Switchings of semifield multiplications

Abstract

Let $B(X,Y)$ be a polynomial over $\mathbb {F}_{q^n}$ which defines an $\mathbb {F}_q$-bilinear form on the vector space $\mathbb {F}_{q^n}$, and let $\xi$ be a nonzero element in $\mathbb {F}_{q^n}$. In this paper, we consider for which $B(X,Y)$, the binary operation $xy+B(x,y)\xi$ defines a (pre)semifield multiplication on ${\mathbb {F}}_{q^n}$. We prove that this question is equivalent to finding $q$-linearized polynomials $L(X)\in \mathbb {F}_{q^n}[X]$ such that ${\mathrm {Tr}}_{q^n/q}(L(x)/x)\ne 0$ for all $x\in \mathbb {F}_{q^n}^*$. For $n\le 4$, we present several families of $L(X)$ and we investigate the derived (pre)semifields. When $q$ equals a prime $p$, we show that if $n>\frac{1}{2}(p-1)(p^2-p+4)$, $L(X)$ must be $a_0 X$ for some $a_0\in \mathbb {F}_{p^n}$ satisfying ${\mathrm {Tr}}_{q^n/q}(a_0)\ne 0$. Finally, we include a natural connection with certain cyclic codes over finite fields, and we apply the Hasse–Weil–Serre bound for algebraic curves to prove several necessary conditions for such kind of $L(X)$.

1 Introduction

A semifield $\mathbb {S}$ is an algebraic structure satisfying all the axioms of a skewfield except (possibly) the associativity. In other words, it satisfies the following axioms:

• (S1) $(\mathbb {S},+)$ is a group, with identity element $0$;

• (S2) $(\mathbb {S}\setminus \{0\},*)$ is a quasigroup;

• (S3) $0*a=a*0=0$ for all $a$;

• (S4) The left and right distributive laws hold, namely for any $a,b,c\in \mathbb {S}$,

  $$\begin{aligned} (a+b)*c= & {} a*c+b*c,\\ a*(b+c)= & {} a*b+a*c; \end{aligned}$$

• (S5) There is an element $e\in \mathbb {S}$ such that $e*x=x*e=x$ for all $x\in \mathbb {S}$.

A finite field is a trivial example of a semifield. Furthermore, if $\mathbb {S}$ does not necessarily have a multiplicative identity, then it is called a presemifield. For a presemifield ${\mathbb {S}}$, $({\mathbb {S}},+)$ is necessarily abelian [17]. A semifield is not necessarily commutative or associative. However, by Wedderburn’s Theorem [27], in the finite case, associativity implies commutativity. Therefore, a non-associative finite commutative semifield is the closest algebraic structure to a finite field. We refer to [18] for a recent and comprehensive survey.

The first family of non-trivial semifields was constructed by Dickson [7] more than a century ago. In [17], Knuth showed that the additive group of a finite semifield $\mathbb {S}$ is an elementary abelian group, and the additive order of the nonzero elements in $\mathbb {S}$ is called the characteristic of $\mathbb {S}$. Hence, any finite semifield can be represented by $({\mathbb {F}}_q, +, *)$, where $q$ is a power of a prime $p$. Here $({\mathbb {F}}_q, +)$ is the additive group of the finite field ${\mathbb {F}}_q$ and $x*y$ can be written as $x*y=\sum _{i,j}a_{ij} x^{p^i}y^{p^j}$, which forms a mapping from ${\mathbb {F}}_q\times {\mathbb {F}}_q$ to ${\mathbb {F}}_q$.

Geometrically speaking, there is a well-known correspondence, via coordinatisation, between (pre)semifields and projective planes of Lenz-Barlotti type V.1, see [5, 13]. In [1], Albert showed that two (pre)semifields coordinatise isomorphic planes if and only if they are isotopic.

Definition 1.1

Let $\mathbb {S}_1=({\mathbb {F}}_p^n, +, *)$ and $\mathbb {S}_2=({\mathbb {F}}_p^n, +, \star )$ be two presemifields. If there exist three bijective linear mappings $L, M, N:{\mathbb {F}}_{p}^n\rightarrow {\mathbb {F}}_p^n$ such that

$$\begin{aligned} M(x)\star N(y)=L(x*y) \end{aligned}$$

for any $x,y\in {\mathbb {F}}_p^n$, then $\mathbb {S}_1$ and $\mathbb {S}_2$ are called isotopic, and the triple $(M,N,L)$ is called an isotopism between $\mathbb {S}_1$ and $\mathbb {S}_2$.

Let ${\mathbb {P}}=({\mathbb {F}}_{p^n}, +, *)$ be a presemifield. We can obtain a semifield from it via isotopisms in several ways, such as the well known Kaplansky’s trick (see [18], p 2]). The following method was recently given by Bierbrauer [2]. Define a new multiplication $\star$ by the rule

$$\begin{aligned} x\star y := B^{-1}(B_1(x)*y), \end{aligned}$$

(1.1)

where $B(x):=1*x$ and $B_1(x)*1=1*x$. We have $x\star 1= B^{-1}(B_1(x)*1)= B^{-1}(1*x)=x$ and $1\star x=B^{-1}(B_1(1)*x)=B^{-1}(1*x)=x$, thus $({\mathbb {F}}_{p^n}, +, \star )$ is a semifield with identity $1$. In particular, when ${\mathbb {P}}$ is commutative, $B_1$ is the identity mapping.

Let ${\mathbb {S}}=({\mathbb {F}}_{p^n},+,*)$ be a semifield. The subsets

$$\begin{aligned} N_l({\mathbb {S}})= & {} \left\{ a\in {\mathbb {S}}: (a*x)*y=a*(x*y) \text { for all }x,y\in {\mathbb {S}}\right\} \!,\\ N_m({\mathbb {S}})= & {} \{a\in {\mathbb {S}}: (x*a)*y=x*(a*y) \text { for all }x,y\in {\mathbb {S}}\},\\ N_r({\mathbb {S}})= & {} \{a\in {\mathbb {S}}: (x*y)*a=x*(y*a) \text { for all }x,y\in {\mathbb {S}}\}, \end{aligned}$$

are called the left, middle and right nucleus of ${\mathbb {S}}$, respectively. It is easy to check that these sets are finite fields. The subset $N({\mathbb {S}})=N_l({\mathbb {S}})\cap N_m({\mathbb {S}}) \cap N_r({\mathbb {S}})$ is called the nucleus of ${\mathbb {S}}$. It is easy to see if ${\mathbb {S}}$ is commutative, then $N_l({\mathbb {S}})=N_r({\mathbb {S}})$ and $N_l({\mathbb {S}})\subseteq N_m({\mathbb {S}})$, therefore $N_l({\mathbb {S}})=N_r({\mathbb {S}})=N({\mathbb {S}})$. In [13], a geometric interpretation of these nuclei is discussed. The subset $\{a\in \mathbb {S}: a*x =x*a\text { for all } x\in \mathbb {S}\}$ is called the commutative center of $\mathbb {S}$ and its intersection with $N(\mathbb {S})$ is called the center of ${\mathbb {S}}$.

Let $G$ be a group and $N$ a subgroup. A subset $D$ of $G$ is called a relative difference set with parameters $(|G|/|N|,|N|,|D|,\lambda )$ if the list of differences of $D$ covers every element in $G\setminus N$ exactly $\lambda$ times, and no element in $N\setminus \{0\}$. We call $N$ the forbidden subgroup.

Jungnickel [15] showed that every semifield $\mathbb {S}$ of order $q$ leads to a $(q,q,q,1)$-relative difference set $D$ in a group $G$ which is not necessarily abelian. Assume that $\mathbb {S}$ is commutative. If $q=p^n$ and $p$ is odd, then $G$ is isomorphic to the elementary abelian group $C_{p}^{2n}$; if $q=2^n$, then $G\cong C_{4}^n$. ($C_m$ is the cyclic group of order $m$.)

Let $p$ be an odd prime. A function $f:\mathbb {F}_{p^n}\rightarrow \mathbb {F}_{p^n}$ is called planar if the mapping

$$\begin{aligned} x\mapsto f(x+a)-f(x) \end{aligned}$$

is a permutation of $\mathbb {F}_{p^n}$ for every $a\in \mathbb {F}_{p^n}^*$. Planar functions were first defined by Dembowski and Ostrom in [6]. It is not difficult to verify that planar functions over $\mathbb {F}_{p^n}$ are equivalent to $(p^n,p^n,p^n,1)$-relative difference sets in $C_{p}^{2n}$. Planar functions over $\mathbb {F}_{2^n}$, introduced recently in [25, 29], has a slightly different definition: A function $f:\mathbb {F}_{2^n}\rightarrow \mathbb {F}_{2^n}$ is called planar, if the mapping

$$\begin{aligned} x\mapsto f(x+a)+f(x)+ax \end{aligned}$$

is a permutation of $\mathbb {F}_{2^n}$ for every $a\in \mathbb {F}_{2^n}^*$. They are equivalent to $(2^n,2^n,2^n,1)$-relative difference sets in $C_{4}^{n}$; see [29], Theorem 2.1].

Let $f$ be a planar function over $\mathbb {F}_{q^n}$, where $q$ is a power of prime. A switching of $f$ is a planar function of the form $f+g\xi$ where $g$ is a mapping from $\mathbb {F}_{q^n}$ to $\mathbb {F}_q$ and $\xi \in \mathbb {F}_{q^n}^*$. Switchings of planar functions over $\mathbb {F}_{p^n}$, where $p$ is an odd prime, were investigated by Pott and the third author in [24]. In [29], it is proved that switchings of the planar function $f(x)=0$ defined over $\mathbb {F}_{2^n}$ can be written as affine polynomials $\sum a_i x^{2^i}+b$, which are equivalent to $f(x)$ itself.

In the present paper, we will investigate the switchings of (pre)semifield multiplications. To be precise, we will consider when the binary operation

$$\begin{aligned} x*y= x\star y+ B(x,y)\xi \end{aligned}$$

on $\mathbb {F}_{q^n}$ defines a (pre)semifield multiplication, where $\star$ is a given (pre)semifield multiplication, $\xi \in \mathbb {F}_{q^n}^*$ and $B(x,y)$ is an $\mathbb {F}_q$-bilinear form from $\mathbb {F}_{q^n}\times \mathbb {F}_{q^n}$ to $\mathbb {F}_q$. (One may identify $\mathbb {F}_{q^n}$ with $\mathbb {F}_q^n$, although it is not necessary.) We call $x*y$ a switching neighbour of $x\star y$. In particular, we will concentrate on the case in which $\star$ is the multiplication of a finite field.

In Sect. 2, we show that finding $B$ such that $x*y:= xy+ B(x,y)\xi$ defines a (pre)semifield multiplication is equivalent to finding $q$-linearized polynomials $L(X)\in \mathbb {F}_{q^n}[X]$ such that ${\mathrm {Tr}}_{q^n/q}(L(x)/x)\ne 0$ for all $x\in \mathbb {F}_{q^n}^*$. For $n\le 4$, we give in Sect. 3 several $q$-linearized polynomials $L(X)\in \mathbb {F}_{q^n}[X]$ satisfying this condition and we discuss the presemifields of the corresponding switchings. In Sect. 4, we prove that when $q=p$ is a prime and $n>(p-1)(p^2-p+4)/2$, the only $L(X)$ satisfying the above condition are those of the form $\beta X$ where ${\text {Tr}}_{p^n/p}(\beta )\ne 0$. In Sect. 5, we explore a connection of the $q$-linearized polynomials $L(X)$ satisfying the above condition with certain cyclic codes over ${\mathbb {F}}_q$. Finally, in Sect. 6 we derive several necessary conditions for the existence of the $q$-linearized polynomials $L(X)$ from the Hasse–Weil–Serre bound for algebraic curves over finite fields.

