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Suppose that $k$ is an algebraically closed field. GL$(n,k)$ is the general linear group. It can be considered as $\{x \in \mathbb{A^{n^2}}: det(x) \neq 0\}$. Clearly, this is an open subset of $\mathbb{A^{n^2}}$. But why is it closed? and why is it connected?

Let T$(n,k)$ denotes the subgroup of GL$(n,k)$ consists of diagonal matrices. It is clearly closed. But how to prove its connectedness?

Many thanks~

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    Dear ShinyaSakai, More generally, any open subset of an irreducible topological space is irreducible (hence connected). Since $\mathbb{A}_k^{n^2}$ is irreducible, so is $GL_n(k)$. For the second question, note that $T(n, k)$ is isomorphic to $k^* \times \dots \times k^*$.2011-05-21

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