Constructions and equivalence of Sidon spaces

Abstract

Sidon spaces have been introduced by Bachoc et al. (in: Mathematical Proceedings of the Cambridge Philosophical Society, Cambridge University Press, 2017) as the q-analogue of Sidon sets. The interest on Sidon spaces has increased quickly, especially after the work of Roth et al. (IEEE Trans Inform Theory 64(6):4412–4422, 2017), in which they highlighted the correspondence between Sidon spaces and cyclic subspace codes. Up to now, the known constructions of Sidon Spaces may be divided in three families: the ones contained in the sum of two multiplicative cosets of a fixed subfield of $\mathbb {F}_{q^n}$, the ones contained in the sum of more than two multiplicative cosets of a fixed subfield of $\mathbb {F}_{q^n}$ and finally the ones obtained as the kernel of subspace polynomials. In this paper, we will mainly focus on the first class of examples, for which we provide characterization results and we will show some new examples, arising also from some well-known combinatorial objects. Moreover, we will give a quite natural definition of equivalence among Sidon spaces, which relies on the notion of equivalence of cyclic subspace codes and we will discuss about the equivalence of the known examples.

1 Introduction

Bachoc et al. [1] introduced the notion of Sidon space as an important object to study in order to prove the linear analogue of Vosper’s theorem, which characterizes the equality in the linear analogue of Cauchy–Davenport inequality proved in [2, 19]. Sidon spaces can be seen as the q-analog of Sidon sets, a well-studied combinatorial object introduced by Simon Szidon, see [14]. The definition of Sidon space is the following. An ${{\mathbb {F}}}_{q}$-subspace V of $\mathbb {F}_{q^n}$ is called a Sidon space if for all nonzero $a,b,c,d \in V$ such that $ab=cd$ then

$$\begin{aligned}\{a {{\mathbb {F}}}_q,b {{\mathbb {F}}}_q\}=\{c {{\mathbb {F}}}_q,d {{\mathbb {F}}}_q\},\end{aligned}$$

where $e{{\mathbb {F}}}_{q}=\{e\lambda :\lambda \in {{\mathbb {F}}}_{q}\}$.

Sidon spaces gained a lot of attention especially because of their connection with cyclic subspace codes, pointed out by Roth, Raviv and Tamo in [40]. Indeed, this connection allowed them to prove, for most of the cases, a conjecture on the existence of cyclic subspace codes by Horlemann-Trautmann, Manganiello, Braun and Rosenthal in [46].

Let n and k be positive integers such that $k \le n$. We denote by $\mathcal {G}_q(n,k)$ the set of k-dimensional ${{\mathbb {F}}}_q$-subspaces of ${{\mathbb {F}}}_{q^n}$. For $U \in \mathcal {G}_q(n,k)$ and $\alpha \in \mathbb {F}_{q^n}$, denote by $\textrm{Orb}(U)=\{\alpha U :\alpha \in \mathbb {F}_{q^n}^*\}$ the orbit of U, where $\alpha U=\{\alpha u :u \in U\}$. We can equip $\mathcal {G}_q(n,k)$ with the subspace metric, that is,

$$\begin{aligned}d(U,V)=\dim _{{{\mathbb {F}}}_q}(U)+\dim _{{{\mathbb {F}}}_q}(V)-2\dim _{{{\mathbb {F}}}_q}(U \cap V),\end{aligned}$$

where $U,V \in \mathcal {G}_{q}(n,k)$. A constant dimension subspace code as a subset $\mathcal {C}$ of $\mathcal {G}_{q}(n,k)$ endowed with the subspace metric. In particular, its minimum distance $d(\mathcal {C})$ is the minimum of distances between two distinct elements in $\mathcal {C}$. Codes in the subspace metric have been investigated especially after the well-celebrated paper of Koetter and Kschischang [21], in which they showed how to use such codes in random network coding. Because of their algebraic structure, the most investigated class is those of cyclic subspace codes, originally introduced by Etzion and Vardy in [15]. A constant dimension subspace code $\mathcal {C} \subseteq \mathcal {G}_q(n,k)$ is said to be cyclic if for every $\alpha \in {{\mathbb {F}}}_{q^n}^*$ and every $V \in C$ then $\alpha V \in C$, that is, $\mathcal {C}$ is the union of orbit of subspaces in $\mathcal {G}_q(n,k)$.

The following result establishes a correspondence between Sidon spaces and cyclic subspace codes.

Theorem 1.1

[40, Lemma 34] Let $U \in \mathcal {G}_q(n,k)$. Then, U is a Sidon space if and only if $|\textrm{Orb}(U)|=\frac{q^n-1}{q-1}$ and $d(\textrm{Orb}(U))=2k-2$, i.e., for all $\alpha \in \mathbb {F}_{q^n}{\setminus } {{\mathbb {F}}}_{q}$

$$\begin{aligned}\dim _{{{\mathbb {F}}}_{q}}(U\cap \alpha U)\le 1.\end{aligned}$$

Note that $|\textrm{Orb}(U)|=\frac{q^n-1}{q-1}$ is equivalent to require that U is strictly ${{\mathbb {F}}}_{q}$-linear; see [34, Theorem 1].

Therefore, equivalently we can talk about k-dimensional Sidon spaces or optimal one-orbit cyclic subspace codes with minimum distance $2k-2$.

As a consequence of [1] (cf. Theorem 4.4), a Sidon space V in $\mathbb {F}_{q^n}$ has dimension at most n/2. Despite the classical case of Sidon sets, there are constructions of Sidon spaces having maximum dimension for any $q\ge 3$ and n even provided in [40].

In this paper, we will use the terminology of Sidon spaces and our aim is to show new constructions of such spaces and to give some insight on the equivalence of these objects. All of our results may be read in terms of cyclic subspace codes via the correspondence given in Theorem 1.1.

Known examples of Sidon spaces can be divided in three families, based on their descriptions:

• I as subspace of the sum of two multiplicative cosets¹ of a fixed subfield of $\mathbb {F}_{q^n}$;

• II as subspace of the sum of more than two multiplicative cosets of a fixed subfield of $\mathbb {F}_{q^n}$;

• III as kernel of subspace polynomials.

In the first family, we find the examples in [40, Constructions 11 and 15], [50, Lemma 3.1], [24, Lemmas 4.1 and 4.2] and [51, Lemma 2.4], in the second one we find examples in [53, Theorems 3.2 and 3.6], [50, Lemma 3.3, Theorems 3.5 and 3.13], [24, Theorems 4.5, 4.7, 4.9, 4.13-18 and 4.20], [49, Lemma 5 and Theorem 6], [52, Theorems 3.1 and 3.2] and [16, Theorem 3.3], and for the last family we can find examples in [9, Corollary 4 and Section IIIA] and [41, Section 7.3].

We also mention an explicit construction of Sidon spaces in $\mathcal {G}_q(n,k)$ that does not belong to the above-described families, whose parameters are n odd, $4k-1 \le n$ and $n \mid (q-1)$, see [53, Theorem 3.8]. Furthermore in [40], existence results of Sidon spaces, for any n and k such that $k \le \lfloor \frac{n-2}{4}\rfloor$ or $\left( {\begin{array}{c}k+1\\ 2\end{array}}\right) \le n$, are provided, see [40, Theorems 19 and 31].

In this paper, we construct and provide characterization results for Sidon spaces in $\mathcal {G}_{q}(n,m)$ with the following parameters:

• $q\ge 2$, $n=rk$, $r\ge 2$ and $m\le k=n/r$ (cf. Sects. 3 and 4);

• $q\ge 3$, $n=rk$, $r\ge 6$ and $m\le 2k$ (cf. Corollary 5.4).

The parameters are already covered by the previously mentioned constructions; however, via the notion of equivalence we introduce, we will show that we have a wide variety of examples with these parameters. This may be also of interest from a cryptographic point of view: In [39], a multivariate public key cryptosystem based on the Sidon spaces found in [40] was developed. It has been recently broken in [11] due to the nice algebraic structure of the chosen Sidon spaces. As the authors of [39] suggested, new examples of Sidon spaces whose ration between its dimension and the degree extension is small may be useful to build up a stronger cryptosystem.

1.1 Main contributions

We deal with Sidon spaces which belong to Family I, that is, k-dimensional Sidon spaces of $\mathbb {F}_{q^{n}}$ contained in $\mathbb {F}_{q^k}+\gamma \mathbb {F}_{q^k}$, for some $k \mid n$ and $\gamma \in \mathbb {F}_{q^{n}}\setminus \mathbb {F}_{q^k}$. We will first give a characterization result of those subspaces which turn out to be Sidon with the aid of some auxiliary subspaces, which also yield us to obtain a description via linearized polynomials for those subspaces having dimension k. As a matter of fact, we will show how to construct examples from scattered polynomials. We will also show examples of linearized polynomials defining Sidon spaces, which are not scattered. This shows that the family of Sidon spaces in Family I is very large. Moreover, we will also investigate the direct sum of Sidon spaces, already exploited in [24], in order to construct Sidon spaces in higher degree field extensions with less assumptions. As a by-product, we obtain new examples of Sidon spaces. Finally, we will deal with the notion of equivalence of Sidon spaces, following the approach used in [54]. Indeed, since many more examples of Sidon spaces are emerging, it is important to give a definition of equivalence for them, which can naturally arise from the classification of isometries for cyclic subspace codes established by Gluesing-Luerssen and Lehmann in [18]. Thanks to the above equivalence, we can see how all the known examples of the Family I arise from a particular scattered polynomial. As a consequence, we will show that the family of examples we found is very large and contains many non-equivalent examples. We will then prove that some of our examples cannot be obtained as the kernel of a subspace trinomial and hence they cannot be obtained from the known examples of the third family.

2 Linearized polynomials and scattered polynomials

A linearized polynomial (or q-polynomial) is a polynomial of the shape

$$\begin{aligned} f(x)=\sum _{i=0}^{t}f_i x^{q^i}, \quad f_i \in {{\mathbb {F}}}_{q^k}. \end{aligned}$$

If f is nonzero, the q-degree of f will be the maximum i such that $f_i \ne 0$.

The set of linearized polynomials forms an ${{\mathbb {F}}}_q$-algebra with the usual addition between polynomials and the composition, defined by

$$\begin{aligned} (f_ix^{q^i})\circ (g_jx^{q^j})=f_ig_j^{q^i}x^{q^{i+j}}, \end{aligned}$$

on q-monomials, and then extended by associativity and distributivity and the multiplication by elements of ${{\mathbb {F}}}_q$. We denote this ${{\mathbb {F}}}_q$-algebra by ${{\mathcal {L}}}_{k,q}$. The elements of the quotient algebra $\tilde{\mathcal {L}}_{k,q}$ obtained from $\mathcal {L}_{k,q}$ modulo the two-sided ideal $(x^{q^k}-x)$ are represented by linearized polynomials of q-degree less than k, i.e.,

$$\begin{aligned} \tilde{\mathcal {L}}_{k,q} = \left\{ \sum _{i=0}^{k-1}f_i x^{q^i}, \quad f_i \in {{\mathbb {F}}}_{q^k} \right\} . \end{aligned}$$

It is well known that the following isomorphism as ${{\mathbb {F}}}_q$-algebra holds

$$\begin{aligned} (\tilde{{{\mathcal {L}}}}_{k,q},+,\circ )\cong (\textrm{End}_{{{\mathbb {F}}}_{q}}({{\mathbb {F}}}_{q^k}),+,\circ ), \end{aligned}$$

where the linearized polynomial f is identified with the endomorphism of ${{\mathbb {F}}}_{q^k}$ defined by its evaluation map.

Thanks to the above isomorphism, we can identify a linearized polynomial with the endomorphism it defines and so we will speak of kernel and rank of a q-polynomial f meaning by this the kernel and rank of the corresponding endomorphism, denoted by $\ker (f)$ and $\textrm{rk}(f)$, respectively.

Consider

$$\begin{aligned}f(x)=a_0x+a_1x^{q^s}+\cdots +a_{k-1}x^{q^{s(k-1)}}+a_kx^{q^{sk}}\in \mathcal {L}_{n,q},\end{aligned}$$

with $k\le n-1$, $\gcd (s,k)=1$ and let $a_k\ne 0$. Then,

$$\begin{aligned} \dim _{{{\mathbb {F}}}_q}(\ker (f))\le k. \end{aligned}$$

(1)

When the equality holds, we say that f is a subspace polynomial. It is clear that every k-dimensional ${{\mathbb {F}}}_{q}$-subspace of $\mathbb {F}_{q^n}$ is the kernel of a unique monic subspace polynomial of q-degree k.

We refer to [25] for more details on this class of polynomials.

In connection with optimal codes in the rank metric, Sheekey in [42] introduced the notion of scattered polynomials.

Definition 2.1

A polynomial $f \in {\mathcal {L}}_{k,q}$ is said to be scattered if for any $a,b \in \mathbb {F}_{q^k}^*$ such that $f(a)/a=f(b)/b$ implies that a and b are ${{\mathbb {F}}}_{q}$-proportional.

