ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 16 (2019) 585-608 https://doi.org/10.26493/1855-3974.1651.e79 (Also available at http://amc-journal.eu) A Carlitz type result for linearized polynomials* Bence Csajbok t MTA-ELTE Geometric and Algebraic Combinatorics Research Group, ELTE Eötvös Loránd University, Budapest, Hungary, Department of Geometry, 1117 Budapest, Pazmany P. stny. 1/C, Hungary Giuseppe Marino Dipartimento di Matematica e Fisica, Universita degli Studi della Campania "Luigi Vanvitelli", Viale Lincoln 5,1-81100 Caserta, Italy and Dipartimento di Matematica e Applicazioni "Renato Caccioppoli", Universita degli Studi di Napoli "Federico II", Via Cintia, Monte S.Angelo I-80126 Napoli, Italy Olga Polverino Dipartimento di Matematica e Fisica, Universita degli Studi della Campania "Luigi Vanvitelli", Viale Lincoln 5, I-81100 Caserta, Italy Received 21 March 2018, accepted 9 October 2018, published online 30 April 2019 For an arbitrary q-polynomial f over Fqn we study the problem of finding those q-polynomials g over Fqn for which the image sets of f (x)/x and g(x)/x coincide. For n < 5 we provide sufficient and necessary conditions and then apply our result to study maximum scattered linear sets of PG(1, q5). Keywords: Linearized polynomial, linear set, direction. Math. Subj. Class.: 11T06, 51E20 * The research was supported by the Italian National Group for Algebraic and Geometric Structures and their Applications (GNSAGA - INdAM). ^The first author is supported by the Janos Bolyai Research Scholarship of the Hungarian Academy of Sciences. The first author acknowledges the support of OTKA Grant No. K 124950. E-mail addresses: csajbokb@cs.elte.hu (Bence Csajbok), giuseppe.marino@unicampania.it, giuseppe.marino@unina.it (Giuseppe Marino), olga.polverino@unicampania.it (Olga Polverino) Abstract ©® This work is licensed under https://creativecommons.org/licenses/by/4.0/ 586 Ars Math. Contemp. 16 (2019) 445-463 1 Introduction Let Fqn denote the finite field of qn elements where q = ph for some prime p. For n > 1 and s | n the trace and norm over Fqs of elements of Fqn are defined as Trqn/qs (x) = x + xqS +-----+ xq" s and Nqn/qs (x) = x1+qS+ hq" s, respectively. When s = 1 then we will simply write Tr(x) and N(x). Every function f: Fqn ^ Fqn can be given uniquely as a polynomial with coefficients in Fqn and of degree at most qn - 1. The function f is Fq -linear if and only if it is represented by a q-polynomial, that is, n-1 f (x) = £ Oixqi (1.1) ¿=0 with coefficients in Fqn. Such polynomials are also called linearized. If f is given as in (1.1), then its adjoint (w.r.t. the symmetric non-degenerate bilinear form defined by (x, y) = Tr(xy)) is n-1 f (x) := £ af"xqn-i, ¿=0 i.e. Tr(xf (y)) = Tr(yf(x)) for any x, y G Fqn. The aim of this paper is to study what can be said about two q-polynomials f and g over Fqn if they satisfy Im (M) = Im (g^) , (1.2) where by Im(f (x)/x) we mean the image of the rational function f (x)/x, i.e. {f (x)/x : x G Fqn }. For a given q-polynomial f, the equality (1.2) clearly holds with g(x) = f (Ax)/A for each A g Fq„. It is less obvious that (1.2) holds also for g(x) = f (Ax)/A, cf. [2, Lemma 2.6] and the first part of [8, Section 3], see also the proof of [18, Theorem 3.3.9]. When one of the functions in (1.2) is a monomial then the answer to the question posed above follows from McConnel's generalization [25, Theorem 1] of a result due to Carlitz [7] (see also Bruen and Levinger [6]). Theorem 1.1 ([25, Theorem 1]). Let p denote a prime, q = ph, and 1 < d a divisor of q — 1. Also, let F: Fq ^ Fq be a function such that F(0) = 0 and F(1) = 1. Then q-1 q-1 (F(x) — F(y)) d = (x — y) d for all x, y G Fq if and only if F (x) = xpj for some 0 < j < h and d | pj — 1. Indeed, when the function F of Theorem 1.1 is Fq -linear, we easily get the following corollary (see Section 2 for the proof, or [16, Corollary 1.4] for the case when q is an odd prime). Corollary 1.2. Let g(x) and f (x) = axq , q = ph, be q-polynomials over Fqn satisfying Condition (1.2). Denote gcd(k, n) by t. Then g(x) = ^xq with gcd(s, n) = t for some ft with Nqn/qt (a) = Nqn/qt (ft). Another