Closing the Permanent-Rank Boundary for the Alon–Jaeger–Tarsi Conjecture

1. Introduction

The Alon–Jaeger–Tarsi conjecture asks whether, for every finite field $\mathbb{F}$ with $\left|\mathbb{F}\right|\ge 4$ and every $M\in {GL}_n\left(\mathbb{F}\right)$, there is a vector $x\in {\left({\mathbb{F}}^{*}\right)}^n$ such that $Mx\in {\left({\mathbb{F}}^{*}\right)}^n$; see [1]. The non-prime finite fields were settled in the original work, and Nagy and Pach recently proved the conjecture for primes $p>61$, $p\ne 79$ [4]. The small prime fields remain the main unresolved range.

For a matrix A, write $prankA$ for the largest order of a square submatrix of A having nonzero permanent. Yu proved

$$prankA\ge {\displaystyle \frac{rankA}{2}}$$

(1)

and used a concatenation argument to establish the Alon–Jaeger–Tarsi conjecture over ${\mathbb{F}}_p$ whenever $n<2^{p-2}$ [5]. For $p=5$ this leaves $n=8$ as the first dimension outside that theorem. A 2026 sequel of Yu still presents the ${\mathbb{F}}_5$ assertion as a conjecture while noting the recent classification of equality in (1) [6].

Our observation is that the strict inequality in Yu’s dimension bound can be removed. If $prankM$ is strictly larger than half the rank, the integer rounding in Yu’s recurrence reaches full permanent rank at the boundary. If equality holds, a rigidity theorem of Kisley and Shader identifies the matrix up to transformations that preserve the algebraic torus ([3] Theorem 4.8).

Theorem 1.1.

Let $p\ge 5$ be prime and $n\le 2^{p-2}$. For every $M\in {GL}_n\left({\mathbb{F}}_p\right)$ there exists $x\in {\left({\mathbb{F}}_p^{*}\right)}^n$ such that $Mx\in {\left({\mathbb{F}}_p^{*}\right)}^n$.

Corollary 1.2.

For every $M\in {GL}_8\left({\mathbb{F}}_5\right)$ there exists $x\in {\left({\mathbb{F}}_5^{*}\right)}^8$ such that $Mx\in {\left({\mathbb{F}}_5^{*}\right)}^8$.

The proof is entirely algebraic. We include the short concatenation and coefficient arguments to make clear exactly where the boundary is gained.

2. A Permanent-Rank Recurrence

For matrices having the same number of rows, $\left[BA\right]$ denotes horizontal concatenation. We first isolate the one-step estimate implicit in Yu’s proof.

Lemma 2.1

(Yu’s recurrence). Let B be an $n\times N$ matrix over a field of characteristic different from 2, and let $A\in {GL}_n\left(\mathbb{F}\right)$. If $r=prankB$, then

$$prank\left[BA\right]\ge ⌈{\displaystyle \frac{r+n}{2}}⌉.$$

Proof. 

The statement is immediate if $r=n$. Otherwise select r columns of B forming an $n\times r$ matrix $D_0$ with full permanent rank. Enlarge $D_0$ by a maximal subset of the columns of A such that the resulting $n\times s$ matrix D has permanent rank s.

If $s=n$, the conclusion is immediate, so assume $s<n$.

Yu’s column-appending lemma [5] states that

$$L\left(D\right)=\{z\in {\mathbb{F}}^n:prank\left[Dz\right]=s\}$$

is a linear subspace of dimension at most s. By maximality, every column of A not selected for D belongs to $L\left(D\right)$. There are $s-r$ selected columns of A, and all the columns of A span ${\mathbb{F}}^n$. Therefore

$$n\le (s-r)+dimL\left(D\right)\le 2s-r.$$

Thus $s\ge (r+n)/2$, and integrality gives the result. □

For $A\in {GL}_n\left(\mathbb{F}\right)$, let $A^{\left[j\right]}=[A\cdots A]$ be the concatenation of j copies of A, and set

$$k_j=prank{A}^{\left[j\right]},d_j=n-k_j.$$

Then Theorem 2.1 gives the useful deficiency recurrence

$$d_{j+1}\le ⌊{\displaystyle \frac{d_j}{2}}⌋.$$

(2)

3. From Full Permanent Rank to a Torus Point

The next lemma is the standard polynomial implication behind Yu’s argument. We give the coefficient calculation because the restriction to $p-2$ copies is essential.

