3 Examples of simple and non-simple linear sets of PG(1, q

Classes and equivalence of linear sets in PG(1, q

)

Bence Csajb´ok, Giuseppe Marino and Olga Polverino^∗

Abstract

The equivalence problem ofFq-linear sets of rank nof PG(1, qⁿ) is investigated, also in terms of the associated variety, projecting configu- rations,Fq-linear blocking sets of R´edei type and MRD-codes. We call anFq-linear setLU of rank nin PG(W,Fqⁿ) = PG(1, qⁿ)simple if for any n-dimensional Fq-subspace V of W, LV is PΓL(2, qⁿ)-equivalent to LU only when U and V lie on the same orbit of ΓL(2, qⁿ). We prove that U = {(x,Tr_qn/q(x)) : x ∈ Fqⁿ} defines a simple Fq-linear set for each n. We provide examples of non-simple linear sets not of pseudoregulus type for n > 4 and we prove that allFq-linear sets of rank 4 are simple in PG(1, q⁴).

1 Introduction

Linear sets are natural generalizations of subgeometries. Let Λ = PG(W,Fqⁿ)

=P G(r−1, qⁿ), whereW is a vector space of dimensionr overFqⁿ. A point setLof Λ is said to be an Fq-linear set of Λ of rankk if it is defined by the non-zero vectors of ak-dimensional Fq-vector subspaceU ofW, i.e.

L=L_U ={hui_Fqn:u∈U \ {0}}.

The maximum field of linearity of an Fq-linear set LU is Fq^t ift | n is the largest integer such thatL_U is anFq^t-linear set. In the recent years, starting from the paper [20] by Lunardon, linear sets have been used to construct or characterize various objects in finite geometry, such as blocking sets and mul- tiple blocking sets in finite projective spaces, two-intersection sets in finite

∗The research was supported by Ministry for Education, University and Research of Italy MIUR (Project PRIN 2012 ”Geometrie di Galois e strutture di incidenza”) and by the Italian National Group for Algebraic and Geometric Structures and their Applications (GNSAGA - INdAM). The first author was partially supported by the J´anos Bolyai Re- search Scholarship of the Hungarian Academy of Sciences and by OTKA Grant No. K 124950.

projective spaces, translation spreads of the Cayley Generalized Hexagon, translation ovoids of polar spaces, semifield flocks and finite semifields. For a survey on linear sets we refer the reader to [27], see also [16].

One of the most natural questions about linear sets is their equivalence.

Two linear setsL_U andL_V of PG(r−1, qⁿ) are said to be PΓL-equivalent (or simplyequivalent) if there is an elementϕin PΓL(r, qⁿ) such thatL^ϕ_U =L_V. In the applications it is crucial to have methods to decide whether two linear sets are equivalent or not. Forf ∈ΓL(r, qⁿ) we have L_Uf =L^ϕ_U^f, where ϕ_f denotes the collineation of PG(W,Fqⁿ) induced by f. It follows that if U andV areFq-subspaces ofW belonging to the same orbit of ΓL(r, qⁿ), then L_U and L_V are equivalent. The above condition is only sufficient but not necessary to obtain equivalent linear sets. This follows also from the fact thatFq-subspaces ofW with different ranks can define the same linear set, for example Fq-linear sets of PG(r−1, qⁿ) of rank k ≥ rn−n+ 1 are all the same: they coincide with PG(r−1, qⁿ). As it was showed recently in [7], ifr = 2, then there existFq-subspaces ofW of the same ranknbut on different orbits of ΓL(2, qⁿ) defining the same linear set of PG(1, qⁿ).

This observation motivates the following definition. AnFq-linear setL_U of PG(W,Fqⁿ) = PG(r−1, qⁿ) with maximum field of linearityFq is called simple if for each Fq-subspace V of W, L_U = L_V only if U and V are in the same orbit of ΓL(r, qⁿ) or, equivalently, if for each Fq-subspace V ofW, LV is PΓL(r, qⁿ)-equivalent to LU only if U and V are in the same orbit of ΓL(r, qⁿ).

Natural examples of simple linear sets are the subgeometries (cf. [19, Theorem 2.6] and [15, Section 25.5]). In [6] it was proved that Fq-linear sets of rank n+ 1 of PG(2, qⁿ) admitting (q+ 1)-secants are simple. This allowed the authors to translate the question of equivalence to the study of the orbits of the stabilizer of a subgeometry on subspaces and hence to obtain the complete classification of Fq-linear blocking sets in PG(2, q⁴).

Until now, the only known examples of non-simple linear sets are those of pseudoregulus type of PG(1, qⁿ) for n≥5 and n6= 6, see [7].

