TITLE: Hartshorne's connectedness theorem
QUESTION [5 upvotes]: A key step in proving Hartshorne's connectedness theorem is the following: 

Let $(R,\mathfrak m)$ be a Noetherian local ring such that $\operatorname{depth}R\geq 2$. Then $U=\operatorname{Spec}R\setminus \{\mathfrak m\}$ is connected.

What does the conclusion of this theorem say algebraically?
Note: The hypothesis has a condition on depth, which is an algebraic condition. Now i know that for a commutative ring with unit, $\operatorname{Spec}R$ is connected iff $R$ has no non-trivial idempotents. But here, we are removing a point, so how does this translate back to the ring in question?

REPLY [3 votes]: Let's assume that $(R,\mathfrak m)$ is a Noetherian local ring with $\dim R\ge 1$. Then $U=\operatorname{Spec}R\setminus \{\mathfrak m\}$ is disconnected if and only if the set of minimal primes of $R$ can be partitioned into two subsets $\{\mathfrak p_1,\dots,\mathfrak p_t\}$ and $\{\mathfrak p_{t+1},\dots,\mathfrak p_n\}$ such that $\sqrt{\mathfrak p_i+\mathfrak p_j}=\mathfrak m$ for $i\le t$ and $j>t$. (This book can be a good reference.)