ARS MATHEMATICA CONTEMPORANEA
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 19 (2020) 125-145 https://doi.org/10.26493/1855-3974.2137.7fa
(Also available at http://amc-journal.eu)
A new family of maximum scattered linear sets
in PG(1, q6)*
Daniele Bartoli ©
Dipartimento di Matematica e Informática, Universita degli Studi di Perugia,
Perugia, Italy
Corrado Zanella ©
Dipartimento di Tecnica e Gestione dei Sistemi Industriali, Universita degli Studi di Padova,Vicenza, Italy
Ferdinando Zullo t ©
Dipartimento di Matematica e Fisica, Universita degli Studi della Campania "Luigi Vanvitelli", Caserta, Italy
Received 5 October 2019, accepted 11 July 2020, published online 12 November 2020
We generalize the example of linear set presented by the last two authors in "Vertex properties of maximum scattered linear sets of PG(1, qn)" (2019) to a more general family, proving that such linear sets are maximum scattered when q is odd and, apart from a special case, they are new. This solves an open problem posed in "Vertex properties of maximum scattered linear sets of PG(1, qn)" (2019). As a consequence of Sheekey's results in "A new family of linear maximum rank distance codes" (2016), this family yields to new MRD-codes with parameters (6,6, q; 5).
Keywords: Scattered linear set, MRD-code, linearized polynomial. Math. Subj. Class. (2020): 51E20, 05B25, 51E22
* The research of all the authors was supported by the Italian National Group for Algebraic and Geometric Structures and their Applications (GNSAGA - INdAM).
ÎThis research of the third author was supported by the project "VALERE: VAnviteLli pEr la RicErca" of the University of Campania "Luigi Vanvitelli" and was partially funded by a fellowship from the Department of Management and Engineering (DTG) of the Padua University.
E-mail addresses: daniele.bartoli@unipg.it (Daniele Bartoli), corrado.zanella@unipd.it (Corrado Zanella), ferdinando.zullo@unicampania.it (Ferdinando Zullo)
Abstract
©® This work is licensed under https://creativecommons.org/licenses/by/4.0/
126
Ars Math. Contemp. 19 (2020) 125-145
1	Introduction
Let A = PG(V, Fqn) = PG(1, qn), where V is a vector space of dimension 2 over Fqn. If U is a k-dimensional Fq-subspace of V, then the Fq-linear set LU is defined as
Lu = {(u>F,„ : u e U \{0}},
and we say that LU has rank k. Two linear sets LU and LW of PG(1, qn) are said to be PrL-equivalent if there is an element ^ in PrL(2, qn) such that LU = LW. Itmay happen that two Fq-linear sets LU and LW of PG(1, qn) are PrL-equivalent even if the Fq-vector subspaces U and W are not in the same orbit of rL(2, qn) (see [5, 12] for further details). In this paper we focus on maximum scattered Fq-linear sets of PG(1, qn), that is, Fq-linear sets of rank n in PG(1, qn) of size (qn — 1)/(q — 1).
If ((0,1)>Fqn is not contained in the linear set LU of rank n of PG(1, qn) (which we can always assume after a suitable projectivity), then U = Uf := {(x, f (x)) : x e Fqn} for some linearized polynomial (or q-polynomial) f (x) = J2n- aixq e Fqn [x]. In this case we will denote the associated linear set by Lf. If Lf is scattered, then f (x) is called a scattered q-polynomial; see [24].
The first examples of scattered linear sets were found by Blokhuis and Lavrauw in [3] and by Lunardon and Polverino in [18] (recently generalized by Sheekey in [24]). Apart from these, very few examples are known, see Section 3.
In [24, Section 5], Sheekey established a connection between maximum scattered linear sets of PG(1, qn) and MRD-codes, which are interesting because of their applications to random linear network coding and cryptography. We point out his construction in the last section. By the results of [1] and [2], it seems that examples of maximum scattered linear sets are rare.
In this paper we will prove that any
fh(x) = hq-1xq — hq2-1xq2 + xq4 + xq5, h e Fq6, hq3+1 = —1, q odd (1.1) is a scattered q-polynomial. This will be done by considering two cases:
Case 1: h e Fq, that is, fh(x) = xq — xq2 + xq4 + xq5; the condition hq3+1 = —1 implies q = 1 (mod 4).
Case 2: h e Fq. In this case h = 1, otherwise h e Fq2 and then we have hq+1 = 1, a contradiction to hq +1 = —1.
Note that in Case 1, this example coincides with the one introduced in [27], where it has been proved that fh is scattered for q = 1 (mod 4) and q < 29. In Corollary 3.11 we will prove that the linear set Lh associated with fh(x) is new, apart from the case of q a power of 5 and h e Fq. This solves an open problem posed in [27].
Finally, in Section 4 we prove that the Fq-linear MRD-codes with parameters (6,6, q; 5) arising from linear sets Lh are not equivalent to any previously known MRD-code, apart from the case h e Fq and q a power of 5; see Theorem 4.1.
2	Lh is scattered
A q-polynomial (or linearized polynomial) over Fqn is a polynomial of the form
f (x) = E
t
i=o
D. Bartoli et al.: A new family of maximum scattered linear sets in PG(1, q6)
127
where a G Fqn and t is a positive integer. We will work with linearized polynomials of degree less than or equal to qn-1. For such a kind of polynomial, the Dickson matrix1 M (f) is defined as
M(f) :=
/ ao ai
q aq
n-i ao
a
a
a
an-i \ q
n-2
/
G Fn„xn,
where aj = 0 for i > t.
Recently, different results regarding the number of roots of linearized polynomials have been presented, see [4, 9, 22, 23, 26]. In order to prove that a certain polynomial is scattered, we make use of the following result; see [4, Corollary 3.5].
Theorem 2.1. Consider the q-polynomial f (x) = En=01 ajxq' over Fqn and, with m as a variable, consider the matrix
m
M(m) :=
a
a1
1
an 1
2
qn— 1
mq
The determinant of the (n — i) x (n — i) matrix obtained by M (m) after removing the first i columns and the last i rows of M(m) is a polynomial Mn-i(m) G Fqn [m]. Then the polynomial f (x) is scattered if and only if M0(m) and M1(m) have no common roots.
