Spectrum is sober

Statement

Topological statement

The spectrum of a commutative unital ring, with the usual topology, is a sober space: in other words, any irreducible closed subset is the closure of a one-point subset.

Ring-theoretic statement

Any irreducible closed subset in the spectrum of a commutative unital ring corresponds to a prime ideal under the Galois correspondece between a ring and its spectrum.

Proof

Given: A commutative unital ring $R$ , its spectrum $Spec(R)$ , an irreducible closed subset $A$ of $Spec(R)$ . Let $\mathcal{Z}(I)$ , for an ideal $I$ , denote the set of prime ideals containing $I$ .

To prove: There is a prime ideal $P$ such that $A$ is precisely the set of all prime ideals containing $P$

Proof: By definition of closed set, there exists a unique radical ideal $P$ such that $A$ is precisely the set of prime ideals containing $P$ . We need to show that $P$ is in fact prime. Suppose not. Then there exist ideals $J$ and $K$ of $R$ such that $JK \subseteq P$ but neither $J$ nor $K$ is contained in $P$ .

We now argue the following:

$\mathcal{Z}(J) \cup \mathcal{Z}(K) = \mathcal{Z}(P) = A$ : If $Q$ is a prime ideal containing $P$ , then $JK \subset Q$ . But primeness of $Q$ forces $J \subset Q$ or $K \subset Q$ , thus forcing $Q \in \mathcal{Z}(J) \cup \mathcal{Z}(K)$ .
Both are closed subsets: This is by definition
Both are proper subsets: If $\mathcal{Z}(J) = \mathcal{Z}(P)$ , then $P$ being a radical ideal, must be the radical of $J$ . But that'd force $J \subset P$ , a contradiction to assumption. Hence $\mathcal{Z}(J)$ must be a proper subset of $A$ . A similar argument holds for $\mathcal{Z}(K)$ .