Bourbaki

Lemma 2

Tag 0DAH, permanent. Cite it as [Bourbaki, Tag 0DAH].

Commutative Algebra, chapter X, § 8. Dualité des modules de longueur finie

Let AA be a ring, α\alpha an ideal of finite type of AA, and MM an AA-module each element of which is annihilated by a power of α\alpha. Let A^\widehat{A} denote the separated completion of AA for the α\alpha-adic topology.

a) There exists on MM a unique structure of A^\widehat{A}-module extending the given structure of AA-module.

b) *The sub-A^\widehat{A}-modules of MM are its sub-AA-modules, and one has Hom⁡A(M,P)=Hom⁡A^(M,P)\operatorname{Hom}_A(M, P) = \operatorname{Hom}_{\widehat{A}}(M, P) for every A^\widehat{A}-module PP.

a) Let us identify A^\widehat{A} with the projective limit of the rings A/αnA/\alpha^n, and endow MM with the discrete topology. Let a=(an)a = (a_n) be an element of A^\widehat{A}, and xx an element of MM. Since xx is annihilated by a power of α\alpha, the sequence (anx)(a_n x) is stationary; let axax denote its limit. The mapping (a,x)↦ax(a, x) \mapsto ax defines on MM a structure of A^\widehat{A}-module extending the given structure of AA-module.

Conversely, suppose given such a structure on MM; let a=(an)a = (a_n) be an element of A^\widehat{A}, xx an element of MM and mm an integer such that αmx=0\alpha^m x = 0. For every integer nn, a−ana - a_n belongs to αn\alpha^n, which is equal to αnA^\alpha^n \widehat{A} (III, § 2, n° 12, cor. 2 of prop. 16); hence one has ax=anxax = a_n x for n≥mn \geq m, whence the assertion of uniqueness.

b) It follows from the preceding that one has Ax=A^xAx = \widehat{A}x for every x∈Mx \in M; the A^\widehat{A}-submodules of MM are therefore its AA-submodules. Finally, let uu be an AA-linear homomorphism of MM into a A^\widehat{A}-module PP. Let a=(an)a = (a_n) be an element of A^\widehat{A}, xx an element of MM, and mm an integer such that αmx=0\alpha^m x = 0; one has αmu(x)=0\alpha^m u(x) = 0. Since a−ama - a_m belongs to αmA^\alpha^m \widehat{A}, one has

u(ax)=u(amx)=amu(x)=au(x), u(ax) = u(a_m x) = a_m u(x) = au(x),

so that uu is A^\widehat{A}-linear.

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

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Translated with machine help, not yet reviewed by a person.