The Stacks project

Proof. In the proof below we will use Lemmas 37.75.4 and 37.75.3 without further mention. We will also use elementary properties of constructible subsets of schemes and topological spaces, see Topology, Section 5.15 and Properties, Section 28.2. Using this the reader sees question is local on $X$ and $Y$; details omitted. Hence we may assume $X$ and $Y$ are affine. If we can find a surjective morphism $Y' \to Y$ of finite presentation such that the level sets of $w$ pull back to locally constructible subsets of $X' = Y' \times _ Y X$, then we conclude by Morphisms, Theorem 29.22.3.

Assume $X$ and $Y$ affine. We may choose an immersion $X \to T$ where $T \to Y$ is finite, see Lemma 37.43.3. By Morphisms, Lemma 29.48.6 after replacing $Y$ by $Y'$ surjective finite locally free over $Y$, replacing $X$ by $Y' \times _ Y X$ and $T$ by a scheme finite locally free over $Y'$ containing $Y' \times _ Y T$ as a closed subscheme, we may assume $T$ is finite locally free over $Y$, contains closed subschemes $T_ i$ mapping isomorphically to $Y$ such that $T = \bigcup _{i = 1, \ldots , n} T_ i$ (set theoretically). Since $T_ i \subset T$ is a constructible closed subset (as the image of a finitely presented morphism $Y \to T$ of schemes), we see that for $I \subset \{ 1, \ldots , n\} $ the intersection $\bigcap _{i \in I} T_ i$ is a constructible closed subset of $T$ and hence maps to a constructible closed subset of $Y$.

For a disjoint union decomposition $\{ 1, \ldots , n\} = I_1 \amalg \ldots \amalg I_ r$ with nonempty parts consider the subset $Y_{I_1, \ldots , I_ r} \subset Y$ consisting of points $y \in Y$ such that $T_ y = \{ x_1, \ldots , x_ r\} $ consists of exactly $r$ points with $x_ j \in T_ i \Leftrightarrow i \in I_ j$. By our remarks above this is a constructible partition of $Y$. There exists an affine scheme $Y'$ of finite presentation over $Y$ such that the image of $Y' \to Y$ is exactly $Y_{I_1, \ldots , I_ r}$, see Algebra, Lemma 10.29.4. Hence we may assume that $Y = Y_{I_1, \ldots , I_ r}$ for some disjoint union decomposition $\{ 1, \ldots , n\} = I_1 \amalg \ldots \amalg I_ r$. In this case $T = T(1) \amalg \ldots \amalg T(r)$ with $T(j) = \bigcap _{i \in I_ j} T_ i$ is a decomposition of $T$ into disjoint closed (and hence open) subsets. Intersecting with the locally closed subscheme $X$ we obtain an analogous decomposition $X = X(1) \amalg \ldots \amalg X(r)$ into open and closed parts. The morphism $X(j) \to Y$ an immersion. Since $w$ is a weighting, it follows that $w|_{X(j)}$ is locally constant¹ and we conclude. $\square$

Proof. By Lemma 37.75.1 formation of the function $\int _ f w$ commutes with arbitrary base change and by Lemma 37.75.3 after base change we still have a weighthing. This means that if we can find $Y' \to Y$ surjective and of finite presentation, then it suffices to prove the result after base change to $Y'$, see Morphisms, Theorem 29.22.3.

The question is local on $Y$ hence we may assume $Y$ is affine. Then $X$ is quasi-compact and quasi-separated (as $f$ is of finite presentation). Suppose that $X = U \cup V$ are quasi-compact open. Then we have

Thus if we know the result for $w|_ U$, $w|_ V$, $w|_{U \cap V}$ then we know the result for $w$. By the induction principle (Cohomology of Schemes, Lemma 30.4.1) it suffices to prove the lemma when $X$ is affine.

