}}$. Then
$\dim_{\mathbb{F}_{q}}(\ker(ax+bx^{\sigma}-x^{\sigma^{d}}))=d$ with
$a,b\in\mathbb{F}_{q^{n}}$ if and only if $a$ and $b$ satisfy the following
system

$\sum_{\begin{subarray}{c}i_{0},\ldots,i_{t}=0\\\
i_{0}+\ldots+i_{t}=d-1-t\end{subarray}}^{d-1-t}\left(\prod_{j_{0}=1}^{i_{0}}a^{\sigma^{d(d-1-j_{0})+g}}\right)\left(\prod_{j_{1}=1}^{i_{1}}a^{\sigma^{d(d-i_{0}-j_{1}-2)+g+1}}\right)\cdot\ldots\cdot$
$\left(\prod_{j_{t-1}=1}^{i_{t-1}}a^{\sigma^{d(d-i_{0}-\ldots-
i_{t-2}-j_{t-1}-t)+g+t-1}}\right)\cdot\left(\prod_{j_{t}=1}^{i_{t}}a^{\sigma^{d(i_{t}-j_{t})+t+g}}\right)\cdot$
$\left(\prod_{u=0}^{t-1}b^{\sigma^{d(d-2-i_{0}-\cdots-i_{u}-u)+u+g}}\right)$
$=\left\\{\begin{array}[]{lll}(-1)^{v}b^{\frac{\sigma^{v}-1}{\sigma-1}}a^{-\frac{\sigma^{v+1}-1}{\sigma-1}}&\text{if}\,\,t=d-g+v,\,\,0\leq
v\leq g-1,\\\
(-1)^{g}a^{-\frac{\sigma^{g}-1}{\sigma-1}}b^{\frac{\sigma^{g}-1}{\sigma-1}}&\text{if}\,\,t=0,\\\
0&\text{otherwise},\end{array}\right.$

As a byproduct, in Corollary 4.1 we also get necessary conditions on the
coefficients of the trinomial to have maximum kernel. Then we use Theorem 1.4
to introduce a new class of $\sigma$-linearized trinomials with maximum kernel
when $q$ is even and we determine such equations when $d=3$ and $d=4$. Finally
we apply these results to investigate rank metric codes introduced in [8], to
construct examples of quasi-subfield polynomials and of cyclic subspace codes.
These constructions of subspace codes yield new solutions to the conjecture
posed in [37].

The paper is organized as follows. Section 2 regards the proof of Theorem 1.3.
Section 3 is devoted to the proof of Theorem 1.4 and in Section 4 we pointed
out some necessary conditions on the coefficients of the trinomial to have
maximum kernel. In Section 5 we apply Theorem 1.4 to obtain a new family of
$\sigma$-linearized trinomials with maximum kernel. In Section 6 we give
explicit conditions on the coefficient of a $\sigma$-linearized trinomials
when the $\sigma$-degree is either $3$ or $4$. Section 7 deals with the
applications in rank metric codes, quasi-subfield polynomials and cyclic
subspace codes. We conclude the paper listing some possible open problems in
Section 8.

##  2 Proof of Theorem 1.3

Let $d,n$ and $s$ be positive integers with $\gcd(s,n)=1$ such that $d\geq 3$.
Let denote by $\sigma$ the automorphism $x\in\mathbb{F}_{q^{n}}\mapsto
x^{q^{s}}\in\mathbb{F}_{q^{n}}$. Let
$L(x)=-x^{\sigma^{d}}+bx^{\sigma}+ax\in\mathcal{L}_{n,q,\sigma}$. When $s=1$,
Theorem 1.3 is [24, Theorem 1.1]. More precisely,

######  Theorem 2.1.

Let $d,n$ be positive integers such that $d\geq 3$.

  1. (a)

If $n\leq(d-1)d$ and $d$ does not divide $n$, then there is no polynomial
$L=x^{q^{d}}-bx^{q}-ax$ with $a,b\in{\mathbb{F}}_{q^{n}}$ that splits
completely over ${\mathbb{F}}_{q^{n}}$.

  2. (b)

Let $n=id$ with $i\in\\{1,\ldots,d-1\\}$. Let
$L=x^{q^{d}}-bx^{q}-ax\in{\mathbb{F}}_{q^{n}}[x]$. Then $L$ has $q^{d}$ roots
in ${\mathbb{F}}_{q^{n}}$ if and only if $a^{1+q^{d}+\cdots+q^{(i-1)d}}$ and
$b=0$.

  3. (c)

If $n=d(d-1)+1$, then $\dim_{\mathbb{F}_{q}}(\ker(ax+bx^{q}-x^{q^{d}}))=d$
with $a,b\in\mathbb{F}_{q^{n}}$ if and only if

$\left\\{\begin{array}[]{lll}\mathrm{N}_{q^{n}/q}(a)=(-1)^{d-1},\\\
b=-a^{qf_{1}}\,\,\text{where}\,\,f_{1}=\sum_{i=0}^{d-1}q^{id},\\\
d-1\,\,\text{is a power of}\,\,p.\end{array}\right.$

Theorem 1.3 is a consequence of Theorem 2.1, Theorem 1.2 and of the following
lemma.

######  Lemma 2.2.

[23, Lemma 3.2] Let $q$ be a prime power, and $\delta,n,s$ positive integers
such that $\gcd(s,n)=1$ and $\delta<n$. Let $U^{\prime}$ be an
${\mathbb{F}}_{q^{s}}$subspace of $\mathbb{F}_{q^{sn}}$ with
$\dim_{{\mathbb{F}}_{q^{s}}}(U^{\prime})=\delta$ and let
$U=U^{\prime}\cap{\mathbb{F}}_{q^{n}}$. Then
$\dim_{{\mathbb{F}}_{q}}(U)\leq\delta$.

_Proof of Theorem 1.3_
Let $q^{\prime}=q^{s}$, denote by $U^{\prime}$ the kernel of $f(x)$ in
$\mathbb{F}_{q^{\prime n}}$ and by $U$ the kernel of $f(x)$ in
$\mathbb{F}_{q^{n}}$, that is $U=U^{\prime}\cap{\mathbb{F}}_{q^{n}}$. By
Theorem 1.1 $\dim_{{\mathbb{F}}_{q^{\prime}}}(U^{\prime})\leq d$ and Lemma 2.2
implies that if $\dim_{{\mathbb{F}}_{q}}(U)=d$ then
$\dim_{{\mathbb{F}}_{q^{\prime}}}(U^{\prime})=d$.

  * (a)

If $n\leq d(d-1)$ and $d\nmid n$, then
$\dim_{{\mathbb{F}}_{q^{\prime}}}(U^{\prime})<d$ by (a) of Theorem 2.1 and
Lemma 2.2 implies $\dim_{{\mathbb{F}}_{q}}(U)<d$.

  * (b)

If $n\leq d(d-1)$, $d\mid n$ and $\dim_{{\mathbb{F}}_{q}}(U)=d$, then
$\dim_{{\mathbb{F}}_{q^{\prime}}}(U^{\prime})=d$ and by (b) of Theorem 2.1
$b=0$ and $\mathrm{N}_{q^{\prime n}/q^{\prime}}(a)=\mathrm{N}_{q^{n}/q}(a)=1$,
since $\gcd(s,n)=1$. The converse trivially holds, see e.g. [9, Corollary
3.5].

  * (c)

Let $n=d(d-1)+1$ and assume that $f(x)$ has maximum kernel, that is
$\dim_{{\mathbb{F}}_{q}}(U)=d$ and hence
$\dim_{{\mathbb{F}}_{q^{\prime}}}(U^{\prime})=d$. By (c) of Theorem 2.1 we
have

$\left\\{\begin{array}[]{lll}\mathrm{N}_{q^{\prime
n}/q^{\prime}}(a)=\mathrm{N}_{q^{n}/q}(a)=(-1)^{d-1},\\\
b=-a^{q^{\prime}\sum_{i=0}^{d-1}q^{\prime id}},\\\ d-1\,\,\text{is a power
of}\,\,p.\end{array}\right.$

Conversely, suppose that

$\left\\{\begin{array}[]{lll}\mathrm{N}_{q^{n}/q}(a)=(-1)^{d-1},\\\
b=-a^{q^{\prime}\sum_{i=0}^{d-1}q^{\prime id}},\\\ d-1\,\,\text{is a power
of}\,\,p,\end{array}\right.$

then by (c) of Theorem 2.1 $\dim_{{\mathbb{F}}_{q^{\prime}}}(U^{\prime})=d$.
Since $f(x)$, seen as an ${\mathbb{F}}_{q^{\prime}}$-linearized polynomial,
has maximum kernel by Theorem 1.2

$A_{n}=C_{L}C_{L}^{q^{\prime}}\cdot\ldots\cdot C_{L}^{q^{\prime n-1}}=I_{d}.$

As $C_{L}\in{\mathbb{F}}_{q^{n}}^{d\times d}$, $A_{n}$ coincides with
$C_{L}C_{L}^{\sigma}\cdot\ldots\cdot C_{L}^{\sigma^{n-1}}$ and from Theorem
1.2 follows that the ${\mathbb{F}}_{q}$-linearized polynomial $f(x)$ has
maximum kernel. ∎

##  3 Proof of Theorem 1.4

We divide two subsections: in the first we prove some preliminary results and
we develop the machinery we use, and in the last subsection we show Theorem
1.4.

###  3.1 Preliminaries results

The $\sigma$-companion matrix $C_{L}$ of $L(x)=-x^{\sigma^{d}}+bx^{\sigma}+ax$
(defined as in Theorem 1.2) is

$C_{L}=\left(\begin{matrix}0&0&\cdots&0&a\\\ 1&0&\cdots&0&b\\\
0&1&\cdots&0&0\\\ \vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&\cdots&1&0\end{matrix}\right)\in\mathbb{F}_{q^{n}}^{d\times d}.$

We define $A_{n}=C_{L}C_{L}^{\sigma}\cdots C_{L}^{\sigma^{n-1}}$, where
$C^{\sigma}$ denotes the matrix $C$ in which $\sigma$ is applied to each of
its entries. By Theorem 1.2, $\dim_{{\mathbb{F}}_{q}}(\ker(L))=d$ if and only
if $A_{n}=I_{d}$.

Let $M_{l,k}$ denote the $(l,d)$-entry of $A_{k}=C_{L}C_{L}^{\sigma}\cdots
C_{L}^{\sigma^{k-1}}$, for $l\in\\{1,\ldots,d\\}$. The recursive relation of
the next result will play a crucial role in the paper.

######  Proposition 3.1.

Set $M_{l,l-d}=1$, and $M_{l,k}=0$ for $k\leq 0$ and $k\neq l-d$.

  * •

For every $k\geq d$, the $(l,j)$-entry of $A_{k}$ is $M_{l,k-d+j}$.

  * •

For every $1\leq l\leq d$ and $k\geq 1$, the following relation holds

$M_{l,k}=M_{l,k-d}a^{\sigma^{k-1}}+M_{l,k-d+1}b^{\sigma^{k-1}}.$ (3)

###### Proof.

First note that when multiplying a row $\mathbf{v}=(v_{1},\ldots,v_{d})$ by a
Frobenius of the $\sigma$-companion matrix, namely $C^{\sigma^{i}}$, we get a
row vector which may be obtained by shifting on the left the entries of
$\mathbf{v}$ and by replacing its $d$-th entry with the element
$v_{1}a^{\sigma^{i}}+v_{2}b^{\sigma^{i}}$, that is

$(v_{1},\ldots,v_{d})\left(\begin{matrix}0&0&\cdots&0&a^{\sigma^{i}}\\\
1&0&\cdots&0&b^{\sigma^{i}}\\\ 0&1&\cdots&0&0\\\
\vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&\cdots&1&0\end{matrix}\right)=(v_{2},\ldots,v_{d},v_{1}a^{\sigma^{i}}+v_{2}b^{\sigma^{i}}).$
(4)

Then we have:

$A_{1}=A=\left(\begin{matrix}0&0&\cdots&0&M_{1,1}\\\ 1&0&\cdots&0&M_{2,1}\\\
\vdots&\vdots&&&\vdots\\\
0&0&\cdots&1&M_{d,1}\end{matrix}\right),A_{2}=\left(\begin{matrix}0&0&\cdots&M_{1,1}&M_{1,2}\\\
0&0&\cdots&M_{2,1}&M_{2,2}\\\ \vdots&\vdots&&&\vdots\\\
0&0&\cdots&M_{d,1}&M_{d,2}\end{matrix}\right),\ldots.$

So, if $k\geq d$, we have that $M_{l,k-1}$ is the $(l,d-1)$-entry of $A_{k}$,
$M_{l,k-2}$ is the $(l,d-2)$ entry of $A_{k}$ etc. In general, $M_{l,k-h}$ is
the $(l,d-h)$-entry of $A_{k}$, for any $h\in\\{0,1,\ldots,d-1\\}$, that is
$M_{l,k-d+j}$ is the $(l,j)$-entry of $A_{k}$, for $j=1,\ldots,d$, that proves
the first part of the statement.

