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
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
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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