 # A new family of maximum scattered linear sets in PG(1,q^6)

We generalize the example of linear set presented by the last two authors in "Vertex properties of maximum scattered linear sets of PG(1,q^n)" (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 are new. This solves an open problem posed in "Vertex properties of maximum scattered linear sets of PG(1,q^n)" (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).

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## 1. Introduction

Let , where

is a vector space of dimension

over . If is a -dimensional -subspace of , then the -linear set is defined as

 LU={⟨u⟩Fqn:u∈U∖{0}},

and we say that has rank . Two linear sets and of are said to be -equivalent if there is an element in such that . It may happen that two –linear sets and of are -equivalent even if the -vector subspaces and are not in the same orbit of (see [11, 5] for further details). In this paper we focus on maximum scattered -linear sets of , that is, -linear sets of rank in of size .

If is not contained in the linear set of rank of (which we can always assume after a suitable projectivity), then for some linearized polynomial (or -polynomial) . In this case we will denote the associated linear set by . If is scattered, then is called a scattered -polynomial; see .

The first examples of scattered linear sets were found by Blokhuis and Lavrauw in  and by Lunardon and Polverino in  (recently generalized by Sheekey in ). Apart from these, very few known examples are known, see Section 3.

In [23, Sect. 5], Sheekey established a connection between maximum scattered linear sets of 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  and , it seems that examples of maximum scattered linear sets are rare.

In this paper we will prove that any

 (1.1) fh(x)=hq−1xq−hq2−1xq2+xq4+xq5,h∈Fq6,hq3+1=−1,q odd

is a scattered -polynomial. This will be done by considering two cases:
Case 1: , that is, ; the condition implies .
Case 2: . In this case , otherwise and then we have , a contradiction to .

Note that in Case 1, this example coincides with the one introduced in , where it has been proved that is scattered for and . In Corollary 3.11 we will prove that the linear set associated with is new, apart from the case of a power of five and . This solves an open problem posed in .

Finally, in Section 4 we prove that the -linear MRD-codes with parameters arising from linear sets are not equivalent to any previously known MRD-code, apart from the case and a power of ; see Theorem 4.1.

## 2. Lh is scattered

A -polynomial (or linearized polynomial) over is a polynomial of the form

 f(x)=t∑i=0aixqi,

where and is a positive integer. We will work with linearized polynomials of degree less than or equal to . For such a kind of polynomial, the Dickson matrix111This is sometimes called autocirculant matrix. is defined as

 M(f):=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝a0a1…an−1aqn−1aq0…aqn−2⋮⋮⋮⋮aqn−11aqn−12…aqn−10⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠∈Fn×nqn,

where for .

Recently, different results regarding the number of roots of linearized polynomials have been presented, see [4, 7, 21, 22, 25]. In order to prove that a certain polynomial is scattered, we make use of the following result; see [4, Theorem 3.4].

###### Theorem 2.1.

Let be the Dickson matrix associated with a nonzero -polynomial over . Denote by the submatrix of obtained by considering the last columns and the first rows of , with . Then if and only if

 det(M0(f))=det(M1(f))=…=det(Mt−1(f))=0, and det(Mt(f))≠0.

### 2.1. Case 1

Let

 f(x)=xq−xq2+xq4+xq5∈Fq6[x].

By Theorem 2.1, is scattered if and only if for each the determinants of the following two matrices do not vanish at the same time

 (2.1) M5(m)=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝1−1011mq1−1011mq21−1011mq31−1011mq41⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠,M6(m)=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝m1−10111mq1−10111mq21−10011mq31−1−1011mq411−1011mq5⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠.
###### Theorem 2.2.

The polynomial is scattered if and only if .

###### Proof.

If is even, then for the matrix has rank two and is not scattered.

Suppose now . Then consider such that . So and, by direct checking,

 det(M5(¯¯¯¯¯m))=(¯¯¯¯¯m2+4)2=0,det(M6(¯m))=−(¯¯¯¯¯m2+4)3=0

and is not scattered.