2 Preliminary discussion

Let ${\mathrm {Tr}}_{q^n/q}$ be the trace function from $\mathbb {F}_{q^n}$ to $\mathbb {F}_q$. We define

$$\begin{aligned} B(x,y):= {\mathrm {Tr}}_{q^n/q}\left( \sum _{i=0}^{n-1}b_i x y^{q^i}\right) ,\qquad x,y\in {\mathbb {F}}_{q^n}, \end{aligned}$$

where $b_i\in {\mathbb {F}}_{q^n}$. It is easy to see that $B(x,y)$ defines an $\mathbb {F}_q$-bilinear form from $\mathbb {F}_{q^n}\times \mathbb {F}_{q^n}$ to $\mathbb {F}_q$, and every such bilinear form can be written in this way.

In the next theorem, we consider the switchings of a finite field multiplication.

Theorem 2.1

Let $x*y:= xy+ B(x,y)\xi$, where $B(x,y):= {\mathrm {Tr}}_{q^n/q}(\sum _{i=0}^{n-1}b_i x y^{q^i})$, $b_i\in {\mathbb {F}}_{q^n}$, and $\xi \in \mathbb {F}_{q^n}^*$. Then $*$ defines a presemifield multiplication on $\mathbb {F}_{q^n}$ if and only if for any $a\in {\mathbb {F}}_{q^n}^*$, ${\mathrm {Tr}}_{q^n/q}(M(a)/a)\ne -1$, where $M(X):= \xi \sum _{i=0}^{n-1}b_i X^{q^i}\in \mathbb {F}_{q^n}[X]$.

Proof

$(\Rightarrow )$ Let $x*y$ be a presemifield multiplication. Assume to the contrary that there is $a\in \mathbb {F}_{q^n}^*$ such that

$$\begin{aligned} {\mathrm {Tr}}_{q^n/q}(M(a)/a)=-1. \end{aligned}$$

We consider the equation $x*a=0$. It has a solution $x$ if and only if there exists $u\in \mathbb {F}_q$ such that

$$\begin{aligned} xa&= \xi u\quad \text {and} \end{aligned}$$

(2.1)

$$\begin{aligned} B(x,a)&= -u. \end{aligned}$$

(2.2)

Plugging (2.1) into (2.2), we have $B(\xi u/a,a)=-u$, which means that

$$\begin{aligned} u {\mathrm {Tr}}_{q^n/q}\left( \xi \sum _{i=0}^{n-1}b_i a^{q^i-1}\right) =-u, \end{aligned}$$

i.e.

$$\begin{aligned} u {\mathrm {Tr}}_{q^n/q}(M(a)/a)=-u, \end{aligned}$$

which holds for any $u\in \mathbb {F}_q$ according to our assumption. Therefore, $x*a=0$ has a nonzero solution. It contradicts our assumption that $*$ defines a presemifield multiplication.

$(\Leftarrow )$ It is easy to see that the left and right distributivity of the multiplication $*$ hold. We only need to show that for any $a\ne 0$, $x*a=0$ if and only if $x=0$. This is achieved by reversing the first part of the proof. $\square$

Let $x*y$ be the multiplication defined in Theorem 2.1. Then it is straightforward to verify that the presemifield $(\mathbb {F}_{q^n},+,*)$ is isotopic to $(\mathbb {F}_{q^n},+,\star )$, where

$$\begin{aligned} x\star y := xy+B'(x,y) \end{aligned}$$

and $B'(x,y)= {\mathrm {Tr}}_{q^n/q}(\xi \sum _{i=0}^{n-1}b_i x y^{q^i})$. Therefore, we can restrict ourselves to the switchings of finite field multiplications with $\xi =1$.

For the switchings

$$\begin{aligned} x\star y+ B(x,y)\xi \end{aligned}$$

of a (pre)semifield multiplication $\star$, it is difficulty to obtain explicit conditions on $B(x,y)$. The reason is that generally we can not explicitly write down the solution of $x\star a=\xi u$ as we did for (2.1).

Let $\alpha$ be an element in $\mathbb {F}_{q^n}$ such that ${\mathrm {Tr}}_{q^n/q}(\alpha )=1$. To find $M(X)$ satisfying the condition in Theorem 2.1, we only need to consider the $q$-linearized polynomial $L(X):=M(X)+\alpha X\in \mathbb {F}_{q^n}[X]$ such that

$$\begin{aligned} {\mathrm {Tr}}_{q^n/q}(L(x)/x)\ne 0 \quad \text {for all } x\in \mathbb {F}_{q^n}^*. \end{aligned}$$

(2.3)

Obviously, when $L(X)=\beta X$, where ${\mathrm {Tr}}_{q^n/q}(\beta )\ne 0$, we have ${\mathrm {Tr}}_{q^n/q}(L(x)/x)\ne 0$ for every nonzero $x$. The question is whether there are other $L$’s. We will give several results concerning this question throughout Sects. 3–6.

The proof of next proposition is also straightforward.

Proposition 2.2

Let $L(X)=\sum _{i=0}^{n-1}a_i X^{q^i}\in \mathbb {F}_{q^n}[X]$. If ${\mathrm {Tr}}_{q^n/q}(L(x)/x)\ne 0$ for all $x\in \mathbb {F}_{q^n}^*$, then the mapping $x\mapsto L(x)$ is a permutation of ${\mathbb {F}}_{q^n}$.

We include several lemmas which will be used later to investigate the commutativity of presemifield multiplications.

Lemma 2.3

Let $x*y:= xy+ B(x,y)$, where $B(x,y):= {\mathrm {Tr}}_{q^n/q}(\sum _{i=0}^{n-1}b_i x y^{q^i})$, $b_i\in {\mathbb {F}}_{q^n}$. Then $*$ is commutative if and only if $b_i=b_{n-i}^{q^i}$ for every $i=1,\dots ,n-1$.

Proof

Clearly, $x*y=y*x$ if and only if $B(x,y)=B(y,x)$, i.e.

$$\begin{aligned} {\mathrm {Tr}}_{q^n/q}\left( \sum _{i=0}^{n-1}b_i x y^{q^i}\right) = {\mathrm {Tr}}_{q^n/q}\left( \sum _{i=0}^{n-1}b_i y x^{q^i}\right) , \end{aligned}$$

which means that

$$\begin{aligned} {\mathrm {Tr}}_{q^n/q}\left( x\sum _{i=1}^{n-1}(b_i-b_{n-i}^{q^i})y^{q^i}\right) =0 \end{aligned}$$

for every $x,y\in \mathbb {F}_{q^n}$. Therefore we complete the proof. $\square$

It is possible that a non-commutative presemifield ${\mathbb {P}}$ is isotopic to a commutative presemifield. We can use the next criterion given by Bierbrauer [2], as a generalization of Ganley’s criterion [8], to test whether this happens.

Lemma 2.4

A presemifield $({\mathbb {P}},+,*)$ is isotopic to a commutative semifield if and only if there is some nonzero $v$ such that $A(v*x)*y=A(v*y)*x$, where $A: \mathbb {F}_{q^n}\rightarrow \mathbb {F}_{q^n}$ is defined by $A(x)*1=x$.

Given an arbitrary presemifield multiplication, it is not easy to get the explicit expression for $A(x)$. However, we can do it for the switchings of multiplications of finite fields.

Lemma 2.5

Let $x*y:= xy+ B(x,y)$ be a switching of $\mathbb {F}_{q^n}$, where $B(x,y):= {\mathrm {Tr}}_{q^n/q}(\sum _{i=0}^{n-1}b_i x y^{q^i})$, $b_i\in {\mathbb {F}}_{q^n}$. Let $A: \mathbb {F}_{q^n}\rightarrow \mathbb {F}_{q^n}$ be such that $A(x)*1=x$ for every $x\in \mathbb {F}_{q^n}$. Then

$$\begin{aligned} A(x)=x+ {\mathrm {Tr}}_{q^n/q}\left( \frac{-tx}{1+ {\mathrm {Tr}}_{q^n/q}(t)}\right) , \end{aligned}$$

(2.4)

where $t=\sum _{i=0}^{n-1}b_i$.

Proof

First, we have

$$\begin{aligned} u*1&= u+B(u,1)\\&= u+ {\mathrm {Tr}}_{q^n/q}\left( \sum _{i=0}^{n-1}b_i u\right) \\&= u+ {\mathrm {Tr}}_{q^n/q}(tu). \end{aligned}$$

It is worth noting that $1*1=1+ {\mathrm {Tr}}_{q^n/q}(t)\ne 0$. Let $s:=-t/(1+ {\mathrm {Tr}}_{q^n/q}(t))$. Replacing $u$ by the expression in (2.4), we have

$$\begin{aligned} A(x)*1&= x+ {\mathrm {Tr}}_{q^n/q}(sx)+ {\mathrm {Tr}}_{q^n/q}\bigl [tx+t {\mathrm {Tr}}_{q^n/q}(sx)\bigr ]\\&= x+ {\mathrm {Tr}}_{q^n/q}\bigl [s(1+ {\mathrm {Tr}}_{q^n/q}(t))x+tx\bigr ]\\&= x. \end{aligned}$$

$\square$

3 Switchings of $\mathbb {F}_{q^n}$ for small $n$

In this section, we investigate the switchings of finite fields $(\mathbb {F}_{q^n},+,\cdot )$ where $n\le 4$.

Lemma 3.1

Let $L(X)=a_1 X^q + a_0 X\in \mathbb {F}_{q^2}[X]$. Then the polynomial

$$\begin{aligned} f(X)= {\mathrm {Tr}}_{q^2/q}(L(X)/X) \end{aligned}$$

has no root in $\mathbb {F}_{q^2}^*$ if and only if the equation $x^{q-1} =y$ has no solution $x\in {\mathbb {F}}_{q^2}^*$ for every $y\in {\mathbb {F}}_{q^2}$ satisfying

$$\begin{aligned} a_1 y^2+ {\mathrm {Tr}}_{q^2/q}(a_0)y+a_1^{q}=0. \end{aligned}$$

(3.1)

Proof

Let $y:= x^{q-1}$, where $x\in \mathbb {F}_{q^2}^*$. Then

$$\begin{aligned} {\mathrm {Tr}}_{q^2/q}(L(x)/x)&= {\mathrm {Tr}}_{q^2/q}(a_1 x^{q-1}+a_0)\\&= {\mathrm {Tr}}_{q^2/q}(a_1 y+a_0)\\&= a_1^q y^q+a_1 y+ {\mathrm {Tr}}_{q^2/q}(a_0)\\&= y^q(a_1 y^2+ {\mathrm {Tr}}_{q^2/q}(a_0)y+a_1^q) \end{aligned}$$

since $y^{q+1}=1$. Therefore, $f$ has a nonzero root if and only if there exists a $(q-1)$th power in $\mathbb {F}_{q^2}^*$ satisfying (3.1). $\square$

Theorem 3.2

Let $L(X)=a_1 X^q + a_0 X\in \mathbb {F}_{q^2}[X]$. Then

$$\begin{aligned} f(X)= {\mathrm {Tr}}_{q^2/q}(L(X)/X) \end{aligned}$$

(3.2)

has no root in $\mathbb {F}_{q^2}^*$ if and only if $g(X)=X^2+ {\mathrm {Tr}}_{q^2/q}(a_0)X+a_1^{q+1}\in \mathbb {F}_q[X]$ has two distinct roots in $\mathbb {F}_q$.

Proof

If $a_1=0$, then $f(X)= {\mathrm {Tr}}_{q^2/q}(a_0)$ and $g(X)=X^2+ {\mathrm {Tr}}_{q^2/q}(a_0)X$. It is clear that $f$ has no nonzero roots if and only if $g$ has two distinct roots.

In the rest of the proof, we assume that $a_1\ne 0$.

$(\Leftarrow )$ Let $a_1 y\in \mathbb {F}_q$ ($y\in \mathbb {F}_{q^2}$) be a root of $g$. By Lemma 3.1, it suffices to show that $y^{q+1}\ne 1$.

