# 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.