Determinantal polynomials and the base polynomial of a square matrix over a finite field

Abstract

Purpose

The author studies forms over finite fields obtained as the determinant of Hermitian matrices and use these determinatal forms to define and study the base polynomial of a square matrix over a finite field.

Design/methodology/approach

The authors give full proofs for the new results, quoting previous works by other authors in the proofs. In the introduction, the authors quoted related references.

Findings

The authors get a few theorems, mainly describing some monic polynomial arising as a base polynomial of a square matrix.

Originality/value

As far as the author knows, all the results are new, and the approach is also new.

Citation

Ballico, E. (2024), "Determinantal polynomials and the base polynomial of a square matrix over a finite field", Arab Journal of Mathematical Sciences, Vol. 30 No. 2, pp. 245-255. https://doi.org/10.1108/AJMS-10-2022-0242

1. Introduction

For any field K, any positive integer m and any number of variables t₁, …, t_m, we call K[t₁, …, t_m] the polynomial ring over K with variables t₁, …, t_m, not the vector space of all polynomial functions Km→K. These two rings are isomorphic if and only if the field K is infinite. If K is a finite field, then the ring of polynomial functions Km→K is isomorphic to K[t1,…,tm]/(t1#K−t1,…,tm#K−tm). In this paper, we are always taking K finite, with either #K = q or #K = q², where q is a fixed prime power.

Fix a prime p and a p-power q. For any M=(mij)∈Mn,n(Fq2), let M^† denote the matrix (mjiq). M is said to be Hermitian if M = M^†. Note that the diagonal elements of a Hermitian matrix are elements of Fq and that the set of all Hermitian n × n matrices forms an Fq vector space of dimension n². We briefly recall the notion of Hermitian geometry for the Galois degree 2 extension Fq2 of Fq. The Frobenius map σ:t→tq is a generator of the Galois group of this degree 2 extension. The Hermitian form (i.e. σ-sesquilinear form) 〈,〉:Fq2n×Fq2n→Fq2 is defined by the formula

〈(u1,…,un),(v1,…,vn)〉≔∑i=1nuiqvi.

Fix positive integers m, n and m n × n Hermitian matrices M1,…,Mm∈Mn,n(Fq2). Set

fM1,…,Mm(t1,…,tm)≔det(t1M1+⋯+tmMm)

and call it the determinantal polynomial of the Hermitian matrices M₁, …, M_m. For m ≥ 2 set

gM1,…,Mm−1(t1,…,tm)≔fM1,…,Mm−1,In×n(t1,…,tm).

We say that gM1,…,Mm−1,In×n(t1,…,tm) is the base polynomial of the Hermitian matrices M₁, …, M_m−1.

All polynomials fM1,…,Mm(t1,…,tm) are homogeneous degree n polynomials with coefficients in Fq (Lemma 1).

The motivation for this paper came from Kippenhahn’s paper on the numerical range, his definition of the base polynomial f(x, y, z) and his use of the dual curve of the plane curve {f(x, y, z) = 0} to characterize the numerical range ([1, 2]), which is even now a source of inspirations ([3, 4]). The numerical range of a matrix is also defined for matrices M∈Mn,n(Fq2) ([5–8]), using a choice of a certain element β∈Fq2\Fq ([5, 6]). With this choice for any M∈Mn,n(Fq2), we get uniquely determined Hermitian matrices M+,M−∈Mn,n(Fq2) such that M = M₊ + βM₋ (see References [1, 2] for more details). The field Fq2 is a degree 2 extension of Fq. First assume q odd. There is α∈Fq, which is a square in Fq2, but not in Fq. We take β∈Fq2 such that β² = α and set M₊≔(M + M^†)/2 and M₋≔(M − M ^†)/2β. Now assume q even. There is ε∈Fq such that the polynomial t² + t + ɛ has no root in Fq. We call β one of its root in Fq2 (the other one is β + 1). We set M₋≔M + M^† and M₊≔(β + 1)M + βM^†.

Using M₊ and M_−, one can use Kippelmahn’s definition of the base polynomial of a square complex matrix and set

bp(M)(x,y,z)=gM+,M−(x,y,z)=det(xM++yM−+zIn×n).

Note that bp(M) is a homogeneous degree n polynomial with zⁿ as one of its monomials and that its coefficient is 1. We call monic such degree n forms. A form f∈Fq[t1,…,tm] is said to be concise if there is no linear change of coordinates such that in the new coordinates f does not depend on all coordinates. For degree 2 forms conciseness is equivalent to the smoothness of their zero-locus (Remark 10).