Case 1. Assume that $q$ is even. Since $g$ has two distinct roots, we have ${\mathrm {Tr}}_{q^2/q}(a_0)\ne 0$. Since

$$\begin{aligned} (a_1y)^{q+1}=(a_1y)^2= {\mathrm {Tr}}_{q^2/q}(a_0)a_1y+a_1^{q+1}, \end{aligned}$$

we have

$$\begin{aligned} y^{q+1}=1+\frac{ {\mathrm {Tr}}_{q^2/q}(a_0)y}{a_1^q}\ne 1. \end{aligned}$$

Case 2. Assume that $q$ is odd. We have $y=\frac{1}{2a_1}(- {\mathrm {Tr}}_{q^2/q}(a_0)+d)$, where $d\in \mathbb {F}_q^*$ and $d^2= {\mathrm {Tr}}_{q^2/q}(a_0)^2-4a_1^{q+1}$. Suppose to the contrary that $y^{q+1}=1$. It follows that

$$\begin{aligned} (- {\mathrm {Tr}}_{q^2/q}(a_0)+d)^{q+1}=4a_1^{q+1}, \end{aligned}$$

which means

$$\begin{aligned} {\mathrm {Tr}}_{q^2/q}(a_0)^2+d^2-2d {\mathrm {Tr}}_{q^2/q}(a_0)=4a_1^{q+1}. \end{aligned}$$

Hence

$$\begin{aligned} 2d^2-2d {\mathrm {Tr}}_{q^2/q}(a_0)=0. \end{aligned}$$

Therefore $d= {\mathrm {Tr}}_{q^2/q}(a_0)$. But then $d^2= {\mathrm {Tr}}_{q^2/q}(a_0)^2\ne {\mathrm {Tr}}_{q^2/q}(a_0)^2-4a_1^{q+1}$, which is a contradiction.

$(\Rightarrow )$ We first show that $g$ is reducible in $\mathbb {F}_q[x]$. Otherwise, let $a_1y\in \mathbb {F}_{q^2}\setminus \mathbb {F}_q$ be a root of $g$. Then $(a_1y)^{q+1}=a_1^{q+1}$, thus $y^{q+1}=1$. By Lemma 3.1, $f$ has nonzero roots.

It remains to show that ${\mathrm {Tr}}_{q^2/q}(a_0)^2-4a_1^{q+1}\ne 0$. Assume to the contrary that ${\mathrm {Tr}}_{q^2/q}(a_0)^2-4a_1^{q+1}=0$.

Case 1. Assume that $q$ is even. It follows that ${\mathrm {Tr}}_{q^2/q}(a_0)=0$. Write $a_1=x^2$, where $x\in \mathbb {F}_{q^2}$, and let $y=x^{q-1}$. Then $a_1y$ is a root of $g$, which leads to a contradiction.

Case 2. Assume that $q$ is odd. Then $a_1y=- {\mathrm {Tr}}_{q^2/q}(a_0)/2$ is a root of $g$, and

$$\begin{aligned} y^{q+1}=\frac{ {\mathrm {Tr}}_{q^2/q}(a_0)^2}{4a_1^{q+1}}=1, \end{aligned}$$

which is impossible by Lemma 3.1. $\square$

Remark

When $n=2$, if there is some $L(X)$ such that (3.2) has no root in $\mathbb {F}_{q^2}^*$, then we can define a presemifield multiplication $*$ over $\mathbb {F}_{q^2}$ via Theorem 2.1. Let $\mathbb {S}=(\mathbb {F}_{q^2},+,\star )$ be a semifield which is isotopic to $(\mathbb {F}_{q^2},+,*)$. We may assume that $\star$ is defined by (1.1) and hence $\mathbb {S}$ has identity $1$. There are $a_{ij}\in \mathbb {F}_{q^2}$ such that $x*y=\sum _{i,j}a_{ij}x^{q^i}y^{q^j}$ for all $x,y\in \mathbb {F}_{q^2}$. Thus there are $b_{ij}\in \mathbb {F}_{q^2}$ such that $x\star y=\sum _{i,j}b_{ij}x^{q^i}y^{q^j}$ for all $x,y\in \mathbb {F}_{q^2}$. It follows that the center of $\mathbb {S}$ contains $\mathbb {F}_q$. (For $x\in \mathbb {F}_q$ and $y\in \mathbb {F}_{q^2}$, we have $x\star y=x(1\star y)=xy$ and $y\star x=x(y\star 1)=xy$. This implies that $\mathbb {F}_q$ is contained in both the commutative center and the nucleus of $\mathbb {S}$.) Due to the classification of two-dimensional finite semifields by Dickson [7], $\mathbb {S}$ is isotopic to a finite field.

Theorem 3.3

Let $q$ be a power of an odd prime and let $L(X)=a_1 X^{q^2} + a_0 X\in \mathbb {F}_{q^4}[X]$ with $a_1\ne 0$. Then ${\mathrm {Tr}}_{q^4/q}(L(X)/X)$ has no root in $\mathbb {F}_{q^4}^*$ if and only if $a_1^{q^2+1}$ is a square in ${\mathbb {F}}_q^*$ and ${\mathrm {Tr}}_{q^4/q}(a_0)=0$.

Proof

Let $b= {\mathrm {Tr}}_{q^4/q}(a_0)$. Let $x\in \mathbb {F}_{q^4}^*$ and set $y:= x^{q^2-1}$ and $z:= a_1 y+a_1^{q^2}/{y}$. Then

$$\begin{aligned} {\mathrm {Tr}}_{q^4/q}(L(x)/x)&= {\mathrm {Tr}}_{q^4/q}(a_1x^{q^2-1}+a_0) \nonumber \\&= a_1y+a_1^qy^q+a_1^{q^2}/{y}+a_1^{q^3}/{y^q}+ {\mathrm {Tr}}_{q^4/q}(a_0)\nonumber \\&= z+z^q+b. \nonumber \\&= \left( z+\frac{b}{2}\right) ^q + \left( z+\frac{b}{2}\right) . \end{aligned}$$

(3.3)

Thus ${\mathrm {Tr}}_{q^4/q}(L(x)/x)=0$ if and only if $(z+\frac{b}{2})^{q-1}=-1$ or $0$, i.e., $z=t-\frac{b}{2}$ for some $t\in T:= \{t\in \mathbb {F}_{q^4}:t^{q}=-t\}\subset {\mathbb {F}}_{q^2}$. Since $z=a_1 y+a_1^{q^2}/{y}$, we see that $z=t-\frac{b}{2}$ if and only if

$$\begin{aligned} a_1 y^2+\left( \frac{b}{2}-t\right) y+a_1^{q^2}=0. \end{aligned}$$

(3.4)

By the proof of Theorem 3.2, we see that $\{x\in \mathbb {F}_{q^4}^*:y=x^{q^2-1}\ \text {satisfies (3.4)}\}\ne \emptyset$ if and only if

$$\begin{aligned} g(X):=X^2+\left( \frac{b}{2}-t\right) X+a_1^{q^2+1} \end{aligned}$$

has two distinct roots in $\mathbb {F}_{q^2}$. Therefore, to sum up, ${\mathrm {Tr}}_{q^4/q}(L(x)/x)$ has no root in $\mathbb {F}_{q^4}^*$ if and only if $g(X)$ has two distinct roots in $\mathbb {F}_{q^2}$ for every $t\in T$. We now proceed to prove the “if” and the “only if” portions of the theorem separately.

$(\Leftarrow )$ Assume $b=0$ and $a_1^{q^2+1}$ is a square in ${\mathbb {F}}_q^*$. Then $a_1^{q^2+1}\ne t^2$ for all $t\in T$. Hence

$$\begin{aligned} \Delta :=\left( \frac{b}{2}-t\right) ^2-4a_1^{q^2+1}=t^2-4a_1^{q^2+1}\in \mathbb {F}_{q}^*. \end{aligned}$$

It follows that $g$ has two distinct roots in ${\mathbb {F}}_{q^2}$.

$(\Rightarrow )$ Assume that ${\mathrm {Tr}}_{q^4/q}(L(X)/X)$ has no root in $\mathbb {F}_{q^4}^*$. We want to show

• R1. $b=0$, and

• R2. $a_1^{q^2+1}$ is a square in $\mathbb {F}_q^*$. Equivalently, $a_1^{q^2+1}$ is in $\mathbb {F}_q$ and there is no $t\in T$ such that $t^2=4 a_1^{q^2+1}$.

Now we assume that $\Delta =\left( \frac{b}{2}-t\right) ^2-4a_1^{q^2+1}\ne 0$ always has a square root in $\mathbb {F}_{q^2}$ for every $t\in T$. Choose an element $\xi$ of $\mathbb {F}_{q^2}\setminus \mathbb {F}_q$ such that $\xi ^{q-1}=-1$. Then every element of $\mathbb {F}_{q^2}$ can be written as $z+w\xi$, where $z$, $w\in \mathbb {F}_q$, and $T=\{x\xi : x \in \mathbb {F}_q\}$. We write $a_1^{q^2+1}=A_1+A_2\xi$. As $\Delta$ is always a square in $\mathbb {F}_{q^2}^*$, the equation

$$\begin{aligned} (z+w\xi )^2=(x\xi -b/2)^2-(A_1+A_2\xi ) \end{aligned}$$

(3.5)

in $(z,w)$ has solutions for every $x\in \mathbb {F}_q$. Expanding (3.5), we have

$$\begin{aligned} z^2+w^2\alpha= & {} x^2\alpha +b^2/4 -A_1, \end{aligned}$$

(3.6)

$$\begin{aligned} 2wz= & {} -xb-A_2, \end{aligned}$$

(3.7)

where $\alpha =\xi ^2\in \mathbb {F}_q$.

If we can show that $b=0$ and $A_2=0$, then the proof is complete (R2 can be easily derived from the condition that $\Delta \ne 0$). Suppose to the contrary that at least one of $b$ and $A_2$ is not $0$. Then there exists at most one $x=x_0\in \mathbb {F}_q$ such that $w=0$ by (3.7). Now assume that $w\ne 0$. From (3.7) we have

$$\begin{aligned} z=-\frac{xb+A_2}{2w}. \end{aligned}$$

Plugging it into (3.6), we get

$$\begin{aligned} \frac{(xb+A_2)^2}{4w^2}+w^2\alpha =x^2\alpha + \frac{b^2}{4}-A_1, \end{aligned}$$

i.e.,

$$\begin{aligned} \alpha (w^2)^2-\left( x^2\alpha +\frac{b^2}{4}-A_1\right) w^2+\frac{(xb+A_2)^2}{4}=0. \end{aligned}$$

For every given $x\in \mathbb {F}_q\setminus \{x_0\}$, this equation always has a solution $w$ in $\mathbb {F}_q$. It follows that

$$\begin{aligned} f(x)=\left( x^2\alpha + \frac{b^2}{4}-A_1\right) ^2-\alpha (xb+A_2)^2 \end{aligned}$$

is always a square in $\mathbb {F}_q$. Let $\psi$ be the multiplicative character of $\mathbb {F}_q$ of order $2$, and for convenience we set $\psi (0)=0$. Then we have

$$\begin{aligned} \sum _{c\in \mathbb {F}_q}\psi (f(c))\ge q-6. \end{aligned}$$

(3.8)

On the other hand, by Theorem 5.41 in [19] (it is routine to verify all the conditions for $f(x)$, because $(b,A_2)\ne (0,0)$ and $(A_1,A_2)\ne (0,0)$), we have

$$\begin{aligned} \sum _{c\in \mathbb {F}_q} \psi (f(c))\le 3\sqrt{q}. \end{aligned}$$

Therefore $q-6\le 3\sqrt{q}$, which means that $q=3,5,7,9,11,13,17,19$. We can use MAGMA [3] to show that $f(x)$ is not always a square for $x\in \mathbb {F}_q\setminus \{x_0\}$ when $q\le 19$. Hence $b=A_2=0$, which completes the proof. $\square$

Theorem 3.4

Let $q$ be a power of an odd prime. Let $a_1\in {\mathbb {F}}_{q^4}^*$ such that $a_1^{q^2+1}$ is a square in ${\mathbb {F}}_q^*$ and let $\tilde{a}_0$ be an element in $\mathbb {F}_{q^4}$ such that ${\mathrm {Tr}}_{q^4/q}(\tilde{a}_0)=-1$. Define

$$\begin{aligned} x*y= xy + {\mathrm {Tr}}_{q^4/q}(a_1xy^{q^2}+\tilde{a}_0xy). \end{aligned}$$

According to Theorems 2.1 and 3.3, $(\mathbb {F}_{q^4}, +, *)$ forms a presemifield. Furthermore, it is isotopic to a commutative semifield.

