# change of rings for faithfully flat modules-Bourbaki's proof

I’m having trouble with a theorem in Bourbaki’s Algebre Commutative, namely

C1.$$\S$$.5 Proposition 8: on change of rings for faithfully flat modules)

The theorem is in 4 parts. I understand the first part and its proof but Bourbaki gives a one-sentence ‘proof ‘ of how the second part follows from the first that I can’t follow. I’ll specialize to the case where the rings involved are commutative with identity and explain Bourbaki’s notation. If $$A$$ and $$B$$ are rings, $$M$$ is a $$B$$-module and $$\rho :A \rightarrow B$$ is a ring homomorphism, then $$\rho_*(M)$$ is the $$A$$-module defined by $$am=\rho (a)m \forall a \in A, m \in M$$. Here are the first two parts of the theorem, in my translation. “Let $$A$$ and $$B$$ be rings and $$\rho :A \rightarrow B$$ a ring homomorphism. Suppose that there exists a $$B$$-module $$E$$ such that $$\rho_*(E)$$ is a faithfully flat $$A$$-module.

(i) For every $$A$$-module $$F$$, the canonical homomorphism $$j:F \rightarrow B \otimes_AF$$ such that $$x \mapsto 1 \otimes x, x \in F$$ is injective. (ii) For every ideal $$\mathfrak a$$ of $$A$$, $$\rho^{-1}(B \mathfrak a)=\mathfrak a .$$ N.B. In (i) by $$B \otimes_AF$$, Bourbaki means, it seems to me, $$\rho_*(B) \otimes_AF .$$ In (ii) by $$B \mathfrak a$$ Bourbaki means, it seems to me,$$B \rho (\mathfrak a)$$, the ideal in $$B$$ generated by $$\rho (\mathfrak a ).$$ Bourbaki says that (ii) follows from (i) by taking $$F=A/ \mathfrak a$$. I don’t see how to go from a statement about $$j$$ that involves tensor products over $$A$$ to a statement about the ring homomorphism $$\rho$$. Approaches I have tried: If we have a monomorphism or exact sequence of modules, we can tensor with a flat module and still have a monomorphism or exact sequence; the converse is also true for a faithfully flat module. If $$M$$ is an $$R$$-module then $$M \otimes_R R$$ is isomorphic to $$M$$. If $$U$$ is an $$A$$- module, $$V$$ is an $$A,B$$-bimodule and $$W$$ is a $$B$$-module, then $$U \otimes_A (V \otimes_B W)$$ is isomorphic to $$(U \otimes_A V) \otimes_B W$$. Since we are working in homological algebra, all we need to do is point to an exact sequence and say “ker =im” at an appropriate place but I can’t see how to do it. Thanks for your help.

For any $$A$$-module $$M$$ and any ideal $$I$$ of $$A$$, the natural map $$M \otimes_{A} A/I \to M/IM$$ sending the simple tensor $$m \otimes [a]$$ to $$[am]$$ is an isomorphism of $$A$$-modules. If $$\rho \colon A \to B$$ is a ring homomorphism and we view $$B$$ as an $$A$$-module via $$\rho$$, then the aforementioned natural map $$B \otimes_{A} A/I \to B/IB$$ is not just an isomorphism of $$A$$-modules, but is in fact a ring isomorphism.
Hence, suppose that (i) holds. Taking $$F = A/I$$ in the statement of (i), we are given that the $$A$$-module homomorphism $$j \colon A/I \to B \otimes_{A} A/I$$ sending $$[a]$$ to $$1 \otimes [a]$$ is injective. In fact, $$j$$ is also a ring homomorphism, and if we compose $$j$$ with the natural isomorphism $$B \otimes_{A} A/I \to B/IB$$, then we obtain the natural ring morphism $$\overline{\rho} \colon A/I \to B/IB, [a] \mapsto [\rho(a)]$$ induced by the composition of $$\rho$$ with the canonical quotient map $$B \to B/IB$$.
To summarize what we have learned so far, (i) tells us that $$\overline{\rho}$$ is injective. The kernel of the composite map $$A \xrightarrow{~\rho~} B \to B/IB$$ is just $$\rho^{-1}(IB)$$, which always contains $$I$$; this is why $$\overline{\rho}$$ is well-defined. The fact that $$\overline{\rho}$$ is injective tells you that $$I$$ contains $$\rho^{-1}(IB)$$; this gives the equality $$\rho^{-1}(IB) = I$$, as desired.