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

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

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

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

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

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.

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.

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.

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 $t∈Fq2$⁠.

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).

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). □

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. □

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. 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 $Fq$⁠.

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 $fM1,…,Mm(t1,…,tm)=fM1,…,Mm(x1,…,xm−1,0)$⁠. □

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

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.

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.

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.

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. □

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.