Lemma 2
Tag 0D0X, permanent. Cite it as [Bourbaki, Tag 0D0X].
Commutative Algebra, chapter X, § 1. Profondeur
Let be a local homomorphism of noetherian local rings, a finitely generated -module and an element of . The following two conditions are equivalent:
(i) the -module is flat and the homothety is injective; (ii) the -module is flat and the homothety is injective.
When they are satisfied, the homothety is injective for every -module .
Assume that the hypotheses of (i) are satisfied, and let us prove (ii) as well as the last assertion. Let be an -module. Since the -module is flat, we deduce from the exact sequence the exact sequences
where and ; it follows that the homothety of ratio is injective in , and bijective in . Assume moreover that the A-module is finitely generated; then the B-module is finitely generated (A, X, p. 107, Prop. 6), hence is zero since belongs to (Nakayama's lemma), which implies that the A-module is flat (A, X, p. 74, Theorem 2).
(ii) (i): assume that the hypotheses of (ii) are satisfied. Consider the two exact sequences of B-modules
(1)
(2)
where and are the canonical homomorphisms. It follows that the homomorphism is surjective, and (since is flat) that the kernel of the homomorphism is isomorphic to . But the mapping , equal to , is injective by hypothesis; it follows that is bijective and injective, and consequently that one has . It follows that the A-module is flat (III, § 5, No. 2, Theorem 1 and No. 4, Proposition 2).
Since and are flat over , the same is true of (exact sequence (2)). One then deduces from exact sequence (1) that is isomorphic to the kernel of , hence is zero; thus the homothety is injective by Nakayama's lemma.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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