The Stacks project

Lemma 49.6.6. Let $A \to B$ be a flat quasi-finite homomorphism of Noetherian rings. The pairing of Remark 49.6.4 induces an isomorphism $J \to \mathop{\mathrm{Hom}}\nolimits _ B(\omega _{B/A}, B)$.

Proof. We first prove this when $A \to B$ is finite and flat. In this case we can localize on $A$ and assume $B$ is finite free as an $A$-module. Let $b_1, \ldots , b_ n$ be a basis of $B$ as an $A$-module and denote $b_1^\vee , \ldots , b_ n^\vee $ the dual basis of $\omega _{B/A}$. Note that $\sum b_ i \otimes c_ i \in J$ maps to the element of $\mathop{\mathrm{Hom}}\nolimits _ B(\omega _{B/A}, B)$ which sends $b_ i^\vee $ to $c_ i$. Suppose $\varphi : \omega _{B/A} \to B$ is $B$-linear. Then we claim that $\xi = \sum b_ i \otimes \varphi (b_ i^\vee )$ is an element of $J$. Namely, the $B$-linearity of $\varphi $ exactly implies that $(b \otimes 1)\xi = (1 \otimes b)\xi $ for all $b \in B$. Thus our map has an inverse and it is an isomorphism.

Let $\mathfrak q \subset B$ be a prime lying over $\mathfrak p \subset A$. We will show that the localization

\[ J_\mathfrak q \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ B(\omega _ B/A, B)_\mathfrak q \]

is an isomorphism. This suffices by Algebra, Lemma 10.23.1. By Algebra, Lemma 10.145.2 we can find an ├ętale ring map $A \to A'$ and a prime ideal $\mathfrak p' \subset A'$ lying over $\mathfrak p$ such that $\kappa (\mathfrak p') = \kappa (\mathfrak p)$ and such that

\[ B' = B \otimes _ A A' = C \times D \]

with $A' \to C$ finite and such that the unique prime $\mathfrak q'$ of $B \otimes _ A A'$ lying over $\mathfrak q$ and $\mathfrak p'$ corresponds to a prime of $C$. Let $J'$ be the annihilator of $\mathop{\mathrm{Ker}}(B' \otimes _{A'} B' \to B')$. By Lemmas 49.2.5, 49.6.2, and 49.6.5 the map $J' \to \mathop{\mathrm{Hom}}\nolimits _{B'}(\omega _{B'/A'}, B')$ is gotten by applying the functor $- \otimes _ B B'$ to the map $J \to \mathop{\mathrm{Hom}}\nolimits _ B(\omega _{B/A}, B)$. Since $B_\mathfrak q \to B'_{\mathfrak q'}$ is faithfully flat it suffices to prove the result for $(A' \to B', \mathfrak q')$. By Lemmas 49.2.7, 49.6.1, and 49.6.5 this reduces us to the case proved in the first paragraph of the proof. $\square$

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