On a conjecture about maximum scattered subspaces of F q 6 × F q 6
aa r X i v : . [ m a t h . C O ] A p r ON A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q DANIELE BARTOLI, BENCE CSAJB ´OK, AND MARIA MONTANUCCI
Abstract.
Maximum scattered subspaces are not only objects of intrinsic interest in finitegeometry but also powerful tools for the construction of MRD-codes, projective two-weightcodes, and strongly regular graphs. In 2018 Csajb´ok, Marino, Polverino, and Zanella introduceda new family of maximum scattered subspaces in F q × F q arising from polynomials of type f b ( x ) = bx q + x q for certain choices of b ∈ F q . Throughout characterizations for f b ( x ) and f b ( x ) giving rise to equivalent maximum scattered subspaces, the authors conjectured that theportion of new and inequivalent maximum scattered subspaces obtained in this way is quitelarge. In this paper first we find necessary and sufficient conditions for b to obtain a maximumscattered subspace. Such conditions were found independently with different techniques also byPolverino and Zullo 2019. Then we prove the conjecture on the number of new and inequivalentmaximum scattered subspaces of this family. Introduction
For a prime power q let V denote an r -dimensional vector space over the finite field F q n .Then V is also an F q -vector space of dimension rn . A t -spread of V is a partition of V \ { } by F q -subspaces of dimension t . In particular D := {h u i F qn \ { } : u ∈ V } is the so calledDesarguesian n -spread of V . An F q -subspace U of V is called scattered with respect to a spread S if U meets each element of S in at most a one-dimensional F q -subspace. In [8] Blokhuis andLavrauw proved that rn/ V w.r.t. aDesarguesian spread. In this paper by a scattered subspaces we will always mean a scatteredsubspace w.r.t. the Desarguesian spread D . When rn is even then scattered subspaces ofdimension rn/ r − , q n ) formed by the lattice of F q n -subspaces of V .For a k -dimensional F q -subspace U of V the linear set of rank k defined by U is L U = {h u i F qn : u ∈ U \ { }} ⊆ Λ . When dealing with linear sets, equivalence is an important issue. Two linear sets L U and L W ofΛ are called PΓL-equivalent if there exists an element ψ ∈ PΓL( r, q n ) such that ψ ( L U ) = L W .It is possible that L U and L W are PΓL-equivalent even if U and W are not ΓL-equivalent, i.e.there is no ϕ ∈ ΓL( r, q n ) such that ϕ ( U ) = W ; see for instance [12, 16]. The second author is supported by the J´anos Bolyai Research Scholarship of the Hungarian Academy ofSciences and partially by OTKA Grant No. PD 132463 and OTKA Grant No. K 124950.
1N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q If U is a (maximum) scattered subspace, then L U is called a (maximum) scattered linearset. Maximum scattered linear sets of Γ are two-intersection sets w.r.t. hyperplanes: theymeet hyperplanes of Γ either in ( q rn − n − / ( q −
1) or in ( q rn − n +1 − / ( q −
1) points; see [8].It follows that they define projective two-weight codes and strongly regular graphs; see [10].After the series of papers [3, 4, 8, 14, 22] it is now known that maximum scattered subspacesexist whenever rn is even.If U i is a maximum scattered subspace in V i = F r i q n , for i = 1 ,
2, then U ⊕ U is a maximumscattered subspace in V ⊕ V ; see [4] where these objects were used to construct small completecaps. Thus, maximum scattered subspaces of F q n can be used to construct maximum scatteredsubspaces in F rq n for r ≥ n -dimensional (i.e. maximum) scattered subspaces of F q n × F q n are important. For every n -dimensional F q -subspace U of F q n × F q n there exist asuitable basis and a q -polynomial f ( x ), that is, a polynomial of the form P i a i x q i , over F q n andof degree less than q n such that U = U f = { ( x, f ( x )) : x ∈ F q n } . Then, L U = L f = ((cid:28)(cid:18) , f ( x ) x (cid:19)(cid:29) F qn : x ∈ F ∗ q n ) ⊆ PG(1 , q n ) . The set of directions determined by the graph { ( x, g ( x )) : x ∈ F q n } ⊂ AG(2 , q n ) of a function g : F q n → F q n is D g := (cid:26) g ( x ) − g ( y ) x − y : x, y ∈ F q n , x = y (cid:27) . If g is additive, then D g = (cid:26) g ( x ) x : x ∈ F ∗ q n (cid:27) and hence | L f | = | D f | . If U is maximum scattered, then | D f | = | L U | = ( | U | − / ( q −
1) =( q n − / ( q − F q n → F q n function whichdetermines less than ( q n + 3) / , q n ) in at least q points.Sheekey in [32] discovered the following relation between maximum rank distance codes andmaximum scattered subspaces of F q n × F q n . If U f is a maximum scattered subspace, then thefollowing set of F q -linear F q n → F q n maps C f := { f a,b : x ax + bf ( x ) } is an F q -subspace of dimension 2 n such that for any two h, g ∈ C f , the rank of h − g (as F q -linear transformation of F q n ) is at lest n −
1. In other words, C f is a linear maximumrank distance code with minimum rank distance n −