In this paper we focus on linear sets of rank n of PG(1, qⁿ). We first introduce a method which can be used to find non-simple linear sets of rank nof PG(1, qⁿ). LetL_U be a linear set of rank nof PG(W,Fqⁿ) = PG(1, qⁿ) and let β be a non-degenerate alternating form of W. Denote by ⊥ the orthogonal complement map induced by Tr_qⁿ_/q◦β on W (considered as an Fq-vector space). Then U and U^⊥ defines the same linear set (cf. Result 2.1) and if U and U^⊥ lie on different orbits of ΓL(W,Fqⁿ), then L_U is non- simple. Using this approach we show that there are non-simple linear sets

of ranknof PG(1, qⁿ) forn≥5, not of pseudoregulus type (cf. Proposition 3.10). Contrary to what we expected initially, simple linear sets are harder to find than non-simple linear sets. We prove that the linear set of PG(1, qⁿ) defined by the trace function is simple (cf. Theorem 3.7). We also show that linear sets of rank nof PG(1, qⁿ) are simple forn≤4 (cf. Theorem 4.5).

Moreover, in PG(1, qⁿ) we extend the definition of simple linear sets and introduce the Z(ΓL)-class and the ΓL-class for linear sets of rank n. In Section 5 we point out the meaning of these classes in terms of equivalence of the associated blocking sets, MRD-codes and projecting configurations.

2 Definitions and preliminary results

2.1 Dual linear sets with respect to a symplectic polarity of a line

For α ∈ Fqⁿ and a divisor h of n we will denote by Tr_qⁿ_/q^h(α) the trace of α over the subfield F_q^h, that is, Tr_qⁿ_/q^h(α) = α +α^qh +. . .+α^qn−h. By N_qn/q^h(α) we will denote the norm of α over the subfield Fq^h, that is, N_qn/q^h(α) =α^{1+qh+...+qn−h}. Since in the paper we will use only norms over Fq, the function N_qⁿ_/q will be denoted simply by N.

Starting from a linear set L_U of PG(r, qⁿ) and using a polarity τ of the space it is always possible to construct another linear set, which is calleddual linear set ofL_U with respect to the polarityτ (see [27]). In particular, letL_U be anFq–linear set of ranknof a line PG(W,Fqⁿ) and letβ:W×W −→Fqⁿ

be a non-degenerate reflexive Fqⁿ–sesquilinear form on the 2-dimensional vector spaceW overFqⁿ determining a polarity τ. The map Tr_qⁿ_/q◦β is a non-degenerate reflexiveFq–sesquilinear form onW, whenW is regarded as a 2n-dimensional vector space over Fq (see [14]).

Let ⊥_β and ⊥⁰_β be the orthogonal complement maps defined by β and Tr_qⁿ_/q◦β on the lattices of the Fqⁿ-subspaces and Fq-subspaces of W, re- spectively. The dual linear set of LU with respect to the polarity τ is the Fq–linear set of ranknof PG(W,Fqⁿ) defined by the orthogonal complement U^⊥0β and it will be denoted byL^τ_U. Also, up to projective equivalence, such a linear set does not depend onτ [27, Proposition 2.5].

For a point P =hzi_Fqn ∈ PG(W,Fqⁿ) the weight of P with respect to the linear setLU is wLU(P) := dimq(hzi_Fqn ∩U).

Result 2.1. From [27, Property 2.6] (withr = 2,s= 0 andt=n) it can be easily seen that ifL_U is anFq–linear set of ranknof a line PG(1, qⁿ) andL^τ_U

is its dual linear set with respect to a polarityτ, then wL^τ_U(P^τ) =wLU(P) for each pointP ∈PG(1, qⁿ). Ifτ is a symplectic polarity of a line PG(1, qⁿ), thenP^τ =P and henceL_U =L^τ_U =L

U^⊥0β. 2.2 F^q-linear sets of PG(1, qⁿ) of class r

In this paper we investigate the equivalence ofFq-linear sets of ranknof the projective line PG(W,Fqⁿ) = PG(1, qⁿ). The first step is to determine the Fq-vector subspaces ofW defining the same linear set. This motivates the definition of theZ(ΓL)-class and ΓL-class of a linear setL_U of PG(1, qⁿ) (cf.

Definitions 2.4 and 2.5). The next proposition relies on the characterization of functions over Fq determining few directions. It states that the Fq-rank ofL_U of PG(1, qⁿ) is uniquely defined when the maximum field of linearity of LU isFq. This will allow us to state our definitions and results without further conditions on the rank of the correspondingFq-subspaces.

For anFq toFq functionf, the set of directions determined byf is D_f :=

f(x)−f(y)

x−y :x, y∈Fqⁿ, x6=y

.

Theorem 2.2 (Ball et al. [3] and Ball [1]). Let f be a function from Fq

to Fq, q =p^h, and let N be the number of directions determined by f. Let s=p^e be maximal such that any line with a direction determined by f that is incident with a point of the graph of f is incident with a multiple of s points of the graph of f. Then one of the following holds.

1. s= 1 and (q+ 3)/2≤N ≤q+ 1, 2. e|h, q/s+ 1≤N ≤(q−1)/(s−1), 3. s=q and N = 1.

Moreover if s >2, then the graph of f is Fs-linear.

Proposition 2.3. LetLU be an Fq-linear set ofPG(W,Fqⁿ) = PG(1, qⁿ)of rank n. The maximum field of linearity ofL_U is Fq^d, where

d= min{w_LU(P) :P ∈LU}.

