Spectrum is sober

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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.


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.