Let M be an A-module and a an ideal of A. The sub-S−1A-modules (S−1a)(S−1M), a(S−1M), (S−1a)j(M) (where j:M→S−1M is the canonical mapping) and S−1(aM) of S−1M are identical. In particular, if a and b are two ideals of A, then
(S−1a)(S−1b)=a(S−1b)=(S−1a)b=S−1(ab).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.