Bourbaki

Exercise 20, § 2

Commutative Algebra, II, 131

Let F\mathscr{F} be a topologizing and idempotent set of left ideals of a ring AA. (a) Suppose that every ideal of F\mathscr{F} contains a finitely generated ideal belonging to F\mathscr{F}. Let MM be an AA-module and (Ni)i∈I(N_i)_{i \in I} a right directed family of submodules of MM, whose union is MM. Show that MFM_{\mathscr{F}} is the union of its sub-AFA_{\mathscr{F}}-modules (Ni)F(N_i)_{\mathscr{F}}. In particular, for every family (Mλ)λ∈L(M_\lambda)_{\lambda \in L} of AA-modules, (⨁λ∈LMλ)F\left( \bigoplus_{\lambda \in L} M_\lambda \right)_{\mathscr{F}} is canonically isomorphic to ⨁λ∈L(Mλ)F\bigoplus_{\lambda \in L} (M_\lambda)_{\mathscr{F}}. (b) Suppose that condition (a) holds and also that, for every surjective AA-module homomorphism u:M→Nu : M \to N, uF:MF→NFu_{\mathscr{F}} : M_{\mathscr{F}} \to N_{\mathscr{F}} is surjective. Then, the canonical mapping AF⊗AM→MFA_{\mathscr{F}} \otimes_A M \to M_{\mathscr{F}} defined by the external law on MFM_{\mathscr{F}} as an AFA_{\mathscr{F}}-module is an AFA_{\mathscr{F}}-module isomorphism (reduce it to the case where MM is free); the ring AFA_{\mathscr{F}} is a flat left AA-module and the left AFA_{\mathscr{F}}-modules coincide with the left AA-modules MM such that the mapping jM:M→MFj_M : M \to M_{\mathscr{F}} is bijective. (c) Suppose that AA is an infinite product ∏t∈IAt\prod_{t \in I} A_t, of rings; let e, denote the unit element of AA, and take F\mathscr{F} to be the set of left ideals of AA containing the two-sided ideal a=⨁t∈IAta = \bigoplus_{t \in I} A_t, (the AtA_t, being canonically identified with ideals of AA). Show that, for every left AA-module MM, MFM_{\mathscr{F}} is identified with the product ∏t∈I(etM)\prod_{t \in I} (e_t M) and in particular AF=AA_{\mathscr{F}} = A. Deduce that, for every surjective AA-homomorphism u:M→Nu : M \to N, uF:MF→NFu_{\mathscr{F}} : M_{\mathscr{F}} \to N_{\mathscr{F}} is surjective, but give examples of AA-modules MM such that jMj_M is not bijective. (d) Suppose that condition (a) holds and also that AA is commutative. Show that, if MM is 9-negligible, M(F)=0M_{(\mathscr{F})} = 0.

Solution

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