Lemma 38.20.5. In Situation 38.20.3. Let $B \to C$ is a local map of local $A$-algebras and $N$ a $C$-module. Denote $F'_{lf} : \mathcal{C} \to \textit{Sets}$ the functor associated to the pair $(C, N)$. If $M \cong N$ as $B$-modules and $B \to C$ is finite, then $F_{lf} = F'_{lf}$.

Proof. Let $A'$ be an object of $\mathcal{C}$. Set $C' = C \otimes _ A A'$ and $N' = N \otimes _ A A'$ similarly to the definitions of $B'$, $M'$ in Situation 38.20.3. Note that $M' \cong N'$ as $B'$-modules. The assumption that $B \to C$ is finite has two consequences: (a) $\mathfrak m_ C = \sqrt{\mathfrak m_ B C}$ and (b) $B' \to C'$ is finite. Consequence (a) implies that

$V(\mathfrak m_{A'}C' + \mathfrak m_ C C') = \left( \mathop{\mathrm{Spec}}(C') \to \mathop{\mathrm{Spec}}(B') \right)^{-1}V(\mathfrak m_{A'}B' + \mathfrak m_ B B').$

Suppose $\mathfrak q \subset V(\mathfrak m_{A'}B' + \mathfrak m_ B B')$. Then $M'_{\mathfrak q}$ is flat over $A'$ if and only if the $C'_{\mathfrak q}$-module $N'_{\mathfrak q}$ is flat over $A'$ (because these are isomorphic as $A'$-modules) if and only if for every maximal ideal $\mathfrak r$ of $C'_{\mathfrak q}$ the module $N'_{\mathfrak r}$ is flat over $A'$ (see Algebra, Lemma 10.39.18). As $B'_{\mathfrak q} \to C'_{\mathfrak q}$ is finite by (b), the maximal ideals of $C'_{\mathfrak q}$ correspond exactly to the primes of $C'$ lying over $\mathfrak q$ (see Algebra, Lemma 10.36.22) and these primes are all contained in $V(\mathfrak m_{A'}C' + \mathfrak m_ C C')$ by the displayed equation above. Thus the result of the lemma holds. $\square$

There are also:

• 2 comment(s) on Section 38.20: Flattening functors

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).