Proposition (ascending union of ideals is an ideal):
Let be a totally ordered set, and let be a chain of left, right or both-sided ideals of a ring . Then
is a left- right- or both-sided ideal of respectively.
Proof: Indeed, let first . Then and for certain . Set , then since we are dealing with a chain, and consequently . The remaining property, closedness under multiplication by , we prove for left ideals, since the proof for right ideals is the same, and the claim for both-sided ideals then follows by combining the other two claims (or is proved directly in the same way). Thus, suppose that the are right ideals of , and let . Then we have whenever , but for each there exists some such that this is the case, hence the claim.
Proposition (existence of prime ideal not intersecting multiplicative set and containing ideal):
Let be a ring, let be a left, right, or both-sided ideal of , and let be a multiplicatively closed subset of so that . Then there exists a left, right or both-sided ideal of respectively such that contains , is prime, does not intersect and is maximal (with respect to inclusion) among all ideals that do not intersect .
(On the condition of the axiom of choice.)
Proof: Define to be the set of all (left, right or both-sided; we'll omit this distinction for brevity in the remainder of the proof) ideals that do not intersect and contain . This is a nonempty set, since . Moreover, given any ascending chain in with respect to inclusion, where is any totally ordered set, we obtain that
indeed, is an ideal, it certainly contains , and finally it does not intersect . Thus, is inductive with respect to inclusion, and we may choose a maximal element by Zorn's lemma, which we denote by . Claim that in fact does have all the required properties. All that is left to show is that is a prime ideal.
TODO: Distinguish between completely prime ideals and prime ideals in noncommutative rings