2.1 Case 1
Let
f (x) = xq — xq + xq + xq G Fq6 [x].
By Theorem 2.1, f (x) is scattered if and only if for each m G Fq6 the determinants of the following two matrices do not vanish at the same time
M5(m) =
Me(m) =
(1	—1	0	1	1	
mq	1	—1	0	1	
1	mq	1	—1	0	,
1	1	mq	1	— 1	
0	1	1	mq	1	
m	1	—1	0	1	1 ^
1	mq	1	—1	0	1
1	1	mq	1	—1	0
0	1	1	mq	1	—1
—1	0	1	1	mq4	1
1	—1	0	1	1	mq J
This is sometimes called autocirculant matrix.
n — 1
n— 1
q
q
a
a
2
0
q
q
a
m
a
n1
n1
q
a
2
1
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Ars Math. Contemp. 19 (2020) 125-145
Theorem 2.2. The polynomial f (x) is scattered if and only if q = 1 (mod 4).
Proof. If q is even, then for m = 0 the matrix M6 (0) has rank two and f (x) is not scattered. Suppose now q = 3 (mod 4). Then let m G Fq2 \ Fq such that m2 = -4. So
m = mq = mq = -mq = -mq = -mq and, by direct checking,
det(M5(m)) = (m2 +4)2 = 0, det(M6 (m)) = -(m2 + 4)3 = 0
and f (x) is not scattered.
Assume q = 1 (mod 4) and suppose that f (x) is not scattered. Then there exists m0 G Fq6 such that
(det(M5(mo)))q =0,
Consider
P1 = det
(det(M6(m0)))q =0, s,t = 0,1, 2, 3, 4, 5.
(2.1)
1	-1	0	1	1
Y	1	-1	0	1
1	Z	1	-1	0
1	1	U	1	-1
0	1	1	V	1
P2 = det
X	1	-1	0	1	1
1	Y	1	-1	0	1
1	1	Z	1	-1	0
0	1	1	U	1	-1
-1	0	1	1	V	1
1	-1	0	1	1	W
(2.2)
Therefore,
X = mo, Y = m0, ..., W = m0 _. p(0) p _. p(0)
0, . . . , W = m0
is a root of P1 =: P|0), P2 =: P2(0) and of the polynomials inductively defined by
P(j)(X, Y, Z, U, V, W) = P(j-1)(Y, Z, U, V, W, X), j = 1, 2, 3,4,5, i = 1, 2 which arise from Equation 2.1. These polynomials satisfy
(2.3)
P
(j-i)
(m0,
q q q q
m0, m0 , m0 , m0 , m
5q
))' = P(j)(m0,<
q q q q
m0, m0 , m , m
).
One obtains a set S of twelve equations in X, Y, Z, U, V, W having a nonempty zero set. The following arguments are based on the fact that taking the resultant R of two polynomials in S with respect to any variable, the equations S U {R} admit the same solutions. We have
P1 = YZUV -YZU - 2YZ + 2YU + 4Y - ZUV + 2ZV - 2UV + 4V +16 = 0. (2.4) Consider the following resultants:
Q1 := ResV (P^, P1) = 2(XY2ZU - XY2ZW + XY2UW + 2XY2W
-	2XYZU + 2XYZW - 2XYUW + 8XYW + 8XY - 8XW + 16X
-	Y2ZUW - 2Y2ZU + 2YZUW - 8YZU - 8YZ + 8YU - 8YW + 8ZU - 16Z + 16U - 16W),
Q2 := ResV(P1(4),P1) = XYZW - XYZ - XYW + 2XZ
-	2XW - 2YZ + 2YW + 4Z + 4W + 16,
Q3 := ResV(P1(5),P1) = XYZU - XYZ - 2XY + 2XZ + 4X - YZU + 2YU - 2ZU + 4U + 16.
4
q
m
D. Bartoli et al.: A new family of maximum scattered linear sets in PG(1, q6)
129
They all must be zero, as well as
Resw(Resu(Qi,Qs), Q2) = 8(YZ - 4)(Y2 + 4)(X - Z)(XZ + 4)(XY - 4). (2.5) We distinguish a number of cases.
1.	Suppose that Y2 = -4. Since q = 1 (mod 4), X = Y = Z = U = V = W .So
P1 = X4 - 2X3 + 8X + 16
and the resultant between X2 +4 and P1 with respect to X is 227 = 0 and then (2.3) is not a root of P1, a contradiction.
2.	Condition YZ = 4 is clearly equivalent to XY = 4. This means that Y = U = W = 4/X, Z = V = X. Therefore, by (2.4) we get X2 + 4 = 0 and we proceed as above.
3.	Case XZ = -4. In this case Z = -4/X, U = -4/Y, V = -4/Z = X, W = Y, X = Z and therefore X2 = -4 and we can proceed as above.
4.	Condition X = Z implies X G Fq2 and so X = Z = V and Y = U = W .By substituting in P1 and P2,
X3Y3 + 3X3Y - 6X2Y2 - 12X2 + 3XY3 + 24XY - 12Y2 - 64 = 0, X2Y2 - X2Y + 2X2 - XY2 - 4XY + 4X + 2Y2 + 4Y + 16 = 0.
Eliminating Y from these two equations one gets
8(X2 +4)6 = 0,
and so X2 + 4 = 0. We proceed as in the previous cases.
This proves that such m0 G Fq6 does not exist and the assertion follows.	□
2.2 Case 2
We apply the same methods as in Section 2.1. In the following preparatory lemmas (and in the rest of the paper) q is a power of an arbitrary prime p.
Lemma 2.3. Let h G Fq6 be such that hq3+1 = -1, h4 = 1. Then
1. hq = -h;
2.	hq2+1 = 1;
3.	hq2+1 = ±hq, if q is odd;
4.	h4q2 +4 + 14h2q2+2q+2 + h4q = 0 implies p = 2 and hq2-q+1 = 1 or q = 32s, s G N*, hq2-q+1 = ±V-1.
