A Large Family of Maximum Scattered Linear Sets of ${{\,\mathrm{{PG}}\,}}(1,q^n)$ and Their Associated MRD Codes

Abstract

Linear sets in projective spaces over finite fields were introduced by Lunardon (Geom Dedic 75(3):245–261, 1999) and they play a central role in the study of blocking sets, semifields, rank-metric codes, etc. A linear set with the largest possible cardinality and rank is called maximum scattered. Despite two decades of study, there are only a limited number of maximum scattered linear sets of a line ${{\,\mathrm{{PG}}\,}}(1,q^n)$. In this paper, we provide a large family of new maximum scattered linear sets over ${{\,\mathrm{{PG}}\,}}(1,q^n)$ for any even $n\ge 6$ and odd q. In particular, the relevant family contains at least

$$\begin{aligned} {\left\{ \begin{array}{ll} \left\lfloor \frac{q^t+1}{8rt}\right\rfloor ,&{} \text { if }t\not \equiv 2\pmod {4};\\ \left\lfloor \frac{q^t+1}{4rt(q^2+1)}\right\rfloor ,&{} \text { if }t\equiv 2\pmod {4}, \end{array}\right. } \end{aligned}$$

inequivalent members for given $q=p^r$ and $n=2t>8$, where $p=\textrm{char}({{\mathbb {F}}}_q)$. This is a great improvement of previous results: for given q and $n>8$, the number of inequivalent maximum scattered linear sets of ${{\,\mathrm{{PG}}\,}}(1,q^n)$ in all classes known so far, is smaller than $q^2\phi (n)/2$, where $\phi$ denotes Euler’s totient function. Moreover, we show that there are a large number of new maximum rank-distance codes arising from the constructed linear sets.

1 Introduction

Let V be a vector space over ${{\mathbb {F}}}_{q^n}$ of dimension r and $\Omega ={{\,\mathrm{{PG}}\,}}(V,q^n)={{\,\mathrm{{PG}}\,}}(e-1,q^n)$. A set of points $L_U$ of $\Omega$ is called an ${{\mathbb {F}}}_q$-linear set of rank k if it consists of the points defined by the non-zero elements of an ${{\mathbb {F}}}_q$-subspace U of V of dimension k, that is,

$$\begin{aligned} L_U=\left\{ \langle \textbf{u} \rangle _{{{\mathbb {F}}}_{q^n}}: \textbf{u}\in U\setminus \{ \textbf{0} \} \right\} . \end{aligned}$$

The term linear was introduced by Lunardon [19] who considered a special kind of blocking sets. In the past two decades after this work, linear sets have been intensively investigated and applied to construct and characterize various objects in finite geometry, such as blocking sets, two-intersection sets, translation spreads of the Cayley generalized Hexagon, translation ovoids of polar spaces, semifields and rank-metric codes. We refer to [1, 16, 25,26,27] and the references therein.

The most interesting linear sets are those satisfying certain extremal properties. Firstly, it is clear that $|L_U|\le \frac{q^k-1}{q-1}$. When the equality is achieved, L is called scattered. A scattered linear set $L_U$ of $\Omega$ with largest possible rank k is called a maximum scattered linear set. In [6], it is proved that the largest possible rank is $k=en/2$ if e is even, and $(en-n)/2\le k\le en/2$ if e is odd. In particular, when $e=2$, i.e. $L_U$ is a maximum scattered linear set over a projective line, its rank k equals n.

For a given linear set $L_U$ of rank n of a projective line, by a suitable collineation of ${{\,\mathrm{{PG}}\,}}(1,q^n)$, we may always assume that the point $\langle (0,1)\rangle _{{{\mathbb {F}}}_{q^n}}$ is not in $L_U$. This means

$$\begin{aligned} U=U_f:=\{(x,f(x)): x\in {{\mathbb {F}}}_{q^n} \}, \end{aligned}$$

for a q-polynomial f(x) over ${{\mathbb {F}}}_{q^n}$; i.e. an element of the set

$$\begin{aligned}{{\mathcal {L}}}_{n,q}[x]=\left\{ \sum _{i=0}^{n-1} c_i x^{q^i}:c_i\in {{\mathbb {F}}}_{q^n} \right\} .\end{aligned}$$

Since polynomials in this set define ${{\mathbb {F}}}_q$-linear maps of ${{\mathbb {F}}}_{q^n}$ seen as an ${{\mathbb {F}}}_q$-vector space, they are also known in the literature as linearized polynomials. Given a q-polynomial f, we use $L_f$ to denote the linear set defined by $U_f$. It is not difficult to show that $L_f$ is scattered if and only if for any $z,y\in {{\mathbb {F}}}_{q^n}^*$ the condition

$$\begin{aligned}\frac{f(z)}{z} = \frac{f(y)}{y}\end{aligned}$$

implies that z and y are ${{\mathbb {F}}}_q$-linearly dependent. Hence, a q-polynomial satisfying this condition is usually called a scattered polynomial (over ${{\mathbb {F}}}_{q^n}$), see [27]. The condition for a q-polynomial f(x) to be scattered can be rephrased by saying that if $f(\gamma x)=\gamma f(x)$, for x and $\gamma \in {{\mathbb {F}}}_{q^n}$ with $x \ne 0$, then $\gamma \in {{\mathbb {F}}}_q$.

Scattered polynomials are also strongly related to maximum rank-distance (MRD, for short) codes. Given a scattered polynomial f, the set of q-polynomials

$$\begin{aligned}{{\mathcal {C}}}_f=\{ ax+bf(x): a,b\in {{\mathbb {F}}}_{q^n} \}\end{aligned}$$

defines a linear MRD code of minimum distance $n-1$ over ${{\mathbb {F}}}_q$. For recent surveys on MRD codes and their relations with linear sets, we refer to [26, 28].

Up to now, there are only three families of maximum scattered linear sets in ${{\,\mathrm{{PG}}\,}}(1,q^n)$ for infinitely many n. We list the corresponding scattered polynomials over ${{\mathbb {F}}}_{q^n}$ below:

• (i) $x^{q^s}$, where $1 \le s \le n-1$ and $\gcd (s,n)=1$; see [6].

• (ii) $\delta x^{q^s}+x^{q^{n-s}}$, where $n \ge 4$,  $N_{q^n/q}(\delta )\notin \{0,1\}$, $\gcd (s,n)=1$ and $N_{q^n/q}:\,x \in {{\mathbb {F}}}_{q^n} \mapsto x^{\frac{q^n-1}{q-1}} \in {{{\mathbb {F}}}_q},$ is the norm function of ${{\mathbb {F}}}_{q^n}$ over ${{\mathbb {F}}}_q$; see [20, 27].

• (iii) $\psi ^{(k)}(x)$, where $\psi (x)= \frac{1}{2}\left( x^q+x^{q^{t-1}} -x^{q^{t+1}}+x^{q^{2t-1}}\right)$, q odd, $n=2t$ and

  • t is even and $\gcd (k,t)=1$, or

  • t is odd, $\gcd (k,2t)=1$, and $q\equiv 1\pmod {4}$; see [18].

For $n\in \{6,8\}$, there are other families of scattered polynomials over ${{\mathbb {F}}}_{q^n}$; see [4, 9, 10, 23, 31]. According to the asymptotic classification results of them obtained in [2, 3, 5, 13], maximum scattered linear sets in ${{\,\mathrm{{PG}}\,}}(1,q^n)$ seem rare.

Two linear sets $L_U$ and $L_{U'}$ of ${{\,\mathrm{{PG}}\,}}(1,q^n)$ are said to be $P\Gamma L$-equivalent (or projectively equivalent) if there exists $\varphi \in P\Gamma L(2,q^n)$ such that $L_U^\varphi =L_{U'}$. For two given q-polynomials, it is well-known that ${{\mathcal {C}}}_f$ is equivalent to ${{\mathcal {C}}}_g$ if and only if $U_f$ and $U_g$ are on the same $\Gamma L(2,q^n)$-orbit (see Theorem 2.1), which further implies that $L_f$ and $L_g$ are $P\Gamma L$-equivalent. However, the converse statement is not true in general. For instance, if $U_f=\left\{ (x, x^q): x\in {{\mathbb {F}}}_{q^n} \right\}$ and $U_g=\left\{ (x, x^{q^s}): x\in {{\mathbb {F}}}_{q^n} \right\}$ with $s\ne 1,n-1$ and $\gcd (s,n)=1$, then $U_f$ and $U_g$ are not $\Gamma L(2,q^n)$-equivalent, but obviously $L_f=\left\{ \langle (1, x^{q-1})\rangle _{{{\mathbb {F}}}_{q^n}}: x\in {{\mathbb {F}}}_{q^n}^* \right\} =L_g$. For more results on the equivalence problems, we refer to [8, 11].

Table 1 Numbers of inequivalent ${{\mathcal {C}}}_f$ and $P\Gamma L(2,q^n)$-inequivalent $L_f$, where f is a scattered polynomial in (i), (ii) or (iii), $q=p^r$ with $p=\textrm{char}({{\mathbb {F}}}_q)$ and $\phi$ denotes Euler’s totient function

In Table 1, we list the numbers of inequivalent ${{\mathcal {C}}}_f$ and $P\Gamma L(2,q^n)$-inequivalent $L_f$, for a scattered polynomial f in each one of the three known families. The proof of what is stated in Table 1 will be provided in Sect. 2.

In this paper, we present a new family of maximum scattered linear sets in ${{\,\mathrm{{PG}}\,}}(1,q^n)$ where $q=p^r$, p is an odd prime and $n=2t\ge 6$; see Theorem 3.1. In particular, when $t>4$, this new family provides at least

$$\begin{aligned} {\left\{ \begin{array}{ll} \left\lfloor \frac{q^t+1}{4rt}\right\rfloor ,&{} \text { if }t\not \equiv 2\pmod {4};\\ \left\lfloor \frac{q^t+1}{2rt(q^2+1)}\right\rfloor ,&{} \text { if }t\equiv 2\pmod {4} \end{array}\right. } \end{aligned}$$

inequivalent number of MRD codes (Corollary 4.3) and at least

$$\begin{aligned} {\left\{ \begin{array}{ll} \left\lfloor \frac{q^t+1}{8rt}\right\rfloor ,&{} \text { if }t\not \equiv 2\pmod {4};\\ \left\lfloor \frac{q^t+1}{4rt(q^2+1)}\right\rfloor ,&{} \text { if }t\equiv 2\pmod {4} \end{array}\right. } \end{aligned}$$

$P\Gamma L(2,q^n)$-inequivalent maximum linear sets (Theorem 5.1). Therefore, the number of maximum scattered linear sets in ${{\,\mathrm{{PG}}\,}}(1,q^n)$ (and hence of MRD codes) grows exponentially with respect to n.

The remaining part of this paper is organized as follows. In Sect. 2, we introduce more results on the equivalence of maximum scattered linear sets in ${{\,\mathrm{{PG}}\,}}(1,q^n)$ and the associated MRD codes, and explain Table 1 in details. In Sect. 3, we exhibit a family of polynomials f over ${{\mathbb {F}}}_{q^n}$, and prove that they are scattered. The equivalence between the MRD codes ${{\mathcal {C}}}_f$ associated to the members of this family are completely determined in Sect. 4, in which we also study the MRD codes derived from the adjoint maps of our scattered polynomials. Based on these results, the $P\Gamma L$-equivalence of the associated maximum linear sets are investigated in Sect. 5.

2 Equivalence of MRD Codes and Linear Sets

A rank-distance code (or RD code for short) ${{\mathcal {C}}}$ is a subset of the set of $m \times n$ matrices ${{\mathbb {F}}}^{m \times n}_q$ over ${{{\mathbb {F}}}_q}$ endowed with the rank distance

$$\begin{aligned} d(A, B) = {{\,\textrm{rk}\,}}(A - B) \end{aligned}$$

for any $A, B \in {{\mathbb {F}}}_q^{m \times n}$. The minimum distance of an RD code ${{\mathcal {C}}}$, $|{{\mathcal {C}}}|\ge 2$, is defined as

$$\begin{aligned} d({{\mathcal {C}}}) = \min _{\underset{M \not = N}{M, N \in {{\mathcal {C}}}}} d(M, N). \end{aligned}$$

A rank-distance code of ${{\mathbb {F}}}_q^{m \times n }$ with minimum distance d has parameters (m, n, q; d). If ${{\mathcal {C}}}$ is an ${{\mathbb {F}}}_q$-linear subspace of ${{\mathbb {F}}}_q^{m \times n}$, then ${{\mathcal {C}}}$ is called ${{\mathbb {F}}}_q$-linear RD code and its dimension $\dim _{{{\mathbb {F}}}_q} {{\mathcal {C}}}$ is defined to be the dimension of ${{\mathcal {C}}}$ as a subspace over ${{\mathbb {F}}}_q$. The Singleton-like bound [12] for an (m, n, q; d) RD-code ${{\mathcal {C}}}$ is

$$\begin{aligned} |{{\mathcal {C}}}| \le q^{\max \{m,n\}(\min \{m,n\}-d+1)}. \end{aligned}$$

If ${{\mathcal {C}}}$ attains this size, then ${{\mathcal {C}}}$ is a called Maximum Rank-Distance code, MRD code for short. In this paper we will consider only the case in which the codewords are square matrices, i.e. $m=n$. Note that if $n = d$, then an MRD code ${{\mathcal {C}}}$ consists of $q^n$ invertible endomorphisms of ${{\mathbb {F}}}_{q^n}$; such ${{\mathcal {C}}}$ is called spread set of ${{\,\textrm{End}\,}}_{{{{\mathbb {F}}}_q}}({{\mathbb {F}}}_{q^n})$. In particular if ${{\mathcal {C}}}$ is ${{{\mathbb {F}}}_q}$-linear, it is called a semifield spread set of ${{\,\textrm{End}\,}}_{{{{\mathbb {F}}}_q}}({{\mathbb {F}}}_{q^n})$, see [15]. Two ${{{\mathbb {F}}}_q}$-linear codes ${{\mathcal {C}}}$ and ${{\mathcal {C}}}'$ are called equivalent if there exist $A,B \in {{\,\mathrm{{GL}}\,}}(n, q)$ and a field automorphism $\sigma$ of ${{{\mathbb {F}}}_q}$ such that

$$\begin{aligned} {{\mathcal {C}}}' = \{AC^\sigma B :C \in {{\mathcal {C}}}\}. \end{aligned}$$

The aforementioned link lies in the fact that rank-distance codes can be described by means of q-polynomials over ${{\mathbb {F}}}_{q^n}$, considered modulo $x^{q^n}-x$. After fixing an ordered ${{\mathbb {F}}}_q$-basis $\{b_1,b_2,\ldots ,b_n\}$ for ${{\mathbb {F}}}_{q^n}$ it is possible to give a bijection $\Phi$ which associates for each matrix $M\in {{\mathbb {F}}}_q^{n\times n}$ a unique q-polynomial $f_M\in {{\mathcal {L}}}_{n,q}[x]$. More precisely, put $\textbf{b}=(b_1,b_2,\ldots ,b_n)\in {{\mathbb {F}}}_{q^n}^n$, then $\Phi (M)=f_M$ where for each ${{\textbf{u}}}=(u_1,u_2,\ldots ,u_n)\in {{\mathbb {F}}}_q^n$ we have $f_M({{\textbf{b}}}\cdot {{\textbf{u}}})={{\textbf{b}}}\cdot {{\textbf{u}}} M$.

For a scattered polynomial f, as the kernel of the ${{\mathbb {F}}}_q$-linear map $z\mapsto az+bf(z)$ is of dimension at most 1 for any $a,b\in {{\mathbb {F}}}_{q^n}$ with $(a,b)\ne (0,0)$, the set of q-polynomials

$$\begin{aligned}{{\mathcal {C}}}_f=\big \{ ax+bf(x): a,b\in {{\mathbb {F}}}_{q^n} \big \}\end{aligned}$$

defines a linear MRD code of minimum distance $n-1$ over ${{\mathbb {F}}}_q$.

