Lemma 49.2.9. Let $A \to B$ be a flat quasi-finite homomorphism of Noetherian rings. Then $\omega _{B/A}$ is a flat $A$-module.

**Proof.**
Let $\mathfrak q \subset B$ be a prime lying over $\mathfrak p \subset A$. We will show that the localization $\omega _{B/A, \mathfrak q}$ is flat over $A_\mathfrak p$. This suffices by Algebra, Lemma 10.39.18. By Algebra, Lemma 10.145.2 we can find an étale 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

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 $C$. By Lemma 49.2.5 and Algebra, Lemma 10.100.1 it suffices to show $\omega _{B'/A', \mathfrak q'}$ is flat over $A'_{\mathfrak p'}$. Since $\omega _{B'/A'} = \omega _{C/A'} \times \omega _{D/A'}$ by Lemma 49.2.7 this reduces us to the case where $B$ is finite flat over $A$. In this case $B$ is finite locally free as an $A$-module and $\omega _{B/A} = \mathop{\mathrm{Hom}}\nolimits _ A(B, A)$ is the dual finite locally free $A$-module. $\square$

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