case for which we know a complete answer to our problem is when f(x) = Tr(x). B. Csajbok et al.: A Carlitz type result for linearized polynomials 587 Theorem 1.3 ([8, Theorem 3.7]). Let f (x) = Tr(x) and let g(x) be a q-polynomial over Fqn such that Im(f (x)/x) = Im(g(x)/x). Then g(x) = Tr(Ax)/A for some A G Fq„. Note that in Theorem 1.3 we have f (x) = f (x) and the only solutions for g are g(x) = f (Ax)/A, while in Corollary 1.2 we have (up to scalars) y(n) different solutions for g, where y is the Euler's totient function. The problem posed in (1.2) is also related to the study of the directions determined by an additive function. Indeed, when f is additive, then Im(f (x)/x) = ( f (x) - f (y) : x = y, x, y G Fqn I x - y is the set of directions determined by the graph of f, i.e. by the point set Gf := {(x, f (x)) : x G Fqn} c AG(2, qn). Hence, in this setting, the problem posed in (1.2) corresponds to finding the Fq-linear functions whose graph determines the same set of directions. The size of Im(f (x)/x) (for any f, not necessarily additive) was studied extensively. When f is Fq-linear the following result holds. Theorem 1.4 ([1, 3]). Let f be a q-polynomial over Fqn, with maximum field of linearity qn — 1 qn-1 + 1 < | Im(f (x)/x)| < Fq. Then q - 1 The classical examples which show the sharpness of these bounds are the monomial functions xqS, with gcd(s, n) = 1, and the Tr(x) function. However, these bounds are also achieved by other polynomials which are not "equivalent" to these examples (see Section 2 for more details). Two Fq-linear polynomials f (x) and h(x) of Fqn [x] are equivalent if the two graphs Gf and Gh are equivalent under the action of the group rL(2, qn), i.e. if there exists an element y G rL(2, qn) such that Gj = Gh. In such a case, we say that f and h are equivalent (via y) and we write h = fv. It is easy to see that in this way we defined an equivalence relation on the set of q-polynomials over Fqn. If f and g are two q-polynomials such that Im(f (x)/x) = Im(g(x)/x), then Im(fv(x)/x) = Im(gv(x)/x) for any admissible y G rL(2, qn) (see Proposition 2.6). This means that the problem posed in (1.2) can be investigated up to equivalence. For n < 4, the only solutions for g in problem (1.2) are the trivial ones, i.e. either g(x) = f (Ax)/x or g(x) = f (Ax)/x (cf. Theorem 2.8). For the case n = 5, in Section 4, we prove the following main result. Theorem 1.5. Let f (x) and g(x) be two q-polynomials over Fq5, with maximum field of linearity Fq, such that Im(f (x)/x) = Im(g(x)/x). Then either there exists y G rL(2, q5) suchthat fv(x) = axqi and gv(x) = ^xqj with N(a) = N(P) for some G {1, 2, 3,4}, or there exists A G F*5 such that g(x) = f (Ax)/A or g(x) = f (Ax)/A. Finally, the relation between Im(f (x)/x) and the linear sets of rank n of the projective line PG(1, qn) will be pointed out in Section 5. As an application of Theorem 1.5 we get a criterium of PrL(2, q5)-equivalence for linear sets in PG(1, q5) and this allows us to prove 588 Ars Math. Contemp. 16 (2019) 445-463 that the family of (maximum scattered) linear sets of rank n and of size (qn - 1)/(q - 1) in PG(1, qn) found by Sheekey in [27] contains members which are not-equivalent to the previously known linear sets of this size. 2 Background and preliminary results Let us start this section by the following immediate corollary of Theorem 1.4. Proposition 2.1. If Im(f (x)/x) = Im(g(x)/x) for two q-polynomials f and g over Fqn, then their maximum fields of linearity coincide. Proof. Let Fqm and Fqk be the maximum fields of linearity of f and g, respectively. Suppose to the contrary m < k. Then | Im(g(x)/x)| < (qn — 1)/(qk — 1) < qn-fc+1 + 1 < qn-m + 1 < | Im(f (x)/x)|, a contradiction. □ Now we are able to prove Corollary 1.2. Proof. The maximum field of linearity of f (x) is Fqt, thus, by Proposition 2.1, g(x) has to be a qf -polynomial as well. Then for t > 1 the result follows from the t = 1 case (after substituting q for qf and n/t for n) and hence we can assume that f (x) and g(x) are strictly Fq-linear. By (1.2), we note that g(1) = azq -1, for some z0 G F*„. Let F(x) := g(x)/g(1), then F is a q-polynomial