Bourbaki

Proposition 17

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

Commutative Algebra, chapter III, § 2. General results on filtered rings and modules

Let AA be a commutative ring and (mλ)λ∈L(m_{\lambda})_{\lambda \in L} a family of ideals of AA, distinct from AA, such that mλm_{\lambda} and mμm_{\mu} are relatively prime for λ≠μ\lambda \neq \mu. For every family s=(s(λ))λ∈Ls = (s(\lambda))_{\lambda \in L} of integers ≥0\geq 0, of finite support, set as=⋂λ∈Lmλs(λ)a_s = \bigcap_{\lambda \in L} m_{\lambda}^{s(\lambda)} (equal to the product of the mλs(λ)m_{\lambda}^{s(\lambda)} for the λ\lambda such that s(λ)≠0s(\lambda) \neq 0; cf. Chapter 11, § 1, no. 2, Propositions 3 and 5); the asa_s form a fundamental system of neighbourhoods of 0 with respect to a topology T\mathscr{T} compatible with the ring structure on AA; let A^\hat{A} be the Hausdorff completion of AA with respect to this topology. On the other hand, for all λ∈L\lambda \in L, let AλA_{\lambda} be the ring AA with the mλm_{\lambda}-adic topology and let A^λ\hat{A}_{\lambda} be its Hausdorff completion. If u:A→∏λ∈LAλu : A \to \prod_{\lambda \in L} A_{\lambda} denotes the diagonal homomorphism, uu is continuous and the corresponding homomorphism u^:\hat{u} :

A^→(∏λ∈LAλ)^=∏λ∈LA^λ \hat{A} \to \left( \prod_{\lambda \in L} A_{\lambda} \right)^{\hat{}} = \prod_{\lambda \in L} \hat{A}_{\lambda}

(General Topology, Chapter III, § 6, no. 5 and Chapter II, § 3, no. 9, Corollary 2 to Proposition 18) is a topological ring isomorphism.

The first assertion follows from General Topology, Chapter III, § 6, no. 3, Example 3. Let us set B=∏λ∈LAλB = \prod_{\lambda \in L} A_{\lambda}; as the topology on AA is finer than each of the mλm_{\lambda}-adic topologies, the mappings prλ∘u\mathrm{pr}_{\lambda} \circ u are continuous and hence uu is continuous. Also, u(as)u(a_s) is the intersection of the diagonal AA of BB and the open set ⋂λ∈Lprλ−1(mλs(λ))\bigcap_{\lambda \in L} \mathrm{pr}_{\lambda}^{-1}(m_{\lambda}^{s(\lambda)}) of BB; it follows that uu is a strict morphism from the additive group AA to BB with image AA. Now Δ\Delta is dense in BB. For let b=(aλ)λ∈Lb = (a_{\lambda})_{\lambda \in L} be an element of BB; every neighbourhood of bb in BB contains a set of the form b+Vb + V, where V=⋂λ∈Lprλ−1(mλs(λ))V = \bigcap_{\lambda \in L} \mathrm{pr}_{\lambda}^{-1}(m_{\lambda}^{s(\lambda)}) for a family s=(s(λ))λ∈Ls = (s(\lambda))_{\lambda \in L} with finite support of integers ≥0\geq 0. As the mλs(λ)m_{\lambda}^{s(\lambda)} are relatively prime in pairs (Chapter II, § 1, no. 2, Proposition 3), there exists x∈Ax \in A such that x≡aλ(modmλs(λ))x \equiv a_{\lambda} \pmod{m_{\lambda}^{s(\lambda)}} for all λ\lambda (loc. cit., Proposition 5) and hence (b+V)∩A≠∅(b + V) \cap A \neq \varnothing. The Hausdorff completion of the group B/ΔB/\Delta is then {0}\{0\}; applying Lemma 2 of no. 12 to the exact sequences 0→A→uB,A→uB→B/Δ0 \to A \xrightarrow{u} B, A \xrightarrow{u} B \to B/\Delta, we see that u^\hat{u} is an isomorphism of A^\hat{A} onto B^\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.