# Constructions and equivalence of Sidon spaces

Chiara Castello, Olga Polverino, Paolo Santonastaso and Ferdinando Zullo The
research was supported by the project “VALERE: VAnviteLli pEr la RicErca” of
the University of Campania “Luigi Vanvitelli” and was partially supported by
the Italian National Group for Algebraic and Geometric Structures and their
Applications (GNSAGA - INdAM).

###### Abstract

Sidon spaces have been introduced by Bachoc, Serra and Zémor in 2017 as the
$q$-analogue of Sidon sets. The interest on Sidon spaces has increased
quickly, especially after the work of Roth, Raviv and Tamo in 2018, 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.

Keywords: Sidon space; cyclic subspace code; linearized polynomial; $q$-analog

MSC2020: Primary 11T99; 11T06. Secondary 11T71; 94B05.

##  1 Introduction

Bachoc, Serra and Zémor in [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, 17]. Sidon spaces can be seen as the
$q$-analog of Sidon sets, a well-studied combinatorial object introduced by
Simon Szidon, see [13]. 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

$\\{a{\mathbb{F}}_{q},b{\mathbb{F}}_{q}\\}=\\{c{\mathbb{F}}_{q},d{\mathbb{F}}_{q}\\},$

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

Sidon spaces gained a lot of attention especially because of their connection
with cyclic subspace codes, pointed out by Roth, Raviv and Tamo in [37].
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 [43].

Let $n$ and $k$ be positive integers such that $k\leq n$. 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 $\mathrm{Orb}(U)=\\{\alpha
U\colon\alpha\in\mathbb{F}_{q^{n}}^{*}\\}$ the orbit of $U$, where $\alpha
U=\\{\alpha u\colon u\in U\\}$. We can equip $\mathcal{G}_{q}(n,k)$ with the
subspace metric, that is

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

where $U,V\in\mathcal{G}_{q}(n,k)$. 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 $d(\operatorname{\mathcal{C}})$ is
the minimum of distances between two distinct elements in
$\operatorname{\mathcal{C}}$. Codes in the subspace metric have been
investigated especially after the well-celebrated paper of Koetter and
Kschischang [19], 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
[14]. 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.

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

$\dim_{{\mathbb{F}}_{q}}(U\cap\alpha U)\leq 1.$

Note that $|\mathrm{Orb}(U)|=\frac{q^{n}-1}{q-1}$ is equivalent to require
that $U$ is strictly ${\mathbb{F}}_{q}$-linear; see [32, Theorem 1].

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

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:

  1. 1.

as subspace of the sum of two multiplicative cosets111In this paper, a
multiplicative coset of a subfield $\mathbb{F}_{q^{k}}$ of
$\mathbb{F}_{q^{n}}$ is $\alpha\mathbb{F}_{q^{k}}$, for some
$\alpha\in\mathbb{F}_{q^{n}}$. of a fixed subfield of $\mathbb{F}_{q^{n}}$;

  2. 2.

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

  3. 3.

as kernel of subspace polynomials.

In the first family we find the examples in [37, Constructions 11 and 15],
[47, Lemma 3.1], [22, Lemmas 4.1 and 4.2] and [48, Lemma 2.4], in the second
one we find examples in [49, Theorems 3.2 and 3.6], [47, Lemma 3.3, Theorems
3.5 and 3.13], [22, Theorems 4.5, 4.7, 4.9, 4.13-18 and 4.20] and [46, Lemma 5
and Theorem 6], and for the last family we can find examples in [9, Corollary
4 and Section IIIA] and [38, Section 7.3].

We will deal with Sidon spaces which belong to the first family, that is
$k$-dimensional Sidon spaces of $\mathbb{F}_{q^{k\ell}}$ contained in
$\mathbb{F}_{q^{k}}+\gamma\mathbb{F}_{q^{k}}$, for some
$\gamma\in\mathbb{F}_{q^{k\ell}}\setminus\mathbb{F}_{q^{k}}$. We will first
give a characterization result and we will also obtain a description with the
aid of linearized polynomials. 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.
Moreover, we will also investigate the direct sum of Sidon spaces, already
exploited in [22], in order to construct Sidon spaces in higher degree field
extensions with less assumptions. This will give new examples of Sidon spaces.
Finally, we will deal with the notion of equivalence of Sidon spaces,
following the approach used in [50]. 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 [16].
Thanks to the above equivalence, we can see how all the known examples of the
first family 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

$f(x)=\sum_{i=0}^{t}f_{i}x^{q^{i}},\quad f_{i}\in{\mathbb{F}}_{q^{k}}.$

If $f$ is nonzero, the $q$-degree of $f$ will be the maximum $i$ such that
$f_{i}\neq 0$.

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

$(f_{i}x^{q^{i}})\circ(g_{j}x^{q^{j}})=f_{i}g_{j}^{q^{i}}x^{q^{i+j}},$

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.

$\tilde{\mathcal{L}}_{k,q}=\left\\{\sum_{i=0}^{k-1}f_{i}x^{q^{i}},\quad
f_{i}\in{\mathbb{F}}_{q^{k}}\right\\}.$

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

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

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 $\mathrm{rk}(f)$, respectively.

Consider

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

with $k\leq n-1$, $\gcd(s,k)=1$ and let $a_{k}\neq 0$. Then

$\dim_{{\mathbb{F}}_{q}}(\ker(f))\leq k.$ (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 [23] for more details on this class of polynomials.

