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Can we find a Zariski open subset of $\mathbb{C}^n$ for some $n\ge 1$ which is not connected?

If $n=1$, this is not possible, since Zariski open subsets of $\mathbb{C}$ are just complements of finite sets.

Definition. A subset $U\subseteq\mathbb{C}^n$ is Zariski-open if its complement $\mathbb{C}^n-U$ is a zero set of polynomials $p_1,\ldots,p_k\in\mathbb{C}[x_1,\ldots,x_n]$.

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No.

In the usual topology, this is because if $U$ is connected of dimension $n$ and $Z \subset U$ is of real codimension at least two then $U \backslash Z$ is connected.

In the Zariski topology, something is stronger : $U$ is irreducible since $\Bbb C^n$ is, in particular connected.

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  • $\begingroup$ Is it true also that if $U$ is a Zariski-open subset of a connected affine variety $X$ then $U$ is also connected? $\endgroup$ – Simon Parker Jun 25 '18 at 13:03
  • $\begingroup$ @SimonParker : no, for example $xy = 0$ is a connected subvariety of $\Bbb C^2$ but is not connected if you remove zero. It is true if $X$ is irreducible. $\endgroup$ – Nicolas Hemelsoet Jun 25 '18 at 13:27

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