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I was wondering if someone could clarify the following. Let $W$ be an affine variety (which I don't assume to be irreducible) in $\mathbb{C}^n$. Let $U$ be an open (in Zariski topology) and suppose $W \cap U$ is non-empty. Does it then follow that $W \cap U$ has dimension $(\dim W)$?

Any comments would be appreciated!

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    $\begingroup$ Remark: It is true if $ W$ is irreducible. $\endgroup$ – HeinrichD Mar 20 '17 at 19:29
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    $\begingroup$ The irreducible case follows purely topologically too. So does the counter example (take the disjoint union of two spaces of different dimensions and take as your open set the smaller space). $\endgroup$ – Asvin Mar 20 '17 at 20:31
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Consider the affine variety, $$W = V(xz, yz) \subset \mathbb C^3.$$ $W$ is the union of the $z$-axis and the plane $z = 0$. The dimension of $W$ is two.

Take $U$ to be the complement of the $z = 0$ plane. So $W \cap U$ is the $z$-axis with the origin removed. This has dimension one.

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