Lemma 35.25.1. Let $f : X \to Y$ be a flat, quasi-compact, surjective monomorphism. Then f is an isomorphism.

## 35.25 Application of fpqc descent of properties of morphisms

The following lemma may seem a bit frivolous but turns out is a useful tool in studying étale and unramified morphisms.

**Proof.**
As $f$ is a flat, quasi-compact, surjective morphism we see $\{ X \to Y\} $ is an fpqc covering of $Y$. The diagonal $\Delta : X \to X \times _ Y X$ is an isomorphism (Schemes, Lemma 26.23.2). This implies that the base change of $f$ by $f$ is an isomorphism. Hence we see $f$ is an isomorphism by Lemma 35.23.17.
$\square$

We can use this lemma to show the following important result; we also give a proof avoiding fpqc descent. We will discuss this and related results in more detail in Étale Morphisms, Section 41.14.

Lemma 35.25.2. A universally injective étale morphism is an open immersion.

**First proof.**
Let $f : X \to Y$ be an étale morphism which is universally injective. Then $f$ is open (Morphisms, Lemma 29.36.13) hence we can replace $Y$ by $f(X)$ and we may assume that $f$ is surjective. Then $f$ is bijective and open hence a homeomorphism. Hence $f$ is quasi-compact. Thus by Lemma 35.25.1 it suffices to show that $f$ is a monomorphism. As $X \to Y$ is étale the morphism $\Delta _{X/Y} : X \to X \times _ Y X$ is an open immersion by Morphisms, Lemma 29.35.13 (and Morphisms, Lemma 29.36.16). As $f$ is universally injective $\Delta _{X/Y}$ is also surjective, see Morphisms, Lemma 29.10.2. Hence $\Delta _{X/Y}$ is an isomorphism, i.e., $X \to Y$ is a monomorphism.
$\square$

**Second proof.**
Let $f : X \to Y$ be an étale morphism which is universally injective. Then $f$ is open (Morphisms, Lemma 29.36.13) hence we can replace $Y$ by $f(X)$ and we may assume that $f$ is surjective. Since the hypotheses remain satisfied after any base change, we conclude that $f$ is a universal homeomorphism. Therefore $f$ is integral, see Morphisms, Lemma 29.45.5. It follows that $f$ is finite by Morphisms, Lemma 29.44.4. It follows that $f$ is finite locally free by Morphisms, Lemma 29.48.2. To finish the proof, it suffices that $f$ is finite locally free of degree $1$ (a finite locally free morphism of degree $1$ is an isomorphism). There is decomposition of $Y$ into open and closed subschemes $V_ d$ such that $f^{-1}(V_ d) \to V_ d$ is finite locally free of degree $d$, see Morphisms, Lemma 29.48.5. If $V_ d$ is not empty, we can pick a morphism $\mathop{\mathrm{Spec}}(k) \to V_ d \subset Y$ where $k$ is an algebraically closed field (just take the algebraic closure of the residue field of some point of $V_ d$). Then $\mathop{\mathrm{Spec}}(k) \times _ Y X \to \mathop{\mathrm{Spec}}(k)$ is a disjoint union of copies of $\mathop{\mathrm{Spec}}(k)$, by Morphisms, Lemma 29.36.7 and the fact that $k$ is algebraically closed. However, since $f$ is universally injective, there can only be one copy and hence $d = 1$ as desired.
$\square$

We can reformulate the hypotheses in the lemma above a bit by using the following characterization of flat universally injective morphisms.

Lemma 35.25.3. Let $f : X \to Y$ be a morphism of schemes. Let $X^0$ denote the set of generic points of irreducible components of $X$. If

$f$ is flat and separated,

for $\xi \in X^0$ we have $\kappa (f(\xi )) = \kappa (\xi )$, and

if $\xi , \xi ' \in X^0$, $\xi \not= \xi '$, then $f(\xi ) \not= f(\xi ')$,

then $f$ is universally injective.

**Proof.**
We have to show that $\Delta : X \to X \times _ Y X$ is surjective, see Morphisms, Lemma 29.10.2. As $X \to Y$ is separated, the image of $\Delta $ is closed. Thus if $\Delta $ is not surjective, we can find a generic point $\eta \in X \times _ S X$ of an irreducible component of $X \times _ S X$ which is not in the image of $\Delta $. The projection $\text{pr}_1 : X \times _ Y X \to X$ is flat as a base change of the flat morphism $X \to Y$, see Morphisms, Lemma 29.25.8. Hence generalizations lift along $\text{pr}_1$, see Morphisms, Lemma 29.25.9. We conclude that $\xi = \text{pr}_1(\eta ) \in X^0$. However, assumptions (2) and (3) guarantee that the scheme $(X \times _ Y X)_{f(\xi )}$ has at most one point for every $\xi \in X^0$. In other words, we have $\Delta (\xi ) = \eta $ a contradiction.
$\square$

Thus we can reformulate Lemma 35.25.2 as follows.