Now, let $k\geq d$, we have

$A_{k}=A_{k-1}C^{\sigma^{k-1}}=$
$=\left(\begin{matrix}M_{1,k-1-d+1}&M_{1,k-1-d+2}&\cdots&M_{1,k-1}\\\
M_{2,k-1-d+1}&M_{2,k-1-d+2}&\cdots&M_{2,k-1}\\\ \vdots&\vdots&\vdots&\vdots\\\
M_{d,k-1-d+1}&M_{d,k-1-d+2}&\cdots&M_{d,k-1}\end{matrix}\right)\left(\begin{matrix}0&0&\cdots&0&a^{\sigma^{k-1}}\\\
1&0&\cdots&0&b^{\sigma^{k-1}}\\\ 0&1&\cdots&0&0\\\
\vdots&\vdots&\ddots&\vdots&\vdots\\\ 0&0&\cdots&1&0\end{matrix}\right)$

and then

$M_{l,k}=M_{l,k-d}a^{\sigma^{k-1}}+M_{l,k-d+1}b^{\sigma^{k-1}},$

that is Equation (3) when $k\geq d$.

Assume that $k\leq d-1$, then

$M_{l,k}=(0,\cdots,0,1,0,\cdots,0,M_{l,1},\cdots,M_{l,k-1})\cdot\left(a^{\sigma^{k-1}},b^{\sigma^{k-1}},0,\ldots,0\right)^{T}$

where the $1$ in the first vector is in the $(l-(k-1))$-position, so that
$M_{l,k}$ is zero when $l-(k-1)\notin\\{1,2\\}$. If either $l-(k-1)=1$ or
$l-(k-1)=2$, that is $k=l$ or $k=l-1$, we have $M_{l,l}=a^{\sigma^{l-1}}$ and
$M_{l,l-1}=b^{\sigma^{l-2}}$. Equation (3) then follows also for $k\leq d-1$.
∎

As a consequence of Proposition 3.1, we get the following result.

######  Proposition 3.2.

For any $i\in\\{0,\ldots,d-3\\}$ $M_{1,j}=0$, for every
$j\in\\{id+2,\ldots,(i+1)d-(i+1)\\}$.

###### Proof.

We prove the claim by induction on $i$. For $i=0$ the assertion easily follows
by (3). Note that by Equation (3), if $M_{1,k}=M_{1,k+1}=0$ then we have also
$M_{1,k+d}=0$. So, if we assume the statement holds true for
$i\in\\{0,\ldots,d-4\\}$, then we have
$M_{1,id+2+d}=0,\ldots,M_{1,(i+1)d-(i+1)+d-1}=0$, that is the statement for
$i+1$. ∎

For $M_{l,j}$’s the following holds.

######  Proposition 3.3.

For any $l\in\\{2,\ldots,d\\}$,

$M_{l,k}=\begin{cases}1&\text{if}\,\,k=l-d,\\\ 0&\text{if}\,\,k\leq
0\,\,\text{and}\,\,k\neq l-d,\\\ b^{\sigma^{l-2}}&\text{if}\,\,k=l-1,\\\
a^{\sigma^{l-1}}&\text{if}\,\,k=l\\\ 0&\text{if}\,\,1\leq k\leq d-1,\,\,k\neq
l\,\,\text{and}\,\,k\neq l-1.\end{cases}$

###### Proof.

The first part coincides with the definition of $M_{l,k}$ when $k\leq 0$. Now,
let $k\geq 0$. As $k\leq d-1$, then by applying Equation (3), $M_{l,k-d+1}\neq
0$ if and only if $k=l-1$. Similarly, $M_{l,k-d}$ is nonzero if and only if
$k=l$. The assertion then follows. ∎

Furthermore, according to (2) of Theorem 1.2, we are interested in determining
the $(l,1)$-entries of $A_{n}$. By Proposition 3.1, $(l,1)$-entries of $A_{n}$
correspond to $M_{l,n-d+1}$ for $l\in\\{1,2,\ldots,d\\}$. So, Theorem 1.2 may
be rephrased in our case as follows.

######  Corollary 3.4.

Let $n\geq 1$, $d\geq 3$, $a,b\in{\mathbb{F}}_{q^{n}}$ and
$L(x)=-x^{\sigma^{d}}+bx^{\sigma}+ax\in\mathcal{L}_{n,q,\sigma}$. Then
$\dim_{\mathbb{F}_{q}}(\ker(L(x))=d$ if and only if

$M_{l,n-d+1}=\begin{cases}1&\mbox{if }l=1,\\\ 0&\mbox{if }l=2,\ldots
d.\end{cases}$

Our aim now is to write $M_{l,k}$ as a combination of $M_{l,i}$’s which are
easier to calculate. This can be done by using the recursive formula (3).
Indeed, by applying (3), after $j$ steps $M_{l,k}$ will depend on
$M_{l,k-jd},\ldots,M_{l,k-jd+j}$, see Figure 1.

$k-3d$$k-3d+1$$k-3d+1$$k-3d+2$$k-3d+1$$k-3d+2$$k-3d+2$$k-3d+3$$k-2d$$k-2d+1$$k-2d+1$$k-2d+2$$k-d$$k-d+1$$k$
Figure 1: Recursive indexes of the $M_{i,j}$’s

Let $j$ and $t$ be two positive integers with $0\leq t\leq j$. The element
$M_{l,k-jd+t}$ appears when we apply $j$ times the recursive formula (3) to
$M_{l,k}$. Taking into account Figure 1, we note that $M_{l,k-jd+t}$ is
obtained $\binom{j}{t}$ times, each of which is obtained going right $t$ times
and going left $j-t$ times in the Figure 1.

So, by applying $j$ times (3) we have

$M_{l,k}=\sum\limits_{t=0}^{j}c^{k}_{j,t}M_{l,k-jd+t},$ (5)

where $c_{j,t}^{k}\in{\mathbb{F}}_{q^{n}}$ denotes the coefficient of
$M_{l,k-jd+t}$, which can be computed as follows. As $c^{k}_{j,t}$ is the
coefficient of $M_{l,k-jd+t}$, this means that in order to obtain
$M_{l,k-jd+t}$ we go right through Figure 1 exactly $t$ times at the steps
$i_{1},\ldots,i_{t}$ with $i_{1}\leq\ldots\leq i_{t}$. Note that when we go
right in Figure 1 we multiply by a power of $b$, otherwise we multiply by a
power of $a$. More precisely, if we first go left $i_{1}$ times in the
expression of $M_{l,k}$ will appear $M_{l,k-i_{1}d}$ multiplied by
$a^{\sigma^{k-1}}\cdot\ldots\cdot a^{\sigma^{k-(i_{1}-1)d-1}}$. Then, if we go
right in the expression of $M_{l,k}$ will appear
$a^{\sigma^{k-1}}\cdot\ldots\cdot
a^{\sigma^{k-(i_{1}-1)d-1}}M_{l,k-(i_{1}+1)d+1}$ multiplied by
$b^{\sigma^{k-i_{1}d-1}}$. Therefore, we apply this procedure $j$ times and
hence

$c^{k}_{j,t}=\sum_{\begin{subarray}{c}i_{0},\ldots,i_{t}=0\\\
i_{0}+\ldots+i_{t}=j-t\end{subarray}}^{j-t}\left(\prod_{j_{0}=1}^{i_{0}}a^{\sigma^{k-(j_{0}-1)d-1}}\right)\left(\prod_{j_{1}=1}^{i_{1}}a^{\sigma^{k-(i_{0}+j_{1})d}}\right)\cdot\ldots\cdot$
(6)
$\left(\prod_{j_{t-1}=1}^{i_{t-1}}a^{\sigma^{k-(i_{0}+\cdots+i_{t-2}+j_{t-1}+t-2)d+t-2}}\right)\cdot\left(\prod_{j_{t}=1}^{i_{t}}a^{\sigma^{k-(j-i_{t}+j_{t}-1)d+t-1}}\right)\cdot$
$\left(\prod_{u=0}^{t-1}b^{\sigma^{k-(i_{0}+\cdots+i_{u}+u)d+u-1}}\right).$

Note that

$c_{j,0}^{k}=\prod_{j_{0}=1}^{j}a^{\sigma^{k-(j_{0}-1)d-1}}\,\,\,\,\text{and}\,\,\,\,c_{j,j}^{k}=\prod_{u=0}^{j-1}b^{\sigma^{k-ud+u-1}}.$

In particular, for $k=n-d+1$, we have

$M_{l,n-d+1}=\sum\limits_{t=0}^{j}c_{j,t}M_{l,n-(j+1)d+t+1},$

where we set $c_{j,t}=c_{j,t}^{n-d+1}$.

Let $n=d(d-1)+g$ with $0\leq g<d$. We have

$M_{l,n-d+1}=\sum\limits_{t=0}^{j}c_{j,t}M_{l,(d-j-2)d+t+1+g},$

and choosing $j=d-1$, we have

$M_{l,n-d+1}=\sum_{t=0}^{d-1}c_{d-1,t}M_{l,-d+t+1+g}.$ (7)

The first case $g=1$ and $\sigma\colon x\in{\mathbb{F}}_{q^{n}}\mapsto
x^{q}\in{\mathbb{F}}_{q^{n}}$ is studied in [24].

When $a$ and $b$ satisfies certain conditions, the $c_{i,j}^{k}$ have a
simpler representation.

######  Lemma 3.5.

Suppose that $a^{\sigma^{d}}b=a^{\sigma}b^{\sigma^{d}}$, then we have

$M_{l,k}=\sum_{i=0}^{j}\binom{j}{i}z_{j,i}^{k}M_{l,k-jd+i},$

for every positive integer $k\geq d+1$ and
$j\in\\{0,\ldots,\lfloor\frac{k-1}{d}+1\rfloor\\}$. Hence, the $c_{j,i}^{k}$
defined in (5) are

$c_{j,i}^{k}=\binom{j}{i}z_{j,i}^{k},$

and the $z^{k}_{j,i}$ are determined by the following recursion:

$z_{j,i}^{k}=\begin{cases}1&\mbox{if }j=i=0,\\\
z_{j-1,0}^{k}a^{\sigma^{k-(j-1)d-1}}&\mbox{if }i=0,\\\
z_{j-1,i}^{k}a^{\sigma^{k-(j-1)d+i-1}}=z_{j-1,i-1}^{k}b^{\sigma^{k-(j-1)d+i-2}}&\mbox{if
}0<i<j,\\\ z_{j-1,j-1}^{k}b^{\sigma^{k-(j-1)d+j-2}}&\mbox{if }i=j.\end{cases}$
(8)

###### Proof.

First observe that when $k\geq d+1$ then
$a^{\sigma^{k-1}}b^{\sigma^{k-d-1}}=a^{\sigma^{k-d}}b^{\sigma^{k-1}}$. We will
prove the statement by induction. The case $j=1$ follows by (3); now we prove
the assertion for $j=2$. Let $k\geq d+1$. By Equation (3),