If we denote by $U_f=\{(x,f(x)):x \in {{\mathbb {F}}}_{q^k}\}\subseteq \mathbb {F}_{q^k}^2$ the k-dimensional ${{\mathbb {F}}}_{q}$-subspace defined by the graph of f, the polynomial f turns out to be scattered if for any $a \in \mathbb {F}_{q^k}^*$ we have

$$\begin{aligned} \dim _{{{\mathbb {F}}}_{q}}(S_{a,f})= 1, \end{aligned}$$

(2)

where $S_{a,f}=\{ \rho \in {{\mathbb {F}}}_{q^k}:\rho (a,f(a)) \in U_f \}$, i.e., $S_{a,f}={{\mathbb {F}}}_{q}$.

3 Characterization of Sidon spaces with n composite

Let $\gamma \in \mathbb {F}_{q^n}$ be a root of an irreducible polynomial of degree $\frac{n}{k}\geqslant 2$ over $\mathbb {F}_{q^k}$, i.e., $\mathbb {F}_{q^n}=\mathbb {F}_{q^k}(\gamma )$ and $[{{\mathbb {F}}}_{q^n}:{{\mathbb {F}}}_{q^k}]=\frac{n}{k}$.

Let U be an ${{\mathbb {F}}}_{q}$-subspace of $\mathbb {F}_{q^k}^2$ of dimension m over ${{\mathbb {F}}}_{q}$, and let define

$$\begin{aligned}V_{U,\gamma }=\lbrace u+v\gamma : (u,v)\in U\rbrace \in \mathcal {G}_q(n,m).\end{aligned}$$

We are going first to characterize the Sidon spaces of the form $V_{U,\gamma }$ in terms of U and $\gamma$ and we will see that in the maximum dimension case we may use linearized polynomials and read the Sidon space property directly on it.

For any $(u,v)\in U$, define

$$\begin{aligned}S_{(u,v)}^\gamma =\lbrace \lambda \in \mathbb {F}_{q^n}| \lambda (u+v\gamma )\in V_{U,\gamma }\rbrace \subseteq \mathbb {F}_{q^n}.\end{aligned}$$

It is easy to see that $S_{(u,v)}^\gamma$ is an ${{\mathbb {F}}}_{q}$-subspace of $\mathbb {F}_{q^n}$ containing ${{\mathbb {F}}}_{q}$.

Definition 3.1

Let U be an ${{\mathbb {F}}}_{q}$-subspace of $\mathbb {F}_{q^k}^2$ of dimension m over ${{\mathbb {F}}}_{q}$, and let $\gamma \in \mathbb {F}_{q^n}$ be a root of an irreducible polynomial of degree $\frac{n}{k}\geqslant 2$ over $\mathbb {F}_{q^k}$. We say that U has the Sidon space property with respect to $\gamma$ if for every ${{\mathbb {F}}}_{q}$-linearly independent $(u,v),(u',v')\in U$ we have that

$$\begin{aligned}S_{(u,v)}^\gamma \cap S_{(u',v')}^\gamma ={{\mathbb {F}}}_{q}.\end{aligned}$$

Our aim is to characterize the Sidon space property of a subspace of the form $V_{U,\gamma }$ via the property above defined on U. In the next examples, we see that this property is strongly related also to $\gamma$ and not only to U.

Example 3.2

Consider the extension $\mathbb {F}_{3^8}/\mathbb {F}_{3^4}$. Let $\mathbb {F}_{3^4}=\mathbb {F}_{3}(g_1)$ and $\mathbb {F}_{3^8}=\mathbb {F}_{3^4}(\gamma )$, where $g_1\in \mathbb {F}_{3^4}$ has $x^4-x^3-1$ as minimal polynomial over $\mathbb {F}_{3}$ and $\gamma \in \mathbb {F}_{3^8}$ has $x^2+g_1^{35}x+g_1$ as minimal polynomial over $\mathbb {F}_{3^4}$. Note that $\mathbb {F}_{3^8}=\mathbb {F}_{3^4}(\gamma )=\mathbb {F}_{3^4}(\gamma ^2)$.

Denoting by

$$\begin{aligned}U=\{(x,x^3) :x \in \mathbb {F}_{3^4}\},\end{aligned}$$

then $V_{U,\gamma }=\lbrace x+x^3\gamma :x\in \mathbb {F}_{3^4}\rbrace$ is a Sidon space in $\mathbb {F}_{3^8}$, whereas $V_{U,\gamma ^2}=\lbrace x+x^3\gamma ^2:x\in \mathbb {F}_{3^4}\rbrace$ is not a Sidon space.

We are going to give a characterization of Sidon spaces which are contained in the sum of two multiplicative cosets of $\mathbb {F}_{q^k}$ via the Sidon space property.

Theorem 3.3

Let $\gamma \in \mathbb {F}_{q^n}$ be a root of an irreducible polynomial of degree $\frac{n}{k}\geqslant 2$ over $\mathbb {F}_{q^k}$. Let U be an ${{\mathbb {F}}}_{q}$-subspace of $\mathbb {F}_{q^k}^2$ of dimension m over ${{\mathbb {F}}}_{q}$, and let $V=V_{U,\gamma }\in \mathcal {G}_q(n,m)$. Then, V is a Sidon space if and only if U has the Sidon space property with respect to $\gamma$.

Proof

Suppose that V is a Sidon space and, by contradiction, suppose that there exist $(u,v),(u',v')\in U \setminus \{(0,0)\}$ which are ${{\mathbb {F}}}_{q}$-linearly independent and there exists $w\in (S_{(u,v)}^\gamma \cap S_{(u',v')}^\gamma ){\setminus }{{\mathbb {F}}}_{q}$. Since $w\in S_{(u,v)}^\gamma \cap S_{(u',v')}^\gamma$, we have that

$$\begin{aligned}{} & {} w (u+v\gamma )=wu+wv\gamma \in V\\{} & {} w (u'+v'\gamma )=wu'+wv'\gamma \in V. \end{aligned}$$

Now, let $a=wu+wv\gamma ,b=u'+v'\gamma ,c=u+v\gamma ,d=w u'+w v'\gamma$. Then, a, b, c, d are nonzero elements of V and it results

$$\begin{aligned} ab=cd. \end{aligned}$$

(3)

Since $w \notin {{\mathbb {F}}}_{q}$ and (u, v) and $(u',v')$ are not ${{\mathbb {F}}}_{q}$-proportional, we get $\lbrace a{{\mathbb {F}}}_{q}, b{{\mathbb {F}}}_{q}\rbrace \ne \lbrace c{{\mathbb {F}}}_{q}, d{{\mathbb {F}}}_{q}\rbrace$, a contradiction. Conversely, suppose that for every $(u,v),(u',v')\in U$ such that $(u,v),(u',v')$ are not ${{\mathbb {F}}}_{q}$-proportional we have that $S_{(u,v)}^\gamma \cap S_{(u',v')}^\gamma ={{\mathbb {F}}}_{q}$ and by contradiction suppose that V is not a Sidon space, i.e., there exist $a=u+v\gamma , b=u'+v'\gamma , c=t+z\gamma , d=t'+z'\gamma \in V$ distinct nonzero elements of V such that

$$\begin{aligned}ab=cd\end{aligned}$$

and $\lbrace a{{\mathbb {F}}}_{q}, b{{\mathbb {F}}}_{q}\rbrace \ne \lbrace c{{\mathbb {F}}}_{q}, d{{\mathbb {F}}}_{q}\rbrace$. As a consequence, we have that

$$\begin{aligned} \frac{a}{c}=\frac{u+v\gamma }{t+z\gamma }=\frac{t'+z'\gamma }{u'+v'\gamma }=\frac{d}{b}=\lambda \in \mathbb {F}_{q^n}\setminus {{\mathbb {F}}}_{q}\end{aligned}$$

and

$$\begin{aligned} \frac{a}{d}=\frac{u+v\gamma }{t'+z'\gamma }=\frac{t+z\gamma }{u'+v'\gamma }=\frac{c}{b}=\mu \in \mathbb {F}_{q^n}\setminus {{\mathbb {F}}}_{q}. \end{aligned}$$

Since $u+v\gamma =\lambda (t+z\gamma )\in V$ and $t'+z'\gamma =\lambda (u'+v'\gamma )\in V$, then $\lambda \in (S_{(u',v')}^\gamma \cap S_{(t,z)}^\gamma ){\setminus }{{\mathbb {F}}}_{q}$. Note that this is possible, by hypothesis, if and only if $(u',v')$ and (t, z) are ${{\mathbb {F}}}_{q}$-proportional; that is, there exists $\eta \in {{\mathbb {F}}}_{q}^*$ such that $(u',v')=\eta (t,z)$. Then, we have $\frac{c}{b}=\frac{t+z\gamma }{u'+v'\gamma }=\mu =\frac{1}{\eta }\in {{\mathbb {F}}}_{q}$ and this is not possible. $\square$

If the degree $[{{\mathbb {F}}}_{q^k}(\gamma ):{{\mathbb {F}}}_{q^k}]$ is greater than two, then the Sidon space property of U does not depend on $\gamma$ and we can replace the subspaces $S_{(u,v)}^{\gamma }$ of $\mathbb {F}_{q^n}$ by $S_{(u,v)}$, which are subspaces of $\mathbb {F}_{q^k}$ and are independent from $\gamma$.

Theorem 3.4

Let U be an ${{\mathbb {F}}}_{q}$-subspace of $\mathbb {F}_{q^k}^2$, and let $\gamma$ be a root of an irreducible polynomial over $\mathbb {F}_{q^k}$ of degree greater than two. Then, U has the Sidon space property with respect to $\gamma$ if and only if

$$\begin{aligned} S_{(u,v)}\cap S_{(u',v')}={{\mathbb {F}}}_{q}, \end{aligned}$$

(4)

for any ${{\mathbb {F}}}_{q}$-linearly independent $(u,v),(u',v')\in U$, where

$$\begin{aligned}S_{(u,v)}=\{\lambda \in \mathbb {F}_{q^k} :\lambda (u,v) \in U\}.\end{aligned}$$

Proof

If U has the Sidon space property with respect to $\gamma$, then (4) trivially holds. Now, suppose that (4) holds for every ${{\mathbb {F}}}_{q}$-linearly independent $(u,v),(u',v')\in U$. Fix any $(u,v),(u',v')\in U$ in such a way that are ${{\mathbb {F}}}_{q}$-linearly independent and suppose by contradiction that there exists

$$\begin{aligned} \mu \in (S_{(u,v)}^\gamma \cap S_{(u',v')}^\gamma )\setminus \mathbb {F}_q, \end{aligned}$$

and $\mu \notin \mathbb {F}_{q^k}$ because of the assumption of $S_{(u,v)}\cap S_{(u',v')}$. Then,

$$\begin{aligned} \mu (u+\gamma v), \mu (u'+\gamma v') \in V_{U,\gamma }, \end{aligned}$$

i.e.,

$$\begin{aligned} \mu (u+\gamma v)=w+\gamma z\,\,\quad \text {and}\quad \,\, \mu (u'+\gamma v')=w'+\gamma z', \end{aligned}$$

(5)

for some $(w,z),(w',z') \in U$. Therefore,

$$\begin{aligned} (u+\gamma v)(w'+\gamma z')=(u'+\gamma v')(w+\gamma z). \end{aligned}$$

(6)

By $[{{\mathbb {F}}}_{q^k}(\gamma ):{{\mathbb {F}}}_{q^k}]>2$ and Eq. (6), it follows

$$\begin{aligned} \left\{ \begin{array}{lll} uw'-u'w=0,\\ vz'-v'z=0,\\ uz'+vw'=u'z+v'w. \end{array} \right. \end{aligned}$$

(7)

Note that since $\mu \notin \mathbb {F}_{q^k}$, the pairs $(u,w),(v,z),(u',w')$ and $(v',z')$ are different from (0, 0). We start by proving that w and $w'$ cannot be zero.

Case 1: $w=0$.

By System (7), it follows that $w'=0$ and $u\ne 0$.

Case 1.1: $v=0$.

In this case, also $v'=0$ and so we obtain

$$\begin{aligned} \frac{\mu }{\gamma }=\frac{z}{u}=\frac{z'}{u'} \in \mathbb {F}_{q^k}, \end{aligned}$$

hence $\frac{\mu }{\gamma } \in S_{(u,0)}\cap S_{(u',0)}$, implying that $\mu /\gamma =\alpha \in {{\mathbb {F}}}_{q}$. Also, $(u,v)=(u,0)$, $(u',v')=(u',0)$ $(0,z/\alpha )=(0,u)$ and $(0,z'/\alpha )=(0,u')$ are in U. Let $\bar{\alpha }=u'/u$, then $\bar{\alpha }\in S_{(u,0)}\cap S_{(0,u)}={{\mathbb {F}}}_{q}$; that is, (u, v) and $(u',v')$ are ${{\mathbb {F}}}_{q}$-proportional, a contradiction.