1. Note that for each map ϕ a : x ax , ϕ a ◦ C f = C f and hence C f has an idealiser isomorphic to F q n which is maximum possible (someauthors call such MRD-codes F q n -linear). By fixing an F q -basis for F q n one can easily obtain a N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q matrix representation of this code, i.e. a subspace of F n × nq . The converse also holds: F q -linearMRD-codes of F n × nq with an idealiser isomorphic to F q n and with minimum distance n − F q n × F q n ; cf. [13, Proposition 6.1]. Moreover,equivalent MRD-codes correspond to equivalent maximum scattered subspaces and viceversa;see [32, Theorem 8]. For a survey on MRD-codes we refer to [33].If q = 2 and U f is maximum scattered in F n × F n then the graph of f together with thecommon point h (0 , , i F n of vertical (affine) lines is a translation oval O of PG(2 , n ) withnucleus (common point of tangents to O ) the common point h (1 , , i F qn of horizontal (affine)lines. The converse holds as well: every translation oval corresponds to a maximum scattered F -subspace of F n × F n . Then it follows from Payne’s classification of translation ovals [29](see also [20, 21] by Hirschfeld) that up to ΓL-equivalence, maximum scattered F -subspacesof F n × F n are all monomials, more precisely: they are of the form U f for some f ( x ) = 2 s ,with gcd( s, n ) = 1. This example can be generalized for every q , but for q > m | n the norm function N q n /q m : F q n → F q m is defined as x x ( q n − / ( q m − . The known maximum scattered F q -subspaces of F q n × F q n are members ofone of the following families of scattered subspaces: • U ,ns := (cid:8)(cid:0) x, x q s (cid:1) : x ∈ F q n (cid:9) , 1 ≤ s ≤ n −
1, gcd( s, n ) = 1 [8, 17]; • U ,ns,δ := n(cid:16) x, δx q s + x q n − s (cid:17) : x ∈ F q n o , n ≥
4, N q n /q ( δ ) = 0 , q = 2, gcd( s, n ) = 1,see [25] for s = 1, [26, 32] for s = 1; • U ,ns,b := n(cid:16) x, bx q s + x q s + n/ (cid:17) : x ∈ F q n o , n ∈ { , } , gcd( s, n/
2) = 1, N q n /q n/ ( b ) = 0 , b , see [13]; • U c := n(cid:16) x, x q + x q + cx q (cid:17) : x ∈ F q o , q odd, c + c = 1, see [15, 27]; • U h := n(cid:16) x, h q − x q − h q − x q + x q + x q (cid:17) : x ∈ F q o , q odd, h q +1 = −
1, see [6, 36].Note that these families may overlap, but each of them contains examples not contained inother families. The only known families of scattered subspaces with examples for each n > U ,ns and U ,ns,δ . By [18], for n ≤ q n /q ( δ ) = 1 is not only sufficientbut also necessary in order to get maximum scattered subspaces of the form U ,ns,δ . Very recentlyin [30] it was proved that for n ≥
10 the family U ,ns,b does not contain scattered subspaces. Apolynomial f ( x ) so that { ( x q t , f ( x )) : x ∈ F q mn } is scattered in F q nm × F q nm for infinitely many m is called exceptional scattered of index t , see [5, 7, 19]. The known exceptional scatteredpolynomials are those defining U ,ns and U ,ns,δ .In this paper we will investigate the subspaces U s,b := U , s,b = n(cid:16) x, bx q s + x q s +3 (cid:17) : x ∈ F q o of F q × F q . The first part of the next result is a special case of [13, Proposition 5.1], the secondpart can be proved by the same arguments: N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q Result 1.1.
Take some b, ¯ b ∈ F ∗ q , N q /q ( b ) = 1 , N q /q (¯ b ) = 1 and s, ¯ s ∈ { , , , } .The subspaces U s,b and U ¯ s, ¯ b are ΓL(2 , q ) -equivalent if and only if • s = ¯ s and N q /q (¯ b ) = N q /q ( b ) σ , or • s + ¯ s = 6 and N q /q (¯ b )N q /q ( b ) σ = 1 ,for some automorphism σ ∈ Aut ( F q ) .The subspaces U s,b and U ¯ s, ¯ b are GL(2 , q ) -equivalent if and only if • s = ¯ s and N q /q (¯ b ) = N q /q ( b ) , or • s + ¯ s = 6 and N q /q (¯ b ) q ¯ s N q /q ( b ) = 1 . Also, by the first paragraph of [13, Section 5], for s ∈ { , , , } the F q -subspaces(1) { ( x, αx q s + βx q s +3 ) : x ∈ F q } and { ( x, α ′ x q s + β ′ x q s +3 ) : x ∈ F q } are GL(2 , q )-equivalent when N q /q ( αβ ′ ) = N q /q ( α ′ β ). It follows that U ,b is equivalent to U ,b and U ,b is equivalent to U ,b . Also, by Result 1.1, U ,b is equivalent to U ,b − q and henceevery subspace U s,b , s ∈ { , , , } , is GL(2 , q )-equivalent to a subspace U b := U ,b = { ( x, bx q + x q ) : x ∈ F q } . In [13] the authors proved that U b is not ΓL(2 , q )-equivalent to subspaces of the form U , s or U , s,δ ; cf. [13, Theorem 6.2]. By [15,27] U b is not ΓL(2 , q )-equivalent to U c and by [6, Proposition3.5] neither to U h . By Result 1.1 two subspaces U b and U c , N q /q ( b ) = 1, N q /q ( c ) = 1, areGL(2 , q )-equivalent if and only if N q /q ( b ) = N q /q ( c ) and they are ΓL(2 , q )-equivalent if andonly if N q /q ( b ) = N q /q ( c ) σ , for some σ ∈ Aut ( F q ). This motivates the definition ofΓ := { N q /q ( b ) : U b is maximum scattered } . The size of Γ is the same as the number of pairwise not GL(2 , q )-equivalent maximumscattered subspaces of the form U b . In [13] the authors proved that Γ = ∅ and conjectured thefollowing. Conjecture 1.2 ( [13, Conjecture 7.4]) . Let q be a prime power. Then | Γ | = ⌊ ( q + q + 1)( q − / ⌋ . The conjecture was proven in [13] for q ≤
32 using GAP. In this paper Conjecture 1.2 is provedfor an arbitrary prime power q . For q odd and even see Theorems 2.9 and 2.19, respectively.Suppose q = p e for some prime p . From our proof, it will turn out that U b is maximum scatteredif and only if U b p is maximum scattered (cf. Theorems 2.4 and 2.12). It follows that w ∈ Γ ifand only if w p ∈ Γ. A consequence of our result is the following.
Theorem 1.3.
