Let ℌ be a class of n×n Hankel matrices HA whose entries, depending on a given matrix A, are linear forms in n variables with coefficients in a finite field 𝔽q. For every matrix in ℌ, it is shown that the varieties specified by the leading minors of orders from 1 to n-1 have the same number qn-1 of points in 𝔽qn. Further properties are derived, which show that sets of varieties, tied to a given Hankel matrix, resemble a set of hyperplanes as regards the number of points of their intersections.
1. Introduction
The representation of hypersurfaces of small degree as determinants is a classical subject. For instance, Hesse [1] discussed the representation of the plane quartic by symmetric determinants, and many different problems have been tackled over the years; see, for example, [2, 3]. An important question, when hypersurfaces are defined over finite fields, is the computation of the number of points. In general this is very difficult, for example, [4], and most frequently only bounds are given. This paper considers hypersurfaces over finite fields, which are defined by determinants of Hankel matrices whose entries are linear forms in the variables. These Hankel matrices are encountered in the proof of certain properties of finite state automata whose state change is governed by tridiagonal matrices [5, 6]. They also occur in the study of some decoding algorithms for error-correcting codes [7, 8].
It is remarkable that, for these determinantal varieties, the exact number of points can in many instances be explicitly found, in terms of the size of the field and the number of variables.
Let p(z)=zn+a1zn-1+a2zn-2+⋯+an-1z+an be an irreducible polynomial of degree n over 𝔽q with root α∈𝔽qn, which is thus an eigenvalue of the companion matrix A which is assumed to have the coefficients of p(z) in the last column, all 1s in the first subdiagonal, and the remaining entries are 0s [9].
The definition of Hankel matrices that we are dealing with uses the Krylov matrices
(1)K(A,x)=(x,Ax,A2x,…,An-1x),K(AT,yT)=(yT,ATyT,(AT)2yT,…,(AT)n-1yT),
where y=(y1,…,yn) is a row vector of n independent variables and xT=(x1,…,xn) is a column vector of n independent variables. Every Krylov matrix is nonsingular unless x and y are all-zero vectors, as will be proved later.
Definition 1.
The class ℌ consists of n×n matrices defined as
(2)HA=K(AT,yT)TK(A,x).
These are Hankel matrices, because the entries
(3)(HA)ij=yAiAjx=yAi+jx
are clearly the same whenever the index sum i+j=h is constant. When the vector y is a fixed element yo of 𝔽qn, the corresponding subclass of ℌ is denoted by ℌ(yo).
Given a polynomial f in the ring 𝔽q[x1,…,xn], the variety 𝒱(f) is defined as the set of points in the affine space 𝔽qn that annihilate f; that is,
(4)𝒱(f)={(a1,a2,…,an)∈𝔽pn∣f(a1,…,an)=0}⊆𝔽pn.
More generally, given s polynomials f1,…,fs∈𝔽p[x1,…,xn] the variety 𝒱(f1,…,fs) is the set of solutions of the system
(5)f1=0,…,fs=0.
Note that 𝒱(f1,…,fs)=⋂i=1s𝒱(fi) is the intersection and 𝒱(f1f2,…,fs)=⋃i=1s𝒱(fi) is the union of the varieties 𝒱(f1),…,𝒱(fs).
The entries in HA are bilinear forms of the entries in y and x. Let Dj(x1,…,xn) denote the leading minor of order j of a given Hankel matrix HA obtained fixing y=(b1,…,bn)∈𝔽qn, and define the determinantal varieties as 𝒱(Dj(x1,…,xn))≔{(a1,…,an)∈𝔽qn:Dj(a1,…,an)=0}. Then, we prove that every polynomial Dj(x1,…,xn) is irreducible over 𝔽¯q (Proposition 10), and obtain the following general result.
Theorem 2.
We have |𝒱(Dj(x1,…,xn))|=qn-1 if j=1,…,n-1 and |𝒱(Dn(x1,…xn))|=1.
While proving this theorem, the cardinality of certain subsets S(i,j,n)⊂𝔽qn is also computed. The sets S(i,j,n) are the zero-loci of all Dh(x1,…,xn)’s with i≤h≤j (Theorem 18). That is, every S(i,j,n) is specified by j+1-i equations of degree higher than 1; nevertheless its cardinality qn+i-1-j is the same as in the case of the intersections in 𝔽qn of j+1-i distinct hyperplanes. In the next section, preliminary notions, properties, and useful lemmas are collected, while the main results are proved in Section 3.
