Corollary 3

Under the hypotheses of Theorem 3, suppose that $\operatorname{Spec}(A_p)$ is irreducible (resp. that $A_p$ is an integral domain) for every $p \in U$; then $\operatorname{Spec}(A_p)$ is irreducible (resp. $A_p$ is an integral domain) for every $p \in \operatorname{Spec}(A)$.

Let $(Y_i)_{i \in I}$ be the (finite) family of irreducible components of $\operatorname{Spec}(A)$. Let $p \in U$; since $\operatorname{Spec}(A_p)$ is irreducible, $p$ contains only one minimal prime ideal of $A$, and therefore belongs to only one of the $Y_i$ (II, § 4, No. 3, Cor. 2 of Prop. 14). The subsets $Y_i \cap U$ are closed subsets of $U$, pairwise disjoint, non-empty since $U$ is dense in $\operatorname{Spec}(A)$, and irreducible by II, § 4, No. 1, Prop. 7; they are therefore the connected components of $U$. Their closures $Y_i$ are the connected components of $\operatorname{Spec}(A)$ (Cor. 1). This proves that the connected components of $\operatorname{Spec}(A)$ are irreducible, and therefore that $\operatorname{Spec}(A_p)$ is irreducible for every $p$ (No. 8).

Suppose that $A_q$ is an integral domain for every $q \in U$. Let $p \in \operatorname{Spec}(A)$. Since $\operatorname{Spec}(A_p)$ is irreducible, the nilradical of $A_p$ is the unique minimal prime ideal of $A_p$; it therefore belongs to $\operatorname{Ass}(A_p)$ (IV, § 1, No. 3, Cor. 1 of Prop. 7), and consequently is equal to $qA_p$, where $q$ is a prime ideal associated with $A$ (IV, § 1, No. 2, Cor. of Prop. 5). One has $q \in U$ (Remark 1) and $qA_q \in \operatorname{Ass}(A_q)$ (loc. cit.); since $A_q$ is an integral domain, $q$ is zero, and therefore $A_p$ is an integral domain.