The Stacks project

Proof. There is an immediate reduction to the case where $X$, $X'$, $Y$, $Y'$ are affine. In other words, we have a cocartesian diagram of Noetherian rings

with $A \to B$ flat and quasi-finite. The base change map $\omega _{B/A} \otimes _ B B' \to \omega _{B'/A'}$ is an isomorphism (Lemma 49.2.10) and maps the trace element $\tau _{B/A}$ to the trace element $\tau _{B'/A'}$ (Lemma 49.4.4). Hence the finite $B$-module $Q = \mathop{\mathrm{Coker}}(\tau _{B/A} : B \to \omega _{B/A})$ satisfies $Q \otimes _ B B' = \mathop{\mathrm{Coker}}(\tau _{B'/A'} : B' \to \omega _{B'/A'})$. Thus $\mathfrak {D}_{B/A}B' \subset \mathfrak {D}_{B'/A'}$ which means we obtain the closed immersion $R' \to R \times _ Y Y'$. Since $R = \text{Supp}(Q)$ and $R' = \text{Supp}(Q \otimes _ B B')$ (Algebra, Lemma 10.40.5) we see that $R' \to R \times _ Y Y'$ is bijective by Algebra, Lemma 10.40.6. The equality $\mathfrak {D}_{B/A}B' = \mathfrak {D}_{B'/A'}$ holds if $B \to B'$ is flat, e.g., if $A \to A'$ is flat, see Algebra, Lemma 10.40.4. Finally, if $\omega _{B/A}$ is invertible, then we can localize and assume $\omega _{B/A} = B \lambda $. Writing $\tau _{B/A} = b\lambda $ we see that $Q = B/bB$ and $\mathfrak {D}_{B/A} = bB$. The same reasoning over $B'$ gives $\mathfrak {D}_{B'/A'} = bB'$ and the lemma is proved. $\square$