The number of pairwise not
ΓL(2 , q ) -equivalent maximum scattered subspacesof the form U b is at least | Γ | = ⌊ ( q + q + 1)( q − / ⌋ e , where q = p e , for some p prime. N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q The proof of the conjecture
Here we analyse more in detail the family U b in order to prove Conjecture 1.2. Doing so, wefollow the same notation and approach as in [13, Section 7]. We have, in particular, that U b isscattered if and only if for each m ∈ F q the q -polynomial r m,b ( x ) := mx + bx q + x q , seen as alinear transformation of F q , has kernel of dimension at most one over F q . In the whole sectionwe will assume b = 0. This is equivalent to require that the associated Dickson matrix(2) M ( m, b ) := m b m q b q m q b q m q b q
00 0 1 0 m q b q b q m q has rank at least five; for different proofs see [24, Section 2.4], [28, Proposition 5], [34, Proposi-tion 4.4]. As the definition of Γ suggests, the norm of b plays a crucial role in the investigationof U b ; in this whole section we will abbreviate it simply as N := N q /q ( b ) = b q +1 ∈ F q . Note that the determinant of the 4 × M ( m, b ) is b q ( N − q and itdoes not vanish when b = 0 and N = 1. In order to compute the elements in Γ we give a precisedescription of the values b giving rise to a maximum scattered linear set U b . We will need thefollowing result. Theorem 2.1 ( [11, Theorem 1.3]) . Let M ( f ) be the n × n Dickson matrix associated with a q -linearized polynomial over F q n . Denote by M r ( f ) the ( n − r ) × ( n − r ) submatrix of M ( f ) obtainedby considering the last n − r columns and the first n − r rows of M ( f ) . Then dim ker f = t ifand only if det( M ( f )) = det( M ( f )) = · · · = det( M t − ( f )) = 0 , and det( M t ( f )) = 0 . The cases q odd and q even will be analyzed separately. The next lemma will be crucial inboth cases. Lemma 2.2. If N = 1 , then U b is not scattered. If N = 1 and U b is not scattered, then (3) φ b ( T ) := b q +1 T + ( − b q + q + q + q + q +1 + 2 b q + q + q +1 − b q +1 − b q + q + b q + q ) T − b q + q ( b q +1 − q + q ∈ F q [ T ] has a root which is a (1 + q + q ) -th power in F ∗ q . N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q Proof.
By Theorem 2.1, U b is not scattered if and only if there exists m ∈ F q such that(4) det m b m q b q m q b q m q b q
00 0 1 0 m q b q b q m q = det b m q b q m q b q m q b q
00 1 0 m q b q = 0 . The first condition is G ( m, m q , m q , m q , m q , m q ) = 0, where G ( X, Y, Z, U, V, W ) :=
U V W XY Z + U W Y + b q +1 U V W + V XZ + b q +1 U V Z + b q + q V W X + b q + q U Y Z + b q + q XY Z + b q + q W XY − b q + q + q + q + q +1 + b q + q + q + q + b q + q + q +1 + b q + q + q +1 − b q + q − b q +1 − b q + q + 1 , while the second condition is F ( m, m q , m q , m q , m q , m q ) = 0, where F ( X, Y, Z, U, V, W ) := − b q Y ZU − bZU V + b q + q + q + q +1 − b q + q +1 − b q + q + q + b q . Clearly, this is equivalent to F ( m, m q , m q , m q , m q , m q ) = 0 = F ( m, m q , m q , m q , m q , m q ) , where F ( X, Y, Z, U, V, W ) := − b q ZU V − b q U V W + b q + q + q + q + q − b q + q + q − b q + q + q + b q ,F ( X, Y, Z, U, V, W ) := − bU V W − b q V W X + b q + q + q + q +1 − b q + q + q − b q + q +1 + b q . Next, we eliminate
X, Y, Z from the above equations. For two multivariate polynomials f, g ∈ F q [ X , . . . , X n ] denote by Res X i ( f, g ) the resultant of f and g considered as polynomials in X i .Then one can verify thatRes Z (Res Y (Res X ( G, F ) , Res X ( G, F )) , Res Y (Res X ( G, F ) , Res X ( G, F ))) = − U V φ b ( U V W ) . It follows that (4) can hold only if m = 0 or φ b ( m q + q + q ) = 0.If m = 0, then (4) reads − ( N − q + q = 0 and b q ( N − q +1 = 0; which holds if and onlyif N = 1. If N = 1, then 0 is not a root of φ b and hence φ b ( m q (1+ q + q ) ) = 0 only if φ b has aroot which is a (1 + q + q )-th power in F ∗ q . (cid:3) N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q The q odd case. In this section we will always assume q odd. First we give sufficient andnecessary conditions on b so that U b is maximum scattered. This condition is equivalent to theone obtained in [31, Theorem 7.3] with different techniques.The discriminant ∆ b of φ b ( T ) will play a crucial role in our investigation:∆ b := (cid:16) b q + q + q + q + q +1 (cid:17) − b q + q +2 q + q + q +2 − b q +2 q + q + q +2 q +1 − b q + q + q +2 q + q +1 + b q +2 + b q +2 q + b q +2 q +8 b q + q + q + q + q +1 − b q + q + q +1 − b q + q + q +1 − b q + q + q + q = N q + q +1) − N q + q +2 + N q +2 q +1 + N q + q +1 ) + N + N q + N q +8 N q + q +1 − N q +1 + N q + q + N q +1 ) . Note that ∆ b belongs to F q . Proposition 2.3.
There is a root of φ b ( T ) which is a ( q + q + 1) -th power in F q if and onlyif one of the following conditions holds: • ∆ b = 0 , • ∆ b is a square in F ∗ q and N ∈ F q , • ∆ b is a non-square in F q .In all cases, each of the roots of φ b ( T ) are (1 + q + q ) -th powers.Proof. Let δ ∈ F q be such that δ = ∆ b . Consider a root t of φ b ( T ), namely t = − B + δ A , where A = b q +1 , (5) B = − b q + q + q + q + q +1 + 2 b q + q + q +1 − b q +1 − b q + q + b q + q . We have that t is a ( q + q +1)-th power in F q if and only if t = 0 or t ( q − q +1) = 1, equivalently,when t q + q = t q +1 , that is, by (5):(6) A δ q +1 + A ( δ + δ q ) + A ( δ q + δ q ) + A δ q + q + A = 0 , N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q where A = b q + q ,A = b q + q + q +2 q + q +1 − b q + q + q + q + q +1 + b q + q + q +1 + b q + q + q + q − b q +2 q ,A = b q + q +2 q + q + q +2 + b q +2 + 2 b q + q + q + q + q +1 − b q + q + q +1 − b q + q + q +1 ,A = − b q +1 ,A = − b q +2 q +3 q +2 q +2 q +3 + 2 b q + q +3 q + q + q +3 − b q +3 + b q +2 q +2 q +3 q +2 q +2 +2 b q +2 q +2 q + q +2 q +2 − b q + q +2 q + q + q +2 + 2 b q +2 q + q +2 + 3 b q +2 q + q +2 − b q +2 q + q +2 q +2 q +1 − b q + q +2 q +1 − b q + q + q +3 q + q +1 + 8 b q + q + q +2 q + q +1 − b q + q +2 q +1 + b q +2 q + q +2 q − b q + q +2 q + q + b q +3 q . We distinguish three cases: • ∆ b = 0. In this case t = − B/ (2 A ) and direct computations show t q + q = t q +1 . Hence t is a ( q + q + 1)-th power in F q . • ∆ b is a square in F ∗ q . This means that δ q i = δ for each i = 1 , . . . ,
