Lemma 49.2.5. If $A \to A_1$ is flat, then the base change map (49.2.3.1) induces an isomorphism $\omega _{B/A} \otimes _ B B_1 \to \omega _{B_1/A_1}$.

**Proof.**
Assume that $A \to A_1$ is flat. By construction of $\omega _{B/A}$ we may assume that $A \to B$ is finite. Then $\omega _{B/A} = \mathop{\mathrm{Hom}}\nolimits _ A(B, A)$ and $\omega _{B_1/A_1} = \mathop{\mathrm{Hom}}\nolimits _{A_1}(B_1, A_1)$. Since $B_1 = B \otimes _ A A_1$ the result follows from More on Algebra, Lemma 15.65.4.
$\square$

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