Lemma 52.14.2. Let $I \subset \mathfrak a$ be ideals of a Noetherian ring $A$. Assume

1. $A$ is $I$-adically complete and has a dualizing complex,

2. if $\mathfrak p \subset A$ is a minimal prime not contained in $V(I)$ and $\mathfrak q \in V(\mathfrak p) \cap V(\mathfrak a)$, then $\dim ((A/\mathfrak p)_\mathfrak q) > \text{cd}(A, I) + 1$,

3. any nonempty open $V \subset \mathop{\mathrm{Spec}}(A)$ which contains $V(I) \setminus V(\mathfrak a)$ is connected1.

Then $V(I) \setminus V(\mathfrak a)$ is either empty or connected.

Proof. We may replace $A$ by its reduction. Then we have the inequality in (2) for all associated primes of $A$. By Proposition 52.12.2 we see that

$\mathop{\mathrm{colim}}\nolimits H^0(V, \mathcal{O}_ V) = \mathop{\mathrm{lim}}\nolimits H^0(T_ n, \mathcal{O}_{T_ n})$

where the colimit is over the opens $V$ as in (3) and $T_ n$ is the $n$th infinitesimal neighbourhood of $T = V(I) \setminus V(\mathfrak a)$ in $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$. Thus $T$ is either empty or connected, since if not, then the right hand side would have a nontrivial idempotent and we've assumed the left hand side does not. Some details omitted. $\square$

 For example if $A$ is a domain.

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