Spec R, Compact

Spec R is Compact

If R is commutative, spec R is compact.

Given an open cover for spec R, replace each open set with a union of base sets. This is still an open cover, with possibly more sets.

since the space is covered, the intersection of the complements of the designated base sets is empty. Let E be the union of the elements that generate these base sets. Thus E is contained in no proper ideal, and E generates R. Write 1 as a finite linear combination of generators g1 g2 g3 … gn from E. The corresponding base open sets og1 through ogn cover spec R. These are brought in by at most n open sets from our original cover. Thus a finite subcover will do, and spec R is compact.

The Subspace defined by a Base Open Set

If T is spec R, let S be the subspace ox, for some element x in R. Now a set is closed in S if it comes from a closed set in T. In other words,vE remains a closed set in S if E misses x. (If E contains x, well, vE doesn't intersect S at all.)

If x is nilpotent then every prime ideal contains x, and S is empty. This isn't very interesting, so assume the powers of x never reach 0.

Let W be the fractions of R by the powers of x. Apply prime correspondence, and the prime ideals of R missing x, which are the points of ox, correspond to the prime ideals in W. This forms a bijection between S and spec W. Let's show it is a homeomorphism.

A closed set in either space is the set of prime ideals that contain a given ideal. Let E be an ideal in R and let F be the corresponding ideal in the fraction ring W. Now the primes containing E correspond to the primes containing F. In other words, our map carries closed sets in spec R to closed sets in spec W. conversely, let F be an ideal in W and take the numerators to find an ideal E in R. Once again the prime ideals containing E correspond to the prime ideals containing F. The map is bicontinuous, and is a homeomorphism.

Since spec W is compact, ox is a compact subspace of spec R.

It follows that an open set in spec R is compact iff it can be represented as a finite union of base open sets.

This is not a typical topology. In the plane, for instance, a nonempty open set is never compact.

Noetherian Criterion

Assume R is commutative and noetherian. As you recall, every ideal of R is finitely generated. Let H be such an ideal, hence vH is the intersection of vg1 vg2 … vgn, for the generators of H. In other words, oH is the finite union of base open sets. This holds for every H, hence every open set in spec R is compact. Of course every closed set in a compact space is also compact.

Actually this theorem is trumped by a stronger result, that is almost a corollary. Since R is noetherian, spec R is noetherian, and in such a space, every set, closed open or otherwise, is compact.

Now - don't assume the converse. Every open set in R could be compact, even though R is not noetherian. See the previous section for an example.