2. Preliminaries
It is direct to check that vα=(1,α,…,αn-1) is a row eigenvector of A, associated with the eigenvalue α; that is, vαA=αvα.
Let σ:𝔽¯q→𝔽¯q denote the q-Frobenius; that is, set σ(x)≔xq for all x. The action of σ is extended to vectors and matrices component-wise. Since σ(A)=A, because the entries of this matrix are in 𝔽q, we have
(6)σ(vαA)=σ(αvα)⟹σ(vα)A=σ(α)σ(vα);
that is, all eigenvectors of A are conjugate vectors under σ. Hence the matrix
(7)B=[vασ(vα)σ2(vα)⋮σn-1(vα)]
reduces A to diagonal form over 𝔽qn; that is,
(8)D=BAB-1=diag(α,…,σn-1(α)),D being the diagonal matrix of the eigenvalues of A.
Observe that, writing (8) as AB-1=B-1D, the columns of B-1 are column eigenvectors of A. Thus there is a column vector u that allows us to write B-1 in the form
(9)B-1=(uα,σ(uα),…,σn-1(uα)).
The following lemma is useful to show that every matrix similar to A gives the same class ℌ. Let GL(n,𝔽q) denote the general linear group of n × n nonsingular matrices with entries in 𝔽q.
Lemma 3.
Matrices of GL(n,𝔽q) that have the same characteristic irreducible polynomial p(z) are 𝔽q-similar.
Proof.
Let α be a root of p(z). To prove the lemma it is sufficient to show that any two matrices A and E of GL(n,𝔽q), having the same characteristic polynomial p(z), are similar. The previous arguments indicate that there are two 𝔽qn-matrices B and S of form (7) such that
(10)BAB-1=D,SES-1=D.
Multiplying the first equation by S-1 on the left, and by S on the right, we have (S-1B)A(S-1B)-1=E. Thus, the lemma is proved by showing that S-1B is a 𝔽q-matrix. Since we may always assume that
(11)S-1=(sα,σ(sα),…,σn-1(sα)),
where sα is a convenient column eigenvector of E and B is of form (7), we have
(12)S-1B=sαvα+σ(sα)σ(vα)+⋯+σn-1(sα)σn-1(vα)=sαvα+σ(sαvα)+⋯+σn-1(sαvα),
which is patently invariant under the action of the automorphism σ; thus S-1B is a 𝔽q-matrix.
Corollary 4.
A and AT are 𝔽q-similar.
2.1. B and z
The equation w=yB-1 defines an 𝔽q-linear mapping ψ from 𝔽qn into 𝔽qnn. Taking the vector y to be the element of 𝔽qn(13)yo=(Tr(1),Tr(α),Tr(α2),…,Tr(αn-1))∈𝔽qn,
we have wo=(1,1,…,1)=ψ(yo). The image Im(ψ) is the 𝔽qn-linear span of (1,1…,1); hence it is a one-dimensional 𝔽qn-linear subspace of 𝔽qnn.
Equation (8) implies that BA=DB; then, introducing the vector
(14)zT=B(x1,x2,…,xn)T=(z1,z2,…,zn),
it is immediate to see that zi=σi-1(z1) for every i, whenever xT∈𝔽qn. The linear forms yoAjxT are transformed into linear forms woDjzT, and matrix HA can be written as
(15)HA=[woTzwoTDzwoTD2z⋯woTDn-1zwoTDzwoTD2zwoTD3z⋯woTDnzwoTD2zwoTD3zwoTD4z⋯woTDn+1z⋮⋮⋮⋯⋮woTDn-1zwoTDnz⋯woTD2n-2z].
Definition 5.
Let Dj(x1,…,xn) denote the leading minor of order j of a given Hankel matrix HA in ℌ. When there is no ambiguity surrounding the variables, this minor is in brief denoted by Dj(1). The determinant of HA is Dn(x1,…,xn), or Dn(1).
Lemma 6.
Let x be a vector of n variables, and let y be a constant vector in 𝔽qn; then we have the following.
The determinant Dn(x1,…,xn) of HA is zero if and only if all variables are set equal to zero.
The matrix HA is a linear combination of n nonsingular matrices, the coefficients of the linear combination being the entries of x.
Any linear combination of the rows of HA is a set of n linearly independent linear forms.