Proof

According to Lemma 2.4, we only have to show that there exists some $v$ such that

$$\begin{aligned} A(v*x)*y=A(v*y)*x \end{aligned}$$

for every $x$, $y\in \mathbb {F}_{q^4}$, where $A$ is given by (2.4).

Using the same notation as in Lemma 2.5, we set $t=a_1+\tilde{a}_0$ and $s=-t/(1+ {\mathrm {Tr}}_{q^4/q}(t))$. Now,

$$\begin{aligned} A(v*x)&= A(vx+ {\mathrm {Tr}}_{q^4/q}(a_1vx^{q^2}+\tilde{a}_0vx))\\&= vx+ {\mathrm {Tr}}_{q^4/q}(a_1vx^{q^2}+\tilde{a}_0vx) + {\mathrm {Tr}}_{q^4/q}\bigl [s(vx+ {\mathrm {Tr}}_{q^4/q}(a_1vx^{q^2}+\tilde{a}_0vx))\bigr ]\\&= vx+(1+ {\mathrm {Tr}}_{q^4/q}(s)) {\mathrm {Tr}}_{q^4/q}(a_1vx^{q^2}+\tilde{a}_0vx) + {\mathrm {Tr}}_{q^4/q}(svx)\\&= vx + \frac{ {\mathrm {Tr}}_{q^4/q}(a_1vx^{q^2}+\tilde{a}_0vx)}{1+ {\mathrm {Tr}}_{q^4/q}(a_1+\tilde{a}_0)}-\frac{ {\mathrm {Tr}}_{q^4/q}((a_1+\tilde{a}_0)vx)}{1+ {\mathrm {Tr}}_{q^4/q}(a_1+\tilde{a}_0)}\\&= vx + \frac{ {\mathrm {Tr}}_{q^4/q}(a_1vx^{q^2}-a_1vx)}{1+ {\mathrm {Tr}}_{q^4/q}(a_1+\tilde{a}_0)}. \end{aligned}$$

For convenience, let $r(x)$ denote $A(v*x)-vx$. Then

$$\begin{aligned} A(v*x)*y =\,&vxy +r(x) y + {\mathrm {Tr}}_{q^4/q}(a_1vxy^{q^2}+\tilde{a}_0vxy) +r(x) {\mathrm {Tr}}_{q^4/q}(a_1 y^{q^2}+\tilde{a}_0 y)\\ =\,&vxy + \frac{ {\mathrm {Tr}}_{q^4/q}(a_1vx^{q^2}-a_1vx)}{1+ {\mathrm {Tr}}_{q^4/q}(a_1+\tilde{a}_0)}(y+ {\mathrm {Tr}}_{q^4/q}(a_1y^{q^2}+\tilde{a}_0 y))\\&+ {\mathrm {Tr}}_{q^4/q}(a_1vxy^{q^2}+\tilde{a}_0vxy). \end{aligned}$$

It is not difficult to see that if $v$ is an element in $\mathbb {F}_{q^4}$ such that $a_1v\in \mathbb {F}_{q^2}$, then $A(v*x)*y=A(v*y)*x$, from which it follows that $(\mathbb {F}_{q^4},+,*)$ is isotopic to a commutative semifield. $\square$

Theorem 3.5

Let $q$ be a power of an odd prime. Let $a_1\in {\mathbb {F}}_{q^4}^*$ such that $a_1^{q^2+1}$ is a square in ${\mathbb {F}}_q^*$ and let $\tilde{a}_0$ be an element in $\mathbb {F}_{q^4}$ such that ${\mathrm {Tr}}_{q^4/q}(\tilde{a}_0)=-1$. Let $x*y$ be defined as in Theorem 3.4, i.e.,

$$\begin{aligned} x*y= xy + {\mathrm {Tr}}_{q^4/q}(a_1xy^{q^2}+\tilde{a}_0xy). \end{aligned}$$

Then the presemifield $(\mathbb {F}_{q^4}, +, *)$ is isotopic to Dickson’s semifield.

Proof

We have already shown in Theorem 3.4 that $(\mathbb {F}_{q^4}, +, *)$ is isotopic to a commutative semifield, which is denoted by $\mathbb {S}$. Next we are going to prove that its middle nucleus $N_m(\mathbb {S})$ is of size $q^2$ and its left nucleus $N_l(\mathbb {S})$ is of size $q$. Furthermore, as $\mathbb {S}$ is commutative, we have $N_r(\mathbb {S})= N_l(\mathbb {S})$ . Due to the classification of semifields planes of order $q^4$ with kernel $\mathbb {F}_{q^2}$ and center $\mathbb {F}_q$ by Cardinali, Polverino and Trombetti in [4], $(\mathbb {F}_{q^4},+,*)$ is isotopic to Dickson’s semifield.

To determine the middle and left nuclei of $\mathbb {S}$, we need to introduce another presemifield multiplication $x\circ y$, which corresponds to the dual spread of the spread defined by $x*y$. (For more details on the dual spread, see [16].) Actually, $x\circ y$ is defined as

$$\begin{aligned} x \circ y := xy + (a_1 y^{q^2}+\tilde{a}_0 y) {\mathrm {Tr}}_{q^4/q}(x). \end{aligned}$$

(3.9)

It is straightforward to verify that ${\mathrm {Tr}}_{q^4/q}(x(z\circ y)-z(x*y))=0$. Let $\mathbb {S}'$ denote a semifield which is isotopic to the presemifield defined by $x\circ y$. According to the interchanging of nuclei of semifields in the so called Knuth orbit ([16] and [18], Sect. 1.4]), we have $N_l(\mathbb {S}')\cong N_m(\mathbb {S})$ and $N_m(\mathbb {S}')\cong N_l(\mathbb {S})$.

To determine $N_l(\mathbb {S}')$ and $N_m(\mathbb {S}')$, we use the connection between certain homology groups as described in [13], Theorem 8.2] and [14], Result 12.4]. To be precise, we want to find every $q$-linearized polynomial $A(X)$ over $\mathbb {F}_{q^4}$ such that for every $y\in \mathbb {F}_{q^4}$, there is a $y'\in \mathbb {F}_{q^4}$ satisfying $A(x)\circ y= x\circ y'$ for every $x\in \mathbb {F}_{q^4}$. The set $\mathcal M(\mathbb {S}')$ of all such $A(X)$ is equivalent to the middle nucleus $N_m(\mathbb {S}')$.

First, it is routine to verify that $A(X)=uX$ with $u\in \mathbb {F}_q$ is in $\mathcal {M}(\mathbb {S}')$. Next we show that there are no other $A(X)$ in $\mathcal {M}(\mathbb {S}')$.

Assume that

$$\begin{aligned} A(x)y+ {\mathrm {Tr}}_{q^4/q}(A(x))(a_1y^{q^2}+\tilde{a}_0y)=xy'+ {\mathrm {Tr}}_{q^4/q}(x)(a_1y'^{q^2}+\tilde{a}_0y') \end{aligned}$$

(3.10)

holds for every $x\in \mathbb {F}_{q^4}$.

Let $x_0\in \mathbb {F}_{q^4}^*$ be such that ${\mathrm {Tr}}_{q^4/q}(x_0)= {\mathrm {Tr}}_{q^4/q}(A(x_0))=0$. Then

$$\begin{aligned} A(x_0)y=x_0y'. \end{aligned}$$

It means that $y'=uy$ holds for each $y\in \mathbb {F}_{q^4}$, where $u=A(x_0)/x_0$. Plugging it into (3.10), we have

$$\begin{aligned} A(x)y+ {\mathrm {Tr}}_{q^4/q}(A(x))(a_1y^{q^2}+\tilde{a}_0y)=uxy+ {\mathrm {Tr}}_{q^4/q}(x)(a_1(uy)^{q^2}+\tilde{a}_0uy). \end{aligned}$$

From this equation we can deduce that

$$\begin{aligned} A(x)-ux+( {\mathrm {Tr}}_{q^4/q}(A(x))- {\mathrm {Tr}}_{q^4/q}(x)u)\tilde{a}_0= & {} 0, \end{aligned}$$

(3.11)

$$\begin{aligned} ( {\mathrm {Tr}}_{q^4/q}(A(x))- {\mathrm {Tr}}_{q^4/q}(x)u^{q^2})a_1= & {} 0. \end{aligned}$$

(3.12)

Since $a_1\ne 0$, from (3.12) we see that

$$\begin{aligned} {\mathrm {Tr}}_{q^4/q}(A(x))=u^{q^2} {\mathrm {Tr}}_{q^4/q}(x) \end{aligned}$$

(3.13)

for every $x\in \mathbb {F}_{q^4}$. From (3.13) it follows that $u\in \mathbb {F}_q$. Therefore, by (3.11), we have $A(x)=ux$ where $u\in \mathbb {F}_q$. Hence $|N_l(\mathbb {S})|=|N_m(\mathbb {S}')|= q$.

Next we determine every $q$-linearized polynomial $A(X)$ over $\mathbb {F}_{q^4}$ such that for every $y\in \mathbb {F}_{q^4}$, there is a $y'\in \mathbb {F}_{q^4}$ satisfying $A(x\circ y)= x\circ y'$ for every $x\in \mathbb {F}_{q^4}$. The set of all such $A(X)$ is equivalent to the left nucleus $N_l(\mathbb {S}')$.

Assume that

$$\begin{aligned} A(xy+ {\mathrm {Tr}}_{q^4/q}(x)(a_1y^{q^2}+\tilde{a}_0y))=xy'+ {\mathrm {Tr}}_{q^4/q}(x)(a_1y'^{q^2}+\tilde{a}_0y'). \end{aligned}$$

(3.14)

It is readily verified that when $A(X)=cX$ for some $c\in \mathbb {F}_{q^2}$, (3.14) holds for all $x$ and $y$ in $\mathbb {F}_{q^4}$ with $y'=cy$. Hence $\mathbb {F}_{q^2}$ is a subfield contained in $N_l(\mathbb {S}')$. On the other hand, $N_l(\mathbb {S}')$ has to be a proper subfield of $\mathbb {F}_{q^4}$, for otherwise $\mathbb {S}'$ would be a finite field, which would lead to a contradiction. Therefore, we have $|N_m(\mathbb {S})|=|N_l(\mathbb {S}')|=q^2$, which completes the proof. $\square$

Theorem 3.6

Let $q$ be a power of a prime and let $u,v$ be elements in $\mathbb {F}_{q^3}^*$ such that $\mathrm{N}_{q^3/q}(-v/u)\ne 1$. For every $\beta \in \mathcal {B}$, where

$$\begin{aligned} {\mathcal {B}}:= \left\{ x\in \mathbb {F}_{q^3}: {\mathrm {Tr}}_{q^3/q}(u^{q^2}v^qx)= u^{q^2+q+1}+v^{q^2+q+1}\right\} , \end{aligned}$$

the equation

$$\begin{aligned} ux^{q^2-1}+vx^{q-1}+\beta =0 \end{aligned}$$

(3.15)

has no solution in $\mathbb {F}_{q^3}^*$. Let $L(X):=u^{q^2}v^q(ua^{q^2-1}X^{q^2}+va^{q-1}X^{q}+\theta X)$, where $\theta \in {\mathcal {B}}$ and $a\in \mathbb {F}_{q^3}^*$. Then the polynomial ${\mathrm {Tr}}_{q^3/q}(L(X)/X)$ has no root in $\mathbb {F}_{q^3}^*$.

Proof

When $\beta =0$, (3.15) becomes $x^{q-1}(ux^{q(q-1)}+v)=0$. If there exists $x\in \mathbb {F}_{q^3}^*$ such that $ux^{q(q-1)}+v=0$, then $\mathrm{N}_{q^3/q}(-v/u)=\mathrm{N}_{q^3/q}(x^{q(q-1)})=1$, which leads to a contradiction.