$\begin{split}M_{l,k}&=M_{l,k-d}a^{\sigma^{k-1}}+M_{l,k-d+1}b^{\sigma^{k-1}}\\\
&=(M_{l,k-2d}a^{\sigma^{k-d-1}}+M_{l,k-2d+1}b^{\sigma^{k-d-1}})a^{\sigma^{k-1}}+\\\
&\hskip
14.22636pt(M_{l,k-2d+1}a^{\sigma^{k-d}}+M_{l,k-2d+2}b^{\sigma^{k-d}})b^{\sigma^{k-1}}\\\
&=M_{l,k-2d}a^{\sigma^{k-d-1}+\sigma^{k-1}}+M_{l,k-2d+1}(a^{\sigma^{k-1}}b^{\sigma^{k-d-1}}+a^{\sigma^{k-d}}b^{\sigma^{k-1}})\\\
&\hskip 14.22636pt+M_{l,k-2d+2}b^{\sigma^{k-d}+\sigma^{k-1}}\end{split}$ (9)

As $a^{\sigma^{k-1}}b^{\sigma^{k-d-1}}=a^{\sigma^{k-d}}b^{\sigma^{k-1}}$,
putting $z^{k}_{2,0}=a^{\sigma^{k-d-1}+\sigma^{k-1}}$,
$z^{k}_{2,1}=2a^{\sigma^{k-d}}b^{\sigma^{k-1}}$ and
$z^{k}_{2,2}=b^{\sigma^{k-d}+\sigma^{k-1}}$, (9) may be rewritten as follows

$M_{l,k}=z^{k}_{2,0}M_{l,k-2d}+2z^{k}_{2,1}M_{l,k-2d+1}+z^{k}_{2,2}M_{l,k-2d+2},$

and the statement for $j=2$ is proved. Now, assume that the statement holds
for any $t<j$, that is

$M_{l,k}=\sum_{i=0}^{t}\binom{t}{i}z_{t,i}^{k}M_{l,k-td+i},$

where $z_{t,i}^{k}$ are defined as in (8). Then by applying (3) to
$M_{l,k-(j-1)d+i}$ we have

$\begin{split}M_{l,k}&=\sum_{i=0}^{j-1}\binom{j-1}{i}z^{k}_{j-1,i}M_{l,k-(j-1)d+i}\\\
&=\sum_{i=0}^{j-1}\binom{j-1}{i}z^{k}_{j-1,i}(M_{l,k-jd+i}a^{\sigma^{k-(j-1)d+i-1}}+M_{l,k-jd+i+1}b^{\sigma^{k-(j-1)d+i-1}})\\\
&=\binom{j-1}{0}z^{k}_{j-1,0}M_{l,k-jd}a^{\sigma^{k-(j-1)d-1}}+\binom{j-1}{j-1}z^{k}_{j-1,j-1}M_{l,k-jd+j}b^{\sigma^{k-(j-1)d+j-2}}\\\
&+\sum_{i=1}^{j-1}M_{l,k-jd+i}\left(\binom{j-1}{i}z^{k}_{j-1,i}a^{\sigma^{k-(j-1)d+i-1}}+\binom{j-1}{i-1}z^{k}_{j-1,i-1}b^{\sigma^{k-(j-1)d+i-2}}\right).\end{split}$

By hypothesis we have

$a^{\sigma^{k-(j-2)d+i-2}}b^{\sigma^{k-(j-1)d+i-2}}=a^{\sigma^{k-(j-1)d+i-1}}b^{\sigma^{k-(j-2)d+i-2}}.$
(10)

Furthermore,

$z_{j-1,i}^{k}a^{\sigma^{k-(j-1)d+i-1}}=z_{j-2,i-1}^{k}a^{\sigma^{k-(j-1)d+i-1}}b^{\sigma^{k-(j-2)d+i-2}}$

and

$z_{j-1,i-1}^{k}b^{\sigma^{k-(j-1)d+i-2}}=z_{j-2,i-1}^{k}a^{\sigma^{k-(j-2)d+i-2}}b^{\sigma^{k-(j-1)d+i-2}}$

and by (10), these two expressions are equal. As
$\binom{j-1}{i}+\binom{j-1}{i-1}=\binom{j}{i}$ we get

$M_{l,k}=\binom{j}{0}z^{k}_{j-1,0}M_{l,k-jd}a^{\sigma^{k-(j-1)d-1}}+\binom{j}{j}z^{k}_{j-1,j-1}M_{l,k-jd+j}b^{\sigma^{k-(j-1)d+j-2}}$
$+\sum_{i=1}^{j-1}\binom{j}{i}z^{k}_{j-1,i}M_{l,k-jd+i}a^{\sigma^{k-(j-1)d+i-1}},$

that is the statement. ∎

###  3.2 Proof of Theorem 1.4

Let $n=(d-1)d+g$ and $1\leq g\leq d-1$. We also remind that in Proposition
3.1, we set $M_{l,l-d}=1$, and $M_{l,k}=0$ for $k\leq 0$ and $k\neq l-d$.

_Proof of Theorem 1.4_
Let start by recalling that, by (7), we have

$M_{l,n-d+1}=\sum_{t=0}^{d-1}c_{d-1,t}M_{l,-d+1+g+t}.$ (11)

By Theorem 1.2, $\dim_{{\mathbb{F}}_{q}}(\ker(L(x)))=d$ if and only if
$M_{l,n-d+1}=\begin{cases}1&\text{if}\,\,l=1,\\\ 0&\text{if}\,\,l\neq
1.\end{cases}$ So, (11) and Proposition 3.3 imply that if $g\leq d-2$

$\begin{cases}1=M_{1,n-d+1}=M_{1,1}c_{d-1,d-g}=ac_{d-1,d-g}\\\
0=M_{2,n-d+1}=M_{2,1}c_{d-1,d-g}+M_{2,2}c_{d-1,d-g+1}\\\ \hskip
67.7177pt=bc_{d-1,d-g}+a^{\sigma}c_{d-1,d-g+1}\\\ \vdots\\\
0=M_{g,n-d+1}=c_{d-1,d-2}M_{g,g-1}+c_{d-1,d-1}M_{g,g}\\\ \hskip
67.7177pt=b^{\sigma^{g-2}}c_{d-1,d-2}+a^{\sigma^{g-1}}c_{d-1,d-1}\\\
0=M_{g+1,n-d+1}=c_{d-1,d-1}M_{g+1,g}+c_{d-1,0}M_{g+1,-d+g+1}\\\ \hskip
78.81425pt=b^{\sigma^{g-1}}c_{d-1,d-1}+c_{d-1,0}\\\
0=M_{g+2,n-d+1}=c_{d-1,1}M_{g+2,-d+1+g+1}=c_{d-1,1}\\\ \vdots\\\
0=M_{d,n-d+1}=c_{d-1,d-g-1}\end{cases}$

and if $g=d-1$

$\begin{cases}1=M_{1,n-d+1}=M_{1,1}c_{d-1,d-g}=ac_{d-1,1}\\\
0=M_{2,n-d+1}=M_{2,1}c_{d-1,1}+M_{2,2}c_{d-1,2}=bc_{d-1,1}+a^{\sigma}c_{d-1,2}\\\
\vdots\\\ 0=M_{d-1,n-d+1}=c_{d-1,d-2}M_{d-1,d-2}+c_{d-1,d-1}M_{d-1,d-1}\\\
\hskip 79.6678pt=b^{\sigma^{d-3}}c_{d-1,d-2}+a^{\sigma^{d-2}}c_{d-1,d-1}\\\
0=M_{d,n-d+1}=c_{d-1,d-1}M_{d,d-1}+c_{d-1,0}M_{d,0}\\\ \hskip
68.28644pt=b^{\sigma^{d-2}}c_{d-1,d-1}+c_{d-1,0},\end{cases}$

which can be written as

$\begin{cases}c_{d-1,d-g}=a^{-1}\\\ c_{d-1,d-g+1}=-ba^{-1}a^{-\sigma}\\\
\vdots\\\ c_{d-1,d-g+u}=(-1)^{u}b^{\sigma^{u-1}}b^{\sigma^{u-2}}\cdots
b^{\sigma}ba^{-1}a^{-\sigma}\cdots a^{-\sigma^{u}}\\\ \vdots\\\
c_{d-1,d-1}=(-1)^{g-1}b^{\sigma^{g-2}}b^{\sigma^{g-3}}\cdots
b^{\sigma}ba^{-1}a^{-\sigma}\cdots a^{-\sigma^{g-1}}\\\ \hskip
41.25641pt=(-1)^{g-1}b^{\frac{\sigma^{g-1}-1}{\sigma-1}}a^{-\frac{\sigma^{g}-1}{\sigma-1}}\\\
c_{d-1,0}=(-1)^{g}b^{\sigma^{g-1}}b^{\sigma^{g-2}}\cdots
b^{\sigma}ba^{-1}a^{-\sigma}\cdots a^{-\sigma^{g-1}}\\\ \hskip
29.87547pt=(-1)^{g}b^{\frac{\sigma^{g}-1}{\sigma-1}}a^{-\frac{\sigma^{g}-1}{\sigma-1}}\\\
c_{d-1,1}=0\\\ \vdots\\\ c_{d-1,d-g-1}=0\end{cases}$

if $g\leq d-2$ and

$\begin{cases}c_{d-1,1}=a^{-1}\\\ c_{d-1,2}=-ba^{-1}a^{-\sigma}\\\ \vdots\\\
c_{d-1,1+u}=(-1)^{u}\prod_{i=0}^{u-1}b^{\sigma^{i}}\prod_{i=0}^{u}a^{\sigma^{i}}\\\
\vdots\\\ c_{d-1,d-1}=(-1)^{d-2}b^{\sigma^{d-3}}b^{\sigma^{d-4}}\cdots
b^{\sigma}ba^{-1}a^{-\sigma}\cdots a^{-\sigma^{d-2}}\\\ \hskip
41.25641pt=(-1)^{d-2}b^{\frac{\sigma^{d-2}-1}{\sigma-1}}a^{-\frac{\sigma^{d-1}-1}{\sigma-1}}\\\
c_{d-1,0}=(-1)^{d-1}b^{\sigma^{d-2}}b^{\sigma^{d-3}}\cdots
b^{\sigma}ba^{-1}a^{-\sigma}\cdots a^{-\sigma^{d-2}}\\\ \hskip
29.59067pt=(-1)^{d-1}b^{\frac{\sigma^{d-1}-1}{\sigma-1}}a^{-\frac{\sigma^{d-1}-1}{\sigma-1}}\\\
\end{cases}$ (12)

if $g=d-1$.

In order to determine the $c_{d-1,i}$’s we use Equation (6) with $j=d-1$ and
$k=n-d+1$ and we get

$c^{k}_{j,t}=\sum_{\begin{subarray}{c}i_{0},\ldots,i_{t}=0\\\
i_{0}+\ldots+i_{t}=j-t\end{subarray}}^{j-t}\left(\prod_{j_{0}=1}^{i_{0}}a^{\sigma^{k-(j_{0}-1)d-1}}\right)\left(\prod_{j_{1}=1}^{i_{1}}a^{\sigma^{k-(i_{0}+j_{1})d}}\right)\cdot\ldots\cdot$
$\left(\prod_{j_{t-1}=1}^{i_{t-1}}a^{\sigma^{k-(i_{0}+\cdots+i_{t-2}+j_{t-1}+t-2)d+t-2}}\right)\cdot\left(\prod_{j_{t}=1}^{i_{t}}a^{\sigma^{k-(j-i_{t}+j_{t}-1)d+t-1}}\right)\cdot$
$\left(\prod_{u=0}^{t-1}b^{\sigma^{k-(i_{0}+\cdots+i_{u}+u)d+u-1}}\right).$

Hence, by looking to the previous systems and the above relation we have

$c_{d-1,t}=\sum_{\begin{subarray}{c}i_{0},\ldots,i_{t}=0\\\
i_{0}+\ldots+i_{t}=d-1-t\end{subarray}}^{d-1-t}\left(\prod_{j_{0}=1}^{i_{0}}a^{\sigma^{d(d-1-j_{0})+g}}\right)\left(\prod_{j_{1}=1}^{i_{1}}a^{\sigma^{d(d-i_{0}-j_{1}-2)+g+1}}\right)\cdot\ldots\cdot$
$\left(\prod_{j_{t-1}=1}^{i_{t-1}}a^{\sigma^{d(d-i_{0}+\cdots-
i_{t-2}-j_{t-1}-t)+g+t-1}}\right)\cdot\left(\prod_{j_{t}=1}^{i_{t}}a^{\sigma^{d(i_{t}-j_{t})+t+g}}\right)\cdot$
$\left(\prod_{u=0}^{t-1}b^{\sigma^{d(d-2-i_{0}-\cdots-i_{u}-u)+u+g}}\right)$
$=\left\\{\begin{array}[]{lll}(-1)^{v}b^{\frac{\sigma^{v}-1}{\sigma-1}}a^{-\frac{\sigma^{v+1}-1}{\sigma-1}}&\text{if}\,\,t=d-g+v,\,\,0\leq
v\leq g-1,\\\
(-1)^{g}a^{-\frac{\sigma^{g}-1}{\sigma-1}}b^{\frac{\sigma^{g}-1}{\sigma-1}}&\text{if}\,\,t=0,\\\
0&\text{otherwise},\end{array}\right.$

which proves the assertion. ∎

######  Remark 3.6.

We point out that the proof of (c) of Theorem 1.3 can be also obtained by
manipulating the conditions of Theorem 1.4.