5. Recalling that δ = ∆ b , one gets from (6)2 δ ( N q − N )( − N + N q − N q + N q + q + δ ) = 0 . Then either N ∈ F q or − N + N q − N q + N q + q ∈ F q . Since N ∈ F q , N ∈ F q follows inboth cases. Then b q = b q +1 /b q and b q = b q +1 /b q . Now, ∆ b reads N ( N − ( N + 3).Let z ∈ F ∗ q be such that z = ( N − N +3), so that t = ( N ( N − + N ( N − z ) / (2 b q +1 ).It can be verified that t q + q = t q +1 in this case. • ∆ b is not a square in F ∗ q . This yields δ q = − δ and (6) reads − A δ − A δ + A = − ∆ b ( A + A ) + A = 0 . (cid:3) Theorem 2.4. U b is a maximum scattered F q -subspace of F q × F q if and only if ∆ b is a squarein F ∗ q .Proof. By Lemma 2.2, if φ b has no roots which are (1 + q + q )-th powers in F q then U b isscattered. Suppose that φ b ( t ) = 0 for some t = m q + q + q where m ∈ F ∗ q . Then m q i + q i +1 + q i +2 = t q i +3 for i = 0 , , . . . , m q + q = m q + q m q + q + q m q + q + q = t q + q t q and m q + q + q = t q +1 t q . N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q Then, by defining G and F as in Lemma 2.2 and using the expressions above, one can eliminate m as follows:det M ( m, b ) = G ( m, m q , m q , m q , m q , m q ) = t − q − q ( b q +1 t q + q +1 − b q + q t q + q (7) + b q +1 t q + q + b q + q t q +2 q − b q +1 t q + q + b q + q t q + q + q + b q + q + q +1 t q + q − b q + q t q + q + b q + q + q + q t q + q + b q + q t q + q + q + b q + q t q + q + q + b q + q + q +1 t q + q − b q + q + q + q + q +1 t q + q + t q +1 + t q + q + t q + q + t q + q + q +1 ) , det M ( m, b ) = F ( m, m q , m q , m q , m q , m q )(8) = − b q + q + q − b q + q +1 + b q + q + q + q +1 − b q t q + b q − bt q . By Proposition 2.3, it is convenient to distinguish four cases: • ∆ b is a square in F ∗ q and N / ∈ F q . By Proposition 2.3 the roots of φ b ( T ) are not( q + q + 1)-th powers and hence U b is scattered. • ∆ b = 0. The solution of φ b ( T ) = 0 is t = − B/ (2 A ), where A, B are as in (5). ByProposition 2.3, t = 0 and hence N = 1, thus either U b is not scattered, or we can write t = m q + q + q for some m ∈ F ∗ q . In the latter case, by substituting t = − B/ (2 A ) in (7)we get zero. Substituting t = − B/ (2 A ) in (8) we get a quantity divisible by ∆ b , andhence again zero. It follows that U b is not scattered. • ∆ b is a square in F ∗ q and N ∈ F q . In this case a root t of φ b ( T ) can be written as(9) t = N ( N − + N ( N − z b q +1 , where z ∈ F ∗ q satisfies z = ( N − N + 3). Then substituting (9) in (8) and recallingthat N ∈ F q , we get det M ( m, b ) = − b − q ( N − N z, which is non-zero since N = 1 would give ∆ b = 0. It follows that U b is scattered. • ∆ b is not a square in F q . In this case a root t of φ b ( T ) can be written as(10) t = − B + δ b q +1 , where δ = ∆ b and δ q = − δ . Note that t = 0, since then we would have N = 1 andhence ∆ b = 0. Therefore, by Proposition 2.3 we can write t = m q + q for some m ∈ F ∗ q .Then by substituting (10) in (7) and (8) we obtain det M ( m, b ) = det M ( m, b ) = 0and thus U b is not scattered. (cid:3) Next we consider the condition ∆ b being a square in F ∗ q . It will be useful to write ∆ b as(11) ∆ b = ( N q + q +1 − N − N q − N q ) + 8 N q + q +1 − N q +1 + N q +1 + N q + q ) . In what follows we will count the exact number of norms N such that ∆ b is a square in F ∗ q . N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q Note that each N ∈ F q \ F q is a root of multiplicity one of the irreducible polynomial(12) T − ( N + N q + N q ) T + ( N q +1 + N q + q + N q +1 ) T − N q + q +1 ∈ F q [ T ] . This fact leads us to consider polynomials F ( T ) := T − ST + RT − P ∈ F q [ T ]satisfying(13) ( S − P ) + 8 P − R = U for some U ∈ F ∗ q . Proposition 2.5.
The number of norms N ∈ F q \ F q such that ∆ b is a square in F ∗ q is threetimes the number of irreducible polynomials F ( T ) = T − ST + RT − P ∈ F q [ T ] satisfying (13) .Proof. It follows from (11) and (12). (cid:3)
Proposition 2.6.
The number of polynomials F ( T ) = T − ST + RT − P ∈ F q [ T ] satisfying (13) is ( q − q ) / .Proof. For each choice of
S, P ∈ F q there are ( q −
1) pairs (
R, U ) such that U = 0 and( S − P ) + 8 P − R = U . This yields q ( q − / S, R, P ) ∈ F q such that( S − P ) + 8 P − R is a square in F ∗ q . (cid:3) Proposition 2.7.
The number of pairs ( t, F ( T )) , where F ( T ) satisfies (13) and t ∈ F q is aroot of F ( T ) is exactly ( q − q )( q + 1) / .Proof. For each such t we have that t − St + Rt − P = 0 and hence P = t − St + Rt . From( S − P ) + 8 P − R = U one gets(14) ( S − t + St − Rt ) + 8( t − St + Rt ) − R = U . • If t = 1, then the above equation defines (in terms of S , R and U ) a hyperbolic quadricwhich has exactly 2 q + 1 ideal points, namely those satisfying (cid:0) − Rt + St + S − U (cid:1) (cid:0) − Rt + St + S + U (cid:1) = 0 . Therefore there exist exactly ( q + 1) − q − q triples ( S, R, U ) satisfying (14).Among them those with U = 0 satisfy( S − t + St − Rt ) + 8( t − St + Rt ) − R = 0and they are q , that is the number of the affine F q -rational points of an irreducibleconic with a unique point at infinity obtained when S = t and R = 1 + t . Thismeans that the number of distinct pairs ( S, R ) ∈ F q satisfying (14) and such that( S − t + St − Rt ) + 8( t − St + Rt ) − R = 0 is exactly ( q − q ) / S, R, − U )and ( S, R, U ) come in pairs.