Now suppose $\beta \ne 0$. Assume to the contrary that (3.15) has a solution $x\in \mathbb {F}_{q^3}^*$. Let $y:= x^{q-1}$. Then we have $uy^{q+1}+vy+\beta =0$. It follows that

$$\begin{aligned} y^q=\frac{-vy-\beta }{uy}, \end{aligned}$$

(3.16)

and

$$\begin{aligned} y^{q^2}=\frac{v^q(vy+\beta )-\beta ^qu y}{-u^q(vy+\beta )}. \end{aligned}$$

Hence

$$\begin{aligned} y^{q^2}y^qy=\frac{v^q(vy+\beta )-\beta ^q u y}{u^{q+1}}, \end{aligned}$$

which is equal to $1$ since $y=x^{q-1}$. Therefore,

$$\begin{aligned} (v^{q+1}-\beta ^q u)y+v^q\beta =u^{q+1}. \end{aligned}$$

(3.17)

Suppose that $u\beta ^q= v^{q+1}$. Then $u^{q^2}v^q \beta =v^{q^2+1}v^q$, and ${\mathrm {Tr}}_{q^3/q}(u^{q^2}v^q \beta )=3v^{q^2+q+1}$. On the other hand, we also have $u^{q+1}=v^q\beta$ from (3.17). It follows that ${\mathrm {Tr}}_{q^3/q}(u^{q^2}v^q \beta )=3u^{q^2+q+1}$. All together with $\beta \in {\mathcal {B}}$, we have that

$$\begin{aligned} u^{q^2+q+1}+v^{q^2+q+1}=3v^{q^2+q+1}=3u^{q^2+q+1}, \end{aligned}$$

which can not holds for $3\not \mid q$. Moreover, if $3\mid q$, then $u^{q^2+q+1}=-v^{q^2+q+1}$ which contradicts the assumption that $\mathrm{N}_{q^3/q}(-v/u)\ne 1$. Hence $u\beta ^q\ne v^{q+1}$.

Since $u\beta ^q\ne v^{q+1}$, from (3.17) we obtain

$$\begin{aligned} y=\frac{u^{q+1}-v^q\beta }{v^{q+1}-\beta ^qu}. \end{aligned}$$

(3.18)

Plugging (3.18) into (3.16), we have

$$\begin{aligned} \frac{u^{q^2+q}-v^{q^2}\beta ^q}{v^{q^2+q}-\beta ^{q^2}u^q}=\frac{vu^q-\beta ^{q+1}}{v^q\beta -u^{q+1}}. \end{aligned}$$

Hence

$$\begin{aligned}&u^{q^2+q}v^q\beta - u^{q^2+2q+1} + u^{q+1}v^{q^2} \beta ^q-v^{q^2+q}\beta ^{q+1}\\ =\,&v^{q^2+q+1}u^q-\beta ^{q^2}vu^{2q}-v^{q^2+q}\beta ^{q+1}+\beta ^{q^2+q+1}u^q. \end{aligned}$$

Dividing it by $u^q$, we have

$$\begin{aligned} \beta ^{q^2+q+1}-(u^{q}v\beta ^{q^2}+uv^{q^2}\beta ^q+u^{q^2}v^q\beta )+u^{q^2+q+1}+v^{q^2+q+1}=0. \end{aligned}$$

It follows from ${\mathrm {Tr}}_{q^3/q}(u^{q^2}v^q\beta )= u^{q^2+q+1}+v^{q^2+q+1}$ that

$$\begin{aligned} \beta ^{q^2+q+1}=0. \end{aligned}$$

Hence $\beta =0$, which is a contradiction. Therefore, (3.15) has no solution in $\mathbb {F}_{q^3}^*$.

Furthermore, if ${\mathrm {Tr}}_{q^3/q}(L(X)/X)$ has a root $x_0\in \mathbb {F}_{q^3}^*$, then $u^{q^2}v^q(u(a x_0)^{q^2-1}+v(ax_0)^{q-1}+\theta )=\gamma$ for some $\gamma \in \mathbb {F}_{q^3}$ satisfying ${\mathrm {Tr}}_{q^3/q}(\gamma )=0$. We write $\gamma$ as $\gamma =u^{q^2}v^q\tau$ for some $\tau \in \mathbb {F}_{q^3}$. Then $\theta -\tau \in {\mathcal {B}}$ and

$$\begin{aligned} u(a x_0)^{q^2-1}+v(ax_0)^{q-1}+\theta -\tau =0, \end{aligned}$$

which contradicts the fact that (3.15) has no solution in $\mathbb {F}_{q^3}^*$. $\square$

For given $u$ and $v$, it is not difficult to see that for different $a$, we obtain isotopic semifields via Theorem 3.6: Let the multiplication corresponding to $a=1$ be $xy+B(x,y)$. Then for other $a\in \mathbb {F}_{q^3}^*$, the semifield multiplication is $xy + B(x/a,ay)$. Furthermore, when $u=v$ and $a=1$, it follows from Lemma 2.3 that the presemifield ${\mathbb {P}}$ derived from $L(x)$ in Theorem 3.6 is commutative. It is worth noting that, up to isotopism, we can obtain non-commutative semifields via Theorem 3.6. For instance, let $q=4$ and let $\xi$ be a primitive element of $\mathbb {F}_{q^3}$ which is a root of $X^6+X^4+X^3+X+1$. Setting $u=\xi ^5$, $v=\xi$ and $\beta =\xi ^{62}$, we can use Lemma 2.4 and computer to show that the presemifield ${\mathbb {P}}$ derived from Theorem 3.6 is not isotopic to a commutative one.

According to the classification of semifields of order $q^3$ with center containing $\mathbb {F}_q$ in [21], the presemifield obtained via Theorem 3.6 is either finite field or generalized twisted field.

Besides all the $L$’s described in this section, we did not find any other examples. Thus we propose the following question:

Question 3.7

For $n>4$, is there a $q$-linearized polynomial $L(X)=\sum _{i=0}^{n-1}a_iX^{q^i}\in \mathbb {F}_{q^n}[X]$ with $(a_1,\dots ,a_{n-1})\ne (0,\dots , 0)$ satisfying (2.3)?

4 Switchings of $\mathbb {F}_{p^n}$ for large $n$

The main result of this section is a negative answer to Question 3.7 when $q=p$ (prime) and $n$ is large.

Theorem 4.1

Let $q=p$, where $p$ is a prime, and assume $n\ge \frac{1}{2}(p-1)(p^2-p+4)$. If $L(X)=\sum _{i=0}^{n-1}a_i X^{p^i}\in {\mathbb {F}}_{p^n}[X]$ satisfies (2.3), i.e.,

$$\begin{aligned} {\mathrm{Tr}}_{p^n/p}\bigl (L(x)/x\bigr )\ne 0\quad \text {for all}\ x\in {\mathbb {F}}_{p^n}^*, \end{aligned}$$

then $a_1=\cdots =a_{n-1}=0$.

In 1971, Payne [22] considered a similar problem which calls for the determination of all $2$-linearized polynomials $L=\sum _{i=0}^{n-1}a_i X^{2^i}\in {\mathbb {F}}_{2^n}[ X]$ such that both $L( X)$ and $L( X)/ X$ are permutation polynomials of ${\mathbb {F}}_{2^n}$. Such linearized polynomials give rise to translation ovoids in the projective plane $\text {PG}(2,{\mathbb {F}}_{2^n})$ [23]. Payne later solved the problem by showing that such linearized polynomials can have only one term [23]. For a different proof of Payne’s theorem, see [11], Sect. 8.5]. For the $q$-ary version of Payne’s theorem, see [12].

4.1 Preliminaries

Let $L(X)=\sum _{i=0}^{n-1}a_i X^{q^i}\in {\mathbb {F}}_{q^n}[ X]$. For $x\in {\mathbb {F}}_{q^n}^*$, we have

$$\begin{aligned} {\text {Tr}}_{q^n/q}\left( \frac{L(x)}{x}\right) ={\text {Tr}}_{q^n/q}\left( \sum _{i=0}^{n-1}a_ix^{q^i-1}\right) =\sum _{0\le i,j\le n-1}a_i^{q^j}x^{q^j(q^i-1)}. \end{aligned}$$

Therefore (2.3) is equivalent to

$$\begin{aligned}&\left[ \;\sum _{0\le i,j\le n-1}a_i^{q^j} X^{q^j(q^i-1)}\right] ^{q-1}\equiv \;{\text {Tr}}_{q^n/q}(a_0)^{q-1}+\left[ 1-{\text {Tr}}_{q^n/q}(a_0)^{q-1}\right] X^{q^n-1}\nonumber \\&\quad \pmod { X^{q^n}- X}. \end{aligned}$$

(4.1)

Let $\Omega =\{0,1,\dots ,q^n-1\}$ and $\Omega _0=\{0,1,\dots ,\frac{q^n-1}{q-1}\}$. For $\alpha ,\beta \in \Omega _0$, define $\alpha \oplus \beta \in \Omega _0$ such that $\alpha \oplus \beta \equiv \alpha +\beta \pmod {\frac{q^n-1}{q-1}}$ and

$$\begin{aligned} \alpha \oplus \beta ={\left\{ \begin{array}{ll} 0&{}\text {if}\ \alpha =\beta =0,\\ \frac{q^n-1}{q-1}&{}\text {if}\ \alpha +\beta \equiv 0\pmod {\frac{q^n-1}{q-1}}\ \text {and}\ (\alpha , \beta )\ne (0,0). \end{array}\right. } \end{aligned}$$

For $d_0,\dots ,d_{n-1}\in {\mathbb {Z}}$, we write

$$\begin{aligned} (d_0,\dots ,d_{n-1})_q=\sum _{i=0}^{n-1}d_iq^i. \end{aligned}$$

When $q$ is clear from the context, we write $(d_0,\dots ,d_{n-1})_q=(d_0,\dots ,d_{n-1})$. For $j,i\in {\mathbb {Z}}$, $i\ge 0$, let

$$\begin{aligned} s(j,i)=(\overset{0}{0}\ \cdots \ 0\ \underbrace{\overset{j}{1}\ \cdots \ 1}_i\ 0\ \cdots \overset{n-1}{0})_q, \end{aligned}$$

where the positions of the digits are labeled modulo $n$ and the string of $1$’s may wrap around. For example, with $n=4$,

$$\begin{aligned} s(1,3)=(0\; 1\; 1\; 1),\qquad s(3,2)=(1\; 0\; 0\; 1). \end{aligned}$$

Note that

$$\begin{aligned} s(j,i)\equiv q^j\frac{q^i-1}{q-1}\pmod {q^n-1}. \end{aligned}$$

For each $\alpha \in \Omega _0$, let $C(\alpha )$ denote the coefficient of $X^{\alpha (q-1)}$ in the left side of (4.1) after reduction modulo $X^{q^n}-X$. Then we have

$$\begin{aligned} C(\alpha )=\sum _{\begin{array}{c} 0\le j_1,i_1,\dots ,j_{q-1},i_{q-1}\le n-1\\ s(j_1,i_1)\oplus \cdots \oplus s(j_{q-1},i_{q-1})=\alpha \end{array}} \prod _{k=1}^{q-1}a_{i_k}^{q^{j_k}}. \end{aligned}$$

(4.2)

Let

$$\begin{aligned} S=\{s(j,i):0\le j\le n-1,\ 1\le i\le n-1\}. \end{aligned}$$

If $C(\alpha )=0$, we can derive from (4.2) useful information about $a_i$’s if we know the possible ways to express $\alpha$ as an $\oplus$ sum of $q-1$ elements (not necessarily distinct) of $S\cup \{0\}$.