In Sections 4 and 5, we study the realizability problem (which monic forms are of the form bp(A) for some A) for 2 × 2 matrices. At the end of Section 4, we collect several questions concerning the base polynomials.

We get some negative results, i.e. many matrices have base polynomials not interesting and unrelated to the numerical range of any non-zero matrix. We prove the following result.

Theorem 1.

Fix A∈Mn,n(Fq2).

(i) Assume either A = A₊ or A = βA₋. Then bp(A) = zⁿ if and only if 0 is the unique eigenvalue of A over F¯q.

(ii) There are q² 2 × 2 matrices A such that A = A₊ (resp. A = βA₋) and bp(A) = z².

(iii) Assume n = 2. Then bp(A) = z² if and only if there are a∈Fq, e∈Fq, c∈Fq2, d∈Fq2 such that

(1) −a2=cq+1,−e2=dq+1,−2ae=cqd+cdq

and A = A₊ + βA₋, where

(2) A+=accq−a

(3) A−=eddq−e

(iv) Assume q even. There are (q − 1)(q² − 1) matrices A∈M2,2(Fq2) such that bp(A) = z², A₊ ≠ 0 and A₋ ≠ 0. Each such A is of the form A = A₊ + βA₋ with A₊ and A₋ as in (2) and (3). Each such matrix A is obtaining taking c∈Fq2\{0}, t∈Fq\{0}, setting d ≔ tc and taking as a and e the only elements of Fq such that a² = c^q+1 and e² = d^q+1.

(v) Take q odd. There are at least q² matrices A∈M2,2(Fq2) such that A₊ ≠ 0, A₋ ≠ 0 and bp(A) = z². Some of them may be obtained taking A₊ as in (2) and taking A = A₊ + βA₊.

Remark 1.

Concerning part (i) of Theorem 1, we have a complete description of the q² matrices. The ones with A = A₊ (resp. A = βA₋) are the ones described in (2) (resp. (3)) with a, c (resp. e, d) as in (1).

We get some positive results (obtaining a monic polynomial as the base polynomial of a square matrix). This is called the reconstruction problem for monic polynomials. We prove the case of 2 × 2 matrices, i.e. we prove the following result.

Proposition 1.

All monic degree 2 forms are realized as a base polynomial.

Definition 1.

Let K be a field. Take f ∈ K[x₁, …, x_n]. We say that f depends on n variables or that it does not depend on <n variables or that it is concise if there is no pair (g, M), where M ∈ M_n−1,n(K), g ∈ K [y₁, …, y_n−1] and f(x₁, …, x_n) = g(y₁, …, y_n−1), where

(y1,…,yn−1)=M(x1,…,xn)t.

We say that the polynomial 0 depends on 0 variables. In Section 3, we study the conciseness of some determinantal polynomial and of some base polynomial, with the main results only for 2 × 2 matrices. We conclude Section 3 with several questions.

We found only a weak connection between the study of our determinantal polynomial and the (in principle) very similar problem of the description of a homogeneous form as a determinant of a matrix of linear forms. A. Beauville wrote the beautiful paper [9], which also contains realization as the determinant of a symmetric matrix of linear forms and as the Pfaffian of an anti-symmetric matrix. We discuss this topic in Section 5 which studies bp(A) for a matrix M∈Mn,n(Fq2) such that M+∈Mn,n(Fq) and M−∈Mn,n(Fq). Of course, it depends on the choice of β∈Fq2\Fq. Section 5 also contains the use of [9] for fM1,…,Mm, mainly for m = 3.

We thank a referee for useful suggestions.

2. Preliminaries

For any matrix M=(aij)∈Mn,n(Fq2) set M(q)=(aijq). Thus, M is Hermitian if and only if M^t = M^(q). Note that (M1+M2)(q)=M1(q)+M2(q) and that (tM)^(q) = t^qM^(q) for all t∈Fq2.

Remark. 2

Assume q = p^e for some e > 0. The field Fq is the set of all z∈F¯p such that z^q = z ([10, page 1], [11, Theorem 2.5]). Fix any a∈Fq\{0}. Since q + 1 is invertible in Fq, the polynomial t ^q+1 − a and its derivative (q + 1)t^q have no common zero. Hence, the polynomial t^q+1 − a has q + 1 distinct roots in F¯q. Fix any one of them, b. Since a^q−1 = 1, bq2−1=1. Thus, b∈Fq2. Thus, for any a∈Fq\{0} there are exactly q + 1 elements c∈Fq2 such that c^q+1 = a. Obviously, 0 is the only element t of Fq2 such that t^q+1 = 0.