Proof. The first three are easy computations. Consider now
h4q2+4 + 14h2q2+2q+2 + h4q = 0.
For p = 2 the equation above implies hq2-q+1 = 1.
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Ars Math. Contemp. 19 (2020) 125-145
Assume now p = 2. Since h = 0, it is equivalent to
(hq2-q+1 )4 + 14(hq2-q+1)2 + 1 = 0,
that is (hq2-q+1)2 = -7 ± 4V3 = (V—3 ± 2V—1)2. Let z = -7 ± 4V3. Note that hq -9+1 = ±vz belongs to F q2. We distinguish two cases.
•	Vz e Fq. Then
-1 = hq3+1 = (hq2-q+1)q+1 = (±vz)q+1 = z = -7 ± 4V3,
a contradiction if p = 3. Also, z = -1, q is an even power of 3, and hq2-q+1 =
±V-1.
•	Vz e Fq .Then
-1 = hq3+1 = (hq2-q+1)q+1 = (±vz)q+1 = -z = 7 + 4V3, a contradiction if p = 2.	□
Lemma 2.4. Let h e Fq6 be such that hq3+1 = -1, h4 = 1. If a root a of the polynomial
hq+1Tq+1 + (hq2+q+2 + h2q2+2)Tq + (h2q2+2 - hq2+1)t
+ hq2+2q+1 + h2q2+q+1 - h2q - hq2+q e Fq6 [T]
belongs to Fq6, then one of the following cases occurs:
•	p = 2, hq2-q+1 = 1; or
•	q = 32s, s > 0, hq2-q+1 = ±V-1; or
•	a = ±(hq2 + hq); or
•	h e Fq.
Proof. First, note that a = 0 would imply hq (hq + h)q(hq2+1 -1) = 0 which is impossible by Lemma 2.3. Therefore a = 0 and aqi = mpX), where
4(X ) = -(hq2+1 - 1)(hq2+1X + h2q + hq2+q ) mt(X ) = h(hq X + hq2+q+1 + h2q2+1)
¿2(x ) = -(hq + h)(2hq2+q+1x + h2q2+q+2 + h3q2+2 + h3q + hq2+2q )
m2(X ) = hq+1(h2q2+2X + h2q X + 2hq2+2q+1 + 2h2q2 +q+1) 4(X ) = (hq + h)q (3h2q2+q+2X + h3q X + h3q2+q+3 + h4q2+3 + 3hq2+3q+1
+ 3h2q2+2q+1)
m3(X ) = hq2+q (h3q2+3X + 3hq2+2q+1X + 3h2q2+2q+2 + 3h3q2+q+2 + h4q + hq2+3q )
D. Bartoli et al.: A new family of maximum scattered linear sets in PG(1, q6)
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) = (h«2+1 - i)(h4«2+4x + 6h2«2+2«+2x + h4« X + 4h3«2+2«+3 + 4h4q2+q+3
+ 4h92+4q+i +4h292+3«+1)
m4(X ) = h«2 (4h3«2+«+3X + 4h«2+3«+1X + h4«2+«+4 + h5«2+4 + 6h2«2+3«+2 + 6h3«2 +2«+2 + h5« + h«2+4« ) 4(X ) = -(h« + h)(h5«2+5X + 10h3«2+2«+3X + 5h«2+4«+1X + 5h4«2+2«+4
+ 5h5«2 +«+4 + 10h2«2 +4«+2 + 10h3«2+3«+2 + h6« + h«2+5« ) m5(X ) = 5h4«2+«+4X + 10h2«2+3«+2X + h5« X + h5«2+«+5 + h6«2+5 + 10h3«2+3«+3 + 10h4«2+2«+3 + 5h«2+5«+1 + 5h2«2+4«+1 4(X ) = (h« + h)« (6h5«2+«+5X + 20h«3+3«+3X + 6Xh«2+5«+1 + h6«2 +«+6 + h7«2+6 + 15h4«2+3«+4 + 15h5«2+2«+4 + 15h2«2+5«+2 + 15h3«2+4«+2 + h7« + h«2+6« ) m6(X ) = h6«2+6X + 15h4«2+2«+4X + 15h2«2 +4«+2X + h«6 X + 6h5«2+2«+5
+ 6h6«2 +«+5 + 20h3«2 +4«+3 + 20h4«2+3«+3 + 6h«2+6«+1 + 6h2«2+5«+1.
Since a«6 = a, in particular
(h2«2+2 + h2«)(h4«2+4 + 14h2«2+2«+2 + h4«)(h«2 - h«)(a + h« + h«2 )(a - h« - h«2 ) = 0. The claim follows from Lemma 2.3.	□
Lemma 2.5. Let h G F«6 be such that h«3+1 = -1, h4 = 1. If a root a of the polynomial
h«+1T«2+1 + (h? + h)«+1 g F«6 [T]
belongs to F«6, then
a = ±(h«2 + h« ).
Proof. If a = 0, then h« + h = 0, a contradiction to Lemma 2.3. So we can suppose a = 0. Then
«2 (h«-1 + 1)«+1 a« =--
a
~3 I „2 _
" 1a
a« = (h«-1 + 1)« +« -«-1c
(h«—1 + 1)95 + 94-93-92+9+1 (h« + h)2« a« = —	—
So, a = ±(h« + h«).	□
Let h G F«6 be such that h«3+1 = -1, h4 = 1. By Theorem 2.1 the polynomial
fh(x) = h«-1x« - (h«2-1 )x«2 + x«4 + x«5
a
a
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Ars Math. Contemp. 19 (2020) 125-145
is scattered if and only if for each m e F«6 the determinant of the following two matrices do not vanish at the same time
Me(m) =
M5(m) =
m	h«-1	—h«2-1	0	1	1
1	m«	h«2-«	h-«-1	0	1
1	1	m«2	—h-«2-1	h-«2-«	0
0	1	1	m«3	h1-«	—h1-«2
h«+1	0	1	1	m«4	h«-«2
V—h«2+1	h«2+«	0	1	1	«5 m«
z'h«-1	-h«2-1	0	1	1 \	
m«	h«2-«	h-«-1	0	1	
1	m«2	—h-«2-1	h-«2-«	0	
1	1	m«3	h1-«	—h1-«2	
0	1	1	m«4	h«-«2	
(2.6)
(2.7)
Theorem 2.6. Let h e F«6, q = 2s, be such that h«3+1 = 1. Then the polynomial
fh(x) = h«-1x« — (h«2-1)x«2 + x«4 + x«5 is not scattered.
