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Let $GL_n$ be the group of all invertible matrices of order $n$ and $D_n$ the subgroup of $GL_n$ consisting of all invertible diagonal matrices of order $n$. How to show that $D_n$ is closed in $GL_n$ with respect to Zariski topology? I think that we have to show that $D_n$ is the set of zeros of some set of polynomials. What are these polynomials? Thank you very much.

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    $\begingroup$ Hint: the function which returns the $(i,j)$-th entry of a matrix is a polynomial function on $\mathrm{GL}_n$. $\endgroup$ – mdp Sep 25 '13 at 9:46
  • $\begingroup$ Do you really mean for your diagonal matrices to have order $n$? $\endgroup$ – Tobias Kildetoft Sep 25 '13 at 9:48
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    $\begingroup$ @MattPressland, thank you very much. I think that $D_n=\{g=(x_{ij}) : f_{ij}(g) = x_{ij}=0, i\neq j, i, j \in \{1, \ldots, n\}\}$. $\endgroup$ – LJR Sep 25 '13 at 9:52
  • $\begingroup$ @TobiasKildetoft, thank you very much. I mean that the diagonal matrices are $n$ by $n$ matrices. $\endgroup$ – LJR Sep 25 '13 at 9:55
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So this question doesn't stay on the unanswered list, I'll copy my hint from the comments to an answer, as it seemed to help.

Hint: the function which returns the $(i,j)$-th entry of a matrix is a polynomial function on $\mathrm{GL}_n$.

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