##  4 Some necessary conditions

In this section we will give some necessary condition on the coefficients of
$L(x)=ax+bx^{\sigma}-x^{\sigma^{d}}\in\mathcal{L}_{n,q,\sigma}$ to have kernel
of maximum dimension. In order to do this, we find the coefficients
$c_{d-1,0}$ and $c_{d-1,d-1}$. Indeed, the coefficient $c_{d-1,0}^{k}$ can be
obtained by applying (6) with $t=0$ and $j=d-1$, that is

$c_{d-1,0}^{k}=a^{\sigma^{k-1}+\sigma^{k-d-1}+\ldots+\sigma^{k-(d-2)d-1}}.$

With $k=n-d+1$, we get

$c_{d-1,0}=c_{d-1,0}^{k}=a^{\sigma^{g}(\sigma^{d(d-2)}+\ldots+\sigma^{d}+1)}=a^{\sigma^{g}e_{1}},$
(13)

where $e_{1}=\frac{\sigma^{d(d-1)}-1}{\sigma^{d}-1}$. Similarly, the
coefficient $c_{d-1,d-1}^{k}$ can be obtained by applying (6) with $t=d-1$ and
$j=d-1$, that is

$c_{d-1,d-1}^{k}=b^{\sigma^{k-1}+\sigma^{k-d}+\ldots+\sigma^{k-(d-2)d+d-3}}.$

With $k=n-d+1$, we get

$c_{d-1,d-1}=c_{d-1,d-1}^{n-d+1}=b^{\sigma^{g-1}(\sigma^{d-1}+\ldots+\sigma^{(d-1)(d-1)})}=b^{\sigma^{g-1}e_{2}},$
(14)

where $e_{2}=\frac{\sigma^{d(d-1)}-\sigma^{d-1}}{\sigma^{d-1}-1}$.

As a consequence of, we get the following necessary conditions.

######  Corollary 4.1.

Let $n=(d-1)d+g$, with $1\leq g\leq d-1$. If $L(x)$ has maximum kernel then

$\begin{cases}b^{\frac{\sigma^{g}-1}{\sigma-1}}a^{-\frac{\sigma^{g}-1}{\sigma-1}}=(-1)^{g}a^{\sigma^{g}e_{1}},\\\
b^{\frac{\sigma^{g-1}-1}{\sigma-1}}a^{-\frac{\sigma^{g}-1}{\sigma-1}}=(-1)^{g-1}b^{\sigma^{g-1}e_{2}}.\end{cases}$

##  5 A new family of $\sigma$-linearized trinomials with maximum kernel

In this section we introduce a new family of $\sigma$-linearized trinomials
with maximum kernel, relying on the results of Section 3. In particular we
prove the following.

######  Theorem 5.1.

Let $L(x)=-x^{\sigma^{d}}+bx^{\sigma}+ax\in\mathcal{L}_{n,q,\sigma}$, with
$n=(d-1)d+d-1$. If

  * •

$q$ is even,

  * •

$d$ is a power of $2$,

  * •

$\mathrm{N}_{q^{n}/q}(a)=1$,

  * •

$b=a^{-\sigma^{d}\frac{\sigma^{d(d-1)}-1}{\sigma^{d}-1}}$,

then $\dim_{{\mathbb{F}}_{q}}(\ker(L))=d$.

######  Lemma 5.2.

If $a,b\in\mathbb{F}_{q^{n}}$ with
$b=a^{-\sigma^{d}\frac{\sigma^{d(d-1)}-1}{\sigma^{d}-1}}$ then

$a^{\sigma^{d}}b=a^{\sigma}b^{\sigma^{d}},$

and hence

$a^{\sigma^{k-1}}b^{\sigma^{k-d-1}}=a^{\sigma^{k-d}}b^{\sigma^{k-1}},$

for every positive integer $k\geq d+1$. Moreover, for every
$r\in\\{1,\ldots,d-1\\}$ and $u\leq r-1$,

$b^{\sigma^{d(r-u)+u-1}}a^{-\sum_{i=1}^{r-u}\sigma^{id+(u-1)}}=a^{-\sigma^{u}}b^{-\sigma^{u-1}}a^{-\sum_{i=1}^{r-u-1}\sigma^{id+u}}.$

###### Proof.

We have

$\begin{split}a^{\sigma^{d}}b&=a^{\sigma^{d}-\sum_{i=1}^{d-1}\sigma^{id}}\\\
&=a^{-\sum_{i=2}^{d-1}\sigma^{id}}\\\
&=a^{-\sum_{i=2}^{d}\sigma^{id}+\sigma^{d^{2}}}\\\
&=a^{-\sum_{i=2}^{d}\sigma^{id}+\sigma}\pmod{a^{\sigma^{n}}-a}\\\
&=a^{\sigma}a^{-\sigma^{d}\sum_{i=2}^{d}\sigma^{(i-1)d}}\\\
&=a^{\sigma}b^{\sigma^{d}}.\\\ \end{split}$

For the second part, using
$b=a^{-\sigma^{d}\frac{\sigma^{d(d-1)}-1}{\sigma^{d}-1}}$,
$\sigma^{n}=\mathrm{id}$ and $n=d^{2}-1$, we have

$\begin{split}b^{\sigma^{d(r-u)+u-1}}&=a^{-\sum_{i=1}^{d-1}\sigma^{d(r-u+i)+u-1}}\\\
&=a^{-\sigma^{u-1}(\sum_{i=1}^{u-r+d-1}\sigma^{d(r-u+i)}+\sum_{i=u-r+d}^{d-1}\sigma^{d(r-u+i)})}\\\
&=a^{-\sigma^{u-1}(\sum_{i=1}^{u-r+d-1}\sigma^{d(r-u+i)})}a^{-\sigma^{u-1}(\sum_{i=u-r+d}^{d-1}\sigma^{d(r-u+i)})},\end{split}$

and note that $d+u-r-1\leq d-2$ and $u+d-r-1\geq u\geq 1$.

Putting together $a^{-\sum_{i=1}^{r-u}\sigma^{id+(u-1)}}$ and
$a^{-\sigma^{u-1}(\sum_{i=1}^{u-r+d-1}\sigma^{d(r-u+i)})}$ we obtain
$b^{\sigma^{u-1}}$.

Now consider $a^{-\sigma^{u-1}(\sum_{i=u-r+d}^{d-1}\sigma^{d(r-u+i)})}$. Then
it is equal to

$a^{-\sigma^{u-1}(\sum_{i=d}^{d-1+r-u}\sigma^{id})}=a^{-\sigma^{u-1}(\sum_{i=0}^{r-u-1}\sigma^{id+1})}=a^{-\sigma^{u}}a^{-\sum_{i=1}^{r-u-1}\sigma^{id+u}}.$

So, the assertion has been proved. ∎

The first part of the previous lemma allows us to apply Lemma 3.5.

######  Remark 5.3.

With $k=n-d+1$ and $n=d(d-1)+d-1=d^{2}-1$, $z_{j,i}^{k}$ in (8), setting
$z_{j,i}=z_{j,i}^{n-d+1}$, becomes:

$z_{j,i}=z_{j,i}^{n-d+1}=\begin{cases}1&\mbox{if }j=i=0,\\\
z_{j-1,0}a^{\sigma^{d(d-j)-1}}&\mbox{for }i=0,\\\
z_{j-1,i}a^{\sigma^{d(d-j)+i-1}}=z_{j-1,i-1}b^{\sigma^{d(d-j)+i-2}}&\mbox{for
}0<i<j,\\\ z_{j-1,j-1}b^{\sigma^{d(d-j)+j-2}}&\mbox{for }i=j.\end{cases}$ (15)

Furthermore, by Lemma 3.5 we have

$c_{j,i}=\binom{j}{i}z_{j,i}.$

In the next lemma we compute the $c_{d-1,i}$’s.

######  Lemma 5.4.

For any $r\in\\{2,\ldots,d-1\\}$,

$c_{d-1,r}=\binom{d-1}{r}\prod_{v=1}^{r-1}b^{\sigma^{(r-v)d+v-1}}a^{-\sum_{i=1}^{r-1}\sigma^{id}}a^{-1}.$

and

$c_{d-1,1}=\binom{d-1}{1}a^{-1}.$ (16)

###### Proof.

By Remark 5.3, for any $r\geq 2$

$\begin{split}c_{d-1,r}&=\binom{d-1}{r}b^{\sigma^{d+r-2}}z_{d-2,r-1}\\\
&=\binom{d-1}{r}b^{\sigma^{d+r-2}}b^{\sigma^{2d+r-3}}z_{d-3,r-2}\\\
&=\binom{d-1}{r}\prod_{v=0}^{r-1}b^{\sigma^{(r-v)d+v-1}}z_{d-(r+1),0}\\\
\end{split}$

and then applying Remark 5.3 to the case $z_{j,0}$ it follows

$\begin{split}c_{d-1,r}&=\binom{d-1}{r}\prod_{v=0}^{r-1}b^{\sigma^{(r-v)d+v-1}}a^{\sigma^{d(d-(d-(r+1)))-1}+\cdots+\sigma^{d(d-1)-1}}\\\
&=\binom{d-1}{r}\prod_{v=0}^{r-1}b^{\sigma^{(r-v)d+v-1}}a^{\sigma^{d(r+1)-1}+\cdots+\sigma^{d(d-1)-1}}\\\
\end{split}$

From $v=0$, in the product appears
$b^{\sigma^{rd-1}}a^{\sigma^{d(r+1)-1}+\cdots+\sigma^{d(d-1)-1}}$ which, using
$b=a^{-\sigma^{d}\frac{\sigma^{d(d-1)}-1}{\sigma^{d}-1}}$, equals to

$a^{-\sigma^{d^{2}-1}-\sigma^{d(d+1)-1}-\cdots-\sigma^{d(d-1+r)-1}}=a^{-1}a^{-\sigma^{d}-\cdots-\sigma^{d(-1+r)}},$

so that

$c_{d-1,r}=\binom{d-1}{r}\prod_{v=1}^{r-1}b^{\sigma^{(r-v)d+v-1}}a^{-\sum_{i=1}^{r-1}\sigma^{id}}a^{-1}.$

When $r=1$

$\begin{split}c_{d-1,1}&=\binom{d-1}{1}b^{\sigma^{d-1}}z_{d-2,0}\\\
&=\binom{d-1}{1}b^{\sigma^{d-1}}a^{\sigma^{2d-1}}a^{\sigma^{3d-1}}\cdots
a^{\sigma^{d(d-1)-1}}\\\
&=\binom{d-1}{1}a^{-\sum_{i=1}^{d-1}\sigma^{(i+1)d-1}+\sigma^{2d-1}+\sigma^{3d-1}+\cdots+\sigma^{d(d-1)-1}}\\\
&=\binom{d-1}{1}a^{-\sigma^{d^{2}-1}}\\\ &=\binom{d-1}{1}a^{-1}.\end{split}$
(17)

∎

Now we are ready to prove Theorem 5.1.