Case 1.2: $v \ne 0$.

Therefore, by the second and third equations of System (7) we get $\beta =u/u'=v/v'=z/z' \in S_{(u',v')}$. Since (0, z) and $(0,z')$ are both in U, then we also obtain that $\beta \in S_{(0,z')}$. Since $u'\ne 0$, then $\beta \in S_{(u',v')} \cap S_{(0,z')}={{\mathbb {F}}}_{q}$, which implies that (u, v) and $(u',v')$ are ${{\mathbb {F}}}_{q}$-proportional.

Case 2: $w'=0$.

Argue as for Case 1.

Case 3: $w,w' \ne 0$.

We have two subcases that need to be analyzed.

Case 3.1: $z,z'\ne 0$.

Let

$$\begin{aligned} \lambda =\frac{u}{w}=\frac{u'}{w'}\in {{\mathbb {F}}}_{q^k}\,\,\,\text {and}\,\,\,\rho =\frac{v}{z}=\frac{v'}{z'}\in {{\mathbb {F}}}_{q^k}. \end{aligned}$$

From the third equation of System (7), we have

$$\begin{aligned} wz'\lambda +zw'\rho =w'z\lambda +z' w\rho , \end{aligned}$$

that is

$$\begin{aligned} (\rho -\lambda ) (zw'-z'w)=0. \end{aligned}$$

If $zw'-z'w \ne 0$, we have $\rho =\lambda$ and $\lambda =\mu ^{-1}$, that is $\mu \in {{\mathbb {F}}}_{q^k}$. This immediately implies that

$$\begin{aligned}\mu \in S_{(u,v)}\cap S_{(u',v')}={{\mathbb {F}}}_{q}, \end{aligned}$$

a contradiction. Hence, $zw'=z'w$, and then, there exists $\xi \in {{\mathbb {F}}}_{q^k}$ such that $(w,z)=\xi (w',z')$; then, by (6) it follows that

$$\begin{aligned} u+\gamma v=\xi (u'+\gamma v'), \end{aligned}$$

and so $\xi \in S_{(u',v')}\cap S_{(w',z')}={{\mathbb {F}}}_{q}$ since $(u',v')$ and $(w',z')$ are not ${{\mathbb {F}}}_{q}$-proportional by (5). Hence, (u, v) and $(u',v')$ are ${{\mathbb {F}}}_{q}$-linearly dependent, again a contradiction.

Case 3.2: either $z=0$ or $z'=0$.

Without loss of generality, let $z=0$. From the second equation of System (7), we get $z'=0$. Note that $(w,0),(w',0) \in U$. From (5), it follows that

$$\begin{aligned} \mu (u+\gamma v)=w\,\,\,\text {and}\,\,\,\mu (u'+\gamma v')=w', \end{aligned}$$

from which we get that

$$\begin{aligned} (u',v')=\frac{w}{w'} (u,v). \end{aligned}$$

Moreover, $w/w' \in S_{(w',0)}$ and so

$$\begin{aligned} \frac{w}{w'}\in S_{(u,v)}\cap S_{(w',0)}={{\mathbb {F}}}_{q}, \end{aligned}$$

since $v\ne 0$. We have again a contradiction since (u, v) and $(u',v')$ turn out to be ${{\mathbb {F}}}_{q}$-proportional. $\square$

Remark 3.5

The above theorem cannot be extended to the case of extension fields of degree two, see, for instance, Example 3.2.

Remark 3.6

By definition, it follows that

$$\begin{aligned} \dim _{{{\mathbb {F}}}_{q}}(S_{(u,v)})=\dim _{{{\mathbb {F}}}_{q}}(U \cap \langle (u,v) \rangle _{\mathbb {F}_{q^k}}), \end{aligned}$$

where $\langle (u,v) \rangle _{\mathbb {F}_{q^k}}=\{a(u,v) :a \in \mathbb {F}_{q^k}\}$.

The above result allows us to give the following definition.

Definition 3.7

Let U be an ${{\mathbb {F}}}_{q}$-subspace of $\mathbb {F}_{q^k}^2$. We say that U has the Sidon space property if for every ${{\mathbb {F}}}_{q}$-linearly independent $(u,v),(u',v')\in U$ we have that

$$\begin{aligned}S_{(u,v)}\cap S_{(u',v')}={{\mathbb {F}}}_{q}.\end{aligned}$$

As a corollary of Theorem 3.4, we have the following.

Corollary 3.8

Let U be an ${{\mathbb {F}}}_{q}$-subspace of $\mathbb {F}_{q^k}^2$. If U has the Sidon space property, then it defines a Sidon space in every extension of ${{\mathbb {F}}}_{q^k}$ of degree greater than two; that is,

$$\begin{aligned} V_{U,\gamma }=\lbrace u+v\gamma : (u,v)\in U\rbrace \end{aligned}$$

is a Sidon space in ${{\mathbb {F}}}_{q^k}(\gamma )$ for every $\gamma$ such that $[{{\mathbb {F}}}_{q^k}(\gamma ):\mathbb {F}_{q^k}]>2$.

When $V_{U,\gamma }$ is a Sidon space, then also the related subspaces $S_{(u,v)}^\gamma$ and $S_{(u,v)}$ are Sidon spaces.

Proposition 3.9

Let $\gamma \in \mathbb {F}_{q^n}$ be a root of an irreducible polynomial of degree $\frac{n}{k}\geqslant 2$ over $\mathbb {F}_{q^k}$. Let U be an ${{\mathbb {F}}}_{q}$-subspace of $\mathbb {F}_{q^k}^2$ of dimension m over ${{\mathbb {F}}}_{q}$ and suppose that $V=V_{U,\gamma }\in \mathcal {G}_q(n,m)$ is a Sidon space. Then,

• $S_{(u,v)}^\gamma$ is a Sidon space in $\mathbb {F}_{q^n}$, for any nonzero vector $(u,v)\in U$;

• if $[{{\mathbb {F}}}_{q^k}(\gamma ):\mathbb {F}_{q^k}]>2$, then $S_{(u,v)}$ is a Sidon space in $\mathbb {F}_{q^k}$, for any nonzero vector $(u,v)\in U$.

Proof

Let’s fix a nonzero vector $(u,v)\in U$. Because of its definition, we have that

$$\begin{aligned} \alpha (u+v\gamma )\in V_{U,\gamma }, \end{aligned}$$

for any $\alpha \in S_{(u,v)}^\gamma$, that is,

$$\begin{aligned} (u+v\gamma )S_{(u,v)}^\gamma \subseteq V_{U,\gamma }. \end{aligned}$$

Since $V_{U,\gamma }$ is a Sidon space and every of its subspaces is a Sidon space as well, we have that $(u+v\gamma )S_{(u,v)}^\gamma$ is a Sidon space and hence $S_{(u,v)}^\gamma$ is a Sidon space. Similar arguments can be performed for $S_{(u,v)}$. $\square$

When the dimension of $V=V_{U,\gamma }$ is k, then it can be represented by means of a linearized polynomial.

Lemma 3.10

Let $\gamma \in \mathbb {F}_{q^n}$ be a root of an irreducible polynomial of degree $\frac{n}{k}\geqslant 2$ over $\mathbb {F}_{q^k}$. Let $U\subseteq \mathbb {F}_{q^k}^2$ be an ${{\mathbb {F}}}_{q}$-subspace such that $\dim _{{{\mathbb {F}}}_{q}}(U)=k$ and $U\cap \langle (0,1)\rangle _{{{\mathbb {F}}}_{q^k}}=\lbrace (0,0)\rbrace$. Let’s consider $V_{U,\gamma }=\lbrace a+b\gamma : (a,b)\in U\rbrace \in \mathcal {G}_q(n,k)$. Then, there exists $f\in \mathcal {L}_{k,q}$ such that

$$\begin{aligned}V_{U,\gamma }=\lbrace u+f(u)\gamma :u\in \mathbb {F}_{q^k}\rbrace .\end{aligned}$$

Proof

Since $\dim _{{{\mathbb {F}}}_{q}}(U)=k$ and $U\cap \langle (0,1)\rangle _{{{\mathbb {F}}}_{q^k}}=\lbrace (0,0)\rbrace$, for any $a \in {{\mathbb {F}}}_{q^k}$ there exists a unique $b \in {{\mathbb {F}}}_{q^k}$ such that $(a,b)\in U$. This implies that the map

$$\begin{aligned} \begin{aligned} f:\mathbb {F}_{q^k}&\rightarrow \mathbb {F}_{q^k}\\ a&\mapsto b: (a,b)\in U \end{aligned} \end{aligned}$$

is well defined and since $V_{U,\gamma }$ is an ${{\mathbb {F}}}_{q}$-subspace, it results that f is an ${{\mathbb {F}}}_{q}$-linear map, that is, $f \in \mathcal {L}_{k,q}$. $\square$

Notation 3.11

Let $f\in \mathcal {L}_{k,q}$ be a linearized polynomial over $\mathbb {F}_{q^k}$ and let $U_{f}=\lbrace (u,f(u)):u\in \mathbb {F}_{q^k}\rbrace \subseteq \mathbb {F}_{q^k}^2$ be the ${{\mathbb {F}}}_{q}$-subspace of $\mathbb {F}_{q^k}^2$ defined by f which has dimension k over ${{\mathbb {F}}}_{q}$. Then, let’s denote by

$$\begin{aligned}{} & {} V_{f,\gamma }=\lbrace u+f(u)\gamma :u\in \mathbb {F}_{q^k}\rbrace \in \mathcal {G}_q(n,k),\\{} & {} S_{(u,f(u))}^\gamma =S_{u,f}^\gamma \,\,\, \text {for every} \,\,\,u\in \mathbb {F}_{q^k}, \end{aligned}$$

and

$$\begin{aligned} S_{(u,f(u))} =S_{u,f}\,\,\, \text {for every} \,\,\,u\in \mathbb {F}_{q^k}. \end{aligned}$$

Therefore, we give the following definition on a linearized polynomial f, which will correspond to say that $V_{f,\gamma }$ is a Sidon space.

Definition 3.12

Let $f\in \mathcal {L}_{k,q}$ be a linearized polynomial over $\mathbb {F}_{q^k}$. We say that f is a Sidon space polynomial with respect to $\gamma$ if for every $x_0,x_1\in \mathbb {F}_{q^k}$ such that $x_0,x_1$ are not ${{\mathbb {F}}}_{q}$-proportional we have that $S_{x_0,f}^\gamma \cap S_{x_1,f}^\gamma ={{\mathbb {F}}}_{q}$. Moreover, we say that f is a Sidon space polynomial if for every $x_0,x_1\in \mathbb {F}_{q^k}$ such that $x_0,x_1$ are not ${{\mathbb {F}}}_{q}$-proportional we have that $S_{x_0,f}\cap S_{x_1,f}={{\mathbb {F}}}_{q}$.

Theorem 3.3 and Corollary 3.8 allow us to characterize Sidon spaces of form $V_{f,\gamma }$.

Corollary 3.13

Let $f\in \mathcal {L}_{k,q}$ be a linearized polynomial over $\mathbb {F}_{q^k}$. Then,

• 1. $V_{f,\gamma }$ is a Sidon space if and only if f is a Sidon space polynomial with respect to $\gamma$, where $\gamma$ is a root of an irreducible polynomial of degree at least two over $\mathbb {F}_{q^k}$;

• 2. Also, $V_{f,\gamma }$ is a Sidon space for every $\gamma$ such that $[{{\mathbb {F}}}_{q^k}(\gamma ):{{\mathbb {F}}}_{q^k}]>2$ if and only f is a Sidon space polynomial.

Remark 3.14

It is interesting to observe that because of Proposition 3.9, f is a Sidon space polynomial and then $\ker (f)$ is a Sidon space of $\mathbb {F}_{q^k}$.

Note that if $f\in \mathcal {L}_{k,q}$ has the property that $\dim _{{{\mathbb {F}}}_{q}}(S_{x,f}^\gamma )\leqslant 2$ for every $x\in \mathbb {F}_{q^k}$, then the Sidon space property is easier to check.

Corollary 3.15

Let $\gamma \in \mathbb {F}_{q^n}$ be a root of an irreducible polynomial of degree $\frac{n}{k}\geqslant 2$ over $\mathbb {F}_{q^k}$, and let $f\in \mathcal {L}_{k,q}$ be such that $\dim _{{{\mathbb {F}}}_{q}} (S_{x,f}^\gamma )\leqslant 2$ for every $x\in \mathbb {F}_{q^k}$. Then, $V_{f,\gamma }$ is a Sidon space if and only if $S_{x_0,f}^\gamma \ne S_{x_1,f}^\gamma$ for every $x_0,x_1\in \mathbb {F}_{q^k}$ such that $x_0$ and $x_1$ are not ${{\mathbb {F}}}_{q}$-proportional and $\dim _{{{\mathbb {F}}}_{q}}(S_{x_0,f}^\gamma )=\dim _{{{\mathbb {F}}}_{q}}(S_{x_1,f}^\gamma )=2$.

