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

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.