over Fqn, with F(0) = 0 and F(1) = 1. Also, from (1.2), for each x G Fq„ there exists z G Fq„ such that F(x) / z ,zo qk-1 This means that for each x G Fq„ we get N (^Xr) = 1. By Theorem 1.1 (applied to the q-polynomial F with d = q - 1 | qn - 1 and using the fact that F is additive) it follows that F(x) = xpj for some 0 < j < nh. Then Proposition 2.1 yields p = qs with gcd(s, n) = 1. We get the first part of the statement by putting ft = g(1). Then from the assumption (1.2) it is easy to deduce N(a) = N(ft). □ We will use the following definition. Definition 2.2. Let f and g be two equivalent q-polynomials over Fqn via the element ( G rL(2, qn) represented by the invertible matrix a b c d and with companion automorphism a of Fqn. Then g(xx)): x G V} = { d)(f (x) : x G V H ( d ^w : x G Fq^ . (2.1) Let Kj(x) = axff + bf (x)ff and Hj(x) = cxff + df (x)ff. B. Csajbok et al.: A Carlitz type result for linearized polynomials 589 Proposition 2.3. Let f and g be q-polynomials over Fqn such that g = fv for some y G rL(2, qn). Then Kj is invertible andg(x) = Hj((Kj)-1(x)). Proof. It easily follows from (2.1). □ From (2.1) it is also clear that fv(x)^ _ f c + dz7 _ ^ f f (x) Im ( — ) H : z G Im ( ^ ^ (2.2) and hence | Im(fj(x)/x)| = | Im(f (x)/x)|. (2.3) From Equation (2.3) and Theorem 1.4 the next result easily follows. Proposition 2.4. If two q-polynomials over Fqn are equivalent, then their maximum fields of linearity coincide. Note that | Im(g(x)/x)| = | Im(f (x)/x)| does not imply the equivalence of f and g. In fact, in the last section we will list the known examples of q-polynomials f which are not equivalent to monomials but the size of Im(f (x)/x) is maximal. To find such functions was also proposed in [16] and, as it was observed by Sheekey, they determine certain MRD-codes [27]. Let us give the following definition. Definition 2.5. An element y G rL(2, qn) represented by the invertible matrix a b c d and with companion automorphism a of Fqn is said to be admissible w.r.t. a given q-polynomial f over Fqn if either b = 0 or — (a/b)CT G Im(f (x)/x). The following results will be useful later in the paper. Proposition 2.6. If Im(f (x)/x) = Im(g(x)/x) for some q-polynomials over Fqn, then Im(fv(x)/x) = Im(gv(x)/x) holds for each admissible y G rL(2, qn). Proof. From Im(f (x)/x) = Im(g(x)/x) it follows that any y G rL(2, qn) admissible w.r.t. f is admissible w.r.t. g as well. Hence Kj and Kj are both invertible and we may construct f and gv as indicated in Proposition 2.3. The statement now follows from Equation (2.2). □ Proposition 2.7. Let f and g be q-polynomials over Fqn and take some y G rL(2, qn) with companion automorphism a. Then gv(x) = fv(ACTx)/ACT for some A G F*„ if and only if g(x) = f (Ax)/A. Proof. First we prove the "if" part. Since g(x) = f (Ax)/A = o f o w>)(x), where wa denotes the scalar map x G Fqn ^ ax G Fqn, direct computations show that Hj = o Hj o and Kj = o Kj o Then gv = o fv o and the first part of the statement follows. The "only if" part follows from the "if" part applied to gj(x) = fj(Aax)/ACT and y-1; and from (fj)j-1 = f and (gj)j-i = g. □ 590 Ars Math. Contemp. 16 (2019) 445-463 Next we summarize what is known about problem (1.2) for n < 4. Theorem 2.8. Suppose Im(f (x)/x) = Im(g(x)/x) for some q-polynomials over Fqn, n < 4, with maximum field of linearity Fq. Then there exist < G GL(2, qn) and X G F*„ such that the following holds. • If n = 2 then fv(x) = xq and g(x) = f (Xx)/X. • If n = 3 then either fv(x) = Tr(x) and g(x) = f (Xx)/X or fv(x) = xq and g(x) = f (Xx)/X or g(x) = f (Xx)/X. • If n = 4 then g(x) = f (Xx)/X or g(x) = f (Xx)/X. Proof. In the n = 2 case f (x) = ax + bxq, b = 0. Let < be represented by the matrix 1 0 \ -a/b 1/b). Then < G GL(2,q2) maps f (x) to xq. Then Proposition 2.6 and Corollary 1.2 give gv (x) = fv (px)and hence Proposition 2.7 gives g(x) = f (Xx)/X for some X G Fqn. If n = 3 then according to [21, Theorem 5] and [8, Theorem 1.3] there exists < G GL(2, q3) such that either fv(x) = Tr(x) or fv(x) = xq. In the former case Proposition 2.6 and Theorem 1.3 give gv(x) = fv(px)/p and the assertion follows from Proposition 2.7. In the