The Stacks project

Proof. Let $x_1, \ldots , x_ n \in X$. Choose a quasi-compact open $U \subset X$ with $x_ i \in U$. Then $U \to Y$ is quasi-affine by Lemma 37.43.2. Hence there exists an affine open $V \subset U$ containing $x_1, \ldots , x_ n$ by Properties, Lemma 28.29.5. $\square$

Proof. If $U_ i \subset X$, $i \in I$ is a collection of opens such that the restrictions $f|_{f^{-1}(U_ i)} : f^{-1}(U_ i) \to U_ i$ are finite, then with $U = \bigcup U_ i$ the restriction $f|_{f^{-1}(U)} : f^{-1}(U) \to U$ is finite, see Morphisms, Lemma 29.44.3. Thus the problem is local on $X$ and we may assume that $X$ is affine.

Assume $X$ is affine. Write $Y = \bigcup _{j = 1, \ldots , m} V_ j$ with $V_ j$ affine. This is possible since $f$ is quasi-finite and hence in particular quasi-compact. Each $V_ j \to X$ is quasi-finite and separated. Let $\eta \in X$ be a generic point of an irreducible component of $X$. We see from Morphisms, Lemmas 29.20.10 and 29.51.1 that there exists an open neighbourhood $\eta \in U_\eta $ such that $f^{-1}(U_\eta ) \cap V_ j \to U_\eta $ is finite. We may choose $U_\eta $ such that it works for each $j = 1, \ldots , m$. Note that the collection of generic points of $X$ is dense in $X$. Thus we see there exists a dense open $W = \bigcup _\eta U_\eta $ such that each $f^{-1}(W) \cap V_ j \to W$ is finite. It suffices to show that there exists a dense open $U \subset W$ such that $f|_{f^{-1}(U)} : f^{-1}(U) \to U$ is finite. Thus we may replace $X$ by an affine open subscheme of $W$ and assume that each $V_ j \to X$ is finite.

Assume $X$ is affine, $Y = \bigcup _{j = 1, \ldots , m} V_ j$ with $V_ j$ affine, and the restrictions $f|_{V_ j} : V_ j \to X$ are finite. Set

This is a nowhere dense closed subset of $V_ j$ because it is the boundary of the open subset $V_ i \cap V_ j$ in $V_ j$. By Morphisms, Lemma 29.48.7 the image $f(\Delta _{ij})$ is a nowhere dense closed subset of $X$. By Topology, Lemma 5.21.2 the union $T = \bigcup f(\Delta _{ij})$ is a nowhere dense closed subset of $X$. Thus $U = X \setminus T$ is a dense open subset of $X$. We claim that $f|_{f^{-1}(U)} : f^{-1}(U) \to U$ is finite. To see this let $U' \subset U$ be an affine open. Set $Y' = f^{-1}(U') = U' \times _ X Y$, $V_ j' = Y' \cap V_ j = U' \times _ X V_ j$. Consider the restriction

of $f$. This morphism now has the property that $Y' = \bigcup _{j = 1, \ldots , m} V'_ j$ is an affine open covering, each $V'_ j \to U'$ is finite, and $V_ i' \cap V_ j'$ is (open and) closed both in $V'_ i$ and $V'_ j$. Hence $V_ i' \cap V_ j'$ is affine, and the map

is surjective. This implies that $Y'$ is separated, see Schemes, Lemma 26.21.7. Finally, consider the commutative diagram

The south-east arrow is finite, hence proper, the horizontal arrow is surjective, and the south-west arrow is separated. Hence by Morphisms, Lemma 29.41.9 we conclude that $Y' \to U'$ is proper. Since it is also quasi-finite, we see that it is finite by Lemma 37.44.1, and we win. $\square$

Proof. Let $n$ be an integer bounding the degree of the fibres of $X \to S$. By Morphisms, Lemma 29.57.5 we see that any base change has degrees of fibres bounded by $n$ also. In particular, all the integers $r$ that occur in the statement of the lemma will be $\leq n$. We will prove the lemma by induction on $n$. The base case is $n = 0$ which is obvious.

