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.
Best Answer
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.