College Math Teaching

January 28, 2015

Prime Ideal Topology

Filed under: Uncategorized — collegemathteaching @ 11:38 am

Note: this is just a watered down version of the Zariski topology on the spectrum of a commutative ring. I got the idea from Steen and Seecbach’s book Counterexamples in Topology.

I am presenting the idea while attempting to use as little ring theory as possible, as some of my students have not had abstract algebra as yet.

Consider the integers Z = \{....-3, -2, -1, 0, 1, 2, 3,... \} . An ideal I is a subset of Z that consists of 0 and all multiples of a given integer. The smallest positive integer k in an ideal is the generator of that ideal; we denote that ideal by (k) .

Examples: (1) =Z since every integer is a multiple of 1, (2) = \{ ...-6, -4, -2, 0, 2, 4, 6, ... \} and (6) = \{ ... -12, -6, 0, 6, 12, 18, ... \} .

An ideal is PRIME if it is generated by a prime number. Now if p is a prime and x \notin (p), y \notin (p) xy \notin (p) because neither has p as a prime factor. So the prime ideals are those ideals whose compliments are multiplicatively closed.

Consider X , the set of prime ideals of Z . That is, X = \{ (0), (2), (3), (5), (7), (11), ... \} The elements (points) of X are prime ideals. Yes, (0) is a prime ideal because if ab  =0 then a = 0 or b = 0 . (1) is not a prime ideal because (1) = Z and a prime ideal cannot be all of Z .

Let’s create a basis for a topology on X : let V_x = \{ I \in X, x \notin I \} . We are indexing subsets of prime ideals by positive integers and zero. Now V_0 = \emptyset as every ideal contains zero. V_1 = X as no prime ideal contains 1 . Now if x \in Z , x has a prime factorization which contains prime factors p_1, p_2, ...p_k , so x \in \cap^k_{i=1} (p_i) so V_x = X -\cup_{i=1}^k (p_i) . That is, the open basis elements are those collections of prime ideals that have a finite complement (those generated by the prime factorization of x .

Examples: V_2 = X - (2), V_8 = X - (2), V_{60} = X - \{(2) \cup (3) \cup (5) \} .

What is a closed set in this topology? A set C \subset X is closed if there exists some ideal I (not necessarily prime) such that C = \{ P \in X | I \subset P \} . For example, C = \{ (2), (3), (5) \} is a closed set because X -C is open. And note (30) \subset (2), (30) \subset (3), (30) \subset (5) . Note: traditionally, the Zariski topology is defined in terms of closed sets.

Clearly a finite union of closed sets is closed and an arbitrary intersection of closed sets is closed.

Now this topology is irreducible, which means that every non-empty pair of open sets intersect as all of them contain (0) . Remember: V_0 = \emptyset as every ideal contains 0. Hence, this topology is not Hausdorff nor is it T_1 ( a topology is T_1 if every two points lie in different open sets, though these sets may intersect each other). This does have the property of being T_0 in that, given two points, there is at least one open set that does not contain both of the points as (p) \notin V_p for p prime.

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