Ideal membership - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T18:10:10Z http://mathoverflow.net/feeds/question/63828 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63828/ideal-membership Ideal membership Mohsen 2011-05-03T17:19:43Z 2011-05-07T15:04:57Z <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>$$X = \begin{pmatrix} x_{11} &amp; x_{12} &amp; \cdots &amp; x_{1n} \\ x_{21} &amp; x_{22} &amp; \cdots &amp; x_{2n} \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ x_{n1} &amp; x_{n2} &amp; \cdots &amp; x_{nn} \\ \end{pmatrix}.$$</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>