_Proof of Theorem 5.1_
By Lemma 5.4, for any $r\geq 2$ it follows

$c_{d-1,r}=\binom{d-1}{r}\prod_{v=1}^{r-1}b^{\sigma^{(r-v)d+v-1}}a^{-\sum_{i=1}^{r-1}\sigma^{id}}a^{-1}.$

Applying $r-1$ times Lemma 5.2 to $c_{d-1,r}$ we have that

$\begin{split}c_{d-1,r}&=\binom{d-1}{r}b^{\sigma^{(r-1)d}}a^{-\sum_{i=1}^{r-1}\sigma^{id}}\prod_{v=2}^{r-1}b^{\sigma^{(r-v)d+v-1}}a^{-1}\\\
&=\binom{d-1}{r}\prod_{v=2}^{r-1}b^{\sigma^{(r-v)d+v-1}}a^{-\sum_{i=1}^{r-2}\sigma^{id+1}}ba^{-\sigma-1}\\\
&=\vdots\\\
&=\binom{d-1}{r}\prod_{i=0}^{r-2}b^{\sigma^{i}}\prod_{i=0}^{r-1}a^{-\sigma^{i}},\end{split}$
(18)

for $r\geq 2$.

Recall that $L(x)$ has maximum kernel if and only if $a$ and $b$ satisfy
System (12). So, substituting (17) and (18) in (12) we obtain

$\begin{cases}\binom{d-1}{1}a^{-1}=a^{-1}\\\ \vdots\\\
\binom{d-1}{r}\prod_{i=0}^{r-2}b^{\sigma^{i}}\prod_{i=0}^{r-1}a^{-\sigma^{i}}=\prod_{i=0}^{r-2}b^{\sigma^{i}}\prod_{i=0}^{r-1}a^{-\sigma^{i}}\\\
\vdots\\\
c_{d-1,0}=b^{\frac{\sigma^{d-1}-1}{\sigma-1}}a^{-\frac{\sigma^{d-1}-1}{\sigma-1}}\\\
\end{cases}$ (19)

From (13) we have

$c_{d-1,0}=a^{\frac{\sigma^{d^{2}-1}-\sigma^{d-1}}{\sigma^{d}-1}},$

so that the last equation of (19), since $\sigma^{d^{2}}=\sigma$, becomes

$a^{\frac{\sigma^{d^{2}-1}-\sigma^{d-1}}{\sigma^{d}-1}}=b^{\frac{\sigma^{d-1}-1}{\sigma-1}}a^{-\frac{\sigma^{d-1}-1}{\sigma-1}},$

that is

$a^{\frac{\sigma^{d^{2}-1}-\sigma^{d-1}}{\sigma^{d}-1}}=a^{-\frac{\sigma^{d^{2}}-\sigma^{d}}{\sigma^{d}-1}\frac{\sigma^{d-1}-1}{\sigma-1}}a^{-\frac{\sigma^{d-1}-1}{\sigma-1}},$

which is equivalent to $\mathrm{N}_{q^{n}/q}(a)=1$ and so the last equation of
System (19) is satisfied.

By Lucas’s Theorem we have

$\binom{d-1}{r}=\prod_{v=0}^{u}\binom{d_{v}}{r_{v}}\pmod{2},$

where $d-1=d_{u}2^{u}+\ldots+d_{1}2+d_{0}$ and
$r=r_{u}2^{u}+\ldots+r_{1}2+r_{0}$. As $d$ is a power of $2$, namely
$d=2^{u+1}$, then

$d-1=2^{u}+\ldots+2+1,$

so that $d_{v}=1$ for each $v\in\\{0,\ldots,u\\}$ and Lucas’s Theorem implies
that $\binom{d-1}{r}=1$. Therefore, the remaining equations of System (19) are
satisfied. ∎

##  6 $\sigma$-degree $3$ and $4$

In this section we will make use of Theorems 1.3 and 1.4 in order to to give
explicit relations on $a,b\in{\mathbb{F}}_{q^{n}}$ ensuring that the
polynomial $L(x)=ax+bx^{\sigma}-x^{\sigma^{d}}$ has maximum kernel when
$d\in\\{3,4\\}$.

######  Proposition 6.1.

Let $L(x)=ax+bx^{\sigma}-x^{\sigma^{3}}\in\mathcal{L}_{n,q,\sigma}$ with
$\sigma$ a generator of
$\mathrm{Gal}({\mathbb{F}}_{q^{n}}\colon{\mathbb{F}}_{q})$.

  * •

If $n\leq 5$ and $n\neq 3$ then $\dim_{{\mathbb{F}}_{q}}(\ker(L(x)))<3$;

  * •

if $n\in\\{3,6\\}$ then $\dim_{{\mathbb{F}}_{q}}(\ker(L(x)))=3$ if and only if
$b=0$ and $\mathrm{N}_{q^{6}/q^{3}}(a)=1$;

  * •

if $n=7$ then $\dim_{{\mathbb{F}}_{q}}(\ker(L(x)))=3$ if and only if

$\begin{cases}\mathrm{N}_{q^{7}/q}(a)=1,\\\
b=-a^{\sigma(1+\sigma^{3}+\sigma^{6})},\\\ q\,\,\text{is even};\end{cases}$

  * •

if $n=8$ then $\dim_{{\mathbb{F}}_{q}}(\ker(L(x)))=3$ if and only if

$\begin{cases}\mathrm{N}_{q^{8}/q}(a)=1,\\\ q\equiv 0,2\pmod{3},\\\
b=\frac{-\alpha}{a^{\sigma^{6}+\sigma^{3}}},\\\ \alpha\,\,\text{is a root
of}\,\,x^{2}+x+1.\end{cases}$

###### Proof.

By Theorem 1.3, the only part that remains to prove is the last point. Let
$n=8$. By Theorem 1.4, $\dim_{{\mathbb{F}}_{q}}(\ker(L(x)))=3$ if and only if

$\begin{cases}b^{\sigma^{5}}a^{\sigma^{3}}+a^{\sigma^{5}}b^{\sigma^{2}}=a^{-1},\\\
b^{\sigma^{5}+\sigma^{3}}=-ba^{-1-\sigma},\\\
a^{\sigma^{5}+\sigma^{2}}=b^{\sigma+1}a^{-1-\sigma},\end{cases}$ (20)

from which clearly we have $a,b\neq 0$. System (20) is equivalent to the
following system

$\begin{cases}b^{\sigma^{5}}a^{\sigma^{3}+1}+a^{\sigma^{5}+1}b^{\sigma^{2}}=1,\\\
b^{\sigma^{5}+\sigma^{3}+\sigma}=-a^{\sigma^{5}+\sigma^{2}},\\\
a^{\sigma^{5}+\sigma^{2}+\sigma+1}=b^{\sigma+1}.\end{cases}$ (21)

Clearly, by the second equation of the above system we also have
$b^{\sigma^{6}+\sigma^{4}+\sigma^{2}}=-a^{\sigma^{6}+\sigma^{3}}$ and
$b^{\sigma^{7}+\sigma^{5}+\sigma^{3}}=-a^{\sigma^{7}+\sigma^{4}}$, and hence
by the third equation of System (21) the equation
$\mathrm{N}_{q^{8}/q}(a)=\mathrm{N}_{q^{8}/q}(b)$. Therefore, System (21) is
equivalent to

$\begin{cases}b^{\sigma^{5}}a^{\sigma^{3}+1}+a^{\sigma^{5}+1}b^{\sigma^{2}}=1,\\\
b^{\sigma^{5}+\sigma^{3}+\sigma}=-a^{\sigma^{5}+\sigma^{2}},\\\
\mathrm{N}_{q^{8}/q}(a)=\mathrm{N}_{q^{8}/q}(b),\end{cases}$

which is equivalent to

$\begin{cases}-b^{\sigma^{5}}b^{\sigma^{3}+\sigma+\sigma^{7}}-b^{1+\sigma^{6}+\sigma^{4}}b^{\sigma^{2}}=1,\\\
b^{\sigma^{5}+\sigma^{3}+\sigma}=-a^{\sigma^{5}+\sigma^{2}},\\\
\mathrm{N}_{q^{8}/q}(a)=\mathrm{N}_{q^{8}/q}(b).\end{cases}$ (22)

By multiplying the second equation by $b^{\sigma^{7}}$ and rewriting the first
equation, System (22) becomes

$\begin{cases}\mathrm{N}_{q^{8}/q^{2}}(b)+\mathrm{N}_{q^{8}/q^{2}}(b)^{\sigma}=-1,\\\
a^{\sigma^{4}+\sigma}b^{\sigma^{6}}=-\mathrm{N}_{q^{8}/q^{2}}(b),\\\
\mathrm{N}_{q^{8}/q^{2}}(a)=\mathrm{N}_{q^{8}/q^{2}}(b).\end{cases}$

We now use that $\mathrm{N}_{q^{8}/q}(a)=1$ because of Theorem 1.1, so that
the above system can be written as follows

$\begin{cases}\mathrm{N}_{q^{8}/q^{2}}(b)+\mathrm{N}_{q^{8}/q^{2}}(b)^{\sigma}=-1,\\\
a^{\sigma^{4}+\sigma}b^{\sigma^{6}}=-\mathrm{N}_{q^{8}/q^{2}}(b),\\\
\mathrm{N}_{q^{8}/q}(b)=\mathrm{N}_{q^{8}/q^{2}}(b)^{\sigma+1}=1,\\\
\mathrm{N}_{q^{8}/q}(a)=1.\end{cases}$ (23)

Assume now that $\mathrm{N}_{q^{8}/q^{2}}(b)\in{\mathbb{F}}_{q}$, the first
equation of the above system implies that $q$ cannot be even and
$\mathrm{N}_{q^{8}/q^{2}}(b)=-\frac{1}{2}$, so that the third equation of
System (23) yields $q$ to be a power of $3$ and System (23) becomes

$\begin{cases}b=\frac{-1}{a^{\sigma^{6}+\sigma^{3}}},\\\
\mathrm{N}_{q^{8}/q}(a)=1.\end{cases}$

Suppose that $\mathrm{N}_{q^{8}/q^{2}}(b)\notin{\mathbb{F}}_{q}$. By the first
and the third equation of System (23) then $\mathrm{N}_{q^{8}/q^{2}}(b)$ and
$\mathrm{N}_{q^{8}/q^{2}}(b)^{\sigma}$ are two distinct roots of

$x^{2}+x+1=0.$

This equation has two roots not in $\mathbb{F}_{q}$ if and only if $q\equiv
2\pmod{3}$ by [18, (xi) Section 1.5]. So, System (23) admits at least a
solution if and only if $q\equiv 2\pmod{3}$ and the solutions are

$\begin{cases}b=\frac{-\alpha}{a^{\sigma^{6}+\sigma^{3}}},\\\
\mathrm{N}_{q^{8}/q}(a)=1.\end{cases}$

This completes the proof. ∎

######  Proposition 6.2.

Let $L(x)=ax+bx^{\sigma}-x^{\sigma^{4}}\in\mathcal{L}_{n,q,\sigma}$ with
$\sigma$ a generator of
$\mathrm{Gal}({\mathbb{F}}_{q^{6}}\colon{\mathbb{F}}_{q})$.