latter case Proposition 2.6 and Corollary 1.2 give gv(x) = axq where i G {1,2} and N(a) = 1. If i = 1, then gv(x) = fv(px)/p where pq-1 = a and the assertion follows from Proposition 2.7. Let now i = 2 and denote by A B C D the matrix of 1. Also, let A denote the determinant of this matrix and recall that fv(x) = xq, with < G GL(2, q3). Then by Proposition 2.3 KJ-1 (x) = Ax + Bxq is invertible and its inverse is the map _Aq+q2 x - Aq2 Bxq + B1+qxq2 V(x) := N(A)+N(B) . Also, by Proposition 2.3 we have (fj)j-i (x) = CV(x) + DV(x)q, which gives f (x) = (fv)v-i (x). Using similar arguments, since N(a) = 1, direct computations show . . . . . . (Aq+q2 C + Bq+q2D)x - Bq2 Aaq2+1xq + AqAaxq2 g(x) = (gj)j-1 (x) =-n(A) + N(B )-, and hence g(x) = f(Xx)/X for each X G F*3 with Xq-1 = A1-q/aq. The case n = 4 is [8, Proposition 4.2]. □ B. Csajbok et al.: A Carlitz type result for linearized polynomials 591 Remark 2.9. Theorem 2.8 yields that there is a unique equivalence class of q-polynomials, with maximum field of linearity Fq, when n = 2. For n = 3 there are two non-equivalent classes and they correspond to the classical examples: Tr(x) and xq. Whereas, for n = 4, from [8, Sec. 5.3] and [5, Table on p. 54], there exist at least eight non-equivalent classes. The possible sizes for the sets of directions determined by these strictly Fq-linear functions are q3 + 1, q3+q2 -q+1, q3+q2 + 1 and q3+q2+q+1 and each of them is determined by at least two non-equivalent q-polynomials. Also, by [13, Theorem 3.4], if f is a q-polynomial over Fq4 for which the set of directions is of maximum size then f is equivalent either to xq or to 6xq + xq3, for some 6 e F*4 with N(6) = 1 (see [23]). 3 Preliminary results about Tr(x) and the monomial q-polynomials over Fq5 Let q be a power of a prime p. We will need the following results. Proposition 3.1. Let f (x) = ^4=0 0jxqi and g(x) = Tr(x) be q-polynomials over Fq5. Then there is an element f e rL(2, q5) such that Im(fv(x)/x) = Im(g(x)/x) if and only if 01020304 = 0, (oi/o2)q = 02/03, (02/03^ = 03/04 and N(ai) = N(o2). Proof. Let f e rL(2, q5) such that Im(fv(x)/x) = Im(g(x)/x). By Proposition 2.4, the maximum field of linearity of f is Fq and by Theorem 1.3 there exists A e F*5 such that fv(x) = Tr(Ax)/A. This is equivalent to the existence of 0, b, c, d, 0d — bc = 0 and a: x ^ xph such that y e Fq4 ={ (° d)(f(xxV): x e Fq5 Then cx? + df (x)? G Fq for each x G Fq5. Let z = x?. Then 4 4 cz + d ^ a?zqi = cqzq + dq ^ a?qzq ¿=0 ¿=0 for each z. As polynomials of z, the left and right-hand sides of the above equation coincide modulo zq — z and hence comparing coefficients yield c + da? - J'0 — d a4 do? — cq + dq a?q, dafc+i =dq afcq, for k =1, 2,3. If d = 0, then c = 0, a contradiction. Since d = 0, if one of ai, a2, a3, a4 is zero, then all of them are zero and hence f is Fq5 -linear. This is not the case, so we have aia2a3a4 =0. Then the last three equations yield q Ol \ 02 02 J a3 q a2 a3 a3 a4 592 Ars Math. Contemp. 16 (2019) 445-463 and by taking the norm of both sides in da2 = dq a^9 we get N(ai ) = N(a2). Now assume that the conditions of the assertion hold. It follows that a3 = a2+1/a1 and a4 = a|+1/a2 = a22+q+1/af. Let a = ai/a1 for i = 0,1, 2, 3, 4. Then a1 = 1, N(a2) = 1, a3 = a2+1 and a4 = a1+q+q2. We have a2 = A9-1 for some A G F*5. If 'a b vc d) = lyl - A1-q4a0/a1 A1-q4/a1 then a b\ f x c d) /(x) : + A1-q4 x9 + Aq-q4 xq2 + Aq2-q4 xq3 + Aq3-q4 x97 VTr(xAq4 )/Aq4 i.e. /v(x) = Tr(Aq4x)/Aq4, where ^ is defined by the matrix ab cd □ Proposition 3.2. Let / (x) = ^4=0 ajxqi, with aia2a3a4 = 0. Then there is an element p G rL(2, q5) such that Im(/v(x)/x) = Im(xq/x) if and only if one of the following holds: 1. (01/02)® = 02/03, (02/03)® = «3/04 and N(ai) = N(a2), or 2. (04/01)® = «1/03, (01/02)® = 03/04 and N(oi) = N(o3). In both cases, if the condition on the norms does not hold, then Im(/v(x)/x) = Im(Tr(x)/x). Proof. We first note that the monomials xq* and xq5 * are equivalent via the map (i1 Hence, by Corollary 1.2, the statement holds if and only if there exist 0, b, c, d, 0d — bc = 0, a: x ^ xp and i G {1,2} such that ) : y G } = {(0 d)(/) : x G }. (3,) holds then let