If the maximum field of linearity ofLU isFq, then the rank ofLU as anFq- linear set is uniquely defined, i.e. for eachFq-subspaceV of W ifL_U =L_V, thendim_q(V) =n.

Proof. We first note that since LU is an Fq-linear set of PG(1, qⁿ) of rank n, then|L_U| ≤(qⁿ−1)/(q−1) and henceL_U 6= PG(1, qⁿ).

Since the action of ΓL(2, qⁿ) preserves the maximum field of linearity and the weight of points, we can assume, up to the action of ΓL(2, qⁿ), that U = {(x, f(x)) : x ∈ Fqⁿ} for some q-polynomial f over Fqⁿ. Since f is linear,|L_U|is the size of the set of directions determined byf. Also, a line` with slopem meets the graph off inq^tpoints, wheret=wLU(h(1, m)i_Fqn), i.e.

z∈F^∗qⁿ:f(z)/z=m =q^t−1.

Let d = min{w_LU(P) : P ∈ LU}. If q = p^e, p prime, then p^de is the largestp-power such that every line with a determined direction that meets the graph off meets the graph of f in a multiple of s=p^de points. Then Theorem 2.2 yields that eithers=qⁿ and f(x) = λxfor some λ∈Fqⁿ, or Fq^d is a proper subfield ofFqⁿ and

q^n−d+ 1≤ |L_U| ≤ qⁿ−1

q^d−1. (1)

Moreover, ifq^d>2, then f is Fq^d-linear. In our case we already know that f isFq-linear, so even in the caseq^d= 2 it follows thatU is anFq^d-subspace ofW and henceLU is anF_q^d-linear set.

We show that Fq^d is the maximum field of linearity of LU. Suppose, contrary to our claim, thatL_U isFq^r-linear of rankz for somer > d. Then LU is also Fq-linear of rank rz. It follows that rz ≤ n since otherwise LU = PG(1, qⁿ). Then for the size ofLU we get|L_U| ≤(q^rz−1)/(q^r−1)≤ (qⁿ−1)/(q^r−1), and this number is less than the lower bound in (1). This showsr=d.

Now suppose that Fq is the maximum field of linearity of LU and let V be an r-dimensional Fq-subspace of W such that L_U =L_V. We cannot have r > n since L_U 6= PG(1, qⁿ). Suppose, contrary to our claim, that r ≤ n−1. Then |L_U| ≤ (qⁿ⁻¹−1)/(q−1) contradicting (1) which gives qⁿ⁻¹+ 1≤ |L_U|. This concludes the proof.

Now we can give the following definitions of classes of anFq-linear set of a line.

Definition 2.4. Let L_U be an Fq-linear set of PG(W,Fqⁿ) = PG(1, qⁿ) of ranknwith maximum field of linearityFq. We say thatL_U is ofZ(ΓL)-class r if r is the largest integer such that there exist Fq-subspaces U1, U2, . . . , Ur

of W with L_Ui = L_U for i ∈ {1,2, . . . , r} and Ui 6= λUj for each λ ∈ F^∗qⁿ

and for eachi6=j,i, j ∈ {1,2, . . . , r}.

Definition 2.5. Let LU be an Fq-linear set of PG(W,Fqⁿ) = PG(1, qⁿ) of rank n with maximum field of linearity Fq. We say that L_U is of ΓL-class sif s is the largest integer such that there exist Fq-subspaces U₁, U₂, . . . , U_s of W withLUi =LU for i∈ {1,2, . . . , s} and there is no f ∈ΓL(2, qⁿ) such thatUi =U_j^f for each i6=j, i, j∈ {1,2, . . . , s}.

Simple linear sets (cf. Section 1) of PG(1, qⁿ) are exactly those of ΓL- class one. The next proposition is easy to show.

Proposition 2.6. Let LU be an Fq-linear set of PG(1, qⁿ) of rank n with maximum field of linearityFq and letϕbe a collineation ofPG(1, qⁿ). Then L_U and L^ϕ_U have the sameZ(ΓL)-class and ΓL-class.

Remark 2.7. Let L_U be an Fq-linear set of rank n of PG(1, qⁿ) with ΓL- class sand letU₁, U₂, . . . , U_s beFq-subspaces belonging to different orbits of ΓL(2, qⁿ) and definingLU. The PΓL(2, qⁿ)-orbit of LU is the set

s

[

i=1

{LU_i^f:f ∈ΓL(2, qⁿ)}.

3 Examples of simple and non-simple linear sets of PG(1, q

)

Let LU be an Fq–linear set of rank n of PG(1, qⁿ). We can always assume (up to a projectivity) thatL_U does not contain the pointh(0,1)i_Fqn. Then U = U_f ={(x, f(x)) :x ∈ Fqⁿ}, for some q-polynomialf(x) =Pn−1

i=0 a_ix^qi overFqⁿ. For the sake of simplicity we will writeLf instead ofLU_f to denote the linear set defined byU_f.

According to Result 2.1 and using the same notations as in Section 2.1 if LU is anFq-linear set of ranknof PG(1, qⁿ) andτ is a symplectic polarity, thenU^⊥0β defines the same linear set as U. Since in generalU^⊥0β and U are not equivalent under the action of the group ΓL(2, qⁿ), simple linear sets of a line are harder to find than non-simple linear sets.