Let $\alpha =(d_0,\dots ,d_{n-1})_q\in \Omega$, where $0\le d_i\le q-1$. If $d_i>d_{i-1}$ ($d_i<d_{i-1}$), where the subscripts are taken modulo $n$, we say that $i$ is an ascending (descending) position of $\alpha$ with multiplicity $|d_i-d_{i-1}|$. The multiset of ascending (descending) positions of $\alpha$ is denoted by $\text {Asc}(\alpha )$ ($\text {Des}(\alpha )$). The multiset cardinality $|\text {Asc}(\alpha )|$ ($=|\text {Des}(\alpha )|$) is denoted by $\text {asc}(\alpha )$. For example, if $\alpha =(2\; 0\; 1\; 1\; 3\; 0)$, then

$$\begin{aligned} \text {Asc}(\alpha )=\{0,0,2,4,4\},\quad \text {Des}(\alpha )=\{1,1,5,5,5\},\quad \text {asc}(\alpha )=5. \end{aligned}$$

Assume that $\alpha \in \Omega$ has $\text {asc}(\alpha )=q-1$. Then $\alpha$ cannot be a sum of less than $q-1$ elements (not necessarily distinct) of $S$. Moreover, if

$$\begin{aligned} \alpha =s(j_1,i_1)+\cdots +s(j_{q-1},i_{q-1}), \end{aligned}$$

where $0\le j_1,\dots ,j_{q-1}\le n-1$ and $1\le i_1,\dots ,i_{q-1}\le n-1$, we must have $\{j_1,\dots ,j_{q-1}\}=\text {Asc}(\alpha )$ and $\{j_1+i_1,\dots ,j_{q-1}+i_{q-1}\}=\text {Des}(\alpha )$, where $j_k+i_k$ is taken modulo $n$.

4.2 Proof of Theorem 4.1

Lemma 4.2

Let $q=p$, where $p$ is a prime, and assume $L=\sum _{i=0}^{n-1}a_i{X}^{p^i}\in {\mathbb {F}}_{p^n}[{X}]$ satisfies (2.3). Then for all $1\le i_1<\cdots <i_{p-1}$ and $0\le t_{p-2}\le \cdots \le t_1$ with $i_{p-1}+t_1\le n-2$, we have

$$\begin{aligned} \sum _\tau \prod _{k=1}^{p-1} a_{i_{p-k}+\tau (p-k)}^{p^{ i_{p-1}-i_{p-k} }}=0, \end{aligned}$$

(4.3)

where $(\tau (1),\dots ,\tau (p-1))$ runs through all permutations of $(t_1,\dots ,t_{p-2},0)$.

Proof

Let

$$\begin{aligned} \alpha= & {} (\overbrace{1\ \cdots \ 1}^{i_{p-1}-i_{p-2}}\ \cdots \ \overbrace{p-2\ \cdots \ p-2}^{i_2-i_1}\ \overbrace{p-1\ \cdots \ p-1}^{i_1}\\&\underbrace{p-2\ \cdots \ p-2}_{t_{p-2}}\ \underbrace{p-3\ \cdots \ p-3}_{t_{p-3}-t_{p-2}}\ \cdots \ \underbrace{1\ \cdots \ 1}_{t_1-t_2}\ \underbrace{0\ \cdots \ 0}_{\begin{array}{c} n-i_{p-1}-t_1\\ \ge 2 \end{array}})\in \Omega _0. \end{aligned}$$

For $1\le k\le p-2$, we have

$$\begin{aligned} \alpha +(k\ \cdots \ k)=(\;\overbrace{k\!+\!1\ \cdots \ k\!+\!1\ \cdots \ p\!-\!1\ \cdots \ p-1\ 0\ 1\ \cdots \ 1\ \cdots }^{i_{p-1}}\ \overbrace{\cdots \ d}^{t_1}\ \overbrace{e\ \underbrace{k\ \cdots \ k}_{\ge 1}}^{n-i_{p-1}-t_1}), \end{aligned}$$

where $e=k+1$ or $k$, depending on whether it receives a carry from the preceding digit. If $e=k+1$, then $\text {asc}(\alpha +(k\;\cdots \; k))\ge p-1-k+k+1=p$. If $e=k$, then $t_1>0$ and $d\ge k+1$, which also implies that $\text {asc}(\alpha +(k\;\cdots \; k))\ge p$. Therefore $\alpha +(k\;\cdots \; k)$ is not a sum of $\le p-1$ elements (not necessarily distinct) of $S$, i.e., not a sum of $p-1$ elements (not necessarily distinct) of $S\cup \{0\}$.

On the other hand, we have $\text {asc}(\alpha )=p-1$ and

$$\begin{aligned} \text {Asc}(\alpha )= & {} \{0,i_{p-1}-i_{p-2},\dots ,i_{p-1}-i_1\},\\ \text {Des}(\alpha )= & {} \{i_{p-1},i_{p-1}+t_{p-2},\dots ,i_{p-1}+t_1\}. \end{aligned}$$

Therefore, the only possible ways to express $\alpha$ as a sum of $p-1$ elements (not necessarily distinct) of $S\cup \{0\}$ are

$$\begin{aligned} \alpha= & {} s(0,i_{p-1}+\tau (p-1))+s(i_{p-1}-i_{p-2},i_{p-2}+\tau (p-2))\\&+\cdots +s(i_{p-1}-i_1,i_1+\tau (1)), \end{aligned}$$

where $(\tau (1),\dots ,\tau (p-1))$ is a permutation of $(t_1,\dots ,t_{p-2},0)$. Together with the fact that for $1\le k\le p-2$, $\alpha +(k\;\cdots \;k)$ is not a sum of $p-1$ elements (not necessarily distinct) of $S\cup \{0\}$, we have proved that

$$\begin{aligned} \alpha =\alpha _1\oplus \cdots \oplus \alpha _{p-1},\quad \alpha _i\in S\cup \{0\}, \end{aligned}$$

if and only if

$$\begin{aligned} \{\alpha _1,\dots ,\alpha _{p-1}\}= & {} \bigl \{s(0,i_{p-1}+\tau (p-1)),s(i_{p-1}-i_{p-2},i_{p-2}+\tau (p-2)),\\&\quad \dots ,s(i_{p-1}-i_1,i_1+\tau (1))\bigr \}, \end{aligned}$$

where $(\tau (1),\dots ,\tau (p-1))$ is a permutation of $(t_1,\dots ,t_{p-2},0)$.

Now we have

$$\begin{aligned} 0\,= & {} C(\alpha )\qquad \text {(by (4.1))}\nonumber \\= & {} (p-1)!\sum _\tau \prod _{k=1}^{p-1} a_{i_{p-k}+\tau (p-k)}^{p^{ i_{p-1}-i_{p-k}}} \qquad \text {(by (4.2))}, \end{aligned}$$

(4.4)

which gives (4.3). $\square$

Proof of Theorem 4.1

$1^\circ$ We first show that for all $1\le k\le p-1$ and

$$\begin{aligned} 1+\sum _{j=0}^{k-1} j\le i_k<\cdots <i_{p-1}\le n-k-1, \end{aligned}$$

we have

$$\begin{aligned} a_{i_k}\cdots a_{i_{p-1}}=0. \end{aligned}$$

We use induction on $k$. When $k=1$, the conclusion follows from Lemma 4.2 with $t_{p-2}=\cdots =t_1=0$. Assume $2\le k\le p-1$. In Lemma 4.2, let $t_1=k-1$, $t_2=k-2,\ \dots ,\ t_{k-1}=1$, $t_k=\cdots =t_{p-2}=0$, $i_{k-1}=i_k-1$, $i_{k-2}=i_k-2,\ \dots ,\ i_1=i_k-(k-1)$, and note that $i_{p-1}+t_1=i_{p-1}+k-1\le n-2$. We have

$$\begin{aligned} \sum _\tau \prod _{j=1}^{p-1}a_{i_j+\tau (j)}^*=0, \end{aligned}$$

(4.5)

where $(\tau (1),\dots ,\tau (p-1))$ runs through all permutations of $(k-1,k-2,\dots ,1,0,\dots ,0)$ and the $*$’s are suitable powers of $p$. (In general, we use a $*$ to denote a positive integer exponent whose exact value is not important.) Multiplying (4.5) by $a_{i_k}\cdots a_{i_{p-1}}$ gives

$$\begin{aligned} a_{i_k}^*\cdots a_{i_{p-1}}^*+\sum _{\begin{array}{c} \tau \\ (\tau (1),\dots ,\tau (k-1))\ne (k-1,\dots ,1) \end{array}} a_{i_k}\cdots a_{i_{p-1}}\prod _{j=1}^{p-1}a_{i_j+\tau (j)}^*=0. \end{aligned}$$

(4.6)

When $(\tau (1),\dots ,\tau (k-1))\ne (k-1,\dots ,1)$, at least one of $i_1+\tau (1),\dots ,i_{p-1}+\tau (p-1)$, say $i_{k-1}'$, is less than $i_k$. Also note that $i_{k-1}'\ge i_1=i_k-(k-1)\ge 1+1+2+\cdots +(k-2)$. Therefore by the induction hypothesis, $a_{i_{k-1}'}a_{i_k}\cdots a_{i_{p-1}}=0$. Thus the $\sum$ in (4.6) equals $0$, which gives $a_{i_k}\cdots a_{i_{p-1}}=0$.

$2^\circ$ Let $k=p-1$ in $1^\circ$. We have

$$\begin{aligned} a_i=0\quad \text {for all}\ 1+\frac{1}{2}(p-2)(p-1)\le i\le n-p. \end{aligned}$$

$3^\circ$ We claim that

$$\begin{aligned} a_i=0\quad \text {for all}\ 1\le i\le \frac{1}{2}(p-2)(p-1). \end{aligned}$$

Assume to the contrary that this is not true. Let $1\le l\le \frac{1}{2}(p-2)(p-1)$ be the largest integer such that $a_l\ne 0$. Let

$$\begin{aligned} \alpha =(\underbrace{\underbrace{1\ \cdots \ 1}_l\ \underbrace{0\ \cdots \ 0}_{p+1}\ \underbrace{1\ \cdots \ 1}_l\ \underbrace{0\ \cdots \ 0}_{p+1}\ \cdots \ \underbrace{1\ \cdots \ 1}_l\ \underbrace{0\ \cdots \ 0}_{p+1}}_{{p}-{\text {1 copies}}}\ 0\ \cdots \ 0 )\in \Omega _0. \end{aligned}$$

(Here we used the assumption that $n\ge (p-1)\bigl [\frac{1}{2}(p-2)(p-1)+p+1\bigr ]$.) For $0\le k\le p-2$, we have $\text {asc}(\alpha +(k\; \cdots \; k))=p-1$ and

$$\begin{aligned} \text {Asc}\bigl (\alpha +(k\; \cdots \; k)\bigr )= & {} \bigl \{0,\,l+p+1,\, 2(l+p+1),\,\dots ,\, (p-2)(l+p+1)\bigr \},\\ \text {Des}\bigl (\alpha +(k\; \cdots \; k)\bigr )= & {} \bigl \{l,\, l+p+1+l,\, 2(l+p+1)+l,\, \dots ,\, (p-2)(l+p+1)+l\bigr \}. \end{aligned}$$

If $\alpha +(k\; \cdots \; k)$ is expressed as a sum of $p-1$ elements (not necessarily distinct) of $S$, the expression must be of the form

$$\begin{aligned} \alpha +(k\; \cdots \; k)=s(0,i_1)+s(l+p+1,i_2)+\cdots +s((p-2)(l+p+1),i_{p-1}), \end{aligned}$$

(4.7)

where $i_1,\dots ,i_{p-1}\in \{1,\dots ,n-1\}$, and in modulus $n$

$$\begin{aligned}&\bigl \{i_1,\, l+p+1+i_2,\, \dots ,\, (p-2)(l+p+1)+i_{p-1}\bigr \}\nonumber \\&\quad =\;\bigl \{l,\, l+p+1+l,\, 2(l+p+1)+l,\, \dots ,\, (p-2)(l+p+1)+l\bigr \}. \end{aligned}$$

(4.8)

We further require $a_{i_1}\cdots a_{i_{p-1}}\ne 0$, which implies that $i_1,\dots ,i_{p-1}\in \{1,\dots ,l\}\cup \{n-p+1,\dots , n-1\}$. It follows from (4.8) that $i_1=\cdots =i_{p-1}=l$. Thus we have

(4.9)

which is a contradiction.