ay = 0j/01 fo a3 = a2+1, a4 = a2+q+q and it turns out that ( 1 0 ) _ + „ +2 1 ~3 , a9 If Condition 1 holds then let aj = aj/a1 for j = 0,1,2, 3,4. So a1 = 1, N(a2) = 1, 2 1+9+92+93 1 ) V-ao 1/aJ /(x) ' 1 "()( x a1+9+92+93 1 J yxq + a2xq2 + a3xq3 + a4xq4y B. Csajbok et al.: A Carlitz type result for linearized polynomials 593 Hence (3.1) is satisfied with i = 1, a : x ^ x and a b c d 1 1+q+q2+q3 2 1 0 Ï ao 1/aW If condition (1.2) holds then let aj = a.j/a3 for j = 0,1, 2, 3,4. So a3 = 1, N(«i) = 1, a2 = «1+q+q , a4 = «1+q and (3.1) is satisfied with i = 2, a: x ^ x and ab cd 1+q+q3+q4 ■1 1 1 10. ao 1/a3J Suppose now that (3.1) holds and put z = xCT. Then 4 i 4 za + b £ aj zqj j=o cz + d£aj zqj j=o for each z g Fq5 and hence, as polynomials in z, the left-hand side and right-hand side of the above equation coincide modulo zq5 - z. The coefficients of z, zqi and zq with i g {1, 2} and k g {1, 2, 3,4} \ {i} give jq = c + da j, bq a . aqi + bqi ajq* = daj, *k-i respectively, where the indices are considered modulo 5. Note that d6 = 0 since otherwise also a = c = 0 and hence ad — 6c = 0. With {r, s, t} = {1,2,3,4} \ {i}, the last three equations yield: at- First assume i = 1. Then we have at q a1 a2 a2 a3 - = - and - = -. «V «3 V «3 / «4 If N(«i) = N(«2), from Proposition 3.1 and Equation (2.3) it follows that | Im(xq/x)| = | Im(Tr(x)/x)|. Since | Im(xq/x)| = (qn - 1)/(q - 1) and | Im(Tr(x)/x)| = qn-1 + 1, we get a contradiction. Now assume i = 2. Then we have («4/«1)q = «1/«3 and «3 a4 (3.2) 4 q a 1 2 q k q a a r—i r a a s—i s q a a si s 2 q 594 Ars Math. Contemp. 16 (2019) 445-463 q2 11 q2 Multiplying these two equations yields and hence a2 — a!+q+q3/a|3+q. (3.3) By (3.2) this implies a4 — a13+1/a|3. (3.4) If N(ai) — N(as), then also N(ai) — N(a2) — N(as) — N(a4). We show that in this case Im(/v(x)/x) — Im(Tr(x)/x), so we must have N(a1) — N(a3). According to Proposition 3.1 it is enough to show (a1/a2)q — a2/a3 and (a2/a3)q — a3/a4. By (3.2) 4 we have (a1/a2)q — (a3/a4)q , which equals a2/a3 if and only if (a2/a3)q — a3/a4, i.e. a1+q — a4a|. Taking into account (3.3) and (3.4), this equality follows from N(a1) — N(a3). □ 4 Proof of the main theorem In this section we prove Theorem 1.5. In order to do this, we use the following two results and the technique developed in [8]. Lemma 4.1 ([8, Lemma 3.4]). Let / and g be two linearized polynomials over Fqn. If Im(/(x) /x) — Im(g(x) /x), then for each positive integer d the following holds ^ f/M^d — ^ ^g(xrd Lemma 4.2 (See for example [8, Lemma 3.5]). For any prime power q and integer d we haveYlxd — —1 if q — 1 | d and Y^xd — 0 otherwise. Proposition 4.3. Let /(x) — ^4=0 ajxqi and g(x) — ^4=0 bjxqi be two q-polynomials over Fq5 such that Im(/(x)/x) — Im(g(x)/x). Then the following relations hold between the coefficients of / and g: ao — bo, (4.1) a1a4 — b1&4, (4.2) 22 a2a3 — 62 b3 , (4.3) a?+1a32 + a2a4+q 2 — bq+1b32 + &2&4+9 2, (4.4) a1a2+q3 + a1+q3 a4 — &1&2+q3 + b1+q3 b% (4.5) 1+q+q2 q3 1 1+q q2+q3 1 q 1+q2+q3 1 q2 q3 q , 1+q+q3 q2 , a^ a2 + a2 a3 + aja^ + aj a2a3 a4 + a^ a4 + aqaq3 a aq2 + a aqaq2 aq3 + a1+q2 aq+q3 + a aq+q2+q3 — a1a2 a3a4 + a1a2a3 a4 + a1 a4 + a3a4 — b1+q+q2 bf + b1+q b32+q3 + bqb!+q2+q3 + bq2 b2b33 64 + b1+q+q3 b42 + bqbq3 b3b42 + b1bqb32 b43 + b1+q2 b4+q3 + b3b4+q2+q3, (4.6) B. Csajbok et al.: A Carlitz type result for linearized polynomials 595 N(ai) + N(a2) + N(a3) + N(a4) + Tr(a?a2 a3 + a?+q a2 a1+q + a?+q2 a23+q4 a4 + a1+q2+q4 a3' a4 + a2a32+q3+q4 a4 + af af^4 a1+q + öo aq a^ ~p ai a0 a/i 2 3 4 1 2 4 (47) N(bi) + N(62) + N(6a) + N(b4) + Tr(61622+q3+q4 63 + 6fq3 624 &1+q2 + 2 3 4 2 4 3 2 3 4 2 3 4 6l+q 62 + 64 + 6l+q 63 64 + 6263 64 + 61 63 6l+q + 62+q3 634 6l+q2 + 6i2 624 64+q+q3). Proof. Equations (4.1)-(4.5) follow from [8, Lemma 3.6]. To prove (4.6) we will use Lemma 4.1 with d = q3 + q2 + q +1. This gives us + al a2 a4+q+q ) = £ ai aq «m an3 £ x rqi-1+qj+1-q+qm+2-q2 + qn+3-q3 = 1 1, then q2 divides q +1 — k and hence k = q +1, or k = q2 + q + 1. In the former case i = j = n = 2 and m = 4, or i = j = 2 and n = m = 3, or i = 3, j = 1, m = 4 and n = 2, or i = 3, j = 1 and m = n = 3, or m =1, i = 2, j = 4 and n = 3. In the latter case i = 3 and n = m = j =4. Then (4.6) follows. To prove (4.7) we follow the previous approach with d = q4 + q3 + q2 + q +1. We obtain £ Oia?am an3 af = £ bib«6^ bf bf, where the summation is on the quintuples (i, j, m, n, r) with elements taken from {1,2, 3, 4} such that Li,j,m,n,r := (qi — 1) + (qj+1 — q) + (qm+2 — q2) + (qn+3 — q3) + (qr+4 — q4) is divisible by q5 — 1. Then Li,j,m,n,r — Ki,j/,m/,n/,r/ (m°d q 1), where Kij,m',n',r' = (q' -1) + (q3 - q) + (qm - q2) + (qn - q3) + (qr - q4), 5 596 Ars Math. Contemp. 