Spectrum is sober

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 $$\mathcal{Z}(K)$$.

Thus, we have expressed $$A$$ as a union of two proper closed subsets, a contradiction. Hence, our original assumption was false, and $$P$$ is prime.