Proof. Consider m = h«2 + h«. So,
m« = ^ + hq2, h
—«4 , 1 m = h +—2,
h«2
-«2 1 1 m« =^- + T, h« h
m
« = h« + h.
By direct checking, in this case, both det(M6(m)) fh(x) is not scattered.
m «
1 1
hF + h«,
det(M5(m)) = 0 and therefore
□
Theorem 2.7. Let h e F«6, q = ps, p > 2, be such that h«3+1 = —1 and h e F«. Then the polynomial fh(x) = h«-1x« — (h« -1)x« + x« + x« is scattered.
Proof. First we note that h4 = 1 since q is odd, he F«, and h«3+1 = —1. Suppose that
f(x) is not scattered. Then det(M6(m0)) = det(M5(m0)) = 0 for some m0 e F«6. Consider
X = mo, Y = m0, Z = m0 , U = m0 , V = m0 , W = m0 .
With a procedure similar to the one in the proof of Theorem 2.2, we will compute resultants starting from the polynomials associated with det(M6(m0)), det(M5(m0))« , and det(M5(mo))«5.
Eliminating W using det(M5(m0))« = 0 and U using det(M5(m0))« = 0, one gets
from det(M6(m0)) = 0
h«2+2«+V1(X, Y)^2(X, Y, Z, VVs(X, Y, Z, V) = 0,
D. Bartoli et al.: A new family of maximum scattered linear sets in PG(1, q6)
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where
<1(X, Y) = hq+1XY + h2q2+2X - hq2+1X + hq2+q+2Y + h2q2+2Y + h«2+2«+1 + h2«2+«+1 - h2« - h«2+«; <2(X, Y, Z, V) = hq2+q+2XYZV - hq2+q+2XYZ - h2XY - hq+1XY
-	h2«2+«+1XZV - h2«2+2XV - h2«2+«+1XV - h«2+2«+3YZ _ h2«2+q+3YZ_h«2+«+2Y - fr2«2+2Y - h^^Y
-	h2«2+«+1Y - h«2+2«+1ZV - h2«2+«+1ZV - h2«2+«+1V
-	h3«2+1V - h2«2+2«V - h3«2+«V + h2«2+«+3 + h3«2+3
+ h2?2+2?+2 + h3q2+q+2 - 2h«2+3+2 - 2h2«2+2 - 2h«2+23+1
-	2h2q2+q+1 + hq+1 + hq2+1 + h2q + hq2 +«; <3(X, Y, Z, V) = hq2+q+2XYZV + hq2+q+2XYZ - h2XY - hq+1XY
+ h2«2+«+1XZV - h2«2+2XV - h2,2+,+1IV - h«2+2«+3YZ
-	h2?2+?+3YZ + h«2+«+2Y + h2«2+2Y + h«2+2«+1Y
+ h2«2+«+1Y - h«2+2«+1ZV - h2«2+«+1ZV + h2«2+«+1V + h3«2+1V + h2«2+2« V + h3«2+« V + h2«2+«+3 + h3«2+3 + h2?2+2?+2 + h3?2+9+2 - 2h«2+3+2 - 2h2«2+2 - 2h«2+23+1
-	2h2q2+q+1 + hq+1 + hq2+1 + h2q + hq2 +«.
• If <1(X, Y) = 0, then by Lemma 2.4 either q = 32s and hq2-q+1 = ±V-1, or X = ±(hq2 + hq). In this last case,
Y	= ±(-h-1 + hq2), Z = ±(-h-q - h-1), U = ± (-h-q2 - h-q)
2 (2.8)
V	= ±(h - h-q ), W = ±(hq + h).
By substituting in det(M5(m0)) one obtains
4(h + hq)q+1(hq2+1 - 1)(hq2+1 - hq) = 0
and
4(h + hq )q+1(hq2+1 - 1)(hq2+1 + hq) = 0, respectively. Both are not possible due to Lemma 2.3.
Consider now the case q = 32s, hq2-q+1 = ±V-1 and X = ±(hq2 + hq). So, using <1(X, Y) = 0 and hq2-q+1 = ±V-1
det(M5(m0)) = 0
hq2+2q+1(hq2 + hq)(hq + h)(hq2+1 - 1)(hq2+q + hq)3(hq2+q - hq)3 •
• (h2q2+2 - hq2+1 + h2q)(X + hq + hq2)2(X - hq - hq2)2 = 0.
By Lemma 2.3 we get
h2«2+2 - h«2+1 + h2« = 0, which yields to a contradiction.
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Ars Math. Contemp. 19 (2020) 125-145
If	=0 and yi (X,Y) = 0, eliminating V in det(M5(m0)) =0 one
gets
2h3q2+2q+1(h?+2yZ - h«2 + 2 - h«2+«+1 + h* + h) •
•	(hXY + hq2+q+1 + h2q2+1 - hq2 - hq) •
•	(hq+1 XZ + hq+1 + hq2+1 + h2q + hq2+q) •
•	(hq+2YZ + hY + hqY - hq2 +q+1Z + hqZ - hq2+2 - hq2+q+1 + hq + h) = 0.
-	If hq+2YZ - hq2+2 - hq2+q+1 + hq + h = 0 then, from
_ hq2+2 + hq2+q+1 - hq - h =	hq+2Y	'
det(M5) = 0 gives
(hq + h)q+1(hY - hq2+1 + 1)(hY + hq2+1 - 1) = 0.
So, (2.8) holds and as in the case (X, Y) = 0 a contradiction arises.