We claim the set of points $s \in S$ with $\deg _{\kappa (s)}(X_ s) = n$ is an open subset $S_ n \subset S$ and that $X \times _ S S_ n \to S_ n$ is finite locally free of degree $n$. Namely, suppose that $s \in S$ is such a point. Choose an elementary étale morphism $(U, u) \to (S, s)$ and a decomposition $U \times _ S X = W \amalg V$ as in Lemma 37.41.6. Since $V \to U$ is finite, flat, and locally of finite presentation, we see that $V \to U$ is finite locally free, see Morphisms, Lemma 29.48.2. After shrinking $U$ to a smaller neighbourhood of $u$ we may assume $V \to U$ is finite locally free of some degree $d$, see Morphisms, Lemma 29.48.5. As $u \mapsto s$ and $W_ u = \emptyset $ we see that $d = n$. Since $n$ is the maximum degree of a fibre we see that $W = \emptyset $! Thus $U \times _ S X \to U$ is finite locally free of degree $n$. By Descent, Lemma 35.23.30 we conclude that $X \to S$ is finite locally free of degree $n$ over $\mathop{\mathrm{Im}}(U \to S)$ which is an open neighbourhood of $s$ (Morphisms, Lemma 29.36.13). This proves the claim.

Let $S' = S \setminus S_ n$ endowed with the reduced induced scheme structure and set $X' = X \times _ S S'$. Note that the degrees of fibres of $X' \to S'$ are universally bounded by $n - 1$. By induction we find a stratification $S' = S_0 \amalg \ldots \amalg S_{n - 1}$ adapted to the morphism $X' \to S'$. We claim that $S = \coprod _{r = 0, \ldots , n} S_ r$ works for the morphism $X \to S$. Let $g : T \to S$ be a morphism of schemes and assume that $X \times _ S T \to T$ is finite locally free of degree $r$. As remarked above this implies that $r \leq n$. If $r = n$, then it is clear that $T \to S$ factors through $S_ n$. If $r < n$, then $g(T) \subset S' = S \setminus S_ d$ (set theoretically) hence $T_{red} \to S$ factors through $S'$, see Schemes, Lemma 26.12.7. Note that $X \times _ S T_{red} \to T_{red}$ is also finite locally free of degree $r$ as a base change. By the universal property of the stratification $S' = \coprod _{r = 0, \ldots , n - 1} S_ r$ we see that $g(T) = g(T_{red})$ is contained in $S_ r$. Conversely, suppose that we have $g : T \to S$ such that $g(T) \subset S_ r$ (set theoretically). If $r = n$, then $g$ factors through $S_ n$ and it is clear that $X \times _ S T \to T$ is finite locally free of degree $n$ as a base change. If $r < n$, then $X \times _ S T \to T$ is a morphism which is separated, flat, and locally of finite presentation, such that the restriction to $T_{red}$ is finite locally free of degree $r$. Since $T_{red} \to T$ is a universal homeomorphism, we conclude that $X \times _ S T_{red} \to X \times _ S T$ is a universal homeomorphism too and hence $X \times _ S T \to T$ is universally closed (as this is true for the finite morphism $X \times _ S T_{red} \to T_{red}$). It follows that $X \times _ S T \to T$ is finite, for example by Lemma 37.44.1. Then we can use Morphisms, Lemma 29.48.2 to see that $X \times _ S T \to T$ is finite locally free. Finally, the degree is $r$ as all the fibres have degree $r$. $\square$

Proof. The question is local on $S$, hence we may assume that $S$ is affine. By Morphisms, Lemma 29.57.9 the fibres of $f$ are universally bounded in this case. Hence the existence of the stratification follows from Lemma 37.45.3.