  * •

If $n\leq 11$ and $n\neq 4,8$ then $\dim_{{\mathbb{F}}_{q}}(\ker(L(x)))<4$;

  * •

if $n\in\\{4,8,12\\}$ then $\dim_{{\mathbb{F}}_{q}}(\ker(L(x)))=4$ if and only
if $b=0$ and $\mathrm{N}_{q^{n}/q^{4}}(a)=1$;

  * •

if $n=13$ then $\dim_{{\mathbb{F}}_{q}}(\ker(L(x)))=4$ if and only if

$\begin{cases}\mathrm{N}_{q^{n}/q}(a)=-1,\\\
b=-a^{\sigma(1+\sigma^{4}+\sigma^{8}+\sigma^{12})},\\\ q\,\,\text{is a power
of}\,\,3;\end{cases}$

  * •

if $n=14$ then $\dim_{{\mathbb{F}}_{q}}(\ker(L(x)))=4$ if and only if

$\begin{cases}ac_{3,2}=1,\\\ a^{1+\sigma}b^{\sigma^{6}+\sigma^{3}}=-b,\\\
a^{1+\sigma+\sigma^{2}+\sigma^{6}+\sigma^{10}}=b^{1+\sigma},\\\
c_{3,1}=0,\end{cases}$

where

$c_{3,2}=a^{\sigma^{10}}b^{\sigma^{6}}b^{\sigma^{3}}+b^{\sigma^{10}}a^{\sigma^{7}}b^{\sigma^{3}}+b^{\sigma^{10}}b^{\sigma^{7}}a^{\sigma^{4}},$

and

$c_{3,1}=a^{\sigma^{10}}a^{\sigma^{6}}b^{\sigma^{2}}+a^{\sigma^{10}}b^{\sigma^{6}}a^{\sigma^{3}}+b^{\sigma^{10}}a^{\sigma^{7}}a^{\sigma^{3}}.$

  * •

if $n=15$ then $\dim_{{\mathbb{F}}_{q}}(\ker(L(x)))=4$ if and only if

$\begin{cases}\mathrm{N}_{q^{15}/q}(a)=1,\\\
b=\frac{1}{a^{\sigma^{4}+\sigma^{8}+\sigma^{12}}},\\\ q\,\,\text{is even}.\\\
\end{cases}$

###### Proof.

Again, by applying Theorem 1.3 we get the first three points. Let $n=15$. By
Theorem 1.4 $L(x)$ has maximum kernel if and only if

$\begin{cases}b^{\sigma^{11}}a^{\sigma^{8}}a^{\sigma^{4}}+a^{\sigma^{11}}b^{\sigma^{7}}a^{\sigma^{4}}+a^{\sigma^{11}}a^{\sigma^{7}}b^{\sigma^{3}}=a^{-1},\\\
a^{\sigma^{11}}b^{\sigma^{7}}b^{\sigma^{4}}+b^{\sigma^{11}}a^{\sigma^{8}}b^{\sigma^{4}}+b^{\sigma^{11}}b^{\sigma^{8}}a^{\sigma^{5}}=-ba^{-1}a^{-\sigma},\\\
b^{\sigma^{5}+\sigma^{8}+\sigma^{11}}=b^{\sigma}ba^{-1}a^{-\sigma}a^{-\sigma^{2}},\\\
a^{\sigma^{3}+\sigma^{7}+\sigma^{11}}=-b^{\sigma^{2}}b^{\sigma}ba^{-1}a^{-\sigma}a^{-\sigma^{2}}.\\\
\end{cases}$ (24)

By the third and the fourth equations of (24) we get
$b^{\sigma^{2}+\sigma^{5}+\sigma^{8}+\sigma^{11}}=-a^{\sigma^{3}+\sigma^{7}+\sigma^{11}}$
and, together with its $\sigma$ and $\sigma^{2}$ power and the fourth equation
of (24), gives $\mathrm{N}_{q^{15}/q}(a)=\mathrm{N}_{q^{15}/q}(b)$. So, (24)
is equivalent to

$\begin{cases}b^{\sigma^{11}}a^{\sigma^{8}}a^{\sigma^{4}}+a^{\sigma^{11}}b^{\sigma^{7}}a^{\sigma^{4}}+a^{\sigma^{11}}a^{\sigma^{7}}b^{\sigma^{3}}=a^{-1},\\\
a^{\sigma^{11}}b^{\sigma^{7}}b^{\sigma^{4}}+b^{\sigma^{11}}a^{\sigma^{8}}b^{\sigma^{4}}+b^{\sigma^{11}}b^{\sigma^{8}}a^{\sigma^{5}}=-ba^{-1}a^{-\sigma},\\\
b^{\sigma^{2}+\sigma^{5}+\sigma^{8}+\sigma^{11}}=-a^{\sigma^{3}+\sigma^{7}+\sigma^{11}},\\\
\mathrm{N}_{q^{15}/q}(a)=\mathrm{N}_{q^{15}/q}(b).\end{cases}$

By replacing the third equation in the first equation of the above system and
by Theorem 1.1 we get

$\begin{cases}\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma^{2}}+\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma}+\mathrm{N}_{q^{15}/q^{3}}(b)=-1,\\\
a^{\sigma^{11}}b^{\sigma^{7}}b^{\sigma^{4}}+b^{\sigma^{11}}a^{\sigma^{8}}b^{\sigma^{4}}+b^{\sigma^{11}}b^{\sigma^{8}}a^{\sigma^{5}}=-ba^{-1}a^{-\sigma},\\\
b^{\sigma^{2}+\sigma^{5}+\sigma^{8}+\sigma^{11}}=-a^{\sigma^{3}+\sigma^{7}+\sigma^{11}},\\\
\mathrm{N}_{q^{15}/q}(a)=\mathrm{N}_{q^{15}/q}(b)=-1.\end{cases}$ (25)

By the third equation we get
$b=\frac{-\mathrm{N}_{q^{15}/q^{3}}(b)}{a^{\sigma^{4}+\sigma^{8}+\sigma^{12}}}$,
so that by replacing such value in the second equation of System (25) we get

$\begin{cases}\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma^{2}}+\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma}+\mathrm{N}_{q^{15}/q^{3}}(b)=-1,\\\
\mathrm{N}_{q^{15}/q^{3}}(b)^{2\sigma}+\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma+\sigma^{2}}+\mathrm{N}_{q^{15}/q^{3}}(b)^{2\sigma^{2}}=\mathrm{N}_{q^{15}/q^{3}}(b),\\\
b=\frac{-\mathrm{N}_{q^{15}/q^{3}}(b)}{a^{\sigma^{4}+\sigma^{8}+\sigma^{12}}},\\\
\mathrm{N}_{q^{15}/q}(a)=\mathrm{N}_{q^{15}/q}(b)=-1.\end{cases}$ (26)

Note that if $\mathrm{N}_{q^{15}/q^{3}}(b)\in{\mathbb{F}}_{q}$, the square of
the first equation and the sum of the second equation with its $\sigma$-powers
of System (26) implies that $q$ is not a power of $3$ and
$\mathrm{N}_{q^{15}/q^{3}}(b)=-1/3$. From the second equation of System (26)
we get $q$ is even and $\mathrm{N}_{q^{15}/q^{3}}(b)=1$. So, in this case
System (26) becomes

$\begin{cases}b=\frac{1}{a^{\sigma^{4}+\sigma^{8}+\sigma^{12}}},\\\
\mathrm{N}_{q^{15}/q}(a)=1.\end{cases}$

Now, assume $\mathrm{N}_{q^{15}/q^{3}}(b)\notin{\mathbb{F}}_{q}$, so that
$|\\{\mathrm{N}_{q^{15}/q^{3}}(b),\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma},\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma^{2}}\\}|=3$.
If $q$ is not a power of $3$, then by using the first equation of System (26),
the second equation may be replaced by the following

$\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma+1}+\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma^{2}+1}+\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma^{2}+\sigma}=1.$

Hence, System (26) is equivalent to

$\begin{cases}\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma^{2}}+\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma}+\mathrm{N}_{q^{15}/q^{3}}(b)=-1,\\\
\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma+1}+\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma^{2}+1}+\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma^{2}+\sigma}=1,\\\
b=\frac{-\mathrm{N}_{q^{15}/q^{3}}(b)}{a^{\sigma^{4}+\sigma^{8}+\sigma^{12}}},\\\
\mathrm{N}_{q^{15}/q}(a)=\mathrm{N}_{q^{15}/q}(b)=-1,\end{cases}$

that is
$\mathrm{N}_{q^{15}/q^{3}}(b),\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma},\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma^{2}}$
are distinct solutions of

$x^{3}+x^{2}+x+1=0$

not lying in ${\mathbb{F}}_{q}$. Hence the contradiction follows by noting
that $-1$ is a root of $x^{3}+x^{2}+x+1$. Let $q$ be a power of $3$. Arguing
as before, one have that System (26) is equivalent to

$\begin{cases}\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma^{2}}+\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma}+\mathrm{N}_{q^{15}/q^{3}}(b)=-1,\\\
\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma+1}+\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma^{2}+1}+\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma^{2}+\sigma}\\\
\hskip
56.9055pt=-1+\mathrm{N}_{q^{15}/q^{3}}(b)^{2}+\mathrm{N}_{q^{15}/q^{3}}(b)^{2\sigma}+\mathrm{N}_{q^{15}/q^{3}}(b)^{2\sigma^{2}},\\\
b=\frac{-\mathrm{N}_{q^{15}/q^{3}}(b)}{a^{\sigma^{4}+\sigma^{8}+\sigma^{12}}},\\\
\mathrm{N}_{q^{15}/q}(a)=\mathrm{N}_{q^{15}/q}(b)=-1.\end{cases}$ (27)

By the third and the fourth equations we have

$\mathrm{N}_{q^{15}/q^{3}}(b)=\frac{-\mathrm{N}_{q^{15}/q^{3}}(b)^{5}}{\mathrm{N}_{q^{15}/q^{3}}(a)^{\sigma^{2}+\sigma+1}}=\mathrm{N}_{q^{15}/q^{3}}(b)^{5},$

that is $\mathrm{N}_{q^{15}/q^{3}}(b)^{2}=\pm 1$, so that by the second
equation of System (27) we have

$\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma+1}+\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma^{2}+1}+\mathrm{N}_{q^{15}/q^{3}}(b)^{\sigma^{2}+\sigma}=-1.$

So, $\mathrm{N}_{q^{15}/q^{3}}(b)$ is a solution of the following system

$\left\\{\begin{array}[]{ll}x^{3}+x^{2}-x+1=0,\\\ x^{2}=\pm
1,\end{array}\right.$

from which we get $\mathrm{N}_{q^{15}/q^{3}}(b)=0$, a contradiction. ∎

##  7 Applications

###  7.1 Rank metric codes

Rank metric codes were introduced by Delsarte [10] in 1978 and they have been
intensively investigated in recent years because of their applications; we
refer to [35] for a recent survey on this topic. The set of $m\times n$
matrices ${\mathbb{F}}_{q}^{m\times n}$ over ${\mathbb{F}}_{q}$ may be endowed
with a metric, called _rank metric_ , defined by

$d(A,B)=\mathrm{rk}\,(A-B).$

A subset $\operatorname{\mathcal{C}}\subseteq{\mathbb{F}}_{q}^{m\times n}$
equipped with the rank metric is called a _rank metric code_ (shortly, a _RM_
-code). The minimum distance of $\operatorname{\mathcal{C}}$ is defined as

$d=\min\\{d(A,B)\colon A,B\in\operatorname{\mathcal{C}},\,\,A\neq B\\}.$

Denote the parameters of a RM-code
$\operatorname{\mathcal{C}}\subseteq{\mathbb{F}}_{q}^{m,n}$ with minimum
distance $d$ by $(m,n,q;d)$. We are interested in ${\mathbb{F}}_{q}$-_linear_
RM-codes, i.e. ${\mathbb{F}}_{q}$-subspaces of ${\mathbb{F}}_{q}^{m\times n}$.
Delsarte showed in [10] that the parameters of these codes must obey a
Singleton-like bound, i.e.

$|\operatorname{\mathcal{C}}|\leq q^{\max\\{m,n\\}(\min\\{m,n\\}-d+1)}.$

When equality holds, we call $\operatorname{\mathcal{C}}$ a _maximum rank
distance_ (_MRD_ for short) code. Examples of ${\mathbb{F}}_{q}$-linear MRD-
codes were first found in [10, 14]. We say that two ${\mathbb{F}}_{q}$-linear
RM-codes $\operatorname{\mathcal{C}}$ and
$\operatorname{\mathcal{C}}^{\prime}$ are equivalent if there exist
$X\in\mathrm{GL}(m,q)$, $Y\in\mathrm{GL}(n,q)$, and $\sigma\in{\rm
Aut}({\mathbb{F}}_{q})$ such that

$\operatorname{\mathcal{C}}^{\prime}=\\{XC^{\sigma}Y\colon
C\in\operatorname{\mathcal{C}}\\}.$

The _left_ and _right_ idealisers of $\operatorname{\mathcal{C}}$ are defined
in [21] as $L(\operatorname{\mathcal{C}})=\\{A\in\mathrm{GL}(m,q)\colon
A\operatorname{\mathcal{C}}\subseteq\operatorname{\mathcal{C}}\\}$ and
$R(\operatorname{\mathcal{C}})=\\{B\in\mathrm{GL}(n,q)\colon\operatorname{\mathcal{C}}B\subseteq\operatorname{\mathcal{C}}\\}$.
They are invariant under the equivalence of rank metric codes, and have been
investigated in [22]; further invariants have been introduced in [15, 27].