Consider the non-degenerate symmetric bilinear form of Fqⁿ over Fq

defined by the following rule

< x, y >:= Tr_qⁿ_/q(xy). (2) Then theadjoint mapfˆof anFq-linear mapf(x) =Pn−1

i=0 a_ix^qi ofFqⁿ (with

respect to the bilinear formh,i) is fˆ(x) :=

n−1

X

i=0

a^q_iⁿ⁻ⁱx^qn−i. (3) Letη:F²qⁿ×F²qⁿ →Fqⁿ be the non-degenerate alternating bilinear form ofF²_qⁿ defined by

η((x, y),(u, v)) =xv−yu. (4) Thenη induces a symplectic polarity on the line PG(1, qⁿ) and

η⁰((x, y),(u, v)) = Tr_qⁿ_/q(η((x, y),(u, v))) (5) is a non-degenerate alternating bilinear form onF²qⁿ, when F²qⁿ is regarded as a 2n-dimensional vector space over Fq. We will always denote in the paper by⊥and⊥⁰ the orthogonal complement maps defined byηand η⁰ on the lattices of the Fqⁿ-subspaces and the Fq-subspaces of F²qⁿ, respectively.

Direct calculation shows that

U_f^⊥0 =U_fˆ. (6)

Result 2.1 and (6) allow us to slightly reformulate [4, Lemma 2.6].

Lemma 3.1 ([4]). Let Lf ={h(x, f(x))i_Fqn:x ∈ F^∗_qⁿ} be an Fq–linear set of PG(1, qⁿ) of rank n, with f(x) a q-polynomial over Fqⁿ, and let fˆ be the adjoint of f with respect to the bilinear form (2). Then for each point P ∈PG(1, qⁿ) we have w_Lf(P) =w_Lˆ

f(P). In particular, L_f =L_fˆand the maps defined by f(x)/xand fˆ(x)/xhave the same image.

Lemma 3.2. Letϕbe anFq-linear map ofFqⁿ and forλ∈F^∗_qⁿ letϕλdenote the Fq-linear map: x 7→ ϕ(λx)/λ. Then for each point P ∈ PG(1, qⁿ) we havew_Lϕ(P) =w_Lϕλ(P). In particular, L_ϕ=L_ϕλ.

Proof. The statements follow fromλUϕλ =Uϕ.

Remark 3.3. The results of Lemmas 3.1 and 3.2 can also be obtained via Dickson matrices. For a q-polynomial f(x) = Pn−1

i=0 a_ix^qi over Fqⁿ let D_f denote the associated Dickson matrix (or q-circulant matrix)

D_f :=











a₀ a₁ . . . an−1

a^q_n−1 a^q₀ . . . a^q_n−2 ... ... ... ... a^q₁ⁿ⁻¹ a^q₂ⁿ⁻¹ . . . a^q₀ⁿ⁻¹









 .

When f(x) = λx for some λ∈ Fqⁿ we will simply write Dλ. The rank of the matrixD_f equals the rank of theFq-linear map f, see for example [29].

We will denote the pointh(1, λ)i_qn by P_λ.

Transposition preserves the rank of matrices and D_f^T =Dfˆ, D^T_λ =Dλ. It follows that

dimqker(D_f −D_λ) = dimqker(D_f−D_λ)^T = dimqker(D_fˆ−D_λ), and hence for each λ∈Fqⁿ we have w_Lf(P_λ) =w_Lˆ

f(P_λ).

Let f_µ(x) =f(xµ)/µ. It is easy to see that D_1/µD_fD_µ=D_fµ and dimqker(D_f −D_λ) = dimqkerD_1/µ(D_f −D_λ)Dµ= dimqker(D_fµ−D_λ), and hence wL_f(P_λ) =wL_fµ(P_λ) for each λ∈Fqⁿ.

From the previous arguments it follows that linear sets L_f withf(x) = f(x) are good candidates for being simple. In the next section we show thatˆ the trace function, which has the previous property, defines a simple linear set. We are going to use the following lemmas which will also be useful later.

Lemma 3.4. Let f and g be two linearized polynomials. If L_f =L_g, then for each positive integer dthe following holds

X

x∈F^∗_qn

f(x) x

d

= X

x∈F^∗_qn

g(x) x

d

.

Proof. IfL_f =L_g=:L, then{f(x)/x:x∈F^∗qⁿ}={g(x)/x:x∈F^∗qⁿ}=:H.

For eachh∈H we have|{x:f(x)/x=h}|=qⁱ−1, whereiis the weight of the pointh(1, h)i_qⁿ ∈Lw.r.t. U_f, and similarly|{x:g(x)/x=h}|=q^j−1, where j is the weight of the point h(1, h)i_qⁿ ∈ L w.r.t. U_g. Because of the characteristic ofFqⁿ, we obtain:

X

x∈F^∗qn

f(x) x

d

=−X

h∈H

h^d= X

x∈F_qn^∗

g(x) x

d

.