N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q • Assume t = 1. In this case P = 1 − S + R and Equation (14) reads( R − S − U + 3)( R − S + U + 3) = 0 . So, any pair (
S, R ) ∈ F q with R − S + 3 = 0 gives rise to one pair (1 , F ( T )) satisfyingour conditions. In this case we obtain q − q such pairs.In total, the number of suitable pairs is( q − q − q q − q ) = ( q − q )( q + 1)2 . (cid:3) Proposition 2.8.
The number of pairs ( B, F ( T )) , where F ( T ) satisfies (13) and F ( T ) =( T − A )( T − B )( T − B q ) for some A ∈ F q , B ∈ F q is exactly (cid:0) q − q + 2 q + 3 (cid:1) . Proof.
In ( S − P ) + 8 P − R = U we have S = A + B + B q , R = AB + AB q + B q +1 and P = AB q +1 . Thus one gets(15) ( B − B q ) + A (8 B q +1 − B q +1 )( B + B q )) + A ( B q +1 − = U . When B q +1 = 1, the discriminant of the left hand side, as a polynomial in A , is(16) 16( B − q +1) B q +1 . • If B = 0, then (15) reads A = U . In this case A can have q − q − , F ( T )) with the given conditions. • If B = 1, then by (15) U = 0 and hence B cannot have this value. • If B = 1 and B q +1 = 1, then the conic in the variables A and U defined by (15) has q affine points and the line U = 0 contains exactly one of them. Thus the number ofsolutions for each such B is ( q − /
2. There are q possible values for B and thus intotal q ( q − / B, F ( T )) in this case. • In F q , there are ( q − / B such that B q +1 is a non-square in F q . In thesecases (15) has no solutions with U = 0, cf. (16). Since the affine part of the conic inthe variables A and U , defined by (15), has q − B we get( q − / A and hence in total ( q − q − / B, F ( T )) with thegiven conditions. • In F q , there are ( q − / − ( q + 1) values of B such that B q +1 is a square in F q \ { , } .In these cases (15) has two solutions with U = 0, and since the affine part of the conicin the variables A and U , defined by (15), has q − B weget ( q − / A and hence in total ( q − q − q − / B, F ( T ))with the given conditions.So in total the number of suitable pairs is14 ( q − (cid:0) q − (cid:1) + 14 ( q − (cid:0) q − q − (cid:1) + 12 ( q − q + q − (cid:0) q − q + 2 q + 3 (cid:1) . N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q (cid:3) We are now in the position to prove Conjecture 1.2 when q is odd. Theorem 2.9. If q is odd then | Γ | = ( q − q − q − .Proof. We will count the exact number of norms N such that ∆ b is a square of F ∗ q . For i ∈ { , , , } denote by γ i the number of polynomials F ( T ) = T − ST + RT − P ∈ F q [ T ]with exactly i distinct roots in F q and satisfying (13). Consider the pairs ( t, F ( T )), where F ( T )is a polynomial satisfying (13) and t ∈ F q is a root of F ( T ). By Proposition 2.7, the numberof such pairs is ( q − q )( q + 1) / γ + 2 γ + 3 γ = ( q − q )( q + 1)2 . By Proposition 2.6, the number of polynomials F ( T ) satisfying (13) is q ( q − /
2, thus(18) γ = q ( q − − γ − γ − γ . Denote by δ q the number of N ∈ F q such that ∆ b is a square in F ∗ q . Note that this numberdepends on q , however, we won’t need its exact value. It is easy to see that δ q is the same asthe number of polynomials F ( T ) satisfying (13) and with a unique 3-fold root in F q . Denoteby A the number of totally reducible polynomials F ( T ) over F q (polynomials which can befactorized into linear factors over F q ) satisfying (13) and with at most 2 distinct roots in F q .Also, denote by A the number of reducible polynomials F ( T ) over F q satisfying (13) and witha unique, one-fold root in F q . Then γ = A + δ q ,γ = A − δ q . By Proposition 2.5 the number of norms N such that ∆ b is a square of F ∗ q is, using also (17)and (18), 3 γ + δ q = 3 q ( q − − γ − γ − γ + δ q = 3 q ( q − − ( q − q )( q + 1)2 − γ − γ + δ q = 3 q ( q − − ( q − q )( q + 1)2 − A − A . In Proposition 2.8, B ∈ F q \ F q and B q define the same polynomial F ( T ) and hence2 A + A = 12 (cid:0) q − q + 2 q + 3 (cid:1) , thus3 γ + δ q = 3 q ( q − − ( q − q )( q + 1)2 − (cid:0) q − q + 2 q + 3 (cid:1) = 12 (cid:0) q − q − q − (cid:1) . (cid:3) N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q The q even case. Similarly to the q odd case, we will need some preparation. Define N and φ b ( T ) in the exact same way as in (3). Since q is even, we have now φ b ( T ) = AT + BT + C ∈ F q [ T ] , where A = b q +1 ,B = N q + q +1 + N + N q + N q ,C = A q ( N + 1) q + q +1 . As before we will assume b = 0 and hence A = 0. Proposition 2.10.
