Lemma 52.14.3. Let $A$ be a Noetherian domain which has a dualizing complex and which is complete with respect to a nonzero $f \in A$. Let $f \in \mathfrak a \subset A$ be an ideal. Assume every irreducible component of $Z = V(\mathfrak a)$ has codimension $> 2$ in $X = \mathop{\mathrm{Spec}}(A)$, i.e., assume every irreducible component of $Z$ has codimension $> 1$ in $Y = V(f)$. Then $Y \setminus Z$ is connected.

Proof. This is a special case of Lemma 52.14.2 (whose proof relies on Proposition 52.12.3). Below we prove it using the easier Proposition 52.12.6.

Set $U = X \setminus Z$. By Proposition 52.12.6 we have an isomorphism

$\mathop{\mathrm{colim}}\nolimits \Gamma (V, \mathcal{O}_ V) \to \mathop{\mathrm{lim}}\nolimits _ n \Gamma (U, \mathcal{O}_ U/f^ n \mathcal{O}_ U)$

where the colimit is over open $V \subset U$ containing $U \cap Y$. Hence if $U \cap Y$ is disconnected, then for some $V$ there exists a nontrivial idempotent in $\Gamma (V, \mathcal{O}_ V)$. This is impossible as $V$ is an integral scheme as $X$ is the spectrum of a domain. $\square$

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