**Proof.**
Proof of (1). We may work locally on $Y$, hence we may assume $\omega _{Y/X}$ is free of rank $1$. Say $\omega _{Y/X} = \mathcal{O}_ Y\lambda $. Then we can write $\tau _{Y/X} = h \lambda $ and then we see that $R$ is defined by $h$, i.e., $R$ is locally principal.

Proof of (2). We may assume $Y \to X$ is given by a finite free ring map $A \to B$ and that $\omega _{B/A}$ is free of rank $1$ as $B$-module. Choose a $B$-basis element $\lambda $ for $\omega _{B/A}$ and write $\text{Trace}_{B/A} = b \cdot \lambda $ for some $b \in B$. Then $\mathfrak {D}_{B/A} = (b)$ and $D_ f$ is cut out by $\det (\text{Trace}_{B/A}(b_ ib_ j))$ where $b_1, \ldots , b_ n$ is a basis of $B$ as an $A$-module. Let $b_1^\vee , \ldots , b_ n^\vee $ be the dual basis. Writing $b_ i^\vee = c_ i \cdot \lambda $ we see that $c_1, \ldots , c_ n$ is a basis of $B$ as well. Hence with $c_ i = \sum a_{il}b_ l$ we see that $\det (a_{il})$ is a unit in $A$. Clearly, $b \cdot b_ i^\vee = c_ i \cdot \text{Trace}_{B/A}$ hence we conclude from the computation in the proof of Lemma 49.9.5 that $\text{Norm}_{B/A}(b)$ is a unit times $\det (\text{Trace}_{B/A}(b_ ib_ j))$.

Proof of (3). In the notation above we see from Lemma 49.9.6 and the assumption that $h$ does not vanish in the associated points of $Y$, which implies that $h$ is a nonzerodivisor. The canonical isomorphism sends $1$ to $\tau _{Y/X}$, see Divisors, Lemma 31.14.10.
$\square$

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