Lemma 31.17.6. Let $\pi : X \to Y$ be a finite locally free morphism of degree $d \geq 1$. Then there exists a canonical norm of degree $d$ whose formation commutes with arbitrary base change.

Proof. Let $V \subset Y$ be an affine open such that $(\pi _*\mathcal{O}_ X)|_ V$ is finite free of rank $d$. Choosing a basis we obtain an isomorphism

$\mathcal{O}_ V^{\oplus d} \cong (\pi _*\mathcal{O}_ X)|_ V$

For every $f \in \pi _*\mathcal{O}_ X(V) = \mathcal{O}_ X(\pi ^{-1}(V))$ multiplication by $f$ defines a $\mathcal{O}_ V$-linear endomorphism $m_ f$ of the displayed free vector bundle. Thus we get a $d \times d$ matrix $M_ f \in \text{Mat}(d \times d, \mathcal{O}_ Y(V))$ and we can set

$\text{Norm}_\pi (f) = \det (M_ f)$

Since the determinant of a matrix is independent of the choice of the basis chosen we see that this is well defined which also means that this construction will glue to a global map as desired. Compatibility with base change is straightforward from the construction.

Property (1) follows from the fact that the determinant of a $d \times d$ diagonal matrix with entries $g, g, \ldots , g$ is $g^ d$. To see property (2) we may base change and assume that $Y$ is the spectrum of a field $k$. Then $X = \mathop{\mathrm{Spec}}(A)$ with $A$ a $k$-algebra with $\dim _ k(A) = d$. If there exists an $x \in X$ such that $f \in A$ vanishes at $x$, then there exists a map $A \to \kappa$ into a field such that $f$ maps to zero in $\kappa$. Then $f : A \to A$ cannot be surjective, hence $\det (f : A \to A) = 0$ as desired. $\square$

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