most recent 30 from http://mathoverflow.net 2013-05-26T05:48:18Z http://mathoverflow.net/feeds/question/57719 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57719/algebraic-closure-of-commuting-pairs-of-matrices spelas 2011-03-07T19:40:33Z 2011-03-07T21:05:51Z

<p>Let $F$ be an arbitrary field of characteristic $0$, $K$ its algebraic closure. Define $M=\{ (x,y)\in M_n(F)×M_n(F) \mid [x,y]=0\}$ and let $N$ be the Zariski closure of $M$ in $K^{2n^2}$.</p> <p>How can one show that $N$ contains the set $\{(axa^{-1},aya^{-1}) \mid (x,y)\in N, a\in \mathrm{GL}(n,K)\}$?</p> <p>Thank you.</p>

http://mathoverflow.net/questions/57719/algebraic-closure-of-commuting-pairs-of-matrices/57730#57730 Guntram 2011-03-07T21:05:51Z 2011-03-07T21:05:51Z

<p>Note that $M$ is invariant under the $GL_n(F)$-action given by $a \cdot (x,y):=(axa^{-1},aya^{-1})$. It follows that its closure $N$ is also invariant under $GL_n(F)$. Since $F$ is infinite and $GL_n$ is reductive, the rational points $GL_n(F)$ are Zariski-dense in $GL_n(K)$ by Borel, Linear Algebraic Groups, Corr. 18.3. (This is probably the piece of information you were missing). </p> <p>Then it follows that $N$ is $GL_n(K)$-invariant, as claimed. </p>