Ideal membership - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T17:42:52Zhttp://mathoverflow.net/feeds/question/63828http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/63828/ideal-membershipIdeal membershipMohsen2011-05-03T17:19:43Z2011-05-07T15:04:57Z
<p>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
<code>$$ X =
\begin{pmatrix}
x_{11} & x_{12} & \cdots & x_{1n} \\
x_{21} & x_{22} & \cdots & x_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
x_{n1} & x_{n2} & \cdots & x_{nn} \\
\end{pmatrix}.
$$</code>
Consider the ideal
<code>$$I=\langle x_{11}^{q-1}-x_{11},x_{12}^{q-1}-x_{12},\ldots,x_{nn}^{q-1}-x_{nn}\rangle$$</code>
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 $f$.</p>