16 (2019) 445-463 such that j ' = j + 1, m' = m + 2, n' = n + 3, r' = r + 4 (mod 5) with j' G {0, 2, 3, 4}, m' G {0,1, 3, 4}, n' G {0,1, 2, 4}, r' G {0,1, 2, 3}. (4.9) For q = 2 and q = 3 we can determine by computer those quintuples (i, j', m', n', r') for which Ki,j/,m/,n/,r/ is divisible by q5 - 1 and hence (4.7) follows. So we may assume q > 3. Then 3 - q2 - q3 - q4 = (q - 1) + (1 - q) + (1 - q2) + (1 - q3) + (1 - q4) < K (q4 - 1) + (q4 - q) + (q4 - q2) + (q4 - q3) + (q3 - q4) = 3q4 - 1 - q - q2, and hence Li,j,m,n,r is divisible by q5 - 1 if and only if Kj m',n>,r> = 0. It follows that qi + qj' + qm' + qn' + qr' = 1 + q + q2 + q3 + q4. (4.10) For h G {0,1, 2, 3,4} let ch denote the number of elements in the multiset {i, j', m', n', r'} which equals h. So 4 £ Chqh = 1 + q + q2 + q3 + q4 h=0 for some 0 < ch < 5 with ^h=0 ch = 5. We cannot have c0 = 5 since q > 1. If ci = 5 for some 1 < i < 4 then the left hand side of (4.10) is not congruent to 1 modulo q, a contradiction. It follows that ch < 4. Thus for q > 3 (4.10) holds if and only if ch = 1 for h = 0,1,2, 3,4 and we have to find those quintuples (i, j', m', n', r') for which i G {1, 2,3,4}, {i, j', m', n', r'} = {0,1, 2, 3,4} and (4.9) are satisfied. This can be done by computer and the 44 solutions yield (4.7). □ 4.1 Proof of Theorem 1.5 Proof. Since f has maximum field of linearity Fq, we cannot have a1 = a2 = a3 = a4 = 0. If three of {a1, a2, a3, a4} are zeros, then f (x) = a0x + aixqi, for some i G {1, 2, 3,4}. Hence with ^ represented by ( 1 0 ^ y-a0/ai 1/ai J we have fv(x) = xqi. Then Proposition 2.6 and Corollary 1.2 give gv(x) = £xqj where N(£) = 1 and j G {1,2,3,4}. Now, we distinguish three main cases according to the number of zeros among {a1, a2, a3, a4}. Two zeros among {a\,a2,a3,a4} Applying Proposition 4.3 we obtain a0 = b0. The two non-zero coefficients can be chosen in six different ways, however the cases a1a2 = 0 and a1a3 = 0 correspond to a3a4 = 0 and a2a4 = 0, respectively, since Im(f (x)/x) = Im(f(x)/x). Thus, after possibly interchanging f with f, we may consider only four cases. First let 4 f (x) = a0x + a1xq + a4xq , a^4 = 0. B. Csajbok et al.: A Carlitz type result for linearized polynomials 597 2 Applying Proposition 4.3 we obtain 0 = 62. Since 6i64 = 0, from (4.4) we get 62 = 63 = 0 and hence (4.7) gives N(ai) + N(a4) = N(bi) + N(b4). Also, from (4.2) we have N(ai) N(«4) = N(6i) N(&4). It follows that either N(ai) = N(61 ) and N(a4) = N(64), or N(ai) = N(64) and N(a4) = N(6i). In the first case 6i = aiAq_i for some A G F*5 and by (4.2) we get g(x) = f (Ax)/A. In the latter case 6i = a^- for some A G F*5 and by (4.2) we get g(x) = f (Ax)/A. Now consider f (x) = aixq + a3xq , aia3 = 0. Applying Proposition 4.3 and arguing as above we have either 62 = 64 = 0 or 6i = 63 = 0. In the first case from (4.6) we obtain a>i+q2+q3 = 6?63+q2+q3 and together with (4.4) this yields N(ai) = N(6i) and N(a3) = N(63). In this case g(x) = 2 4 f (Ax)/A for some A G F*5 .If 6i = 63 = 0, then in g (x) the coefficients of xq and xq are zeros thus applying the result obtained in the former case we get Ag(x) = f (Ax) and hence after substituting y = Ax and taking the adjoints of both sides we obtain g(y) = f (^y)/^, where ^ = A_i. The cases 2 2 3 f (x) = aixq + a2xq and f (x) = a2xq + a3xq can be handled in a similar way, applying Equations (4.2) - (4.7) of Proposition 4.3. One zero among {a^, a2, a3, a4} Since Im(f (x)/x) = Im(f (x)/x), we may assume ai =0 or a2 = 0. First suppose a1 = 0. Then by (4.2) either b1 =0 or b4 = 0. In the former case putting together Equations (4.3), (4.4), (4.5) we get N(a*) = N(6j) for