Given two scattered polynomials f and g over ${{\mathbb {F}}}_{q^n}$, the corresponding MRD codes ${{\mathcal {C}}}_f$ and ${{\mathcal {C}}}_g$ are equivalent if and only if there exists a triple $(L_1,L_2,\sigma )$, with $L_1$, $L_2\in {{\mathcal {L}}}_{n,q}[x]$ permuting ${{\mathbb {F}}}_{q^n}$ and $\sigma \in {{\,\mathrm{{Aut}}\,}}({{\mathbb {F}}}_{q})$ such that

$$\begin{aligned} L_1\circ \varphi ^\sigma \circ L_2 \in {{\mathcal {C}}}_g \text { for all }\varphi \in {{\mathcal {C}}}_f,\end{aligned}$$

where $\circ$ stands for the composition of maps and $\varphi ^\sigma (x)= \sum a_i^\sigma x^{q^i}$ for $\varphi (x)=\sum a_i x^{q^i}$. If $f=g$, the set of the triples defined as above is a group which is called the automorphism group of the code ${{\mathcal {C}}}_f$ and it is denoted by $\textrm{Aut}({{\mathcal {C}}}_f)$.

The following result concerning the equivalence of MRD codes associated with scattered polynomials is proved in [27].

Theorem 2.1

Let f and g be two scattered polynomials over ${{\mathbb {F}}}_{q^n}$, respectively. The MRD-codes ${{\mathcal {C}}}_f$ and ${{\mathcal {C}}}_g$ are equivalent if and only if $U_f$ and $U_g$ are $\Gamma L(2,q^n)$-equivalent.

For two scattered polynomials f and g, by Theorem 2.1 and the definition of $P\Gamma L$-equivalence of linear sets, if ${{\mathcal {C}}}_f$ and ${{\mathcal {C}}}_g$ are equivalent, then $L_f$ and $L_g$ are also $P\Gamma L$-equivalent. The converse does not hold, see [11].

For this paper, we only need the necessary and sufficient conditions in Theorem 2.1 to interpret the equivalence problem in Sect. 4. However, in general, the equivalence problem for MRD codes could be more complicated. We refer to the surveys [26, 28] for its precise definition and related results. See [7] for the hardness of testing the equivalence between rank-distance codes. The left idealizer and the right idealizer of any given rank-distance code ${{\mathcal {C}}}$ are invariant under equivalence. These two concepts were introduced in [17], and in [21] with different names. If a rank-distance code ${{\mathcal {C}}}$ is given as a subset of ${{\mathcal {L}}}_{n,q}[x]$, then its left idealizer and right idealizer are defined as

$$\begin{aligned}I_L({{\mathcal {C}}}) =\{\varphi \in {{\mathcal {L}}}_{n,q}[x]: \varphi \circ f\in {{\mathcal {C}}}\text { for all }f\in {{\mathcal {C}}}\}, \end{aligned}$$

and

$$\begin{aligned}I_R({{\mathcal {C}}}) =\{\varphi \in {{\mathcal {L}}}_{n,q}[x]: f\circ \varphi \in {{\mathcal {C}}}\text { for all }f\in {{\mathcal {C}}}\}, \end{aligned}$$

respectively. When ${{\mathcal {C}}}$ is an MRD-code, it is well known that all nonzero elements in $I_L({{\mathcal {C}}})$ and $I_R({{\mathcal {C}}})$ are invertible and each of the idealizers must be a subfield of ${{\mathbb {F}}}_{q^n}$. In particular, if ${{\mathcal {C}}}$ is an ${{\mathbb {F}}}_{q^n}$-subspace of ${{\mathcal {L}}}_{n,q}[x]$, then $I_L({{\mathcal {C}}})$ is isomorphic to ${{\mathbb {F}}}_{q^n}$ and ${{\mathcal {C}}}$ is said to be an ${{\mathbb {F}}}_{q^n}$-linear MRD code.

Obviously, for every MRD code ${{\mathcal {C}}}_f$ associated with a scattered polynomial f over ${{\mathbb {F}}}_{q^n}$, its left idealizer

$$\begin{aligned} I_L({{\mathcal {C}}}_f)=\big \{ \alpha x: x \in {{\mathbb {F}}}_{q^n}\big \}, \end{aligned}$$

(1)

clearly isomorphic to ${{\mathbb {F}}}_{q^n}$.

For a q-polynomial $f(x)=\sum _{i=0}^{n-1}{a_i}x^{q^i}$ over ${{\mathbb {F}}}_{q^n}$, the adjoint map $\hat{f}$ of it with respect to the bilinear form $\langle x,y \rangle =\textrm{Tr}_{q^n/q} (xy)$ is

$$\begin{aligned} \hat{f}(x) = a_0 x+\sum _{i=1}^{n-1} a_i^{q^{n-i}} x^{q^{n-i}}. \end{aligned}$$

For a given scattered polynomial f over ${{\mathbb {F}}}_{q^n}$, its adjoint ${\hat{f}}$ is a scattered polynomial over ${{\mathbb {F}}}_{q^n}$ as well, and $U_f$ and $U_{\hat{f}}$ (and hence ${{\mathcal {C}}}_f$ and ${{\mathcal {C}}}_{{\hat{f}}}$) are not necessarily equivalent. However, they define exactly the same linear set of ${{\,\mathrm{{PG}}\,}}(1,q^n)$; see [1, Lemma 2.6] and [8, Lemma 3.1]. This also implies that ${{\mathcal {C}}}_f$ and ${{\mathcal {C}}}_{{\hat{f}}}$ are both MRD-codes.

To investigate the $P\Gamma L$-equivalence among linear sets of a line, we need thefollowing result

Lemma 2.2

[8, Lemma 3.6] Let $f(x)=\sum _{i=0}^{n-1} \alpha _i x^{q^i}$ and $g(x)=\sum _{i=0}^{n-1} \beta _i x^{q^i}$ be two q-polynomials over ${{\mathbb {F}}}_{q^n}$ such that $L_f=L_g$. Then

• (a) $\alpha _0=\beta _0$,

• (b) $\alpha _k \alpha _{n-k} ^{q^k} = \beta _k \beta _{n-k}^{q^k}$ for $k=1,2,\cdots , n-1$,

• (c) $\alpha _1 \alpha _{k-1}^q\alpha _{n-k}^{q^k} + \alpha _k \alpha _{n-1}^{q} \alpha _{n-k+1}^{q^k}= \beta _1 \beta _{k-1}^q\beta _{n-k}^{q^k} + \beta _k \beta _{n-1}^{q} \beta _{n-k+1}^{q^k}$, for $k=2,3,\cdots , n-1$.

In the following we provide details on the estimates stated in Table 1 for the three families.

Family (i) The number of inequivalent MRD codes defined in (i) comes from the well-known results on the equivalence of Delsarte–Gabidulin codes and their generalizations; see [22, Theorem 4.4] for instance. Moreover all monomials $f(x)=x^{q^s}$ with $\gcd (s,n)=1$, determine the same linear set in ${{\,\mathrm{{PG}}\,}}(1,q^n)$; in fact,

$$\begin{aligned}L_f:=\left\{ \big \langle \big (1, x^{q^{s}-1}\big ) \big \rangle _{{{\mathbb {F}}}_{q^n}}: x\in {{\mathbb {F}}}_{q^n}^* \right\} ;\end{aligned}$$

the so-called linear set of pseudo-regulus type.

Family (ii) For the number of inequivalent MRD defined in (ii), we need more complicated analysis. Firstly, we state the following result which is a special case of [22, Theorem 4.4].

Proposition 2.3

For $\theta ,\eta \in {{\mathbb {F}}}_{q^n}$ such that $N_{q^n/q}(\theta ), N_{q^n/q}(\eta )\notin \{0,1\}$, with $1\le s,t \le \frac{n-1}{2}$ satisfying $\gcd (s,n)=1$, let $f(x) = \eta x^{q^s}+x^{q^{n-s}}$ and $g(x)=\theta x^{q^t}+x^{q^{n-t}}$. Then ${{\mathcal {C}}}_f$ and ${{\mathcal {C}}}_g$ are equivalent if and only if $s=t$ and

$$\begin{aligned} \theta =\eta ^{\tau } z^{q^{2s}-1} \end{aligned}$$

for some $\tau \in {{\,\mathrm{{Aut}}\,}}({{\mathbb {F}}}_{q^n})$ and $z\in {{\mathbb {F}}}_{q^n}^*$.

By Hilbert’s Theorem 90, if m|n, $\{x\in {{\mathbb {F}}}^*_{q^n}: N_{q^n/q^{m}}(x)=1 \} = \{ y^{q^{m}-1}: y\in {{\mathbb {F}}}^*_{q^n} \}$. As

$$\begin{aligned}\gcd (q^{2s}-1,q^n-1) = q^{\gcd (2s,n)}-1=q^{\gcd (2,n)}-1= {\left\{ \begin{array}{ll} q^2-1, &{}2\mid n,\\ q-1, &{} 2\not \mid n, \end{array}\right. } \end{aligned}$$

if $N_{q^n/q^{\gcd (2,n)}}(\theta )= N_{q^n/q^{\gcd (2,n)}}(\eta )$, then we can always find $z\in {{\mathbb {F}}}_{q^n}^*$ such that $\theta = \eta z^{q^{2s}-1}$. Hence, under the maps $\eta \mapsto \eta z^{q^{2s}-1}$ for $z\in {{\mathbb {F}}}_{q^n}^*$, the elements in ${{\mathbb {F}}}_{q^n}^*$ are partitioned into $q^{\gcd (2,n)}-1$ orbits. Moreover, $\theta$ and $\eta$ are in the same orbit under the maps $\eta \mapsto \eta ^\tau z^{q^{2s}-1}$ for $z\in {{\mathbb {F}}}_{q^n}^*$ and $\tau \in {{\,\mathrm{{Aut}}\,}}({{\mathbb {F}}}_{q^n})$ if and only if $N_{q^n/q^{\gcd (2,n)}}(\theta )= \left( N_{q^n/q^{\gcd (2,n)}}(\eta )\right) ^{\tau '}$ for some $\tau '\in {{\,\mathrm{{Aut}}\,}}({{\mathbb {F}}}_{q^{\gcd (2,n)}})$, which implies that $N_{q^n/q^{\gcd (2,n)}}(\theta )$ and $N_{q^n/q^{\gcd (2,n)}}(\eta )$ must belong either to ${{\mathbb {F}}}_p\setminus \{0,1\}$ or to F(m) for some $m\, |\, r\cdot \gcd (2,n)$, where $m>1$ and

$$\begin{aligned}F(m)=\big \{ x\in {{\mathbb {F}}}_{p^m}: x\notin {{\mathbb {F}}}_{p^k} \text { with }k<m, k\mid m\big \}.\end{aligned}$$

Therefore, the total number $\Theta (q,n)$ of inequivalent MRD codes in family (ii) for given $q=p^r$ and n, where $p=\textrm{char}({{\mathbb {F}}}_q),$ is

$$\begin{aligned} \Theta (q,n)=\left( \sum _{j\mid r\cdot \gcd (2,n), j\ne 1} \frac{|F(j)|}{j} + (p-2)\right) \frac{\phi (n)}{2}. \end{aligned}$$

By computation, we have

$$\begin{aligned} \sum _{j\mid r\cdot \gcd (2,n), j\ne 1} \frac{|F(j)|}{j} + (p-2) \le&{\left\{ \begin{array}{ll} \sum _{j\mid r, j\ne 1}|F(j)| + (p-2), &{} 2\not \mid n,\\ \frac{1}{2}\sum _{j\mid 2r, j\ne 1}|F(j)| + (p-2), &{} 2\mid n, \end{array}\right. } \\ =&{\left\{ \begin{array}{ll} q-2, &{} 2\not \mid n,\\ \frac{1}{2}\left( q^2-p \right) +p-2, &{} 2\mid n. \end{array}\right. } \end{aligned}$$

As a consequence, we can derive the following upper bound for $\Theta (q,n)$:

$$\begin{aligned} \Theta (q,n)\le {\left\{ \begin{array}{ll} (q-2)\frac{\phi (n)}{2}, &{} 2\not \mid n,\\ \frac{q^2+p-4}{2}\frac{\phi (n)}{2}, &{} 2\mid n. \end{array}\right. } \end{aligned}$$

(2)

One can also derive a lower bound

$$\begin{aligned} \Theta (q,n)\ge {\left\{ \begin{array}{ll} \frac{q-2}{r}\frac{\phi (n)}{2}, &{} 2\not \mid n,\\ \frac{q^2-q}{2r}\frac{\phi (n)}{2}, &{} 2\mid n. \end{array}\right. } \end{aligned}$$

Note that for q large most choices of $\theta$ satisfying $N_{q^n/q}(\theta )\notin \{0,1\}$, $N_{q^n/q^{\gcd (s,n)}}(\theta )$ is not in any proper subfield of ${{\mathbb {F}}}_{q^{\gcd (s,n)}}$. Therefore, $\Theta (n,q)$ is asymptotically close to the above lower bound when q is getting large. By (2), we get an upper bound on the number of inequivalent $\mathcal {C}_f$ in family (ii) in Table 1.

Let $\Lambda (n,q)$ be the number of $P\Gamma L$-inequivalent Lunardon–Polverino linear sets over ${{\,\mathrm{{PG}}\,}}(1,q^n)$. Clearly, the upper bound for $\Theta (n,q)$ is an upper bound for $\Lambda (n,q)$, but, actually, this could be smaller. In [10, Section 3], $\Lambda (n,q)$ is determined for $n=6,8$. The precise value of $\Lambda (n,q)$ for all $n\ge 3$ can be found in a recent preprint [30, Section 4]. The value of $\Lambda (n,q)$ is approximately $\frac{q \phi (n)}{4r}$ for odd n and $\frac{{q^2}\phi (n)}{8r}$ for even n provided that q is large enough.

Family (iii) Regarding (iii) in Table 1, for fixed q and n, there are exactly $\phi (n)/2$ inequivalent MRD-codes derived from this family of scattered polynomials; see [18, Theorem 5.4]. The precise number of $P\Gamma L$-inequivalent linear sets provided by it is still unknown, but it is obviously smaller than or equal to $\phi (n)/2$.

3 Construction

Inspired by the results obtained in [18] where polynomials $\psi ^{(k)}(x)$ defined in (iii) were introduced, and their scatteredness proven, we investigate here a similar problem for a natural generalization of these examples.

Precisely, set $n=2t$, $t \ge 3$ and let q be an odd prime power. We consider the following q-polynomials:

$$\begin{aligned} \psi _{h,t}(x)= x^{q}+x^{q^{t-1}}-h^{1-q^{t+1}}x^{q^{t+1}}+h^{1-q^{2t-1}}x^{q^{2t-1}} \in {{\mathbb {F}}}_{q^{n}}[x] \end{aligned}$$

(3)

where $h \in {{\mathbb {F}}}_{q^n} \setminus {{\mathbb {F}}}_{q^t}$ and $h^{q^t +1 } =-1$.

First we note that if $t=3$ in (3), then $\psi _{h,3}(x)$ are the scattered polynomials and they are the adjoint polynomials of those constructed by Bartoli et al. in [4].

Furthermore, if we allow $h \in {{{\mathbb {F}}}_{q^t}}$, since $-1=h^{q^t+1}=h^2$, then $h \in {{\mathbb {F}}}_{q^2}$. Then we may distinguish two cases:

• (a) $q \equiv 1 \pmod 4$. In this case $h \in {{{\mathbb {F}}}_q}$ and $\psi _{h,t}(x)$ becomes

  $$\begin{aligned} x^{q}+x^{q^{t-1}}-x^{q^{t+1}}+x^{q^{2t-1}}. \end{aligned}$$

  This polynomial was proven to be scattered for each $t \ge 3$ in [18].

• (b) $q \equiv 3 \pmod 4$. In this case $h\in {{\mathbb {F}}}_{q^2}$ and $h^q=-h$; hence, t must be even and $\psi _{h,t}(x)$ becomes

  $$\begin{aligned} x^{q}+x^{q^{t-1}}+x^{q^{t+1}}-x^{q^{2t-1}}. \end{aligned}$$

  Also, this polynomial was proven to be scattered in [18].

In this section, our goal is to prove the following main result.