In connection with optimal codes in the rank metric, Sheekey in [39]
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 $U_{f}=\\{(x,f(x))\colon
x\in{\mathbb{F}}_{q^{k}}\\}\subseteq\mathbb{F}_{q^{k}}^{2}$ 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

$\dim_{{\mathbb{F}}_{q}}(S_{a,f})=1,$ (2)

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

##  3 Characterization of Sidon spaces

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

$V_{U,\gamma}=\\{u+v\gamma:(u,v)\in U\\}\in\mathcal{G}_{q}(n,m).$

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

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

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
$(u,v),(u^{\prime},v^{\prime})\in U$ we have that

$S_{(u,v)}^{\gamma}\cap
S_{(u^{\prime},v^{\prime})}^{\gamma}={\mathbb{F}}_{q}.$

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

$U=\\{(x,x^{3})\colon x\in\mathbb{F}_{3^{4}}\\},$

then $V_{U,\gamma}=\\{x+x^{3}\gamma\colon x\in\mathbb{F}_{3^{4}}\\}$ is a
Sidon space in $\mathbb{F}_{3^{8}}$, whereas
$V_{U,\gamma^{2}}=\\{x+x^{3}\gamma^{2}\colon x\in\mathbb{F}_{3^{4}}\\}$ 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 consider $V=V_{U,\gamma}\in\mathcal{G}_{q}(n,m)$.
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 $(u,v),(u^{\prime},v^{\prime})\in U\setminus\\{(0,0)\\}$ which are
${\mathbb{F}}_{q}$-linearly independent and there exists
$w\in(S_{(u,v)}^{\gamma}\cap
S_{(u^{\prime},v^{\prime})}^{\gamma})\setminus{\mathbb{F}}_{q}$. Since $w\in
S_{(u,v)}^{\gamma}\cap S_{(u^{\prime},v^{\prime})}^{\gamma}$, we have that

$w(u+v\gamma)=wu+wv\gamma\in V$

$w(u^{\prime}+v^{\prime}\gamma)=wu^{\prime}+wv^{\prime}\gamma\in V$.

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

$ab=cd.$ (3)

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

$ab=cd$

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

$\frac{a}{c}=\frac{u+v\gamma}{t+z\gamma}=\frac{t^{\prime}+z^{\prime}\gamma}{u^{\prime}+v^{\prime}\gamma}=\frac{d}{b}=\lambda\in\mathbb{F}_{q^{n}}\setminus{\mathbb{F}}_{q}$
(4)

and

$\frac{a}{d}=\frac{u+v\gamma}{t^{\prime}+z^{\prime}\gamma}=\frac{t+z\gamma}{u^{\prime}+v^{\prime}\gamma}=\frac{c}{b}=\mu\in\mathbb{F}_{q^{n}}\setminus{\mathbb{F}}_{q}.$
(5)

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

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

$S_{(u,v)}\cap S_{(u^{\prime},v^{\prime})}={\mathbb{F}}_{q},$ (6)

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

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

###### Proof.

If $U$ has the Sidon space property with respect to $\gamma$, then (6)
trivially holds. Now, suppose that (6) holds for every
${\mathbb{F}}_{q}$-linearly independent $(u,v),(u^{\prime},v^{\prime})\in U$.
Fix any $(u,v),(u^{\prime},v^{\prime})\in U$ in such a way that are
${\mathbb{F}}_{q}$-linearly independent and suppose by contradiction that
there exists

$\mu\in(S_{(u,v)}^{\gamma}\cap
S_{(u^{\prime},v^{\prime})}^{\gamma})\setminus\mathbb{F}_{q},$

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

$\mu(u+\gamma v),\mu(u^{\prime}+\gamma v^{\prime})\in V_{U,\gamma},$

i.e.

$\mu(u+\gamma v)=w+\gamma z\,\,\,\text{and}\,\,\,\mu(u^{\prime}+\gamma
v^{\prime})=w^{\prime}+\gamma z^{\prime},$ (7)

for some $(w,z),(w^{\prime},z^{\prime})\in U$. Therefore,

$(u+\gamma v)(w^{\prime}+\gamma z^{\prime})=(u^{\prime}+\gamma
v^{\prime})(w+\gamma z),$ (8)

and hence since $[{\mathbb{F}}_{q^{k}}(\gamma):{\mathbb{F}}_{q^{k}}]>2$ it
follows

$\left\\{\begin{array}[]{lll}uw^{\prime}-u^{\prime}w=0,\\\
vz^{\prime}-v^{\prime}z=0,\\\
uz^{\prime}+vw^{\prime}=u^{\prime}z+v^{\prime}w.\end{array}\right.$ (9)

Note that since $\mu\notin\mathbb{F}_{q^{k}}$ the pairs
$(u,w),(v,z),(u^{\prime},w^{\prime})$ and $(v^{\prime},z^{\prime})$ are
different from $(0,0)$. We start by proving that $w$ and $w^{\prime}$ cannot
be zero.
Case 1: $w=0$.
By System (9) it follows that $w^{\prime}=0$ and $u\neq 0$.
Case 1.1: $v=0$.
In this case, also $v^{\prime}=0$ and so we obtain

$\frac{\mu}{\gamma}=\frac{z}{u}=\frac{z^{\prime}}{u^{\prime}}\in\mathbb{F}_{q^{k}},$

hence $\frac{\mu}{\gamma}\in S_{(u,0)}\cap S_{(u^{\prime},0)}$, implying that
$\mu/\gamma=\alpha\in{\mathbb{F}}_{q}$. Also, $(u,v)=(u,0)$,
$(u^{\prime},v^{\prime})=(u^{\prime},0)$ $(0,z/\alpha)=(0,u)$ and
$(0,z^{\prime}/\alpha)=(0,u^{\prime})$ are in $U$. Let
$\bar{\alpha}=u^{\prime}/u$, then $\bar{\alpha}\in S_{(u,0)}\cap
S_{(0,u)}={\mathbb{F}}_{q}$, that is $(u,v)$ and $(u^{\prime},v^{\prime})$ are
${\mathbb{F}}_{q}$-proportional, a contradiction.
Case 1.2: $v\neq 0$.
Therefore, by the second and third equations of System (9) we get
$\beta=u/u^{\prime}=v/v^{\prime}=z/z^{\prime}\in S_{(u^{\prime},v^{\prime})}$.
Since $(0,z)$ and $(0,z^{\prime})$ are both in $U$, then we also obtain that
$\beta\in S_{(0,z^{\prime})}$. Since $u^{\prime}\neq 0$ then $\beta\in
S_{(u^{\prime},v^{\prime})}\cap S_{(0,z^{\prime})}={\mathbb{F}}_{q}$, which
implies that $(u,v)$ and $(u^{\prime},v^{\prime})$ are
${\mathbb{F}}_{q}$-proportional.
Case 2: $w^{\prime}=0$.
Argue as for Case 1.
Case 3: $w,w^{\prime}\neq 0$.
We have two subcases that need to be analyzed.
Case 3.1: $z,z^{\prime}\neq 0$.
Let