Proof.
Dn(x1,…,xn) is not zero, because it is the product of two determinants that are different from zero
(16)Dn(x1,…,xn)=det(K(AT,yT)TK(A,x))=det(K(AT,yT)T)det(K(A,x)).
In particular, det(K(A,x))=0 if and only if x is the all-zero vector; the same observation holds for y. This proves point (1).
Point (2) is proved by writing
(17)K(AT,yT)TK(A,x)=∑i=1nxiMi,
where the matrices Mi have constant entries that depend on y, and taking xj=1 and xi=0 for every i≠j, that is, x=ej. When x=ej, we have
(18)Dn(0,…,1,…,0)=det(K(AT,yT)T)det(K(A,ej))=det(Mj).
This implies that det(Mj)≠0.
Point (3) is proved by noting that Dn(x1,…,xn)=0 has only one solution, namely, x1=⋯=xn=0, and Dn(x1,…,xn)=0 identifies linear combinations of the rows of HA. It follows that every linear combination of the rows should have only the all-zero solution; therefore the n entries in every row must be linearly independent, by a theorem of Rouché-Capelli.
By correspondence (14), every Dj(x1,x2,…,xn) is transformed into a polynomial Qj(z1,…,zn) in the variables zis with the coefficients in 𝔽qn.
2.2. Auxiliary Results
Let V(a1,a2,…,aj) be a Vandermonde determinant of order j identified by the j-tuple (a1,a2,…,aj).
Definition 7.
For every triple of integers j, i, and t such that t≥2i-1≥2j-1>0, the subset S(j,i,t) of 𝔽qt is defined as
(19)S(j,i,t)≔{(a1,…,at)∈𝔽qt:Dh(a1,…,at)=0}mmmmmmmmmmmmmmmmm∀h∈{j,…,i}.
Definition 8.
The set ℭjn is defined to be the collection of (nj) subsets, where each subset consists of the unordered collection of j distinct integers from the set {1,2,…,n}.
Every subset 𝔥i={h1,…,hj} defines a mapping τi(ℓ)=hℓ from the set {1,2,…,j} into {1,2,…,n}.
Lemma 9.
Consider a Hankel matrix HA, as defined in (2) with y=yo∈𝔽qn; the leading minors Dj(x1,…,xn), j=1,…,n are multivariate homogeneous polynomials of degree j, which may be written over 𝔽qn, in the form
(20)Dj(x1,…,xn)=Qj(z1,…,zn)=∑{h1,h2,…,hj}∈CjnV(σh1-1(α),…,σhj-1(α))2zh2⋯zhj,
where the summation is extended to all combinations of the n integers {1,2,…,n}, taking j at a time, and the coefficients of the monomials zh1⋯zhj are squares of Vandermonde determinants.
Proof.
In matrix (15), the bilinear forms wDi+j-2z have the explicit expression
(21)∑h=1nzhσh-1(αi+j-2)=∑h=1nzhσh-1(αi-1)σh-1(αj-1),
where i is the row index and j is the column index. Each column is a linear combination of columns with coefficients zh such that all columns with the same coefficient zh are proportional. Matrix (15) can be written as a sum of the form
(22)HA=∑h=1nzhσh-1(Γ),
where Γ is the n×n matrix
(23)[1α⋯αn-1αα2⋯αn⋮⋮⋮⋮αn-1αn⋯α2n-1],
which has rank 1, since every row is proportional to the first row, and the same holds for the columns. The leading minor Dj(x1,…,xn) is computed by writing the determinant as a sum of nj determinants, which contain a single variable zk in every column, determinants with repeated variables are 0, because of the previous observation that their corresponding columns are proportional, and in the remaining determinants the corresponding variable is collected from each column.
The coefficient of the monomial zh1zh2⋯zhj is obtained as follows. Let ui=σhi-1(α). Then the coefficient of zh1,…,zhj is equal to
(24)∑τ∈Sj|1uτ(2)⋯uτ(j)uτ(1)uτ(2)2⋯uτ(j)j⋮⋮⋮⋮uτ(1)j-1uτ(2)j⋯uτ(j)2j-1|=∑τ∈Sjsgn(τ)∏ℓ=1juτ(ℓ)ℓ-1|11⋯1u1u22⋯ujj⋮⋮⋮⋮u1j-1u2j⋯uj2j-1|.