Theorem 3.1

Let $n=2t$, $t\ge 3$ and let q be an odd prime power. For each $h \in {{\mathbb {F}}}_{q^n}\setminus {{{\mathbb {F}}}_{q^t}}$ such that $h^{q^t+1}=-1$, the ${{{\mathbb {F}}}_q}$-linearized polynomial $\psi _{h,t}(x)$ is scattered.

Now we note that polynomials described in (3) can be rewritten in the following fashion:

$$\begin{aligned} \psi _{h,t}(x)=L(x)+M(x), \end{aligned}$$

(4)

where $L(x)=x^q-h^{1-q^{t+1}}x^{q^{t+1}}$ and $M(x)=x^{q^{t-1}}+h^{1-q^{2t-1}}x^{q^{2t-1}}$.

It is straightforward to see that L(x) and M(x) are ${{\mathbb {F}}}_{q^t}$-semilinear maps of ${{\mathbb {F}}}_{q^{n}}$ with companion automorphisms $x \mapsto x^q$ and $x \mapsto x^{q^{t-1}}$, respectively. Moreover, we have that

$$\begin{aligned} \ker L = \Big \{ x \in {{\mathbb {F}}}_{q^n}:\, x-h^{q^{2t-1}-q^t}x^{q^t}= 0\Big \} \end{aligned}$$

(5)

and similarly

$$\begin{aligned} \ker M = \Big \{x \in {{\mathbb {F}}}_{q^n}:\, x+h^{q^{t+1}-q^t}x^{q^t}= 0\Big \}. \end{aligned}$$

(6)

In addition, since $h^{q^t+1}=-1$, we have

$$\begin{aligned} L(x)^{q^t}&=\Big (x^q-h^{1-q^{t+1}}x^{q^{t+1}}\Big )^{q^t}=x^{q^{t+1}}-h^{q^t-q}x^q\\&= -h^{q^t-q}\Big (x^q-h^{1-q^{t+1}}x^{q^{t+1}}\Big )=-h^{q^t-q}L(x) \end{aligned}$$

and similarly, we may prove that $M(x)^{q^t}=h^{q^t-q^{t-1}}M(x)$. Hence, we obtain that

$$\begin{aligned} \text {im} \, L = \big \{ z \in {{\mathbb {F}}}_{q^{2t}}:\, z^{q^t}+h^{q^t-q}z= 0 \big \} \end{aligned}$$

(7)

and

$$\begin{aligned} \text {im} \, M = \big \{ z \in {{\mathbb {F}}}_{q^{2t}}: \, z^{q^t}-h^{q^t-q^{t-1}}z= 0 \big \}. \end{aligned}$$

(8)

Clearly, the sets in (5), (6), (7) and (8) are 1-dimensional ${{{\mathbb {F}}}_{q^t}}$-subspaces of ${{\mathbb {F}}}_{q^n}$.

Proposition 3.2

Let $n=2t$, $t\ge 1$ and let $h \in {{\mathbb {F}}}_{q^{2t}}$ be such that $h^{q^t+1}=-1$. Then $h^{q^2+1}\ne 1$ and $h^{q^{t-2}}\ne -h$.

Proof

First, as q is odd, $\gcd (q^2+1, q^{2t}-1)=2$ if t is odd, and

$$\begin{aligned} \gcd \big (q^2+1, q^t+1\big )&={\left\{ \begin{array}{ll} 2, &{} t\equiv 0 \pmod {4};\\ q^2+1, &{} t\equiv 2 \pmod {4}. \end{array}\right. } \end{aligned}$$

Assume on the contrary that $h^{q^2+1}=1$. Together with $h^{q^t+1}=-1$ and the above GCD conditions, we deduce the following results.

If t is odd, then $h^2=1$ which contradicts $h^{q^t+1}=-1$. If $4\mid t$, then $h^2=-1$ which implies $h^{q^2+1}=-1$ contradicting the assumption. If $2\mid t$ but $4\not \mid t$, then $h^{q^t+1}=1$ contradicting $h^{q^t+1}=-1$.

From $h^{q^t+1}=-1$ and $h^{q^2+1}\ne 1$, we finally derive $h^{q^{t-2}}\ne -h$ directly. $\square$

Proposition 3.3

Let $n=2t$ with $t \ge 3$. The finite field ${{\mathbb {F}}}_{q^n}$, seen as ${{{\mathbb {F}}}_{q^t}}$-vector space, is both the direct sum of $\ker L$ and $\ker M$, and of $\text {im} \,L$ and $\text {im} \,M$.

Proof

Since $\ker L$ and $\ker M$ are 1-dimensional ${{{\mathbb {F}}}_{q^t}}$-subspaces of ${{\mathbb {F}}}_{q^n}$, it is enough to prove that $\ker L \cap \ker M = \{0\}$. In this regard, let $u \in \ker L \cap \ker M$. By (5) and (6), we get $h^{q^{2t-1}}=-h^{q^{t+1}}$, i.e. $(h^{q^{t-2}})^{q^{t+1}}=-h^{q^{t+1}}$. Since $h^{{q^t}+1}=-1$, we get $h^{q^{t-2}}=-h$, contradicting Proposition 3.2.

Taking into account (7) and (8), a similar argument shows that the additive group of ${{\mathbb {F}}}_{q^n}$, seen as ${{{\mathbb {F}}}_{q^t}}$-vector space, can be also written as $\text {im} \, L \oplus \text {im} \, M$. $\square$

Consider now the following ${{{\mathbb {F}}}_{q^t}}$-linear maps of ${{\mathbb {F}}}_{q^n}$

$$\begin{aligned} R(x)=x^{q^t}+h^{q^{t-1}-q}x \quad \text {and}\quad T(x)=x^{q^t}+h^{q-q^{t-1}}x. \end{aligned}$$

(9)

It is straightforward to see that $\dim _{{{{\mathbb {F}}}_{q^t}}} \ker R = \dim _{{{{\mathbb {F}}}_{q^t}}}\ker T= 1$; moreover, $\ker T = h^{q^{t-1}-q} \ker R$, moreover by the same argument that we use above, one gets that

$$\begin{aligned}\text {im} \,R =\big \{z \in {{\mathbb {F}}}_{q^n}: z^{q^t}-h^{q-q^{t-1}}z=0\big \}\end{aligned}$$

and

$$\begin{aligned}\text {im} \,T =\big \{z \in {{\mathbb {F}}}_{q^n}: z^{q^t}-h^{q^{t-1}-q}z=0\big \},\end{aligned}$$

obtaining that $\text {im} \, T=h^{q-q^{t-1}}\text {im} \,R$.

Lemma 3.4

Let $\rho , \tau \in {{\mathbb {F}}}^*_{q^n}$, $n=2t$ and $t \ge 3$ be such that $\rho \in \ker R$ and $\tau \in \ker T$. Then

• (i) $\{1,\rho \}$ and $\{1,\tau \}$ are ${{{\mathbb {F}}}_{q^t}}$-bases of ${{\mathbb {F}}}_{q^n}$.

• (ii) If $\tau =h^{q^{t-1}-q} \rho$ and an element $\gamma \in {{\mathbb {F}}}_{q^n}$ has components ($\lambda _1$,$\mu _1$) in the ordered ${{\mathbb {F}}}_{q^t}$-basis $\{1,\rho \}$, then the components of $\gamma$ in the ordered ${{{\mathbb {F}}}_{q^t}}$-basis $\{1,\tau \}$ are

  $$\begin{aligned} \left( \lambda _1+ \mu _1\rho \left( 1-h^{q^{t-1}-q}\right) , \mu _1\right) . \end{aligned}$$

  (10)

Proof

(i) It is enough showing that $\rho$ and $\tau$ are not in ${{{\mathbb {F}}}_{q^t}}$. We will show that $\rho \not \in {{{\mathbb {F}}}_{q^t}}$. A similar argument can be applied to $\tau$ as well. Suppose that $\rho \in {{{\mathbb {F}}}_{q^t}}$, then $\rho ^{q^t-1}=1$. Then, by hypothesis,

$$\begin{aligned} 1=\rho ^{q^t-1}=-h^{q^{t-1}-q}. \end{aligned}$$

Hence $h^{q^{t-2}}=-h$ which, by Proposition 3.2, is not the case.

(ii) Let $\gamma \in {{\mathbb {F}}}_{q^n}$ and suppose that $\gamma =\lambda _1+\mu _1\rho$ with $\lambda _1,\mu _1 \in {{{\mathbb {F}}}_{q^t}}$. Also, denote by $\lambda _2$ and $\mu _2$ the components of $\gamma$ in the ${{\mathbb {F}}}_{q^t}$-basis $\{1,\tau \}$. Of course, we have

$$\begin{aligned} \lambda _2+\mu _2\tau =\lambda _1+\mu _1\rho \end{aligned}$$

(11)

Raising (11) to the $q^t$-th power, and taking into account that $\rho \in \ker R$ and $\tau \in \ker T$, we get the following linear system in the unknowns $\lambda _2$ and $\mu _2$

$$\begin{aligned} {\left\{ \begin{array}{ll} \lambda _2+\mu _2\tau =\lambda _1+\mu _1\rho \\ \lambda _2 -\mu _2h^{q-q^{t-1}} \tau =\lambda _1-\mu _1h^{q^{t-1}-q}\rho . \end{array}\right. } \end{aligned}$$

Clearly, this linear system has a unique solution; i.e.,

$$\begin{aligned} \lambda _2=\lambda _1+\mu _1\rho \left( 1-h^{q^{t-1}-q}\right) \quad \text {and} \quad \mu _2=\mu _1. \end{aligned}$$

Hence, the assertion follows. $\square$

Proposition 3.5

For any nonzero vectors $u \in \ker L$, $v \in \ker M$ and any $a \in {{\mathbb {F}}}_{q^n}$, the following statements are equivalent:

• (i) $a \in \ker R$;

• (ii) $av \in \ker L$;

• (iii) $a M(u) \in \text {im} \,L$.

Proof

Clearly, if a is zero the statement is trivially verified. Suppose that $a \in {{\mathbb {F}}}_{q^n}^*$. Let $\rho$ be a nonzero vector in $\ker R$ which means $\ker R=\langle \rho \rangle _{q^t}$.

$(i) \Rightarrow (ii)$. Let $a \in \langle \rho \rangle _{q^t}$, then there exists $\lambda \in {{{\mathbb {F}}}_{q^t}}$ such that $a=\lambda \rho$. Then

$$\begin{aligned} L(av)=\lambda ^qL(\rho v)=\lambda ^q \left( (\rho v)^q - h^{1-q^{t+1}}(\rho v)^{q^{t+1}} \right) . \end{aligned}$$

Since $\rho \in \ker R$ and $v \in \ker M$, by (6) and (9), we get

$$\begin{aligned} (\lambda \rho v)^q\left( 1-h^{-q^{t+2}+1+q^t-q^2}\right) . \end{aligned}$$

Moreover, since $h^{q^t+1}=-1$, the latter expression is equal to 0; hence, $av \in \ker L$.

$(ii)\Rightarrow (iii)$ Let $v \in \ker M$. Since $av \in \ker L$,

$$\begin{aligned}0=L(av)=(av)^q-h^{1-q^{t+1}}(av)^{q^{t+1}}=v^q\left( a^q+h^{1-q^{t+2}}a^{q^{t+1}}\right) ,\end{aligned}$$

that is

$$\begin{aligned} a+h^{q^{2t-1}-q^{t+1}}a^{q^t}=0. \end{aligned}$$

(12)

We will prove that $aM(u) \in \text {im} L$. So, putting $z=M(u)$, by (7), this is equivalent to prove that

$$\begin{aligned} (az)^{q^t}+h^{q^t-q}(az)=0. \end{aligned}$$

By Proposition 3.3, since $u\in \ker L$, $z=M(u)\ne 0$. Also, since $h^{q^t+1}=-1$, by (8), we have

$$\begin{aligned} (az)^{q^t}+h^{q^t-q}(az)=zh^{q^t-q}\cdot \left( a^{q^t}h^{q^{2t-1}-q^{t+1}}+a\right) \end{aligned}$$

and by (12) the last expression equals 0, proving the result.

$(iii) \Rightarrow (i)$ As before, by Proposition 3.3, $z=M(u)$ is a nonzero element of $\text {im} \,M$. Since $az \in \text {im} \,L$, by (7) and (8), we obtain

$$\begin{aligned} 0{} & {} =(az)^{q^t}+h^{q^t-q}(az)=z\left( a^{q^t}h^{q^t-q^{t-1}}+h^{q^t-q}a\right) \\{} & {} =h^{q^t-q^{t-1}}z\left( a^{q^t}+h^{q^{t-1}-q}a\right) , \end{aligned}$$

which implies $a^{q^t}+h^{q^{t-1}-q}a=0$. Then, by (9), $a \in \ker R$. Finally, since $\ker R$ is a 1-dimensional ${{{\mathbb {F}}}_{q^t}}$-subspace of ${{\mathbb {F}}}_{q^n}$, $a=\lambda \rho$ for some $\lambda \in {{{\mathbb {F}}}_{q^t}}$. $\square$

Using similar techniques as in the previous proposition we can show the following result

Proposition 3.6

For any nonzero vectors $u \in \ker L$, $v \in \ker M$ and any $b \in {{\mathbb {F}}}_{q^n}$, the following statements are equivalent:

• (i) $b \in \ker T$;

• (ii) $b\,u \in \ker M$;

• (iii) $b L(v) \in \text {im} \, M$.

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

Proof of Theorem 3.1

Let $\psi (x):=\psi _{h,t}(x)$, we want to prove that for each $x\in {{\mathbb {F}}}_{q^n}^*$ and for each $\gamma \in {{\mathbb {F}}}_{q^n}$ such that

$$\begin{aligned} \psi (\gamma x)=\gamma \psi (x) \end{aligned}$$

(13)

we get $\gamma \in {{{\mathbb {F}}}_q}$. Recall that

$$\begin{aligned}\psi (x)=L(x)+M(x)\end{aligned}$$

as in (4). Also, by Proposition 3.3, any $x \in {{\mathbb {F}}}_{q^n}$ can be uniquely written as $x=x_1+x_2$, where $x_1 \in \ker L$ and $x_2 \in \ker M$. Similarly, by Lemma 3.4, if $\gamma \in {{\mathbb {F}}}_{q^n}$ there are exactly two elements $\lambda _1,\mu _1 \in {{{\mathbb {F}}}_{q^t}}$ and two elements $\lambda _2, \mu _2 \in {{{\mathbb {F}}}_{q^t}}$ such that

$$\begin{aligned} \lambda _1+\mu _1\rho =\gamma = \lambda _2+\mu _2\tau \end{aligned}$$

where $\rho \in \ker R$ and $\tau = h^{q^{t-1}-q}\rho$. It has been already noted that $\tau \in \ker T$. Putting $a=\mu _1\rho$ and $b=\mu _2\tau$, which imply $a\in \ker R$ and $b\in \ker T$, Condition (13) may be re-written as follows

$$\begin{aligned} \begin{aligned} L((\lambda _1+a)(x_1+x_2))+M((\lambda _2+b)(x_1+x_2))&= (\lambda _2+b)L(x_1+x_2)\\&\quad +(\lambda _1+a)M(x_1+x_2). \end{aligned} \end{aligned}$$

(14)

Also, since $x_1 \in \ker L$, $x_2 \in \ker M$, L(x) and M(x) are ${{{\mathbb {F}}}_{q^t}}$-semilinear maps and by (ii) of Proposition 3.5 and (ii) of Proposition 3.6, Eq. (14) is equivalent to

$$\begin{aligned} \begin{aligned} L(\lambda _1 x_2)+L(a x_1)+M(\lambda _2x_1) +M(b x_2)&=\lambda _2L(x_2)+b L(x_2)\\&\quad +\lambda _1M(x_1)+a M(x_1). \end{aligned} \end{aligned}$$

and hence

$$\begin{aligned} \begin{aligned} \lambda _1 ^qL(x_2)+L(a x_1)-\lambda _2L(x_2)-a M(x_1)&= b L(x_2)+\lambda _1M(x_1)\\&\quad -\lambda _2^{q^{t-1}}M(x_1)-M(b x_2). \end{aligned} \end{aligned}$$

(15)

Now, since the image spaces of the maps L(x) and M(x) are ${{{\mathbb {F}}}_{q^t}}$-spaces, taking (iii) of Proposition 3.5 and (iii) of Proposition 3.6 into account, the expressions on left and right hand sides of (15) belong to $\text {im} \, L$ and $\text {im} \, M$, respectively. By Proposition 3.3, both sides of (15) must be equal to zero an hence we obtain the following system

$$\begin{aligned} {\left\{ \begin{array}{ll} L(ax_1)-aM(x_1)=\big (\lambda _2-\lambda ^q_1\big )L(x_2)\\ bL(x_2)-M(bx_2)=\Big (\lambda _2^{q^{t-1}}-\lambda _1\Big )M(x_1). \end{array}\right. } \end{aligned}$$

Raising to the q-th power the second equation, we get

$$\begin{aligned} {\left\{ \begin{array}{ll} L(ax_1)-aM(x_1)=\big (\lambda _2-\lambda ^q_1\big )L(x_2)\\ b^qL(x_2)^q-M(bx_2)^q=\big (\lambda _2-\lambda _1^q\big )M(x_1)^q. \end{array}\right. } \end{aligned}$$

(16)

Since $a=\mu _1 \rho$, $b=\mu _2\tau$ and $\tau =h^{q^{t-1}-q}\rho$, from Lemma 3.4 it follows

$$\begin{aligned}\lambda _1+\mu _1\rho =\mu _1\rho \big (1-h^{q^{t-1}-q}\big )+\mu _1\tau =\lambda _2+\mu _2\tau \end{aligned}$$

and, since $\{1,\tau \}$ is an ${{\mathbb {F}}}_{q^t}$-basis of ${{\mathbb {F}}}_{q^n}$, we get $\mu _1=\mu _2$ and $b=h^{q^{t-1}-q}a$.