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

From the third equation of System (9), we have

$wz^{\prime}\lambda+zw^{\prime}\rho=w^{\prime}z\lambda+z^{\prime}w\rho,$

that is

$(\rho-\lambda)(zw^{\prime}-z^{\prime}w)=0.$

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

$\mu\in S_{(u,v)}\cap S_{(u^{\prime},v^{\prime})}={\mathbb{F}}_{q},$

a contradiction. Hence, $zw^{\prime}=z^{\prime}w$, then there exists
$\xi\in{\mathbb{F}}_{q^{k}}$ such that $(w,z)=\xi(w^{\prime},z^{\prime})$,
then by (8) it follows that

$u+\gamma v=\xi(u^{\prime}+\gamma v^{\prime}),$

and so $\xi\in S_{(u^{\prime},v^{\prime})}\cap
S_{(w^{\prime},z^{\prime})}={\mathbb{F}}_{q}$ since $(u^{\prime},v^{\prime})$
and $(w^{\prime},z^{\prime})$ are not ${\mathbb{F}}_{q}$-proportional by (7).
Hence, $(u,v)$ and $(u^{\prime},v^{\prime})$ are ${\mathbb{F}}_{q}$-linearly
dependent, again a contradiction.
Case 3.2: either $z=0$ or $z^{\prime}=0$.
Without loss of generality, let $z=0$. From the second equation of System (9),
we get $z^{\prime}=0$. Note that $(w,0),(w^{\prime},0)\in U$. From (7) it
follows that

$\mu(u+\gamma v)=w\,\,\,\text{and}\,\,\,\mu(u^{\prime}+\gamma
v^{\prime})=w^{\prime},$

from which we get that

$(u^{\prime},v^{\prime})=\frac{w}{w^{\prime}}(u,v).$

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

$\frac{w}{w^{\prime}}\in S_{(u,v)}\cap S_{(w^{\prime},0)}={\mathbb{F}}_{q},$

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

######  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

$\dim_{{\mathbb{F}}_{q}}(S_{(u,v)})=\dim_{{\mathbb{F}}_{q}}(U\cap\langle(u,v)\rangle_{\mathbb{F}_{q^{k}}}).$

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 $(u,v),(u^{\prime},v^{\prime})\in U$ we have that

$S_{(u,v)}\cap S_{(u^{\prime},v^{\prime})}={\mathbb{F}}_{q}.$

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

$V_{U,\gamma}=\\{u+v\gamma:(u,v)\in U\\}$

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

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

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

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

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 $S_{(u,v)}$. ∎

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
$U\cap\langle(0,1)\rangle_{{\mathbb{F}}_{q^{k}}}=\\{(0,0)\\}$. Let consider
$V_{U,\gamma}=\\{a+b\gamma:(a,b)\in U\\}\in\mathcal{G}_{q}(n,k)$. Then there
exists $f\in\mathcal{L}_{k,q}$ such that

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

###### Proof.

Since $\dim_{{\mathbb{F}}_{q}}(U)=k$ and
$U\cap\langle(0,1)\rangle_{{\mathbb{F}}_{q^{k}}}=\\{(0,0)\\}$, 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

$\displaystyle f\colon\mathbb{F}_{q^{k}}$ $\displaystyle\to\mathbb{F}_{q^{k}}$
$\displaystyle a$ $\displaystyle\mapsto b:(a,b)\in U$

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
$f\in\mathcal{L}_{k,q}$. ∎

######  Notation 3.11.

Let $f\in\mathcal{L}_{k,q}$ be a linearized polynomial over
$\mathbb{F}_{q^{k}}$ and let
$U_{f}=\\{(u,f(u)):u\in\mathbb{F}_{q^{k}}\\}\subseteq\mathbb{F}_{q^{k}}^{2}$
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 denote by

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

and

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

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. 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. 2.

Also, $V_{f,\gamma}$ is a Sidon space for every $\gamma$ such that
$[{\mathbb{F}}_{q^{k}}(\gamma)\colon{\mathbb{F}}_{q^{k}}]>2$ 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 then $\ker(f)$ is a Sidon space of $\mathbb{F}_{q^{k}}$.

Let observe 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
$S_{x_{0},f}^{\gamma}\neq S_{x_{1},f}^{\gamma}$ 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. ∎

######  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

$S_{x_{0},f}=\\{\lambda\in\mathbb{F}_{q^{k}}\colon\lambda(x_{0},f(x_{0}))=(y_{0},f(y_{0})),\,\,\text{for
some }y_{0}\in\mathbb{F}_{q^{k}}\\}$ $=\\{\lambda\in\mathbb{F}_{q^{k}}\colon
f(\lambda x_{0})-\lambda f(x_{0})=0\\}.$

Therefore,

$\dim_{{\mathbb{F}}_{q}}(S_{x_{0},f})=\dim_{{\mathbb{F}}_{q}}(\ker(f(\lambda
x_{0})-\lambda f(x_{0}))),$

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

$\dim_{{\mathbb{F}}_{q}}(S_{x_{0},f})=\dim_{{\mathbb{F}}_{q}}(\ker(f(x)-(f(x_{0})/x_{0})x)).$

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

We start this section 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 $x_{0}\in\mathbb{F}_{q^{k}}^{*}$

$\dim_{{\mathbb{F}}_{q}}(S_{x_{0},f})=1.$

Hence, $f$ trivially satisfies the Sidon space property. ∎

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

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

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

$UV=\langle uv\colon u\in U,v\in
V\rangle_{{\mathbb{F}}_{q}}\,\,\,\text{and}\,\,\,U^{2}=UU.$

Then we recall also the following theorem by Bachoc, Serra and Zémor in [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

$\dim_{{\mathbb{F}}_{q}}(V^{2})\geqslant 2k.$

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
$\mathrm{N}_{q^{k}/q}(a)=a^{\frac{q^{k}-1}{q-1}}$.