Proof

Suppose that $x_0$ and $x_1$ are not ${{\mathbb {F}}}_{q}$-proportional and $\dim _{{{\mathbb {F}}}_{q}}( S_{x_0,f}^\gamma )=\dim _{{{\mathbb {F}}}_{q}}( S_{x_1,f}^\gamma )=2$, then $1\leqslant \dim _{{{\mathbb {F}}}_{q}} (S_{x_0,f}^\gamma \cap S_{x_1,f}^\gamma )\leqslant 2$. If $\dim _{{{\mathbb {F}}}_{q}}(S_{x_0,f}^\gamma \cap S_{x_1,f}^\gamma )=2$, then $S_{x_0,f}^\gamma \cap S_{x_1,f}^\gamma =S_{x_0,f}^\gamma =S_{x_1,f}^\gamma$ and this is not possible because of the assumptions. Therefore, $\dim _{{{\mathbb {F}}}_{q}} (S_{x_0,f}^\gamma \cap S_{x_1,f}^\gamma )=1$ and by Corollary 3.13 we have that $V_{f,\gamma }$ is a Sidon space. $\square$

Remark 3.16

From the above corollary, we also have that when $n/k>2$, we can replace $S_{x,f}^\gamma$ by $S_{x,f}$ and we can remove the dependence on $\gamma$.

Remark 3.17

When using linearized polynomials, we can explicitly find $S_{x_0,f}$ as the kernel of a certain linearized polynomial related to f. Indeed, we have

$$\begin{aligned} S_{x_0,f}= & {} \{ \lambda \in \mathbb {F}_{q^k}:\lambda (x_0,f(x_0))=(y_0,f(y_0)),\,\,\text {for some } y_0 \in \mathbb {F}_{q^k}\} \\= & {} \{ \lambda \in \mathbb {F}_{q^k}:f(\lambda x_0)-\lambda f(x_0)=0 \}. \end{aligned}$$

Therefore,

$$\begin{aligned} \dim _{{{\mathbb {F}}}_{q}}(S_{x_0,f})=\dim _{{{\mathbb {F}}}_{q}}(\ker (f(\lambda x_0)-\lambda f(x_0))), \end{aligned}$$

where $f(\lambda x_0)-\lambda f(x_0)$ is seen as a linearized polynomial in $\lambda$. Moreover, by Remark 3.6 we can compute this dimension also as follows

$$\begin{aligned} \dim _{{{\mathbb {F}}}_{q}}(S_{x_0,f})=\dim _{{{\mathbb {F}}}_{q}}(\ker (f(x)-(f(x_0)/x_0) x)). \end{aligned}$$

The polynomial version of our characterization allows us to provide new constructions of Sidon spaces that we will explore in the next section.

4 Constructions of Sidon spaces

In the previous section, we have seen a characterization of Sidon spaces in terms of polynomials, so in this section we will show some examples of Sidon space polynomials. We start by proving that if $f\in \mathcal {L}_{k,q}$ is scattered, then the Property 2 of Corollary 3.13 holds and f is a Sidon space polynomial.

Theorem 4.1

Let $f\in \mathcal {L}_{k,q}$ be a scattered polynomial. Then, f is a Sidon space polynomial.

Proof

By (2), we know that for any $x_0 \in \mathbb {F}_{q^k}^*$

$$\begin{aligned}\dim _{{{\mathbb {F}}}_{q}}(S_{x_0,f})=1.\end{aligned}$$

Hence, f trivially satisfies the Sidon space property. $\square$

As a consequence of Corollary 3.13, we have examples of Sidon spaces.

Corollary 4.2

Let $\gamma \in \mathbb {F}_{q^n}$ be a root of an irreducible polynomial of degree $\frac{n}{k}> 2$ over $\mathbb {F}_{q^k}$ and $f\in \mathcal {L}_{k,q}$ scattered, then

$$\begin{aligned} V_{f,\gamma }=\lbrace u+f(u)\gamma :u\in \mathbb {F}_{q^k}\rbrace \end{aligned}$$

is a Sidon space in $\mathbb {F}_{q^n}$.

Remark 4.3

In the case of $n=2k$, we cannot get the same conclusion. This seems not a limit of this argument; indeed, $f(x)=x^q$ is a scattered polynomial but Example 3.2 provides examples of 2-dimensional extension of $\mathbb {F}_{q^k}$ in which this polynomial does not define a Sidon space.

We now show a characterization for monomials defining a Sidon space polynomial when $n=2k$, which clearly involves also $\gamma$. To this aim, we recall that if U and V are two ${{\mathbb {F}}}_{q}$-subspaces of $\mathbb {F}_{q^n}$, then

$$\begin{aligned} UV=\langle uv :u\in U, v \in V\rangle _{{{\mathbb {F}}}_{q}}\,\,\,\text {and}\,\,\, U^2=UU. \end{aligned}$$

Then, we recall also the following theorem by Bachoc et al. [1] which gives a lower bound on the dimension of $V^2$ of a Sidon space V.

Theorem 4.4

[1, Theorem 18] Let $V\in \mathcal {G}_q(n,k)$ be a Sidon space of dimension $k\geqslant 3$, then

$$\begin{aligned} \dim _{{{\mathbb {F}}}_{q}}(V^2)\geqslant 2k. \end{aligned}$$

Thus, we are able to prove the following characterization result. Note that the norm over ${{\mathbb {F}}}_{q}$ of an element $a \in \mathbb {F}_{q^k}$ is $\textrm{N}_{q^k/q}(a)=a^{\frac{q^k-1}{q-1}}$.

Theorem 4.5

Let $\gamma \in \mathbb {F}_{q^{2k}}\setminus \mathbb {F}_{q^k}$ and let $s\in \mathbb {N}$ such that $\gcd (s,k)=1$. Then, $V_{x^{q^s},\gamma }$ is a Sidon space if and only if $\textrm{N}_{q^{2k}/q}(\gamma )\ne 1$.

Proof

Denote by $x^2+bx+c\in \mathbb {F}_{q^k}[x]$ the minimal polynomial of $\gamma$ over $\mathbb {F}_{q^k}$. Note that $c=\gamma ^{q^k+1}=\gamma ^{\frac{q^{2k}-1}{q^k-1}}$ and so

$$\begin{aligned} \textrm{N}_{q^{k}/q}(c)=\textrm{N}_{q^{2k}/q}(\gamma ). \end{aligned}$$

Therefore,

$$\begin{aligned} \textrm{N}_{q^{k}/q}(c)=1 \,\, if\,\, and \,\, only\,\, if\,\, \textrm{N}_{q^{2k}/q}(\gamma )=1. \end{aligned}$$

If $\textrm{N}_{q^{2k}/q}(\gamma )\ne 1$, then $\textrm{N}_{q^{k}/q}(c)\ne 1$ and by [40, Theorem 16], it follows that $V_{x^{q^s},\gamma }$ is a Sidon space.

Conversely, suppose that $\textrm{N}_{q^{2k}/q}(\gamma )=\textrm{N}_{q^{k}/q}(c)=1$. Then,

$$\begin{aligned} \begin{aligned} V_{x^{q^s},\gamma }^2&=\langle (u+u^{q^s}\gamma )(v+v^{q^s}\gamma ):u,v\in \mathbb {F}_{q^k}\rangle _{{{\mathbb {F}}}_{q}}\\&=\langle (uv-(uv)^{q^s}c)+(uv^{q^s}+u^{q^s}v-b(uv)^{q^s})\gamma :u,v\in \mathbb {F}_{q^k}\rangle _{{{\mathbb {F}}}_{q}}. \end{aligned} \end{aligned}$$

(8)

Therefore, by (8) it follows that

$$\begin{aligned} V_{x^{q^s},\gamma }^2\subseteq \textrm{Im}(T)\oplus \gamma \mathbb {F}_{q^k}\end{aligned}$$

where $T:y\in \mathbb {F}_{q^k}\mapsto y-y^{q^s}c\in \mathbb {F}_{q^k}$.

Since $\textrm{N}_{q^{k}/q}(c)=1$, then $\dim _{{{\mathbb {F}}}_{q}}(\textrm{Im}(T))=k-1$ and so $V_{x^{q^s},f}^2$ is contained in an ${{\mathbb {F}}}_{q}$-subspace of dimension $2k-1$ of $\mathbb {F}_{q^{2k}}$. Therefore, by Theorem 4.4 it follows that $V_{x^{q^s},\gamma }$ cannot be a Sidon space. $\square$

We resume in Table 1 the list of the known examples of scattered polynomials belonging to $\mathcal {L}_{k,q}$, and different entries of the tables correspond to $\Gamma \textrm{L}(2,q^k)$-inequivalent subspaces²$U_f=\{(x,f(x)) \,:\, x\in \mathbb {F}_{q^k}\}$.

Table 1 Known examples of scattered polynomials f

Remark 4.6

From Table 1, we get (up to equivalence, as we will see later) all the known examples of Sidon spaces contained in the sum of two multiplicative cosets of $\mathbb {F}_{q^k}$; for instance, [40, Construction 11] belongs to Family i) of Table 1.

Remark 4.7

In the monomial case, being scattered corresponds to being a Sidon space polynomial. Indeed, $f(x)=ax^{q^s} \in \mathcal {L}_{k,q}$ with $\gcd (s,k)=1$ is scattered and hence a Sidon space polynomial. Suppose that $\gcd (s,k)>1$ and write $s=s' t$, where $t=\gcd (k,s)>1$. Then, $V_{f,\gamma }$ is an ${{\mathbb {F}}}_{q^t}$-subspace of ${{\mathbb {F}}}_{q^k}(\gamma )$ and hence it cannot be a Sidon space because of Theorem 1.1.

We can give sufficient conditions in order to ensure that the binomial $f(x)=x^{q^i}+\delta x^{q^j}$ is a Sidon space polynomial.

Proposition 4.8

Let $f(x)=x^{q^i}+\delta x^{q^j}\in \mathcal {L}_{k,q}$, with $\gcd (k,j-i)=1$ and $j>i\ge 1$. Then, f is a Sidon space polynomial.

Proof

Suppose by contradiction that f is not a Sidon space polynomial, i.e., there exist $x_0,x_1\in \mathbb {F}_{q^k}$ such that $x_0,x_1$ are not ${{\mathbb {F}}}_{q}$-proportional and $\dim _{{{\mathbb {F}}}_{q}}(S_{x_0,f}\cap S_{x_1,f})\ge 2$. Let $\lambda \in (S_{x_0,f}\cap S_{x_1,f})\setminus {{\mathbb {F}}}_{q}$, then

$$\begin{aligned} f(\lambda x_0)=\lambda f(x_0)\,\,\,\text {and}\,\,\,f(\lambda x_1)=\lambda f(x_1), \end{aligned}$$

that is

$$\begin{aligned}(\lambda ^{q^i}-\lambda )x_0^{q^i}=\delta x_0^{q^j}(\lambda -\lambda ^{q^j})\,\,\,\text {and}\,\,\,(\lambda ^{q^i}-\lambda )x_1^{q^i}=\delta x_1^{q^j}(\lambda -\lambda ^{q^j}).\end{aligned}$$

If $\lambda -\lambda ^{q^i}=0$, then $\lambda ^{q^j}-\lambda =0$, i.e., $\lambda ^{q^{j-i}}=\lambda$ and so $\lambda \in \mathbb {F}_{q^k}\cap \mathbb {F}_{q^{j-i}}={{\mathbb {F}}}_{q}$ since $\gcd (k,j-i)=1$ and this is not possible. Therefore, $\lambda -\lambda ^{q^i}\ne 0$ and so we get

$$\begin{aligned}x_0^{q^i-q^j}=\delta \frac{\lambda -\lambda ^{q^j}}{\lambda ^{q^i}-\lambda } \,\,\,\text {and}\,\,\, x_1^{q^i-q^j}=\delta \frac{\lambda -\lambda ^{q^j}}{\lambda ^{q^i}-\lambda }.\end{aligned}$$

Thus, we have

$$\begin{aligned}x_0^{q^i-q^j}=x_1^{q^i-q^j},\end{aligned}$$

, i.e., $\frac{x_0}{x_1}\in \mathbb {F}_{q^k}\cap \mathbb {F}_{q^{j-i}}={{\mathbb {F}}}_{q}$, that is, a contradiction. $\square$

Proposition 4.8 gives a sufficient condition to determine if a linearized binomial defines a Sidon space polynomial; however, this is not also necessary since the examples iv) and vi) in Table 1 have the property that the difference of the exponents of the q-th powers that appear divides k and since they are scattered, by Theorem 4.1, they are also Sidon space polynomials. We will now show that in the case in which $(i-j,k)\ne 1$, a binomial as in Proposition 4.8 gives a Sidon space polynomial if and only if the binomial is also scattered.

