Lemma 49.2.3. Let $A \to B$ be a quasi-finite map of Noetherian rings.
If $A \to B$ factors as $A \to A_ f \to B$ for some $f \in A$, then $\omega _{B/A} = \omega _{B/A_ f}$.
If $g \in B$, then $(\omega _{B/A})_ g = \omega _{B_ g/A}$.
If $f \in A$, then $\omega _{B_ f/A_ f} = (\omega _{B/A})_ f$.
Proof. Say $A \to B' \to B$ is a factorization with $A \to B'$ finite and $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(B')$ an open immersion. In case (1) we may use the factorization $A_ f \to B'_ f \to B$ to compute $\omega _{B/A_ f}$ and use Algebra, Lemma 10.10.2. In case (2) use the factorization $A \to B' \to B_ g$ to see the result. Part (3) follows from a combination of (1) and (2). $\square$
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