Lemma 35.25.4. Let $f : X \to Y$ be a morphism of schemes. Let $X^0$ denote the set of generic points of irreducible components of $X$. If

$f$ is étale and separated,

for $\xi \in X^0$ we have $\kappa (f(\xi )) = \kappa (\xi )$, and

if $\xi , \xi ' \in X^0$, $\xi \not= \xi '$, then $f(\xi ) \not= f(\xi ')$,

then $f$ is an open immersion.

Lemma 35.25.5. Let $f : X \to Y$ be a morphism of schemes which is locally of finite type. Let $Z$ be a closed subset of $X$. If there exists an fpqc covering $\{ Y_ i \to Y\} $ such that the inverse image $Z_ i \subset Y_ i \times _ Y X$ is proper over $Y_ i$ (Cohomology of Schemes, Definition 30.26.2) then $Z$ is proper over $Y$.

**Proof.**
Endow $Z$ with the reduced induced closed subscheme structure, see Schemes, Definition 26.12.5. For every $i$ the base change $Y_ i \times _ Y Z$ is a closed subscheme of $Y_ i \times _ Y X$ whose underlying closed subset is $Z_ i$. By definition (via Cohomology of Schemes, Lemma 30.26.1) we conclude that the projections $Y_ i \times _ Y Z \to Y_ i$ are proper morphisms. Hence $Z \to Y$ is a proper morphism by Lemma 35.23.14. Thus $Z$ is proper over $Y$ by definition.
$\square$

Lemma 35.25.6. Let $f : X \to S$ be a morphism of schemes. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $\{ g_ i : S_ i \to S\} _{i \in I}$ be an fpqc covering. Let $f_ i : X_ i \to S_ i$ be the base change of $f$ and let $\mathcal{L}_ i$ be the pullback of $\mathcal{L}$ to $X_ i$. The following are equivalent

$\mathcal{L}$ is ample on $X/S$, and

$\mathcal{L}_ i$ is ample on $X_ i/S_ i$ for every $i \in I$.

**Proof.**
The implication (1) $\Rightarrow $ (2) follows from Morphisms, Lemma 29.37.9. Assume $\mathcal{L}_ i$ is ample on $X_ i/S_ i$ for every $i \in I$. By Morphisms, Definition 29.37.1 this implies that $X_ i \to S_ i$ is quasi-compact and by Morphisms, Lemma 29.37.3 this implies $X_ i \to S$ is separated. Hence $f$ is quasi-compact and separated by Lemmas 35.23.1 and 35.23.6.

This means that $\mathcal{A} = \bigoplus _{d \geq 0} f_*\mathcal{L}^{\otimes d}$ is a quasi-coherent graded $\mathcal{O}_ S$-algebra (Schemes, Lemma 26.24.1). Moreover, the formation of $\mathcal{A}$ commutes with flat base change by Cohomology of Schemes, Lemma 30.5.2. In particular, if we set $\mathcal{A}_ i = \bigoplus _{d \geq 0} f_{i, *}\mathcal{L}_ i^{\otimes d}$ then we have $\mathcal{A}_ i = g_ i^*\mathcal{A}$. It follows that the natural maps $\psi _ d : f^*\mathcal{A}_ d \to \mathcal{L}^{\otimes d}$ of $\mathcal{O}_ X$ pullback to give the natural maps $\psi _{i, d} : f_ i^*(\mathcal{A}_ i)_ d \to \mathcal{L}_ i^{\otimes d}$ of $\mathcal{O}_{X_ i}$-modules. Since $\mathcal{L}_ i$ is ample on $X_ i/S_ i$ we see that for any point $x_ i \in X_ i$, there exists a $d \geq 1$ such that $f_ i^*(\mathcal{A}_ i)_ d \to \mathcal{L}_ i^{\otimes d}$ is surjective on stalks at $x_ i$. This follows either directly from the definition of a relatively ample module or from Morphisms, Lemma 29.37.4. If $x \in X$, then we can choose an $i$ and an $x_ i \in X_ i$ mapping to $x$. Since $\mathcal{O}_{X, x} \to \mathcal{O}_{X_ i, x_ i}$ is flat hence faithfully flat, we conclude that for every $x \in X$ there exists a $d \geq 1$ such that $f^*\mathcal{A}_ d \to \mathcal{L}^{\otimes d}$ is surjective on stalks at $x$. This implies that the open subset $U(\psi ) \subset X$ of Constructions, Lemma 27.19.1 corresponding to the map $\psi : f^*\mathcal{A} \to \bigoplus _{d \geq 0} \mathcal{L}^{\otimes d}$ of graded $\mathcal{O}_ X$-algebras is equal to $X$. Consider the corresponding morphism

It is clear from the above that the base change of $r_{\mathcal{L}, \psi }$ to $S_ i$ is the morphism $r_{\mathcal{L}_ i, \psi _ i}$ which is an open immersion by Morphisms, Lemma 29.37.4. Hence $r_{\mathcal{L}, \psi }$ is an open immersion by Lemma 35.23.16 and we conclude $\mathcal{L}$ is ample on $X/S$ by Morphisms, Lemma 29.37.4. $\square$

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