Lemma 49.9.5. Let $f : Y \to X$ be a finite flat morphism of Noetherian schemes. Then $\text{Norm}_ f : f_*\mathcal{O}_ Y \to \mathcal{O}_ X$ maps $f_*\mathfrak {D}_ f$ into the ideal sheaf of the discriminant $D_ f$.

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

\begin{align*} \text{Norm}_{B/A}(b) & = \det (b_ i^\vee (bb_ j)) \\ & = \det ( (b \cdot b_ i^\vee )(b_ j)) \\ & = \det ((c_ i \cdot \text{Trace}_{B/A})(b_ j)) \\ & = \det (\text{Trace}_{B/A}(c_ ib_ j)) \\ & = \det (a_{il}) \det (\text{Trace}_{B/A}(b_ l b_ j)) \end{align*}

which proves the lemma. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).