Corollary 4

Let $A$ be a noetherian ring whose spectrum is connected. Suppose there exists an integer $d \geq 1$ such that one has $\operatorname{prof}(A_p) \geq 2$ for every prime ideal $p$ of $A$ of height $> d$.

a) For every closed subset $Z$ of $\operatorname{Spec}(A)$ of codimension $> d$, the space $\operatorname{Spec}(A) - Z$ is connected.

b) Let $Y$ and $Y'$ be irreducible components of $\operatorname{Spec}(A)$. There exists a sequence $X_1, X_2, \ldots, X_n$ of irreducible components of $\operatorname{Spec}(A)$ such that one has $X_1 = Y$, $X_n = Y'$ and, for $i = 1, \ldots, n-1$

Let $Z \subset \operatorname{Spec}(A)$ be a closed subset of codimension $> d$. For every $p \in Z$, one has $\dim(A_p) > d$, hence $\operatorname{prof}(A_p) \geq 2$, which implies that $\operatorname{prof}_Z(A)$ is $\geq 2$ (No. 5, Prop. 8) and therefore that $\operatorname{Spec}(A) - Z$ is connected (Theorem 3).

Let us prove b). Denote by $Z$ the union of the sets $X' \cap X''$ where $(X', X'')$ runs through the (finite) set of pairs of irreducible components of $\operatorname{Spec}(A)$ such that $\operatorname{codim}(X' \cap X'', \operatorname{Spec}(A)) > d$. By a), the set $\operatorname{Spec}(A) - Z$ is connected. All the irreducible components of $\operatorname{Spec}(A)$ meet $\operatorname{Spec}(A) - Z$; their traces on $\operatorname{Spec}(A) - Z$ are the irreducible components of $\operatorname{Spec}(A) - Z$ (II, § 4, No. 1, Prop. 7). Since $\operatorname{Spec}(A) - Z$ is connected, there exists a sequence $X_1, \ldots, X_n$ of irreducible components of $\operatorname{Spec}(A)$ such that one has $X_1 - Z = Y - Z$, $X_n - Z = Y' - Z$ and $(X_i - Z) \cap (X_{i+1} - Z) \neq \varnothing$ for $1 \leq i \leq n-1$; in other words one has $X_1 = Y$, $X_n = Z$ and $\operatorname{codim}(X_i \cap X_{i+1}) \leq d$.