# 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 , its spectrum , an irreducible closed subset of . Let , for an ideal , denote the set of prime ideals containing .

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

*Proof*: By definition of closed set, there exists a unique radical ideal such that is precisely the set of prime ideals containing . We need to show that is in fact prime. Suppose not. Then there exist ideals and of such that but neither nor is contained in .

We now argue the following:

- : If is a prime ideal containing , then . But primeness of forces or , thus forcing .
- Both are closed subsets: This is by definition
- Both are proper subsets: If , then being a radical ideal, must be the radical of . But that'd force , a contradiction to assumption. Hence must be a proper subset of . A similar argument holds for .

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