## 52.14 Application to connectedness

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].

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

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

2. $c$ 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. $\square$

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 52.12.3 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. $\square$

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. $\square$

[1] For example if $A$ is a domain.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).