The Stacks project

Proof. As the Dedekind different for $A \to B$ is defined we can apply Lemma 49.14.1 to see that $Y \to X$ satisfies condition (1) of Remark 49.14.2. Recall that there is a canonical isomorphism $c : \mathcal{L}_{B/A} \to \omega _{B/A}$, see Lemma 49.8.2. Let $K = Q(A)$ and $L = K \otimes _ A B$ as above. By construction the map $c$ fits into a commutative diagram

where the right vertical arrow sends $x \in L$ to the map $y \mapsto \text{Trace}_{L/K}(xy)$ and the lower horizontal arrow is the base change map (49.2.3.1) for $\omega _{B/A}$. We can factor the lower horizontal map as

Since all associated points of $\omega _{V/X}$ map to associated primes of $A$ (Lemma 49.2.8) we see that the second map is injective. The element $\tau _{V/X}$ maps to $\text{Trace}_{L/K}$ in $\mathop{\mathrm{Hom}}\nolimits _ K(L, K)$ by the very definition of trace elements (Definition 49.4.1). Thus $\tau$ as in condition (2) of Remark 49.14.2 exists if and only if $1 \in \mathcal{L}_{B/A}$ and then $\tau = c(1)$. In this case, by Lemma 49.8.1 we see that $\mathfrak {D}_{B/A} \subset B$. Finally, the agreement of $\mathfrak {D}_ f$ with $\mathfrak {D}_{B/A}$ is immediate from the definitions and the fact $\tau = c(1)$ seen above. $\square$