If $a=0$, we have $\mu _1=0$ and hence $\gamma =\lambda _1=\lambda _2\in {{\mathbb {F}}}_{q^t}$. Also, from (16), if $\lambda _2\ne \lambda _1^q$, then $L(x_2)=M(x_1)=0$. By Proposition 3.3, $x=x_1=x_2=0$, a contradiction. Then $\lambda _1=\lambda _2=\lambda _1^q$, which gives $\lambda _1\in {{\mathbb {F}}}_q$, i.e. $\gamma \in {{\mathbb {F}}}_q$.

In the remainder of the proof, we are going to show that $a\ne 0$, i.e. $\gamma \in {{\mathbb {F}}}_{q^{2t}}\setminus {{\mathbb {F}}}_{q^t}$ leads to contradictions. Depending on the value of $x_1$ and $x_2$, we separate the proof into three cases.

Case 1 $x_1=0$. The system in (16) is reduced to

$$\begin{aligned} {\left\{ \begin{array}{ll} \left( \lambda _2-\lambda ^q_1\right) L(x_2)=0\\ b^qL(x_2)^q-M(bx_2)^q=0. \end{array}\right. } \end{aligned}$$

By the second equation, taking into that $b\in \ker T$, $x_2\in \ker M$ and $h^{q^t+1}=-1$, we get

$$\begin{aligned} b^{q-1}=\frac{h^{q-1}}{\left( x_2\left( 1+h^{q-q^{t-1}}\right) \right) ^{q^2-1}}. \end{aligned}$$

Then, there exists $\lambda \in {{\mathbb {F}}}_{q}^*$ such that

$$\begin{aligned} b=\lambda \cdot \frac{h}{\left( x_2\left( 1+h^{q-q^{t-1}}\right) \right) ^{q+1}}. \end{aligned}$$

Since $b \in \ker T$, then $b^{q^t}+h^{q-q^{t-1}}b=0$ and we get

$$\begin{aligned} \frac{h^{q^t-1}}{\left( x_2\left( 1+h^{q-q^{t-1}}\right) \right) ^{q^t(q+1)}} + \frac{h^{q-q^{t-1}}}{\left( x_2\left( 1+h^{q-q^{t-1}}\right) \right) ^{q+1}}=0, \end{aligned}$$

whence, since $x_2 \in \ker M$, $x_2^{q^t}=-h^{q^{t}-q^{t+1}}x_2$ and hence

$$\begin{aligned} \Biggl ( \frac{1+h^{q-q^{t-1}}}{ h^{q-1}\left( 1+h^{q^{t-1}-q}\right) } \Biggr )^{q+1}=-h^{1+q-q^{t-1}-q^t}. \end{aligned}$$

This is equivalent to

$$\begin{aligned} \Biggl ( \frac{h^{q^{t-1}-1}\big (1+h^{q-q^{t-1}}\big )}{ h^{q-1}\big (1+h^{q^{t-1}-q}\big )} \Biggr )^{q+1}=-1, \end{aligned}$$

whence we have $1^{q+1}=-1$, a contradiction.

Case 2 $x_2=0$. The system in (16) is reduced to

$$\begin{aligned} {\left\{ \begin{array}{ll} L(ax_1)-aM(x_1)=0\\ \big (\lambda _2-\lambda _1^q\big )M(x_1)^q=0. \end{array}\right. } \end{aligned}$$

By the first equation, taking into account that $a\in \ker R$, $x_1 \in \ker L$ and $h^{q^t+1}=-1$, we obtain

$$\begin{aligned} a^{q-1}=\big (x^{q}_1\big )^{q^{t-2}-1}\cdot \frac{1+h^{1-q^{t-2}}}{1+h^{q^{t+2}-1}}=\left( x^{q}_1\big (1+h^{q^{t+2}-1}\big )\right) ^{q^{t-2}-1}. \end{aligned}$$

Then there exists $\lambda \in {{{\mathbb {F}}}_q}^*$ such that

$$\begin{aligned} a=\lambda \left( x^q_1\big (1+h^{q^{t+2}-1}\big )\right) ^\nu , \end{aligned}$$

where $\nu =(q^{t-2}-1)/(q-1)$.

By (9), since $a \in \ker R$, then

$$\begin{aligned} \left( x_1^{q^{t+1}}\big (1+h^{q^2-q^{t}}\big )\right) ^\nu +h^{q^{t-1}-q}\left( x_1^q\big (1+h^{q^{t+2}-1}\big )\right) ^\nu =0. \end{aligned}$$

Moreover, since $x_1 \in \ker L$, then

$$\begin{aligned} \Biggl (\ \frac{h^{q^{t+1}-1} \big (1+h^{q^2-q^t}\big )}{1+h^{q^t-q^2}} \Biggr )^\nu =-h^{q^{t-1}-q}. \end{aligned}$$

The last expression is equivalent to

$$\begin{aligned} \Biggl (\ \frac{h^{q^{t}-q} \big (1+h^{q^2-q^t}\big )}{h^{q^2-q}\big (1+h^{q^t-q^2}\big )} \Biggr )^\nu =-1, \end{aligned}$$

whence $1^\nu =-1$, leading to a contradiction.

Case 3 $x_1,x_2 \ne 0$. Recall that $a \in \ker R$, $b=h^{q^{t-1}-q}a$, $\lambda _2 = \lambda _1+(1-h^{q^{t-1}-q})a$, $x_1 \in \ker L$ and $x_2 \in \ker M$. Then, by (16), a turns out to be a nonzero solution of the following linear system

$$\begin{aligned} {\left\{ \begin{array}{ll} x_1^q\big (1+h^{q^{t}-q^2}\big )a^q-\left( M(x_1)+\big (1-h^{q^{t-1}-q}\big )L(x_2)\right) a\\ \quad =\big (\lambda _1-\lambda ^q_1\big )L(x_2)\\ h^{q^t-q^2}L(x_2)^qa^q+\left( x_2^{q^t}\big (1+h^{q^{t-1}-q}\big )-\big (1-h^{q^{t-1}-q}\big )M(x_1)^q\right) a\\ \quad =\big (\lambda _1-\lambda _1^q\big )M(x_1)^q. \end{array}\right. } \end{aligned}$$

(17)

By $x_1\in \ker L$ and $x_2\in \ker M$, we obtain the following two equations which will be frequently used later,

$$\begin{aligned} L(x_2)&= x_2^q\Big (1+h^{1-q^{t+2}}\Big ),\\ M(x_1)&= x_1^{q^{t-1}} \Big (1+h^{1-q^{t-2}}\Big ). \end{aligned}$$

- Case 3.1 First of all, suppose that $\lambda _1 \in {{{\mathbb {F}}}_q}$, then System (17) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} x_1^q\Big (1+h^{q^{t}-q^2}\Big )a^q-\left( M(x_1)+\Big (1-h^{q^{t-1}-q}\Big )L(x_2)\right) a=0\\ h^{q^t-q^2}L(x_2)^qa^q+\left( x_2^{q^t}\Big (1+h^{q^{t-1}-q}\Big )-\Big (1-h^{q^{t-1}-q}\Big )M(x_1)^q\right) a=0. \end{array}\right. } \end{aligned}$$

(18)

and since a is a nonzero solution then

$$\begin{aligned}{} & {} x_1^q\Big (1+h^{q^{t}-q^2}\Big )\left( x_2^{q^t}\Big (1+h^{q^{t-1}-q}\Big )-\Big (1-h^{q^{t-1}-q}\Big )M(x_1)^q\right) \nonumber \\{} & {} \quad =-h^{q^t-q^2}L(x_2)^q\left( M(x_1)+\Big (1-h^{q^{t-1}-q}\Big )L(x_2)\right) . \end{aligned}$$

(19)

Since $L(x_2) \ne 0 \ne M(x_1)$, from (18) we get

$$\begin{aligned}{} & {} M(x_1)^q\left( x_1^q\Big (1+h^{q^{t}-q^2}\Big )a^q-M(x_1)a\right) \\{} & {} \quad =L(x_2)\left( h^{q^t-q^2}L(x_2)^q a^q+x_2^{q^t}\Big (1+h^{q^{t-1}-q}\Big )a\right) \end{aligned}$$

whence

$$\begin{aligned}{} & {} \left( x_1^q M(x_1)^q\big (1+h^{q^{t}-q^2}\big ) - h^{q^t-q^2}L(x_2)L(x_2)^q\right) a^q \nonumber \\{} & {} \quad =\left( M(x_1)M(x_1)^q+x_2^{q^t}(1+h^{q^{t-1}-q})L(x_2)\right) a. \end{aligned}$$

(20)

Next we want to show that the coefficient of $a^q$ in (20) cannot be 0. By way of contradiction, suppose that

$$\begin{aligned} x_1^q\left( 1+h^{q^{t}-q^2}\right) M(x_1)^q=h^{q^t-q^2}L(x_2)L(x_2)^q, \end{aligned}$$

(21)

from (19) it follows

$$\begin{aligned} x_1^q x_2^{q^t}\left( 1+h^{q^{t}-q^2}\right) \left( 1+h^{q^{t-1}-q}\right) =-h^{q^t-q^2}L(x_2)^qM(x_1). \end{aligned}$$

(22)

Since $x_1 \in \ker L$ and $x_2 \in \ker M$, this is equivalent to

$$\begin{aligned} x_1^{q^{t-2}-1}=-\big (x_2^q\big )^{q^{t-2}-1}\frac{\big (1+h^{q-q^{t-1}}\big )\big (1+h^{q^{t-2}-1}\big )}{\big (1+h^{1-q^{t+2}}\big )\big (1+h^{q^{2t-1}-q^{t-3}}\big )}. \end{aligned}$$

(23)

This formula is equivalent to

$$\begin{aligned} x_1^{q^{t-2}-1}=-\Biggl (x_2^q\cdot \frac{1+h^{1-q^{t+2}}}{\big (1+h^{q-q^{t-1}} \big )h^{q^{t-1}}} \Biggr )^{q^{t-2}-1}. \end{aligned}$$

(24)

Since

$$\begin{aligned} d=\gcd \left( 2t,t-2\right) =\gcd (4,t-2)=\left\{ \begin{array}{lll} 1, &{}\text{ if } t \text{ odd }\\ 2, &{}\hbox { if}\ t \equiv 0 \pmod 4\\ 4, &{}\hbox { if}\ t \equiv 2 \pmod 4 \end{array} \right. \end{aligned}$$

there exists a solution of the equation $x^{q^{t-2}-1}=-1$ for any $t \ge 3$ and $t \not \equiv 2 \pmod 4$ in ${{\mathbb {F}}}_{q^n}$. Thus, for $t\equiv 2 \pmod 4$, Eq. (24) gives a contradiction. In the remaining cases,

$$\begin{aligned} x_1= \omega x^q_2 \cdot \frac{1+h^{1-q^{t+2}}}{\big (1+h^{q-q^{t-1}} \big )h^{q^{t-1}}}, \end{aligned}$$

for some $\omega \in {{\mathbb {F}}}_{q^n}^*$ satisfying $\omega ^{q^{t-2}-1}=-1$. By substituting this expression in (21), since $x_1\in \ker L$, we get

$$\begin{aligned} \omega ^{q+1}=-1, \end{aligned}$$

(25)

and hence $\omega \in {{\mathbb {F}}}_{q^2}$.

If t is odd, since $\omega ^{q^{t-2}}=-\omega$, then $\omega ^q=-\omega$ and from (25), we get $\omega =\pm 1$. If $t \equiv 0 \pmod 4$, since $\omega =\omega ^{q^{{t-2}}}=-\omega$. In both cases we get a contradiction.

Then, by (20), we get

$$\begin{aligned} \begin{aligned} a^{q-1}&=\frac{M(x_1)M(x_1)^q-h^{q^t-q^{t+1}}x_2L(x_2)(1+h^{q^{t-1}-q})}{x_1^q(1+h^{q^{t}-q^2})M(x_1)^q-h^{q^t-q^2}L(x_2)L(x_2)^q}\\&=h^{q-1} \cdot \frac{1+h^{q^{t-1}-q}}{1+h^{q^t-q^2}} \cdot \frac{x_1M(x_1)-x_2L(x_2)}{x_1^qM(x_1)^q-x_2^qL(x_2)^q}\\&=\Biggl ( \frac{h}{(1+h^{q^{t-1}-q})(x_1M(x_1)-x_2L(x_2))} \Biggr )^{q-1}, \end{aligned} \end{aligned}$$

(26)

whence

$$\begin{aligned} a=\lambda \cdot \frac{h}{(1+h^{q^{t-1}-q})(x_1M(x_1)-x_2L(x_2))} \end{aligned}$$

for some $\lambda \in {{\mathbb {F}}}_{q}^*$. Since $a \in \ker R$, then

$$\begin{aligned}{} & {} \frac{h^{q^t}}{\left( (1+h^{q^{t-1}-q})(x_1M(x_1)-x_2L(x_2))\right) ^{q^t}}+ h^{q^{t-1}-q}\cdot \\{} & {} \quad \qquad \frac{h}{(1+h^{q^{t-1}-q})(x_1M(x_1)-x_2L(x_2))}=0. \end{aligned}$$

Recalling that $x_1 \in \ker L$ and $x_2 \in \ker M$, we get

$$\begin{aligned} x_1M(x_1)-x_2L(x_2) = x_1^{q^{t-1}+1}\left( 1+h^{1-q^{t-2}}\right) -x_2^{q+1}\left( 1+h^{1-q^{t+2}}\right) . \end{aligned}$$

This implies that

$$\begin{aligned} -\frac{1}{h^{q^t}(1+h^{q-q^{t-1}})}+\frac{h^{q^{t-1}-q}\cdot h}{1+h^{q^{t-1}-q}}=0, \end{aligned}$$

which means $h^{q^t+1}=1$, a contradiction. Then $\lambda _1$ may not belong to ${{{\mathbb {F}}}_q}$.

- $\textit{Case 3.2}$ Let $\lambda _1\notin {{\mathbb {F}}}_q$ and let a be a nonzero solution of System (17). If this system admits more than one solution, then each $2\times 2$ minor of the associated matrix of (17) is zero. In particular Eqs. (21) and (22) hold true, obtaining a contradiction as in the previous case.

