Bourbaki

Corollary 1

Tag 04GM, permanent. Cite it as [Bourbaki, Tag 04GM].

Commutative Algebra, chapter III, § 3. m-adic topologies on Noetherian rings

Let AA be a commutative Noetherian ring, mm an ideal of AA, EE a finitely generated AA-module and FF and GG two submodules of EE. Let A,E,F,GA, E, F, G be given the mm-adic topologies and let ii be the canonical mapping from EE to ee. Then:

F^=A^.i(F),(F+G)∧=F^+G^,(F∩G)∧=F^∩G^, \hat{F} = \hat{A}.i(F), \quad (F + G)^{\wedge} = \hat{F} + \hat{G}, \quad (F \cap G)^{\wedge} = \hat{F} \cap \hat{G},
(F:G)∧=F^:G^. (F : G)^{\wedge} = \hat{F} : \hat{G}.

Moreover, if aa and bb are two ideals of AA and c=ab,c^=a^b^c = ab, \hat{c} = \hat{a}\hat{b}.

By Theorem 3, E^,F^,G^\hat{E}, \hat{F}, \hat{G} are canonically identified with A^⊗AE,A^⊗AF,A^⊗AG\hat{A} \otimes_A E, \hat{A} \otimes_A F, \hat{A} \otimes_A G, which establishes the first two formulae. The third and fourth follow respectively from Chapter I, § 2, no. 6, Proposition 6 and no. 10, Proposition 12. Finally, as a^=A^i(a),b^=A^i(b),c^=A^i(c)\hat{a} = \hat{A}i(a), \hat{b} = \hat{A}i(b), \hat{c} = \hat{A}i(c),

c^=A^i(ab)=A^i(a)i(b)=a^b^. \hat{c} = \hat{A}i(ab) = \hat{A}i(a)i(b) = \hat{a}\hat{b}.

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

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Transcribed from the printed volume. Not translated.