Prime Ideals (Pt. III)
Point of Post: This is a continuation of this post.
So, why is this lemma so important? It clearly isn’t a ‘practical’ method to show that certain ideals are prime. I mean, theoretically it shows that and are prime in and (for an integral domain ) but it is much harder to show that they are maximal (in terms of containment) ideals disjoint from the multiplicative sets and . Indeed, in both these cases the primality of the ideals follows from the fact that the multiplicative set isn’t in their complement–it is their complement. Moreover, it’s clear that even if we had some multiplicative set it doesn’t seem likely that we would actually (in most cases) find the maximal (in terms of containment) ideal of . But, if we are able to show that always has a maximal (in terms of containment) ideal, then we may conclude that has ‘a lot’ of prime ideal–at least one, in certain cases as we shall see below.
Theorem: Let be a commutative unital ring and a multiplicative subset of . Then the set of all ideals disjoint from , ordered by containment, is either empty or has a maximal element.
Proof: Let be a chain in . We claim then that is an element of . In particular, we claim that is an ideal and it is disjoint from . To see that really is an ideal we merely note that if then by definition we have that and for , but since is a chain we may assume without loss of generality that and so and thus . Next, if and then for some and so . Thus, is an ideal as claimed. The fact that is basic set theory since to assume that is to assume that for some contrary to assumption.
Since was arbitrary we may conclude from Zorn’s Lemma.
Remark: To see why may be empty consider the multiplicative set sitting inside .
From this and the previous theorem we may conclude the following:
Theorem: Let be a commutative unital ring and a multiplicative subset of . If there exists an ideal disjoint from then there exists a prime ideal disjoint from .
Or, in the more user-friendly version:
Theorem: Let be a commutative unital ring and a multiplicative subset of . If is an ideal disjoint from then there exists a prime ideal disjoint from .
This existence theorem is only the tip of the iceberg concerning the consequences of the innocuous lemma. The next consequence we’d like to discuss shall seem somewhat contrived at first, but hopefully the usefulness will become readily apparent soon after. Namely, it’s not hard to see from the above discussion that the focus of our recent discussion has been more on multiplicative properties of rings and less on the additive properties. What I have in mind is that we defined a prime ideal to be a subgroup of with the property that if and only if or . In particular, has the absorption property of an ideal and has a multiplicative complement. Well, what if we forewent the seemingly cumbersome subgroup structure and just wondered about sets with the second property. Namely, we want to see if we can somehow characterize subsets of a commutative unital ring with the property that if and only if or . Let’s start with (as always) and work our way up. The place to begin (once again, as always) is with the primes and inevitably. Namely, if is of the type described, where does ‘put the primes and ‘? Well, the first thing we note is that just like in the case where is actually a subgroup, so that it would be an ideal, one has that if and only if . Thus, where goes becomes an almost afterthought and we are left with the real meat–the primes. To see where they ‘go’ we begin by noticing that if a prime then for every . Moreover, it’s clear that if for some and prime then since, inductively we see that implies or , etc. Thus, a dichotomy emerges for primes to powers. Namely we have that or . But, from the absorption property we see that this is equivalent to or . Therefore, we may conclude that for the case when we have that is precisely the union of where . Seeing that sets of this form are, in fact, the type of sets we wish to study we may conclude that:
Theorem: A set has the property that if and only if or if and only or if it is of the form
where is a set of (some) primes.
Recalling our earlier characterization of we may phrase this in language more amenable to our current goals:
Theorem: For a set the following are equivalent
This gives us the conviction to state and prove the following:
Theorem: Let be a commutative unital ring, then the following are equivalent:
Proof: Clearly the case is simple, so we assume that . Evidently if is a union of prime ideals then it satisfies property . Conversely, suppose that satisfies property and let be arbitrary. Since is multiplicative and is disjoint from we have that there exists some prime ideal with the property that . Thus, evidently we have that
from where the conclusion follows.
1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.
2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.
3. Reid, Miles. Undergraduate Commutative Algebra. Cambridge: Cambridge UP, 1995. Print.