Remark 3.

Note that (−1)^q = −1 in Fq. Since (u + v)^q = u^q + v^q and (u − v)^q = u^q + (−1)^qv^q = u^q − v^q for all u,v∈Fq2, det(M^(q)) = det(M)^q for all M∈Mn,n(Fq2). Now assume that M is Hermitian, i.e. assume M = M^†. Thus, det(M)=det((M(q))t)=det(M(q))=det(M)q. Hence, det(M)∈Fq by Remark 2.

Remark 4.

For any two Hermitian matrices A,B∈Mn,n(Fq2), there is a unique M∈Mn,n(Fq2) such that A = M₊ and B = M₋, the matrix M = A + βB.

Remark 5.

Take A,B∈Mn,n(Fq2) and a,b∈Fq. We have (aA + bB)₊ = aA₊ + bB₊ and (aA + bB)₋ = aA₋ + bB₋. Usually these equalities fail if a∈Fq2\Fq. For instance, if A is Hermitian, A ≠ 0 and a = β, then (aA)₊ = 0, while (aA)₋ = A.

For any A=(aij)∈Mn,n(Fq2) and any B=(bij)∈Mm,m(Fq2) let A ⊕ B denote the matrix (cij)∈Mn+m,n+m(Fq2) such that c_ij = a_ij if 1 ≤ i ≤ n and 1 ≤ j ≤ n, c_ij = 0 if either i > n and j ≤ n or i ≤ n and j > n, c_ij = b_i−a,j−n if i > n and j > n. The matrix A ⊕ B is called the unitary direct sum of A and B. Since (A ⊕ B)₊ = A₊ ⊕ B₊ and (A ⊕ B)₋ = A₋⊕ B₋, bp(A ⊕ B) = bp(A)bp(B).

Lemma 1.

Fix positive integers m, n and take m n × n Hermitian matrices M1,…,Mm∈Mn,n(Fq2). Then fM1,…,Mm(t1,…,tm)∈Fq[t1,…,tm]

Proof. Since Mi∈Mn,n(Fq2) for all i, fM1,…,Mm(t1,…,tm)∈Fq2[t1,…,tm]. Thus to prove that fM1,…,Mm(t1,…,tm)∈Fq[t1,…,tm], it is sufficient to prove that all its coefficients are preserved by the Frobenius map x↦x^q. Let α∈Fq2 be the coefficient of t1e1⋯tmem. Since the Frobenius map is additive, αqt1e1⋯tmem is a monomial of fM1q,…,Mmq(t1,…,tm). Recall that det(Mi)q=det(Mi(q)) (Remark 3). Since det(Mi(q))=det((Mi(q))t) and Mi†=(Miq)t, then α^q = α. Hence, α∈Fq (Remark 2). □

Lemma 2.

Take M∈M2,2(Fq2) such that M = M^†. The matrix M has 0 as its unique eigenvalue in F¯q if and only if there are a∈Fq and c∈Fq2 such that

M=accq−a,where

(4) −a2=cq+1

Moreover, there are exactly q² such matrices.

Proof. A 2 × 2 matrix over a field K has 0 as its unique eigenvalue over the algebraic closure of K if and only if its traces and determinant are 0. Since M = M^†, these are exactly the conditions on the entries of M stated in the lemma. For any a∈Fq\{0}, there are q + 1 elements c∈Fq2 such that c^q+1 = −a² (Remark 2). 0 is the unique c∈Fq2 such that c^q+1 = 0. Since #(Fq\{0})=q−1, there are 1 + (q − 1)(q + 1) = q² such matrices. □

Remark 6.

The definition of bp(A) depends on the definitions of A₊ and A₋, which depend on the choice of a suitable β∈Fq2\Fq. We explore the dependency of A₊, A₋ and bp(A) for different choices of β if q is odd. Assume q odd. Take a different choice and call it γ. We write A₊(β), A₋(β), bp(A)_β, A₊(γ), A₋(γ) and bp(A)_γ for the matrices and polynomials obtained from these two choices. Since q is odd, A₊(β) = A₊(γ) and A−(γ)=γβA−(β). Thus, bp(A)γ(x,y,z)=bp(A)β(x,γβy,z).

