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Let $X$ be a scheme (e.g. of finite type over $\mathbb C$, but it does not matter) and let $Z\subset X$ be an irreducible component of $X$. Suppose we have an open subscheme $U\subset Z$.

How to characterize when $U$ will still be open in $X$?

The answer is always if the irreducible components are the connected components of $X$, which happens for regular schemes. If two irreducible components meet (hence $X$ is not regular), I did some examples (so I know the answer is not always), but still I cannot figure out a pattern. However, I feel this should be well established.

Thanks for any help.

Brenin
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    At least if $U$ does not intersect any irreducible component different from $Z$, then $U$ will be open in $X$. I think we can't say more. – paf Aug 11 '16 at 14:42

1 Answers1

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The subset $U\subset Z\subset X$ is open in $X$ if and only it is open in $Z$ and if it does not meet any of the other components. In fact, if $U\subset Z$ is open in $Z$ and doesn't have a point in common with any component different from $Z$, $U\subset X$ is open by basic point-set topology.

Conversely, suppose that $U\subset X$ is open. Then $U\cap Y\subset Y$ is open for each irreducible component $Y\subset X$; the claim is now equivalent to the fact that $Y\cap U\not=\emptyset$ can only happen for $Y=Z$. Indeed, if $Y\cap U$ is open in $Y$ and non-empty, it has to contain the generic point of $Y$. But as $U\subset Z$, also $Z$ contains the generic point of $Y$ and, thus, all of $Y$. But this is impossible unless $Y=Z$.

Ben
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  • For those of you who don't want to think or talk about generic points: if $Y\cap U\subset Y$ is open and non-empty, then it's dense in $Y$. Thus, $Y = \overline{U\cap Y}\subset \overline{U} \subset Z$. – Ben Aug 11 '16 at 15:25
  • although It is an old question, may I ask you something? I think in the first part of proof , we need "$X$ has a finitely many irreducible component" . (ex. $X$ noetherian) . If not, how can I show this? I don't understand "by basic point-set topology" part. If $X$ has a infinitely many irreducible component, how can I show $U$ is open in $X$? – hew Oct 13 '20 at 12:24
  • You’re right, I must have been assuming that implicitly. – Ben Oct 14 '20 at 19:01