Then, System (17) must admits a unique nonzero solution $(a,a^q) \in {{\mathbb {F}}}_{q^n}^2$. Bycomputing the ratio $a^{q-1}$ of its components, we get

$$\begin{aligned} \begin{aligned} a^{q-1}&=\frac{\begin{vmatrix} L(x_2)&-M(x_1) \\ M(x_1)^q&x_2^{q^t}(1+h^{q^{t-1}-q}) \end{vmatrix}}{\begin{vmatrix} x_1^q(1+h^{q^{t}-q^2})&L(x_2) \\ h^{q^t-q^2}L(x_2)^q&M(x_1)^q \end{vmatrix}}\\&=\frac{M(x_1)M(x_1)^q-h^{q^t-q^{t+1}}x_2L(x_2)(1+h^{q^{t-1}-q})}{x_1^q(1+h^{q^{t}-q^2})M(x_1)^q-h^{q^t-q^2}L(x_2)L(x_2)^q}. \end{aligned} \end{aligned}$$

This is again Eq. (26). Repeating the arguments as in Case 3.1 we get a contradiction. $\square$

4 A New Family of MRD Codes

Let start by the following preliminary general result.

Lemma 4.1

Let f(x) be a scattered polynomial and let ${{\mathcal {C}}}_f$ denote the associated MRD code. Then ${{\,\mathrm{{Aut}}\,}}({{\mathcal {C}}}_f)$ consists of elements $(\alpha x^{q^m},L_2,\sigma )\in {{\mathcal {L}}}_{n,q}[x]\times {{\mathcal {L}}}_{n,q}[x]\times \textrm{Aut}({{\mathbb {F}}}_q)$ with $L_2$ an invertible map, $\alpha \in {{\mathbb {F}}}_{q^n}^*$ and $m \in \{0,1,\ldots , n-1\}$ such that

$$\begin{aligned}{{\mathcal {C}}}_{f^{\sigma q^m}} \circ x^{q^m}\circ L_2= {{\mathcal {C}}}_f.\end{aligned}$$

Also, for any $\alpha \in {{\mathbb {F}}}_{q^n}^*$ and $m \in \{0,1,\ldots ,n-1\}$, there is a bijection between each $L_2$ such that $(\alpha x^{q^m}, L_2, \sigma ) \in {{\,\mathrm{{Aut}}\,}}(\mathcal {C}_f)$ and each ${{\,\mathrm{{GL}}\,}}(2,q^n)$-equivalence map from $U_f$ to $U_{f^{\sigma q^m}}$, where $U_f =\{ (x,f(x)): x \in {{\mathbb {F}}}_{q^n} \}$. In particular, the multiplicative group $I_R({{\mathcal {C}}}_f) \setminus \{0\}$ and the ${{\,\mathrm{{GL}}\,}}(2,q^n)$-automorphism group of $U_f$ are isomorphic.

Proof

Suppose that $\varphi \in I_L({{\mathcal {C}}}_f)^*:=I_L({{\mathcal {C}}}_f)\setminus \{0\}$ and $(L_1, L_2,\sigma )\in {{\,\mathrm{{Aut}}\,}}({{\mathcal {C}}}_f)$ where $L_1,L_2$ are invertible ${{{\mathbb {F}}}_q}$-polynomials and $x^\sigma =x^{p^\ell }$ with $0 \le \ell \le r-1$. Then for any $g\in {{\mathcal {C}}}_f$, there exists an element $g'\in {{\mathcal {C}}}_f$ such that

$$\begin{aligned} \varphi \circ (L_1 \circ g^\sigma \circ L_2) = L_1 \circ g'^{\sigma } \circ L_2 \end{aligned}$$

which means

$$\begin{aligned} \varphi \circ (L_1 \circ x^{p^\ell } \circ g \circ x^{p^{nr-\ell }} \circ L_2) = L_1 \circ x^{p^\ell } \circ g' \circ x^{p^{nr-\ell }} \circ L_2 \\ ( L_1\circ x^{p^\ell })^{-1} \circ \varphi \circ (L_1 \circ x^{p^\ell }) \circ g=g'\in {{\mathcal {C}}}_f, \end{aligned}$$

and hence, $( L_1\circ x^{p^\ell })^{-1} \circ \varphi \circ (L_1 \circ x^{p^\ell }) \in I_L({{\mathcal {C}}}_f)$. By (1),

$$\begin{aligned} ( L_1\circ x^{p^\ell })^{-1} \circ \varphi \circ (L_1 \circ x^{p^\ell })=\gamma x \end{aligned}$$

for some $\gamma \in {{\mathbb {F}}}^*_{q^n}$ which is equivalent to say that $L_1^{-1} \circ \varphi \circ L_1 \in I_{L}({{\mathcal {C}}}_f)^*$. Thus, $L_1$ is in the normalizer of $I_L({{\mathcal {C}}}_f)^*$ in ${{\,\mathrm{{GL}}\,}}(n,q)$, and hence, by [17, p. 362], it is isomorphic to $({{\mathbb {F}}}^*_{q^n},\cdot )\rtimes {{\,\textrm{Gal}\,}}({{\mathbb {F}}}_{q^n}/ {{\mathbb {F}}}_q)$. Then, $L_1(x)=\alpha x^{q^m}$ for some $\alpha \in {{\mathbb {F}}}_{q^n}^*$ and $m \in \{0,1,\ldots , n-1\}$. So any element in ${{\,\mathrm{{Aut}}\,}}({{\mathcal {C}}}_f)$ is as in the statement and

$$\begin{aligned}{{\mathcal {C}}}_f= \alpha x^{q^m} \circ {{\mathcal {C}}}^\sigma _{f}\circ L_2={{\mathcal {C}}}_{f^{\sigma q^m}} \circ x^{q^m} \circ L_2.\end{aligned}$$

This completes the first part of the proof.

As the identity map is in ${{\mathcal {C}}}_{f^{\sigma q^m}}$, $h(x):=x^{q^m} \circ L_2\in {{\mathcal {C}}}_f$, that is $h(x)=ax+bf(x)$ for some $a,b\in {{\mathbb {F}}}_{q^n}$. Furthermore, ${{\mathcal {C}}}_{f^{\sigma q^m}} \circ h = {{\mathcal {C}}}_f$ also implies the existence of $c,d\in {{\mathbb {F}}}_{q^n}$ such that

$$\begin{aligned} cx+df(x)=f^{\sigma q^m}\left( ax+bf(x)\right) , \end{aligned}$$

(27)

for all $x\in {{\mathbb {F}}}_{q^n}$. By setting $y=ax+bf(x)$, (27) is equivalent to

$$\begin{aligned} \begin{pmatrix} a&{}b\\ c&{}d \end{pmatrix} \begin{pmatrix} x\\ f(x) \end{pmatrix}= \begin{pmatrix} y\\ f^{\sigma q^m }(y) \end{pmatrix}. \end{aligned}$$

(28)

Also, the matrix $\begin{pmatrix} a&{}b\\ c&{}d \end{pmatrix}$ is invertible. Indeed, if there exists $\lambda \in {{\mathbb {F}}}_{q^n}^*$ such that $(c,d)=\lambda (a,b)$ then by (27) we get

$$\begin{aligned}\lambda h(x)=f^{\sigma q^m}(h(x))=x^{\sigma q^m}\circ f\circ x^{-\sigma q^m}\circ h,\end{aligned}$$

and hence $f(k(x))=\mu k(x)$ where $k(x)=x^{-\sigma q^{m}} \circ h(x)$ and $\mu =\lambda ^{-\sigma q^m}\in {{\mathbb {F}}}_{q^n}^*$. Since k(x) is invertible, then $f(x)=\mu x$ contradicting the fact that $L_f$ is scattered. This means that $U_f$ is ${{\,\mathrm{{GL}}\,}}(2,q^n)$-equivalent to $U_{f^{\sigma q^m}}$. Also from (27), it can be easily seen that (c, d) is uniquely determined by (a, b). Therefore, there is a 1–1 correspondence between every $L_2$ and every matrix $\begin{pmatrix} a&{}b\\ c&{}d \end{pmatrix}$ mapping $U_f$. In particular, when $f=f^{\sigma q^m}$, all such $\begin{pmatrix} a&{}b\\ c&{}d \end{pmatrix}$ form the ${{\,\mathrm{{GL}}\,}}(2,q^n)$-automorphism group of $U_f$, denoted by $G_f$. As each nonzero element of $I_R({{\mathcal {C}}}_f)$ is invertible (see [21, Corollary 5.6]), it is straightforward to see that the map described above determines an isomorphism between the groups $G_f$ and $I_R({{\mathcal {C}}}_f){\setminus } \{0\}$. Hence, the result follows. $\square$

Let $\psi _{h,t}$ be defined as in (3) and

$$\begin{aligned} {{\mathcal {C}}}_{h,t}=\{ax+b\psi _{h,t}(x): a,b \in {{\mathbb {F}}}_{q^n}\}. \end{aligned}$$

(29)

By the argument stated in [27, Section 5], ${{\mathcal {C}}}_{h,t}$ is an ${{\mathbb {F}}}_{q^n}$-linear MRD code.

The following result is about the equivalence among ${{\mathcal {C}}}_{h,t}$’s for different h and the automorphism group of ${{\mathcal {C}}}_{h,t}$.

Theorem 4.2

Let $n=2t$ with $t>4$. For each $h,k\in {{\mathbb {F}}}_{q^n}$ satisfying $h^{q^t+1}=k^{q^t+1}=-1$, the following hold

• (a) If $t\not \equiv 2\pmod {4}$, then ${{\mathcal {C}}}_{h,t}$ and ${{\mathcal {C}}}_{k,t}$ are equivalent if and only if $h=\pm k^\rho$ where $\rho \in {{\,\mathrm{{Aut}}\,}}({{\mathbb {F}}}_{q^n})$;

• (b) If $t\equiv 2\pmod {4}$, then ${{\mathcal {C}}}_{h,t}$ and ${{\mathcal {C}}}_{k,t}$ are equivalent if and only if $h=\ell k^\rho$ where $\ell ^{q^2+1}=1$ and $\rho \in {{\,\mathrm{{Aut}}\,}}({{\mathbb {F}}}_{q^n})$.

The full automorphism group ${{\,\mathrm{{Aut}}\,}}({{\mathcal {C}}}_{h,t})$ is isomorphic to $({{\mathbb {F}}}_{q^n}^*,\cdot )\times ({{\mathbb {F}}}_{q^2}^*,\cdot )\rtimes H$, where $H=\{\rho \in {{\,\mathrm{{Aut}}\,}}({{\mathbb {F}}}_{q^n}): h=\pm h^\rho \}$ if $t\not \equiv 2\pmod {4}$, and $H=\{\rho \in {{\,\mathrm{{Aut}}\,}}({{\mathbb {F}}}_{q^n}): (h^\rho /h)^{q^2+1}=1 \}$ if $t\equiv 2\pmod {4}$.

Proof

Let $U_h=\{(x,\psi _{h,t}(x)): x\in {{\mathbb {F}}}_{q^n}\}$ and $U_k=\{(x,\psi _{k,t}(x)): x\in {{\mathbb {F}}}_{q^n}\}$. By Theorem 2.1, we only have to consider the $\Gamma L(2,q^n)$-equivalence between $U_h$ and $U_k$. Since $U_k$ is $\textrm{GL}(2,q^n)$-equivalent to $U_{k^{\rho }}$ for any $\rho$ in $\textrm{Aut}({\mathbb {F}}_{q^n})$, this is equivalent to study the $\textrm{GL}(2,q^n)$-equivalence between $U_h$ and $U_{k^\rho }$. This also means that, to get (a) and (b), we can replace $k^\rho$ by k and we just need to show that a necessary and sufficient condition for that $U_h$ is ${{\,\mathrm{{GL}}\,}}(2,q^n)$-equivalent to $U_k$ is

$$\begin{aligned} h={\left\{ \begin{array}{ll} \pm k, &{} t\not \equiv 2\pmod {4};\\ \ell k, &{} t\equiv 2\pmod {4}, \end{array}\right. } \end{aligned}$$

(30)

where $\ell \in {{\mathbb {F}}}_{q^n}$ satisfies $\ell ^{q^2+1}=1$.

Hence, we only have to consider the existence of invertible matrix $\begin{pmatrix} a&{}b\\ c&{}d \end{pmatrix}$ over ${{\mathbb {F}}}_{q^n}$ such that for each $x\in {{\mathbb {F}}}_{q^n}$ there exists $y\in {{\mathbb {F}}}_{q^n}$ satisfying

$$\begin{aligned} \begin{pmatrix} a&{}b\\ c&{}d \end{pmatrix} \begin{pmatrix} x\\ \psi _{h,t}(x) \end{pmatrix}= \begin{pmatrix} y\\ \psi _{k,t}(y) \end{pmatrix}. \end{aligned}$$

This is equivalent to

$$\begin{aligned} cx+d\psi _{h,t}(x)=\psi _{k,t}\left( ax+b\psi _{h,t}(x)\right) , \end{aligned}$$

(31)

for all $x\in {{\mathbb {F}}}_{q^n}$. The right-hand-side of (31) is

$$\begin{aligned}&\left( b^q+k^{1-q^{t+1}}b^{q^{t+1}} h^{q^{t+1}-q^2}\right) x^{q^2} + \left( b^{q^{t-1}}h^{q^{t-1}-q^{t-2}} + b^{q^{2t-1}}k^{1-q^{2t-1}}\right) x^{q^{t-2}} \\&\quad +\left( b^q+b^{q^{t-1}}-h^{q^{t+1}-q^t}k^{1-q^{t+1}}b^{q^{t+1}} -k^{1-q^{2t-1}}h^{q^{2t-1}-q^t}b^{q^{2t-1}}\right) x^{q^t}\\&\quad +\left( b^{q^{t-1}}+k^{1-q^{2t-1}}h^{q^{2t-1}-q^{2t-2}}b^{q^{2t-1}} \right) x^{q^{2t-2}} - \left( h^{q-q^{t+2}}b^q+k^{1-q^{t+1}}b^{q^{t+1}}\right) x^{q^{t+2}}\\&\quad +\left( b^qh^{q-1} - b^{q^{t-1}}h^{q^{t-1}-1}-k^{1-q^{t+1}}b^{q^{t+1}}+k^{1-q^{2t-1}}b^{q^{2t-1}}\right) x+\psi _{k,t}(ax). \end{aligned}$$

As $t>4$, it is easy to see that the coefficients of $x^{q^2}$, $x^{q^{t-2}}$, $x^{q^t}$, $x^{q^{t+2}}$ and $x^{q^{2t-2}}$ in the right-hand-side of (31) must be 0. Depending on whether the value of b equals 0 or not, we separate the proof into two cases.

Case 1 $b\ne 0$. By the coefficient of $x^{q^2}$ (or equivalently, by the coefficient of $x^{q^{t+2}}$), we get

$$\begin{aligned} b^{q^{t+1}-q}=-k^{-q-1}h^{q^2+q}. \end{aligned}$$

(32)

Similarly, by the coefficient of $x^{q^{t-2}}$ (or equivalently, by the coefficient of $x^{q^{2t-2}}$), we get

$$\begin{aligned} b^{q^{t+1}-q}=-k^{-q+q^2}h^{q-1}. \end{aligned}$$

(33)

By (32) and (33), we obtain

$$\begin{aligned} h^{q^2+1}=k^{q^2+1}.\end{aligned}$$

Let $\ell =h/k$. Then $\ell ^{q^2+1}=1$. By the assumption that $h^{q^t+1}=k^{q^t+1}=-1$, we have $\ell ^{q^t+1}=1$.

If $t\not \equiv 2 \pmod {4}$, then $\ell =\pm 1$; if $t\equiv 2 \pmod {4}$, then we obtain $\ell ^{q^2+1}=1$, which implies $\ell \in {{\mathbb {F}}}_{q^4}$.