For the sake of completeness we give a proof of the following well-known result.

Lemma 3.5. For any prime power q and integerdwe have P

x∈F^∗qx^d=−1 if q−1|d andP

x∈F^∗qx^d= 0 otherwise.

Proof. Letg denote a primitive element of Fq and put s=Pn−2

j=0g^id. Then sg^d=s and hence eithers= 0, org^d= 1. In the latter caseq−1|dsince g was a primitive element and hencex^d= 1 for eachx∈F^∗q.

Lemma 3.6. Let f(x) = Pn−1

i=0 a_ix^qi and g(x) = Pn−1

i=0 b_ix^qi be two q- polynomials over Fqⁿ, such that L_f =L_g. Then

a0 =b0, (7)

and for k= 1,2, . . . , n−1 it holds that

a_ka^q_n−k^k =b_kb^q_n−k^k , (8) for k= 2,3, . . . , n−1 it holds that

a1a^q_k−1a^q_n−k^k +a_ka^q_n−1a^q_n−k+1^k =b1b^q_k−1b^q_n−k^k +b_kb^q_n−1b^q_n−k+1^k . (9) Proof. We are going to use Lemma 3.5 together with Lemma 3.4 with dif- ferent choices ofd.

Withd= 1 we have X

x∈F^∗_qn n−1

X

i=0

a_ix^qi−1 = X

x∈F^∗_qn n−1

X

i=0

b_ix^qi−1,

and hence

n−1

X

i=0

a_i X

x∈F^∗_qn

x^qi−1 =

n−1

X

i=0

b_i X

x∈F^∗_qn

x^qi−1.

Sinceqⁿ−1 cannot divideqⁱ−1 withi= 1,2, . . . , n−1,a0 =b0 =:cfollows.

Letϕdenote theFqⁿ-linear map which fixes (0,1) and maps (1,0) to (1,−c).

Then U_f^ϕ =U_f⁰ and U_g^ϕ = U_g⁰ with f⁰ =Pn−1

i=1 a_ix^qi, g⁰ = Pn−1

i=1 b_ix^qi and of course withLf⁰ =Lg⁰. It follows that we may assume c= 0.

First we show that (8) holds. With d=q^k+ 1, 1≤k≤n−1 we obtain X

1≤i,j≤n−1

a_ia^q_j^k X

x∈F^∗_qn

x^{qi−1+qj+k−qk} = X

1≤i,j≤n−1

b_ib^q_j^k X

x∈F^∗_qn

x^{qi−1+qj+k−qk}.

P

x∈F^∗_qnx^{qi−1+qj+k−qk} =−1 if and only ifqⁱ+q^j+k≡q^k+ 1 (modqⁿ−1), and zero otherwise. Suppose that the former case holds.

First consider j+k≤n−1. Thenqⁱ+q^j+k ≤qⁿ⁻¹+qⁿ⁻¹ < q^k+ 1 + 2(qⁿ−1) hence one of the following holds.

• Ifqⁱ+q^j+k =q^k+ 1, then the right hand side is not divisible by q, a contradiction.

• Ifqⁱ+q^j+k=q^k+1+(qⁿ−1) =qⁿ+q^k, thenj+k=n, a contradiction.

Now consider the case j+k≥n. Then qⁱ+q^j+k≡qⁱ+q^j+k−n≡q^k+ 1 (modqⁿ−1). Sincej+k≤2(n−1), we have qⁱ+q^j+k−n≤qⁿ⁻¹+qⁿ⁻² <

q^k+ 1 + 2(qⁿ−1), hence one of the following holds.

• Ifqⁱ+q^j+k−n=q^k+ 1, then j+k=nand i=k.

• Ifqⁱ+q^j+k−n=q^k+ 1 + (qⁿ−1) =qⁿ+q^k, then there is no solution sincej+k−n /∈ {k, n}.

Hence (8) follows. Now we show that (9) also holds. Note that in this casen≥3, otherwise there is nokwith 2≤k≤n−1. Withd=q^k+q+ 1, we obtain

X

1≤i,j,m≤n−1

a_ia^q_ja^q_m^k X

x∈F^∗_qn

x^{qi−1+qj+1−q+qm+k−qk} =

X

1≤i,j,m≤n−1

b_ib^q_jb^q_m^k X

x∈F^∗_qn

x^{qi−1+qj+1−q+qm+k−qk}.

P

x∈F^∗_qnx^{qi−1+qj+1−q+qm+k−qk} =−1 if and only ifqⁱ+q^j+1+q^m+k≡q^k+q+1 (modqⁿ−1), and zero otherwise. Suppose that the former case holds.

First considerm+k≤n−1. Thenqⁱ+q^j+1+q^m+k≤qⁿ⁻¹+qⁿ+qⁿ⁻¹ <

q^k+q+ 1 + 2(qⁿ−1) hence one of the following holds.

• Ifqⁱ+q^j+1+q^m+k=q^k+q+ 1, then the right hand side is not divisible byq, a contradiction.

