The Stacks project

Proof. Clearly we may assume $Y$ and hence $X$ affine. Since $\pi $ is finite the fibre $\pi ^{-1}(\{ y\} )$ over $y$ is finite. Since $X$ is affine, we can pick $s \in \Gamma (X, \mathcal{L})$ not vanishing in any point of $\pi ^{-1}(\{ y\} )$. This follows from Properties, Lemma 28.29.7 but we also give a direct argument. Namely, we can pick a finite set $E \subset X$ of closed points such that every $x \in \pi ^{-1}(\{ y\} )$ specializes to some point of $E$. For $x \in E$ denote $i_ x : x \to X$ the closed immersion. Then $\mathcal{L} \to \bigoplus _{x \in E} i_{x, *}i_ x^*\mathcal{L}$ is a surjective map of quasi-coherent $\mathcal{O}_ X$-modules, and hence the map

is surjective (as taking global sections is an exact functor on the category of quasi-coherent $\mathcal{O}_ X$-modules, see Schemes, Lemma 26.7.5). Thus we can find an $s \in \Gamma (X, \mathcal{L})$ not vanishing at any point specializing to a point of $E$. Then $X_ s \subset X$ is an open neighbourhood of $\pi ^{-1}(\{ y\} )$. Since $\pi $ is finite, hence closed, we conclude that there is an open neighbourhood $V \subset Y$ of $y$ whose inverse image is contained in $X_ s$ as desired. $\square$

Proof. We will use the correspondence between isomorphism classes of invertible $\mathcal{O}_ X$-modules and elements of $H^1(X, \mathcal{O}_ X^*)$ given in Cohomology, Lemma 20.6.1 without further mention. We explain how to take the norm of an invertible $\mathcal{O}_ X$-module $\mathcal{L}$. Namely, by Lemma 31.17.1 there exists an open covering $Y = \bigcup V_ j$ such that $\mathcal{L}|_{\pi ^{-1}V_ j}$ is trivial. Choose a generating section $s_ j \in \mathcal{L}(\pi ^{-1}V_ j)$ for each $j$. On the overlaps $\pi ^{-1}V_ j \cap \pi ^{-1}V_{j'}$ we can write

for a unique $u_{jj'} \in \mathcal{O}^*_ X(\pi ^{-1}V_ j \cap \pi ^{-1}V_{j'})$. Thus we can consider the elements

These elements satisfy the cocycle condition (because the $u_{jj'}$ do and $\text{Norm}_\pi $ is multiplicative) and therefore define an invertible $\mathcal{O}_ Y$-module. We omit the verification that: this is well defined, additive on Picard groups, and satisfies the property $\text{Norm}_\pi (\pi ^*\mathcal{N}) \cong \mathcal{N}^{\otimes d}$ for all invertible $\mathcal{O}_ Y$-modules $\mathcal{N}$. $\square$

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. $\square$

Proof. Let $\mathcal{L}$ be the ample invertible sheaf on $X$ given to us by assumption. We will prove that $\mathcal{N} = \text{Norm}_\pi (\mathcal{L})$ is ample on $Y$.

Since $X$ is quasi-compact (Properties, Definition 28.26.1) and $X \to Y$ surjective (by the existence of $\text{Norm}_\pi $) we see that $Y$ is quasi-compact. Let $y \in Y$ be a point. To finish the proof we will show that there exists a section $t$ of some positive tensor power of $\mathcal{N}$ which does not vanish at $y$ such that $Y_ t$ is affine. To do this, choose an affine open neighbourhood $V \subset Y$ of $y$. Choose $n \gg 0$ and a section $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ such that

by Properties, Lemma 28.29.6. Then $t = \text{Norm}_\pi (s)$ is a section of $\mathcal{N}^{\otimes n}$ which does not vanish at $x$ and with $Y_ t \subset V$, see Lemma 31.17.3. Then $Y_ t$ is affine by Properties, Lemma 28.26.4. $\square$

Proof. By Properties, Lemma 28.27.1 we see that $\mathcal{O}_ X$ is an ample invertible sheaf on $X$. The proof of Lemma 31.17.4 shows that $\text{Norm}_\pi (\mathcal{O}_ X) = \mathcal{O}_ Y$ is an ample invertible $\mathcal{O}_ Y$-module. Hence Properties, Lemma 28.27.1 shows that $Y$ is quasi-affine. $\square$

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. $\square$

Proof. Let $\mathop{\mathrm{Spec}}(B) \subset Y$ be an affine open subset and let $\mathop{\mathrm{Spec}}(A) \subset X$ be its inverse image. Then $A$ and $B$ are domains. Let $K$ be the fraction field of $A$ and $L$ the fraction field of $B$. Picture:

Since $K/L$ is a finite extension, there is a norm map $\text{Norm}_{K/L} : K^* \to L^*$ of degree $d = [K : L]$; this is given by mapping $f \in K$ to $\det _ L(f : K \to K)$ as in the proof of Lemma 31.17.6. Observe that the characteristic polynomial of $f : K \to K$ is a power of the minimal polynomial of $f$ over $L$; in particular $\text{Norm}_{K/L}(f)$ is a power of the constant coefficient of the minimal polynomial of $f$ over $L$. Hence by Algebra, Lemma 10.38.6 $\text{Norm}_{K/L}$ maps $A$ into $B$. This determines a compatible system of maps on sections over affines and hence a global norm map $\text{Norm}_\pi $ of degree $d$.

Property (1) is immediate from the construction. To see property (2) let $f \in A$ be contained in the prime ideal $\mathfrak p \subset A$. Let $f^ m + b_1 f^{m - 1} + \ldots + b_ m$ be the minimal polynomial of $f$ over $L$. By Algebra, Lemma 10.38.6 we have $b_ i \in B$. Hence $b_0 \in B \cap \mathfrak p$. Since $\text{Norm}_{K/L}(f) = b_0^{d/m}$ (see above) we conclude that the norm vanishes in the image point of $\mathfrak p$. $\square$

Proof. Let $A$ be a Noetherian ring with $pA = 0$. Let $I \subset A$ be the ideal of nilpotent elements. Then $I^ n = 0$ for some $n$ (Algebra, Lemma 10.32.5). Pick $e$ such that $p^ e \geq n$. Then

is well defined. This produces a norm of degree $p^ e$ for $\mathop{\mathrm{Spec}}(A/I) \to \mathop{\mathrm{Spec}}(A)$. Now if $X$ is obtained by glueing some affine schemes $\mathop{\mathrm{Spec}}(A_ i)$ then for some $e \gg 0$ these maps glue to a norm map for $X_{red} \to X$. Details omitted. $\square$

Proof. Case (1) follows from a combination of Lemmas 31.17.6 and 31.17.4. Case (3) follows from a combination of Lemmas 31.17.8 and 31.17.4. In case (2) we first replace $X$ by an irreducible component of $X$ which dominates $Y$ (viewed as a reduced closed subscheme of $X$). Then we can apply Lemma 31.17.7. $\square$

Proof. Case (1) follows from a combination of Lemmas 31.17.6 and 31.17.5. In case (2) we first replace $X$ by an irreducible component of $X$ which dominates $Y$ (viewed as a reduced closed subscheme of $X$). Then we can apply Lemma 31.17.7. $\square$