Bourbaki

Theorem 1

Tag 0465, permanent. Cite it as [Bourbaki, Tag 0465].

Commutative Algebra, chapter II, § 2. Rings and modules of fractions

The ring S−1AS^{-1}A is a flat AA-module (Chapter I, § 2, no. 1, Definition 2).

If u:M′→Mu : M' \to M is an injective homomorphism of AA-modules, it is necessary to establish that S−1u:S−1M′→S−1MS^{-1}u : S^{-1}M' \to S^{-1}M is injective. Now, if m′/sm'/s (m′∈M′m' \in M', s∈Ss \in S) is such that u(m′)/s=0u(m')/s = 0, this implies the existence of an s′∈Ss' \in S such that s′u(m′)=0s'u(m') = 0 (no. 2, Proposition 4) or u(s′m′)=0u(s'm') = 0; as uu is injective, it follows that s′m′=0s'm' = 0, whence m′/s=0m'/s = 0.

The fact that S−1AS^{-1}A is a flat AA-module allows us to apply to it the results of Chapter I, § 2. In particular: (1) If MM is an AA-module and NN a submodule of MM, S−1NS^{-1}N is canonically identified with a submodule of S−1MS^{-1}M generated by iMS(N)i_M^S(N) (Chapter I. § 2, no. 3, Remark 2); with this identification, S−1(M/N)S^{-1}(M/N) is identified with (S−1M)/(S−1N)(S^{-1}M)/(S^{-1}N) and, if PP is a second submodule of MM, then

S−1(N+P)=S−1N+S−1P,S−1(N∩P)=S−1N∩S−1P S^{-1}(N + P) = S^{-1}N + S^{-1}P, \quad S^{-1}(N \cap P) = S^{-1}N \cap S^{-1}P

(Chapter I, § 2, no. 6, Proposition 2). (2) If MM is a finitely generated AA-module, then

S−1Ann⁡(M)=Ann⁡(S−1M) S^{-1} \operatorname{Ann}(M) = \operatorname{Ann}(S^{-1}M)

(Chapter I, § 2, no. 10, Corollary 2 to Proposition 12).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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