Case 2 $b=0$. If (31) holds, then $c=0$ and

$$\begin{aligned} {\left\{ \begin{array}{ll} d=a^q=a^{q^{t-1}}\\ dh^{1-q^{t+1}} = k^{1-q^{t+1}} a^{q^{t+1}}\\ dh^{1-q^{2t-1}} =k^{1-q^{2t-1}}a^{q^{2t-1}}. \end{array}\right. } \end{aligned}$$

(34)

The first equation in (34) implies that $a \in {{\mathbb {F}}}_{q^{\gcd (2t,t-2)}}$ which means $a\in {{\mathbb {F}}}_{q^{\gcd (t-2, 4)}}$. Let $\ell =h/k$. The last two equations in (34) become

$$\begin{aligned} d\ell ^{1-q^{t+1}}=a^{q^3}=d\ell ^{1-q^{2t-1}}. \end{aligned}$$

(35)

Thus $\ell ^{q^{2t-1}-q^{t+1}}=1$. This means $\ell \in {{\mathbb {F}}}_{q^{\gcd (t-2, 2t)}}={{\mathbb {F}}}_{q^{\gcd (t-2, 4)}}$. By the assumption that $h^{q^t+1}=k^{q^t+1}=-1$, we have $\ell ^{q^t+1}=1$. If $\gcd (t-2,4)\in \{1,2\}$ i.e. $t\not \equiv 2\pmod {4}$, then $\ell =\pm 1$ which means $\psi _{k,t}=\psi _{h,t}$ and ${{\mathcal {C}}}_{h,t}$ and ${{\mathcal {C}}}_{k,t}$ are the same; if $\gcd (t-2,4)=4$ i.e. $t\equiv 2\pmod {4}$, then we obtain $\ell ^{q^2+1}=1$.

Next, let us further consider the case $t\equiv 2\pmod {4}$. By (35),

$$\begin{aligned} a^{q^3-q}=\ell ^{1-q^{t+1}}=\ell ^{1-q^3}=\ell ^{(1-q)(q^2+q+1)}=\ell ^{q-q^2}=\ell ^{q+1}. \end{aligned}$$

For a given $\ell \in {{\mathbb {F}}}_{q^4}$, we can always find $a\in {{\mathbb {F}}}_{q^4}$ satisfying the above equation. Moreover, it is routine to verify that such a satisfies (34) provided that $h=\ell k$ with $\ell ^{q^2+1}=1$; note here that by (34) d depends on a. This complete the proof ofequivalence between ${{\mathcal {C}}}_{h,t}$ and ${{\mathcal {C}}}_{k,t}$.

In the final part of this proof, we determine the automorphism group of ${{\mathcal {C}}}_{h,t}$. By the first part of Lemma 4.1, ${{\,\mathrm{{Aut}}\,}}({{\mathcal {C}}}_{h,t})$ consists of $(\alpha x^{q^m}, L_2, \sigma )\in {{\mathcal {L}}}_{n,q}[x] \times {{\mathcal {L}}}_{n,q}[x] \times {{\,\mathrm{{Aut}}\,}}({{\mathbb {F}}}_{q})$ such that

$$\begin{aligned} {{\mathcal {C}}}_{h,t} = \left\{ f^{\sigma q^m} \circ x^{q^m} \circ L_2:f\in {{\mathcal {C}}}_{h,t} \right\} . \end{aligned}$$

Note that $\alpha$ can be any element in ${{\mathbb {F}}}_{q^n}^*$, because we always have $\alpha (a+b\psi _{h,t}(x))\in {{\mathcal {C}}}_{h,t}$ for any $a,b\in {{\mathbb {F}}}_{q^{n}}$.

By the second part of Lemma 4.1, for a given m, there is a bijection between each $L_2$ and each invertible matrix $\begin{pmatrix} a &{} b\\ c&{} d\end{pmatrix}$ such that (28) holds for $f=\psi _{h,t}$. More precisely, following the proof of Lemma 4.1,

$$\begin{aligned} L_2(x)=(ax+b\psi _{h,t}(x))^{q^{n-m}}. \end{aligned}$$

(36)

Write $\rho =\sigma q^m$. Consequently, we only have to determine all the $\rho \in {{\,\mathrm{{Aut}}\,}}({{\mathbb {F}}}_{q^n})$ and all the elements in ${{\,\mathrm{{GL}}\,}}(2,q^n)$ mapping $U_h$ to $U_{h^\rho }$. By (a) and (b), we see that $\rho$ must satisfy the condition $h^{\rho }=\pm h$ for $t\not \equiv 2\pmod {4}$, and $(h^\rho /h)^{q^2+1}=1$ for $t\equiv 2\pmod {4}$. All such $\rho$ form a subgroup $H\subseteq {{\,\mathrm{{Aut}}\,}}({{\mathbb {F}}}_{q^n})$.

To accomplish the proof, we just need to continue the computation of the first part for $k=h$ and determine which matrix $\begin{pmatrix} a&{}b\\ c&{}d \end{pmatrix}$ defines an equivalence map from $U_h$ to itself. Indeed, all such maps provide a subgroup K, which is normal in ${{\,\mathrm{{Aut}}\,}}(C_{h,t})$,

where

$$\begin{aligned} K=\left\{ (\alpha x, L_2, \textrm{id}): \alpha \in {{\mathbb {F}}}_{q^{n}}^*, L_2\in {{\mathcal {L}}}_{n,q}[x] \text { invertible,\ } {{\mathcal {C}}}_{h,t}= \{\alpha f(L_2(x)): f \in {{\mathcal {C}}}_{h,t}\} \right\} . \end{aligned}$$

Moreover, every other element in ${{\,\mathrm{{Aut}}\,}}({{\mathcal {C}}}_{h,t})$ which corresponds to a map from $U_h$ to $U_{h^\rho }$ with $\rho \ne \textrm{id}$ belongs to a coset of K in ${{\,\mathrm{{Aut}}\,}}({{\mathcal {C}}}_{h,t})$, and ${{\,\mathrm{{Aut}}\,}}({{\mathcal {C}}}_{h,t})/K \cong H$.

Depending on whether $b=0$ or not, we consider two cases to determine the elements in K.

When $b=0$, we only have to let $k=h$ in (34). From there we derive that $d=a^q=a^{q^{t-1}}=a^{q^{t+1}}$. Therefore $a\in {{\mathbb {F}}}_{q^{\gcd (t,2)}}^*$.

When $b\ne 0$, by letting $k=h$, we see that the coefficient of $x^{q^t}$ in the right-hand-side of (31) is

$$\begin{aligned} b^q+b^{q^{t-1}}-h^{1-q^t}b^{q^{t+1}} - h^{1-q^t}b^{q^{2t-1}} \end{aligned}$$

which must be 0. Plugging (32) into it and taking into account that $h^{q^t}=-1/h$, we get

$$\begin{aligned}b^q\big (1+h^{q^2-q^t}\big ) + b^{q^{t-1}}\big (1+h^{1-q^{t-2}}\big )=0.\end{aligned}$$

Raising it to the $q^2$-th power and plugging (32) again, we obtain

$$\begin{aligned} b^{q^3} \big (1-h^{q^4+q^2}\big ) - b^q\big (h^{q^2-1} - h^{2q^2}\big )=0,\end{aligned}$$

which means

$$\begin{aligned}b^q\big (h^{-1}-h^{q^2}\big ) =\left( b^q\big (h^{-1}-h^{q^2}\big ) \right) ^{q^2}.\end{aligned}$$

By Proposition 3.2, $h^{-1}\ne h^{q^2}$. Thus

$$\begin{aligned}b=\frac{-\delta }{h^{-q^{2t-1}}-h^q}=\frac{\delta }{h^{q^{t-1}}+h^q}\end{aligned}$$

for some $\delta \in {{\mathbb {F}}}_{q^2}$. Substitute it into (32),

$$\begin{aligned}\frac{\delta ^{q^t}}{h^{q^{2t-1}}+h^{q^{t+1}}} - h^{q+q^{t-1}}\frac{\delta }{h^{q^{t-1}}+h^q}=0,\end{aligned}$$

which means

$$\begin{aligned} \delta ^{q^t}+\delta =0. \end{aligned}$$

(37)

When t is even, it follows that $2\delta =0$. Thus b must be 0 which contradicts the assumption that $b\ne 0$. Thus, by (36), the assumption $m=0$ and the above discussions for $b=0$ and $b\ne 0$, we have determined the subgroup

$$\begin{aligned}K=\left\{ (\alpha , ax, \textrm{id}): \alpha \in {{\mathbb {F}}}_{q^{n}}^*, a\in {{\mathbb {F}}}_{q^2}^* \right\} . \end{aligned}$$

When t is odd, (37) implies $\delta ^{q}=-\delta$. Plugging this value back into (31), we see that the coefficients of $x^{q^2}$, $x^{q^{t-2}}$, $x^{q^t}$, $x^{q^{t+2}}$ and $x^{q^{2t-2}}$ on the right-hand-side of (31) are all 0. Consequently, (31) becomes

$$\begin{aligned} cx+d\psi _{h,t}(x){} & {} =\left( b^qh^{q-1} - b^{q^{t-1}}h^{q^{t-1}-1}-h^{1-q^{t+1}}b^{q^{t+1}}+h^{1-q^{2t-1}}b^{q^{2t-1}}\right) x\\{} & {} \quad +\psi _{h,t}(ax). \end{aligned}$$

By simply setting c equal the coefficient of x in the right-hand-side of the above equation, which means c is determined by b, we only have to guarantee $d\psi _{h,t}(x)=\psi _{h,t}(ax)$. It reduces to the same computation for $b=0$, in which we get $d=a^q=a^{q^{t-1}}=a^{q^{t+1}}$ with $a\in {{\mathbb {F}}}_{q}$, because $\gcd (t,2)=1$. The only difference is that, under the assumption that $b\ne 0$, we allow a to be 0. Therefore,

$$\begin{aligned} K= & {} \left\{ (\alpha x, ax+b\psi _{h,t}(x), \textrm{id}): \alpha \in {{\mathbb {F}}}_{q^{n}}^*, a\in {{\mathbb {F}}}_{q}, \right. \\{} & {} \qquad \, \left. b=\frac{\delta }{h^{q^{t-1}}+h^q}\text { with }\delta ^q=-\delta , (a,b)\ne (0,0) \right\} . \end{aligned}$$

One may check directly that all such $\begin{pmatrix} a &{} b\\ c &{} d \end{pmatrix}$ form a cyclic group of order $q^2-1$, which also follows from the fact that the nonzero elements of the right idealizer

$$\begin{aligned} \nonumber I_R({{\mathcal {C}}}_{h,t})=&\{\varphi \in {{\mathcal {L}}}_{n,q}[x]: f\circ \varphi \in {{\mathcal {C}}}_{h,t} \text { for all }f\in {{\mathcal {C}}}_{h,t} \}\\ =&\left\{ ax+b\psi _{h,t}(x): a\in {{\mathbb {F}}}_{q}, b=\frac{\delta }{h^{q^{t-1}}+h^q} \text { with }\delta ^q=-\delta \right\} \end{aligned}$$

(38)

form the multiplicative group of a finite field.

Hence, no matter t is even or odd, K is isomorphic to $({{\mathbb {F}}}_{q^n}^*,\cdot )\times ({{\mathbb {F}}}_{q^2}^*,\cdot )$. Therefore, ${{\,\mathrm{{Aut}}\,}}({{\mathcal {C}}}_{h,t})\cong ({{\mathbb {F}}}_{q^n}^*,\cdot )\times ({{\mathbb {F}}}_{q^2}^*,\cdot )\rtimes H$. $\square$

Theorem 4.2 shows that our construction provides a big family of inequivalent MRD codes.

Corollary 4.3

Let p be an odd prime number and let r, t be positive integers with $t>4$ and $q=p^r$. The total number N of inequivalent MRD codes ${{\mathcal {C}}}_{h,t}$ is

$$\begin{aligned} N\ge {\left\{ \begin{array}{ll} \left\lfloor \frac{q^t+1}{4rt}\right\rfloor ,&{} \text { if }t\not \equiv 2\pmod {4};\\ \left\lfloor \frac{q^t+1}{2rt(q^2+1)}\right\rfloor ,&{} \text { if }t\equiv 2\pmod {4}. \end{array}\right. } \end{aligned}$$

Proof

In the proof of Theorem 4.2, we have obtained a necessary and sufficient condition (30) for the ${{\,\mathrm{{GL}}\,}}(2,q^n)$-equivalence between $U_{h}$ and $U_{k}$, $n=2t$. For a given h satisfying $h^{q^t+1}=-1$, let $\xi _h$ denote the number of k for which $U_k$ is${{\,\mathrm{{GL}}\,}}(2,q^n)$-equivalent to $U_h$. The value of $\xi _h$ is independent of h and

$$\begin{aligned}\xi _h={\left\{ \begin{array}{ll} 2, &{} t\not \equiv 2\pmod {4};\\ q^2+1, &{} t\equiv 2\pmod {4}. \end{array}\right. }\end{aligned}$$

As there are at most $|{{\,\mathrm{{Aut}}\,}}({{\mathbb {F}}}_{q^n})|=rn$ different $\rho$ such that $U_h$ is ${{\,\mathrm{{GL}}\,}}(2,q^n)$-equivalent to $U_{k^\sigma }$ for a given h, there are at most $nr\xi _h$ choices of k for which $U_{h}$ is $\Gamma L(2,q^n)$-equivalent to $U_{k}$. Therefore, we have obtained the lower bound for N which concludes the proof. $\square$

By Corollary 4.3, we can prove the following result.

Theorem 4.4

Let $n=2t$ with $t>4$ and let q be an odd prime power. The family of ${{\mathbb {F}}}_{q^n}$-linear MRD codes of minimum distance $n-1$

$$\begin{aligned} {{\mathcal {C}}}_{h,t}=\{ ax+b\psi _{h,t}(x): a,b\in {{\mathbb {F}}}_{q^n} \}, \end{aligned}$$

where $\psi _{h,t}(x)= x^{q}+x^{q^{t-1}}-h^{1-q^{t+1}}x^{q^{t+1}}+h^{1-q^{2t-1}}x^{q^{2t-1}}\in {{\mathbb {F}}}_{q^n}[x]$ and h is any element of ${{\mathbb {F}}}_{q^n}$ such that $h^{q^t+1}=-1$, contains examples which are not equivalent to any known ones in Table 1.

Proof

As q can be any odd prime power and n can be any even integer larger than 6, by comparing Corollary 4.3 with the numbers of known inequivalent constructions of ${{\mathbb {F}}}_{q^n}$-linear MRD codes of minimum distance $n-1$ in Table 1, our family must be new. $\square$

Remark 4.5

In a recent paper [24], our family ${{\mathcal {C}}}_{h,t}$ of MRD codes has been generalized by Neri, Santonastaso and Zullo with an extra field automorphism $\sigma$ as follows

$$\begin{aligned} {{\mathcal {C}}}_{h,t,\sigma } = \{ax+b\psi _{h,t,\sigma }(x): a,b\in {{\mathbb {F}}}_{q^{n}} \} \end{aligned}$$

with

$$\begin{aligned} \psi _{h,t,\sigma }(x)=x^\sigma +x^{\sigma ^{t-1}} + h \sigma (h) x^{\sigma ^{t+1}} + h \sigma ^{-1}(h^{-1}) x^{\sigma ^{2t-1}}, \end{aligned}$$

where $\sigma$ is a generator of ${{\,\textrm{Gal}\,}}({{\mathbb {F}}}_{q^n}|{{\mathbb {F}}}_q)$, $n=2t$ with $t\ge 3$ and $h\in {{\mathbb {F}}}_{q^n}$ such that $h^{q^t+1}=-1$. If we take $\sigma (x) =x^q$, then ${{\mathcal {C}}}_{h,t,\sigma }$ coincides with ${{\mathcal {C}}}_{h,t}$. In [24, Section 4], a strengthened version of Theorem 4.4 has been proved: any member of the family ${{\mathcal {C}}}_{h,t,\sigma }$ is not equivalent to the MRD codes in family (i) for $t\ge 3$ or (ii) for $t>4$ in Table 1. The first part of Theorem 4.2 for the members in ${{\mathcal {C}}}_{h,t,\sigma }$ has been also proved in [24, Section 4]. Instead of considering the equivalence map between subspaces $U_h$ and $U_{k^\rho }$ as we did in the proof of Theorem 4.2, they prove it using the rank-metric code equivalence directly. However, if we wanted to further determine the automorphism group of $\textrm{Aut}(\mathcal {C}_{h,t})$, then we would need to find every map preserving $\mathcal {C}_{h,t}$. Nonetheless, Propositions 3.8 and 3.9 in [29] do not seem to be enough toward this aim. This is the reason why we proved Lemma 4.1 and Theorem 4.2 (the second part about automorphism groups) without checking the definition of MRD-code equivalence directly.