$4^\circ$ Finally, we claim that

$$\begin{aligned} a_i=0\quad \text {for all}\ n-p+1\le i\le n-1. \end{aligned}$$

For $x\in \mathbb {F}_{p^n}^*$,

$$\begin{aligned} {\mathrm {Tr}}_{p^n/p}\left( L(x^{-1})/x^{-1}\right)&= {\mathrm {Tr}}_{p^n/p}\left( \sum _{i=0}^{n-1}a_ix^{1-p^i}\right) = {\mathrm {Tr}}_{p^n/p}\left( \sum _{i=0}^{n-1}a_{i}^{p^{n-i}}x^{p^{n-i}-1}\right) \\&= {\mathrm {Tr}}_{p^n/p}\left( \sum _{i=0}^{n-1}a_{n-i}^{p^i} x^{p^i-1}\right) , \end{aligned}$$

where $a_n=a_0$. Thus $L_1(X):=\sum _{i=0}^{n-1}a_{n-i}^{p^i}X^{p^i}$ also satisfies (2.3). By $2^\circ$ and $3^\circ$, $a_{n-i}=0$ for all $1\le i\le n-p$, i.e., $a_i=0$ for all $p\le i\le n-1$. Since $p\le n-p-1$, the claim is proved. $\square$

It appears that the assumption that $n\ge \frac{1}{2}(p-1)(p^2-p+4)$ in Theorem 4.1 may be weakened. On the other hand, when $q$ is not a prime, the proofs of Lemma 4.2 and Theorem 4.1 fail for the following reason: In (4.4) and (4.9), $(p-1)!$ is replaced by $(q-1)!$, which is $0$ in ${\mathbb {F}}_q$. When $q=p^e$, (4.1) becomes

$$\begin{aligned} \left[ \;\prod _{k=0}^{e-1}\sum _{0\le i,j\le n-1}a_1^{p^kq^j}X^{p^kq^j(q^i-1)}\;\right] ^{p-1}\equiv \;&{\text {Tr}}_{q^n/q}(a_0)^{q-1}+\Bigl [1-{\text {Tr}}_{q^n/q}(a_0)^{q-1}\Bigr ]X^{q^n-1}\\&\pmod {X^{q^n}-X}. \end{aligned}$$

The question is how to decipher this equation.

5 A connection to some cyclic codes for general $\mathbb {F}_q$

In this section we prove certain necessary conditions for a $q$-linearized polynomials $L(X) \in \mathbb {F}_{q^n}[X]$ to satisfy ${\mathrm {Tr}}_{q^n/q}(L(x)/x) \ne 0$ for all $x \in \mathbb {F}_{q^n}^*$, where $q$ is a prime power. In particular, we give a natural connection to some cyclic codes. There is also a connection of such cyclic codes to some algebraic curves. In the next section, we will use this connection to algebraic curves to get some necessary conditions for such $q$-linearized polynomials $L(X) \in \mathbb {F}_{q^n}[X]$.

If $L(X)=a_0X \in \mathbb {F}_{q^n}[X]$, then ${\mathrm {Tr}}_{q^n/q}(L(x)/x) \ne 0$ for all $x \in \mathbb {F}_{q^n}^*$ if and only if ${\mathrm {Tr}}_{q^n/q}(a_0) \ne 0$. Hence we assume that $L(X)=a_0X + a_1X^{q} + \cdots +a_{n-1}X^{q^{n-1}} \in \mathbb {F}_{q^n}[X]$ with $(a_1,a_2, \ldots , a_{n-1}) \ne (0,0, \ldots , 0)$.

First we recall some notation and basic facts from coding theory (see, for example, [20]). Let $N=q^n-1$. A code of length $N$ over $\mathbb {F}_q$ is just a nonempty subset of $\mathbb {F}_q^N$. It is called a linear code if it is a vector space over $\mathbb {F}_q$. The set $C^\bot$ of all $N$-tuples in $\mathbb {F}_q^N$ orthogonal to all codewords of a linear code $C$ with respect to the usual inner product on $\mathbb {F}_q^N$ is called the dual code of $C$. The Hamming weight of an arbitrary $N$-tuple $\mathbf u =(u_0,u_1, \ldots ,u_{N-1}) \in \mathbb {F}_q^N$ is

$$\begin{aligned} ||\mathbf u ||=|\{0 \le i \le N-1: u_i \ne 0\}|. \end{aligned}$$

A cyclic code of length $N$ over $\mathbb {F}_q$ is an ideal $C$ of the quotient ring $R=\mathbb {F}_q[X]/\langle X^N-1\rangle$. Here a codeword $(c_0,c_1, \ldots , c_{N-1})\in \mathbb {F}_q^N$ of $C$ corresponds to an element $c_0 + c_1X + \cdots + c_{N-1}X^{N-1}+\langle X^N-1 \rangle \in C$. All ideals of $R$ are principal. The monic polynomial $g(X)$ of the least degree such that $C=\langle g(X)\rangle /\langle X^N-1 \rangle$ is called the generator polynomial of $C$. The dual $C^\bot$ is cyclic with generator polynomial $X^{\deg h}h(X^{-1})/h(0)$, where $h(X)=(X^N-1)/g(X)$.

If $\theta \in \mathbb {F}_{q^n}$ is a root of $g(X)$, then so is $\theta ^q$. A set $B \subset \mathbb {F}_{q^n}$ is called a basic zero set of $C$ if both of the following conditions are satisfied:

• $\{\theta ^{q^i}: \theta \in B, 0 \le i \le n-1\}$ is the set of the roots of $g(X)$.

• If $\theta _1, \theta _2 \in B$ with $\theta _1^{q^i}=\theta _2$ for some integer $i$, then $\theta _1=\theta _2$.

The following proposition gives a natural connection to some cyclic codes. Some arguments in its proof will also be used in the next section.

Proposition 5.1

Let $\gamma$ be a primitive element of $\mathbb {F}_{q^n}^*$. Let $C$ be the cyclic code of length $N=q^n-1$ over $\mathbb {F}_q$ whose dual code $C^{\perp }$ has

$$\begin{aligned} \left\{ 1, \gamma ^{q-1}, \gamma ^{q^2-1}, \ldots , \gamma ^{q^{n-1}-1}\right\} \end{aligned}$$

as a basic zero set. We have the following: There exists a $q$-linearized polynomial $L(X)=a_0X + a_1 X^{q} + \cdots + a_{n-1}X^{q^{n-1}} \in \mathbb {F}_{q^n}[X]$ with $(a_1, a_2, \ldots , a_{n-1}) \ne (0,0, \ldots ,0)$ such that ${\mathrm {Tr}}_{q^n/q}(L(x)/x) \ne 0$ for all $x \in \mathbb {F}_{q^n}^*$ if and only if the cyclic code $C$ has a codeword $(c_0,c_1, \ldots , c_{N-1})$ of Hamming weight $N$ such that $(c_0,c_1, \ldots , c_{N-1}) \ne u(1,1,\ldots ,1)$ for any $u \in \mathbb {F}_q^*$. Moreover the dimension of $C$ over $\mathbb {F}_q$ is $n^2-n+1$.

Proof

We first show that $\{1,\gamma ^{q-1}, \gamma ^{q^2-1}, \ldots , \gamma ^{q^{n-1}-1}\}$ is a basic zero set. This means that the exponents $0,q-1,q^2-1, \ldots , q^{n-1}-1$ are in distinct $q$-cyclotomic cosets modulo $q^n-1$. For $0 \le d < q^n-1$, let $\psi (d)$ be the base $q$ digits of $d$, i.e., $\psi (d)=(d_0,d_1, \ldots , d_{n-1})$, where $0 \le d_i \le q-1$ are integers such that $d=\sum _{i=0}^{n-1}d_iq^i$. Let $\overline{0}, \overline{q-1}, \overline{q^2-1}, \ldots , \overline{q^{n-1}-1}$ denote the $q$-cyclotomic cosets of $0,q-1, q^2-1, \ldots , q^{n-1}-1$ modulo $q^n-1$. Their images under $\psi$ are

Note that the elements in each row are obtained via cyclic shifts of the first element of the row. This proves that $0,q-1,q^2-1, \ldots , q^{n-1}-1$ are in distinct $q$-cyclotomic cosets modulo $q^n-1$. Moreover the cardinality of the union of their $q$-cyclotomic cosets modulo $q^n-1$ is

$$\begin{aligned} 1+(n-1)n=n^2-n+1. \end{aligned}$$

Therefore the dimensions of $C$ is $n^2-n+1$. Finally using Delsarte’s Theorem [26], Theorem 9.1.2] we obtain that the codewords of $C$ in $\mathbb {F}_q^N$ are

$$\begin{aligned} C=\left\{ \left( {\mathrm {Tr}}_{q^n/q}\Big (a_0 + a_1x^{q-1} + \cdots + a_{n-1}x^{q^{n-1}-1}\Big )\right) _{x \in \mathbb {F}_{q^n}^*}: a_0,a_1, \ldots , a_{n-1} \in \mathbb {F}_{q^n}\right\} . \end{aligned}$$

Note that ${\mathrm {Tr}}_{q^n/q}(L(x)/x) = u$ for all $x \in \mathbb {F}_{q^n}^*$ if and only if ${\mathrm {Tr}}_{q^n/q}(L(X)/X)\equiv u \pmod {X^{q^n}-X}$, from which it follows that $(a_1, a_2, \ldots , a_{n-1}) = (0,0, \ldots ,0)$. This completes the proof. $\square$

6 Some conditions via the Hasse–Weil–Serre bound for general $\mathbb {F}_q$

In this section we obtain some necessary conditions for the $q$-linearized polynomials $L(X) \in \mathbb {F}_{q^n}[X]$ such that ${\mathrm {Tr}}_{q^n/q}(L(x)/x) \ne 0$ for all $x \in \mathbb {F}_{q^n}^*$.

The Hasse–Weil–Serre bound for algebraic curves over finite fields implies upper and lower bounds on the Hamming weights of codewords of cyclic codes (see [10, 28]). Using this method we obtain Theorem 6.1.

First we introduce further notations. Let $\mathrm{Res\,}: {\mathbb {Z}}\rightarrow \{0,1, \ldots , q^n-2\}$ be the map such that $\mathrm{Res\,}(j) \equiv j \pmod {q^n-1}$. Put $q=p^m$ with $m \ge 1$, where $p$ is the characteristic of $\mathbb {F}_q$. Let ${\mathrm{Lead}}:\{0,1, \ldots , p^{mn}-2\} \rightarrow \{0,1, \ldots , p^{mn}-2\}$ be the map sending $j$ to the smallest integer $k$ in $\{0,1, \ldots , p^{mn-2}\}$ such that $k \equiv jp^u \pmod {p^{mn}-1}$ for some integer $u\ge 0$. In other words, ${\mathrm{Lead}}(j)$ is the smallest nonnegative integer in the $p$-cyclotomic coset of $j$ modulo $p^{mn}-1$. It is important to note that if $0 < j< p^{mn}-1$, then ${\mathrm{Lead}}(j)$ is a nonnegative integer which is coprime to $p$.

Theorem 6.1

Let $L(X)=a_0X+a_1X^q+ \cdots +a_{n-1}X^{q^{n-1}} \in \mathbb {F}_{q^n}[X]$ be a $q$-linearized polynomial with $(a_1,\ldots , a_{n-1}) \ne (0, \ldots , 0)$. For each $1 \le j \le q^n-2$ with $\gcd (j,q^n-1)=1$, let

$$\begin{aligned} \ell (j)=\max \{{\mathrm{Lead}}(\mathrm{Res\,}(j(q^i-1))): 1 \le i \le n-1 \; \text{ and } \; a_i \ne 0\}. \end{aligned}$$

Moreover, let

$$\begin{aligned} \ell =\min _j \ell (j), \end{aligned}$$

(6.1)

where the minimum is over all integers $1 \le j \le q^n-2$ with $\gcd (j,q^n-1)=1$. Then we have the following:

• Case ${\mathrm {Tr}}_{q^n/q}(a_0) \ne 0$: If

  $$\begin{aligned} q^n+1 - \frac{(q-1)(\ell -1)}{2}\lfloor 2 q^{n/2} \rfloor >1, \end{aligned}$$

  (6.2)

  then it is impossible that ${\mathrm {Tr}}_{q^n/q}(L(x)/x) \ne 0$ for all $x \in \mathbb {F}_{q^n}^*$.

