Lemma 49.14.3. Assume the Dedekind different is defined for $A \to B$. Set $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$. The generalization of Remark 49.14.2 applies to the morphism $f : Y \to X$ if and only if $1 \in \mathcal{L}_{B/A}$ (e.g., if $A$ is normal, see Lemma 49.8.1). In this case $\mathfrak {D}_{B/A}$ is an ideal of $B$ and we have
as coherent ideal sheaves on $Y$.
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. $\square$
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)