Finally, in [14], the authors complete the study of the equivalence issue of the MRD codes ${{\mathcal {C}}}_{h,t,\sigma }$ for $t \in \{3,4\}$.

The following result is a direct consequence of (38) in the proof of Theorem 4.2.

Corollary 4.6

Let ${{\mathcal {C}}}_{h,t}$ be defined as in (29), with $t>4$. Then $I_R({{\mathcal {C}}}_{h,t})\cong {{\mathbb {F}}}_{q^{2}}$.

Finally, let us investigate the adjoint of $\psi _{h,t}$ and the associated MRD code.

Theorem 4.7

The ${{\mathbb {F}}}_{q^n}$-linear MRD code $\hat{{{\mathcal {C}}}}_{h,t}=\{ ax+b\hat{\psi }_{h,t}(x): a,b\in {{\mathbb {F}}}_{q^n} \}$, where $\hat{\psi }_{h,t}$ is the adjoint map of $\psi _{h,t}$ is equivalent to ${{\mathcal {C}}}_{h,t}$.

Proof

Consider the polynomial

$$\begin{aligned}g(x):=h\hat{\psi }_{h,t}(x/h) =x^q - x^{q^{t-1}} + h^{1-q^{t+1}} x^{q^{t+1}}+ h^{1-q^{2t-1}}x^{q^{2t-1}},\end{aligned}$$

we investigate the equivalence between ${{\mathcal {C}}}_{g}$ and ${{\mathcal {C}}}_{h,t}$.

To prove that ${{{\mathcal {C}}}}_{g}$ is equivalent to ${{\mathcal {C}}}_{h,t}$, as in the proof of Theorem 4.2, we only have to consider the existence of invertible matrix $\begin{pmatrix} a&{}b\\ c&{}d \end{pmatrix}$ over ${{\mathbb {F}}}_{q^n}$ such that for each $x\in {{\mathbb {F}}}_{q^n}$ there exists $y\in {{\mathbb {F}}}_{q^n}$ satisfying

$$\begin{aligned} \begin{pmatrix} a&{}b\\ c&{}d \end{pmatrix} \begin{pmatrix} x\\ g(x) \end{pmatrix}= \begin{pmatrix} y\\ \psi _{h,t}(y) \end{pmatrix}. \end{aligned}$$

Straightforward computation show that this happens for $a=d=0$, $b=\left( \frac{h}{h^{q^2+1}-1}\right) ^{q^{2t-1}}$ and $c=-(h^q+h^{q^{t-1}})$. $\square$

5 Equivalence of the Associated Linear Sets

Let $\psi _{h,t}$ be the scattered polynomial over ${{\mathbb {F}}}_{q^n}$, $n=2t$, defined in Theorem (3). Let

$$\begin{aligned} L_{h,t}:= \left\{ \langle (x,\psi _{h,t}(x)) \rangle _{{{\mathbb {F}}}_{q^n}}: x\in {{\mathbb {F}}}^*_{q^n}\right\} , \end{aligned}$$

(39)

which is a maximum scattered linear set of ${{\,\mathrm{{PG}}\,}}(1,q^n)$.

In this part, we consider the $P\Gamma L$-equivalence between $L_{h,t}$ and $L_{k,t}$. Our main result shows that there is a large number of inequivalent maximum scattered linear sets associated with our family of scattered linear sets.

Theorem 5.1

Let p be an odd prime number and let r, t be positive integers with $t>4$ and $q=p^r$. The total number M of inequivalent maximum scattered linear sets $L_{h,t}$ of ${{\,\mathrm{{PG}}\,}}(1,q^n)$, $n=2t$, satisfies

$$\begin{aligned} M\ge {\left\{ \begin{array}{ll} \left\lfloor \frac{q^t+1}{8rt}\right\rfloor ,&{} \text { if }t\not \equiv 2\pmod {4};\\ \left\lfloor \frac{q^t+1}{4rt(q^2+1)}\right\rfloor ,&{} \text { if }t\equiv 2\pmod {4}. \end{array}\right. } \end{aligned}$$

Remark 5.2

We can provide an asymptotic estimation on the number of scattered linear sets we found over the number of known families. As our new family contains (iii) in Table 1 and (i) has only one member, we only have to know the number of inequivalent members of the Lunardon–Polverino family over ${{\mathbb {F}}}_{q^n}$, which has been recently determined in [30]. Let us denote this number by $\Lambda (n,q)$, whose precise value is quite complicated; see [30, Theorem 4.4]. However, for given n, the value of $\Lambda (n,q)$ is approximately $\frac{q \phi (n)}{4r}$ for odd n and $\frac{{q^2}\phi (n)}{8r}$ for even n provided that q is large enough. Together with Theorem 5.1, it follows that the number of our scattered linear sets over the number of known ones for given $n=2t>8$ is approximately at least

$$\begin{aligned} {\left\{ \begin{array}{ll} \dfrac{q^{n/2-2}}{\frac{n}{2}\phi (n)}, &{} \frac{n}{2}\not \equiv 2 \pmod {4};\\ \dfrac{q^{n/2-4}}{\frac{n}{4}\phi (n)}, &{} \frac{n}{2}\equiv 2 \pmod {4}, \end{array}\right. } \end{aligned}$$

provided that q is large enough.

To prove Theorem 5.1, we first restrict to the equivalence of linear sets under ${{\,\mathrm{{PGL}}\,}}(2,q^n)$ and consider two cases which will be handled in Lemmas 5.4 and 5.5, respectively. Then we will consider the $P\Gamma L(2,q^n)$-equivalence and present the proof of Theorem 5.1.

Let $f(x)=\sum _{i=0}^{n-1} \alpha _i x^{q^i}$ and $g(x)=\sum _{i=0}^{n-1} \beta _i x^{q^i}$ be two scattered polynomials over ${{\mathbb {F}}}_{q^n}$ with $\alpha _0=\beta _0=0$. The associated linear sets $L_f$ and $L_g$ are ${{\,\mathrm{{PGL}}\,}}(2,q^n)$-equivalent if and only if there exists an invertible matrix $\begin{pmatrix} a &{} b\\ c &{} d \end{pmatrix}$ over ${{\mathbb {F}}}_{q^n}$ such that

$$\begin{aligned} \left\{ \frac{f(x)}{x}:x\in {{\mathbb {F}}}_{q^n}^* \right\} =\left\{ \frac{cx+dg(x)}{ax+bg(x)}:x\in {{\mathbb {F}}}_{q^n}^* \right\} . \end{aligned}$$

(40)

Remark 5.3

Note that $ax+bg(x)$ has no nonzero solution in ${{\mathbb {F}}}_{q^n}$, otherwise the linear set $L_f$ would contain the point $\langle (0,1) \rangle _{{{\mathbb {F}}}_{q^n}}$.

Depending on whether the value of b equals 0 or not, we may consider the equivalence in two cases.

If $b=0$, then we can assume that $a=1$, and (40) becomes

$$\begin{aligned} \left\{ \sum _{i=0}^{n-1} \alpha _i x^{q^i-1}:x\in {{\mathbb {F}}}_{q^n}^* \right\} = \left\{ c+d\sum _{i=0}^{n-1} \beta _i x^{q^i-1}:x\in {{\mathbb {F}}}_{q^n}^* \right\} , \end{aligned}$$

i.e.

$$\begin{aligned} \left\{ \sum _{i=0}^{n-1} \alpha _i x^{q^i}:x\in {{\mathbb {F}}}_{q^n}^* \right\} = \left\{ cx+d\sum _{i=0}^{n-1} \beta _i x^{q^i}:x\in {{\mathbb {F}}}_{q^n}^* \right\} .\end{aligned}$$

By (a) of Lemma 2.2, $\alpha _0=c+d\beta _0$. As $\alpha _0=\beta _0=0$, c must be 0. Hence

$$\begin{aligned} \left\{ \sum _{i=1}^{n-1} \alpha _i x^{q^i-1}:x\in {{\mathbb {F}}}_{q^n}^* \right\} = \left\{ d\sum _{i=1}^{n-1} \beta _i x^{q^i-1}:x\in {{\mathbb {F}}}_{q^n}^* \right\} . \end{aligned}$$

(41)

Lemma 5.4

Suppose $f=\psi _{h,t}$, $g=\psi _{k,t}$ with $t > 4$. If there exists d such that (41) holds, then $(h/k)^{q^2+1}=1$ which means $U_h$ is $\Gamma L(2,q^n)$-equivalent to $U_k$.

Proof

By (b) of Lemma 2.2,

$$\begin{aligned}\alpha _j\alpha _{n-j}^{q^j} = d^{q^j+1}\beta _k\beta _{n-j}^{q^j}\end{aligned}$$

for $j=1,2,\cdots , n-1$. Plugging the coefficients of $\psi _{h,t}$ and $\psi _{k,t}$, we get

$$\begin{aligned} \big (h^{1-q^{2t-1}}\big )^q=d^{q+1} \big (k^{1-q^{2t-1}}\big )^q \text { and } \big (-h^{1-q^{t+1}}\big )^{q^{t-1}}=d^{q^{t-1}+1} \big (-k^{1-q^{t+1}}\big )^{q^{t-1}}.\end{aligned}$$

By setting $\ell = h/k$, we get

$$\begin{aligned} {\left\{ \begin{array}{ll} \ell ^{q-1}=d^{q+1},\\ \ell ^{q^{t-1}-1} = d^{q^{t-1}+1}. \end{array}\right. } \end{aligned}$$

(42)

From (42), we derive

$$\begin{aligned} d^{(q+1)(q^{t-2}+q^{t-3}+\cdots +1)} = \ell ^{q^{t-1}-1 }= d^{q^{t-1}+1},\end{aligned}$$

which means $d^{2q(q^{t-3}+\cdots +1)}=1$. Hence $d^{q^{t-2}-1}= d^{2(q^{t-3}+\cdots +1)\cdot \frac{q-1}{2}}=1$. It follows that $\ell ^{q^{t-1}-1} =d^{q^{t-1}+1}=d^{q+1}=\ell ^{q-1}$, whence $\ell ^{q^{t-2}}=\ell$.

As $h^{q^t+1}=k^{q^t+1}=-1$, $\ell ^{q^t}=1/\ell$. By $\ell ^{q^{t-2}}=\ell$, we get $\ell ^{q^2+1}=1$. By Theorem 4.2, $U_h$ is $\Gamma L(2,q^n)$-equivalent to $U_k$. $\square$

Next we consider the case $b\ne 0$. Without loss of generality, we assume that $b=1$. Then

$$\begin{aligned} \frac{cx+dg(x)}{ax+g(x)} = \frac{(c-da)x+d(ax+g(x))}{ax+g(x)}=\frac{\bar{c}x}{ax+g(x)}+d,\end{aligned}$$

for any $x\in {{\mathbb {F}}}_{q^n}^*$, where $\bar{c}=c-da\ne 0$. Noting that $ax+g(x)=0$ must have no nonzero solution,

$$\begin{aligned}\frac{\bar{c}x}{ax+g(x)}+d=\frac{\bar{c}\bar{g}(y)}{y}+d, \end{aligned}$$

where $\bar{g}(y)=\sum _{i=0}^{n-1}\gamma _i y^{q^i}$ is the inverse of the map $x\mapsto ax+g(x)$.

Furthermore, (40) becomes

$$\begin{aligned} \left\{ \sum _{i=1}^{n-1} \alpha _i x^{q^i-1}:x\in {{\mathbb {F}}}_{q^n}^* \right\} = \left\{ \bar{c}\sum _{i=1}^{n-1}\gamma _i x^{q^i-1}+d+\bar{c}\gamma _0:x\in {{\mathbb {F}}}_{q^n}^* \right\} . \end{aligned}$$

By (a) of Lemma 2.2,

$$\begin{aligned} d+\bar{c}\gamma _0=0. \end{aligned}$$

(43)

Thus

$$\begin{aligned} \left\{ \frac{1}{\bar{c}} \sum _{i=1}^{n-1} \alpha _i x^{q^i-1}:x\in {{\mathbb {F}}}_{q^n}^* \right\} = \left\{ \sum _{i=1}^{n-1}\gamma _i x^{q^i-1}:x\in {{\mathbb {F}}}_{q^n}^* \right\} . \end{aligned}$$

(44)

Lemma 5.5

Let $f=\psi _{h,t}$, $g=\psi _{k,t}$ with $t >4$. Suppose that there exist a, c and d such that (44) holds.

• (a) If t is even, then a must be 0.

• (b) If t is odd and $a\ne 0$, then $f=g$.

In particular, when $a=0$, $\gamma _0=d=0$.

The proof of the above lemma is one of the most technical parts in this paper. Let us first give a sketch of it. The difficulty lies in the unknown coefficients $\gamma _i$ for $i=1,\ldots , n-1$. Thus we should derive necessary conditions on them as many as possible.

First, as $f=\psi _{h,t}$, there are only four terms with nonzero coefficients in the polynomials of the left-hand-side of (44). Hence, by Lemma 2.2 (b) and (c), all the other terms with zero coefficients can provide many equations on $\gamma _i$; see (45) to (48).

Secondly, by definition, $\gamma _i$ is defined by the inverse of $\psi _{k,t}$ which provides us another set of restrictions on $\gamma _i$; see (51) to (57).

By all these restrictions on $\gamma _i$ and some further computation into four different cases, we can prove that $a=0$ or a has to satisfy $\frac{1}{a^{q^t}} = -k^{q-q^{2t-1}} \frac{1}{a}$; see (59). By plugging (59) into some other necessary conditions derived from (44), we can finish the proof.

Proof

By the second identity in Lemma 2.2, we know that

$$\begin{aligned} \gamma _j \gamma _{2t-j}=0, \end{aligned}$$

(45)

for $j\in \{1,2,\cdots , 2t-1\} \setminus \{1, t-1, t+1, 2t-1\}$,

$$\begin{aligned} \gamma _1\gamma _{2t-1}^q= \left( \frac{1}{\bar{c}} \right) ^{q+1}\tau ^q, \end{aligned}$$

(46)

and

$$\begin{aligned} \gamma _{t-1} \gamma _{t+1}^{q^{t-1}}= \left( \frac{1}{\bar{c}} \right) ^{1+q^{t-1}}\theta ^{q^{t-1}}, \end{aligned}$$

(47)

where $\tau = h^{1-q^{2t-1}}$ and $\theta =-h^{1-q^{t+1}}=h^{1+q}$.

By the third identity in Lemma 2.2, we have

$$\begin{aligned} \gamma _1 \gamma _{j-1}^q \gamma _{2t-j}^{q^j} + \gamma _{j} \gamma _{2t-1}^q \gamma _{2t-j+1}^{q^j}=0, \end{aligned}$$

(48)

for $j\in \{2,3,\cdots , n-1\}$. Letting $j=2$, we obtain

$$\begin{aligned}\gamma _1 \gamma _1^q \gamma _{2t-2}^{q^2} + \gamma _2 \gamma _{2t-1}^q \gamma _{2t-1}^{q^2}=0. \end{aligned}$$

As $\gamma _2\gamma _{2t-2}=0$ and $\gamma _1\gamma _{2t-1}\ne 0$, we derive $\gamma _2=\gamma _{2t-2}=0$. Similarly, by letting $j=t-1$, we have

$$\begin{aligned}\gamma _1 \gamma _{t-2}^q \gamma _{t+1}^{q^{t-1}} + \gamma _{t-1} \gamma _{2t-1}^q \gamma _{t+2}^{q^{t-1}}=0. \end{aligned}$$

Since $\gamma _{t-2}\gamma _{t+2}=0$ and $\gamma _1,\gamma _{t+1}, \gamma _{t-1}, \gamma _{t+1}\ne 0$, from the above equation we deduce

$$\begin{aligned} \gamma _{t-2}=\gamma _{t+2}=0. \end{aligned}$$

Moreover, by (45), (48) and replacing $j-1$ by j in (48),

$$\begin{aligned} \gamma _j \ne 0 \Rightarrow \gamma _{2t-j}=\gamma _{2t-j+1}=\gamma _{2t-j-1}=0, \end{aligned}$$

(49)

for $j\in \{1,2,\cdots , 2t-1\} \setminus \{1, t-1, t+1, 2t-1\}$.