Theorem 4.9

Let $f(x)=x^{q^i}+\delta x^{q^j}\in \mathcal {L}_{k,q}$, with $\gcd (k,j-i)>1$, $\delta \ne 0$ and $j>i\ge 1$. Let $j=i+st$, where $s,t \in \mathbb {N}$ are such that $\gcd (s,k)=1$, $t>1$ and $t \mid k$. Then,

• (i) f is a Sidon space polynomial if and only if f is scattered;

• (ii) if $\gcd (i,t)\ne 1$, then f is not a Sidon space polynomial;

• (iii) if $\textrm{N}_{q^k/q^t}(\delta )=(-1)^{k/t}$, then f is not a Sidon space polynomial.

Proof

i) We need to prove that if f is a Sidon space polynomial, then f is scattered. By contradiction, assume that f is not scattered, hence there exists $x_0 \in \mathbb {F}_{q^k}^*$ such that

$$\begin{aligned} \dim _{{{\mathbb {F}}}_{q}}(S_{x_0,f})>1. \end{aligned}$$

Consider $\rho \in {{\mathbb {F}}}_{q^t} \setminus {{\mathbb {F}}}_{q}$ and let $x_1=\rho x_0$, then it is easy to check that

$$\begin{aligned} S_{x_0,f}=\{ \lambda \in \mathbb {F}_{q^k}:-\lambda (x_0^{q^i}+\delta x_0^{i+st})+\lambda ^{q^i}x_0^{q^i}+\delta \lambda ^{q^{i+st}}x_0^{q^{i+st}}=0 \} \end{aligned}$$

and

$$\begin{aligned} S_{x_1,f}=\{ \lambda \in \mathbb {F}_{q^k}:-\lambda (x_1^{q^i}+\delta x_1^{i+st})+\lambda ^{q^i}x_1^{q^i}+\delta \lambda ^{q^{i+st}}x_1^{q^{i+st}}=0 \}\end{aligned}$$

coincide and hence $S_{x_0,f}\cap S_{x_1,f}=S_{x_0,f} \supset {{\mathbb {F}}}_{q}$, a contradiction to the fact that f is a Sidon space polynomial.

ii) If $\ell =\gcd (i,t)\ne 1$, then $\ell \mid j$ and so $V_{f,\gamma }$ is an ${{\mathbb {F}}}_{q^\ell }$-subspace of $\mathbb {F}_{q^n}$ and hence by Theorem 1.1 it is not a Sidon space.

iii) If $\textrm{N}_{q^k/q^t}(\delta )=(-1)^{k/t}$, then the equation $x^{q^i(1-q^{ts})}=-\delta$ admits a solution $x_0$ in ${{\mathbb {F}}}_{q^k}$ and the set of its solutions is $\{ x_0 \epsilon :\epsilon \in {{\mathbb {F}}}_{q^t}^* \}$. This implies that $\ker (f)=\{x \in {{\mathbb {F}}}_{q^k} :x^{q^i}=-\delta x^{q^{i+ts}}\}=x_0 {{\mathbb {F}}}_{q^t}$ and hence f is not a Sidon space polynomial by Remark 3.14. $\square$

Remark 4.10

In the previous result, we characterize the property of a binomial of being a Sidon space polynomial in terms of scattered polynomials, when $i-j$ is not coprime with k. Therefore, classification results on binomials that are scattered can be given now for Sidon space polynomials. A special family considered in the scattered polynomials framework is

$$\begin{aligned} f(x)=x^{q^i}+\delta x^{q^{i+t}}, \end{aligned}$$

where $\gcd (i,k)=1$ and $k=2t$, originally introduced in [12]. In [4, 38], there is an explicit condition on $\delta$ in order to characterize the scattered f when $t=3$. When $t=4$ in [44, Theorem 1.1], f is scattered if and only if $\delta ^{q^4+1}=-1$ when q is a large odd prime power. Moreover, in [37, Theorem 1.1] it has been proved that if k is larger than $4i+2$, then f is not scattered.

Thanks to Proposition 4.8, we are able to introduce new examples of Sidon spaces via some linearized polynomials which are not scattered.

Corollary 4.11

Let $f(x)=x^{q^s}+\delta x^{q^{2s}} \in \mathcal {L}_{k,q}$ with $\gcd (s,k)=1$, then f is a Sidon space polynomial.

Remark 4.12

When $s=1$ and $k\ge 5$, $f(x)=x^{q}+\delta x^{q^{2}}$ with $\delta \ne 0$ is not scattered; see [47, Remark 3.11] and also [31].

Remark 4.13

The polynomials considered in Proposition 4.8 can be very far from being scattered; that is, it may exist $x_0$ for which the dimension of $S_{x_0,f}$ is large. To this aim, recall from [30, Theorem 1.1] (see also [41, Theorem 1.3]) that if $k=(d-1)d+1$ and $q=p^h$ (for some prime p and $h \in \mathbb {N}$), then $\dim _{{{\mathbb {F}}}_{q}}(\ker (ax+bx^q-x^{q^d}))=d$ if and only if

$$\begin{aligned} \left\{ \begin{array}{lll} \textrm{N}_{q^k/q}(a)=(-1)^{d-1},\\ b=-a^{qf_1}\,\,\,\text {where}\,\,\, f_1=\sum _{i=0}^{d-1} q^{id},\\ d-1\,\, \text {is a power of }p. \end{array} \right. \end{aligned}$$

Therefore, choosing $d=p^\ell +1$, $q=p^h$, $k=(d-1)d+1$, $j=d$ and $i=1$ by the above result we have the existence of a linearized polynomial of the form $mx+ x^q+\delta x^{q^{d}}$ such that

$$\begin{aligned} \dim _{{{\mathbb {F}}}_{q}}(\ker (mx+ x^q+\delta x^{q^{d}}))=d, \end{aligned}$$

and by Remark 3.17 this means that there exists $x_0 \in \mathbb {F}_{q^k}^*$ such that

$$\begin{aligned} \dim _{{{\mathbb {F}}}_{q}}(S_{x_0,f})=d, \end{aligned}$$

where $f(x)=x^q+\delta x^{q^{d}}$ and satisfies the assumptions of Proposition 4.8.

Let now see another example which regards the Lunardon–Polverino polynomials which, under certain assumptions, are Sidon space polynomials also in the case in which they are not scattered.

Corollary 4.14

Let $f(x)=x^{q^s}+\delta x^{q^{s(k-1)}}\in \mathcal {L}_{k,q}$ with $\gcd (s,k)=1$. Then, f is a Sidon space polynomial if and only if either $\textrm{N}_{q^k/q}(\delta )\ne 1$ or k is odd.

Proof

By [47, Theorem 3.4], f is a scattered polynomial if and only if $\textrm{N}_{q^k/q}(\delta )\ne 1$. So, if $\textrm{N}_{q^k/q}(\delta )\ne 1$, then Theorem 4.1 implies that f is a Sidon space polynomial.

If k is odd, since $\gcd (k,s-s(k-1))=1$, Proposition 4.8 implies that f is a Sidon space polynomial. If k is even, then Theorem 4.9 implies that f is a Sidon space polynomial if and only if f is scattered, and hence, if and only if $\textrm{N}_{q^k/q}(\delta )\ne 1$. $\square$

5 Direct sum of Sidon spaces

In [24, Lemma 3.1], the authors proved that, under certain circumstances, the sum of two Sidon spaces is a Sidon space as well. So, [24, Lemma 3.1] can be stated as follows: If U and V are two Sidon spaces of $\mathbb {F}_{q^n}$ satisfying the following two properties:

• $(U+V)^2=U^2\oplus UV\oplus V^2$;

• $\dim _{{{\mathbb {F}}}_{q}}(U \cap \alpha V) \le 1$ for each $\alpha \in {{\mathbb {F}}}_{q^n}^*$,

then $U\oplus V$ is a Sidon space of $\mathbb {F}_{q^n}$. The main problem is that it is not easy to find two Sidon spaces satisfying both these assumptions. In this section, we will weaken the assumptions and we will prove a similar result, but the direct sum will be contained in an extension of $\mathbb {F}_{q^n}$.

Theorem 5.1

Let $V_1 \in \mathcal {G}_q(n,k_1)$, $V_2 \in \mathcal {G}_q(n,k_2)$ be two distinct Sidon spaces. Suppose that $V_1$ and $V_2$ satisfy

$$\begin{aligned} \dim _{{{\mathbb {F}}}_{q}}(V_1 \cap \alpha V_2) \le 1\,\,\, \text {for each}\,\,\, \alpha \in {{\mathbb {F}}}_{q^n}^*. \end{aligned}$$

(9)

Let m be a multiple of n such that $\frac{m}{n}>2$. Let $\delta$ be a root of an irreducible polynomial over ${{\mathbb {F}}}_{q^n}$ of degree $\frac{m}{n}$. Then,

$$\begin{aligned}V=\{v_1+\delta v_2:v_1 \in V_1,v_2 \in V_2\}\end{aligned}$$

is a Sidon space of $\mathcal {G}_q(m,k_1+k_2)$.

Proof

We start by noting that

$$\begin{aligned} V=V_{V_1\times V_2,\delta }. \end{aligned}$$

By Theorem 3.4, it is enough to prove that $V_1\times V_2 \subseteq \mathbb {F}_{q^n}\times \mathbb {F}_{q^n}$ has the Sidon space property, that is, for any $(u_1,u_2),(u_1',u_2') \in V_1\times V_2$ which are not ${{\mathbb {F}}}_{q}$-proportional we have

$$\begin{aligned} S_{(u_1,u_2)}\cap S_{(u_1',u_2')}={{\mathbb {F}}}_{q}. \end{aligned}$$

Let fix $(u_1,u_2),(u_1',u_2') \in V_1\times V_2$ which are not ${{\mathbb {F}}}_{q}$-proportional and suppose that there exists $\rho \in {{\mathbb {F}}}_{q^n}{\setminus } {{\mathbb {F}}}_{q}$ which is in $S_{(u_1,u_2)}\cap S_{(u_1',u_2')}$. This implies the existence of $(w_1,w_2),(w_1',w_2') \in V_1\times V_2$ such that

$$\begin{aligned} \rho (u_1,u_2)=(w_1,w_2)\,\,\,\text {and}\,\,\,\rho (u_1',u_2')=(w_1',w_2'), \end{aligned}$$

so

$$\begin{aligned} \left\{ \begin{array}{llll} \rho u_1=w_1,\\ \rho u_2=w_2,\\ \rho u_1'=w_1',\\ \rho u_2'=w_2'. \end{array} \right. \end{aligned}$$

(10)

and from last equation we get that either $u_2'=w_2'=0$ or $u_2',w_2'$ are both nonzero.

Case 1: Suppose that $u_2'$ and $w_2'$ are both nonzero. Then, from System (10) we get

$$\begin{aligned} \left\{ \begin{array}{lll} \frac{u_1}{u_2'}=\frac{w_1}{w_2'}=\lambda ,\\ \frac{u_2}{u_2'}=\frac{w_2}{w_2'}=\mu ,\\ \frac{u_1'}{u_2'}=\frac{w_1'}{w_2'}=\xi . \end{array} \right. \end{aligned}$$

(11)

which implies that

$$\begin{aligned} \begin{aligned}&u_1,w_1\in V_1\cap \lambda V_2,\\&u_2,w_2\in V_2\cap \mu V_2\\&u_1',w_1'\in V_1\cap \xi V_2. \end{aligned} \end{aligned}$$

(12)

If $(u_1',w_1')=(u_1,w_1)=(0,0)$, then $(u_2,w_2)\ne (0,0)$. Moreover, if $\mu \in {{\mathbb {F}}}_{q}$, then from System (11) it follows that $(u_1,u_2)=(0,u_2)=\mu (0,u_2')=\mu (u_1',u_2')$, i.e., $(u_1,u_2)$ and $(u_1',u_2')$ are ${{\mathbb {F}}}_{q}$-proportional, a contradiction. If $\mu \notin {{\mathbb {F}}}_{q}$, since $V_2$ is a Sidon space, from Equation (12) it follows that $u_2$ and $w_2$ are ${{\mathbb {F}}}_{q}$-proportional, i.e., $\rho \in {{\mathbb {F}}}_{q}$, a contradiction.

Therefore, $(u_1',w_1')\ne (0,0)$ or $(u_1,w_1)\ne (0,0)$; from System (11) this implies that $\xi \ne 0$ or $\lambda \ne 0$.

If $\xi \ne 0$, then by Condition (9) on $V_1$ and $V_2$ we have that $u_1'$ and $w_1'$ are ${{\mathbb {F}}}_{q}$-proportional, i.e., $\rho \in {{\mathbb {F}}}_{q}$, a contradiction.

If $\lambda \ne 0$, then by Condition (9) on $V_1$ and $V_2$ we have that $u_1$ and $w_1$ are ${{\mathbb {F}}}_{q}$-proportional, i.e., $\rho \in {{\mathbb {F}}}_{q}$, a contradiction.

