Bourbaki

Exercise 27, § 2

Commutative Algebra, III

Bourbaki marks this one as harder than the rest.

a) Let AA be a commutative ring endowed with a linear topology. For every multiplicative subset SS of AA, denote by A{S−1}A \left\{ S^{-1} \right\} the separated completion of the ring S−1AS^{-1}A endowed with the topology whose fundamental system of neighbourhoods of 00 is formed by the ideals S−1UλS^{-1}U_\lambda, where (Uλ)(U_\lambda) is a fundamental system of neighbourhoods of 00 in AA formed by ideals of AA. Now prove that A{S−1}A \left\{ S^{-1} \right\} is canonically isomorphic to the projective limit of the rings Sλ−1(A/Uλ)S_\lambda^{-1}(A/U_\lambda) where SλS_\lambda is the canonical image of SS in A/UλA/U_\lambda. If A^\hat{A} is the separated completion of AA, A{S−1}A \left\{ S^{-1} \right\} is canonically isomorphic to A^{S′−1}\hat{A} \left\{ {S'}^{-1} \right\}, where S′S' is the canonical image of SS in A^\hat{A}.

b) In order that A{S−1}A \left\{ S^{-1} \right\} be reduced to 00, it is necessary and sufficient that, in AA, 00 be adherent to SS. Deduce an example where AA is separated and complete, but where this is not so for S−1AS^{-1}A for the topology defined in a).

c) For every continuous homomorphism of AA into a linearly topologized ring BB, separated and complete, such that u(S)u(S) is formed by invertible elements in BB, now prove that u=u′∘ju = u' \circ j, where j:A→A{S−1}j : A \to A \left\{ S^{-1} \right\} is the canonical mapping, and u′u' is continuous; moreover u′u' is determined in a unique way.

d) Let S1,S2S_1, S_2 be two multiplicative subsets of AA, and let S2′S'_2 be the canonical image of S_2 in A\{S^{-1}\}; define a canonical isomorphism of A\{(S_1S_2)^{-1}\} onto A\{S_1^{-1}\}\{S_2'^{-1}\}.

e) Let aa be an open ideal of A, and let a{S−1}a\{S^{-1}\} be the separated completion of S−1aS^{-1}a for the topology induced by that of S−1AS^{-1}A; now prove that a{S−1}a\{S^{-1}\} identifies canonically with an open ideal of A{S−1}A\{S^{-1}\}, and that the discrete ring A{S−1}/a{S−1}A\{S^{-1}\}/a\{S^{-1}\} is isomorphic to S−1(A/a)S^{-1}(A/a). Conversely, if a′a' is an open ideal of A{S−1}A\{S^{-1}\}, its inverse image aa in A is an open ideal such that a′=a{S−1}a' = a\{S^{-1}\}. In particular, the mapping p→p{S−1}p \to p\{S^{-1}\} is an increasing bijection from the set of open prime ideals of A not meeting S onto the set of open prime ideals of A{S−1}A\{S^{-1}\}.

f) Let pp be an open prime ideal in A, and let S=A−pS = A - p. Now prove that A{S−1}A\{S^{-1}\} is a local ring, whose residue field is isomorphic to the field of fractions of A/pA/p.

g) For every element f∈Af \in A, we denote by A{f}A\{f\} the ring A{Sf−1}A\{S_f^{-1}\}, where SfS_f is the multiplicative set of the fn(n≥0)f^n(n \geq 0). When ff runs through a multiplicative subset S of A, the A{f}A\{f\} form an inductive filtered system of rings, whose inductive limit (not provided with a topology) is denoted by A{s}A\{s\}; define a canonical homomorphism

A{s}→A{S−1}. A\{s\} \to A\{S^{-1}\}.

When S=A−pS = A - p, where pp is a open prime ideal of A, now prove that A{s}A\{s\} is a local ring, that the canonical homomorphism A{s}→A{S−1}A\{s\} \to A\{S^{-1}\} is local and that the residue fields of A{s}A\{s\} and A{S−1}A\{S^{-1}\} are canonically isomorphic (cf. chap. II, § 3, exerc. 16).

h) Suppose that the topology of A is the mm-adic topology for an ideal mm of A, that A is separated and complete, and that m/m2m/m^2 is a (A/m)(A/m)-module of type finite. Prove that if m′=m{S−1}m' = m\{S^{-1}\}, the topology of A′=A{S−1}A' = A\{S^{-1}\} is the m′m'-adic topology, one has m′=mA′m' = mA' and m′/m′2m'/{m'}^2 is a (A′/m′)(A'/m')-module of type finite (use prop. 14 of No. 10). If A is noetherian, the same is true of A′A'.

Solution

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Machine translation of the English, by gpt-5-6. Not checked by a person.