######  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 $\mathrm{N}_{q^{2k}/q}(\gamma)\neq 1$.

###### 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

$\mathrm{N}_{q^{k}/q}(c)=\mathrm{N}_{q^{2k}/q}(\gamma).$

Therefore

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

If $\mathrm{N}_{q^{2k}/q}(\gamma)\neq 1$, then $\mathrm{N}_{q^{k}/q}(c)\neq 1$
and by [37, Theorem 16], it follows that $V_{x^{q^{s}},\gamma}$ is a Sidon
space.
Conversely, suppose that
$\mathrm{N}_{q^{2k}/q}(\gamma)=\mathrm{N}_{q^{k}/q}(c)=1$. Then

$\displaystyle V_{x^{q^{s}},\gamma}^{2}$
$\displaystyle=\langle(u+u^{q^{s}}\gamma)(v+v^{q^{s}}\gamma)\colon
u,v\in\mathbb{F}_{q^{k}}\rangle_{{\mathbb{F}}_{q}}$ (10)
$\displaystyle=\langle(uv-(uv)^{q^{s}}c)+(uv^{q^{s}}+u^{q^{s}}v-b(uv)^{q^{s}})\gamma\colon
u,v\in\mathbb{F}_{q^{k}}\rangle_{{\mathbb{F}}_{q}}.$

Therefore, by (10) it follows that

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

where $T:y\in\mathbb{F}_{q^{k}}\mapsto y-y^{q^{s}}c\in\mathbb{F}_{q^{k}}$.
Since $\mathrm{N}_{q^{k}/q}(c)=1$ then
$\dim_{{\mathbb{F}}_{q}}(\mathrm{Im}(T))=k-1$ 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
$V_{x^{q^{s}},\gamma}^{2}$ cannot be a Sidon space. ∎

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

|  $k$ | $f(x)$ | Conditions | References
---|---|---|---|---
i) |  | $x^{q^{s}}$ | $\gcd(s,k)=1$ | [10]
ii) |  | $x^{q^{s}}+\delta x^{q^{s(k-1)}}$ | $\begin{array}[]{cc}\gcd(s,k)=1,\\\ \mathrm{N}_{q^{k}/q}(\delta)\neq 1\end{array}$ | [26, 20]
iii) | $2\ell$ | $\begin{array}[]{cc}x^{q^{s}}+x^{q^{s(\ell-1)}}+\\\ \delta^{q^{\ell}+1}x^{q^{s(\ell+1)}}+\delta^{1-q^{2\ell-1}}x^{q^{s(2\ell-1)}}\end{array}$ | $\begin{array}[]{cc}q\hskip 2.84544pt\text{odd},\\\ \mathrm{N}_{q^{2\ell}/q^{\ell}}(\delta)=-1,\\\ \gcd(s,\ell)=1\end{array}$ | $\begin{array}[]{cc}\text{\cite[cite]{[\@@bibref{}{BZZ,longobardi2021large}{}{}]}}\\\ \text{\cite[cite]{[\@@bibref{}{longobardizanellascatt,neri2022extending,ZZ}{}{}]}}\end{array}$
iv) | $6$ | $x^{q}+\delta x^{q^{4}}$ | $\begin{array}[]{cc}q>4,\\\ \text{certain choices of}\,\delta\end{array}$ | [11, 4, 36]
v) | $6$ | $x^{q}+x^{q^{3}}+\delta x^{q^{5}}$ | $\begin{array}[]{cccc}q\hskip 2.84544pt\text{odd},\\\ \delta^{2}+\delta=1\end{array}$ | [12, 27]
vi) | $8$ | $x^{q}+\delta x^{q^{5}}$ | $\begin{array}[]{cc}q\,\text{odd},\\\ \delta^{2}=-1\end{array}$ | [11]
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 [37, 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 $s=s^{\prime}t$, 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 $j>i\geq 1$. 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 $\dim_{{\mathbb{F}}_{q}}(S_{x_{0},f}\cap
S_{x_{1},f})\geq 2$. Let $\lambda\in(S_{x_{0},f}\cap
S_{x_{1},f})\setminus{\mathbb{F}}_{q}$, then

$f(\lambda x_{0})=\lambda f(x_{0})\,\,\,\text{and}\,\,\,f(\lambda
x_{1})=\lambda f(x_{1}),$

that is

$(\lambda^{q^{i}}-\lambda)x_{0}^{q^{i}}=\delta
x_{0}^{q^{j}}(\lambda-\lambda^{q^{j}})\,\,\,\text{and}\,\,\,(\lambda^{q^{i}}-\lambda)x_{1}^{q^{i}}=\delta
x_{1}^{q^{j}}(\lambda-\lambda^{q^{j}}).$

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
$\lambda-\lambda^{q^{i}}\neq 0$ and so we get

$x_{0}^{q^{i}-q^{j}}=\delta\frac{\lambda-\lambda^{q^{j}}}{\lambda^{q^{i}}-\lambda}\,\,\,\text{and}\,\,\,x_{1}^{q^{i}-q^{j}}=\delta\frac{\lambda-\lambda^{q^{j}}}{\lambda^{q^{i}}-\lambda}.$

Thus we have

$x_{0}^{q^{i}-q^{j}}=x_{1}^{q^{i}-q^{j}},$

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. ∎

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 $(i-j,k)\neq 1$, 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
$\gcd(k,j-i)>1$, $\delta\neq 0$ and $j>i\geq 1$. 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 $\gcd(i,t)\neq 1$ then $f$ is not a Sidon space polynomial;

  * iii)

if $\mathrm{N}_{q^{k}/q^{t}}(\delta)=(-1)^{k/t}$ 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