Remark 7.

For all integers d ≥ 0 and any field K, let K[x,y,z]_d denote the set of all homogeneous degree d polynomials in the variables x, y, z with coefficients in K. The set K[x,y,z]_d is a K-vector space of dimension d+22. Fix M∈Mn,n(Fq2). We have bp(M)∈Fq[x,y,z]n for every M∈Mn,n(Fq2) (Lemma 1).

Lemma 3.

Take f(x,y,z)∈Fq[x,y,z]n such that f(x, y, z) = (z + ax + by)ⁿ for some a,b∈F¯q. Then a,b∈Fq.

Proof. Since Fq is a perfect field, the plane {z + ax + by = 0} is defined over Fq. Thus, there is c∈F¯q, c ≠ 0, such that c(z+ax+by)∈Fq[x,y,z]1. Since c ≠ 0, we first get c∈Fq and then a,b∈Fq. □

Proof of Theorem 1.

:Assume A = A₊, i.e. assume A₋ = 0. Thus, bp(A)=det(Ax+In×nz)∈Fq[x,z]. Since the eigenvalues of A are the roots of the polynomial det(A−tIn×n), we get that bp(A) = zⁿ if and only if all eigenvalues of A are 0, i.e. we get part (i) for A = A₊. If A = βA₋, then just note that bp(A) = bp(A₋) up to changing the names of the variables.

Now assume n = 2. Part (ii) follows from Lemma 2. Part (iii) follows from part (ii) and the explicit computation of the coefficient of xy in the base polynomial bp(A).

Now assume n = 2 and q even. Since q is a 2-power, −2ae = 0 in Fq. Let U denote the set of all (c,d)∈(Fq2\{0})2 such that c^qd + cd^q = 0. Since q is even, (c,d)∈U if and only if c, d are non-zero elements of Fq2 and (dc)q−1=1. By Remark 2, the set Fq\{0} is the set of all t∈Fq2 such that t^q−1 = 1. Thus for every c∈Fq2\{0}, there are exactly q − 1 elements d∈Fq2 such that (c,d)∈U, the elements {tc}t∈Fq\{0}. Take (c,d)∈U. Since Fq is a perfect field and q is even, for every z∈Fq there is a unique w∈Fq such that w² = z. Thus for all (c,d)∈U, there are unique a, e such that c, d, a, e satisfy (1).

Now we prove part (v). Assume n = 2 and q odd. Take a, c satisfying the first equation of (1) and set e≔a and d ≔ c. Note that all equations in (1) are satisfied. □

3. Conciseness of determinantal polynomials

Remark 8.

Fix a field K and f∈K[x1,…,xn]d\{0}. The form f is concise over K¯ if and only if the degree d hypersurface {f=0}⊂Pn−1(K¯) is not a cone. Note that this criterion gives the same answer if we take the irreducible components of the hypersurface f = 0 with their multiplicity or not.

Lemma 4.

Fix fields K⊆L⊆K¯ and f∈K[x1,…,xn]d, d ≥ 2, f ≠ 0. Assume that K is perfect. The form f is concise over L if and only if it is concise over K.

