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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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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
    
This is exercise 14(b), page 143, from Bourbaki, Commutative Algebra. Try to look there because Bourbaki gives a hint in part (a) of the exercise. – user26857 Nov 6 '12 at 17:40
    
@navigetor23 i am the same messi, now i am curious on how to get back to my old account. I came to that conclusion from an example given in fossum's book. I will give exact reference later as i have just made a move and dont have my books handy. – messi Nov 7 '12 at 4:53

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