We will show that $U_ r = S \setminus Z_ r \to S$ is quasi-compact for each $r \geq 0$. This will prove the final statement by elementary topology. Since a composition of quasi-compact maps is quasi-compact it suffices to prove that $U_ r \to U_{r - 1}$ is quasi-compact. Choose an affine open $W \subset U_{r - 1}$. Write $W = \mathop{\mathrm{Spec}}(A)$. Then $Z_ r \cap W = V(I)$ for some ideal $I \subset A$ and $X \times _ S \mathop{\mathrm{Spec}}(A/I) \to \mathop{\mathrm{Spec}}(A/I)$ is finite locally free of degree $r$. Note that $A/I = \mathop{\mathrm{colim}}\nolimits A/I_ i$ where $I_ i \subset I$ runs through the finitely generated ideals. By Limits, Lemma 32.8.8 we see that $X \times _ S \mathop{\mathrm{Spec}}(A/I_ i) \to \mathop{\mathrm{Spec}}(A/I_ i)$ is finite locally free of degree $r$ for some $i$. (This uses that $X \to S$ is of finite presentation, as it is locally of finite presentation, separated, and quasi-compact.) Hence $\mathop{\mathrm{Spec}}(A/I_ i) \to \mathop{\mathrm{Spec}}(A) = W$ factors (set theoretically) through $Z_ r \cap W$. It follows that $Z_ r \cap W = V(I_ i)$ is the zero set of a finite subset of elements of $A$. This means that $W \setminus Z_ r$ is a finite union of standard opens, hence quasi-compact, as desired. $\square$

Proof. The statement simply means that the collection of points where the degree of the fibre is $\geq d$ is open. Thus we can work locally on $S$ and assume $S$ is affine. In this case, for every $W \subset X$ quasi-compact open, the set of points $U_ d(W)$ where the fibres of $W \to S$ have degree $\geq d$ is open by Lemma 37.45.4. Since $U_ d = \bigcup _ W U_ d(W)$ the result follows. $\square$

Proof. We may assume $X$ and $S$ affine. We obtain a filtration $\emptyset = Z_{-1} \subset Z_0 \subset Z_1 \subset Z_2 \subset \ldots \subset Z_ n = S$ as in Lemmas 37.45.3 and 37.45.4. Let $T \subset X$ be the scheme theoretic support of the finite $\mathcal{O}_ X$-module $\mathop{\mathrm{Im}}(g : \mathcal{O}_ X \to \mathcal{O}_ X)$. Note that $T$ is the support of $g$ as a section of $\mathcal{O}_ X$ (Modules, Definition 17.5.1) and for any open $V \subset X$ we have $g|_ V \not= 0$ if and only if $V \cap T \not= \emptyset $. Let $r$ be the smallest integer such that $f(T) \subset Z_ r$ set theoretically. Let $\xi \in T$ be a generic point of an irreducible component of $T$ such that $f(\xi ) \not\in Z_{r - 1}$ (and hence $f(\xi ) \in Z_ r$). We may replace $S$ by an affine neighbourhood of $f(\xi )$ contained in $S \setminus Z_{r - 1}$. Write $S = \mathop{\mathrm{Spec}}(A)$ and let $I = (a_1, \ldots , a_ m) \subset A$ be a finitely generated ideal such that $V(I) = Z_ r$ (set theoretically, see Algebra, Lemma 10.29.1). Since the support of $g$ is contained in $f^{-1}V(I)$ by our choice of $r$ we see that there exists an integer $N$ such that $a_ j^ N g = 0$ for $j = 1, \ldots , m$. Replacing $a_ j$ by $a_ j^ r$ we may assume that $Ig = 0$. For any $A$-module $M$ write $M[I]$ for the $I$-torsion of $M$, i.e., $M[I] = \{ m \in M \mid Im = 0\} $. Write $X = \mathop{\mathrm{Spec}}(B)$, so $g \in B[I]$. Since $A \to B$ is flat we see that