Proof. If f is concise over a field K′ ⊃ K, then f is concise over K. Thus, it is sufficient to prove that if f is not concise over K¯, then it is not concise over K. Assume that f is not concise over K¯, i.e, that the closed hypersurface X(K¯) of Pn−1(K¯) with f as its equation is a cone with, say, vertex E(K¯); in the definition of X(K¯), we allow the multiplicities of the indecomposable factors of f (Remark 8). The set E(K¯) is a non-empty K¯ linear subspace of Pn−1(K¯). The decomposition of f in its irreducible factors and the linear subspace E(K¯) are defined over a finite extension K′ of K. Since K[x₁, …, x_n] is UFD, we reduce to the case in which f is irreducible over K. Since K is perfect, each indecomposable factor of f over K¯ has multiplicity 1 and hence, up to a non-zero multiplicative constant, f is uniquely determined by the set X(K¯) (no multiplicity is required). Since K is perfect, there is a finite extension L of K′ such that L is a Galois extension of K, say with Galois group G. The finite group G acts on X(K¯). Set e≔dimE(K¯). Let v the minimsl dimension of a K¯ linear subspace of Pn(K¯) contained in X(K¯) and containing E(K¯). Let S be the set of all v-dimensional K¯ linear subspace of Pn−1(K¯) contained in X(K¯). Since X(K¯) is a cone with vertex E(K¯), v > 0, ∪L∈SL=X(K¯) and ∩L∈SL=E(K¯). Since the embedding of X(K¯) in Pn−1(K¯) is defined over K, G acts linearly on Pn−1(K¯) and hence it acts on S, i.e. each g ∈ G induces a permutation of S. Thus, g(∩L∈SL)=∩L∈Sg(L) for all g ∈ G. Since each g ∈ G induces a permutation of S and ∩L∈SL=E(K¯), we get g(E(K¯))=E(K¯) for all g ∈ G. Thus E(K¯) is defined over K. Since E(K¯) is defined over K, there are n − e linear forms y0,…,yn−e−1∈K[x1,…,xn]1 such that E(K¯)={y0=⋯=yn−e−1=0}. Since E(K¯) is defined over K, there are yn−e,…,yn∈K[x1,…,xn]1 such that y₀, …, y_n is a new system of coordinates of Pn−1(K¯) and y_n−e, …, y_n are the homogeneous coordinates of E(K¯). Set W≔{y_n−e = ⋯ = y_n = 0}. Note that W is a linear subspace of Pn−1 defined over K, W(K¯)∩E(K¯)=∅, dimW(K¯)+dimE(K¯)=n−2 and y₀, …, y_n−e−1 are homogeneous coordinates of W. Call W̃ the linear subspace of K¯n associated to W. Set u≔f|W̃∈K¯[y0,…,yn−e−1]d. Since X(K¯)≠Pn−1(K¯) and X(K¯) is a cone with vertex E(K¯), X(K¯)∩W(K¯)≠W(K¯), i.e. u ≠ 0. Since f and W are defined over K, u∈K[y0,…,yn−e−1]d. Since X(K¯) is a cone with vertex E(K¯), u (as an element of K[y0,…,yn]d) is an equation of X(K¯). Thus, f is not concise over K. □

For each prime power q and each n ≥ 2, let m(q, n) be the maximal integer m such that there are m Hermitian matrices M1,…,Mm∈Mn,n(Fq2) such that the degree n form fM1,…,Mm(t1,…,tm)∈Fq[t1,…,tm]n is concise over F¯q. By Lemma 4, we get the same integer m(q, n) if we prescribe that fM1,…,Mm(t1,…,tm)∈Fq[t1,…,tm]n is concise over Fq.

Remark 9.

Fix any q. Let Mi∈Mn,n(Fq2), 1 ≤ i ≤ n, be the Hermitian matrix with 1 at (i, i) and 0 elsewhere. Since fM1,…,Mm(t1,…,tm)=∏i=1nti, Remark 8 and Lemma 4 give m(q, n) ≥ n.

Lemma 5.

Take Hermitian matrices M1,…,Mm∈Mn,n(Fq2) which are linearly dependent over Fq. Then fM1,…,Mm(t1,…,tm) is not concise over Fq.

Proof. Suppose for instance that M_m = c₁M₁ + ⋯ + c_m−1M_m−1 for some ci∈Fq. Take the new variables x_i = t_i + c_it_m, 1 ≤ i ≤ m − 1, and x_m = t_m. Note that fM1,…,Mm(t1,…,tm)=fM1,…,Mm(x1,…,xm−1,0). □

Proposition 2.

For any prime power q we have m(q, 2) = 4.

Proof. The set of all Hermitian M∈Mn,n(Fq2) is an n²-dimensional vector space over Fq. Thus, Lemma 5 gives m(q, 2) ≤ 4. Hence, it is sufficient to prove that m(q, 2) ≥ 4. If q is even fix any c∈Fq2\Fq. If q is odd fix any c∈Fq2\Fq such that c^4q − 2c^2q+2 + c⁴ ≠ 0. Set

M1≔1000,M2≔0001,M3≔0ccq0,M4≔0cqc0.

First assume q even. Since c∉Fq, then c^q−1 ≠ 1 and c ≠ 0. Thus c^q + c ≠ 0. Consider the degree 2 binary form h(t3,t4)≔cq+1t32+cq+1t42+(c2+c2q)t3t4. Since the coefficients of t32 and t42 in h(t₃, t₄) are the same and the coefficient of t₃t₄ is non-zero, h(t₃, t₄) is not a square. Thus, h(t₃, t₄) is concise. The binary form t₁t₂ in the variables t₁ and t₂ is concise. The quaternary form fM1,M2,M3,M4(t1,t2,t3,t4)=t1t2+cq+1t32+cq+1t42+(c2+c2q)t3t4 is concise, because the binary forms t₁t₂ and h(t₃, t₄) are concise.

