Bourbaki

Corollary

Tag 046U, permanent. Cite it as [Bourbaki, Tag 046U].

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

Let MM be an AA-module and aa an ideal of AA. The sub-S−1AS^{-1}A-modules (S−1a)(S−1M)(S^{-1}a)(S^{-1}M), a(S−1M)a(S^{-1}M), (S−1a)j(M)(S^{-1}a)j(M) (where j:M→S−1Mj : M \to S^{-1}M is the canonical mapping) and S−1(aM)S^{-1}(aM) of S−1MS^{-1}M are identical. In particular, if aa and bb are two ideals of AA, then

(S−1a)(S−1b)=a(S−1b)=(S−1a)b=S−1(ab). (S^{-1}a)(S^{-1}b) = a(S^{-1}b) = (S^{-1}a)b = S^{-1}(ab).

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

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Cites

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Transcribed from the printed volume. Not translated.