Bourbaki

Proposition 9

Tag 0463, permanent. Cite it as [Bourbaki, Tag 0463].

Commutative Algebra, chapter II, § 2. Rings and modules of fractions

Let I be a right directed preordered set, (Sα)α∈I(S_\alpha)_{\alpha \in I} an increasing family of multiplicative subsets of a ring A and S=⋃α∈ISαS = \bigcup_{\alpha \in I} S_\alpha. We write ρβα=iASβ,SαSβ,Sα\rho_{\beta \alpha} = i^{S_\beta, S_\alpha}_{A^{S_\beta, S_\alpha}} for α≤β\alpha \leq \beta, ρα=iSα,Sα\rho_\alpha = i^{S_\alpha, S_\alpha}. Then (Sα−1A,ρβα)(S_\alpha^{-1}A, \rho_{\beta \alpha}) is a direct system of Zings and, if, for all α∈I\alpha \in I, ρα′\rho'_\alpha is the canonical mapping of Sα−1AS_\alpha^{-1}A to lim⁡→Sα−1A\lim \rightarrow S_\alpha^{-1}A, there exists a unique isomorphism jj from lim⁡→Sα−1A\lim \rightarrow S_\alpha^{-1}A to S−1AS^{-1}A such that j∘ρα′=ραj \circ \rho'_\alpha = \rho_\alpha for all α∈I\alpha \in I.

For α≤β≤γ\alpha \leq \beta \leq \gamma, ργα=ργβ∘ρβα\rho_{\gamma \alpha} = \rho_{\gamma \beta} \circ \rho_{\beta \alpha} (no. 1, Corollary 3 to Proposition 2), hence (Sα−1A,ρβα)(S_\alpha^{-1}A, \rho_{\beta \alpha}) is a direct system. We write A′=lim⁡→Sα−1AA' = \lim \rightarrow S_\alpha^{-1}A; as ρα=ρβ∘ρβα\rho_\alpha = \rho_\beta \circ \rho_{\beta \alpha} for α≤β\alpha \leq \beta (no. 1, Corollary 3 to Proposition 2), (ρα)(\rho_\alpha) is a direct system of homomorphisms and j=lim⁡→ραj = \lim \rightarrow \rho_\alpha is the unique homomorphism from A′A' to S−1AS^{-1}A such that j∘ρα′=ραj \circ \rho'_\alpha = \rho_\alpha for all a∈Ia \in I. The homomorphisms ρα′∘iAαS:A→A′\rho'_\alpha \circ i^{S}_{A^\alpha} : A \to A' are all equal, for ρβα∘iAαS=iAβS\rho_{\beta \alpha} \circ i^{S}_{A^\alpha} = i^{S}_{A^\beta} for a≤βa \leq \beta; let uu be their common value. Clearly the elements of u(S)u(S) are invertible in A′A', which shows that there exists a homomorphism h:S−1A→A′h : S^{-1}A \to A' such that h∘iAS=uh \circ i^{S}_{A} = u (no. 1, Proposition 1). Then

j∘h∘iSA=j∘u=j∘ρα′∘iSαAα=ρα∘iAαS=iAS j \circ h \circ i^{A}_{S} = j \circ u = j \circ \rho'_\alpha \circ i^{A^\alpha}_{S^\alpha} = \rho_\alpha \circ i^{S}_{A^\alpha} = i^{S}_{A}

for all α∈I\alpha \in I and consequently j∘hj \circ h is the identity automorphism of S−1AS^{-1}A. On the other hand, for all α∈I\alpha \in I,

h∘j∘ρα′∘iAαS=h∘ρα∘iAαS=h∘iAS=u=ρα′∘iAαS, h \circ j \circ \rho'_\alpha \circ i^{S}_{A^\alpha} = h \circ \rho_\alpha \circ i^{S}_{A^\alpha} = h \circ i^{S}_{A} = u = \rho'_\alpha \circ i^{S}_{A^\alpha},

whence h∘j∘ρα′=ρα′h \circ j \circ \rho'_\alpha = \rho'_\alpha for all α∈I\alpha \in I; it follows that h∘jh \circ j is the identity automorphism of A′A' and consequently jj is an isomorphism.

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

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Cites

Transcribed from the printed volume. Not translated.