Now, we will use the fact that $\bar{g}(ax+g(x))=x$, namely,

$$\begin{aligned} \bar{g}(ax+ x^q + x^{q^{t-1}}+u x^{q^{t+1}} +vx^{q^{2t-1}} )=x, \end{aligned}$$

(50)

for all $x\in {{\mathbb {F}}}_{q^n}$, where $u=-k^{1-q^{t+1}}$ and $v=k^{1-q^{2t-1}}$. The coefficient of $x^{q^j}$ in the left-hand-side of (50) is

$$\begin{aligned} a^{q^j} \gamma _j+\gamma _{j-1} +\gamma _{j+t+1} + u^{q^{j+t-1}}\gamma _{j+t-1} + v^{q^{j+1}}\gamma _{j+1}. \end{aligned}$$

(51)

By letting $j=0, 1$, $2t-1$, $t+1$, $t-1$ and t in (51) and comparing it with the right-hand-side of (50), we get

$$\begin{aligned} a\gamma _0 + \gamma _{2t-1} + \gamma _{t+1} + u^{q^{t-1}}\gamma _{t-1} + v^q\gamma _{1}&=1, \end{aligned}$$

(52)

$$\begin{aligned} a^q\gamma _1 + \gamma _0&= 0, \end{aligned}$$

(53)

$$\begin{aligned} a^{q^{2t-1}}\gamma _{2t-1} + v\gamma _0&= 0, \end{aligned}$$

(54)

$$\begin{aligned} a^{q^{t+1}}\gamma _{t+1} +u\gamma _0&= 0, \end{aligned}$$

(55)

$$\begin{aligned} a^{q^{t-1}} \gamma _{t-1} +\gamma _0&= 0, \end{aligned}$$

(56)

$$\begin{aligned} \gamma _{t-1} +\gamma _1 +u^{q^{2t-1}}\gamma _{2t-1}+v^{q^{t+1}}\gamma _{t+1}&=0. \end{aligned}$$

(57)

Here we have used the result that $\gamma _t=\gamma _2=\gamma _{2t-2}=\gamma _{t+2}=\gamma _{t-2}=0$.

It is clear that if $a=0$, then $\gamma _0$ must be 0. By (43), $d=0$.

Assume that $a\ne 0$. By (53), (54), (55) and (56) into (52) and (57), we see that $\gamma _0\ne 0$ is completely determined by a, and

$$\begin{aligned} -\gamma _0\left( \frac{1}{a^{q^{t-1}}} + \frac{1}{a^{q}} + \frac{u^{q^{2t-1}}v}{a^{q^{2t-1}}} + \frac{v^{q^{t+1}}u}{a^{q^{t+1}}} \right) =0, \end{aligned}$$

respectively. Recall that $u=-k^{1-q^{t+1}}$, $v=k^{1-q^{2t-1}}$ and $k^{q^t}=-1/k$, from the above equation we deduce

$$\begin{aligned}\frac{1}{a^{q^{t-1}}} + \frac{1}{a^{q}} + \frac{k^2}{a^{q^{2t-1}}} + \frac{k^2}{a^{q^{t+1}}}=0. \end{aligned}$$

Therefore

$$\begin{aligned} \frac{1}{a^{q^{t-1}}} + \frac{1}{a^{q}} = - k^2\left( \frac{1}{a^{q^{t-1}}} + \frac{1}{a^{q}}\right) ^{q^t}. \end{aligned}$$

(58)

Our goal of the next step is to prove

$$\begin{aligned} \frac{1}{a^{q^t}} = -k^{q-q^{2t-1}} \frac{1}{a}, \end{aligned}$$

(59)

always holds.

Let $j=2, t+2, 2t-2, t-2$ in (51), we have

$$\begin{aligned} \gamma _1 + \gamma _{t+3} + u^{q^{t+1}} \gamma _{t+1} + v^{q^3} \gamma _3&= 0, \end{aligned}$$

(60)

$$\begin{aligned} \gamma _{t+1} + \gamma _{3} + u^{q} \gamma _{1} + v^{q^{t+3}} \gamma _{t+3}&= 0, \end{aligned}$$

(61)

$$\begin{aligned} \gamma _{2t-3} + \gamma _{t-1} + u^{q^{t-3}} \gamma _{t-3} + v^{q^{2t-1}} \gamma _{2t-1}&= 0, \end{aligned}$$

(62)

$$\begin{aligned} \gamma _{t-3} + \gamma _{2t-1} + u^{q^{2t-3}} \gamma _{2t-3} + v^{q^{t-1}} \gamma _{t-1}&= 0. \end{aligned}$$

(63)

Depending on the value of $\gamma _3$, $\gamma _{t+3}$, $\gamma _{t-3}$ and $\gamma _{2t-3}$, we separate the proof of (59) into four different cases.

Case (i) $\gamma _3=\gamma _{t+3}=0$. By (53), (55) and (60),

$$\begin{aligned} -\frac{u}{a^{q^{t+1}}} \gamma _0=\gamma _{t+1}=-u^q\gamma _1 =\frac{u^q}{a^q} \gamma _0, \end{aligned}$$

which means $\frac{1}{a^{q^t}} = -u^{1-q^{2t-1}} \frac{1}{a} = -k^{q-q^{2t-1}} \frac{1}{a}$.

Case (ii) $\gamma _{t-3}=\gamma _{2t-3}=0$. By a similar computation of (62) as in Case (i), we get (59) again.

Case (iii) $\gamma _3\ne 0$ and $\gamma _{t-3}\ne 0$. By (49), $\gamma _{2t-3}=\gamma _{2t-4}=\gamma _{t+3}=\gamma _{t+4}=0$. Now, (60) and (61) become

$$\begin{aligned} \gamma _1 + k^{-q^2-q}\gamma _{t+1} + k^{q^3-q^2} \gamma _3&=0,\\ \gamma _3 + \gamma _{t+1} + k^{q^2+q} \gamma _1&=0. \end{aligned}$$

Canceling $\gamma _3$, we get

$$\begin{aligned}(1-k^{q^3+q}) \gamma _{t+1} = - (1-k^{q^3+q})k^{q^2+q}\gamma _1. \end{aligned}$$

By Proposition 3.2, $k^{q^2+1}\ne 1$. Hence, $\gamma _{t+1} = - k^{q^2+q}\gamma _1$. By plugging (53) and (55) into it, we derive (59).

Case (iv) $\gamma _{t+3}\ne 0$ and $\gamma _{2t-3}\ne 0$. By (49), $\gamma _{t-3}=\gamma _{t-4}=\gamma _{3}=\gamma _{4}=0$. As in Case (iii), by canceling $\gamma _{t+3}$ using (60) and (61), we obtain (59) again.

By (49), we have covered all possible cases. Therefore, (59) is proved.

Now we are ready to prove (a) and (b). Our strategy is to give a precise expression for a, which is strong enough to prove (a). Then we further use (46) to get more restrictions on the value of h which leads to (b).

Plugging (59) in (58), we have

$$\begin{aligned} \frac{1}{a^{q^{t-1}}} \left( 1-k^{1+q^{2t-2}}\right) + \frac{1}{a^q}\left( 1-k^{1+q^2}\right) =0,\end{aligned}$$

that is

$$\begin{aligned} \left( \frac{1}{a^q}\left( 1-k^{1+q^2}\right) \right) ^{q^{t-2}} -k^{-1-q^{2t-2}} \frac{1}{a^q}\left( 1-k^{1+q^2}\right) =0.\end{aligned}$$

By $k^{q^{t-2}} = -k^{-q^{2t-2}}$, we have

$$\begin{aligned} \left( \frac{k^{-1}}{a^q}\left( 1-k^{1+q^2}\right) \right) ^{q^{t-2}} +\frac{k^{-1}}{a^q}\left( 1-k^{1+q^2}\right) =0. \end{aligned}$$

Therefore,

$$\begin{aligned} a=k^{-q^{2t-1}} \left( 1-k^{q+q^{2t-1}}\right) \eta , \end{aligned}$$

(64)

where $\eta$ satisfies $\eta ^{q^{t-2}}+\eta = 0$. Since $0=(\eta ^{q^{t-2}}+\eta )^{q^{t+2}} = \eta +(\eta ^{q^{t-2}})^{q^4}=\eta -\eta ^{q^4}$, we get $\eta \in {{\mathbb {F}}}_{q^4}$. Moreover,

$$\begin{aligned} \eta = -\eta ^{q^{t-2}}= {\left\{ \begin{array}{ll} -\eta ^{q^3}=-\eta ^q, &{} t\equiv 1\pmod {4},\\ -\eta =0, &{} t\equiv 2\pmod {4},\\ -\eta ^q, &{} t\equiv 3\pmod {4},\\ -\eta ^{q^2}, &{} t\equiv 0\pmod {4}. \end{array}\right. } \end{aligned}$$

Hence

$$\begin{aligned} \eta ^{q^2-1}= {\left\{ \begin{array}{ll} 1, &{} t\equiv 1\pmod {4},\\ 0, &{} t\equiv 2\pmod {4},\\ 1, &{} t\equiv 3\pmod {4},\\ -1, &{} t\equiv 0\pmod {4}. \end{array}\right. } \end{aligned}$$

(65)

Substitute a in (59) by (64),

$$\begin{aligned} -k^{q-q^{2t-1}}k^{-q^{t-1}}(1-k^{q^{t+1}+q^{t-1}}) \eta ^{q^{t}} = k^{-q^{2t-1}}(1-k^{q+q^{2t-1}})\eta , \end{aligned}$$

which equals

$$\begin{aligned} k^{q}(1-k^{-q-q^{2t-1}}) \eta ^{q^{t}} = k^{-q^{2t-1}}(1-k^{q+q^{2t-1}})\eta . \end{aligned}$$

It implies

$$\begin{aligned} \eta ^{q^t} = -\eta . \end{aligned}$$

(66)

Together with $\eta ^{q^{t-2}}+\eta = 0$, we deduce $\eta ^{q^2} =\eta$. It contradicts (65) when $t\equiv 0\pmod {4}$. Therefore, when t is even, a must be 0 and (a) is proved.

Finally, let us plugging (53) and (54) into (46), we have

$$\begin{aligned} ( \bar{c} \gamma _0)^{q+1}=\frac{\tau ^q}{v^q} a^{q+1}=\ell ^{q-1}a^{q+1}, \end{aligned}$$

(67)

where $\ell =h/k$ which means $\ell ^{q^t}=1/\ell$.

Similarly, by (56) and (55) into (47)

$$\begin{aligned} ( \bar{c} \gamma _0)^{q^{t-1}+1}=\frac{\theta ^{q^{t-1}}}{u^{q^t-1}} a^{q^{t-1}+1}=\ell ^{q^{t-1}-1}a^{q^{t-1}+1}. \end{aligned}$$

(68)

Raising (67) to its $\frac{q^{t-1}+1}{2}$-th power and (68) to its $\frac{q+1}{2}$-th power and canceling $(\bar{c}\gamma _0)^{\frac{1}{2}(q+1)(q^{t-1}+1)}$ and $a^{\frac{1}{2}(q+1)(q^{t-1}+1)}$, we obtain

$$\begin{aligned} \ell ^{q^{t-2}-1}=1. \end{aligned}$$

Again, by $\ell ^{q^t+1}=1$, we have $\ell ^{q^2+1}=1$ which means $\ell ^{\gcd (q^2+1, q^t+1)}=1$. Therefore, since t is odd, we get $\ell ^2=1$, and (b) is proved. $\square$

Now we are ready to prove Theorem 5.1.

Proof of Theorem 5.1

Suppose that $f=\psi _{h,t}$, $g=\psi _{k,t}$, and $L_f$ is $P\Gamma L(2,q^n)$-equivalent to $L_g$. Then there exists an invertible matrix $\begin{pmatrix} a &{} b\\ c &{} d \end{pmatrix}$ over ${{\mathbb {F}}}_{q^n}$ and $\sigma \in {{\,\mathrm{{Aut}}\,}}({{\mathbb {F}}}_{q^n})$ such that

$$\begin{aligned} \left\{ \frac{f(x)}{x}:x\in {{\mathbb {F}}}_{q^n}^* \right\} =\left\{ \frac{cx^\sigma +d(g(x))^\sigma }{ax^\sigma +b(g(x))^\sigma }:x\in {{\mathbb {F}}}_{q^n}^* \right\} =\left\{ \frac{cx+d\bar{g}(x)}{ax+b\bar{g}(x)}:x\in {{\mathbb {F}}}_{q^n}^* \right\} , \end{aligned}$$

where $\bar{g}(x)=\psi _{k^\sigma , t}(x)$.

For a given h satisfying $h^{q^t+1}=-1$, let $\varepsilon _h$ denote the number of k for which $U_k$ is $\Gamma L(2,q^n)$-equivalent to $U_h$.

Depending on the value of b, we separate the remainder part of the proof into two cases:

Case (a) $b=0$. By Lemma 5.4, $\left( h/k^\sigma \right) ^{q^2+1}=1$ which means that $U_h$ is $\Gamma L(2,q^n)$-equivalent to $U_{k^\sigma }$. Thus there are exactly $\varepsilon _h$ choices of k for which $L_{k,t}$ is equivalent to $L_{h,t}$.

Case (b) $b\ne 0$. Without loss of generality, we assume $b=1$ and $f\ne \bar{g}$. By Lemma 5.5, $d=a=0$. Thus

$$\begin{aligned} \left\{ \frac{f(x)}{x}:x\in {{\mathbb {F}}}_{q^n}^* \right\} =\left\{ \frac{cx}{\bar{g}(x)}:x\in {{\mathbb {F}}}_{q^n}^* \right\} . \end{aligned}$$

Suppose that there is another $\tilde{g}(x)=\psi _{\tilde{k},t}(x)$ for certain $\tilde{k}\in {{\mathbb {F}}}_{q^{2t}}$ satisfying $\tilde{k}^{q^t+1}=-1$ and

$$\begin{aligned} \left\{ \frac{f(x)}{x}:x\in {{\mathbb {F}}}_{q^n}^* \right\} =\left\{ \frac{\tilde{c}x}{\tilde{g}(x)}:x\in {{\mathbb {F}}}_{q^n}^* \right\} \end{aligned}$$

for some $\tilde{c}\in {{\mathbb {F}}}_{q^{2t}}$. Then

$$\begin{aligned} \left\{ \frac{\bar{g}(x)}{x}:x\in {{\mathbb {F}}}_{q^n}^* \right\} =\left\{ \frac{c}{\tilde{c}} \cdot \frac{\tilde{g}(x)}{x}:x\in {{\mathbb {F}}}_{q^n}^* \right\} . \end{aligned}$$

By Lemma 5.4, $(\tilde{k}/k^\sigma )^{q^2+1}=1$ which means that $U_{k^\sigma }$ is $\Gamma L(2,q^n)$-equivalent to $U_{\tilde{k}}$. Hence, for the case $b\ne 0$, there are exactly $\varepsilon _{\tilde{k}}$ choices of k for which $L_{k,t}$ is equivalent to $L_{h,t}$.

Finally we combine Case (a) and Case (b), for a given $L_{h,t}$. Noting that $|\{ h\in {{\mathbb {F}}}_{q^n}: h^{q^t+1}=-1 \} |=q^t+1$, there are exactly

$$\begin{aligned}M=\frac{q^t+1}{\varepsilon _h}+\frac{q^t+1}{\varepsilon _{\tilde{k}}} \end{aligned}$$

inequivalent $L_{h,t}$ defined by (39). Recall that

$$\begin{aligned}\varepsilon _h\le n r\xi _h= {\left\{ \begin{array}{ll} 4r t, &{} t\not \equiv 2\pmod {4};\\ 2(q^2+1)rt, &{}t\equiv 2\pmod {4}, \end{array}\right. } \end{aligned}$$

for every possible choice of h; see the proof of Corollary 4.3. We obtain the lower bound of M. $\square$

Remark 5.6

By Theorem 5.1, Family (39) contains much more inequivalent elements compared with the known constructions for infinitely many n listed in Table 1.Therefore, this family must be new.