Lemma 49.2.6. Let $A \to B \to C$ be quasi-finite homomorphisms of Noetherian rings. There is a canonical map $\omega _{B/A} \otimes _ B \omega _{C/B} \to \omega _{C/A}$.

Proof. Choose $A \to B' \to B$ with $A \to B'$ finite such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(B')$ is an open immersion. Then $B' \to C$ is quasi-finite too. Choose $B' \to C' \to C$ with $B' \to C'$ finite and $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(C')$ an open immersion. Then the source of the arrow is

$\mathop{\mathrm{Hom}}\nolimits _ A(B', A) \otimes _{B'} B \otimes _ B \mathop{\mathrm{Hom}}\nolimits _ B(B \otimes _{B'} C', B) \otimes _{B \otimes _{B'} C'} C$

which is equal to

$\mathop{\mathrm{Hom}}\nolimits _ A(B', A) \otimes _{B'} \mathop{\mathrm{Hom}}\nolimits _{B'}(C', B) \otimes _{C'} C$

This indeed comes with a canonical map to $\mathop{\mathrm{Hom}}\nolimits _ A(C', A) \otimes _{C'} C = \omega _{C/A}$ coming from composition $\mathop{\mathrm{Hom}}\nolimits _ A(B', A) \times \mathop{\mathrm{Hom}}\nolimits _{B'}(C', B) \to \mathop{\mathrm{Hom}}\nolimits _ A(C', A)$. $\square$

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