• Case ${\mathrm {Tr}}_{q^n/q}(a_0) = 0$: If

  $$\begin{aligned} q^n+1 - \frac{(q-1)(\ell -1)}{2} \lfloor 2 q^{n/2} \rfloor >q+1, \end{aligned}$$

  (6.3)

  then it is impossible that ${\mathrm {Tr}}_{q^n/q}(L(x)/x) \ne 0$ for all $x \in \mathbb {F}_{q^n}^*$.

Proof

If $\gamma$ is a primite element of $\mathbb {F}_{q^n}^*$, then $\gamma ^j$ is also a primitive element of $\mathbb {F}_{q^n}^*$ for all $1 \le j \le q^n-2$ with $\gcd (j,q^n-1)=1$. Note that

$$\begin{aligned} {\mathrm {Tr}}_{q^n/q}(L(x)/x)= {\mathrm {Tr}}_{q^n/q}\Big (a_0 + a_1 x^{q-1} + \cdots + a_{n-1}x^{q^{n-1}-1}\Big ) \ne 0 \;\; \hbox {for all } x \in \mathbb {F}_{q^n}^*, \end{aligned}$$

if and only if

$$\begin{aligned} {\mathrm {Tr}}_{q^n/q}(L(x^j)/x^j)= {\mathrm {Tr}}_{q^n/q}(a_0 + a_1 x^{j(q-1)} + \cdots + a_{n-1}x^{j(q^{n-1}-1)}) \ne 0 \;\; \text{ for } \text{ all } x \in \mathbb {F}_{q^n}^*. \end{aligned}$$

Moreover, $x^{j(q^i-1)}=x^{\mathrm{Res\,}(j(q^i-1))}$ for $x \in \mathbb {F}_{q^n}^*$, $1 \le i \le n-1$ and $1 \le j \le q^n-2$.

Recall that $\ell$ is defined in (6.1). We choose and fix an integer $1 \le j \le q^n-2$ with $\gcd (j,q^n-1)=1$ such that $\ell =\ell (j)$.

Let $a_{t_1},\dots ,a_{t_s}$ be the nonzero coefficients among $a_1,\dots ,a_{n-1}$. (Note that $s\ge 1$ since $(a_1,\dots ,a_{n-1})\ne (0,\dots ,0)$.) Since $0,q^{t_1}-1,\dots ,q^{t_s}-1$ belong to different $p$-cyclotomic cosets modulo $q^n-1$ and $\text {gcd}(j,q^n-1)=1$, we have that $0,j(q^{t_1}-1),\dots ,j(q^{t_s}-1)$ belong to different $p$-cyclotomic cosets modulo $q^n-1$. Thus $\mathrm{Res\,}(j(q^{t_i}-1))=j_ip^{u_i}$, where $u_i\ge 0$, $p\not \mid j_i$, $1\le i\le s$, and $j_1,\dots ,j_s$ are distinct. We may assume $0<j_1<j_2<\cdots <j_s=\ell$. We have

$$\begin{aligned} a_0 + a_1 X^{\mathrm{Res\,}(j(q-1))} + \cdots + a_{n-1}X^{\mathrm{Res\,}(j(q^{n-1}-1))}=a_0 + b_{1} X^{j_1p^{u_1}} + \cdots + b_{s} X^{j_sp^{u_s}}, \end{aligned}$$

where $b_i=a_{t_i}$, $1\le i\le s$.

Let $\chi$ be the Artin-Shreier type algebraic curve over $\mathbb {F}_{q^n}$ given by

$$\begin{aligned} \chi : Y^q-Y=a_0 + b_1X^{j_1p^{u_1}} + \cdots + b_s X^{j_sp^{u_s}}\!. \end{aligned}$$

Let $S \subset \mathbb {F}_{p^{mn}}^*$ be a complete set of coset representatives of $\mathbb {F}_p^*$ in $\mathbb {F}_{p^{mn}}^*$. For $\mu \in S$, let $\chi _\mu$ be the Artin-Shreier type algebraic curve over $\mathbb {F}_{q^n}$ given by

$$\begin{aligned} \chi _\mu : Y^p-Y=\mu \left( a_0 + b_1X^{j_1p^{u_1}} + \cdots + b_s X^{j_sp^{u_s}}\right) . \end{aligned}$$

Note that $\chi _\mu$ is a degree $p$ covering of the projective line. Using [9], Theorem 2.1] the genus $g(\chi )$ of $\chi$ is computed in terms of the genera of $\chi _\mu$ as

$$\begin{aligned} g(\chi )=\sum _{\mu \in S} g(\chi _\mu ). \end{aligned}$$

(6.4)

Now we determine the genus $g(\chi _\mu )$ of $\chi _\mu$. We choose and fix $\mu \in S$. Let $c_1,c_2, \ldots , c_s \in \mathbb {F}_{p^{mn}}^*$ be such that

$$\begin{aligned} c_1^{p^{u_1}}=\mu b_1, \; c_2^{p^{u_2}}=\mu b_2,\; \ldots ,\; c_s^{p^{u_s}}=\mu b_s. \end{aligned}$$

Let $\chi ^\prime _\mu$ be the Artin-Schreier type algebraic curve over $\mathbb {F}_{q^n}$ given by

$$\begin{aligned} \chi ^\prime _\mu : Y^p-Y=\mu a_0 + c_1X^{j_1} + \cdots + c_s X^{j_s}. \end{aligned}$$

We observe that $\chi _\mu$ and $\chi ^\prime _\mu$ are birationally isomorphic and hence the genera $g(\chi _\mu )$ and $g(\chi ^\prime _\mu )$ are the same. Indeed, if $u_1 \ge 1$, then

$$\begin{aligned} \begin{array}{rcl} Y^p-Y &{} =&{} \mu a_0 + c_1^{p^{u_1}}X^{j_1 p^{u_1}} + c_2^{p^{u_2}}X^{j_2p^{u_2}} + \cdots + c_s^{p^{u_s}}X^{j_sp^{u_s}} \\ \\ &{}=&{} \mu a_0 + \left( c_1^{p^{u_1-1}}X^{j_1p^{u_1-1}}\right) ^p + c_2^{p^{u_2}}X^{j_2p^{u_2}} + \cdots + c_s^{p^{u_s}}X^{j_sp^{u_s}} \end{array} \end{aligned}$$

and hence

$$\begin{aligned} \begin{array}{l} \displaystyle \left[ Y-\left( c_1^{p^{u_1-1}}X^{j_1p^{u_1-1}}\right) \right] ^p - \left[ Y - \left( c_1^{p^{u_1-1}}X^{j_1p^{u_1-1}}\right) \right] \\ \\ =\mu a_0 + c_1^{p^{u_1-1}}X^{j_1p^{u_1-1}} + c_2^{p^{u_2}}X^{j_2p^{u_2}} + \cdots + c_s^{p^{u_s}}X^{j_sp^{u_s}}. \end{array} \end{aligned}$$

This gives a birational isomorphism between $\chi _\mu$ and the curve given by

$$\begin{aligned} Y^p-Y=\mu a_0 + c_1^{p^{u_1-1}}X^{j_1p^{u_1-1}} + c_2^{p^{u_2}}X^{j_2p^{u_2}} + \cdots + c_s^{p^{u_s}}X^{j_sp^{u_s}}. \end{aligned}$$

By induction on $u_1$ we obtain a birational isomorphism between $\chi _\mu$ and the curve given by

$$\begin{aligned} Y^p-Y=\mu a_0 + c_1X^{j_1} + c_2^{p^{u_2}}X^{j_2p^{u_2}} + \cdots + c_s^{p^{u_s}}X^{j_sp^{u_s}}. \end{aligned}$$

Applying the same method to the monomials $c_2^{p^{u_2}}X^{j_2p^{u_2}}, \ldots , c_s^{p^{u_s}}X^{j_sp^{u_s}}$ we conclude that the curves $\chi _\mu$ and $\chi ^\prime _\mu$ are birationally isomorphic.

Recall that the integers $0,j_1, \ldots , j_s$ are in distinct $p$-cyclotomic cosets modulo $q^n-1$. As $c_s \ne 0$ and $\gcd (j_s,p)=1$ we obtain that $\chi _\mu ^\prime$ is absolutely irreducible over $\mathbb {F}_{q^n}$. Moreover $s \ge 1$ and $j_s=\ell$. Hence by [26], Proposition 3.7.8] we have

$$\begin{aligned} g(\chi _\mu )=g(\chi _\mu ^\prime )=(p-1)(\ell -1)/2, \end{aligned}$$

which is independent from the choice of $\mu \in S$. Using (6.4) for the genus $g(\chi )$ of $\chi$ we obtain that

$$\begin{aligned} g(\chi )=\sum _{\mu \in S} g(\chi _\mu )= |S| (p-1)(\ell -1)/2=(q-1)(\ell -1)/2. \end{aligned}$$

Assume that ${\mathrm {Tr}}_{q^n/q}(L(x)/x) \ne 0$ for all $x \in \mathbb {F}_{q^n}^*$. The number $N(\chi )$ of $\mathbb {F}_{q^n}$-rational points of $\chi$ is

$$\begin{aligned} \begin{array}{cl} N(\chi )= 1 + q |\{x \in \mathbb {F}_{q^n}: {\mathrm {Tr}}(L(x)/x)=0\}|=\left\{ \begin{array}{cl} 1 &{} \text{ if } \quad {\mathrm {Tr}}_{q^n/q}(a_0) \ne 0, \\ q+1 &{} \text{ if } \quad {\mathrm {Tr}}_{q^n/q}(a_0) = 0. \end{array}\right. \end{array} \end{aligned}$$

(6.5)

The Hasse–Weil–Serre lower bound on $N(\chi )$ (see, for example, [26], Theorem 5.3.1]) implies that

$$\begin{aligned} N(\chi ) \ge q^n+1 - \frac{(q-1)(\ell -1)}{2} \lfloor 2q^{n/2} \rfloor . \end{aligned}$$

(6.6)

Combining (6.2), (6.3), (6.5) and (6.6), we complete the proof. $\square$

The following corollary, which is a restatement of Theorem 6.1, shows that the distribution of the nonzero coefficients of a $q$-linearized polynomial $L$ satisfying ${\mathrm {Tr}}_{q^n/q}(L(x)/x) \ne 0$ for all $x \in \mathbb {F}_{q^n}^*$ is subject to certain restrictions.

Corollary 6.2

Let $L(X)=a_0X+a_1X^q+ \cdots +a_{n-1}X^{q^{n-1}} \in \mathbb {F}_{q^n}[X]$ be a $q$-linearized polynomial with $(a_1,\ldots , a_{n-1}) \ne (0, \ldots , 0)$. Assume that ${\mathrm {Tr}}_{q^n/q}(L(x)/x) \ne 0$ for all $x \in \mathbb {F}_{q^n}^*$. Then for each integer $1 \le j \le q^n-2$ with $\gcd (j,q^n-1)=1$ we have the following:

• (i) If ${\mathrm {Tr}}_{q^n/q}(a_0) \ne 0$, there exits $1 \le i \le n-1$ such that $a_i \ne 0$ and

  $$\begin{aligned} {\mathrm{Lead}}(\mathrm{Res\,}(j(q^i-1))) \ge 1 + \left\lceil \frac{2q^n}{(q-1) \lfloor 2q^{n/2} \rfloor } \right\rceil . \end{aligned}$$

• (ii) If ${\mathrm {Tr}}_{q^n/q}(a_0) = 0$, there exits $1 \le i \le n-1$ such that $a_i \ne 0$ and

  $$\begin{aligned} {\mathrm{Lead}}(\mathrm{Res\,}(j(q^i-1))) \ge 1 + \left\lceil \frac{2(q^n-q)}{(q-1) \lfloor 2q^{n/2} \rfloor } \right\rceil . \end{aligned}$$