Bourbaki

Proposition 20

Tag 046X, permanent. Cite it as [Bourbaki, Tag 046X].

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

Let A,A′A, A' be two rings, p:A→A′p : A \to A' a homomorphism, SS a multiplicative subset of AA, S′=ρ(S)S' = \rho(S) and ρ′:S−1A→S′−1A′\rho' : S^{-1}A \to {S'}^{-1}A' the homomorphism corresponding to pp (no. 1, Proposition 2).

(i) For every A′A'-module M′M' there exists a unique S−1AS^{-1}A-isomorphism

j:S−1ρ∗(M′)→ρ∗′(S′−1M′) j : S^{-1}\rho_*(M') \to \rho'_*({S'}^{-1}M')
such that j(m′/s)=m′/ρ(s)j(m'/s) = m'/\rho(s) for all m′∈M′,s∈Sm' \in M', s \in S.

(ii) For every AA-module MM, there exists a unique isomorphism

j′:(S−1M)⊗S−1A(S′−1A′)→S′−1(M⊗AA′) j' : (S^{-1}M) \otimes_{S^{-1}A} ({S'}^{-1}A') \to {S'}^{-1}(M \otimes_A A')
of S′−1A′{S'}^{-1}A'*-modules* such that j′((m/s)⊗(a′/s′))=(m⊗a′)/(ρ(s)s′)j'((m/s) \otimes (a'/s')) = (m \otimes a')/(\rho(s)s').

(i) If we consider S′−1M′{S'}^{-1}M' as an AA-module by means of the composite homomorphism iM′S′∘pi^{S'}_{M'} \circ p, the homotheties induced by the elements of SS are bijective, hence there exists a unique homomorphism jj with the stated property (no. 2, Proposition 3). As ρ(S)=S′\rho(S) = S', jj is surjective; moreover, if m′∈M′m' \in M', s∈Ss \in S, m′/ρ(s)=0m'/\rho(s) = 0, there exists t′∈S′t' \in S' such that t′m′=0t'm' = 0; as there exists t∈St \in S such that ρ(t)=t′,t′.m′=0\rho(t) = t', t'.m' = 0 in ρ∗(M′)\rho_*(M'), hence m′/s=0m'/s = 0 in S−1ρ∗(M′)S^{-1}\rho_*(M').

(ii) As (S−1M)⊗S−1A(S′−1A′)=(M⊗AS−1A)⊗S−1A(S′−1A′)(S^{-1}M) \otimes_{S^{-1}A} ({S'}^{-1}A') = (M \otimes_A S^{-1}A) \otimes_{S^{-1}A} ({S'}^{-1}A') and

S′−1(M⊗AA′)=(M⊗AA′)⊗A′(S′−1A′), {S'}^{-1}(M \otimes_A A') = (M \otimes_A A') \otimes_{A'} ({S'}^{-1}A'),
the existence of j′j' follows from the associativity of tensor products.

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.