Assume $X$ and $Y$ are affine. We may choose an open immersion $X \to T$ where $T \to Y$ is finite, see Lemma 37.43.3. Because we may still base change with a suitable $Y' \to Y$ we can use Morphisms, Lemma 29.48.6 to reduce to the case where all residue field extensions induced by the morphism $T \to Y$ (and a foriori induced by $X \to Y$) are trivial. In this situation $\int _ f w$ is just taking the sums of the values of $w$ in fibres. The level sets of $w$ are locally constructible in $X$ (Lemma 37.76.1). The function $w$ only takes a finite number of values by Properties, Lemma 28.2.7. Hence we conclude by Morphisms, Theorem 29.22.3 and some elementary arguments on sums of integers. $\square$

Proof. Let $x \in X$ with image $y \in Y$. Choose an étale neighbourhood $(V, v) \to (Y, y)$ and an open $U \subset X_ V$ such that $\pi : U \to V$ is finite and there is a unique point $u \in U$ mapping to $v$ with $\kappa (u)/\kappa (v)$ purely inseparable. See Lemma 37.41.3. Then $(\int _\pi w|_ U)(v) = w(u)$. It follows from Definition 37.75.2 that after replacing $V$ by a neighbourhood of $v$ we we have $w|_ U(u') \leq w|_ U(u) = w(x)$ for all $u' \in U$. Namely, $w|_ U(u')$ occurs as a summand in the expression for $(\int _\pi w|_ U)(\pi (u'))$. This proves the lemma because the étale morphism $U \to X$ is open. $\square$

Proof. Let $y \in Y$. Let $x_1, \ldots , x_ r \in X$ be the points lying over $y$. Apply Lemma 37.41.5 to get an étale neighbourhood $(U, u) \to (Y, y)$ and a decomposition

as in locus citatus. Observe that $(\int _ f w)(y) = \sum w(v_{i, j})$ where $w(v_{i, j}) = w(x_ i)$. Since $\int _{V_{i, j} \to U} w|_{V_{i, j}}$ is locally constant by definition, we may after shrinking $U$ assume these functions are constant with value $w(v_{i, j})$. We conclude that

This is $\geq (\int _ f w)(y)$ and we conclude because $U \to Y$ is open and formation of the integral commutes with base change (Lemma 37.75.1). $\square$

Proof. It follows from Lemma 37.76.1 and Properties, Lemma 28.2.7 that $w$ only takes a finite number of values on $X$. It follows from Morphisms, Lemma 29.57.9 that $X \to Y$ has bounded geometric fibres. This shows that $\int _ f w$ is bounded. $\square$

Proof. Observe that $Y_ d$ is open by Lemma 37.76.4. Let $y \in Y_ d$. Say $x_1, \ldots , x_ n$ are the points of $X$ lying over $y$. Apply Lemma 37.41.5 to get an étale neighbourhood $(U, u) \to (Y, y)$ and a decomposition

as in locus citatus. Observe that $d = \sum w(v_{i, j})$ where $w(v_{i, j}) = w(x_ i)$. Since $\int _{V_{i, j} \to U} w|_{V_{i, j}}$ is locally constant by definition, we may after shrinking $U$ assume these functions are constant with value $w(v_{i, j})$. We conclude that

This is $\geq (\int _ f w)(y) = d$ and we conclude that $W$ must be the emptyset. Thus $U \times _ Y X \to U$ is finite. By Descent, Lemma 35.23.23 this implies that $X \to Y$ is finite over the image of the open morphism $U \to Y$. In other words, we see that $f$ is finite over an open neighbourhood of $y$ as desired. $\square$

Proof. Choose $b_1, \ldots , b_ n \in B$ generating $B$ as an $A$-module. For each $i$ choose a monic $P_ i \in A[x]$ such that $P_ i(b_ i) = 0$ in $B$, see Algebra, Lemma 10.36.3. Thus $B$ is a quotient of the finite free $A$-algebra $B' = A[x_1, \ldots , x_ n]/(P_1(x_1), \ldots , P_ n(x_ n))$. Let $J \subset B'$ be the kernel of the surjection $B' \to B$. Then $J =(f_1, \ldots , f_ m)$ is finitely generated as $B$ is a finitely generated $A$-algebra, see Algebra, Lemma 10.6.2. Choose an $A$-basis $b'_1, \ldots , b'_ N$ of $B'$. Consider the algebra

where $I$ is the ideal generated by the coefficients in $A[z_1, \ldots , z_ n, y_1, \ldots , y_ m]$ of the basis elements $b'_1, \ldots , b'_ N$ of the expression