Case 2: Suppose that $(u_2',w_2')=(0,0)$, then $u_1',w_1'\ne 0$, then from System (10) (as done for (11)) we get

$$\begin{aligned} \left\{ \begin{array}{ll} \frac{u_1}{u_1'}=\frac{w_1}{w_1'}=\sigma ,\\ \frac{u_2}{u_1'}=\frac{w_2}{w_1'}=l. \end{array} \right. \end{aligned}$$

(13)

from which we derive

$$\begin{aligned} \begin{aligned}&u_1,w_1\in V_1\cap \sigma V_1,\\&u_2,w_2\in V_2\cap l V_1. \end{aligned} \end{aligned}$$

(14)

If $l=0$, then $(u_2,w_2)=(0,0)$ and $u_1,w_1 \ne 0$, and from Eq. (14), if $\sigma \in {{\mathbb {F}}}_{q}$, then $(u_1,u_2)=(u_1,0)$ and $(u_1',u_2')=(u_1',0)$ are ${{\mathbb {F}}}_{q}$-proportional, a contradiction. If $\sigma \notin {{\mathbb {F}}}_{q}$, since $V_1$ is a Sidon space, we have that $u_1$ and $w_1$ are ${{\mathbb {F}}}_{q}$-proportional, implying that $\rho \in {{\mathbb {F}}}_{q}$.

If $l\ne 0$, we can argue as before and get a contradiction also in this case. $\square$

Remark 5.2

The above result can also be proved by using [24, Lemma 3.1] by taking as the two subspaces $V_1$ and $\delta V_2$; however, this version comes quite naturally from our previous discussions.

Remark 5.3

Let observe that if $V_1=V_2=V$, then $V+\delta V$ is not a Sidon space. Indeed, let $\alpha \in \mathbb {F}_{q^n}{\setminus } {{\mathbb {F}}}_{q}$ such that $T=V\cap \alpha V$ has dimension 1 (which always exists if we require that U is a strictly ${{\mathbb {F}}}_{q}$-subspace; that is, U is not $\mathbb {F}_{q^i}$-linear for any $i>1$). Consider now $T+\delta T$, which has dimension 2 and is contained in $(V+\delta V)\cap \alpha (V+\delta V)$, and hence, $V+\delta V$ cannot be a Sidon space.

As a consequence of Theorem 5.1, we can show some examples of Sidon spaces, which are contained in the sum of two multiplicative cosets of $\mathbb {F}_{q^n}$ (so belonging to Family 1 of the Introduction), but different from the one we constructed by using linearized polynomials in the previous section.

In order to use Theorem 5.1, we need to find Sidon spaces which satisfy Condition (9). For instance, we can consider the subspaces of [40, Construction 37]. They prove that the subspaces in [40, Construction 37] satisfy the conditions of Theorem 5.1 when $n=2k$, but it is easy to see that the proof still works when replacing $n/k>2$ and $\gamma _0$ with any element in $\mathbb {F}_{q^n}\setminus \mathbb {F}_{q^k}$.

Corollary 5.4

Let $q\ge 3$, n, k be two positive integers with $k\mid n$ and $n/k\ge 2$, let w be a primitive element of ${{\mathbb {F}}}_{q^k}$, and let $\gamma \in \mathbb {F}_{q^n}\setminus \mathbb {F}_{q^k}$. (If $n=2k$, then see [40, Construction 37] for the assumptions on $\gamma$.) Let m be a multiple of n such that $\frac{m}{n}>2$, and let $\delta$ be a root of an irreducible polynomial over ${{\mathbb {F}}}_{q^n}$ of degree $\frac{m}{n}$. Then,

$$\begin{aligned}V=\{(x+x^q\gamma )+\delta (y+y^q\gamma w): x,y \in \mathbb {F}_{q^k}\}\end{aligned}$$

is a Sidon space in ${{\mathbb {F}}}_{q^m}$.

In particular, if m is odd, the examples of the above corollary are contained in ${{\mathbb {F}}}_{q^n}+\delta {{\mathbb {F}}}_{q^n}$ and has dimension 2k, but ${{\mathbb {F}}}_{q^{2k}}$ is not a subfield of ${{\mathbb {F}}}_{q^m}$ and hence it is different from the construction exhibited in the previous section.

6 Equivalence of Sidon spaces

As already explained in Sect. 1, since Sidon spaces and cyclic subspace codes with a certain minimum distance are equivalent objects, it is quite natural to give a definition of equivalence for Sidon spaces arising from the equivalence of subspace codes; see [54]. The study of the equivalence for subspace codes was initiated by Trautmann in [45], and the case of cyclic subspace codes has been investigated in [18] by Gluesing-Luerssen and Lehmann. Therefore, motivated by [18, Theorem 2.4], we say that two cyclic subspace codes $\textrm{Orb}(U)$ and $\textrm{Orb}(V)$ are linearly equivalent if there exists $i \in \{0,\ldots ,n-1\}$ such that

$$\begin{aligned} \textrm{Orb}(U)=\textrm{Orb}(V^{q^i}), \end{aligned}$$

where $V^{q^i}=\{ v^{q^i} :v \in V \}$. This happens if and only if $U=\alpha V^{q^i}$, for some $\alpha \in \mathbb {F}_{q^n}^*$. We can replace the action of the Frobenius maps $x\in \mathbb {F}_{q^n}\mapsto x^{q^i}\in \mathbb {F}_{q^n}$ with any automorphism in $\textrm{Aut}(\mathbb {F}_{q^n})$, since this will still preserve the metric properties of the codes. Therefore, we say that two cyclic subspace codes $\textrm{Orb}(U)$ and $\textrm{Orb}(V)$ are semilinearly equivalent if there exists $\sigma \in \textrm{Aut}(\mathbb {F}_{q^n})$ such that

$$\begin{aligned} \textrm{Orb}(U)=\textrm{Orb}(V^{\sigma }), \end{aligned}$$

where $V^{\sigma }=\{ v^\sigma :v \in V \}$ and this happens if and only if $U=\alpha V^{\sigma }$, for some $\alpha \in \mathbb {F}_{q^n}^*$. Therefore, we can give the following definition of equivalence among two Sidon spaces.

Definition 6.1

Let U and V be two Sidon spaces in $\mathbb {F}_{q^n}$. Then, we say that U and V are semilinearly equivalent if the associated codes $\textrm{Orb}(U)$ and $\textrm{Orb}(V)$ are semilinearly equivalent; that is, there exist $\sigma \in \textrm{Aut}(\mathbb {F}_{q^n})$ and $\alpha \in \mathbb {F}_{q^n}^*$ such that $U=\alpha V^{\sigma }$. In this case, we will also say that they are equivalent under the action of $(\alpha , \sigma )$.

We can characterize the equivalence between two Sidon spaces which are contained in the sum of two multiplicative cosets of a subfield of $\mathbb {F}_{q^n}$ as follows.

Theorem 6.2

Let k and n be two positive integers such that $k \mid n$. Let U, W be two m-dimensional ${{\mathbb {F}}}_{q}$-subspaces of $\mathbb {F}_{q^k}^2$. Consider

$$\begin{aligned}{} & {} V_{U,\gamma }=\lbrace u+u'\gamma : (u,u')\in U\rbrace \\{} & {} V_{W,\xi }=\lbrace w+w'\xi : (w,w')\in W\rbrace \end{aligned}$$

where $\gamma ,\xi \in \mathbb {F}_{q^n}$ are such that $\lbrace 1,\gamma \rbrace$ and $\lbrace 1,\xi \rbrace$ are $\mathbb {F}_{q^k}$-linearly independent and suppose that they are not contained in any multiplicative coset of ${{\mathbb {F}}}_{q^k}$. Then, $V_{U,\gamma }$ and $V_{W,\xi }$ are semilinearly equivalent under the action of $(\lambda ,\sigma )\in \mathbb {F}_{q^n}^*\times \textrm{Aut}(\mathbb {F}_{q^n})$ if and only if there exists $A=\left( \begin{aligned} \begin{matrix} c &{} d \\ a &{} b \end{matrix} \end{aligned}\right) \in \textrm{GL}(2,\mathbb {F}_{q^k})$ such that $\xi =\frac{a+b\gamma ^\sigma }{c+d\gamma ^\sigma }$, $\lambda =\frac{1}{c+d\gamma ^\sigma }$ and $U^{\sigma }=\{w A :w \in W\}=W \cdot A$.

Proof

Suppose that $V_{U,\gamma }$ and $V_{W,\xi }$ are semilinearly equivalent, i.e., there exist $\lambda \in \mathbb {F}_{q^n}^*$ and $\sigma \in \textrm{Aut}(\mathbb {F}_{q^n})$ such that $\lambda V_{U,\gamma }^{\sigma }=V_{W,\xi }$. Note that

$$\begin{aligned}\lambda V_{U,\gamma }^{\sigma }=\lbrace \lambda u+\lambda u'\gamma ^{\sigma }: (u,u')\in U\rbrace \subseteq \lambda \mathbb {F}_{q^k}+\lambda \gamma ^\sigma \mathbb {F}_{q^k}\end{aligned}$$

and

$$\begin{aligned}V_{W,\xi }=\lbrace w+w'\xi :(w,w')\in W\rbrace \subseteq \mathbb {F}_{q^k}+\xi \mathbb {F}_{q^k}.\end{aligned}$$

Since $V_{W,\xi }=\lambda V_{U,\gamma }^{\sigma }$, then

$$\begin{aligned}V_{W,\xi }=\lambda V_{U,\gamma }^{\sigma }\subseteq (\lambda \mathbb {F}_{q^k}+\lambda \gamma ^{\sigma }\mathbb {F}_{q^k})\cap (\mathbb {F}_{q^k}+\xi \mathbb {F}_{q^k}).\end{aligned}$$

If $\dim _{{{\mathbb {F}}}_{q^k}}((\lambda \mathbb {F}_{q^k}+\lambda \gamma ^{\sigma }\mathbb {F}_{q^k})\cap (\mathbb {F}_{q^k}+\xi \mathbb {F}_{q^k}))=1$, we have $\lambda V_{U,\gamma }^{\sigma }=V_{W,\xi }\subseteq (\lambda \mathbb {F}_{q^k}+\lambda \gamma ^{\sigma }\mathbb {F}_{q^k})\cap (\mathbb {F}_{q^k}+\xi \mathbb {F}_{q^k})=\alpha {{\mathbb {F}}}_{q^k}$, for some $\alpha \in \mathbb {F}_{q^n}$, this is not possible because $V_{W,\xi }$ cannot be contained in a multiplicative coset of ${{\mathbb {F}}}_{q^k}$. Therefore, $\dim _{{{\mathbb {F}}}_{q^k}}(\lambda \mathbb {F}_{q^k}+\lambda \gamma ^{\sigma }\mathbb {F}_{q^k})\cap (\mathbb {F}_{q^k}+\xi \mathbb {F}_{q^k})=2$ and $\lambda \mathbb {F}_{q^k}+\lambda \gamma ^{\sigma }\mathbb {F}_{q^k}=\mathbb {F}_{q^k}+\xi \mathbb {F}_{q^k}$, i.e.,

$$\begin{aligned}\lambda \langle 1,\gamma ^{\sigma }\rangle _{\mathbb {F}_{q^k}}=\langle 1,\xi \rangle _{\mathbb {F}_{q^k}}.\end{aligned}$$

Then,

$$\begin{aligned}\xi =\lambda (a+b\gamma ^{\sigma })\,\,\, \text {with}\,\,\, a,b\in \mathbb {F}_{q^k}\end{aligned}$$

and

$$\begin{aligned}1=\lambda (c+d\gamma ^{\sigma })\,\,\, \text {with}\,\,\,c,d\in \mathbb {F}_{q^k}.\end{aligned}$$

Hence, $\lambda =\frac{1}{c+d\gamma ^{\sigma }}$, $\xi =\lambda (a+b\gamma ^{\sigma })=\frac{a+b\gamma ^{\sigma }}{c+d\gamma ^{\sigma }}$ and $A=\left( \begin{aligned} \begin{matrix} c &{} d \\ a &{} b \end{matrix} \end{aligned}\right)$ is invertible since $\xi \notin {{\mathbb {F}}}_{q^k}$. Moreover, since $\lambda V_{U,\gamma }^{\sigma }=V_{W,\xi }$ and $1, \gamma ^{\sigma }$ are $\mathbb {F}_{q^k}$-linearly independent, it follows that

$$\begin{aligned} \begin{aligned} \lambda ^{-1}V_{W,\xi }&=\lbrace \lambda ^{-1}w+\lambda ^{-1}w'\xi : (w,w')\in W\rbrace \\&=\left\{ (c+d\gamma ^{\sigma })w+(c+d\gamma ^{\sigma })w'\frac{a+b\gamma ^{\sigma }}{c+d\gamma ^{\sigma }}: (w,w')\in W\right\} \\&=\lbrace (c+d\gamma ^{\sigma })w+(a+b\gamma ^{\sigma })w':(w,w')\in W\rbrace ,\\ \end{aligned} \end{aligned}$$

(15)

since $\lambda V_{U,\gamma }^{\sigma }=V_{W,\xi }$, we get

$$\begin{aligned}U^{\sigma }=W \cdot A\end{aligned}$$

and hence we obtain the first part of the assertion.