When $m=n$, an ${\mathbb{F}}_{q}$-linear rank metric code of
${\mathbb{F}}_{q}^{n\times n}$ can be seen an ${\mathbb{F}}_{q}$-subspace of
$\mathcal{L}_{n,q}$ (or $\mathcal{L}_{n,q,\sigma}$) where
$d(f(x),g(x))=\mathrm{rk}(f(x)-g(x))=n-\dim_{{\mathbb{F}}_{q}}(\ker(f(x)-g(x)))$
for any $f(x),g(x)\in\mathcal{L}_{n,q}$. In [8], the authors study those rank
metric codes of $\mathcal{L}_{n,q}$ with both the idealisers isomorphic to
${\mathbb{F}}_{q^{n}}$. In particular, in [8, Theorem 2.2] they proved that
any ${\mathbb{F}}_{q}$-linear rank metric code of $\mathcal{L}_{n,q}$ with
both the idealisers isomorphic to ${\mathbb{F}}_{q^{n}}$ is equivalent to the
${\mathbb{F}}_{q^{n}}$-span of some linearized monomials. Then they classify
these codes when $n\leq 9$, [8, Theorem 1.1], introducing a new class of MRD-
codes

$\operatorname{\mathcal{C}}_{3,n}=\langle
x,x^{q},x^{q^{3}}\rangle_{{\mathbb{F}}_{q^{n}}},$

when $n=7$ and $q$ is odd, and $n=8$ and $q\equiv 1\pmod{3}$. For $n=7$ the
property of $\operatorname{\mathcal{C}}_{3,n}$ of being MRD or not may be
obtained as a consequence of (c) of Theorem 2.1, whereas the case $n=8$ may be
obtained by using Proposition 6.1. Moreover we can determine the number of
minimum weight codewords of

$\operatorname{\mathcal{C}}_{3,n,\sigma}=\langle
x,x^{\sigma},x^{\sigma^{3}}\rangle_{{\mathbb{F}}_{q^{n}}}$

where $\sigma$ is a generator of
$\mathrm{Gal}({\mathbb{F}}_{q^{n}}\colon{\mathbb{F}}_{q})$, when
$\operatorname{\mathcal{C}}_{3,n,\sigma}$ is not an MRD-code and
$n\in\\{7,8\\}$. Note that the minimum distance of
$\operatorname{\mathcal{C}}_{3,n,\sigma}$ is greater than or equal to $n-3$
and the number of minimum weight codewords of an MRD-code (that is those of
weight $n-2$) depends only on its parameters, see [31].

######  Corollary 7.1.

Let $q$ be a prime power, $n$ be a positive integer and let $\sigma$ is a
generator of $\mathrm{Gal}({\mathbb{F}}_{q^{n}}\colon{\mathbb{F}}_{q})$.
Denote by $D$ the number of codewords of weight $n-3$ in
$\operatorname{\mathcal{C}}_{3,n,\sigma}$.

  * •

If $n\in\\{4,5\\}$ then $D=0$, and if $n=6$ then $D=(q^{6}-1)(q^{3}+1)$.

  * •

If $n=7$, then $D$ is either $0$ when $q$ is odd, or
$\frac{(q^{7}-1)^{2}}{q-1}$ when $q$ is even. In particular
$\operatorname{\mathcal{C}}_{3,7,\sigma}$ is an MRD-code if and only if $q$ is
odd.

  * •

If $n=8$, then $D$ is either $0$ when $q\equiv 1\pmod{3}$, or
$\frac{(q^{8}-1)^{2}}{q-1}$ when $q\equiv 0\pmod{3}$, or
$2\frac{(q^{8}-1)^{2}}{q-1}$ when $q\equiv 2\pmod{3}$. In particular
$\operatorname{\mathcal{C}}_{3,8,\sigma}$ is an MRD-code if and only if
$q\equiv 1\pmod{3}$.

###### Proof.

The number of codewords in $\operatorname{\mathcal{C}}_{3,n,\sigma}$ having
weight $n-3$ coincides with the number of polynomials
$\operatorname{\mathcal{C}}_{3,n,\sigma}$ with kernel of dimension $3$. So, by
Theorem 1.3 (see also Theorem 2.1) the first point follows. Let $n=7$. By
Proposition 6.1, the polynomials in $\operatorname{\mathcal{C}}_{3,7,\sigma}$
with kernel of dimension $3$ exists only when $q$ is even and are of the
following form

$\eta(x^{\sigma^{3}}-bx^{\sigma}-ax),$

with $\mathrm{N}_{q^{7}/q}(a)=1$, $b=-a^{\sigma(1+\sigma^{3}+\sigma^{6})}$ and
$\eta\in{\mathbb{F}}_{q^{7}}^{*}$. Similarly, for the case $n=8$ by
Proposition 6.1 the polynomials having kernel of dimension $3$ in
$\operatorname{\mathcal{C}}_{3,8,\sigma}$ exists only if $q\equiv 0,2\pmod{3}$
and have the following shape

$\eta(x^{\sigma^{3}}-bx^{\sigma}-ax),$

with $\mathrm{N}_{q^{8}/q}(a)=1$,
$b=-\frac{\alpha}{a^{\sigma^{3}+\sigma^{6}}}$, where $\alpha$ is a root of
$x^{2}+x+1$ and $\eta\in{\mathbb{F}}_{q^{8}}^{*}$. The assertion then follows
by noting that the equation $x^{2}+x+1=0$ has one root if $q\equiv 0\pmod{3}$
and it has two distinct roots if $q\equiv 1\pmod{3}$. ∎

Now consider $\sigma$ a generator of
$\mathrm{Gal}({\mathbb{F}}_{q^{n}}/{\mathbb{F}}_{q})$, $d$ a positive integer
with $d\leq n$ and let

$\operatorname{\mathcal{C}}_{d,n,\sigma}=\langle
x,x^{\sigma},x^{\sigma^{d}}\rangle_{{\mathbb{F}}_{q^{n}}}.$

The minimum distance of $\operatorname{\mathcal{C}}_{d,n,\sigma}$ is greater
than or equal to $n-d$. Denote by $D$ the number of codewords of weight $n-d$
in $\operatorname{\mathcal{C}}_{d,n,\sigma}$.

Similarly to $\operatorname{\mathcal{C}}_{3,n}$, as a corollary of Proposition
6.2 we determine the number of codewords in
$\operatorname{\mathcal{C}}_{4,n,\sigma}$ with weight $n-4$.

######  Corollary 7.2.

Let $q$ be a prime power, $n$ be a positive integer, $\sigma$ be a generator
of $\mathrm{Gal}({\mathbb{F}}_{q^{n}}/{\mathbb{F}}_{q})$ and let

$\operatorname{\mathcal{C}}_{4,n,\sigma}=\langle
x,x^{\sigma},x^{\sigma^{4}}\rangle_{{\mathbb{F}}_{q^{n}}}.$

  * •

If $5\leq n\leq 11$ and $n\neq 8$ then $D=0$, and if $n\in\\{8,12\\}$ then
$D=\frac{(q^{n}-1)^{2}}{q^{4}-1}$.

  * •

If $n=13$ then

$D=\left\\{\begin{array}[]{ll}0,&\text{if}\,\,q\not\equiv 0\pmod{3},\\\
\frac{(q^{n}-1)^{2}}{q-1},&\text{otherwise}.\end{array}\right.$

  * •

If $n=15$ then

$D=\left\\{\begin{array}[]{ll}0,&\text{if}\,\,q\,\,\text{is odd},\\\
\frac{(q^{n}-1)^{2}}{q-1},&\text{otherwise}.\end{array}\right.$

Using Theorems 1.3, 1.4 and 5.1 and arguing as in Corollary 7.1 we obtain
results on $D$ for $\operatorname{\mathcal{C}}_{d,n,\sigma}$.

######  Corollary 7.3.

Let $q$ be a prime power, $n$ be a positive integer, $\sigma$ be a generator
of $\mathrm{Gal}({\mathbb{F}}_{q^{n}}/{\mathbb{F}}_{q})$.

  * •

If $n\leq d(d-1)$ and $d\nmid n$, then $D=0$.

  * •

If $n\leq d(d-1)$ and $d\mid n$, then $D=\frac{(q^{n}-1)^{2}}{q^{d}-1}$.

  * •

If $n=d(d-1)+1$ then

$D=\left\\{\begin{array}[]{ll}0,&\text{if}\,\,d-1\,\,\text{is not a power of
the characteristic of}\,\,{\mathbb{F}}_{q},\\\
\frac{(q^{n}-1)^{2}}{q-1},&\text{otherwise}.\end{array}\right.$

  * •

If $n=d^{2}-1$, $d$ is a power of $2$, $q$ is even, then
$D\geq\frac{(q^{n}-1)^{2}}{q-1}$.

###  7.2 Quasi-subfield polynomials

A _quasi-subfield polynomial_ is defined as a polynomial of the form
$x^{q^{d}}-\lambda(x)\in{\mathbb{F}}_{q^{n}}[x]$ which divides $x^{q^{n}}-x$
and for which $\log_{q}(\deg(\lambda(x)))<d^{2}/n$. Those polynomials were
recently introduced in [19] in order to solve the Elliptic Curve Discrete
Logarithm Problem (ECDLP).

As for the family of subspace polynomials introduced in [24], we show that the
polynomials in Theorem 5.1 with $\sigma\colon x\in\mathbb{F}_{q^{n}}\mapsto
x^{q}\in{\mathbb{F}}_{q^{n}}$ are quasi-subfield polynomials.

######  Proposition 7.4.

Let $L(x)=x^{q^{d}}-bx^{q}-ax\in\mathcal{L}_{n,q}$, with $n=(d-1)d+d-1$. If

  * •

$q$ is a power of 2,

  * •

$d$ is a power of 2,

  * •

$\mathrm{N}_{q^{n}/q}(a)=1$,

  * •

$b=a^{-q^{d}\frac{q^{d(d-1)}-1}{q^{d}-1}}$,

then $L(x)$ is a quasi-subfield polynomial.

###### Proof.

Clearly, $\log_{q}(\deg(\lambda(x)))=1$ and $d^{2}>n=d^{2}-1$, so that the
condition $\log_{q}(\deg(\lambda(x)))<d^{2}/n$ is satisfied. By Theorem 5.1
$\dim_{{\mathbb{F}}_{q}}(\ker(L(x)))=d$, i.e. $L$ splits completely over
${\mathbb{F}}_{q^{n}}$ and so it divides $x^{q^{n}}-x$. ∎

In [19], the authors developed an algorithm to solve ECDLP over the field
${\mathbb{F}}_{q^{n}}$ using a quasi-subfield polynomial. The complexity of
the algorithm presented in [19] in which $n=d^{2}-1$ and with the use of
$L(x)=x^{q^{d}}-bx^{q}-ax\in\mathcal{L}_{d^{2}-1,q}$ as in Proposition 7.4, is
less than the complexity of a brute force algorithm, but greater than the
generic algorithms (Pollard Rho or Baby-Step-Giant-Step).

See also [12, 13].