-	If hXY + hq2+q+1 + h2q2+1 - hq2 - hq = 0 then, from
_ -hq2 +q+1 - h2q2+1 + hq2 + hq =	hX	'
the equation det(M5(m0)) = 0 yields
2 -i 2 2 (hq + h)(hq +1 - 1)(X - hq - hq)(X + hq + hq)=0.
So, (2.8) holds and as in the case (X, Y) = 0, a contradiction.
-	If hq+1XZ + hq+1 + hq2+1 + h2q + hq2+q = 0 then by Lemma 2.5
22 (X - hq - hq)(X + hq + hq) = 0,
again a contradiction as before.
-	If hq+2 YZ + hY + hqY - hq2+q+1Z + hqZ - hq2+2 - hq2+q+1 + hq + h = 0 then
(hq + h)Y - hq2+2 - hq2+q+1 + hq + h
Z=
hq+2Y - hq2+q+1 + hq
So, substituting U = Zq, V = Zq , W = Zq , X = Zq in det(M5(m0)) = 0 we get
(h - 1)q+1(h + 1)q+1(hq + h)q+1(hq2+1 - 1) • • (hY - hq2+1 + 1)2(hY + hq2+1 - 1)2 = 0.
By Lemma 2.3, (hY - hq2+1 + 1)(hY + hq2+1 - 1) = 0. Since Y = ±(hq2 -1/h) then (2.8) holds and a contradiction arises as in the case (X, Y) = 0.
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• If ^3(X, Y,Z, V) = 0 and <^(X, Y) = 0, eliminating U from det(M5(m0)) = 0 = det(M5(m0))q and then eliminating V using y3(X, Y, Z, V) = 0 one gets
2h3q2+q+1(hq + h)q(hq+2YZ - hq2+2 - hq2+q+1 + hq + h)2 •
•	(hXY + hq2 +q+1 + h2q2+1 - hq2 - hq) •
•	(hq+1XZ + hq+1 + hq2+1 + h2q + hq2+q ) = 0.
A contradiction follows as in the case (X, Y, Z, V) = 0 and (X, Y) = 0. □
3 The equivalence issue
We will deal with the linear sets Lh = Lfh associated with the polynomials defined in (1.1). Note that when h G Fq, such a linear set coincide with the one introduced in [27, Section 5].
3.1 Preliminary results
We start by listing the non-equivalent (under the action of rL(2, q6)) maximum scattered subspaces of F2e, i.e. subspaces defining maximum scattered linear sets.
■ q
Example 3.1.
1.	U1 := {(x,xq) : x G Fq6}, defining the linear set of pseudoregulus type, see [ , 11];
2.	U2 := {(x,Jxq + xq5) : x G Fq6}, Nq6/q(J) G {0,1}, defining the linear set of LP-type, see [16, 18, 20, 24];
3.	Uf := {(x, xq + Jxq ) : x G Fq6}, Nq6/q3 (J) G {0,1}, satisfying further conditions on J and q, see [6, Theorems 7.1 and 7.2] and [23]2;
4.	U4 := {(x, xq + xq3 + Jxq5 ) : x G Fq6}, q odd and J2 + J = 1, see [10,21].
In order to simplify the notation, we will denote by L1 and L^ the Fq-linear set defined by U1 and Urespectively. We will also use the following notation:
Remark 3.2. Consider the non-degenerate symmetric bilinear form of Fq6 over Fq defined by
(x,y) = Tfq6/q (xy^
for each x, y G Fq6. Then the adjoint f of the linearized polynomial f (x) = J25=0 a®xqi G
£6,q with respect to the bilinear form (, ) is
5
f (x) = £ a x-
0
i.e.
^V/q (xf (y)) = T^q6/q (yf (x))
for any x, y G Fq6.
2Here q > 2, otherwise it is not scattered.
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In [10, Propositions 3.1, 4.1 and 5.5] the following result has been proved.
Lemma 3.3. Let Lf be one of the maximum scattered of PG(1, q6) listed before. Then a linear set Lu of PG(1, q6) is PTL-equivalent to Lf ifandonly if U is TL-equivalent either to Uf or to Uf Furthermore, Lu is PTL-equivalent to L3 if and only if U is TL-equivalent to Uf.
We will work in the following framework. Let x0,..., x5 be the homogeneous coordinates of PG(5, q6) and let
E = {((x,xq,...,xq5))F 6 : x € Fq6}
be a fixed canonical subgeometry of PG(5, q6). The collineation a of PG(5, q6) defined by ((x0,..., x5))jJ e = ((x|, x0,..., x4))F6 fixes precisely the points of E. Note that if a is a collineation of PG(5, q6) such that Fix(a) = E, then a = as, with s € {1, 5}.
Let r be a subspace of PG(5, q6) of dimension k > 0 such that r n E = 0, and dim(r n Ta) > k - 2. Let r be the least positive integer satisfying the condition
dim(r n Ta n rff n • • • n rffr) > k - 2r.
(3.1)
Then we will call the integer r the intersection number of r w.r.t. a and we will denote it by intnCT (r); see [27].
Note that if a is as above, then intn^ (r) = intn^ (r) for any r.
As a consequence of the results of [11, 27] we have the following result.
Result 3.4. Let L be a scattered linear set of A = PG(1, q6) which can be realized in PG(5, q6) as the projection of E = Fix(a) from r ~ PG(3, q6) over A. If intnCT(r) = 1,2, then L is not equivalent to any linear set neither ofpseudoregulus type nor ofLP-type.
3.2 Lh is new in most of the cases
The linear set Lh can be obtained by projecting the canonical subgeometry
from
E = {((x,xq,xq ,xq ,xq ,xq ))Wq6 : x € F*6}
r: x0 =0 2 ,
hq Xxi — hq Xx2 + x4 + x5 =0
to
A:
xi =0
x2 =0
x3 = 0 x4 = 0.
Then
F*: xi2=0	1
hq -qx2 + h-q-1x3 + x5 + x0 = 0
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and
Therefore,
r-2 • r2 = 0 2 2
—h-1-q x3 + h-q -qx4 + x0 + x1 = 0.
rnr°
0
0
q2 1
— hq X2 + X4 + X5 =0
2
hq -qx2 + h-q-1x3 + x5 =0
and
r n r° n r0
xo = 0 xi =0 X2 =0 X4 + X5 =0
h-q-1x3 + x5 =0
2 1	2
-1x3 + h-q -qx4 = 0.
