Let $n=2t$ be an even number. Let $F$ denote a finite field where $F=q$. Let $A_{1}, A_{2},\ldots, A_{t}$ and $B_{1},B_{2},\ldots,B_{t}$ be distinct matrices in $M_{n}(F)$. Let $$ X = \begin{pmatrix} x_{11} & x_{12} & \cdots & x_{1n} \\ x_{21} & x_{22} & \cdots & x_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ x_{n1} & x_{n2} & \cdots & x_{nn} \\ \end{pmatrix}. $$ ` Consider the ideal $$I=\langle x_{11}^{q1}x_{11},x_{12}^{q1}x_{12},\ldots,x_{nn}^{q1}x_{nn}\rangle$$ and the polynomial $$f(X)=f(x_{11},x_{12},\ldots,x_{nn})= \prod_{i=1}^{t}\det(XA_{i})\prod_{i=1}^{t}(\det(XB_{i})^{q1}1).$$ I'm looking for some condition on $F$ such that $f(X) \notin I$. Actually I think that $f(X) \notin I$ if $F$ is sufficiently large. In fact I know that, if $F>n^{2}$, then $\prod_{i=1}^{t}\det(XA_{i})\notin I$ and $\prod_{i=1}^{t}(\det(XB_{i})^{q1}1) \notin I$, but I can't find similar result about $f$.