$\dim_{{\mathbb{F}}_{q}}(S_{x_{0},f})>1.$

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

$S_{x_{0},f}=\\{\lambda\in\mathbb{F}_{q^{k}}\colon-\lambda(x_{0}^{q^{i}}+\delta
x_{0}^{i+st})+\lambda^{q^{i}}x_{0}^{q^{i}}+\delta\lambda^{q^{i+st}}x_{0}^{q^{i+st}}=0\\}$

and

$S_{x_{1},f}=\\{\lambda\in\mathbb{F}_{q^{k}}\colon-\lambda(x_{1}^{q^{i}}+\delta
x_{1}^{i+st})+\lambda^{q^{i}}x_{1}^{q^{i}}+\delta\lambda^{q^{i+st}}x_{1}^{q^{i+st}}=0\\}$

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 $\ell=\gcd(i,t)\neq 1$ 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 $\mathrm{N}_{q^{k}/q^{t}}(\delta)=(-1)^{k/t}$, 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
$\\{x_{0}\epsilon\colon\epsilon\in{\mathbb{F}}_{q^{t}}^{*}\\}$. This implies
that $\ker(f)=\\{x\in{\mathbb{F}}_{q^{k}}\colon x^{q^{i}}=-\delta
x^{q^{i+ts}}\\}=x_{0}{\mathbb{F}}_{q^{t}}$ and hence $f$ is not a Sidon space
polynomial by Remark 3.14. ∎

######  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

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

where $\gcd(i,k)=1$ and $k=2t$, originally introduced in [11]. In [4] and
[36], there is an explicit condition on $\delta$ in order to characterize the
scattered $f$ when $t=3$. When $t=4$ in [41, 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 [35, Theorem 1.1] it has been proved that if $k$ is larger than
$4i+2$ then $f$ is not scattered.

Thanks to the 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 $k\geq 5$, $f(x)=x^{q}+\delta x^{q^{2}}$ with $\delta\neq 0$ is
not scattered; see [44, Remark 3.11] and also [29].

######  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 [28, Theorem 1.1] (see also
[38, 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

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

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

$\dim_{{\mathbb{F}}_{q}}(\ker(mx+x^{q}+\delta x^{q^{d}}))=d,$

and by Remark 3.17 this means that there exists
$x_{0}\in\mathbb{F}_{q^{k}}^{*}$ such that

$\dim_{{\mathbb{F}}_{q}}(S_{f,x_{0}})=d,$

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
$\mathrm{N}_{q^{k}/q}(\delta)\neq 1$ or $k$ is odd.

###### Proof.

By [44, Theorem 3.4], $f$ is a scattered polynomial if and only if
$\mathrm{N}_{q^{k}/q}(\delta)\neq 1$. So, if $\mathrm{N}_{q^{k}/q}(\delta)\neq
1$ 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 $\mathrm{N}_{q^{k}/q}(\delta)\neq 1$. ∎

##  5 Direct sum of Sidon spaces

In [22, Lemma 3.1], the authors proved that, under certain circumstances, the
sum of two Sidon spaces is a Sidon space as well. So, [22, 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:

  * •

$(U+V)^{2}=U^{2}\oplus UV\oplus V^{2}$;

  * •

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

then $U+V$ 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

$\dim_{{\mathbb{F}}_{q}}(V_{1}\cap\alpha V_{2})\leq 1\,\,\,\text{for
each}\,\,\,\alpha\in{\mathbb{F}}_{q^{n}}^{*}.$ (11)

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

$V=\\{v_{1}+\delta v_{2}:v_{1}\in V_{1},v_{2}\in V_{2}\\}$

is a Sidon space of $\mathcal{G}_{q}(m,k_{1}+k_{2})$.

###### Proof.

We start by noting that

$V=V_{V_{1}\times V_{2},\delta}.$

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 $(u_{1},u_{2}),(u_{1}^{\prime},u_{2}^{\prime})\in
V_{1}\times V_{2}$ which are not ${\mathbb{F}}_{q}$-proportional we have

$S_{(u_{1},u_{2})}\cap S_{(u_{1}^{\prime},u_{2}^{\prime})}={\mathbb{F}}_{q}.$

Let fix $(u_{1},u_{2}),(u_{1}^{\prime},u_{2}^{\prime})\in V_{1}\times V_{2}$
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
$S_{(u_{1},u_{2})}\cap S_{(u_{1}^{\prime},u_{2}^{\prime})}$. This implies the
existence of $(w_{1},w_{2}),(w_{1}^{\prime},w_{2}^{\prime})\in V_{1}\times
V_{2}$ such that

$\rho(u_{1},u_{2})=(w_{1},w_{2})\,\,\,\text{and}\,\,\,\rho(u_{1}^{\prime},u_{2}^{\prime})=(w_{1}^{\prime},w_{2}^{\prime}),$

so

$\left\\{\begin{array}[]{llll}\rho u_{1}=w_{1},\\\ \rho u_{2}=w_{2},\\\ \rho
u_{1}^{\prime}=w_{1}^{\prime},\\\ \rho
u_{2}^{\prime}=w_{2}^{\prime}.\end{array}\right.$ (12)

and from last equation we get that either $u_{2}^{\prime}=w_{2}=0$ or
$u_{2}^{\prime},w_{2}^{\prime}$ are both non zero.
Case 1: Suppose that $u_{2}^{\prime}$ and $w_{2}^{\prime}$ are both nonzero.
Then from System (12) we get

$\left\\{\begin{array}[]{lll}\frac{u_{1}}{u_{2}^{\prime}}=\frac{w_{1}}{w_{2}^{\prime}}=\lambda,\\\
\frac{u_{2}}{u_{2}^{\prime}}=\frac{w_{2}}{w_{2}^{\prime}}=\mu,\\\
\frac{u_{1}^{\prime}}{u_{2}^{\prime}}=\frac{w_{1}^{\prime}}{w_{2}^{\prime}}=\xi.\end{array}\right.$
(13)

which implies that

$\displaystyle u_{1},w_{1}\in V_{1}\cap\lambda V_{2},$ (14) $\displaystyle
u_{2},w_{2}\in V_{2}\cap\mu V_{2}$ $\displaystyle
u_{1}^{\prime},w_{1}^{\prime}\in V_{1}\cap\xi V_{2}.$