There is a root of φ b ( T ) which is a ( q + q + 1) -th power in F q if and onlyif one of the following conditions holds • B = 0 , • B = 0 , and N ∈ F q \ { , } , • B = 0 , N ∈ F q \ F q , and Tr q / ( AC/B ) = 1 .In all cases, each of the roots of φ b ( T ) are ( q + q + 1) -th powers.Proof. First note that a field element t ∈ F q is a ( q + q + 1)-th power if and only if t q +1 ∈ F q .Let us deal first with the case B = 0. In this case the root t of φ b ( T ) satisfies t = C/A .Therefore t q +1) = C q +1 A q +1 = ( N + 1) q + q +1) ∈ F q and so t q +1 ∈ F q . From now on we assume B = 0. Then t is a root of φ b ( T ) if and only if s := tA/B is a rootof ψ b ( T ) := T + T + AC/B ∈ F q [ T ] . Note that, since
AC/B ∈ F q , we haveTr q / ( AC/B ) = Tr q /q (cid:0) Tr q / ( AC/B ) (cid:1) = 0 , and therefore the roots of ψ b ( T ) (and hence of φ b ( T )) belong to F q ; see [21, page 8]. Also, B ∈ F q and C/A q ∈ F q . • First suppose Tr q / ( AC/B ) = 1. Then the roots of ψ b ( T ) are some s, s q ∈ F q \ F q ,thus s q +1 = AC/B . For a root t = sB/A of φ b ( T ) t q +1 = (cid:18) sBA (cid:19) q +1 = B A q +1 s q +1 = CA q ∈ F q . • Suppose now Tr q / ( AC/B ) = 0. In this case the roots of ψ b ( T ) belong to F q . Wedistinguish two cases depending on N ∈ F q or not. Note that N / ∈ { , } since B = 0. N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q – If N ∈ F q \ { , } then AC/B = 1 / ( N + 1) ∈ F q and hence the roots of ψ b ( T ) aresome s, s + 1 ∈ F q . So, for a root t = sB/A of φ b ( T ) t q +1 = (cid:18) sBA (cid:19) q +1 = B N s ∈ F q . – If N ∈ F q \ F q then first observe AC/B ∈ F q \ F q . Indeed, AC/B = N q +1 ( N +1) q + q +1 / ( N q + q +1 + N + N q + N q ) ∈ F q would yield N q +1 ∈ F q and hence N ∈ F q ∩ F q = F q , a contradiction. Then also the roots of ψ b ( T ) are some s, s + 1 ∈ F q \ F q . Suppose, contrary to our claim, that with t = sB/A (19) t q +1 = (cid:18) BsA (cid:19) q +1 = B A q +1 s ∈ F q . Since s = s + AC/B , it follows that B A q +1 ( s + AC/B ) = B A q +1 s + C/A q ∈ F q . Here
C/A q ∈ F q and hence B A q s ∈ F q . Then dividing by (19) gives s ∈ F q , acontradiction. (cid:3) For a polynomial f ( X ) = P i a i X i ∈ F [ X ] and σ ∈ Aut( F ) we will denote by f σ ( X ) thepolynomial P i a σi X i ∈ F [ X ]. Lemma 2.11. If Tr q / ( AC/B ) = 1 and t is a root of φ q b ( T ) then t q = b q − t + b q − ( N +1) q +1 .Proof. Substitution of s := b q − t + b q − ( N + 1) q +1 in φ q b ( T ) gives 0 and hence s = t q or s = t q + B q /A q = t q + b − − q B . In the latter case(20) t q = b q − t + b q − ( N + 1) q +1 + b − q − B, we will show that this is a contradiction. Note that( tA q /B ) + ( tA q /B ) = ( AC ) q /B . Denote the right hand side above by E , then(21) t q A q /B + ( tA q /B ) = E + E + . . . + E q/ . Expressing t q from (21) and comparing with (20) yields b q − t + b q − ( N + 1) q +1 + b − q − B = tA q − q + BA q ( E + E + . . . + E q/ ) . Note that A q − q = b q − and hence we can eliminate t from the equation above. Then,multiplying by A q /B = b q /B gives N q ( N + 1) q +1 + BB = E + E + . . . + E q/ . N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q The left hand side above is (( N + N q + q ) + (( N + N q + q )) q ) /B . It follows that taking Tr q /q ofboth sides yields 0 = Tr q / ( E ) = Tr q / ( AC/B ) = 1 , and this contradiction proves the claim. (cid:3) Theorem 2.12. U b is a maximum scattered F q -subspace of F q × F q if and only if B = 0 , N ∈ F q \ { , } , Tr q / ( AC/B ) = 0 . (22) Proof.
We follow the same argument as in Theorem 2.4. By Lemma 2.2, if φ b has no roots whichare (1 + q + q )-th powers then U b is scattered. According to Proposition 2.10 it is convenientto distinguish the following cases. • Suppose that B = 0. In this case the solution of φ b ( T ) = 0 is t = p C/A = 0 andby Proposition 2.10 we can find m ∈ F ∗ q such that m q + q + q = p C/A . We will showdet M ( m, b ) = det M ( m, b ) = 0. By B = 0, i.e. Tr q /q ( N ) = N q /q ( N ), we obtain t = c/a = b q + q − q − ( b q + q + q +1 + b q + q + q + q + b q + q + q +1 + 1) . As in (7) and (8), substitution in det M ( m, b ) = G ( m, m q , m q , m q , m q , m q ) and indet M ( m, b ) = F ( m, m q , m q , m q , m q , m q ) , cf. Lemma 2.2, and applying B = 0several times, yields det M ( m, b ) = det M ( m, b ) = 0 and hence U b is not scattered. • Suppose that B = 0 , N ∈ F q \ F q , Tr q / ( AC/B ) = 0 . By Proposition 2.10 the roots of φ b ( T ) are not ( q + q + 1)-th powers in F q and so U b is scattered by Lemma 2.2. • Suppose that B = 0 , N ∈ F q \ { , } , Tr q / ( AC/B ) = 1 . By Proposition 2.10 the roots of φ b ( T ) are ( q + q + 1)-th powers in F q . Let γ bea root of φ q b ( T ) and consider m ∈ F ∗ q such that m q + q +1 = γ . Our aim is to prove G ( m, m q , m q , m q , m q , m q ) = F ( m, m q , m q , m q , m q , m q ) = 0. To do this, it will beuseful to express γ, γ q , . . . , γ q with respect to only one of these conjugated elements.By Lemma 2.11,(23) γ q = b q − ( N + 1) q +1 + b q − γ q . Taking the q -th power in (23) and applying again (23) yields γ q = b q − q ( N q + q + N q + N q + 1) + b q − q γ q = ( N q + q + N q +1 + N + N q + b q + q γ q ) /b q +1 . In a similar way we obtain γ q = ( N q + q +1 + N + N q + N q + b q + q γ q ) /b q + q ,γ q = ( N q +1 + N + N q + q + N q + b q + q γ q ) /b q + q , N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q γ = b q − q ( N + 1) q + b q − q γ q . Now, in (7) after substituting t q i with γ q i +3 , we obtain b q + q G ( m, m q , m q , m q , m q , m q ) = φ q b ( γ q )( φ q b ( γ q ) + γ q N q ( N + N q ) + b q + q ( N + 1) q +1 ( N + N q )) = 0 . Similarly, substitution in (8) yields F ( m, m q , m q , m q , m q , m q ) = 0 and hence U b isnot scattered. • Suppose that B = 0 , N ∈ F q \ { , } , Tr q / ( AC/B ) = 0 . By Proposition 2.10 the roots of φ b ( T ) are ( q + q +1)-powers in F q . Let t = m q + q + q ∈ F ∗ q be a root of φ b ( T ). Then, as in Proposition 2.10, there exists s ∈ F q , a root of ψ b ( T ), such that tA = sB ∈ F q . Then t q i = tA/A q i for i = 0 , , . . . ,
