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I am thinking about the following:

on a locally noetherian scheme X,$p$ and $q$ are generic points of two distinct irreducible components,do there exist two open sets $U,V$ of X with $p\in U,q\in V,U\cap V=\varnothing$?

when the irreducible components corresponding to p and q intersect at Q,we can just pick up a open affine near Q $Spec A$ where A is noetherian.

Spec A must contain p and q who correspond to minimal ideals in A.there are finitely many minimal prime ideal in A since A is noetherian.denote the intersection of minimal prime ideals except $p$ and $q$ as $I$,pick up a $f\in (p\cap I)\cap q^c$,$g\in (q\cap I)\cap p^c$(it’s easy to verify those sets are non-empty),consider $D(f),D(g)$ then it’s done.

So my questions are:

1.is my argument correct above?

2.how to deal with the situation when two irreducible components intersect not?

3.if it doesn’t hold,any simple counterexample?

Thanks for your help.

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I agree with your argument. One can also argue as follows. Given a topological space $X$ and an open subset $Y \subset X$, there is a bijective correspondence between the irreducible components of $Y$ and the irreducible components of $X$ that intersect $Y$. Thus, after replacing $X$ by a quasi-compact open subset containing $p,q$, we may assume that $X$ is a Noetherian scheme. Then $X$ has finitely many irreducible components $C_{1},\dotsc,C_{n}$ with corresponding generic points $p_{1},\dotsc,p_{n}$. For each $i$, set $U_{i} := X \setminus \bigcup_{\ell \ne i} C_{\ell}$. Then $U_{i}$ is an open subset of $X$ containing $p_{i}$, and $U_{i_{1}} \cap U_{i_{2}} = \emptyset$ if $i_{1} \ne i_{2}$ (since $U_{i} \subset C_{i}$).

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  • $\begingroup$ the topological fact you mentioned is very helpful,thank you. $\endgroup$ – XiaYu Dec 7 '17 at 5:16

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