Now assume q odd. We show that we may take c∈Fq2\Fq such that c^4q − 2c^2q+2 + c⁴ ≠ 0, i.e. c^4q−4 − 2c^2q−2 + 1 ≠ 0. If q ≥ 5, it is sufficient to use that #(Fq2\Fq)=q2−q>4q−4. Now assume q = 3. Each c∈F9, c ≠ 0, satisfies c⁸ = 1 and hence it is sufficient to take c such that c⁴ ≠ − 1, i.e. c⁴ = 1. The quaternary form fM1,M2,M3,M4(t1,t2,t3,t4)=t1t2−cq+1t32−cq+1t42+(c2+c2q)t3t4 is concise if and only if the binary form u(t3,t4)≔−cq+1t32−cq+1t42+(c2+c2q)t3t4 in the variables t₃, t₄ is concise. The binary form u(t₃, t₄) is concise, because it has degree 2, −c^q+1 ≠ 0, and the polynomial −c^q+1t² + (c² + c^2q)t − c^q+1 has 2 distinct roots over F¯q by our assumptions on c. □

We ask the following question.

Question 1.

Fix n ≥ 2 and a prime power q. Set m ≔ m(q, n). Is it possible to find m Hermitian matrices M₁, …, M_m such that fM1,…,Mm defines a smooth hypersurface (smooth at all points of Pm−1(F¯q))?

Remark 10.

Recall that a form f in n variables is concise if and only if the hypersurface {f = 0} is not a cone (Remark 8 and Lemma 4). For n = 2, Question 1 is trivially true, because for quadric hypersurfaces not to be a cone is equivalent to smoothness ([10, Lemma 5.1.1]).

Remark 11.

Obviously m(q, n + 1) ≥ m(q, n) for all q and n. We do not know the rate of growth of m(q, n) for a fixed q and n ≫ 0. We have m(q, n) ≤ n² for all n (Lemma 5), but we do not know the values of lim supn→+∞m(q,n)/n2 and lim infn→+∞m(q,n)/n2.

4. Realization of homogeneous polynomials

In this section, we consider the realization problem, i.e. we ask for which homogeneous polynomial f∈Fq[t1,…,tm]n there are Hermitian matrices M1,…,Mm∈Mn,n(Fq2) such that f=fM1,…,Mm. The interested reader should consider the problem of the descriptions of the m-ples (M₁, …, M_m) such that f=fM1,…,Mm.

We only consider the cases m = 1 and m = 2 and the case m = 3 with M3=In×n, i.e. the case of base polynomials, and prove Proposition 1.

4.1 Forms in m ≤ 2 variables

Remark 12.

Since det(M1t1)=det(M1)t1n and for each a∈Fq there is a Hermitian M₁ such that det(M₁) = a (even with M₁ diagonal), the realization problem is trivially satisfied for m = 1.

Remark 13.

Here we observe that the set of all binary n-forms realized by some fM1,M2 is invariant for the action of GL(2,Fq) on the variables x, y. For instance, fM1,M2(y,x)=fM2,M1(x,y) and fM1,M2(x+ay,y)=fM1,aM1+M2(x,y) for any a∈Fq. Use that these transformations generate the group of projective transformations acting on binary forms.

Now take m = 2. We are looking to the realization of binary n-forms, and we call x and y the two variables and M₁ and M₂ the two Hermitian matrices.

Proposition 3.

Take f∈Fq[x,y]. Then there are Hermitian 2 × 2 matrices M₁, M₂ such that f=fM1,M2.

Proof. By Remark 13, it is sufficient to realize at least one element for each orbit for the action of GL(2,Fq).

The binary form 0 is realized by M₁ = M₂ = 0. The binary form x² is realized taking M1=I2×2 and M₂ = 0. The binary form x(x + y) is (up to an Fq linear transformation of Fq2) the only one with 2 distinct roots over Fq. This form is realized taking M1=I2×2 and M₂ = (b_ij), where b₁₁ = b₁₂ = b₂₁ = 0 and b₂₂ = 1.