i G {2,3,4} and hence there exists A g F*5 such that g(x) = f (Ax)/A. If a1 = b4 = 0, then in F,„ : u G U \{0}}. One of the most natural questions about linear sets is their equivalence. Two linear sets Ly and LV of PG(1, qn) are said to be PrL-equivalent (or simply equivalent) if there is an element in PrL(2, qn) mapping Ly to LV. In the applications it is crucial to have methods to decide whether two linear sets are equivalent or not. This can be a difficult problem and some results in this direction can be found in [8, 12]. If Ly and LV are two equivalent Fq-linear sets of rank n in PG(1, qn) and f is an element of rL(2, qn) which induces a collineation mapping Ly to LV, then Lyv = LV. Hence the first step to face with the equivalence problem for linear sets is to determine which Fq-subspaces can define the same linear set. For any q-polynomial f (x) = J2"=-0 ajxqi over Fqn, the graph Gf = {(x,f (x)): x G Fqn} is an Fq-vector subspace of the 2-dimensional vector space V = Fqn x Fqn and the point set Lf := LGf = {((x,f (x))>F,n : x G Fgn } is an Fq-linear set of rank n of PG(1, qn). In this context, the problem posed in (1.2) corresponds to find all Fq-subspaces of V of rank n (cf. [8, Proposition 2.3]) defining the linear set Lf. The maximum field of linearity of f is the maximum field of linearity of Lf, and it is well-defined (cf. Proposition 2.1 and [8, Proposition 2.3]). Also, by the Introduction (Section 1), for any q-polynomial f over Fqn, the linear sets Lf, Lf (with fA(x) := f (Ax)/A for each A G F*„) and Lf- coincide (cf. [2, Lemma 2.6] and the first part of [8, Section 3]). If f and g are two equivalent q-polynomials over Fqn, i.e. Gf and Gg are equivalent w.r.t. the action of the group rL(2, qn), then the corresponding Fq-linear sets Lf and Lg of PG(1, qn) are PrL(2, qn)-equivalent. The converse does not hold (see [12] and [8] for further details). The relation between the problem posed in (1.2) and the equivalence problem of linear sets of the projective line is summarized in the following result. Proposition 5.1. Let Lf and Lg be two Fq-linear sets of rank n of PG(1, qn). Then Lf and Lg are PrL(2, qn)-equivalent if and only if there exists an element f G rL(2, qn) such that Im(fv(x)/x) = Im(g(x)/x). □ B. Csajbok et al.: A Carlitz type result for linearized polynomials 605 Linear sets of rank n of PG(1, qn) have size at most (qn - 1)/(q - 1). A linear set Ly of rank n whose size achieves this bound is called maximum scattered. For applications of these objects we refer to [26] and [19]. Definition 5.2 ([15, 22]). A maximum scattered Fq -linear set Ly of rank n in PG(1, qn) is of pseudoregulus type if it is PrL(2, qn)-equivalent to Lf with f (x) = xq or, equivalently, if there exists an element ( G GL(2, qn) such that lu* = i((x,xq)>Fq„ : x G Fq„}. By Proposition 5.1 and Corollary 1.2, it follows Proposition 5.3. An Fq-linear set Lf of rank n of PG(1, qn) is of pseudoregulus type if and only if f (x) is equivalent to xqi for some i with gcd(i, n) = 1. □ For the proof of the previous result see also [20]. The known pairwise non-equivalent families of q-polynomials over Fqn which define maximum scattered linear sets of rank n in PG(1, qn) are 1. fs (x) = xqS, 1 < s < n - 1, gcd(s, n) = 1 ([4, 11]), 2. gSi<5(x) = JxqS + xqn-s, n > 4, Nqn/q(J) / {0,1}1, gcd(s, n) = 1 ([23] for s = 1, [24, 27] for s = 1), 3. hSj(5(x) = JxqS + xqs+n/2, n G {6, 8}, gcd(s,n/2) = 1, Nq„/q„/2 (J) G {0,1}, for the precise conditions on J and q see [9, Theorems 7.1 and 7.2]2, 4. kb(x) = xq + xq3 + bxq5, n = 6, with b2 + b =1, q = 0, ±1 (mod 5) ([10]). Remark 5.4. All the previous polynomials in cases 2, 3, and 4 above are examples of functions which are not equivalent to monomials but the set of directions determined by their graph has size (qn - 1)/(q - 1), i.e. the corresponding linear sets are maximum scattered. The existence of such linearized polynomials is briefly discussed also in [16, p. 132]. For n = 2 the maximum scattered Fq-linear sets coincide with the Baer sublines. For n = 3 the maximum scattered linear sets are all of pseudoregulus type and the corresponding q-polynomials are all