• Ifqⁱ+q^j+1+q^m+k=q^k+q+ 1 + (qⁿ−1) =qⁿ+q^k+q, thenm+k=n, j+ 1 =k and i= 1, a contradiction.

Now consider the case m+k ≥ n. Then qⁱ +q^j+1 +q^m+k ≡ qⁱ + q^j+1+q^m+k−n≡q^k+q+ 1 (modqⁿ−1). We have qⁱ+q^j+1+q^m+k−n ≤ qⁿ⁻¹+qⁿ+qⁿ⁻²< q^k+q+ 1 + 2(qⁿ−1) hence one of the following holds.

• Ifqⁱ+q^j+1+q^m+k−n=q^k+q+ 1, thenj+ 1 =k,i= 1 andm+k=n.

• If qⁱ +q^j+1 +q^m+k−n = q^k+q + 1 + (qⁿ−1) = qⁿ+q^k+q, then j+ 1 =n,i=kand m+k=n+ 1.

This concludes the proof.

3.1 Linear sets defined by the trace function

We show that there exists at least one simpleFq-linear set in PG(1, qⁿ) for each q and n. Let V = {(x,Tr_qⁿ_/q(x)) : x ∈Fqⁿ}. We show thatL_U =L_V occurs for an Fq-subspace U of W if and only ifV =λU for some λ∈F^∗qⁿ, i.e. LV is of Z(ΓL)-class one and hence simple.

Theorem 3.7. Let V = {(x,Tr_qⁿ_/q(x)) : x ∈ Fqⁿ}, then the Fq-linear set LV of PG(1, qⁿ) is of Z(ΓL)-class one.

Proof. Suppose L_Uf = L_V with U_f = {(x, f(x)) : x ∈ Fqⁿ} and f(x) = Pn−1

i=0 a_ix^qi. We are going to use Lemma 3.6 withg(x) = Tr_qn/q(x). The co- efficientsb0, b1, . . . , bn−1ofg(x) are 1, hencea0 = 1, and fork= 1,2, . . . , n− 1

a_ka^q_n−k^k = 1, (10)

fork= 2,3, . . . , n−1

a₁a^q_k−1a^q_n−k^k +a_ka^q_n−1a^q_n−k+1^k = 2. (11) Note that (10) impliesai 6= 0 for i= 1,2, . . . , n−1. First we prove

ai =a^1+q+...+q₁ ⁱ⁻¹ (12)

by induction onifor each 0< i < n. The assertion holds fori= 1. Suppose that it holds for some integer i−1 with 1 < i < n. We prove that it also holds for i. Then (11) with k=i gives

a₁a^q_i−1a^q_n−iⁱ +a_ia^q_n−1a^q_n−i+1ⁱ = 2. (13) Also, (10) withk=i,k=i−1 andk= 1, respectively, gives

a^q_n−iⁱ = 1/a_i, a^q_n−i+1ⁱ = 1/a^q_i−1,

a^q_n−1 = 1/a1. Then (13) gives

a1a^q_i−1/ai+ai/ a1a^q_i−1

= 2. (14)

It follows that a1a^q_i−1/ai = 1 and hence the induction hypothesis on ai−1

yieldsai =a^1+q+...+q₁ ⁱ⁻¹.

Finally we show N(a1) = 1. First consider n even. Then (10) with k = n/2 gives a^q_n/2^n/2+1 = 1. Applying (12) yields N(a₁) = 1. If n is odd, then (10) with k = (n−1)/2 gives a(n−1)/2a^q_(n+1)/2^(n−1)/2 = 1. Applying (12) yields N(a1) = 1. It follows that a1 = λ^q−1 for some λ ∈ F^∗_qⁿ and hence f(x) =Pn−1

i=0 λ^qi−1x^qi. Then λUf ={(x,Tr_qⁿ_/q(x)) : x∈F^∗_qⁿ}.

Remark 3.8. We point out that in the above theorem we do not have any assumption on the weight of points ofLU. In the special case whenLU =LV

andL_U has a point of weightn−1, then theGL(2, qⁿ)-equivalence of U and V can be deduced also from [8, Theorem 2.3].

3.2 Non-simple linear sets

An Fq-linear set of pseudoregulus type of PG(1, qⁿ) is any linear set equiv- alent to {h(x, x^q)i_Fqn:x ∈ F^∗_qⁿ}. In [7] it was proved that the ΓL-class of such linear sets isϕ(n)/2, hence they are non-simple forn= 5 and n >6.

So far, these are the only known non-simple linear sets of PG(1, qⁿ). Here we show that Fq-linear sets L_f of PG(1, qⁿ) introduced by Lunardon and Polverino, which are not of pseudoregulus type ([23, Theorems 2 and 3]), are non-simple as well. Let us start by proving the following preliminary result.

Proposition 3.9. Let f(x) =Pn−1

i=0 a_ix^qi. There is an Fqⁿ-semilinear map between U_f and U_fˆif and only if the following system of n equations has a solution A, B, C, D∈Fqⁿ, AD−BC6= 0,σ =p^k:

C+Da^σ₀ −a0A=

n−1

X

i=0

(Baia^σ_i)^qn−i, . . .