Hence, dimF 6 (mri) = 1 anddimF 6 (mrinP^) = -1, since qisoddand hq3+1 = 1. So, intnCT (r) = 3 and hence, by Result 3.4 it follows that Lh is not equivalent neither to L1 nor to ¿2.
Generalizing [27, Propositions 5.4 and 5.5] we have the following two propositions.
Proposition 3.5. The linear set is not PrL-equivalent to L3.
Proof. By Lemma 3.3, we have to check whether Uh and U3 are TL-equivalent, with Nq6/q3 (J) G {0,1}. Suppose that there exist p G Aut(Fq6) and an invertible matrix (C d) such that for each x g Fq6 there exists z G Fq6 satisfying
a b
c d) I hp(q-1)xpq — hp(q2-1)xpq2 + xpq4 + xpq5
zq + ¿zq
Equivalently, for each x G Fq6 we have3
cxp + d(hq-1xpq — hq2-1xpq2 + xpq4 + xpq5 ) =
aq xpq + bq (hq2-q xpq2 + h-q-1xpq3 + xpq5 + xp) + ¿[aq4 xpq4 + bq4 (h-q2+q xpq5 — hq+1xp + xpq2 + xpq3 )].
This is a polynomial identity in xp and hence we have the following relations:

c = bq + ¿hq+1bq4
dhq-1 = aq
—dhq2-1 = hq2-q bq + ¿bq4
0 = h-1-q bq + ¿bq 4
d = ¿aq
24
d = bq + ¿hq-q bq .
(3.2)
3We may replace hp by h, since hq +1 = —1 if and only if (hp)q +1 = -1.
z
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From the second and the fifth equations, if a = 0 then Jhq-1 = aq-q4 and Nq6/q3 (J) = 1, which is not possible and so a = d = 0 and b, c = 0. By the last equation, we would get Nq6/q3 (J) = 1, a contradiction.	□
Proposition 3.6. The linear set is PTL-equivalent to L44 (with J2 + J = 1) if and only if there exist a, b, c, d G Fq6 and p G Aut(Fq6 ) such that ad — bc = 0 and either
bq - Jkq2+1bq5
-kq+1bq4 - ¿qbq2
C =
a
d = k-q+1bq3 + ¿bq5 ^ bq3 + (kq-1 + ¿kq+q2 )bq5 = 0 kq2-q bq + (1 + kq2-q )bq3 + ¿kq2-1bq5 = 0 ^-£bq + (k-q+1 + ¿2k1-q2 )bq3 + ¿bq5 = 0
(3.3)
or
c = ¿bq - kq + 1bq
a = -¿qkq+1bq4 - bq2
d = k-q+1bq3 + bq5
¿bq3 + (kq-1 - ¿kq2+q)bq5
0
¿kq -qbq + (kq-q + 1)bq + kq-1bq =( ¿2 b? + (k-9+1 + ¿2k-92+1)b«3 + b?5 = 0,
where k = hp.
(3.4)
Proof. By Lemma 3.3 we have to check whether Uh is equivalent either to UJ4 or to (UJ4)L. Suppose that there exist p G Aut(Fq6) and an invertible matrix (C bd) such that for each x G Fq6 there exists z G Fq6 satisfying
a bW	xp	\ = /	z
c dl \hp(q-1)xpq - hp(q2-1)xpq2 + xpq4 + xpq5/ = \zq + zq3 + ¿zq5
Equivalently, for each x G Fq6 we have
cp + d(kq-1xpq - kq2-1xpq2 + xpq4 + xpq5)
x
-.q xpq
a*x"' + bq(kq -qxpq + k-1-qxpq + xpq + xp)
pq + xp)
+ aq3 xpq3 + bq3 (k-q+1xpq4 - k-q2+1xpq5 + xpq + xpq2 ) + ¿[aq5xpq5 + bq5 (-k1+q2 xp + kq +qxpq + xpq + xpq )].
This is a polynomial identity in xp which yields to the following equations
c = bq - ¿kq2+1bq5
dkq-
aq + bq + ¿kq+q bq
-dkq -1 = kq -qbq + bq 0 = k-q-1bq + aq3 + ¿bq5 d = k-q+1bq3 + ¿bq5 d = bq - k-q2+1bq3 + ¿aq5
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which can be written as (3.3).
Now, suppose that there exist p G Aut(Fq6) and an invertible matrix (ac d) such that for each x G Fq6 there exists z G Fq6 satisfying
a b\i	xp	\ _ /	z
c d) \hp(q-1)xpq - hp(q2-1)xpq2 + xpq4 + xpqJ _ \Szq + zq3 + zq5
Equivalently, for each x G Fq6 we have
cxp + d(kq-1xpq - kq2-1xpq2 + xpq4 + xpq5) _
S[aq xpq + bq (kq2-q xpq2 + k-1-q xpq3 + xpq5 + xp)] + aq3 xpq3 + bq3 (k-q+1xpq4 - k-q2+1xpq5 + xpq + xpq2) + aq5 xpq5 + bq5 (-k1+q2 xp + kq2+q xpq + xpq3 + xpq4).
This is a polynomial identity in xp which yields to the following equations
c = Sbq - kq2+1bq5 dkq-1 = Saq + bq3 + kq+q2 bq5 -dkq2-1 = Skq2-q bq + bq3 0 = Sk-q-1bq + aq3 + bq5 d = k-q+16q3 + bq5 ^d = Sbq - k-q2+16q3 + aq5
which can be written as (3.4).	□
We are now ready to prove that when h G Fq2, Lh is new.
Proposition 3.7. If h G Fq2, then Lh is not WL-equivalent to L4 (with S2 + S = 1).
