Lemma 49.8.2. If the Dedekind different of $A \to B$ is defined, then there is a canonical isomorphism $\mathcal{L}_{B/A} \to \omega _{B/A}$.
Proof. Recall that $\omega _{B/A} = \mathop{\mathrm{Hom}}\nolimits _ A(B, A)$ as $A \to B$ is finite. We send $x \in \mathcal{L}_{B/A}$ to the map $b \mapsto \text{Trace}_{L/K}(bx)$. Conversely, given an $A$-linear map $\varphi : B \to A$ we obtain a $K$-linear map $\varphi _ K : L \to K$. Since $K \to L$ is finite étale, we see that the trace pairing is nondegenerate (Lemma 49.3.1) and hence there exists a $x \in L$ such that $\varphi _ K(y) = \text{Trace}_{L/K}(xy)$ for all $y \in L$. Then $x \in \mathcal{L}_{B/A}$ maps to $\varphi $ in $\omega _{B/A}$. $\square$
Post a comment
Your email address will not be published.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)