Classification conjecture for quadratic permutation systems over finite fields of odd characteristic

About 1 year old · traced to 2025

Let $q$ be an odd prime power and let $n\geq 3$ . A quadratic system is a map

F=(f_1,\dots,f_n):\mathbb{F}_q^n\to\mathbb{F}_q^n,

where each $f_i\in\mathbb{F}_q[x_1,\dots,x_n]$ has degree at most $2$ . Classification conjecture. Such a system induces a permutation of $\mathbb{F}_q^n$ if and only if $F\sim(x_1,\dots,x_n)$ . Here $\sim$ denotes the equivalence relation introduced in the paper. The conjecture extends the complete classification of bivariate quadratic permutation systems to all dimensions $n\geq 3$ , and the paper presents it as an open problem for quadratic systems in several variables over finite fields of odd characteristic.

References

Primary source

Xuan Pang, Yangcheng Li, Pingzhi Yuan and Yuanpeng Zeng, “Determination of Some Types of Permutations over F_q^2 with Low-Degree”, arXiv:2508.01143 (2025).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

An unverified posted construction claims the conjecture is false in every dimension three or higher, while the published work only proves the two-dimensional case.

Pang, Li, Yuan, and Zeng (2025) completely classify quadratic permutation systems over $\mathbb{F}_q^2$ in odd characteristic and conjecture that the same identity classification holds for all $n\geq 3$ .

Known results

Every bivariate quadratic permutation system over $\mathbb{F}_q$ of odd characteristic is equivalent to $(x,y)$ (Pang, Li, Yuan, and Zeng, 2025).

Posted attempt

An undated posted attempt claims a counterexample for every odd prime power $q$ and every $n\geq 3$ : $F(x,y,z,\ldots)=(x+zy,\ y+\delta zx,\ z,\ldots)$ with nonsquare $\delta$ . It argues that $F$ is bijective on each $z$ -fiber, but $\det J_F=1-\delta z^2$ is nonconstant, unlike the identity’s determinant; the claimed invariant would therefore exclude equivalence to the identity. The complete counterexample claim has not been independently verified.

Current status (as of August 2026): The two-dimensional classification is settled, while a posted but unverified counterexample claim would refute the conjecture for all $n\geq 3$ if its equivalence-invariance argument is correct.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

The conjecture is false for EVERY odd prime power $q$ and EVERY $n\ge3$ , under precisely the polynomial-system equivalences defined in the source.

Let $K=\mathbb F_q$ , and choose a nonsquare $\delta\in K^\times$ . Define

F(x,y,z,x_4,\ldots,x_n) =\bigl(x+zy,\ y+\delta zx,\ z,\ x_4,\ldots,x_n\bigr).

For each fixed $z\in K$ , the transformation on $(x,y)$ has matrix

B_z= \begin{pmatrix} 1&z\\ \delta z&1 \end{pmatrix}, \qquad \det B_z=1-\delta z^2.

Since $\delta$ is a nonsquare, $1-\delta z^2\ne0$ for every $z\in K$ . Thus each $z$ -fiber is mapped bijectively to itself, and the remaining coordinates are fixed. Hence $F$ permutes $K^n$ .

Its FORMAL Jacobian determinant, however, is

\det J_F=1-\delta z^2,

which is a nonconstant polynomial.

Having a nonzero constant formal Jacobian determinant is invariant under both equivalences in Definitions 3.1 and 3.2. Indeed, if

G=\rho\circ F\circ\sigma

for nonsingular linear transformations $\rho,\sigma$ , the chain rule gives

\det J_G(\mathbf x) =(\det\rho)(\det\sigma)\det J_F(\sigma\mathbf x).

An invertible linear substitution preserves whether this polynomial is a nonzero constant.

For a coordinate-shift equivalence, the two systems have the source's exact forms

F_0=(f_1(x_1,\ldots,x_k),\ldots,f_k(x_1,\ldots,x_k), x_{k+1},\ldots,x_n)

and

G_0=(f_1(x_1,\ldots,x_k),\ldots,f_k(x_1,\ldots,x_k), x_{k+1}+h_{k+1}(x_1,\ldots,x_k),\ldots, x_n+h_n(x_1,\ldots,x_{n-1})).

Their Jacobian matrices are block lower triangular with the SAME upper-left $k\times k$ block. The lower-right block is respectively the identity or a unit lower-triangular matrix. Therefore

\det J_{G_0}=\det J_{F_0}

as formal polynomials. Coordinate permutations preserve the same constant-determinant property.

The identity system has Jacobian determinant one, so EVERY polynomial system equivalent to it under the specified operations has nonzero constant Jacobian determinant. Our permutation $F$ has nonconstant determinant $1-\delta z^2$ , and therefore cannot be equivalent to the identity.

Already over the smallest admissible field, take $q=3$ , $\delta=2=-1$ , and $n=3$ :

F(x,y,z)=(x+zy,\ y-zx,\ z), \qquad \det J_F=1+z^2.

For $z=0,1,2$ , the fiber determinants are $1,2,2$ . Hence $F$ permutes all 27 points of $\mathbb F_3^3$ but is not equivalent to the identity. This refutes the conjecture at its smallest permitted field and dimension.