The Stacks project

Proof. We will use Lemma 59.82.4 to prove this. First observe that the underlying topological space of $X$ is spectral by Properties, Lemma 28.2.4. Let $Y \subset X$ be an irreducible closed subscheme. To finish the proof we show that $Y \cap Z = Y \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A/I)$ is connected. Replacing $X$ by $Y$ we may assume that $X$ is irreducible and we have to show that $Z$ is connected. Let $X \to \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ be the Stein factorization of $f$ (More on Morphisms, Theorem 37.53.5). Then $A \to B$ is integral and $(B, IB)$ is a henselian pair (More on Algebra, Lemma 15.11.8). Thus we may assume the fibres of $X \to \mathop{\mathrm{Spec}}(A)$ are geometrically connected. On the other hand, the image $T \subset \mathop{\mathrm{Spec}}(A)$ of $f$ is irreducible and closed as $X$ is proper over $A$. Hence $T \cap V(I)$ is connected by More on Algebra, Lemma 15.11.16. Now $Y \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A/I) \to T \cap V(I)$ is a surjective closed map with connected fibres. The result now follows from Topology, Lemma 5.7.5. $\square$