5. Substituting t q = tb q +1 − q − q and t q = tb q − q in det M ( m, b ) = F ( m, m q , m q , m q , m q , m q ) gives b q ( N + 1) which cannot be zero and hence U b is scattered. (cid:3) From now on, our aim is to calculate the number of norms N so that Conditions (22) aresatisfied. One can easily check that(24) Tr q / ( AC/B ) = Tr q/ (cid:0) Tr q /q ( AC/B ) (cid:1) =Tr q/ ( N q +1 + N q +1 + N q + q )( N + N q + N q + N q +1 + N q +1 + N q + q + N q + q +1 + 1)( N q + q +1 + N + N q + N q ) ! . Each N ∈ F q \ F q is a root of multiplicity one of the irreducible polynomial T + ( N + N q + N q ) T + ( N q +1 + N q + q + N q +1 ) T + N q + q +1 ∈ F q [ T ] . This leads us to consider polynomials(25) F ( T ) := T + ST + RT + P ∈ F q [ T ] , satisfying(26) P = S and Tr q/ (cid:18) R ( S + P + R + 1)( P + S ) (cid:19) = 0 , or equivalently, polynomials of the form (25) so that(27) P = S and U + U + R ( S + P + R + 1)( P + S ) ∈ F q [ U ] has two roots in F q . Note that if P = N q + q and S = N + N q + N q , then P = S yields N / ∈ { , } . We proceedsimilarly to the case q odd. After rearranging (27), it is clear that we have to study 4-tuples( U, S, P, R ) satisfying(28) P = S and ( P + S ) U + ( P + S ) U + R ( S + P + R + 1) = 0 . N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q Note that ( u, s, p, r ) satisfies the equation above if and only if ( u + 1 , s, p, r ) satisfies it and forfixed s, p, r there cannot be further solutions. Proposition 2.13.
The number of polynomials F ( T ) = T + ST + RT + P ∈ F q [ T ] satisfying (27) is ( q − q ) / .Proof. For each choice of the pair (
P, S ), P = S , there are q pairs ( R, U ) such that (28) holds,these pairs correspond to affine points of a conic. Then the number of quadruples (
U, S, R, P )satisfying (28) is q − q . The number of polynomials F ( T ) satisfying (27) is the number ofdistinct triples ( S, R, P ) ∈ F q such that (28) is satisfied, which is ( q − q ) / (cid:3) Proposition 2.14.
The number of pairs ( t, F ( T )) , where F ( T ) satisfies (27) and t ∈ F q is aroot of F ( T ) is exactly ( q − q ) / .Proof. We will frequently use P = t + St + Rt to eliminate P .If t = 0 then P = 0 and from (28) one gets ( U + U ) S + R ( S + R + 1) = 0. • If U / ∈ { , } then there are precisely q − S, R ) ∈ F q such that ( U + U ) S + R ( S + R + 1) = 0, two of them with S = 0. So in total q − P = S . • If U = 0 or U = 1 then there are 2 q − S, R ), two of them with S = 0. So intotal 2 q − P = S .Summing up, if t = 0 then the number of 5-tuples ( t, U, S, P, R ) such that F ( T ) satisfies (27)is ( q − q −
3) + 2(2 q −
3) = q − q .If t = 1, then P = 1 + S + R and from (28) one gets U ( U + 1)( R + 1) = 0. • If U / ∈ { , } then R = 1 and hence P = S , so there are no 5-tuples with the requiredproperties. • If U ∈ { , } then there are q − q pairs ( S, P ) with P = S , and each of these pairsuniquely determine R .Summing up, if t = 1 then the number of 5-tuples ( t, U, S, P, R ) such that F ( T ) satisfies (27)is 2( q − q ).Suppose now t / ∈ { , } . • If U ∈ { , } then the equation in (28) reads ( t + 1) R ( t + t + 1 + S ( t + 1) + R ) = 0which is satisfied exactly by 2 q − R, S ), corresponding to affine points of twointersecting affine line. We have P = S if and only if t + S ( t + 1) + Rt = 0 and hencewhen ( S, R ) ∈ { ( t, , ( t / ( t + 1) , } . Thus the number of 5-tuples ( t, U, S, P, R ) suchthat F ( T ) satisfies (27) is 2( q − q − • If U / ∈ { , } then the equation in (28) defines the affine part of a conic C in the variables S =: S ′ /Z and R =: R ′ /Z . The homogenous equation in S ′ , R ′ , Z of C is R ′ (1 + t + t U + t U ) + S ′ ( U + t U + U + t U )+ R ′ S ′ (1 + t ) + Z ( t U + t U ) + R ′ Z (1 + t ) . Its partial derivatives are zero only at the point (1 + t + t , , t ). Substituting thesevalues in the equation above yields (1 + t ) U ( U + 1) which is non-zero and hence C isirreducible. N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q Substituting Z = 0 yields( S ′ ( U t + U ) + R ′ ( U t + 1))( S ′ ( U t + U + t + 1) + R ′ ( U t + t + 1)) = 0 , and hence C has two distinct F q -rational points at infinity.It remains to calculate the number of affine points of C on the affine line t + S ( t +1) + R = 0 (whose points correspond to solutions with P = S ). Substitution of R in(28) gives( t + 1)( t + S ( t + 1))( t U + t U + t + 1 + S ( t U + t U + t + U + U )) = 0 . It is easy to see that the two solutions in S are distinct and hence in this case thenumber of 5-tuples ( t, U, S, P, R ) such that F ( T ) satisfies (27) is ( q − q − q − t, F ( T )) is the number of 4-tuples ( t, S, P, R ) such that F ( T )satisfies (27), which is half of the number of 5-tuples ( t, U, S, P, R ) with the same property, i.e.12 (( q − q ) + 2( q − q ) + 2( q − q −
3) + ( q − q − q − q − q . (cid:3) Lemma 2.15.