Collecting the common factor, the remaining summation is exactly the same determinant; thus we have
(25)|11⋯1u1u22⋯ujj⋮⋮⋮⋮u1j-1u2j⋯uj2j-1|2=V(σh1-1(α),…,σhj-1(α))2,
which gives
(26)Qj(z1,…,zn)=∑𝔥∈ℭjnV(σh1-1(α),…,σhj-1(α))2zh1zh2⋯zhj,
with the summation extended to every subset 𝔥={h1,…,hj} of ℭjn, and this concludes the proof.
Proposition 10.
The product ∏i=1n-1Qi(z1,…,zn) is not identically zero over 𝔽qn.
Furthermore, the leading minors Dj(x1,…,xn), j=1,…n, are irreducible degree-j polynomials over 𝔽¯q.
Proof.
As a consequence of (26), every Qj(z1,…,zn) is irreducible over 𝔽qn. Further, observing that each variable zi occurs at degree 1 in any Qj(z1,…,zn), it has maximum degree n-1 in the product polynomial 𝔇=∏j=1n-1Qj(z1,…,zn). Therefore 𝔇 is not identically zero in 𝔽qnn, because n-1 is certainly less than qn for any q.
To prove the second statement, fix j∈{1,…,n-1}. In this step it is checked that g≔Dj(x1,…,xn)∈𝔽q[x1,…,xn] is irreducible over 𝔽¯q. It is only necessary to use the fact that g∈𝔽q[x1,…,xn] is a homogeneous polynomial of degree j<n which is irreducible over 𝔽qn. Assume that g is reducible over 𝔽¯q and call h∈𝔽¯q an irreducible factor of minimal degree x<j. Let 𝔽qe be the minimal extension of 𝔽q in which h is defined. Since g is irreducible, the polynomials σi(h), 1≤i<e, obtained by applying the Frobenius σi to h are nonproportional irreducible factors of g. Hence deg(g)≥(e-1)x≥nx≥n, which is a contradiction.
Remark 11.
The determinant Dn(x1,…,xn) is found to be
(27)Dn(x1,…,xn)=Δ2∏u=1nzu=p(1)2δ2∏u=1n[σu-1(vα)x],
where δ is the discriminant of p(z), and the product involving x can be seen as a norm in the field 𝔽qn; therefore Dn(x1,…,xn) is irreducible over 𝔽q.
Lemma 12.
The variety 𝒱(D1(1))∩𝒱(D2(1))∩⋯∩𝒱(Dn-1(1)) has cardinality q over 𝔽q.
Proof.
Equation (17) shows that any Dj(1) is a polynomial of degree j with coefficients in 𝔽q; furthermore, every entry is a linear form with coefficients in 𝔽q. Hence D1(1)=0 implies u1=0; in turn D2(1)=0 implies u2=0, given that u1=0 and arguing recursively; finally, Dn-1(1)=0 implies un-1=0, while the variable un is free and may assume q values. The conclusion follows.
Lemma 13.
Let bi≠0, 1≤i≤j, and assume n≥2j-1. Then |𝒱(D1(1)-b1)∩⋯∩𝒱(Dj(1)-bj)|=qn-j.
Proof.
We use induction on j, the case j=1 being obvious. The inductive assumption in 𝔽q2j-3 gives |𝒱(D1(1)-b1)∩⋯∩𝒱(Dj-1(1)-bj-1)|=qj-2. Fix (a1,…,a2j-3)∈𝔽q2j-3 with Di(1)=bi for all 1≤i≤j-1. Since Dj-1(a1,a2,…,a2j-3)≠0, for all a2j-2,c∈𝔽q there is a unique a2j-1∈𝔽q such that Dj(a1,…,a2j-1)=c. Take c=bj. This completes the proof.
3. Main ResultsProposition 14.
The equality |V(Dj(1))|=|V(Dn-j(1))| holds for every 1≤j≤n-1.
Proof.
In the proof of Lemma 9, it was shown that
(28)Dj(1)=Dj(x1,…,xn)=Qj(z1,…,zn),
with zi=∑j=1nσj-1(α)xj. Further, it was noted in that lemma that zi=σi-1(z1) for every i, whenever every xj∈𝔽q.