Conversely, suppose that there exists $A=\left( \begin{aligned} \begin{matrix} c &{} d \\ a &{} b \end{matrix} \end{aligned}\right) \in \textrm{GL}(2,\mathbb {F}_{q^k})$ such that $\xi =\frac{a+b\gamma ^\sigma }{c+d\gamma ^\sigma }$ and $U^{\sigma }=W\cdot A$. Let $\lambda =\frac{1}{c+d\gamma ^\sigma }$. Then, it follows that

$$\begin{aligned} \begin{aligned} V_{W,\xi }&=\lbrace w+w'\xi : (w,w')\in W\rbrace \\&=\left\{ w+w'\frac{a+b\gamma ^{\sigma }}{c+d\gamma ^{\sigma }}: (w,w')\in W\right\} \\&=\lambda \lbrace (c+d\gamma ^{\sigma })w+(a+b\gamma ^{\sigma })w':(w,w')\in W\rbrace \\&=\lambda \lbrace (cw+aw')+(dw+bw')\gamma ^{\sigma }:(w,w')\in W\rbrace \\&=\lambda \lbrace u^{\sigma }+u'^{\sigma }\gamma ^{\sigma }: (u,u')\in U\rbrace \\&=\lambda V_{U,\gamma }^{\sigma }. \end{aligned} \end{aligned}$$

(16)

Hence, $V_{U,\gamma }$ and $V_{W,\xi }$ are semilinearly equivalent under the action of $(\lambda ,\sigma )$. $\square$

Remark 6.3

A first consequence of Theorem 6.2 is that all the known examples ( [40, Constructions 11 and 15], [50, Lemma 3.1], [24, Lemmas 4.1 and 4.2] and [51, Lemma 2.4]) of Sidon spaces contained in the sum of two multiplicative cosets of a field are equivalent to $V_{x^{q^s},\gamma }$. For instance, consider the example shown in [51, Lemma 2.4], that is, $V_{U,\gamma }\subseteq \mathbb {F}_{q^n}$ where

$$\begin{aligned}U=\{ (x,(x^{q^\ell }+ax)b) :x \in \mathbb {F}_{q^k}\},\end{aligned}$$

for some $a,b \in \mathbb {F}_{q^k}$ with $b\ne 0$, $\gamma \in \mathbb {F}_{q^n}{\setminus } \mathbb {F}_{q^k}$ and $\gcd (\ell ,k)=1$. Then, it is easy to see that

$$\begin{aligned} U \begin{pmatrix} 1 &{} -a \\ 0 &{} b^{-1} \end{pmatrix} =U_{x^{q^\ell }}. \end{aligned}$$

Therefore, by Theorem 6.2 the subspace $V_{U,\gamma }$ is equivalent to $V_{x^{q^{\ell }},\gamma '}$ where $\gamma ' =\frac{b^{-1}\gamma }{1-a\gamma }$.

Another important consequence is that every Sidon space of dimension k in an extension of $\mathbb {F}_{q^k}$ which is contained in the sum of two multiplicative cosets of $\mathbb {F}_{q^k}$ is equivalent to a Sidon space defined by a linearized polynomial in $\mathcal {L}_{k,q}$.

Corollary 6.4

Let k and n be two positive integers such that $k \mid n$. Let U be a k-dimensional ${{\mathbb {F}}}_{q}$-subspace of $\mathbb {F}_{q^k}^2$. Consider $V=V_{U,\gamma }$, where $\gamma \in \mathbb {F}_{q^n}{\setminus } {{\mathbb {F}}}_{q^k}$. Then, V is semilinearly equivalent to $V_{f,\xi }$ for some $f\in \mathcal {L}_{k,q}$ and $\xi \in \mathbb {F}_{q^n}{\setminus } {{\mathbb {F}}}_{q^k}$.

Proof

Note that since $\dim _{{{\mathbb {F}}}_{q}}(U)=k$, then [36, Lemma 4.6] there exists at least one $(a,b)\in {{\mathbb {F}}}_{q^k}\times {{\mathbb {F}}}_{q^k}{\setminus } \{(0,0)\}$ such that

$$\begin{aligned} U\cap \langle (a,b)\rangle _{{{\mathbb {F}}}_{q^k}}=\{(0,0)\}. \end{aligned}$$

Consider any $\phi \in \textrm{G L}(2,q^k)$ such that $\varphi ((a,b))=(0,1)$. In this way,

$$\begin{aligned} \phi (U)\cap \langle (0,1)\rangle _{{{\mathbb {F}}}_{q^k}}=\{(0,0)\}, \end{aligned}$$

and Lemma 3.10 implies the assertion. $\square$

Also, from Theorem 6.2 it follows that if $U,W \subseteq \mathbb {F}_{q^k}^2$ are ${{\mathbb {F}}}_{q}$-subspaces of dimension k satisfying the Sidon space property which are $\mathrm {\Gamma L}(2,q^k)$-inequivalent, then $V_{U,\gamma }$ and $V_{W,\gamma }$ are semilinearly inequivalent for any $\gamma \in \mathbb {F}_{q^n}\setminus \mathbb {F}_{q^k}$. Therefore, scattered polynomials in Table 1 can be used to construct semilinearly inequivalent examples of Sidon spaces (via Theorem 4.1), since different entries of the table give rise to $\mathrm {\Gamma L}(2,q^k)$-inequivalent subspaces. Moreover, every entry of Table 1 may contain $\mathrm {\Gamma L}(2,q^k)$-inequivalent subspaces. More precisely, denote by $\Lambda (f)$ the number of $\mathrm {\Gamma L}(2,q^k)$-inequivalent subspaces to $U_f$ in the corresponding family in Table 1. Family (i) in Table 1 is known as pseudoregulus type and

$$\begin{aligned}\Lambda (x^{q^s})=\varphi (k)/2,\end{aligned}$$

where $\varphi$ is the Euler’s totient function. Polynomials in Family (ii) in Table 1 are known as Lunardon–Polverino polynomials and

$$\begin{aligned}\Lambda (x^{q^s}+\delta x^{q^{s(k-1)}})\le \left\{ \begin{array}{ll} \varphi (k)\frac{q-2}{2} &{} \text {if}\,\, 2 \not \mid k,\\ &{} \\ \varphi (k)\frac{(q+1)(q-2)}{4} &{} \text {if}\,\, 2 \mid k, \end{array}\right. \end{aligned}$$

see [26, Section 2.2] and for further bounds see [43, Theorem 4.3]. For the Family iii), we have

$$\begin{aligned} \Lambda (x^{q^s}+x^{q^{s(\ell -1)}}+\delta ^{q^\ell +1}+\delta ^{1-q^{2\ell -1}}x^{q^{s(2\ell -1)}})= \left\lfloor \dfrac{\varphi (\ell )(q^\ell +1)}{4r\ell (q^2+1)}\right\rfloor \end{aligned}$$

if $\ell \equiv 2 \pmod {4}$, and

$$\begin{aligned} \Lambda (x^{q^s}+x^{q^{s(\ell -1)}}+\delta ^{q^\ell +1}+\delta ^{1-q^{2\ell -1}}x^{q^{s(2\ell -1)}})= \left\lfloor \dfrac{\varphi (\ell )(q^\ell +1)}{8r\ell }\right\rfloor \end{aligned}$$

if $\ell \not \equiv 2 \pmod {4}$, where $q=p^r$ and p is prime, see [26, Corollary 4.3] and [33, Theorem 4.12].

Also, by Theorem 6.2, from the same polynomial f we can get inequivalent Sidon spaces $V_{f,\gamma }$ and $V_{f,\xi }$ fixing $\gamma ,\xi \in \mathbb {F}_{q^n}\setminus \mathbb {F}_{q^k}$ such that $(1,\gamma ),(1,\xi ) \in \mathbb {F}_{q^n}\times \mathbb {F}_{q^n}$ are $\mathrm {\Gamma L}(2,q^k)$-inequivalent. Now, we give a lower bound on the number $\Gamma (\gamma )$ of elements $\xi \in \mathbb {F}_{q^n}\setminus \mathbb {F}_{q^k}$ giving inequivalent Sidon spaces of the form $V_{f,\xi }$, where f is fixed. The elements $\xi \in \mathbb {F}_{q^n}\setminus \mathbb {F}_{q^k}$ which gives equivalent Sidon spaces to $V_{f,\gamma }$ are of the form

$$\begin{aligned}\xi = \frac{a+b\gamma ^\sigma }{c+d\gamma ^\sigma },\end{aligned}$$

for some $(a,b,c,d)\in \mathbb {F}_{q^k}$ such that $ad-bc \ne 0$ and $\sigma \in \textrm{Aut}(\mathbb {F}_{q^n})$. Therefore, this number is greater than or equal to the size of L minus one, where

$$\begin{aligned}L=\{ \langle ( a+b\gamma ^\sigma ,c+d\gamma ^\sigma ) \rangle _{\mathbb {F}_{q^n}} :a,b,c,d \in \mathbb {F}_{q^k}, (a,b,c,d)\ne (0,0,0,0) \}.\end{aligned}$$

Then, L is an $\mathbb {F}_{q^k}$-linear set of the form of those studied in [20] and hence $|L|=q^{3k}+1$ when $n/k>3$; we refer to [23, 35] for surveys on linear sets. Hence,

$$\begin{aligned} \Gamma (\gamma ) \ge q^n-q^{3k}|\textrm{Aut}(\mathbb {F}_{q^n})|. \end{aligned}$$

For instance, the number of inequivalent Sidon spaces we can get from Family i) is

$$\begin{aligned} \Lambda (x^{q^s})\Gamma (\gamma )\ge \frac{\varphi (k)}{2} (q^n-q^{3k}|\textrm{Aut}(\mathbb {F}_{q^n})|). \end{aligned}$$

6.1 Sidon spaces as kernel of subspace polynomials

In this section, we will show that the examples of Sidon spaces we found in this paper are not equivalent to the Sidon spaces described as the kernel of a subspace trinomial, and hence, they are not the ones described in [34, 41].

In the next proposition, we show how to read the semilinear equivalence of ${{\mathbb {F}}}_{q}$-subspaces on the related subspace polynomial. We just recall that if $F(x)=\sum _{i=0}^k a_i x^{q^i}\in \mathcal {L}_{n,q}$ and $\sigma \in \textrm{Aut}(\mathbb {F}_{q^n})$, then $F^{\sigma }(x)=\sum _{i=0}^k a_i^{\sigma } x^{q^i}$.

Proposition 6.5

Let V be a k-dimensional ${{\mathbb {F}}}_{q}$-subspace of $\mathbb {F}_{q^n}$, and let $F \in \mathcal {L}_{n,q}$ be the associated subspace polynomial; that is, F is a monic polynomial of q-degree k and $\ker (F)=V$. Every ${{\mathbb {F}}}_{q}$-subspace semilinearly equivalent to V is the kernel of

$$\begin{aligned} \lambda ^{q^k}F^\sigma (\lambda ^{-1}x) \end{aligned}$$

for some $\lambda \in \mathbb {F}_{q^n}$ and $\sigma \in \textrm{Aut}(\mathbb {F}_{q^n})$.

Proof

Any ${{\mathbb {F}}}_{q}$-subspace semilinearly equivalent to V is of the form $\lambda V^{\sigma }$, for some $\lambda \in \mathbb {F}_{q^n}^*$ and $\sigma \in \textrm{Aut}(\mathbb {F}_{q^n})$. Then, it is easy to see that $\lambda ^{q^k}F^\sigma (\lambda ^{-1}x)$ is monic, and it has q-degree k and $\lambda ^{q^k}F^\sigma (\lambda ^{-1}u)=0$, for every $u \in \lambda V^{\sigma }$. $\square$

We now prove that the examples found in [40] are not semilinearly equivalent to the examples found in [34, 41].

Theorem 6.6

Let $\gamma \in \mathbb {F}_{q^n}$ be a root of an irreducible polynomial of degree $\frac{n}{k}\geqslant 2$ over $\mathbb {F}_{q^k}$. Let $\delta =1-(-1)^{k-2}\gamma ^{q^{k-1}+\dots +1}$,

$$\begin{aligned} a_0=-\delta ^q,\,\,\, a_1=-(\gamma ^{q^k}-\gamma ),\,\,\, a_k=\delta \end{aligned}$$

and

$$\begin{aligned}a_i=(-1)^i(\gamma ^{q^{i-1}+\dots +q})(\gamma ^{q^k}-\gamma ),\end{aligned}$$

for $i\in \lbrace 2,\dots ,k-1\rbrace$.

Let $F(x)=\displaystyle \sum _{i=0}^k a_ix^{q^i}\in \mathcal {L}_{n,q}$. Then, $\ker (F)=V_{x^q,\gamma }=\lbrace u+u^q\gamma :u\in \mathbb {F}_{q^k}\rbrace$.