If $(u_{1}^{\prime},w_{1}^{\prime})=(u_{1},w_{1})=(0,0)$ then
$(u_{2},w_{2})\neq(0,0)$. Moreover, if $\mu\in{\mathbb{F}}_{q}$ then from
System (13) it follows that
$(u_{1},u_{2})=(0,u_{2})=\mu(0,u_{2}^{\prime})=\mu(u_{1}^{\prime},u_{2}^{\prime})$,
i.e. $(u_{1},u_{2})$ and $(u_{1}^{\prime},u_{2}^{\prime})$ are
${\mathbb{F}}_{q}$-proportional, a contradiction. If
$\mu\notin{\mathbb{F}}_{q}$, since $V_{2}$ is a Sidon space, from Equation
(14) it follows that $u_{2}$ and $w_{2}$ are ${\mathbb{F}}_{q}$-proportional,
i.e. $\rho\in{\mathbb{F}}_{q}$, a contradiction.
Therefore $(u_{1}^{\prime},w_{1}^{\prime})\neq(0,0)$ or
$(u_{1},w_{1})\neq(0,0)$, from System (13) this implies that $\xi\neq 0$ or
$\lambda\neq 0$.
If $\xi\neq 0$ then by Condition (11) on $V_{1}$ and $V_{2}$ we have that
$u_{1}^{\prime}$ and $w_{1}^{\prime}$ are ${\mathbb{F}}_{q}$-proportional,
i.e. $\rho\in{\mathbb{F}}_{q}$, a contradiction.
If $\lambda\neq 0$ then by Condition (11) 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 $(u_{2}^{\prime},w_{2}^{\prime})=(0,0)$, then
$u_{1}^{\prime},w_{1}^{\prime}\neq 0$, then from System (12) (as done for
(13)) we get

$\left\\{\begin{array}[]{ll}\frac{u_{1}}{u_{1}^{\prime}}=\frac{w_{1}}{w_{1}^{\prime}}=\sigma,\\\
\frac{u_{2}}{u_{1}^{\prime}}=\frac{w_{2}}{w_{1}^{\prime}}=l.\end{array}\right.$
(15)

from which we derive

$\displaystyle u_{1},w_{1}\in V_{1}\cap\sigma V_{1},$ (16) $\displaystyle
u_{2},w_{2}\in V_{2}\cap lV_{1}.$

If $l=0$ then $(u_{2},w_{2})=(0,0)$ and $u_{1},w_{1}\neq 0$ and from Equation
(16), if $\sigma\in{\mathbb{F}}_{q}$ then $(u_{1},u_{2})=(u_{1},0)$ and
$(u_{1}^{\prime},u_{2}^{\prime})=(u_{1}^{\prime},0)$ are
${\mathbb{F}}_{q}$-proportional, a contradiction. If
$\sigma\notin{\mathbb{F}}_{q}$, since $V_{1}$ is a Sidon space, we have
$u_{1}$ and $w_{1}$ are ${\mathbb{F}}_{q}$-proportional, implying that
$\rho\in{\mathbb{F}}_{q}$.
If $l\neq 0$, we can argue as before and get a contradiction also in this
case. ∎

######  Remark 5.2.

The above result can also be proved by using [22, 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). 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 (11). For instance, we can consider the subspaces of [37,
Construction 37]. They prove that the subspaces in [37, 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 $q\geq 3$, $n,k$ be two positive integers with $k\mid n$ and $n/k\geq 2$,
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
[37, 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

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

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

In particular, if $m$ is odd, the examples of the above corollary is 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 the Introduction, 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 [50]. The study of the equivalence
for subspace codes was initiated by Trautmann in [42] and the case of cyclic
subspace codes has been investigated in [16] by Gluesing-Luerssen and Lehmann.
Therefore, motivated by [16, Theorem 2.4], we say that two cyclic subspace
codes $\mathrm{Orb}(U)$ and $\mathrm{Orb}(V)$ are linearly equivalent if there
exists $i\in\\{0,\ldots,n-1\\}$ such that

$\mathrm{Orb}(U)=\mathrm{Orb}(V^{q^{i}}),$

where $V^{q^{i}}=\\{v^{q^{i}}\colon v\in V\\}$. 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
$\mathrm{Aut}(\mathbb{F}_{q^{n}})$, since this will still preserve the metric
properties of the codes. Therefore, we say that two cyclic subspace codes
$\mathrm{Orb}(U)$ and $\mathrm{Orb}(V)$ are semilinearly equivalent if there
exists $\sigma\in\mathrm{Aut}(\mathbb{F}_{q^{n}})$ such that

$\mathrm{Orb}(U)=\mathrm{Orb}(V^{\sigma}),$

where $V^{\sigma}=\\{v^{\sigma}\colon v\in V\\}$ 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
$\mathrm{Orb}(U)$ and $\mathrm{Orb}(V)$ are semilinearly equivalent, that is
there exist $\sigma\in\mathrm{Aut}(\mathbb{F}_{q^{n}})$ 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

$V_{U,\gamma}=\\{u+u^{\prime}\gamma:(u,u^{\prime})\in U\\}$

$V_{W,\xi}=\\{w+w^{\prime}\xi:(w,w^{\prime})\in W\\}$

where $\gamma,\xi\in\mathbb{F}_{q^{n}}$ are such that $\\{1,\gamma\\}$ and
$\\{1,\xi\\}$ 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
$(\lambda,\sigma)\in\mathbb{F}_{q^{n}}^{*}\times\mathrm{Aut}(\mathbb{F}_{q^{n}})$
if and only if there exists $A=\left(\begin{aligned} \begin{matrix}c&d\\\
a&b\end{matrix}\end{aligned}\right)\in\mathrm{GL}(2,\mathbb{F}_{q^{k}})$ such
that $\xi=\frac{a+b\gamma^{\sigma}}{c+d\gamma^{\sigma}}$,
$\lambda=\frac{1}{c+d\gamma^{\sigma}}$ and $U^{\sigma}=\\{wA\colon w\in
W\\}=W\cdot A$.

