Spec R, Dimension of Space and Ring

An Irreducible Spectrum

Recall that a space is irreducible if every pair of nonempty open sets intersects. Equivalently, every pair of proper closed sets fails to cover the entire space. Is spec R irreducible?

Spec R is not irreducible iff we can find two proper closed subspaces that cover all of spec R. this happens iff the ideals A and B are not in rad(0), so that vA and vB are proper, yet every prime ideal contains either A or B, contains A ∩B, and contains AB. In other words, rad(0) contains AB, but neither A nor B.

In summary, spec R is irreducible iff rad(0) is prime.

Is the closed subspace vE irreducible, for a given ideal E? Again, this space is not irreducible iff we can find vA and vB that cover vE. We may as well assume A and B contain E, since our subspace is restricted to those prime ideals that contain E. Neither A nor B lies in rad(E), else one of our closed sets is the entire space. At the same time, AB lies in rad(E), else vA and vB would miss a prime ideal containing E. Once again rad(E) is not prime.

In summary, vE is irreducible iff rad(E) is prime.

The Dimension of Spec R

The dimension of a space is one less then the longest chain of irreducible closed sets, and the dimension of a ring is one less than the longest chain of prime ideals. Use the above to show that a chain of irreducible closed sets corresponds to a chain of prime ideals. Therefore the dimension of spec R, as a topological space, equals the dimension of R as a ring.

An irreducible closed set is called an irreducible component if it is maximal, i.e. at the top of its chain. This happens iff the corresponding ideal is a minimal prime. The correspondence is one to one. Since every closed set lies in an irreducible component, every radical ideal descends to a minimal prime ideal. (You can see a more direct proof of this fact here.)