The Stacks project

Proof. This is clear from Lemmas 49.8.1 and 49.8.3. $\square$

Proof. Consider the map

The image of this map corresponds to the Noether different on affine opens, see Lemma 49.6.7. Hence the result follows from the elementary fact that given an invertible module $\omega $ and a global section $\tau $ the image of $\tau : \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\omega , \mathcal{O}) = \omega ^{\otimes -1} \to \mathcal{O}$ is the same as the annihilator of $\mathop{\mathrm{Coker}}(\tau : \mathcal{O} \to \omega )$. $\square$

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. $\square$

Proof. The norm map is constructed in Divisors, Lemma 31.17.6 and the discriminant of $f$ in Section 49.3. The question is affine local, hence we may assume $X = \mathop{\mathrm{Spec}}(A)$, $Y = \mathop{\mathrm{Spec}}(B)$ and $f$ given by a finite locally free ring map $A \to B$. Localizing further we may assume $B$ is finite free as an $A$-module. Choose a basis $b_1, \ldots , b_ n \in B$ for $B$ as an $A$-module. Denote $b_1^\vee , \ldots , b_ n^\vee $ the dual basis of $\omega _{B/A} = \mathop{\mathrm{Hom}}\nolimits _ A(B, A)$ as an $A$-module. Since the norm of $b$ is the determinant of $b : B \to B$ as an $A$-linear map, we see that $\text{Norm}_{B/A}(b) = \det (b_ i^\vee (bb_ j))$. The discriminant is the principal closed subscheme of $\mathop{\mathrm{Spec}}(A)$ defined by $\det (\text{Trace}_{B/A}(b_ ib_ j))$. If $b \in \mathfrak {D}_{B/A}$ then there exist $c_ i \in B$ such that $b \cdot b_ i^\vee = c_ i \cdot \text{Trace}_{B/A}$ where we use a dot to indicate the $B$-module structure on $\omega _{B/A}$. Write $c_ i = \sum a_{il} b_ l$. We have

which proves the lemma. $\square$

Proof. Since $f$ is of finite presentation and flat, we see that it is étale at a point if and only if it is unramified at that point. Moreover, the formation of the locus of ramified points commutes with base change. See Morphisms, Section 29.36 and especially Morphisms, Lemma 29.36.17. By Lemma 49.9.4 the formation of $R$ commutes set theoretically with base change. Hence it suffices to prove the lemma when $X$ is the spectrum of a field. On the other hand, the construction of $(\omega _{Y/X}, \tau _{Y/X})$ is local on $Y$. Since $Y$ is a finite discrete space (being quasi-finite over a field), we may assume $Y$ has a unique point.

Say $X = \mathop{\mathrm{Spec}}(k)$ and $Y = \mathop{\mathrm{Spec}}(B)$ where $k$ is a field and $B$ is a finite local $k$-algebra. If $Y \to X$ is étale, then $B$ is a finite separable extension of $k$, and the trace element $\text{Trace}_{B/k}$ is a basis element of $\omega _{B/k}$ by Fields, Lemma 9.20.7. Thus $\mathfrak {D}_{B/k} = B$ in this case. Conversely, if $\mathfrak {D}_{B/k} = B$, then we see from Lemma 49.9.5 and the fact that the norm of $1$ equals $1$ that the discriminant is empty. Hence $Y \to X$ is étale by Lemma 49.3.1. $\square$

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. $\square$