in $B'[z_1, \ldots , z_ N, y_1, \ldots , y_ m]$. By construction the element $\sum z_ j b'_ j$ maps to an idempotent $e_{univ}$ in the algebra $B \otimes _ A A_{univ}$. Moreover, if $e \in B \otimes _ A A'$ is an idempotent, then we can lift $e$ to an element of the form $\sum b'_ j \otimes a'_ j$ in $B' \otimes _ A A'$ and we can find $a''_ k \in A'$ such that

is zero in $B' \otimes _ A A'$. Hence we get an $A$-algebra map $A_{univ} \to A$ sending $z_ j$ to $a'_ j$ and $y_ k$ to $a''_ k$ mapping $e_{univ}$ to $e$. This finishes the proof. $\square$

Proof. Recall that a finite type morphism is quasi-finite if and only if it has relative dimension $0$, see Morphisms, Lemma 29.29.5. By Lemma 37.34.9 applied with $d = 0$ we reduce to the case where $X$ and $Y$ are Noetherian. We may choose an open immersion $X \to X'$ such that $X' \to Y$ is finite, see Algebra, Lemma 10.123.14. Note that if we have $Y' \to Y$ and $U'$ as in (2), then

is open immersion between schemes finite over $Y'$ and hence is closed as well. We conclude that $U'$ corresponds to an idempotent in

whose corresponding open and closed subset is contained in the open $Y' \times _ Y X$. Let $Y'_{univ} \to Y$ and idempotent

be the pair constructed in Lemma 37.76.7 for the ring map $\Gamma (Y, \mathcal{O}_ Y) \to \Gamma (X', \mathcal{O}_{X'})$ (here we use that $Y$ is Noetherian to see that $X'$ is of finite presentation over $Y$). Let $U'_{univ} \subset Y'_{univ} \times _ Y X'$ be the corresponding open and closed subscheme. Then we see that

is a closed subset of $U'_{univ}$ and hence has closed image $T \subset Y'_{univ}$. If we set $Y_{univ} = Y'_{univ} \setminus T$ and $U_{univ}$ the restriction of $U'_{univ}$ to $Y_{univ} \times _ Y X$, then we see that the lemma is true. $\square$

Proof. By Lemma 37.76.1 the level sets of $w$ are constructible subsets $E_ k$ of $X$. This implies the function $w$ only takes a finite number of values by Properties, Lemma 28.2.7. Thus there exists an $i$ such that $E_ k$ descends to a construcible subset $E_{i, k}$ in $X_ i$ for all $k$; moreover, we may assume $X_ i = \coprod E_{i, k}$. This follows as the topological space of $X$ is the limit in the category of topological spaces of the spectral spaces $X_ i$ along a directed system with spectral transition maps. See Limits, Section 32.4 and Topology, Section 5.24. We define $w_ i : X_ i \to \mathbf{Z}$ such that its level sets are the constructible sets $E_{i, k}$.

Choose $Y_{i, univ} \to Y_ i$ and $U_{i, univ} \subset Y_{i, univ} \times _{Y_ i} X_ i$ as in Lemma 37.76.8. By the universal property of the construction, in order to show that $w_ i$ is a weighting, it would suffice to show that

is locally constant on $Y_{i, univ}$. By Lemma 37.76.2 this function has constructible level sets but it may not (yet) be locally constant. Set $Y_{univ} = Y_{i, univ} \times _{Y_ i} Y$ and let $U_{univ} \subset Y_{univ} \times _ Y X$ be the inverse image of $U_{i, univ}$. Then, since the pullback of $w$ to $Y_{univ} \times _ Y X$ is a weighting for $Y_{univ} \times _ Y X \to Y_{univ}$ (Lemma 37.75.3) we do have that

is locally constant on $Y_{univ}$. Thus the level sets of $\tau $ are open and closed. Finally, we have $Y_{univ} = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} Y_{i', univ}$ and the level sets of $\tau $ are the inverse limits of the level sets of $\tau _{i'}$ (similarly defined). Hence the references above imply that for sufficiently large $i'$ the level sets of $\tau _{i'}$ are open as well. For such an index $i'$ we conclude that $w_{i'}$ is a weighting of $f_{i'}$ as desired. $\square$