Multiplicative sets

This problem is taken from Atiyah-Macdonald.

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

xy \in S \Leftrightarrow x \in S \mbox{ and } y \in S.

Prove that

(i) S is saturated \Leftrightarrow A-S is a union of prime ideals.

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

Solution:

(i) This is trivial if 0 \in S, since then S=A and A is a union of prime ideals. So suppose otherwise. Let a \in A-S; we will show that a is contained in a prime ideal contained in A-S. Let \Sigma denote the collection of ideals containing (a) and contained in A-S, partially ordered by inclusion. \Sigma is nonempty since (a) \in \Sigma; indeed, since a \notin S and S is saturated, ca \notin S for every c \in A. Moreover, it is clear that every chain in \Sigma has an upper bound — simply take the union of all ideals in the chain. So, by Zorn’s theorem, \Sigma has a maximal element P. We will show that P is prime. Suppose otherwise; then there exist u, v \in A-P such that uv \in P. The ideals (P, u) and (P, v) strictly contain P, so they do not lie in \Sigma by the maximality of P. So there exists s \in S such that s=m+ux, with m \in P, and s' \in S such that s'=m'+ux', where m' \in P. Multiplying these two equations, we have ss'=mm'+mux'+m'vx+uvxx'. This lies in P since uv \in P, and it lies in S since S is multiplicatively closed; this is a contradiction. Thus P is a prime ideal.

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