The relation z=B-1x establishes a one-to-one correspondence between 𝔽qn and a subspace of dimension 1 of 𝔽qnn, and further x=Bz. There thus exists a one-to-one correspondence between the zeros of Dj(1) in 𝔽qn and the zeros of Qj(z1,…,zn) in the one-dimensional subspace of 𝔽qnn, which is the image of 𝔽qn. Referring to (26), which yields the representation Qj(z1,…,zn) of Dj(1), assuming z1≠0, considering the change of variables
(29)zi=1ti∏ℓ≠i-1n-1(σi-1(α)-σℓ(α))2=1𝔡iti,
and recalling that the coefficients of the monomials are squares of Vandermonde determinants V(σh1-1(α),σh2-1(α),…,σhj-1(α))2, we obtain
(30)Qj(z1,…,zn)=1δ2∏i=1ntiQn-j(t1,…,tn),
where δ is the discriminant of the polynomial with root α. The variety 𝒱(Dj(1)) is obtained by considering t1=v(α)xT and the other variables as ti=σi-1(t1), i=2,…,n; thus ∏ti=0 only when every ti=0, ∀i; further δ≠0. Finally, we have the chain of bijections
(31)x⟷zT=B-1xT⟷z1⟷t1={1𝔡1z1ifz1≠00ifz1=0⟷yT⟷x~T=BtT.
In conclusion, this equation shows an explicit one-to-one mapping between the zeros (a1,…,an) of Dj(1) and the zeros (a~1,…,a~n) of Dn-j(1), which implies |𝒱(Dj(1))|=|𝒱(Dn-j(1))|.
In the following example, the procedure for obtaining a point of 𝒱(Dn-j(1)) from a point of 𝒱(Dj(1)) is explicitly illustrated.
Example 15.
Consider the irreducible polynomial p7(z)=z7+z4+z3+z2+1 of degree n=7 over 𝔽3 with the transpose companion matrix
(32)A7=[010000000100000001000000010000000100000001-10-1-1-100].
Taking y=[1,0,0,0,0,0,0], and xT=[x1,x2,x3,x4,x5,x6,x7], the Hankel matrix (17) becomes
(33)HA7=[x1x2x3x4x5x6x7x2x3x4x5x6x7ξ8x3x4x5x6x7ξ8ξ9x4x5x6x7ξ8ξ9ξ10x5x6x7ξ8ξ9ξ10ξ11x6x7ξ8ξ9ξ10ξ11ξ12x7ξ8ξ9ξ10ξ11ξ12ξ13],
where
(34)ξ8=-x1-x3-x4-x5,ξ9=-x2-x4-x5-x6,ξ10=-x3-x5-x6-x7,ξ11=x1+x3+x5-x6-x7,ξ12=x1+x2+x3-x4+x5+x6-x7,ξ13=x1+x2-x3-x4+x5+x6-x7.
The forms D3(1) and D4(1) of degrees 3 and 4, respectively, are
(35)D3(1)=x1x3x5-x1x42-x22x5-x2x3x4-x33,D4(1)=x1x3x5x7-x1x3x62-x1x42x7-x1x4x5x6-x1x53D4(1)-x22x5x7+x22x62-x2x4x3x7+x2x42x6+x2x5x3x6D4(1)-x2x4x52-x33x7-x32x4x6+x32x52+x44.
Given a point xT=[0,1,-1,-1,0,0,1]∈𝔽37 which is a zero of D3(1), a zero of D4(1) is obtained as follows.
Compute the vector z=B7-1xT whose first component is z1=1+α2+α3+α6∈𝔽377, and the remaining entries are obtained as σℓ(z1)=1+α3ℓ-12+α3ℓ-13+α3ℓ-16, ℓ=1,…,6; then compute
(36)𝔡2=∏i=16(α-α3i)2=α+α3+α4-α5,t1=1z1𝔡2=-1-α+α2+α4-α5,
and construct the vector t=[t1,σ(t1),…,σ6(t1)]∈𝔽377. Finally, a zero of D4(1) is obtained as
(37)B7tT=[1,-1,0,1,1,-1,-1]T∈𝔽37,
where B7 is the matrix whose columns are the eigenvectors of A7 in 𝔽377(38)B7=[1111111αα3α32α33α34α35α36α2α2·3α2·32α2·33α2·34α2·35α2·36α3α3·3α3·32α3·33α3·34α3·35α3·36α4α4·3α4·32α4·33α4·34α4·35α4·36α5α5·3α5·32α5·33α5·34α5·35α5·36α6α6·3α6·32α6·33α6·34α6·35α6·36].
Remark 16.