Proof. By Proposition 3.6 we have to show that there are no a, b, c and d in Fq6 such that ad - bc = 0 and (3.3) or (3.4) are satisfied. Note that b = 0 in (3.3) and (3.4) yields a = c = d =0, a contradiction. So, suppose b = 0. Since h G Fq2 then k G Fq2. We start by proving that the last three equations of (3.3), i.e.
1
Eq2 Eqs
Eq1 --------- —
kq2-q bq + (1 + kq2-q )bq3 + Skq2-1bq5 = 0 -Sbq + (k-q+1 + S2k1-q2 )bq3 + Sbq5 = 0,
yield a contradiction. As in the above section, we will consider the q-th powers of Eq1,
i j i
Eq2 and Eq3 replacing bq , kq , and Sq (respectively) by Xj, Yj, and Z^ with i, j G {0,1,2,3,4, 5} and l G {0,1}. Consider the set S of polynomials in the variables Xj, Yj, and Z^
S := {Eqf, Eqf, EqqY : a, 7 G {0,1, 2, 3,4, 5}}.
q q4	q3
By eliminating from S the variables X5, X4, X3, and X2 using Eq1, Eqj, Eqj , and Eqj respectively we obtain
X0Y1 (Z1 Yq2 Y2 — Z1Y0Y22 — Z1Y0 + Z1Y2 — Z0Z2 — Z2) =
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By the conditions on b and k, X0Yi = 0 and therefore
P := ZiYo2Y2 - Z1Y0Y2 - Z1Y0 + ZiY2 - Zo2Z2 - Z2 =0. We eliminate Z1 in S using P, obtaining, w.r.t. b, k, and
bkq2+1(k - kq)(k + kq)(kq2+1 - 1)(kq2+1 + 1) = 0,
a contradiction to k G Fq2.
Consider now the last three equations of (3.4), i.e.
Eq1 : ¿bq3 + (kq-1 - ¿kq2+q)bq5 = 0 Eq2 : ¿kq2-qbq + (kq2-q + 1)bq3 + kq2-1bq5 = 0 Eq3 : ¿2bq + (k-q+1 + ¿2k-q2+1 )bq3 + bq5 = 0.
As before, we will consider the q-th powers of Eq1, Eq2, and Eq3 replacing bqi, kqj, and (respectively) by Xj, Yj, and Z£ with i, j G {0,1,2,3,4, 5} and l G {0,1}. Consider the set S of polynomials in the variables Xj, Yj and Z^
S := {Eqf, Eq2^, Eqf : a,£,7 G {0,1, 2, 3,4, 5}}.
4	3
We eliminate in S the variables X5, X4, X3, and X2 using Eq1, Eqf, Eqf , and Eqf respectively, and we get
yoxo(z1yo2y22 +2z1yoy2y2 + 2Z1Y0Y2 + Z1Y2 - Y,2Y2 - yoy2y; - Y0Y2 - Y2) = 0.
Since b = 0 and k G Fq2, X0 Y0 = 0 and therefore
P := Z^Y2 + 2z1yoy12y2 + 2Z1Y0Y2 + Z1Y2 - Y2Y2 - Y0Y2 Y> - Y0Y2 - Y2 = 0.
Once again we consider the resultants of the polynomials in S and P w.r.t. Z1 and we obtain
bkq2+2q(k - kq)(k + kq)(kq2+1 - 1)(kq2+1 + 1) = 0, a contradiction to k G Fq2.	□
As a consequence of the above considerations and Propositions 3.5 and 3.7, we have the following.
Corollary 3.8. If h G Fq2, then is not PTL-equivalent to any known scattered linear set in PG(1,q6).
3.3 Lh may be defined by a trinomial
Suppose that h G Fq2, then the condition on h becomes hq+1 = -1. For such h we can prove that the linear set can be defined by the q-polynomial (h-1 - 1)xq + xq + (h - 1)xq5.
Proposition 3.9. If h G Fq2, then the linear set is PTL-equivalent to
Ltri := {<(*, (h-1 - 1)xq + xq3 + (h - 1)xq5)>Fq6 : x G F^}.
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Proof. Let A = (acbd) e GL(2, q6) with a = -h + h-1,b = 1,c = h-1 - 1 -h3 + h2 and d = h - h2 - 1. Straightforward computations show that the subspaces Uh and U(h_1_1)xq +(h-1)x,5 are rL(2, q6)-equivalent under the action of the matrix A. Hence, the linear sets Lh and Ltri are PrL-equivalent.	□
The fact that Lh can also be defined by a trinomial will help us to completely close the equivalence issue for Lh when h e Fq2. Indeed, we can prove the following:
Proposition 3.10. If h e Fq2, then the linear set Lh is PrL-equivalent to some L| (S2 + S =1) if and only if h e Fq and q is a power of 5.
Proof. Recall that by [27, Proposition 5.5] if h e Fq and q is a power of 5, then Lh is PrL-equivalent to some L|. As in the proof of Proposition 3.6, by Lemma 3.3 we have to check whether U(h-1-1)xq +xq3 +(h-1)x,5 is rL-equivalent either to UJ4 or to (Uj1)^. Suppose that there exist p e Aut(Fq6) and an invertible matrix (ad) such that for each x e Fq6 there exists z e Fq6 satisfying
a b\ (	xp
c d) \(hrp - 1)xpq + xpq3 + (hp - 1)xpqV \zq + zq3 +
Let k = hp, for which kq+1 = — 1. As in Proposition 3.5, we obtain a polynomial identity, whence
'c = bq (kq — 1) + bq3 + ¿6q5 (k-q — 1) d(k-1 — 1) = aq
0 = bq (k-q — 1) + bq3 (kq — 1) + bq5 J
J q3	(3.5)
d = aq
0 = bq + bq3 (k-q — 1) + bq5 (kq — 1)J ^d(k — 1) = Jaq5.
