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. $\square$

Comment #7168 by Xuande Liu on

I have a problem about the last assertion. By definition we know $(1)\Rightarrow(2)$. But I think it might happen that $Norm_\pi(f_j)=0$ while $f_j\neq 0$. Could you please tell me how to deduce that $(2)\Rightarrow(1)$?

Comment #7170 by on

OK, suppose that $f$ is a regular function on $X$ which does not vanish at any point lying over $y$. Then after shrinking $Y$ we may assume $f$ doesn't vanish at all. Then $f$ is a unit so we have $f g = 1$ for some other regular function $g$ on $X$. Then $1 = Norm(fg) = Norm(f) Norm(g)$ and hence $Norm(f)$ doesn't vanish.

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