Da^σ_m−(an−mA)^qm =

n−1

X

i=0

(Baia^σ_i+m)^qn−i,with m= 1, . . . , n−2, . . .

Da^σ_n−1−(a1A)^qn−1 =

n−1

X

i=0

(Baia^σ_i+n−1)^qn−i, where the indices are taken modulo n.

Proof. Because of cardinality reasons the conditionAD−BC 6= 0 is neces- sary. Then

x f(x)ˆ

:x∈Fqⁿ

=

A B

C D

x^σ f(x)^σ

:x∈Fqⁿ

holds if and only if

Cx^σ+D

n−1

X

j=0

a^σ_jx^σqj =

n−1

X

i=0

a^q_n−iⁱ



Ax^σ +B

n−1

X

j=0

a^σ_jx^σqj





qⁱ

for each x ∈ Fqⁿ. After reducing modulo x^qn −x, this is a polynomial equation of degree at mostqⁿ⁻¹in the variablex^σ. It follows that it holds for eachx∈Fqⁿ if and only if it is the zero polynomial. Comparing coefficients on both sides yields the assertion.

We are able to prove the following.

Proposition 3.10. Consider a polynomial of the form f(x) =δx^q+x^qn−1, where q >4 is a power of the prime p. Ifn >4, then for each generator δ of the multiplicative group ofFqⁿ the linear set L_f is not simple.

Proof. Lemma 3.1 yields L_f = L_fˆ thus it is enough to show the existence of δ such that there is no Fqⁿ-semilinear map between U_f and U_fˆ. In the equations of Proposition 3.9 we have a1 = δ, an−1 = 1 and a0 = a2 = . . . = an−2 = 0. If n >4 then the first two and the last two equations of Proposition 3.9 give

C = (Bδ^σ+1)^qn−1 +B^q, Dδ^σ−A^q = 0,

0 = (Bδ)^qn−1, D−(δA)^qn−1 = 0,

whereσ=p^kfor some integerk. If there is a solution, thenB =C = 0 and (δA)^qn−1δ^σ =A^q. Taking q-th powers on both sides yield

δ^σq+1 =A^q2−1 (15)

and hence

δ

(σq+1)(qn−1)

q−1 = 1. (16)

For eachσ letGσ be the set of elementsδofFqⁿ satisfying (16). For eachσ, G_σ is a subgroup of the multiplicative groupM ofFqⁿ. We show that these are proper subgroups ofM. We haveG_pk =M if and only ifqⁿ−1 divides

(p^kq+1)(qⁿ−1)

q−1 , i.e. when q−1 divides p^kq+ 1. Since gcd(p^w+ 1, p^v−1) is always 1,2, orp^gcd(w,v)+ 1, it follows that forq >4 we cannot haveq−1 as a divisor ofp^kq+ 1.

It follows that for any generator δ of M we have δ /∈ ∪_jG_p^j and hence δ^σq+1 6=A^q2−1 for each σ and for each A.

Remark 3.11. If q = 4, then (15) with k = 2(n−1) + 1 asks for the solution of δ³ =A¹⁵. When n is odd, then{x³:x∈F4ⁿ}={x¹⁵:x∈F4ⁿ} and hence for each δ there exists A such that δ³=A¹⁵.

If q = 3, then (15) with k = n−1 asks for the solution of δ² = A⁸. When n is odd, then {x²: x ∈F3ⁿ} ={x⁸: x ∈F3ⁿ} and hence for each δ there existsA such that δ² =A⁸.

If q = 2, then (15) with k = 0 asks for the solution of δ³ = A³. This equation always has a solution.

4 Linear sets of rank 4 of PG(1, q

)

Fq-linear sets of rank two of PG(1, q²) are the Baer sublines, which are equivalent. As we have mentioned in the introduction, subgeometries are simple linear sets, in fact they haveZ(ΓL)-class one (cf. [19, Theorem 2.6]

and [15, Section 25.5]). There are two non-equivalentFq-linear sets of rank 3 of PG(1, q³), the linear sets of size q² +q + 1 and those of size q² + 1.

Linear sets in both families are equivalent, since the stabilizer of a q-order subgeometry Σ of Σ^∗ = PG(2, q³) is transitive on the set of those points of Σ^∗\Σ which are incident with a line of Σ and on the set of points of Σ^∗ not incident with any line of Σ (cf. Section 5.2 and [18]). In the first case we have the linear sets of pseudoregulus type with ΓL-class 1 andZ(ΓL)-class 2 (cf. Remark 5.6 and Example 5.1). In the second case we have the linear sets defined by Tr_q³_/q with ΓL-class andZ(ΓL)-class 1 (cf. Theorem 3.7, see also [12, Corollary 6]).