Assume and suppose that is not scattered. Then there exists such that

 (detM5(m0))qs=0, (detM6(m0))qt=0, s,t=0,1,2,3,4,5.

Consider

 (2.2) P1=det⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝1−1011Y1−1011Z1−1011U1−1011V1⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠,P2=det⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝X1−10111Y1−10111Z1−10011U1−1−1011V11−1011W⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠.

Therefore,

 (2.3) X=m0, Y=mq0, …, W=mq50

is a root of , and of the polynomials inductively defined by

 P(j)i=P(j−1)i(Y,Z,U,V,W,X),j=1,2,3,4,5,i=1,2.

One obtains a set of twelve equations in having a nonempty zero set. The following arguments are based on the fact that taking the resultant of two polynomials in with respect to any variable, the equations admit the same solutions.

We have

 (2.4) P1=YZUV−YZU−2YZ+2YU+4Y−ZUV+2ZV−2UV+4V+16=0.

Consider the following resultants:

 Q1 := ResV(P(3)1,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(P(4)1,P1)=XYZW−XYZ−XYW+2XZ −2XW−2YZ+2YW+4Z+4W+16, Q3 := ResV(P(5)1,P1)=XYZU−XYZ−2XY+2XZ +4X−YZU+2YU−2ZU+4U+16.

They all must be zero, as well as

 (2.5) ResW(ResU(Q1,Q3),Q2)=8(YZ−4)(Y2+4)(X−Z)(XZ+4)(XY−4).

We distinguish a number of cases.

1. Suppose that . Since , . So

 P1=X4−2X3+8X+16

and the resultant between and with respect to is and then (2.3) is not a root of , a contradiction.

2. Condition is clearly equivalent to . This means that , . Therefore, by (2.4) we get and we proceed as above.

3. Case . In this case , , , , and therefore and we can proceed as above.

4. Condition implies and so and . By substituting in and ,

 X3Y3+3X3Y−6X2Y2−12X2+3XY3+24XY−12Y2−64 =0, X2Y2−X2Y+2X2−XY2−4XY+4X+2Y2+4Y+16 =0.

Eliminating from these two equations one gets

 8(X2+4)6=0,

and so . We proceed as in the previous cases.

This proves that such does not exist and the assertion follows. ∎

### 2.2. Case 2

We apply the same methods of Section 2.1. In the following preparatory lemmas (and in the rest of the paper) is a power of an arbitrary prime .

###### Lemma 2.3.

Let be such that , . Then

1. ;

2. ;

3. , if is odd;

4. implies and or , , .

###### Proof.

The first three are easy computations. Consider now

 h4q2+4+14h2q2+2q+2+h4q=0.

If the equation above implies .

Assume now . Since , it is equivalent to

 (hq2−q+1)4+14(hq2−q+1)2+1=0,

that is . Let . Note that belongs to . We distinguish two cases.

• . Then

 −1=hq3+1=(hq2−q+1)q+1=(±√z)q+1=z=−7±4√3,

a contradiction if . Also, , is an even power of , and .

• . Then

 −1=hq3+1=(hq2−q+1)q+1=(±√z)q+1=−z=7∓4√3,

###### Lemma 2.4.

Let be such that , . If a root 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∈Fq6[T]

belongs to then one of the following cases occurs:

• , ; or

• , , ; or

• ; or

• .

###### Proof.

First, note that would imply which is impossible by Lemma 2.3. Therefore and , where

