The Stacks project

Lemma 31.17.3. Let $\pi : X \to Y$ be a finite morphism of schemes. Assume there exists a norm of degree $d$ for $\pi $. For any $\mathcal{O}_ X$-linear map $\varphi : \mathcal{L} \to \mathcal{L}'$ of invertible $\mathcal{O}_ X$-modules there is an $\mathcal{O}_ Y$-linear map

\[ \text{Norm}_\pi (\varphi ) : \text{Norm}_\pi (\mathcal{L}) \longrightarrow \text{Norm}_\pi (\mathcal{L}') \]

with $\text{Norm}_\pi (\mathcal{L})$, $\text{Norm}_\pi (\mathcal{L}')$ as in Lemma 31.17.2. Moreover, for $y \in Y$ the following are equivalent

  1. $\varphi $ is zero at a point of $x \in X$ with $\pi (x) = y$, and

  2. $\text{Norm}_\pi (\varphi )$ is zero at $y$.

Proof. We choose an open covering $Y = \bigcup V_ j$ such that $\mathcal{L}$ and $\mathcal{L}'$ are trivial over the opens $\pi ^{-1}V_ j$. This is possible by Lemma 31.17.1. Choose generating sections $s_ j$ and $s'_ j$ of $\mathcal{L}$ and $\mathcal{L}'$ over the opens $\pi ^{-1}V_ j$. Then $\varphi (s_ j) = f_ js'_ j$ for some $f_ j \in \mathcal{O}_ X(\pi ^{-1}V_ j)$. Define $\text{Norm}_\pi (\varphi )$ to be multiplication by $\text{Norm}_\pi (f_ j)$ on $V_ j$. An simple calculation involving the cocycles used to construct $\text{Norm}_\pi (\mathcal{L})$, $\text{Norm}_\pi (\mathcal{L}')$ in the proof of Lemma 31.17.2 shows that this defines a map as stated in the lemma. The final statement follows from condition (2) in the definition of a norm map of degree $d$. Some details omitted. $\square$


Comments (0)


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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BCZ. Beware of the difference between the letter 'O' and the digit '0'.