In this section we discuss Grothendieck's connectedness theorem and variants; the original version can be found as [Exposee XIII, Theorem 2.1, SGA2]. There is a version called Faltings' connectedness theorem in the literature; our guess is that this refers to [Theorem 6, Faltings-some]. Let us state and prove the optimal version for complete local rings given in [Theorem 1.6, Varbaro].

reference
Lemma 52.14.1. Let $(A, \mathfrak m)$ be a Noetherian complete local ring. Let $I$ be a proper ideal of $A$. Set $X = \mathop{\mathrm{Spec}}(A)$ and $Y = V(I)$. Denote

$d$ the minimal dimension of an irreducible component of $X$, and

$c$ the minimal dimension of a closed subset $Z \subset X$ such that $X \setminus Z$ is disconnected.

Then for $Z \subset Y$ closed we have $Y \setminus Z$ is connected if $\dim (Z) < \min (c, d - 1) - \text{cd}(A, I)$. In particular, the punctured spectrum of $A/I$ is connected if $\text{cd}(A, I) < \min (c, d - 1)$.

**Proof.**
Let us first prove the final assertion. As a first case, if the punctured spectrum of $A/I$ is empty, then Local Cohomology, Lemma 51.4.10 shows every irreducible component of $X$ has dimension $\leq \text{cd}(A, I)$ and we get $\min (c, d - 1) - \text{cd}(A, I) < 0$ which implies the lemma holds in this case. Thus we may assume $U \cap Y$ is nonempty where $U = X \setminus \{ \mathfrak m\} $ is the punctured spectrum of $A$. We may replace $A$ by its reduction. Observe that $A$ has a dualizing complex (Dualizing Complexes, Lemma 47.22.4) and that $A$ is complete with respect to $I$ (Algebra, Lemma 10.96.8). If we assume $d - 1 > \text{cd}(A, I)$, then we may apply Lemma 52.11.3 to see that

\[ \mathop{\mathrm{colim}}\nolimits H^0(V, \mathcal{O}_ V) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{O}_ U/I^ n\mathcal{O}_ U) \]

is an isomorphism where the colimit is over opens $V \subset U$ containing $U \cap Y$. If $U \cap Y$ is disconnected, then its $n$th infinitesimal neighbourhood in $U$ is disconnected for all $n$ and we find the right hand side has a nontrivial idempotent (here we use that $U \cap Y$ is nonempty). Thus we can find a $V$ which is disconnected. Set $Z = X \setminus V$. By Local Cohomology, Lemma 51.4.10 we see that every irreducible component of $Z$ has dimension $\leq \text{cd}(A, I)$. Hence $c \leq \text{cd}(A, I)$ and this indeed proves the final statement.

We can deduce the statement of the lemma from what we just proved as follows. Suppose that $Z \subset Y$ closed and $Y \setminus Z$ is disconnected and $\dim (Z) = e$. Recall that a connected space is nonempty by convention. Hence we conclude either (a) $Y = Z$ or (b) $Y \setminus Z = W_1 \amalg W_2$ with $W_ i$ nonempty, open, and closed in $Y \setminus Z$. In case (b) we may pick points $w_ i \in W_ i$ which are closed in $U$, see Morphisms, Lemma 29.16.10. Then we can find $f_1, \ldots , f_ e \in \mathfrak m$ such that $V(f_1, \ldots , f_ e) \cap Z = \{ \mathfrak m\} $ and in case (b) we may assume $w_ i \in V(f_1, \ldots , f_ e)$. Namely, we can inductively using prime avoidance choose $f_ i$ such that $\dim V(f_1, \ldots , f_ i) \cap Z = e - i$ and such that in case (b) we have $w_1, w_2 \in V(f_ i)$. It follows that the punctured spectrum of $A/I + (f_1, \ldots , f_ e)$ is disconnected (small detail omitted). Since $\text{cd}(A, I + (f_1, \ldots , f_ e)) \leq \text{cd}(A, I) + e$ by Local Cohomology, Lemmas 51.4.4 and 51.4.3 we conclude that

\[ \text{cd}(A, I) + e \geq \min (c, d - 1) \]

by the first part of the proof. This implies $e \geq \min (c, d - 1) - \text{cd}(A, I)$ which is what we had to show.
$\square$

**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.
$\square$

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