GL(2, q3)-equivalent to xq (cf. [21]). For n = 4 there are two families of maximum scattered linear sets. More precisely, if Lf is a maximum scattered linear set of rank 4 of PG(1, q4), with maximum field of linearity Fq, then there exists ( G GL(2, q4) such that either fv(x) = xq or f(x) = Jxq + xq3, for some J G Fq4 with Nq4/q(J) G {0,1} (cf. [13]). It is easy to see that Lf = Lfs for any s with gcd(s, n) = 1, and fj is equivalent to f if and only if j G {i, n - i}. Also, the graph of gSja is GL(2, qn)-equivalent to the graph of gn_s,5-i. In [23, Theorem 3] Lunardon and Polverino proved that lSm and Lf1 are not PrL(2, qn)-equivalent when q > 3, n > 4. This was extended also for q = 3 [10, Theorem 3.4]. Also in [10], it has been proven that for n = 6,8 the linear sets Lf1, , , s, and Lkb are pairwise non-equivalent for any choice of s, s', J, J', b. In this section we prove that one can find for each q > 2 a suitable J such that LS2,5 of PG(1, q5) is not equivalent to the linear sets LS1^ of PG(1, q5) for each ^ g Fq5, with Nq5/q (m) G {0,1}. In order to do this, we first reformulate Theorem 1.5 as follows. 1 This condition implies q = 2. 2Also here q > 2, otherwise the linear set defined by hs,g is never scattered. 606 Ars Math. Contemp. 16 (2019) 445-463 Theorem 5.5 (Theorem 1.5). Let f (x) and g(x) be two q-polynomials over Fq5 such that Lf = Lg. Then either Lf = Lg is of pseudoregulus type or there exists some A G F*5 such that g(x) = f (Ax)/A or g(x) = f (Ax)/A holds. From [27, Theorem 8] and [24, Theorem 4.4] it follows that the family of Fq-subspaces , s G {1,n - 1}, gcd(s, n) = 1, contains members which are not rL-equivalent to the previously known Fq-subspaces defining maximum scattered linear sets of PG(1, qn). Our next result shows that the corresponding family LSs,5 of linear sets contains (at least for n = 5) examples which are not PrL-equivalent to the previously known maximum scattered linear sets. Theorem 5.6. Let (x) = Sxq2 + xq3 for some S G F*5 with N(S)5 = 1. Then LS2,5 is not PrL(2, q5)-equivalent to any linear set Lg and hence it is a new maximum scattered linear set. Proof. Suppose, contrary to our claim, that is PrL(2, q5)-equivalent to a linear set . From Proposition 5.1 and Theorem 5.5, taking into account that is not of pseudoregulus type, it follows that there exist ^ G rL(2, q5) and A g F*5 such that either (g2,5)v(x) = 5i,m(Ax)/A or (g2,a)v(x) = gi,M(Ax)/A. This is equivalent to say that there exist a, ft, A, B, C, D G Fq5 with AD — BC = 0 and a field automorphism t of Fq5 such that ABA A xT A _ 1 lY z . : x G F„^ = M q ^ q4 : z G F„5 C D J ^ (x)Ty ' q5J-\ V«zq + ftzqV • q where N(a) = N(ft) and aft = 0. We may substitute xT by y, then 2 3 2 3 4 2 3 a(Ay + B£T yq + Byq )q + ft (Ay + B£T yq + Byq )q = Cy + D£T yq + Dyq for each y G Fq5. Comparing coefficients yields C = 0 and aAq + ftBq T = 0, (5.1) 4 ftBq = D£T, (5.2) aBq ¿qT = D, (5.3) 4 aBq + ftAq = 0. (5.4) Conditions (5.2) and (5.3) give Bq4-q = ¿(q+1)T a/ft. (5.5) On the other hand from (5.4) we get Aq = —Bq3 aq2/ftq2 and substituting this into (5.1) we have Bq3-q4 = ¿«4t ftq2+1/aq2+1. (5.6) Equations (5.5) and (5.6) give N(ft/a) = N(¿)2t and N(a/ft)2 = N(¿)T, respectively. It follows that N(¿)5t = 1 and hence N(¿)5 = 1, a contradiction. □ B. Csajbok et al.: A Carlitz type result for linearized polynomials 607 6 Open problems We conclude the paper by the following open problems. 1. Is it true also for n > 5 that for any pair of q-polynomials f (x) and g(x) of Fqn [x], with maximum field of linearity Fq, if Im(f (x)/x) = Im(g(x)/x) then either there exists y G rL(2, qn) such that fv(x) = axqi and gv(x) = £xqj with N(a) = N(£) and gcd(i, n) = gcd(j, n) = 1, or there exists A G Fq„ such that g(x) = f (Ax)/A or g(x) = f(Ax)/A? 2. Is it possible, at least for small values of n > 4, to classify, up to equivalence, the q-polynomials f (x) G Fqn [x] such that | Im(f (x)/x)| = (qn — 1)/(q — 1)? Find new examples! 3. Is it possible, at least for small values of n, to classify, up to equivalence, the q-polynomials f(x) G Fqn [x] such that | Im(f(x)/x)| = qn-1 + 1? 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