From [6, Proposition 2.3] it follows that Fq-linear sets of rank 5 in PG(W, q⁴) = PG(2, q⁴) are simple. The orbits of 5-dimensionalFq-subspaces of W under ΓL(3, q⁴) are also determined (cf. [6, pg. 54]). The results related to R´edei type blocking sets allow to determine all the orbits of 4- dimensional Fq-subspaces of a two-dimensional Fq⁴-space under the group ΓL(2, q⁴). The aim of this section is to prove thatFq-linear sets of rank 4 in

PG(1, q⁴), with maximum field of linearityFq, are simple (cf. Theorem 4.5), since this does not follow from the above mentioned simplicity ofFq-linear blocking sets. As a corollary, a list of orbits under PΓL(2, q⁴) of Fq-linear sets of rank 4 in PG(1, q⁴) can be deduced from [6, pg. 54].

4.1 Subspaces defining the same linear set Lemma 4.1. Let f(x) = P3

i=0aix^qi and g(x) = P3

i=0bix^qi be two q- polynomials over Fq⁴, such that L_f =L_g. Then

N(a1) + N(a2) + N(a3) +a^1+q₁ ²a^q+q₃ ³ +a^q+q₁ ³a^1+q₃ ²+ Trq⁴/q

a1a^q+q₂ ²a^q₃³

= N(b1) + N(b2) + N(b3) +b^1+q₁ ²b^q+q₃ ³ +b^q+q₁ ³b^1+q₃ ²+ Trq⁴/q

b1b^q+q₂ ²b^q₃³

. Proof. We are going to follow the proof of Lemma 3.6. As in that proof, we may assumea₀ =b₀ = 0. In Lemma 3.4 taked= 1 +q+q²+q³. We obtain

X

1≤i,j,k,m≤3

a_ia^q_ja^q_k²a^q_m³ X

x∈F^∗_q4

x^{qi−1+qj+1−q+qk+2−q2+qm+3−q3} =

X

1≤i,j,k,m≤3

b_ib^q_jb^q_k²b^q_m³ X

x∈F^∗_q4

x^{qi−1+qj+1−q+qk+2−q2+qm+3−q3}.

P

x∈F^∗_q4 x^{qi−1+qj+1−q+qk+2−q2+qm+3−q3} =−1 if and only if

qⁱ+q^j+1+q^k+2+q^m+3≡qⁱ+q^j+1+q^k+2+q^m−1 ≡1+q+q²+q³ (modq⁴−1), and zero otherwise. Suppose that the former case holds.

First considerk= 1. Thenqⁱ+q^j+1+q^k+2+q^m−1 ≤q³+q⁴+q³+q² <

1 +q+q²+q³+ 2(q⁴−1) hence one of the following holds.

• Ifqⁱ+q^j+1+q^k+2+q^m−1 = 1 +q+q²+q³, thenm=i=j=k= 1.

• Ifqⁱ+q^j+1+q^k+2+q^m−1 = 1 +q+q²+q³+q⁴−1 =q+q²+q³+q⁴, then {i, j+ 1, k+ 2, m−1} ={1,2,3,4}, hence one of the following holds

i= 1, j= 3, k= 1, m= 3, i= 2, j= 3, k= 1, m= 2.

Now consider the casek≥2. Thenqⁱ+q^j+1+q^k+2+q^m−1 ≡qⁱ+q^j+1+ q^k−2+q^m−1 ≤q³+q⁴+q+q² <1 +q+q²+q³+ 2(q⁴−1) hence one of the following holds.

• Ifqⁱ+q^j+1+q^k−2+q^m−1 = 1+q+q²+q³, then{i, j+1, k−2, m−1}= {0,1,2,3}, hence one of the following holds

i= 1, j= 2, k= 2, m= 3, i= 2, j= 2, k= 2, m= 2, i= 2, j= 2, k= 3, m= 1, i= 3, j= 1, k= 2, m= 2, i= 3, j= 1, k= 3, m= 1.

• Ifqⁱ+q^j+1+q^k−2+q^m−1 = 1 +q+q²+q³+q⁴−1 =q+q²+q³+q⁴, theni=j =k=m= 3.

Proposition 4.2. Let f(x) and g(x) be two q-polynomials over Fq⁴ such thatLf =Lg. If the maximum field of linearity of f is Fq, then

g(x) =f(λx)/λ, or

g(x) = ˆf(λx)/λ.

Proof. By Proposition 2.3, the maximum field of linearity of g is also Fq. First note thatLg =Lf when g is as in the assertion (cf. Lemmas 3.1 and 3.2). Letf(x) =P3

i=0a_ix^qi and g(x) =P3

i=0b_ix^qi.

First we are going to use Lemma 3.6. From (7) we havea0 =b0. From (8) withn= 4 andk= 1,2 we havea₁a^q₃ =b₁b^q₃ anda^1+q₂ ² =b^1+q₂ ², respectively.

From (9) withn= 4 andk= 2 we obtain

a^q+1₁ a^q₂² +a₂a^q+q₃ ² =b^q+1₁ b^q₂²+b₂b^q+q₃ ². (17) Note thata₁a^q₃=b₁b^q₃ implies

N(b1) N(b3) = N(a1) N(a3). (18) Multiplying (17) by b2 and applying a^1+q₂ ² =b^1+q₂ ² yields:

b²₂b^q₃^2+q−b₂(a^q+1₁ a^q₂²+a₂a^q₃^2+q) +b^q+1₁ a^q₂²⁺¹ = 0. (19)