Lemma 3.1.

Let $A\in M_n\left({\mathbb{F}}_p\right)$, where p is prime. If $prank{A}^{[p-2]}=n$, then there exists $x\in {\left({\mathbb{F}}_p^{*}\right)}^n$ such that $Ax\in {\left({\mathbb{F}}_p^{*}\right)}^n$.

Proof. 

Write $A=\left(a_{ij}\right)$ and consider the homogeneous polynomial

$$g(X_1,\dots ,X_n)={\displaystyle \prod _{i=1}^n}\left({\displaystyle \sum _{j=1}^n}{a}_{ij}{X}_j\right).$$

There is an $n\times n$ submatrix N of $A^{[p-2]}$ with $perN\ne 0$. Let $e_j$ be the multiplicity with which the jth column of A occurs in N. Then

$$0\le e_j\le p-2,{\displaystyle \sum _{j=1}^n}{e}_j=n.$$

Labeling the repeated columns in the permanent expansion gives

$$perN=\left({\displaystyle \prod _{j=1}^n}{e}_j!\right)[X_1^{e_1}\dots X_n^{e_n}]g.$$

(3)

The factorials are nonzero in ${\mathbb{F}}_p$, so the coefficient on the right-hand side is nonzero. Apply the Combinatorial Nullstellensatz [2] to the sets $S_j={\mathbb{F}}_p^{*}$. Since $|S_j|=p-1>e_j$, there exists $x\in {\prod }_j{S}_j$ with $g\left(x\right)\ne 0$. Equivalently, all coordinates of both x and $Ax$ are nonzero. □

4. The Equality Case and the Boundary

Let

$$P_2=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right).$$

Two square matrices are monomially equivalent if one is obtained from the other by multiplying on the left and right by invertible monomial matrices. Such matrices permute and rescale coordinates, hence biject the algebraic torus ${\left({\mathbb{F}}^{*}\right)}^n$ with itself and preserve the nowhere-zero property.

Kisley and Shader proved that, over a field of characteristic different from 2, every invertible matrix A satisfying $prankA=rank\left(A\right)/2$ is monomially equivalent to a direct sum of copies of $P_2$ [3]. For $p\ge 5$ the vector ${(1,2)}^T$ is a nowhere-zero input for $P_2$, because

$$P_2\left({\displaystyle \genfrac{}{}{0pt}{}{1}{2}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{3}{-1}}\right)\in {\left({\mathbb{F}}_p^{*}\right)}^2.$$

Consequently every invertible equality-case matrix has a nowhere-zero point.

Proof of Theorem 1.1. 

Let $k_1=prankM$. By (1), $k_1\ge n/2$.

If $k_1=n/2$, the Kisley–Shader classification and the block construction above give the desired point.

Suppose $k_1>n/2$. Then the integer $d_1=n-k_1$ satisfies

$$d_1<{\displaystyle \frac{n}{2}}\le 2^{p-3},$$

and hence $d_1\le 2^{p-3}-1$. Iterating (2) through the remaining $p-3$ copies gives

$$d_{p-2}\le ⌊{\displaystyle \frac{d_1}{2^{p-3}}}⌋=0.$$

Thus $prank{M}^{[p-2]}=n$, and Theorem 3.1 finishes the proof. □

Proof of Theorem 1.2. 

Here $p=5$ and $n=8=2^{p-2}$. More explicitly, if $prankM=4$, the equality classification applies. If $prankM\ge 5$, then Theorem 2.1 yields

$$prank\left[MM\right]\ge ⌈{\displaystyle \frac{5+8}{2}}⌉=7,prank\left[MMM\right]\ge ⌈{\displaystyle \frac{7+8}{2}}⌉=8,$$

so Theorem 3.1 applies. □

Remark 4.1.

The hypothesis $p\ge 5$ is used in the equality case. Over ${\mathbb{F}}_3$, the two ratios $\pm 1$ exhaust ${\mathbb{F}}_3^{*}$, and $P_2$ itself has no nowhere-zero input.