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Let $R$ be a commutative Noetherian ring with unit. Is it true that two prime ideals $p$ and $q$ are in the same connected component of $\text{Spec} R$ iff there exists a series of minimal primes ideals $p_1,\dots,p_n$ with $p\in \overline{p}_1$, $q\in\overline{p}_n$, $\overline{p}_i\cap\overline{p}_{i+1}\neq \emptyset$ for $1\leq i\leq n-1$. (Here $\overline{p}$ denotes the closure of $p$ in $\text{Spec} R$, that is, the set of primes $q$ with $p\subseteq q$.)

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. –  Julian Kuelshammer Nov 5 '12 at 9:06
    
actually i have used this site and asked many questions here, but some reformatting was done on my computer and i forgot my username and password. The above question is not homework. –  messi Nov 5 '12 at 9:10
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And what did you try to do? Can you prove one of the two directions? Where did you get stuck? And what does it mean $p \in p_1$? It seems that $p$ and $p_1$ are both prime ideals of $R$, is it not the case? –  Giovanni De Gaetano Nov 5 '12 at 9:42
    
The condition $p_i\cap p_{i+1}\neq \emptyset$ is empty because the intersection of two ideals is never empty. –  Georges Elencwajg Nov 5 '12 at 9:56
    
You are right, I meant $V(p_i)\cap V(p_{i+1})\neq \emptyset$ for $1\leq i\leq n-1$. Sorry about that. But i think Navigetor23 noted this. –  messi Nov 5 '12 at 14:56

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