 ℓ1(X) = −(hq2+1−1)(hq2+1X+h2q+hq2+q) m1(X) = h(hqX+hq2+q+1+h2q2+1) ℓ2(X) = −(hq+h)(2hq2+q+1X+h2q2+q+2+h3q2+2+h3q+hq2+2q) m2(X) = hq+1(h2q2+2X+h2qX+2hq2+2q+1+2h2q2+q+1) ℓ3(X) = (hq+h)q(3h2q2+q+2X+h3qX+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) ℓ4(X) = (hq2+1−1)(h4q2+4X+6h2q2+2q+2X+h4qX+4h3q2+2q+3+4h4q2+q+3+4hq2+4q+1+4h2q2+3q+1) m4(X) = hq2(4h3q2+q+3X+4hq2+3q+1X+h4q2+q+4+h5q2+4+6h2q2+3q+2+6h3q2+2q+2+h5q+hq2+4q) ℓ5(X) = −(hq+h)(h5q2+5X+10h3q2+2q+3X+5hq2+4q+1X+5h4q2+2q+4+5h5q2+q+4+10h2q2+4q+2 +10h3q2+3q+2+h6q+hq2+5q) m5(X) = 5h4q2+q+4X+10h2q2+3q+2X+h5qX+h5q2+q+5+h6q2+5+10h3q2+3q+3+10h4q2+2q+3 +5hq2+5q+1+5h2q2+4q+1 ℓ6(X) = (hq+h)q(6h5q2+q+5X+20hq3+3q+3X+6Xhq2+5q+1+h6q2+q+6+h7q2+6+15h4q2+3q+4 +15h5q2+2q+4+15h2q2+5q+2+15h3q2+4q+2+h7q+hq2+6q) m6(X) = h6q2+6X+15h4q2+2q+4X+15h2q2+4q+2X+hq6X+6h5q2+2q+5+6h6q2+q+5+20h3q2+4q+3 +20h4q2+3q+3+6hq2+6q+1+6h2q2+5q+1.

Since , in particular

 (h2q2+2+h2q)(h4q2+4+14h2q2+2q+2+h4q)(hq2−hq)(σ+hq+hq2)(σ−hq−hq2)=0.

The claim follows from Lemma 2.3. ∎

###### Lemma 2.5.

Let be such that , . If a root of the polynomial

 hq+1Tq2+1+(hq+h)q+1∈Fq6[T]

belongs to then

 σ=±(hq2+hq).
###### Proof.

If then , a contradiction to Lemma 2.3. So we can suppose . Then

 σq2 = −(hq−1+1)q+1σ σq4 = (hq−1+1)q3+q2−q−1σ σq6 = −(hq−1+1)q5+q4−q3−q2+q+1σ=(hq+h)2qσ.

So, . ∎

Let be such that , . By Theorem 2.1 the polynomial

 fh(x)=hq−1xq−(hq2−1)xq2+xq4+xq5

is scattered if and only if for each the determinant of the following two matrices do not vanish at the same time

 (2.6) M6(m)=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝mhq−1−hq2−10111mqhq2−qh−q−10111mq2−h−q2−1h−q2−q0011mq3h1−q−h1−q2hq+1011mq4hq−q2−hq2+1hq2+q011mq5⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠,
 (2.7) M5(m)=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝hq−1−hq2−1011mqhq2−qh−q−1011mq2−h−q2−1h−q2−q011mq3h1−q−h1−q2011mq4hq−q2⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠.
###### Theorem 2.6.

Let , , be such that . Then the polynomial is not scattered.

###### Proof.

Consider . So,

 ¯¯¯¯¯mq=1/h+hq2,¯¯¯¯¯mq2=1/hq+1/h,¯¯¯¯¯mq3=1/hq2+1/hq,¯¯¯¯¯mq4=h+1/hq2,¯¯¯¯¯mq5=hq+h.

By direct checking, in this case, both and therefore is not scattered. ∎

###### Theorem 2.7.

Let , , , be such that and . Then the polynomial is scattered.

###### Proof.

First we note that since is odd, , and . Suppose that is not scattered. Then for some . Consider

 X=m0,Y=mq0,Z=mq20,U=mq30,V=mq40,W=mq50.

With a procedure similar to the one in the proof of Theorem 2.2, we will compute resultants starting from the polynomials associated with , , and . Eliminating using and using , one gets from

 hq2+2q+1φ1(X,Y)φ2(X,Y,Z,V)φ3(X,Y,Z,V)=0,

where

 φ1(X,Y) = hq+1XY+h2q2+2X−hq2+1X+hq2+q+2Y+h2q2+2Y+hq2+2q+1+h2q2+q+1−h2q−hq2+q; φ2(X,Y,Z,V)