## 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.)