###### Proof.

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

$\lambda V_{U,\gamma}^{\sigma}=\\{\lambda u+\lambda
u^{\prime}\gamma^{\sigma}:(u,u^{\prime})\in
U\\}\subseteq\lambda\mathbb{F}_{q^{k}}+\lambda\gamma^{\sigma}\mathbb{F}_{q^{k}}$

and

$V_{W,\xi}=\\{w+w^{\prime}\xi:(w,w^{\prime})\in
W\\}\subseteq\mathbb{F}_{q^{k}}+\xi\mathbb{F}_{q^{k}}.$

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

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

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.

$\lambda\langle 1,\gamma^{\sigma}\rangle_{\mathbb{F}_{q^{k}}}=\langle
1,\xi\rangle_{\mathbb{F}_{q^{k}}}.$

Then

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

and

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

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

$\displaystyle\lambda^{-1}V_{W,\xi}$
$\displaystyle=\\{\lambda^{-1}w+\lambda^{-1}w^{\prime}\xi:(w,w^{\prime})\in
W\\}$ (17)
$\displaystyle=\left\\{(c+d\gamma^{\sigma})w+(c+d\gamma^{\sigma})w^{\prime}\frac{a+b\gamma^{\sigma}}{c+d\gamma^{\sigma}}:(w,w^{\prime})\in
W\right\\}$
$\displaystyle=\\{(c+d\gamma^{\sigma})w+(a+b\gamma^{\sigma})w^{\prime}:(w,w^{\prime})\in
W\\},$

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

$U^{\sigma}=W\cdot A$

hence we obtain the first part of the assertion.
Conversely, suppose that there exists $A=\left(\begin{aligned}
\begin{matrix}c&d\\\
a&b\end{matrix}\end{aligned}\right)\in\mathrm{GL}(2,\mathbb{F}_{q^{k}})$ 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

$\displaystyle V_{W,\xi}$ $\displaystyle=\\{w+w^{\prime}\xi:(w,w^{\prime})\in
W\\}$ (18)
$\displaystyle=\left\\{w+w^{\prime}\frac{a+b\gamma^{\sigma}}{c+d\gamma^{\sigma}}:(w,w^{\prime})\in
W\right\\}$
$\displaystyle=\lambda\\{(c+d\gamma^{\sigma})w+(a+b\gamma^{\sigma})w^{\prime}:(w,w^{\prime})\in
W\\}$
$\displaystyle=\lambda\\{(cw+aw^{\prime})+(dw+bw^{\prime})\gamma^{\sigma}:(w,w^{\prime})\in
W\\}$
$\displaystyle=\lambda\\{u^{\sigma}+u^{\prime\sigma}\gamma^{\sigma}:(u,u^{\prime})\in
U\\}$ $\displaystyle=\lambda V_{U,\gamma}^{\sigma}.$

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

######  Remark 6.3.

A first consequence of Theorem 6.2 is that all the known examples ([37,
Constructions 11 and 15], [47, Lemma 3.1], [22, Lemmas 4.1 and 4.2] and [48,
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 [48, Lemma 2.4], that is
$V_{U,\gamma}\subseteq\mathbb{F}_{q^{n}}$ where

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

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

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

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

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 [34, 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

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

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

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

and Lemma 3.10 implies the assertion. ∎

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 the Table 1. Family i) in Table 1 is known as
pseudoregulus type and

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

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

$\Lambda(x^{q^{s}}+\delta
x^{q^{s(k-1)}})\leq\left\\{\begin{array}[]{cc}\varphi(k)\frac{q-2}{2}&\text{if}\,\,2\nmid
k,\\\ &\\\ \varphi(k)\frac{(q+1)(q-2)}{4}&\text{if}\,\,2\mid
k,\end{array}\right.$

see [24, Section 2.2] and for further bounds see [40, Theorem 4.3]. For the
Family iii) we have

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

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

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

if $\ell\not\equiv 2\pmod{4}$, where $q=p^{r}$ and $p$ is prime, see [24,
Corollary 4.3] and [31, 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

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

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

$L=\\{\langle(a+b\gamma^{\sigma},c+d\gamma^{\sigma})\rangle_{\mathbb{F}_{q^{n}}}\colon
a,b,c,d\in\mathbb{F}_{q^{k}},(a,b,c,d)\neq(0,0,0,0)\\}.$

Then $L$ is an $\mathbb{F}_{q^{k}}$-linear set of the form of those studied in
[18] and hence $|L|=q^{3k}+1$ when $n/k>3$; we refer to [21, 33] for surveys
on linear sets. Hence,

$\Gamma(\gamma)\geq q^{n}-q^{3k}|\mathrm{Aut}(\mathbb{F}_{q^{n}})|.$

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

$\Lambda(x^{q^{s}})\Gamma(\gamma)\geq\frac{\varphi(k)}{2}(q^{n}-q^{3k}|\mathrm{Aut}(\mathbb{F}_{q^{n}})|).$

###  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, hence they are not the ones described in [32, 38].

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
$\sigma\in\mathrm{Aut}(\mathbb{F}_{q^{n}})$, 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

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

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

###### 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
$\sigma\in\mathrm{Aut}(\mathbb{F}_{q^{n}})$. Then it is easy to see that
$\lambda^{q^{k}}F^{\sigma}(\lambda^{-1}x)$ is monic, it has $q$-degree $k$ and
$\lambda^{q^{k}}F^{\sigma}(\lambda^{-1}u)=0$, for every $u\in\lambda
V^{\sigma}$. ∎

We now prove that the examples found in [37] are not semilinearly equivalent
to the examples found in [32, 38].

######  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
$\delta=1-(-1)^{k-2}\gamma^{q^{k-1}+\dots+1}$,

$a_{0}=-\delta^{q},\,\,\,a_{1}=-(\gamma^{q^{k}}-\gamma),\,\,\,a_{k}=\delta$

and

$a_{i}=(-1)^{i}(\gamma^{q^{i-1}+\dots+q})(\gamma^{q^{k}}-\gamma),$

for $i\in\\{2,\dots,k-1\\}$.
Let $F(x)=\displaystyle\sum_{i=0}^{k}a_{i}x^{q^{i}}\in\mathcal{L}_{n,q}$. Then
$\ker(F)=V_{x^{q},\gamma}=\\{u+u^{q}\gamma:u\in\mathbb{F}_{q^{k}}\\}$.

