Posted by Jason Polak on 04. January 2018 · Write a comment · Categories: commutative-algebra

For a commutative ring, what does the partially ordered set (=poset) of primes look like? I already talked a little about totally ordered sets of primes, but what about in general?

For a general partially ordered set $S$ there are two immediate questions that come to mind:

  1. Does there exist a commutative ring whose poset of primes is $S$?
  2. Does there exist a commutative ring whose poset of primes contains an embedded copy of $S$?

For example, consider this partially ordered set:

I draw the partially ordered sets so that “higher” is larger. This partially ordered set can be embedded into the poset of prime ideals of the integers

What about the totally ordered set $\Z$ itself? It cannot exist in any poset of primes, because it has no minimal or maximal element, whereas the both the intersection and union of a chain of primes are also primes.

Can the closed interval $[0,1]$ be embedded in a poset of primes? Alas, no. Even though $[0,1]$ now has a lower and upper bound, it is a dense ordered set, and a poset of primes cannot contain a “dense part”. More precisely, suppose that $P\subset Q$ are two distinct prime ideals and let $\{P_i\}$ be a maximal chain of prime ideals between $P$ and $Q$. Let $x\in Q – P$ and let
$$P’ = \cup \{ P_i : x\not\in P_i\}\\
Q’ = \cap \{ P_i : x\in P_i\}$$
Then $P’$ and $Q’$ are two distinct prime ideals such that $P’\subset Q’$ and such that there is no prime between $P’$ and $Q’$. So, $[0,1]$ indeed cannot appear in any poset of prime ideals of a commutative ring.

Leave a Reply

Your email address will not be published. Required fields are marked *