Bourbaki

Exercise 16, § 4

Commutative Algebra, II, 143

Bourbaki marks this one as harder than the rest.

(a) Let AA be an absolutely flat (commutative) ring (Chapter I, § 2, Exercise 17). Show that X=Spec⁡(A)X = \operatorname{Spec}(A) is a totally disconnected compact space, that every prime ideal qq of AA is maximal and that AqA_q is canonically isomorphic to the field A/pA/p (show first that, for all f∈Af \in A, V(f)V(f) is both open and closed in XX; on the other hand use § 3, Exercise 9).

(b) Show that the mapping a↦V(a)a \mapsto V(a) is a bijection of the set of ideals of AA onto the set of closed subsets of XX. Moreover, the following conditions are equivalent: (aa) V(a)V(a) is open; (β\beta) aa is finitely generated; (y) aa is generated by an idempotent; (δ\delta) A/aA/a is a projective AA-module.

(c) Let A~\tilde{A} be the structure sheaf of rings over XX. Show that every A~\tilde{A}-Module F\mathscr{F} is of the form E~\tilde{E}, where EE is an AA-module defined up to a unique isomorphism (take E=Γ(X,F)E = \Gamma(X, \mathscr{F})) and use the fact that every point of XX admits a fundamental system of neighbourhoods which are both open and closed). For every ideal aa of AA, show that Hom⁡A(a,E)\operatorname{Hom}_A(a, E) is identified with Γ(X−V(a),E~)\Gamma(X - V(a), \tilde{E}) (note that X−V(a)X - V(a) is the union of the XeX_e, where ee runs through the set of idempotents e∈ae \in a). Deduce that, for EE to be an injective AA-module, it is necessary and sufficient that E~\tilde{E} be flabby.

(d) Let AA be a ring and N\mathfrak{N} its nilradical. The following properties are equivalent:

(a) A/NA/\mathfrak{N} is absolutely flat. (β\beta) X=Spec⁡(A)X = \operatorname{Spec}(A) is Hausdorff. (y) Every point of XX is closed (in other words, every prime ideal of AA is maximal). (Use Exercises 9 of §3.)

Solution

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