Proposition 8

Let A be a ring and S, T two multiplicative subsets of A such that $S \subset T$. The following properties are equivalent:

(a) The homomorphism $i^{T,S}_A : S^{-1}A \to T^{-1}A$ is bijective. (b) For every A-module M, the homomorphism $i^{T,S}_A : S^{-1}M \to T^{-1}M$ is bijective. (c) For all $t \in T$, there exists $a \in A$ such that $at \in S$ (in other words, every element of T divides an element of S). (d) Every prime ideal which meets T meets S.

It has been seen above that $i^{T,S}_A = 1_M \otimes i^{T,S}_A$, which immediately proves the equivalence of (a) and (b). Set $T' = i^S_A(T)$; then (Proposition 7) $T^{-1}A$ is identified with ${T'}^{-1}(S^{-1}A)$ and (a) is equivalent to saying that the elements of $T'$ are invertible in $S^{-1}A$ (no. 1, Remark 5). Now, to say that $(t/1)(a/s) = 1/1$ ($t \in T$, $a \in A, s \in S$) means that there exists $s' \in S$ such that $tas' = ss'$, which shows the equivalence of (a) and (c). We show that (d) implies (c). Let $t$ be an element of T and suppose that $t/1$ is not invertible in $S^{-1}A$; then there exists a maximal ideal $m'$ of $S^{-1}A$ containing $t/1$ (Algebra, Chapter I, § 8, no. 7, Theorem 2) and $p = (i^S_A)^{-1}(m')$ is a prime ideal of A containing $t$ and not meeting S (since the image under $i^S_A$ of an element of S is invertible). Conversely, if there exists a prime ideal $p$ which meets T without meeting S, then no element of $p \cap T$ can divide an element of S; this proves that (c) implies (d) and completes the proof.

It follows from Proposition 8 that, amongst the multiplicative subsets T of A containing S and satisfying the equivalent conditions of Proposition 8, there exists a greatest, consisting of all the elements of A which divide an element of S (cf. Exercise 1).