For α, β, γ ∈ F q , α = 0 , the number of pairs ( X, Y ) ∈ F q , X = γ , such that Y + Y = α ( X + 1)( X + β ) / ( X + γ ) and (29) α ( β + 1) = ( γ + β )( γ + 1) , is either q or q − depending on Tr q/ ( α ) = 1 or Tr q/ ( α ) = 0 , respectively.Proof. First note that α ( X + 1)( X + β )( X + γ ) = α (cid:18) β + 1 X + γ + ( γ + β )( γ + 1) X + γ (cid:19) . Introduce the variable W := 1 / ( X + γ ). Then our equation becomes Y + Y = α + α ( β + 1) W + α ( γ + β )( γ + 1) W , which can be considered as the affine part of a conic C in the variables W =: W ′ /Z and Y =: Y ′ /Z . The partial derivatives of the homogenous equation of C are zero only at the point( a (1 + b ) , , C because of the assumption (29). It is also easy to seethat C has a unique point at infinity which is F q -rational. The number of affine F q -rationalpoints of C with W = 0 is 2 when Tr q/ ( α ) = 0 and zero otherwise. For each solution ( W, Y )with W = 0 there corresponds a unique solution ( X, Y ) with X = 1 /W + γ and hence theresult follows. (cid:3) Proposition 2.16.
The number of polynomials F ( T ) , where F ( T ) satisfies (27) and F ( T ) =( T + K )( T + L )( T + L q ) for some K ∈ F q , L ∈ F q \ F q , is exactly q − q + 4 q . N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q Proof.
Since F ( T ) = T + ST + RT + P , we have S = K + L + L q , R = KL + KL q + L q +1 and P = KL q +1 . Note that P = S is equivalent to K ( L q +1 + 1) = ( L + L q ) and in this casethe equation in (27) reads(30) U + U = ( L + 1) q +1 ( K + 1)(( L + L q ) K + L q +1 )(( L q +1 + 1) K + L + L q ) . First suppose L q +1 = 1. Since L / ∈ F q , in this case S = P holds and (30) reads( L + L q ) ( U + U ) = ( K + 1)(( L + L q ) K + 1) . Each solution (
K, U ) corresponds to an affine point of an irreducible conic with a unique idealpoint and hence there are q such pairs which yields q/ K . Since L can bechosen in q different ways, in total we have q / K, L ) such that F ( T ) satisfies(27).From now on, assume L q +1 = 1. Then, since L / ∈ F q , we can define α = ( L +1) q +1 ( L + L q )( L q +1 +1) , β = L q +1 L q + L and γ = L + L q L q +1 +1 to apply Lemma 2.15. Condition (29) is satisfied and clearly α = 0.Note that S = P is equivalent to γ = K and hence to determine the number of pairs ( K, U ) ∈ F q satisfying (30) we can apply Lemma 2.15. Note thatTr q/ ( α ) = Tr q/ (cid:18) ( L q +1 + 1 + L q + L )( L + L q )( L q +1 + 1) (cid:19) = Tr q/ (cid:18) L + L q L q +1 + 1 + ( L + L q ) ( L q +1 + 1) (cid:19) = 0 . It follows that for each L there are q − K, U ) satisfying (30) and hence ( q − / K . There are q − q choices for L in F q \ F q such that L q +1 = 1. In total we have( q − q − q ) / K, L ) such that F ( T ) satisfies (27).Since ( K, L, L q ) and ( K, L q , L ) give rise to the same polynomial F ( T ) = ( T + K )( T + L )( T + L q ), in total the number of polynomials F ( T ) satisfying (27) is12 (cid:18) q q − q )( q − (cid:19) = q − q + 4 q . (cid:3) Proposition 2.17.
The number of polynomials F ( T ) = ( T + K ) , K ∈ F q , satisfying (27) is q − n, . Proof.
Since F ( T ) = T + ST + RT + P , we have S = K , R = K , and P = K . Note that K / ∈ { , } otherwise S = P . Then (27) reads(31) U + U = 1 K + 1 , which has two roots in U for ( q − / K ∈ F q \ { , } if n is even, and( q − / K ∈ F q \ { , } if n is odd. (cid:3) N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q Proposition 2.18.
The number of polynomials F ( T ) = ( T + K )( T + L ) , K, L ∈ F q , K = L ,which satisfy (27) is q − q + 2 + 2 gcd( n, . Proof.
Since F ( T ) = T + ST + RT + P , we have S = K , R = L , and P = KL . Note that K = 0 and L = 1 otherwise S = P . Then from the equation in (28) one gets(32) K ( L + 1)( U + U ) = L ( K + 1) . If U ∈ { , } , then K = 1 and L ∈ F q \ { } , or L = 0 and K ∈ F q \ { } . The pair ( K, L ) = (1 , q − K, L ) satisfying K = 0, L = 1and (32).From now on suppose U / ∈ { , } . We have L = 1 so we can put V := L / ( L + 1). Then V = 1 and L = V / ( V + 1). Each solution ( K, V ) of K ( U + U ) = V ( K + 1) corresponds toan affine point of an irreducible conic with 2 F q -rational points at infinity. There is one affinesolution with K = 0 and 2 affine solutions with V = 1. It follows that for each U there are q − K = 0 and V = 1, hence the same number of pairs of solutions( K, L ) with K = 0 and L = 1.For U and U + 1 there corresponds the same pair of solution ( K, L ) and hence in total thereare 2(2 q −
3) + ( q − q − q − q + 22suitable pairs ( K, L ). By Proposition 2.18, ( q − n, / K = L . Aftersubtracting this number we obtain the number of polynomials F ( T ) satisfying (27). (cid:3) Theorem 2.19. If q is even then | Γ | = ( q − q − q − .Proof. Finally, we will count the number of norms satisfying (22). For i ∈ { , , , } , denoteby γ i the number of polynomials F ( T ) = T + ST + RT + P ∈ F q [ T ] with exactly i distinctroots in F q and satisfying (27). By Proposition 2.14, γ + 2 γ + 3 γ = q − q . By Propositions 2.16 and 2.17, γ = q − q + 4 q q − n, , and by Proposition 2.18, γ = q − q + 2 + 2 gcd( n, . Combining these results, we obtain γ = q − q + 4 q − n, − . N A CONJECTURE ABOUT MAXIMUM SCATTERED SUBSPACES OF F q × F q By Proposition 2.13 the number of polynomials satisfying (27) is ( q − q ) / γ = q − q − ( γ + γ + γ ) = q − q − q + 2 gcd( n, − . The number of values of N such that (22) is satisfied equals 3 times the number of irreduciblepolynomials F ( T ) satisfying (27) (corresponding to solutions with N ∈ F q \ F q ) plus thenumber of polynomials F ( T ) with a unique 3-fold root in F q and satisfying (27) (correspondingto solutions with N ∈ F q ). Thus this number is3 γ + q − n, q − q − q − . (cid:3) References [1]
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