By our choice of $Z_ r$, the $A/I$-module $B/IB$ is finite locally free of rank $r$. Hence after replacing $S$ by a smaller affine open neighbourhood of $f(\xi )$ we may assume that $B/IB \cong (A/IA)^{\oplus r}$ as $A/I$-modules. Choose a map $\psi : A^{\oplus r} \to B$ which reduces modulo $I$ to the isomorphism of the previous sentence. Then we see that the induced map

is an isomorphism. The lemma follows by taking $\mathcal{F}$ the quasi-coherent sheaf associated to the $A$-module $A[I]$ and the map $\mathcal{F}^{\oplus r} \to \pi _*\mathcal{O}_ V$ the one corresponding to $A[I]^{\oplus r} \subset A^{\oplus r} \to B$. $\square$

Proof. Since $X$ is quasi-compact, there exist finitely many affine opens $U_ i \subset U$ such that $U' = \coprod U_ i \to X$ is surjective. After replacing $U$ by $U'$, we see that we may assume $U$ is affine. In particular $U \to X$ is separated (Schemes, Lemma 26.21.15). Then there exists an integer $d$ bounding the degree of the geometric fibres of $U \to X$ (see Morphisms, Lemma 29.57.9). We will prove the lemma by induction on $d$ for all quasi-compact and separated schemes $U$ mapping surjective and étale onto $X$. If $d = 1$, then $U = X$ and the result holds with $Y = U$. Assume $d > 1$.

We apply Lemma 37.43.2 and we obtain a factorization

with $\pi $ integral and $j$ a quasi-compact open immersion. We may and do assume that $j(U)$ is scheme theoretically dense in $Y$. Note that

where the first summand is the image of $U \to U \times _ X Y$ (which is closed by Schemes, Lemma 26.21.10 and open because it is étale as a morphism between schemes étale over $Y$) and the second summand is the (open and closed) complement. The image $V \subset Y$ of $W$ is an open subscheme containing $Y \setminus U$.

The étale morphism $W \to Y$ has geometric fibres of cardinality $< d$. Namely, this is clear for geometric points of $U \subset Y$ by inspection. Since $U \subset Y$ is dense, it holds for all geometric points of $Y$ for example by Lemma 37.45.3 (the degree of the fibres of a quasi-compact separated étale morphism does not go up under specialization). Thus we may apply the induction hypothesis to $W \to V$ and find a surjective integral morphism $Z \to V$ with $Z$ a scheme, which Zariski locally factors through $W$. Choose a factorization $Z \to Z' \to Y$ with $Z' \to Y$ integral and $Z \to Z'$ open immersion (Lemma 37.43.2). After replacing $Z'$ by the scheme theoretic closure of $Z$ in $Z'$ we may assume that $Z$ is scheme theoretically dense in $Z'$. After doing this we have $Z' \times _ Y V = Z$. Finally, let $T \subset Y$ be the induced reduced closed subscheme structure on $Y \setminus V$. Consider the morphism

This is a surjective integral morphism by construction. Since $T \subset U$ it is clear that the morphism $T \to X$ factors through $U$. On the other hand, let $z \in Z'$ be a point. If $z \not\in Z$, then $z$ maps to a point of $Y \setminus V \subset U$ and we find a neighbourhood of $z$ on which the morphism factors through $U$. If $z \in Z$, then we have a neighbourhood $\Omega \subset Z$ which factors through $W \subset U \times _ X Y$ and hence through $U$. This proves existence.

Assume we have found $Y \to X$ integral and surjective which Zariski locally factors through $U$. Choose a finite affine open covering $Y = \bigcup V_ j$ such that $V_ j \to X$ factors through $U$. We can write $Y = \mathop{\mathrm{lim}}\nolimits Y_ i$ with $Y_ i \to X$ finite and of finite presentation, see Limits, Lemma 32.7.3. For large enough $i$ we can find affine opens $V_{i, j} \subset Y_ i$ whose inverse image in $Y$ recovers $V_ j$, see Limits, Lemma 32.4.11. For even larger $i$ the morphisms $V_ j \to U$ over $X$ come from morphisms $V_{i, j} \to U$ over $X$, see Limits, Proposition 32.6.1. This finishes the proof. $\square$