Since the n forms in the first row of (17) are linearly independent, by Lemma 6, a change of variables from x1,…,xn to u1,…,un takes a matrix HA to the form
(39)HA=[u1u2u3⋯unu2u3⋯unℓ1u3⋯unℓ1ℓ2⋮unℓ1ℓ2⋯ℓn-1],
where the n variables uis are free, and every ℓh is a linear form in the uis.
Fix the integers k≥1 and j≥1, and let Dj(k) denote the j × j determinant of a Hankel matrix with free variable entries ui, i=k,…,2j-2+k(40)|ukuk+1⋯uk+j-2uk+j-1uk+1uk+2⋯uk+j-1uk+j⋮⋮⋮uk+j-1uk+j⋯uk+2j-3uk+2j-2|.
And set D0(1)=1 by definition.
Proposition 17.
Let m,k be natural integers. Let H(2k+2m-1) be a (2k+2m-1)×(2k+2m-1) Hankel matrix with first row (u1,…,u2k+2m-1). Let E(k,m) be the set of points (b1,…,b2k+2m-1)∈𝔽q2k+2m-1 with the same first 2k-1 coordinates bi=ai, i=1,…,2k-1 such that the minor Dk(b1,…,b2k+2m-1)≠0, and the minors Di(b1,…,b2k+2m-1)=0 for all i∈{k+1,…,k+m}. Then E(k,m) has cardinality qm.
Proof.
Observe that the first row (u1,…,u2k+2m-1) of the Hankel matrix H(2k+2m-1) completely specifies the leading (k+m)×(k+m) Hankel submatrix H(k+m), and consequently also every minor Di(1) for i=1,…,k+m.
Let Rj(u1,…,u2k+2m-1) denote the jth row of H(k+m). Let A(k+1,m) be the subset of 𝔽q2m consisting of all (b2k,…,b2k+2m-1)∈𝔽q2m such that Rk+1(a1,…,a2k-1,b2k,…,b2k+2m-1) is linearly dependent on R1(b1,…,a2k-1,b2k,…,b2k+2m-1),…, Rk(b1,…,a2k-1,b2k,…,b2k+2m-1).
The case m=1 is easily settled. Consider the identity
(41)Dk+1(a1,…,a2k-1,u2k,u2k+1,…,u2k+2m-1)=u2k+1Dk(a1,…,a2k-1,u2k,u2k+1,…,u2k+2m-1)+B(a1,…,a2k-1,u2k),
for some B(u1,…,u2k)∈𝔽q[u1,…,u2k], and take u2k=a2k∈𝔽q; it follows that
(42)Dk+1(a1,…,a2k-1,a2k,a2k+1,u2k+2,…,u2k+2m-1)=0,
for a unique a2k+1∈𝔽q because Dk(1)≠0 by hypothesis. Since u2k is any element a2k∈𝔽q (i.e., it may assume q values in 𝔽q, while u2k+1 is uniquely specified), the assertion |E(k,1)|=q is proved.
Now, assume m≥2, and note that row k+1 is uniquely determined up to position k+1 as a linear combination of the above rows up to the same position k+1. Extend this linear combination to uniquely determine the remaining elements of the H(k+m) Hankel matrix.
The assertion |E(k,m)|=qm is a consequence of the following claims.
Claim 1. One has |A(k+1,m)|=qm.
Consider a vector (b2k,…,b2k+2m-1)∈𝔽q2m, which belongs to A(k+1,m) if and only if there are ci∈𝔽q, 1≤i≤k, such that
(43)Rk+1(a1,…,a2k-1,b2k,…,b2k+2m-1)=∑i=1kciRi(a1,…,a2k-1,b2k,…,b2k+2m-1),
since Dk(a1,…,a2k-1,b2k,…,b2k+2m-1)≠0, and this same condition implies that the coefficients c1,…,ck are uniquely determined by the entries of the vector (a1,…,a2k-1) and by the entry b2k=a2k in row Rk+1(a1,…,a2k-1,b2k,…,b2k+2m-1).
We know that for each a2k∈𝔽q there is a unique a2k+1 such that Dk+1(a1,…,a2k-1,a2k,a2k+1)=0.
Fix u2k=a2k and hence fix c1,…,ck, and u2k+1=a2k+1. The values of u2k+i=a2k+i, i=2,…,m are uniquely specified by the linear combination condition, jointly with the Hankel matrix properties. Since the remaining u2k+m+i, i=1,…,m-1 are free, the cardinality of A(k+1,m) is precisely qm.