Proof

Let’s start by observing that since $\dim _{{{\mathbb {F}}}_{q}}(\ker (F))\leqslant k$ and moreover $\dim _{{{\mathbb {F}}}_{q}}(V_{x^q,\gamma })=k$, it is sufficient to show that $V_{x^q,\gamma }\subseteq \ker (F)$. Consider $u+u^q\gamma \in V_{x^q,\gamma }$, then

$$\begin{aligned} \begin{aligned} F(u)=&-\delta ^qu-(\gamma ^{q^k}-\gamma )u^q+\gamma ^q(\gamma ^{q^k}-\gamma )u^{q^2}\\&-\gamma ^{q^2+q}(\gamma ^{q^k}-\gamma )u^{q^3}+\gamma ^{q^3+q^2+q}(\gamma ^{q^k}-\gamma )u^{q^4}\\&\vdots \\&+(-1)^{k-2}\gamma ^{q^{k-3}+\dots +q}(\gamma ^{q^k}-\gamma )u^{q^{k-2}}\\&+(-1)^{k-1}\gamma ^{q^{k-2}+\dots +q}(\gamma ^{q^k}-\gamma )u^{q^{k-1}}\\&+\delta u \end{aligned} \end{aligned}$$

(17)

and

$$\begin{aligned} \begin{aligned} F(u^q\gamma )=&-\delta ^q\gamma u^q-(\gamma ^{q^k}-\gamma )\gamma ^qu^{q^2}\\&+\gamma ^q(\gamma ^{q^k}-\gamma )\gamma ^{q^2}u^{q^3}-\gamma ^{q^2+q}(\gamma ^{q^k}-\gamma )\gamma ^{q^3}u^{q^4}\\&\vdots \\&+(-1)^{k-3}\gamma ^{q^{k-4}+\dots +q}(\gamma ^{q^k}-\gamma )\gamma ^{q^{k-3}}u^{q^{k-2}}\\&+(-1)^{k-2}\gamma ^{q^{k-3}+\dots +q}(\gamma ^{q^k}-\gamma )\gamma ^{q^{k-2}}u^{q^{k-1}}\\&+(-1)^{k-1}\gamma ^{q^{k-2}+\dots +q}(\gamma ^{q^k}-\gamma )\gamma ^{q^{k-1}}u\\&+\delta \gamma ^{q^k}u^q, \end{aligned} \end{aligned}$$

(18)

since $u \in \mathbb {F}_{q^k}$. Since $F(u+u^q\gamma )=F(u)+F(u^q\gamma )$, (17) and (18) imply

$$\begin{aligned} F(u+u^q\gamma )= & {} (-\delta ^q+(-1)^{k-1}\gamma ^{q^{k-1}+q^{k-2}+\dots +q}(\gamma ^{q^k}-\gamma )+\delta )u\\ {}{} & {} +(-(\gamma ^{q^k}-\gamma )-\delta ^q\gamma +\delta \gamma ^{q^k})u^q, \end{aligned}$$

and by using that $\delta =1-(-1)^{k-2}\gamma ^{q^{k-1}+\dots +1}$ we get $F(u+u^q\gamma )=0$. $\square$

Remark 6.7

Since $\gamma \notin {{\mathbb {F}}}_{q^k}$, then all the q-th powers of x up to q-degree k appear in F, and hence, F is far from being a trinomial when $n>6$.

Corollary 6.8

Let $\gamma \in \mathbb {F}_{q^n}$ be a root of an irreducible polynomial of degree $\frac{n}{k}\geqslant 2$ over $\mathbb {F}_{q^k}$. Then, $V_{x^q,\gamma }=\lbrace u+u^q\gamma :u\in \mathbb {F}_{q^k}\rbrace$ is not semilinearly equivalent to the examples in [34, 41].

Proof

By Proposition 6.5 and Theorem 6.6, we have that every subspace semilinearly equivalent to $V_{x^q,\gamma }$ cannot be the kernel of a subspace trinomial. $\square$

In the next result, we show that the class of examples found in Proposition 4.8, as for the monomial, cannot be the kernel of a linearized trinomial of q-degree k and having kernel the Sidon spaces of Proposition 4.8.

Proposition 6.9

Let $\gamma \in \mathbb {F}_{q^n}$ be a root of an irreducible polynomial of degree $\frac{n}{k}\geqslant 2$ over $\mathbb {F}_{q^k}$. Let $V_{g,\gamma }=\lbrace u+g(u)\gamma :u\in \mathbb {F}_{q^k}\rbrace$ be an ${{\mathbb {F}}}_{q}$-subspace of $\mathbb {F}_{q^n}$ where $g(x)= x^{q^i} + \delta x^{q^j} \in \mathcal {L}_{k,q}$ with $0<i<j<k$ and $\delta \ne 0$. If $F\in \mathcal {L}_{n,q}$ is a subspace polynomial with $\ker (F)=V_{g,\gamma }$, then F is not a trinomial.

Proof

Suppose that $F(x)=ax+bx^{q^\ell }+x^{q^k} \in \mathcal {L}_{n,q}$, with $0<\ell <k$, is a subspace polynomial with $\ker (F)=V_{g,\gamma }$, that is,

$$\begin{aligned} F(u+\gamma g(u))=0, \end{aligned}$$

(19)

for any $u \in \mathbb {F}_{q^k}$. Note also that b cannot be zero as $V_{g,\gamma }$ is not contained in any multiplicative coset of $\mathbb {F}_{q^k}$. (19) implies that

$$\begin{aligned} G(u)= & {} (a+1)u+(a\gamma +\gamma ^{q^k})u^{q^i}+(a\gamma \delta +\delta \gamma ^{q^k})u^{q^j}+b u^{q^\ell }\nonumber \\{} & {} \quad +b\gamma ^{q^\ell }u^{q^{i+\ell }}+b\gamma ^{q^\ell }\delta ^{q^\ell }u^{q^{\ell +j}}=0, \end{aligned}$$

(20)

for every $u \in \mathbb {F}_{q^k}$. Hence, the left-hand side G(u) of (20) can be seen as a polynomial in u and (20) implies that this polynomial is the zero polynomial when is reduced modulo $u^{q^k}-u$. To this aim, denote by

$$\begin{aligned}\mathcal {I}=\{0,i,j\}\,\,\,\text {and}\,\,\,\mathcal {J}=\{\ell ,i+\ell ,j+\ell \},\end{aligned}$$

which are seen both as subsets of $\mathbb {Z}/k\mathbb {Z}$. We divide the proof in cases:

Case 1: $|\mathcal {I}\cap \mathcal {J}|\le 2$.

In this case, we have that at least one among $\ell ,i+\ell$ and $j+\ell$ is not in $\mathcal {I}$ and hence the relative coefficient in G(u) is zero, that is, $b=0$, a contradiction.

Case 2: $\mathcal {I}=\mathcal {J}$.

Note that since $0<\ell <k$, then $i\ne \ell +i$ and $j\ne \ell +j$. Therefore, we only have two cases: either

$$\begin{aligned}i+\ell =0, j+\ell =i \,\,\,\text {and}\,\,\,\ell =j,\end{aligned}$$

or

$$\begin{aligned}i+\ell =j, j+\ell =0 \,\,\,\text {and}\,\,\, \ell =i.\end{aligned}$$

Case 2.1: $i+\ell =0$, $j+\ell =i$ and $\ell =j$.

From (20), the coefficients of G are

$$\begin{aligned} \left\{ \begin{array}{lll} a+1+b \gamma ^{q^\ell }=0,\\ a\gamma +\gamma ^{q^k}+b\gamma ^{q^\ell }\delta ^{q^\ell }=0,\\ a\gamma \delta +\delta \gamma ^{q^k}+b=0. \end{array} \right. \end{aligned}$$

(21)

Note that $a+1 \ne 0$; otherwise, the first equation of the above system would imply $b=0$. By multiplying the first equation of System (21) by $\delta ^{q^{\ell }}$ and the third one by $\gamma ^{q^\ell }\delta ^{q^\ell }$, we get

$$\begin{aligned} \left\{ \begin{array}{lll} (a+1)\delta ^{q^\ell }+b \gamma ^{q^\ell } \delta ^{q^\ell }=0,\\ a\gamma +\gamma ^{q^k}+b\gamma ^{q^\ell }\delta ^{q^\ell }=0,\\ a\gamma ^{1+q^\ell }\delta ^{1+q^\ell }+\delta ^{1+q^\ell }\gamma ^{q^k+q^\ell }+b\gamma ^{q^\ell } \delta ^{q^\ell }=0, \end{array} \right. \end{aligned}$$

which can be rewritten as follows

$$\begin{aligned} \left\{ \begin{array}{lll} b \gamma ^{q^\ell } \delta ^{q^\ell }=-(a+1)\delta ^{q^\ell },\\ a\gamma +\gamma ^{q^k}-(a+1)\delta ^{q^\ell }=0,\\ a\gamma ^{1+q^\ell }\delta ^{1+q^\ell }+\delta ^{1+q^\ell }\gamma ^{q^k+q^\ell }-(a+1)\delta ^{q^\ell }=0, \end{array} \right. \end{aligned}$$

and from the difference between the second equation multiplied by $\gamma ^{q^\ell }\delta ^{q^\ell +1}$ and the third equation we obtain

$$\begin{aligned} a \gamma ^{q^\ell }\delta ^{2q^\ell +1}+\gamma ^{q^\ell }\delta ^{2q^\ell +1}=a \delta ^{q^\ell }+\delta ^{q^\ell }, \end{aligned}$$

and hence

$$\begin{aligned} \gamma ^{q^\ell }\delta ^{2q^\ell +1}=\delta ^{q^\ell }, \end{aligned}$$

a contradiction as $\gamma \notin \mathbb {F}_{q^k}$.

Case 2.2: $i+\ell =j, j+\ell =0 \,\,\,\text {and}\,\,\, \ell =i$.

Similar arguments to those of Case 2.1 can be performed to get again the contradiction. $\square$

Arguing as in Proposition 6.9, we obtain the following.

Corollary 6.10

Let $\gamma \in \mathbb {F}_{q^n}$ be a root of an irreducible polynomial of degree $\frac{n}{k}\geqslant 2$ over $\mathbb {F}_{q^k}$. Let $V_{g,\gamma }=\lbrace u+g(u)\gamma :u\in \mathbb {F}_{q^k}\rbrace$ be an ${{\mathbb {F}}}_{q}$-subspace of $\mathbb {F}_{q^n}$ where $g(u)= u^{q^l} + B u^{q^m}$, $B\in \mathbb {F}_{q^k}$ and $\gcd (k,m-l)=1$ with $0<l<m<k$. Then, $V_{g,\gamma }$ is not semilinearly equivalent to the examples in [34, 41].

7 Conclusions and open problems

In this paper, we continued the study of Sidon spaces which are contained in the sum of two multiplicative cosets of a subfield $\mathbb {F}_{q^k}$ of the ambient space $\mathbb {F}_{q^n}$. We give characterization results, which allowed us to check the Sidon space property in the smaller field $\mathbb {F}_{q^k}$ instead of the entire field $\mathbb {F}_{q^n}$, under the assumption that $[\mathbb {F}_{q^n}:\mathbb {F}_{q^k}]>2$. As a consequence, we showed examples arising from scattered spaces/polynomials and others with a wide variety of algebraic behavior. Then, we conclude the paper by analyzing the natural equivalence that can be given on Sidon spaces via the connection with cyclic subspace codes.

We point out some questions/open problems that we think might be of interest:

• In Definition 3.12, we defined linearized polynomials satisfying the Sidon space property. An important task would be to investigate the exceptionality of this property that is to find families of linearized polynomials satisfying the Sidon space property for infinitely many extensions; see [3] for a nice overview. As done for scattered polynomials, see, e.g., [5, 6, 8, 17], a possible strategy is to translate the Sidon space property in terms of algebraic curves/function fields and then try to use heavy algebraic geometry machinery.

• In Corollary 5.4, we find an example of Sidon space with the aid of the direct sum. With the notation of Corollary 5.4, it is clear that if $2k \not \mid m$, this example cannot be the examples described in Sect. 4. When $2k \mid m$, it is not hard to show that it cannot be a Sidon space defined by a scattered polynomial, but we could not exclude that it can be the example defined by the binomial in Proposition 4.8. A possible way to show this is the following: Up to equivalence and with the notation of Corollary 5.4, we may write the example of Corollary 5.4 as $V_{g,\delta '}$ for some $\delta ' \in {{\mathbb {F}}}_{q^m}$ and g as in [32, Theorem 4.20] and then a direct check of the equivalence with the examples in Proposition 4.8 can be done (with non-trivial computations).

• Another interesting step toward the study of Sidon spaces is to investigate those subspaces that are not contained in the sum of two multiplicative cosets of a field. Some results can be readapted, but they need more discussions and we are currently working on it.