###  7.3 Cyclic subspace codes

Let $k$ be a non-negative integer with $k\leq n$, the set of all
$k$-dimensional ${\mathbb{F}}_{q}$-subspaces of ${\mathbb{F}}_{q^{n}}$, viewed
as ${\mathbb{F}}_{q}$-vector space, forms a _Grassmanian space_ over
${\mathbb{F}}_{q}$, which is denoted by $\mathcal{G}_{q}(n,k)$. A _constant
dimension subspace code_ is a subset $C$ of $\mathcal{G}_{q}(n,k)$ endowed
with the metric defined as follows

$d(U,V)=\dim_{{\mathbb{F}}_{q}}(U)+\dim_{{\mathbb{F}}_{q}}(V)-2\dim_{{\mathbb{F}}_{q}}(U\cap
V),$

where $U,V\in C$. The interest in subspace codes has recently increased
because of their application to error correction in random network coding, see
[20]. One of the most studied family of subspace codes was introduced in [11]
and is knows as _cyclic subspace codes_. Let $V\in\mathcal{G}_{q}(n,k)$ and
$\alpha\in{\mathbb{F}}_{q^{n}}^{*}={\mathbb{F}}_{q^{n}}\setminus\\{0\\}$, the
_cyclic shift_ of $V$ by $\alpha$ is defined as $\alpha V=\\{\alpha v:v\in
V\\}$. Note that a cyclic shift of $V$ is an ${\mathbb{F}}_{q}$-subspace of
${\mathbb{F}}_{q^{n}}$ of the same dimension of $V$. So, a subspace code
$C\subseteq\mathcal{G}_{q}(n,k)$ is said to be _cyclic_ if for every
$\alpha\in{\mathbb{F}}_{q^{n}}^{*}$ and every $V\in C$ then $\alpha V\in C$.
Let $V\in\mathcal{G}_{q}(n,k)$, the _orbit_ of $V$ is the set $\\{\alpha
V:\alpha\in{\mathbb{F}}_{q^{n}}^{*}\\}$, and its cardinality is
$(q^{n}-1)/(q^{t}-1)$, for some $t$ which divides $n$, see e.g. [29, Theorem
1]. In particular, every orbit of a subspace $V\in\mathcal{G}_{q}(n,k)$
defines a cyclic subspace code of size $(q^{n}-1)/(q^{t}-1)$, for some $t\mid
n$. Assume $k>1$. It is easy to see that a cyclic subspace code generated by
an orbit of a subspace $V$ with size $(q^{n}-1)/(q-1)$ (that is the maximum
possible) has minimum distance at most $2k-2$. In [37] the following
conjecture arises.

######  Conjecture 7.5.

For every positive integers $n,k$ such that $k\leq n/2$, there exists a cyclic
code of size $\frac{q^{n}-1}{q-1}$ in $\mathcal{G}_{q}(n,k)$ and minimum
distance $2k-2$.

Ben-Sasson et al. [4] used subspace polynomials to generate cyclic subspace
codes with size $\frac{q^{n}-1}{q-1}$ and minimum distance $2k-2$. Precisely,
they proved that Conjecture 7.5 holds true for any given $k$ and infinitely
many values of $n$. Such result was then improved in [29]. However, the
relation between $k$ and $n$ was not immediately clear, as $n$ depends on the
degree of the irreducible factors of $x^{q^{k}}+x^{q}+x$ in
${\mathbb{F}}_{q}$. Finally, Conjecture 7.5 was solved in [33] for most of the
cases, by using _Sidon spaces_.

In this section we construct cyclic subspace codes via subspace polynomials
with the same parameters of the codes of Conjecture 7.5, in which the relation
of $n$ and $k$ is given explicitly, covering some cases of Conjecture 7.5 not
covered before.

In order to do this, we need the concept of _subspace polynomial_ (see [4,
Definition 2]), that is a monic $q$-polynomial with maximum kernel.

Let $V\in\mathcal{G}_{q}(n,k)$, the polynomial $P_{V}(x)=\prod_{v\in V}(x-v)$
is the unique subspace polynomial whose set of roots is $V$. Clearly, if
$P(x)$ is a subspace polynomial having $q$-degree $k$ then $P(x)=P_{V}(x)$,
where $V$ is the kernel of $P$.

######  Definition 7.6.

[4, Definition 3] Let $V\in\mathcal{G}_{q}(n,k)$ and
$P_{V}(x)=x^{q^{k}}+\sum_{j=0}^{i}\alpha_{j}x^{q^{j}}\in\mathcal{L}_{n,q}$,
such that $\alpha_{i}\neq 0$. We define the gap of $V$ as $gap(V)=k-i$.

The gap of a subspace polynomial plays an important role for determining the
distance between two subspace. Indeed, the following holds.

######  Proposition 7.7.

[4, Corollary 2, Lemma 5 and Corollary 3]

  1. 1.

If $U,V\in\mathcal{G}_{q}(n,k)$, then $d(U,V)\geq 2\min(gap(U),gap(V))$.

  2. 2.

If $V\in\mathcal{G}_{q}(n,k)$ and $\alpha\in{\mathbb{F}}_{q^{n}}^{*}$, then
$P_{\alpha V}(x)=\alpha^{q^{k}}P_{V}(\alpha^{-1}x)$. In particular, $V$ and
$\alpha V$ have the same gap.

  3. 3.

Let $V\in\mathcal{G}_{q}(n,k)$ and
$P_{V}(x)=x^{q^{k}}+\sum_{j=0}^{i}\alpha_{j}x^{q^{j}}$. If $\alpha_{h}\neq 0$
for some $h\in\\{1,\ldots,i\\}$ with $\gcd(h,n)=t$, then $V$ has at least
$\frac{q^{n}-1}{q^{t}-1}$ distinct cyclic shifts.

Now we are able to exhibit two families of cyclic subspace codes.

######  Proposition 7.8.

Let $k$ be a positive integer such that $k-1$ is a power characteristic of
${\mathbb{F}}_{q}$. Let $a,b\in\mathbb{F}_{q^{n}}$, with $n=k(k-1)+1$, such
that $\mathrm{N}_{q^{n}/q}(a)=(-1)^{k-1}$ and $b=-a^{qf_{1}}$ where
$f_{1}=\sum_{i=0}^{k-1}q^{ik}$. Let $V$ be the kernel of $P(x)$, with
$P(x)=x^{q^{k}}-bx^{q}-ax$. Then $C=\\{\alpha
V:\alpha\in{\mathbb{F}}_{q^{N}}^{*}\\}$ is a cyclic subspace code in
$\mathcal{G}_{q}(N,k)$, with size $\frac{q^{N}-1}{q-1}$ and minimum distance
$2k-2$, for every $N=nh$ with $h\in\mathbb{N}$.

###### Proof.

Let $P(x)=x^{q^{k}}-bx^{q}-ax$, with $n,k,a$ and $b$ be as in the assumptions.
By (c) of Theorem 2.1, the polynomial $P(x)$ splits completely over
${\mathbb{F}}_{q^{n}}$, so that $P(x)=P_{V}(x)$. By 3. of Proposition 7.7, the
code $C=\\{\alpha
V:\alpha\in{\mathbb{F}}_{q^{n}}^{*}\\}\subseteq\mathcal{G}_{q}(n,k)$ has size
$\frac{q^{n}-1}{q-1}$. Moreover, 1. and 2. of Proposition 7.7, the minimum
distance of $C$ is $2k-2$. Clearly, the result holds in each extension of
${\mathbb{F}}_{q^{n}}$. ∎

Similarly, as a corollary of Theorem 5.1 and Propositions 6.1 one can prove
the following.

######  Proposition 7.9.

Let $k$ and $q$ be powers of $2$. Let $n=k^{2}-1$ and
$a,b\in\mathbb{F}_{q^{n}}$ such that $\mathrm{N}_{q^{n}/q}(a)=(-1)^{k-1}$ and
$b=a^{-\frac{q^{k^{2}}-q}{q^{k}-1}}$. Let $V$ be the kernel of
$P(x)=x^{q^{k}}-bx^{q}-ax$. Then $C=\\{\alpha
V:\alpha\in{\mathbb{F}}_{q^{N}}^{*}\\}$ is a cyclic subspace code in
$\mathcal{G}_{q}(N,k)$, with size $\frac{q^{N}-1}{q-1}$ and minimum distance
$2k-2$, for every $N=nh$ with $h\in\mathbb{N}$.

######  Proposition 7.10.

Let $n=8$, $q\equiv 0,2\pmod{3}$, $a\in{\mathbb{F}}_{q^{8}}$ with
$\mathrm{N}_{q^{8}/q}(a)=1$, $b=\frac{-\alpha}{a^{q^{6}+q^{3}}}$, where
$\alpha$ is a root of $x^{2}+x+1$. Let $V$ be the kernel of
$P(x)=x^{q^{3}}-bx^{q}-ax$. Then $C=\\{\alpha
V:\alpha\in{\mathbb{F}}_{q^{N}}^{*}\\}$ is a cyclic subspace code in
$\mathcal{G}_{q}(N,3)$, with size $\frac{q^{N}-1}{q-1}$ and minimum distance
$4$, for every $N=8h$ with $h\in\mathbb{N}$.

Conjecture 7.5 was solved in [33, Corollary 35.] for the following parameters:

  * •

Any $q,n$ and $k$ with $k$ up to the largest divisor of $n$ that is smaller
than $n/2$;

  * •

For any $q\geq 3$ and even $n\geq 2k$.

So that the results in [33] do not cover the cases in which $n$ is a prime
number. Therefore, as a consequence of Proposition 7.8, we have the following
result.

######  Corollary 7.11.

Let $q=p^{h}$ and $k=p^{\ell}+1$, where $p$ is a prime number and
$h,\ell\in\mathbb{N}$. If $n=(k-1)k+1$ is a prime then the parameters
$(q,n,k)$ of the construction in Proposition 7.8 are new solutions of
Conjecture 7.5.

If $k=p^{\ell}+1$, then $n=p^{2\ell}+p^{\ell}+1$. Up to our knowledge, it is
not known whether or not there infinitely many prime $p$ such that $n$ is a
prime. This problem is related to a well-known conjecture, see Bunyakovsky
conjecture [6]. If $\ell=1$ or $\ell=2$, the values of $n$ are considered in
the OEIS sequence A053182 and A066100, respectively, [28].

##  8 Conclusions and open problems

In this paper we have provided closed formulas for the coefficients of a
$\sigma$-trinomial $f$ over ${\mathbb{F}}_{q^{n}}$ ensuring that $f$ has
maximum kernel. As a consequence, we construct explicit examples of
$\sigma$-linearized trinomials with maximum kernel and characterize those
having $\sigma$-degree $3$ and $4$. Then we applied these results to rank
metric codes, quasi-subfield polynomials and cyclic subspace codes. We
strongly believe that these results may be applied in further contexts.
Indeed, we are currently working on applying these results for the list
decodability of certain rank metric codes, see [34]. However still many open
problems remain and we conclude the paper by listing some of them.

  * •

By the results contained in [2], when $q$ is large enough, there always exists
a $\sigma$-linearized polynomial with maximum kernel of form

$-x^{\sigma^{d}}+a_{d-2}x^{\sigma^{d-2}}+\ldots+a_{0}x\in\mathcal{L}_{n,q,\sigma}.$

Thus, there could be room for $\sigma$-linearized polynomials with maximum
kernel and having some coefficients of high $\sigma$-degree equal to zero.

  * •

When $n=d^{2}-1$, there are others $\sigma$-linearized trinomials with maximum
kernel different from those found in Theorem 5.1, as suggested by Proposition
6.1. Determine the remaining family of $\sigma$-linearized polynomials with
maximum kernel when $n=d^{2}-1$.

  * •

More generally, is it possible to manipulate the equations in Theorem 1.4 to
get some more examples of linearized trinomials with maximum kernel also when
$g\in\\{2,\ldots,d-2\\}$ and $d\geq 4$?

##  9 Acknowledgements

We would like to thank Professor Robert Israel for pointing out Bunyakovsky
conjecture and the OEIS sequences A053182 and A066100. We also thank the
anonymous referee and Bence Csajbók, which helped us to significantly shorten
the proof of Theorem 1.3. This research was supported by the Italian National
Group for Algebraic and Geometric Structures and their Applications (GNSAGA -
INdAM). The second author is also supported by the project “VALERE: VAnviteLli
pEr la RicErca” of the University of Campania “Luigi Vanvitelli”.

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Paolo Santonastaso and Ferdinando Zullo
Dipartimento di Matematica e Fisica,
Università degli Studi della Campania “Luigi Vanvitelli”,
Viale Lincoln, 5
I– 81100 Caserta, Italy
<EMAIL_ADDRESS>