Spec R, An Introduction

Introduction

The prime ideals of a ring R can be assigned a topology, with open and closed sets. This is the Zariski (biography) topology, sometimes called the spectrum of R, or spec R for short. Since maximal ideals are prime, spec R is nontrivial, i.e. it has at least one point.

Let's define the closed sets in this topology; take complements for the open sets. For any set of elements E in R, the collection of prime ideals containing E, denoted v_E, is a closed set. The complement o_E is open. Note that v₁ is the empty set, and v₀ is every prime ideal of R. Thus the empty set and the entire space are closed and open, as they should be.

Recall that the radical of E, written rad(E), is the intersection of the prime ideals containing E. If H = rad(E), then v_H = v_E. We can always ratchet E up to its radical ideal.

Of course we must prove this is a valid topology. Let E₁ E₂ E₃ etc be sets of elements in the ring R, and let U be their union. If a prime ideal P contains U it contains each E_i, and lives in each of the corresponding closed sets. Conversely, if P lives in each of the closed sets then it contains every E_i, and it contains their union. Therefore the intersection of arbitrarily many closed sets is closed, namely v_U. Turn this around and the union of open sets is open, as it should be.

Next consider the union of two closed sets v_E and v_F. Ratchet E and F up to ideals, and let H = E∩F. Let P be a prime ideal in v_E∪v_F. Now P contains E or F, hence it contains H. The intersection contains the product, hence P contains EF. Conversely, a prime ideal P containing EF contains E or F. Put this all together and v_E∪v_F = v_EF. The finite union of closed sets is closed, and the finite intersection of open sets is open. The topology is valid.

Infact this topology has a base, namely o_x, the complement of v_x, for any element x in R. Consider an arbitrary open set o_E with complement v_E. We showed above that v_E is the intersection of v_x for every x in E. Thus o_E is the union over o_x. Every open set is the union of base sets.

We also showed the intersection of two open sets is open, hence o_x∩o_y is open, and covered by base open sets. We have a valid base for the topology.

If R is commutative, then o_x is empty iff x is nilpotent, and o_x is everything iff x is a unit.