Lemma 38.20.6. In Situation 38.20.3 suppose that $B \to C$ is a flat local homomorphism of local rings. Set $N = M \otimes _ B C$. Denote $F'_{lf} : \mathcal{C} \to \textit{Sets}$ the functor associated to the pair $(C, N)$. Then $F_{lf} = F'_{lf}$.

Proof. Let $A'$ be an object of $\mathcal{C}$. Set $C' = C \otimes _ A A'$ and $N' = N \otimes _ A A' = M' \otimes _{B'} C'$ similarly to the definitions of $B'$, $M'$ in Situation 38.20.3. Note that

$V(\mathfrak m_{A'}B' + \mathfrak m_ B B') = \mathop{\mathrm{Spec}}( \kappa (\mathfrak m_ B) \otimes _ A \kappa (\mathfrak m_{A'}) )$

and similarly for $V(\mathfrak m_{A'}C' + \mathfrak m_ C C')$. The ring map

$\kappa (\mathfrak m_ B) \otimes _ A \kappa (\mathfrak m_{A'}) \longrightarrow \kappa (\mathfrak m_ C) \otimes _ A \kappa (\mathfrak m_{A'})$

is faithfully flat, hence $V(\mathfrak m_{A'}C' + \mathfrak m_ C C') \to V(\mathfrak m_{A'}B' + \mathfrak m_ B B')$ is surjective. Finally, if $\mathfrak r \in V(\mathfrak m_{A'}C' + \mathfrak m_ C C')$ maps to $\mathfrak q \in V(\mathfrak m_{A'}B' + \mathfrak m_ B B')$, then $M'_{\mathfrak q}$ is flat over $A'$ if and only if $N'_{\mathfrak r}$ is flat over $A'$ because $B' \to C'$ is flat, see Algebra, Lemma 10.39.9. The lemma follows formally from these remarks. $\square$

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