Lemma 49.2.10. If $A \to B$ 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.**
If $A \to B$ is finite flat, then $B$ is finite locally free as an $A$-module. In this case $\omega _{B/A} = \mathop{\mathrm{Hom}}\nolimits _ A(B, A)$ is the dual finite locally free $A$-module and formation of this module commutes with arbitrary base change which proves the lemma in this case. In the next paragraph we reduce the general (quasi-finite flat) case to the finite flat case just discussed.

Let $\mathfrak q_1 \subset B_1$ be a prime. We will show that the localization of the map at the prime $\mathfrak q_1$ is an isomorphism, which suffices by Algebra, Lemma 10.23.1. Let $\mathfrak q \subset B$ and $\mathfrak p \subset A$ be the prime ideals lying under $\mathfrak q_1$. 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 $C$. Set $A'_1 = A' \otimes _ A A_1$ and consider the base change maps (49.2.3.1) for the ring maps $A \to A' \to A'_1$ and $A \to A_1 \to A'_1$ as in the diagram

where $B' = B \otimes _ A A'$, $B_1 = B \otimes _ A A_1$, and $B_1' = B \otimes _ A (A' \otimes _ A A_1)$. By Lemma 49.2.4 the diagram commutes. By Lemma 49.2.5 the vertical arrows are isomorphisms. As $B_1 \to B'_1$ is étale and hence flat it suffices to prove the top horizontal arrow is an isomorphism after localizing at a prime $\mathfrak q'_1$ of $B'_1$ lying over $\mathfrak q$ (there is such a prime and use Algebra, Lemma 10.39.17). Thus we may assume that $B = C \times D$ with $A \to C$ finite and $\mathfrak q$ corresponding to a prime of $C$. In this case the dualizing module $\omega _{B/A}$ decomposes in a similar fashion (Lemma 49.2.7) which reduces the question to the finite flat case $A \to C$ handled above. $\square$

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