###### 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

$\displaystyle F(u)=$
$\displaystyle-\delta^{q}u-(\gamma^{q^{k}}-\gamma)u^{q}+\gamma^{q}(\gamma^{q^{k}}-\gamma)u^{q^{2}}+$
(19)
$\displaystyle-\gamma^{q^{2}+q}(\gamma^{q^{k}}-\gamma)u^{q^{3}}+\gamma^{q^{3}+q^{2}+q}(\gamma^{q^{k}}-\gamma)u^{q^{4}}-$
$\displaystyle\vdots$
$\displaystyle+(-1)^{k-2}\gamma^{q^{k-3}+\dots+q}(\gamma^{q^{k}}-\gamma)u^{q^{k-2}}+$
$\displaystyle+(-1)^{k-1}\gamma^{q^{k-2}+\dots+q}(\gamma^{q^{k}}-\gamma)u^{q^{k-1}}+$
$\displaystyle+\delta u$

and

$\displaystyle F(u^{q}\gamma)=$ $\displaystyle-\delta^{q}\gamma
u^{q}-(\gamma^{q^{k}}-\gamma)\gamma^{q}u^{q^{2}}+$ (20)
$\displaystyle+\gamma^{q}(\gamma^{q^{k}}-\gamma)\gamma^{q^{2}}u^{q^{3}}-\gamma^{q^{2}+q}(\gamma^{q^{k}}-\gamma)\gamma^{q^{3}}u^{q^{4}}+$
$\displaystyle\vdots$
$\displaystyle+(-1)^{k-3}\gamma^{q^{k-4}+\dots+q}(\gamma^{q^{k}}-\gamma)\gamma^{q^{k-3}}u^{q^{k-2}}$
$\displaystyle+(-1)^{k-2}\gamma^{q^{k-3}+\dots+q}(\gamma^{q^{k}}-\gamma)\gamma^{q^{k-2}}u^{q^{k-1}}+$
$\displaystyle+(-1)^{k-1}\gamma^{q^{k-2}+\dots+q}(\gamma^{q^{k}}-\gamma)\gamma^{q^{k-1}}u+$
$\displaystyle+\delta\gamma^{q^{k}}u^{q},$

since $u\in\mathbb{F}_{q^{k}}$. Since $F(u+u^{q}\gamma)=F(u)+F(u^{q}\gamma)$,
(19) and (20) imply

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

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

∎

######  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 $F$, 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
$V_{x^{q},\gamma}=\\{u+u^{q}\gamma:u\in\mathbb{F}_{q^{k}}\\}$ is not
semilinearly equivalent to the examples in [32, 38].

###### 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.
∎

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
$V_{g,\gamma}=\\{u+g(u)\gamma:u\in\mathbb{F}_{q^{k}}\\}$ 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 $0<i<j<k$. 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

$F(u+\gamma g(u))=0,$ (21)

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
$\mathbb{F}_{q^{k}}$. (21) implies that

$\begin{array}[]{ll}G(u)=(a+1)u+(a\gamma+\gamma^{q^{k}})u^{q^{i}}+(a\gamma\delta+\delta\gamma^{q^{k}})u^{q^{j}}+bu^{q^{\ell}}+\\\
b\gamma^{q^{\ell}}u^{q^{i+\ell}}+b\gamma^{q^{\ell}}\delta^{q^{\ell}}u^{q^{\ell+j}}=0,\end{array}$
(22)

for every $u\in\mathbb{F}_{q^{k}}$. Hence, the left hand-side $G(u)$ of (22)
can be seen as a polynomial in $u$ and (22) implies that this polynomial is
the zero polynomial when is reduced modulo $u^{q^{k}}-u$. To this aim, denote
by

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

which are seen both as subsets of $\mathbb{Z}/k\mathbb{Z}$. We divide the
proof in cases:
Case 1: $|\mathcal{I}\cap\mathcal{J}|\leq 2$.
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 $i\neq\ell+i$ and $j\neq\ell+j$. Therefore,
we only have two cases: either

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

or

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

Case 2.1: $i+\ell=0$, $j+\ell=i$ and $\ell=j$.
From (22) the coefficients of $G$ are

$\left\\{\begin{array}[]{lll}a+1+b\gamma^{q^{\ell}}=0,\\\
a\gamma+\gamma^{q^{k}}+b\gamma^{q^{\ell}}\delta^{q^{\ell}}=0,\\\
a\gamma\delta+\delta\gamma^{q^{k}}+b=0.\end{array}\right.$ (23)

Note that $a+1\neq 0$, otherwise the first equation of the above system would
imply $b=0$. By multiplying the first equation of System (23) by
$\delta^{q^{\ell}}$ and the third one by $\gamma^{q^{\ell}}\delta^{q^{\ell}}$
we get

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

which can be rewritten as follows

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

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

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

and hence

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

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. ∎

Arguing as in Corollary 6.10, 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
$V_{g,\gamma}=\\{u+g(u)\gamma:u\in\mathbb{F}_{q^{k}}\\}$ be an
${\mathbb{F}}_{q}$-subspace of $\mathbb{F}_{q^{n}}$ where
$g(u)=u^{q^{l}}+Bu^{q^{m}}$ and $\gcd(k,m-l)=1$ with $0<l<m<k$. Then
$V_{g,\gamma}$ is not semilinearly equivalent to the examples in [32, 38].

##  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 behaviour. 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 our 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. [8, 6, 15,
5], 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 $2k\nmid m$ this
example cannot be the examples described in Section 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 $V_{g,\delta^{\prime}}$ for some
$\delta^{\prime}\in{\mathbb{F}}_{q^{m}}$ and $g$ as in [30, 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 towards 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.

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Chiara Castello, Olga Polverino, Paolo Santonastaso and Ferdinando Zullo,
Dipartimento di Matematica e Fisica,
Università degli Studi della Campania “Luigi Vanvitelli”,
Viale Lincoln, 5,
I– 81100 Caserta, Italy
E-mail: {chiara.castello,olga.polverino,
<EMAIL_ADDRESS>

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