Claim 2. Since the first k+1 rows of the Hankel matrix H(k+m) are linearly dependent, it follows that Dj(a1,…,a2k+m,b2k+m+1,…,b2k+2m-1)=0 for every j∈{k+2,…,k+m}.
To conclude the proof, it remains to show that the (k+1)th row, constructed as above, is the only possible (k+1)th row that leads to a Hankel matrix satisfying the hypotheses of the proposition. This property is the third claim.
Claim 3. If b2k+i≠a2k+i, for every i=2,…,m, then Dx+k(a1,…,a2k-1,a2k,a2k+1,b2k+2,…,b2k+2m-1)≠0 for every x∈{2,…,m}.
Let x≥2 be the smallest integer such that the (k+x)×(k+x) Hankel matrix H(k+x), with leading minor Dk(1)≠0, has the whole (k+1)th row that is not a linear combination of the above rows: this means that the entry bk+x is different from ak+x.
Let c1,…,ck be the coefficients of the linear combination of the first k rows of H(k+1+x) yielding the row (ak,…,ak+1+x).
From every hth row of the matrix H(k+x), with h≥k+1, the linear combination of the first k rows may be subtracted to get a row whose entries with index y are zero for every y=1,…,h+k+1. The counter-diagonal entries between row k+1 and the bottom row are b2k+x-a2k+x. The determinant of H(k+x) and that of the modified matrix are the same; using the generalized Laplace formula for the expansion of a determinant with respect to the last x+1 rows, we get Dk+1+x(1)=Dk(1)(a2k+x-b2k+x)x≠0.
The contradiction forces b2k+x=a2k+x, which concludes the proof.
Theorem 18.
For all integers n, j, and i such that n≥2j-1≥2i-1>0 we have
(44)|S(j,i,n)|=qn-1-j+i.
Proof.
For all integers n≥t≥2j-1 we have |S(j,i,n)|=|S(j,i,t)|·qn-t, because in each determinant Dh(1), h≤j, the variables xe, e≥2j, do not occur; hence Lemma 12 gives the case i=1 for all j. We may thus assume i≥2. Induction will be applied to j, the case j=2 being obvious. The inductive assumption gives
(45)|S(h,h,2h-1)∖S(h,h-1,2h-1)|=(q-1)q2h-3,
for all h<j. Notice that S(j,i,n)=S(j,1,n)⊔{⊔h=2i(S(j,h,n)∖S(j,h-1,n))}. Lemma 9 gives |S(j,h,n)∖S(j,h-1,n)|=(q-1)q2h-3qn-2j+1+j-h. Hence
(46)|S(j,i,n)|=qn-j+(q-1)∑h=2iqn-j+h-2=qn-j-1+i.
Remark 19.
Take integers n, j, and i such that n≥2j-1≥2i-1≥3. Applying Theorem 18, first for (n,j,i) and then for (n,j,i-1), gives |S(j,i,n)∖S(j,i+1,n)|=qn-j-2+i(q-1).
Proof of Theorem 2.
We know by Lemma 6 that |𝒱(Dj(1))|=|𝒱(Dn-j(1))|, for every 1≤j≤n-1; the proof is completed by showing that |𝒱(Dj(1))|=qn-1 for every 1≤j≤⌊n/2⌋. This is true by the case i=j of Theorem 18. 𝒱(Dn(1)) has only one point, because Dn(1) is an irreducible polynomial over 𝔽q.
Corollary 20.
Given j<n, if gcd{j,q-1}=1, the varieties 𝒱(Dj(1)-b), ∀b∈𝔽q have cardinality qn-1.
Proof.
Performing the substitution xi=txi′ gives the equation tjDj(x1′,…,xn′)-b=0. By hypothesis gcd{j,q-1}=1, the equation tj=b always has a solution in 𝔽q, since we have t=bμ with μj=1 mod q-1. Thus all varieties with b≠0 have the same cardinality, say nb=|𝒱(Dj(1)-1)|, and the equation
(47)qn-1+(q-1)nb=qn
implies nb=qn-1.
Note that, when j has some factor in common with q-1, the cardinalities of 𝒱(Dj(1)-b) are close to qn-1 but depend on b. It is an interesting problem to determine how close these cardinalities are to q(n-1).
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
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