This problem is taken from Atiyah-Macdonald.

A multiplicatively closed subset of a ring is said to be *saturated* if

.

Prove that

(i) is saturated is a union of prime ideals.

(ii) If is any multiplicatively closed subset of , there is a unique smallest saturated multiplicatively closed subset containing , and that is the complement in of the union of the prime ideals which do not meet . ( is called the saturation of .)

*Solution*:

(i) This is trivial if , since then and is a union of prime ideals. So suppose otherwise. Let ; we will show that is contained in a prime ideal contained in . Let denote the collection of ideals containing and contained in , partially ordered by inclusion. is nonempty since \in ; indeed, since and is saturated, for every . Moreover, it is clear that every chain in has an upper bound — simply take the union of all ideals in the chain. So, by Zorn’s theorem, has a maximal element . We will show that is prime. Suppose otherwise; then there exist such that . The ideals and strictly contain , so they do not lie in by the maximality of . So there exists such that , with , and such that , where . Multiplying these two equations, we have . This lies in since , and it lies in since is multiplicatively closed; this is a contradiction. Thus is a prime ideal.

(ii) Let denote the intersection of all saturated, multiplicatively closed subsets of containing . This is nonempty since for example is such a set. It is easy to see that is saturated and multiplicatively closed, and clearly it is the smallest saturated, multiplicatively closed subset of containing . Since is a union of prime ideals by (i), and since , we see thatÂ is a union of prime ideals contained in , i.e. a union of prime ideals not intersecting . Since is minimal, must in fact be the union of all such prime ideals.