Now we consider binary forms which split over Fq2, but not over Fq.

First assume q odd. Up to an Fq linear transformation it is sufficient to realize the form x² − ay² with a not a square in Fq. Take c∈Fq2 such that c^q+1 = a (Remark 2). Take A=I2×2 and B = (b_ij), where b₁₁ = b₂₂ = 0, b₁₂ = c and b₂₁ = c^q.

Now assume q = 2^e even. Since every element of Fq is a square, the form x² + cy² splits and hence up to an Fq linear transformation, it is sufficient to realize the form x² + xy + δy², where δ∈Fq\0 has non-zero absolute trace D(δ), where D(u)=∑i=0e−1u2i for any u∈Fq ([10, p. 3]). Fix any δ∈Fq and take c∈Fq2 such that c^q+1 = δ. Take A=I2×2 and B = (b_ij), where b₁₁ = 1, b₁₂ = c, b₂₁ = c^q and b₂₂ = 0. □

4.2 Base polynomials

Now we take m = 3, M3=In×n, M₁ = A₊, M₂ = A₋ for some A∈Mn,n(Fq2). By Remark 4, it is not restrictive to the existence of a matrix A such that M₁ = A₊ and M₂ = A₋. We call x, y and z the variables. Every degree n base polynomial contains the monomial zⁿ with degree 1. We call monic such forms.

Question 2.

Are there other restrictions?

Remark 14.

Let R denote the set of all polynomials bp(A) with A∈M3,3(Fq2). Take any a,b∈Fq and any A∈M3,3(Fq2). Since a∈Fq, we have (A+aA−)+=A++aA− and (A+aA−)−=A−. Thus, bp(A + aA₋)(x, y, z) = bp(A)(x + ay, y, z). Hence, R is invariant for the linear transformations x↦x + ay, y↦y, z↦z. Since a∈Fq, we have (A+aβA+)+=A+ and (A+aβA+)−=A−+aA+. Thus, bp(A + aβA₊)(x, y, z) = bp(A)(x, ax + y, z). Hence, R is invariant for the linear transformations x↦x, y↦ax + y, z↦z. Since a,b∈Fq, we have (A+(a+βb)In×n)+=A++aIn×n and (A+(a+βb)In×n)−=A−+bIn×n. Thus, bp(A+(a+βb)In×n)(x,y,z)=bp(A)(x,y,z+ax+by). Thus, R is invariant for the linear transformations x↦x, y↦y, z↦z + ax + by. Thus, the set R is invariant for all changes of coordinates (gij)∈GL(3,Fq) such that g₃₃ = 1.

Remark 15.

Take a monic f(x,y,z)∈Fq[x,y,z]n such that f = gh for some monic g, h and 0 < a≔ deg(g) < d. Assume g = bp(A) and h = bp(B) for some A∈Ma,a(Fq2), B∈Mn−a,n−a(Fq2). Then f = bp(A ⊕ B). In particular, if f splits over Fq as a product of n monic linear forms (we allow multiple linear forms), then f = bp(M) for some M∈Mn,n(Fq2). Now assume that f is the product of n linear forms over Fq, say f = L₁⋯L_n with L_i = c_iz_i + a_ix + b_iy, but we allow that some of the forms are not monic. We get ∏i=1nci=1, and hence f is the product of the n monic linear forms z+aicix+biciy.

Proof of Proposition 1.

:By Remark 14, it is sufficient to realize at least one form for each orbit for the action of the subgroup of GL(3,Fq) described in Remark 15. The plane conics over Fq are classified in Ref. [10] in terms of their rank.

There is a unique rank 1 monic conic, z². The binary form z² is realized as a base polynomial taking M₁ = M₂ = 0.

Rank 2 monic conics form 2 orbits, the ones union of 2 lines defined over Fq and the one induced by a form indecomposable over Fq, but decomposable over Fq2. We first check that all rank 2 monic conics which splits over Fq are realized as a base polynomial. For any q, we realize the polynomial (z + x)(z + y) taking the matrix A = A₊ + βA₋ = (a_ij) with a₁₂ = a₂₁ = 0, a₁₁ = 1 and a₂₂ = β.

There is, up to a projective transformation, another rank 2 conic ([10, Th. 5.1.6 for q odd, Th. 5.1.7 for q even]).

