# Ideal membership

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 $$I=\langle x_{11}^{q-1}-x_{11},x_{12}^{q-1}-x_{12},\ldots,x_{nn}^{q-1}-x_{nn}\rangle$$ and the polynomial $$f(X)=f(x_{11},x_{12},\ldots,x_{nn})= \prod_{i=1}^{t}\det(X-A_{i})\prod_{i=1}^{t}(\det(X-B_{i})^{q-1}-1).$$ 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 $\prod_{i=1}^{t}\det(X-A_{i})\notin I$ and $\prod_{i=1}^{t}(\det(X-B_{i})^{q-1}-1) \notin I$, but I can't find similar result about $f$.

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Could you explain the relevance of the tag "algebraic geometry"? –  Martin Brandenburg May 4 '11 at 9:30
Is this supposed to hold for any $A_i$ $B_i$, or just for generic matrices? –  J.C. Ottem May 7 '11 at 18:51
I mean for any distinct $A_{i},B_{i}$. –  Moh514 May 8 '11 at 5:47