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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 $K^{2n^2}$.

How can one show that $N$ contains the set $\{(axa^{-1},aya^{-1}) \mid (x,y)\in N, a\in \mathrm{GL}(n,K)\}$?

Thank you.

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    $\begingroup$ It seems to me that $M$ itself is Zariski closed. $\endgroup$
    – Johannes Ebert
    Commented Mar 7, 2011 at 19:46
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    $\begingroup$ From the way this problem is phrased, it seems to be designed to test understanding of basic algebraic geometry. In the absence of other information, I think it very likely that it is coursework. Voting to close. $\endgroup$
    – Victor Protsak
    Commented Mar 7, 2011 at 20:01
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    $\begingroup$ @Victor: Sorry, I was reading this in an article where it was left as "easy to check". $\endgroup$
    – spelas
    Commented Mar 7, 2011 at 20:18
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    $\begingroup$ My comment was wrong. If $F \neq K$, then $M$ is not Zariski closed. $\endgroup$
    – Johannes Ebert
    Commented Mar 7, 2011 at 20:33

1 Answer 1

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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).

Then it follows that $N$ is $GL_n(K)$-invariant, as claimed.

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