By subtracting the fifth equation from the third equation raised to q2, we get
bq = bq5 (kq — 1),
i.e. either b = 0 or kq — 1 = (bq)q4-1, whence we get either b = 0 or Nq6/q2 (kq — 1) = 1. If b = 0, since k — 1 G Fq2 and Nq6/q2 (k — 1) = (k — 1)3 = 1, then
k3 — 3k2 + 3k — 2 = 0
and, since Nq6/q2 (kq — 1) = 1 and kq = —1/k,
2k3 + 3k2 + 3k +1 = 0,
from which we get
9k2 — 3k + 5 = 0.	(3.6)
If k G Fq then k and kq are the solutions of (3.6) and
—1 = kq+1 = 5,
9'
which holds if and only if q is a power of 7. By (3.6) it follows that k G Fq, a contradiction.
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(3.7)
• If k € Fq, then k2 = -1 and by (3.6) we have k = -4/3, which is possible if and only if q is a power of 5.
Hence, if either k € Fq or k € Fq with q not a power of 5, we have that b = 0 and hence c = 0, a = 0 and d = 0.
By combining the second and the fourth equation of (3.5), we get Nq6/q2 (k-1 - 1) = 1 and, since kq = — 1/k, Nq6/q2 (kq + 1) = -1. Arguing as above, we get a contradiction whenever k € Fq or k € Fq with q not a power of 5.
Now, suppose that there exist p € Aut(Fq6) and an invertible matrix (C bb) such that for each x € Fq6 there exists z € Fq6 satisfying
b\ /	xp	\ = /	z
d)	- 1)xpq + xpq3 + (hp - 1)xpq5y = + zq3 + zq5
Let k = hp. As before, we get the following equations
'c = ¿bq(kq - 1) + bq3 + bq5 (k-q - 1) d(k-1 - 1) = ¿aq 0 = ¿bq (k-q - 1) + bq3 (kq - 1) + bq5 d = aq3
0 = ¿bq + bq3 (k-q - 1) + bq5 (kq - 1) kd(k - 1) = aq5.
By subtracting the fifth equation from the third raised to q2 of the above system we get
bq = bq3 (k-q - 1).
If b = 0, then Nq6/q2 (k-q - 1) = 1. Hence, arguing as above, we get that b = 0 and hence c = 0, a, d = 0. By combining the fourth equation with the second and the fifth equation of (3.7) we get Nq6/q2 (k - 1) = 1, which yields again to a contradiction when k € Fq or k € Fq with q not a power of 5.	□
So, as a consequence of Corollary 3.8 and of the above proposition, we have the following result.
Corollary 3.11. Apart from the case h € Fq and q a power of 5, the linear set is not PTL-equivalent to any known scattered linear set in PG(1, q6).
By Proposition 3.9, when h € Fq2, is a linear set of the family presented in [23, Section 7]. Also, we get an extension of [21, Table 1], where it is shown examples of scattered linear sets which could generalize the family presented in [10]. We do not know whether the linear set Lh, for each h € Fq6 \ Fq2 with hq +1 = -1, may be defined by a trinomial or not.
4 New MRD-codes
Delsarte in [13] (see also [14]) introduced in 1978 rank metric codes as follows. A rank metric code (or RM-code for short) C is a subset of the set of m x n matrices F^*" over Fq equipped with the distance function
d(A, B) = rk (A - B)
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for A, B G F^". The minimum distance of C is
d = min{d(A, B) : A, B G C, A = B}.
We will say that a rank metric code of F^" with minimum distance d has parameters (m, n, q; d). When C is an Fq-subspaceof F^", we say that C is Fq-linear. In the same paper, Delsarte also showed that the parameters of these codes fulfill a Singleton-like bound, i.e.
|C| qmax{m,n}(min{ml"}-d+1)
When the equality holds, we call C a maximum rank distance (MRD for short) code. We will consider only the case m = n and we will use the following equivalence definition for codes of F^™. Two Fq-linear RM-codes C and C' are equivalent if and only if there exist two invertible matrices A, B G F^™ and a field automorphism a such that {ACaB : C GC} = C', or {ACB : C gC} = C', where T denotes transposition. Also, the left and right idealisers of C are L(C) = {A G GL(m, q) : AC C C} and R(C) = {B G GL(m, q) : CB C C} [17, 19]. They are important invariants for linear rank metric codes, see also [15] for further invariants.
In [24, Section 5] Sheekey showed that scattered Fq-linear sets of PG(1, q") of rank n yield Fq-linear MRD-codes with parameters (n, n, q; n - 1) with left idealiser isomorphic to Fqn; see [7, 8, 25] for further details on such kind of connections. We briefly recall here the construction from [24]. Let Uf = {(x, f (x)) : x G Fqn} for some scattered q-polynomial f (x). After fixing an Fq-basis for Fqn we can define an isomorphism between the rings End(Fqn, Fq) and F"x". In this way the set
Cf := {x ^ af (x) + bx : a, b G Fqn}
corresponds to a set of n x n matrices over Fq forming an Fq-linear MRD-code with parameters (n, n, q; n - 1). Also, since Cf is an Fqn-subspace of End(Fqn, Fq) its left idealiser L(Cf) is isomorphic to Fqn. For further details see [6, Section 6].
Let Cf and Ch be two MRD-codes arising from maximum scattered subspaces Uf and Uh of Fqn x Fqn. In [24, Theorem 8] the author showed that there exist invertible matrices A, B and a G Aut(Fq) such that ACJB = Ch if and only if Uf and Uh are rL(2, q")-equivalent
Therefore, we have the following.
Theorem 4.1. The Fq-linear MRD-code Cfh arising from the Fq-subspace Wh has parameters (6,6, q; 5) and left idealiser isomorphic to Fq6, and is not equivalent to any previously known MRD-code, apart from the case h G Fq and q a power of 5.
Proof. From [6, Section 6], the previously known Fq-linear MRD-codes with parameters (6,6, q; 5) and with left idealiser isomorphic to Fq6 arise, up to equivalence, from one of the maximum scattered subspaces of Fq6 x Fq6 described in Section 3. From Corollaries 3.8 and 3.11 the result then follows.	□
ORCID iDs
Daniele Bartoli t> https://orcid.org/0000-0002-5767-1679 Corrado Zanella © https://orcid.org/0000-0002-5031-1961 Ferdinando Zullo © https://orcid.org/0000-0002-5087-2363
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