First assume q odd. We need to represent the equation dx² + z² with d∈Fq and d not a square. Take A = (a_ij) with a₁₁ = d, a₂₂ = 1 and a₁₂ = a₂₁ = 0 (so that A₊ = A and A₋ = 0).

Now assume q even, say q = 2^e for some e > 0. Since every element of Fq is a square, the form z² + cy² splits and hence up to an Fq linear transformation it is sufficient to realize as a base polynomial the form z² + zy + δy², where δ∈Fq\0 has non-zero absolute trace D(δ), where D(u)=∑i=0e−1u2i for any u∈Fq ([10, p. 3]). Fix any δ∈Fq and take c∈Fq2 such that c^q+1 = δ. Take A = 0 and B = (b_ij), where b₁₁ = 1, b₁₂ = c, b₂₁ = c^q and b₂₂ = 0.

For any finite field up to a projective transformation, there is a unique smooth projective conic ([10, Theorems 5.1.6 and 5.1.7]), and we may take z(z + x) − y² as its equation. Use the matrix C = (c_ij) with c₁₁ = 1, c₁₂ = c₂₁ = β and c₂₂ = 0, which have bp(C) = z(z + x) − y² (any q). □

Remark 16.

Remark 15 and Proposition 1 gives that every reducible monic f∈Fq[x,y,z]3 is a base polynomial.

5. M+,M−∈Mn,n(Fq)

A. Beauville studied the realization over a finite field of a form as the determinant of a matrix with entries linear forms ([9]). In this section, we use [9] for matrices M∈Mn,n(Fq2) such that M+∈Mn,n(Fq) and M−∈Mn,n(Fq). Obviously this very strong assumption depends on the choice of β∈Fq2\Fq. For any q, it requires that M+M†∈Mn,n(Fq), but it is stronger.

Remark 17.

Take symmetric matrices A,B∈Mn,n(Fq). Set M ≔ A + βB. Since A, B are symmetric and with coefficients in Fq, they are Hermitian. Thus, M₊ = A and M₋ = B. The matrix Ax+By+zIn,n is symmetric, hence in this case bp(M) is the determinant of a symmetric matrix of linear forms. Conversely, any symmetric matrix of linear forms over Fq with z appearing only in the diagonal and with all coefficients 1 is obtained in this way for some symmetric matrices.

Proposition 4.

Assume q≥(n−1)(n−2)/2+(n−1)(n−2)q. Then every smooth plane curve of degree n defined over Fq is of the form {fM1,M2,M3=0} for some M1,M2,M3∈Mn,n(Fq).

Proof. Let X be a smooth plane curve of degree n defined over Fq. The curve X has genus g≔(n − 1)(n − 2)/2. To get a determinantal equation of X over Fq, it is necessary and sufficient to find a degree g − 1 line bundle L on X defined over Fq and such that h⁰(L) = 0 ([9, Proposition 3.1]). Assume q≥g+2gq. Any smooth projective curve C of genus g defined over Fq satisfies #C(Fq)≥g+1 by the Hasse–Weil theorem ([12, Theoren 9.18]). A theorem proved in Refs. [13, 14] and quoted in [15, Proposition 2.2] says that any smooth genus γ curve C such that #C(Fq)≥γ+1 has a degree γ − 1 line bundle L defined over Fq and with h⁰(L) = h¹(L) = 0. □

The lower bound on q in Proposition 4 is not sharp. The existence of a line bundle L as in the proof of Proposition 4 is related to the computational complexity of the multiplication in finite extensions of a finite field ([13–17]).

The paper [18] and its references gives better information on the number of points of smooth plane curves with a fixed degree and large q. Hasse–Weil bound and related tools may also be used for singular plane curves ([19–21]). See Ref. [22] for results on #Pic0(C)(Fq).

Note that given any f∈Fq[x,y,z]n, f ≠ 0, it is computationally easy to check (a system with the coefficients of f and its partial derivatives) if the plane curve {f = 0} is smooth (smooth at all points, not only at its Fq points). It is also very easy to check when a trivariate polynomial is monic with respect to z. We do not have an always working (or always working for large q) criterion to realize a monic polynomial as bp(A) for some A∈Mn,n(Fq2), but Remark 17 is sufficient if the monic polynomial is the determinant of a symmetric matrix. If q is odd, this is the content of [9, Proposition 4.2].

Funding: The author was partially supported by MIUR and GNSAGA of INdAM (Italy).

Conflict of interests and funding declaration: The author has no conflict of interest and this research received no funding.