The Stacks project

Proof. Let us first prove this in the quasi-compact case, since it is perhaps the most interesting case. In this case we produce inductively a sequence of blowups

and finite sets of effective Cartier divisors $\{ D_{n, i}\} _{i \in I_ n}$. At each stage these will have the property that any triple intersection $D_{n, i} \cap D_{n, j} \cap D_{n, k}$ is empty. Moreover, for each $n \geq 0$ we will have $I_{n + 1} = I_ n \amalg P(I_ n)$ where $P(I_ n)$ denotes the set of pairs of elements of $I_ n$. Finally, we will have

We conclude that for each $n \geq 0$ we have $(b_0 \circ \ldots \circ b_ n)^{-1}(D_ i)$ is a nonnegative integer combination of the divisors $D_{n + 1, j}$, $j \in I_{n + 1}$.

To start the induction we set $X_0 = X$ and $I_0 = I$ and $D_{0, i} = D_ i$.

Given $(X_ n, \{ D_{n, i}\} _{i \in I_ n})$ let $X_{n + 1}$ be the blowup of $X_ n$ in the closed subscheme $Z_ n = \bigcup _{\{ i, i'\} \in P(I_ n)} D_{n, i} \cap D_{n, i'}$. Note that the closed subschemes $D_{n, i} \cap D_{n, i'}$ are pairwise disjoint by our assumption on triple intersections. In other words we may write $Z_ n = \coprod _{\{ i, i'\} \in P(I_ n)} D_{n, i} \cap D_{n, i'}$. Moreover, in a Zariski neighbourhood of $D_{n, i} \cap D_{n, i'}$ the morphism $b_ n$ is equal to the blowup of the scheme $X_ n$ in the closed subscheme $D_{n, i} \cap D_{n, i'}$, and the results of Lemma 115.23.8 apply. Hence setting $D_{n + 1, \{ i, i'\} } = b_ n^{-1}(D_ i \cap D_{i'})$ we get an effective Cartier divisor. The Cartier divisors $D_{n + 1, \{ i, i'\} }$ are pairwise disjoint. Clearly we have $b_ n^{-1}(D_{n, i}) \supset D_{n + 1, \{ i, i'\} }$ for every $i' \in I_ n$, $i' \not= i$. Hence, applying Divisors, Lemma 31.13.8 we see that indeed $b^{-1}(D_{n, i}) = D_{n + 1, i} + \sum \nolimits _{i' \in I_ n, i' \not= i} D_{n + 1, \{ i, i'\} }$ for some effective Cartier divisor $D_{n + 1, i}$ on $X_{n + 1}$. In a neighbourhood of $D_{n + 1, \{ i, i'\} }$ these divisors $D_{n + 1, i}$ play the role of the primed divisors of Lemma 115.23.8. In particular we conclude that $D_{n + 1, i} \cap D_{n + 1, i'} = \emptyset $ if $i \not= i'$, $i, i' \in I_ n$ by part (6) of Lemma 115.23.8. This already implies that triple intersections of the divisors $D_{n + 1, i}$ are zero.

OK, and at this point we can use the quasi-compactness of $X$ to conclude that the invariant

is finite, since after all each $D_ i$ has at most finitely many irreducible components. We claim that for some $n$ the invariant $\epsilon (X_ n, \{ D_{n, i}\} _{i \in I_ n})$ is zero. Namely, if not then by Lemma 115.23.8 we have a strictly decreasing sequence

of positive integers which is a contradiction. Take $n$ with invariant $\epsilon (X_ n, \{ D_{n, i}\} _{i \in I_ n})$ equal to zero. This means that there is no integral closed subscheme $Z \subset X_ n$ and no pair of indices $i, i' \in I_ n$ such that $\epsilon _ Z(D_{n, i}, D_{n, i'}) > 0$. In other words, $\dim _\delta (D_{n, i}, D_{n, i'}) \leq d - 2$ for all pairs $\{ i, i'\} \in P(I_ n)$ as desired.

Next, we come to the general case where we no longer assume that the scheme $X$ is quasi-compact. The problem with the idea from the first part of the proof is that we may get and infinite sequence of blowups with centers dominating a fixed point of $X$. In order to avoid this we cut out suitable closed subsets of codimension $\geq 3$ at each stage. Namely, we will construct by induction a sequence of morphisms having the following shape

Each of the morphisms $j_ n : U_ n \to X_ n$ will be an open immersion. Each of the morphisms $b_ n : X_{n + 1} \to U_ n$ will be a proper birational morphism of integral schemes. As in the quasi-compact case we will have effective Cartier divisors $\{ D_{n, i}\} _{i \in I_ n}$ on $X_ n$. At each stage these will have the property that any triple intersection $D_{n, i} \cap D_{n, j} \cap D_{n, k}$ is empty. Moreover, for each $n \geq 0$ we will have $I_{n + 1} = I_ n \amalg P(I_ n)$ where $P(I_ n)$ denotes the set of pairs of elements of $I_ n$. Finally, we will arrange it so that

We start the induction by setting $X_0 = X$, $I_0 = I$ and $D_{0, i} = D_ i$.

Given $(X_ n, \{ D_{n, i}\} )$ we construct the open subscheme $U_ n$ as follows. For each pair $\{ i, i'\} \in P(I_ n)$ consider the closed subscheme $D_{n, i} \cap D_{n, i'}$. This has “good” irreducible components which have $\delta $-dimension $d - 2$ and “bad” irreducible components which have $\delta $-dimension $d - 1$. Let us set

and similarly

Then $D_{n, i} \cap D_{n, i'} = \text{Bad}(i, i') \cup \text{Good}(i, i')$ and moreover we have $\dim _\delta (\text{Bad}(i, i') \cap \text{Good}(i, i')) \leq d - 3$. Here is our choice of $U_ n$:

By our condition on triple intersections of the divisors $D_{n, i}$ we see that the union is actually a disjoint union. Moreover, we see that (as a scheme)

where $Z_{n, i, i'}$ is $\delta $-equidimensional of dimension $d - 1$ and $G_{n, i, i'}$ is $\delta $-equidimensional of dimension $d - 2$. (So topologically $Z_{n, i, i'}$ is the union of the bad components but throw out intersections with good components.) Finally we set

and we let $b_ n : X_{n + 1} \to X_ n$ be the blowup in $Z_ n$. Note that Lemma 115.23.8 applies to the morphism $b_ n : X_{n + 1} \to X_ n$ locally around each of the loci $D_{n, i}|_{U_ n} \cap D_{n, i'}|_{U_ n}$. Hence, exactly as in the first part of the proof we obtain effective Cartier divisors $D_{n + 1, \{ i, i'\} }$ for $\{ i, i'\} \in P(I_ n)$ and effective Cartier divisors $D_{n + 1, i}$ for $i \in I_ n$ such that $b_ n^{-1}(D_{n, i}|_{U_ n}) = D_{n + 1, i} + \sum \nolimits _{i' \in I_ n, i' \not= i} D_{n + 1, \{ i, i'\} }$. For each $n$ denote $\pi _ n : X_ n \to X$ the morphism obtained as the composition $j_0 \circ \ldots \circ j_{n - 1} \circ b_{n - 1}$.

Claim: given any quasi-compact open $V \subset X$ for all sufficiently large $n$ the maps

are all isomorphisms. Namely, if the map $\pi _ n^{-1}(V) \leftarrow \pi _{n + 1}^{-1}(V)$ is not an isomorphism, then $Z_{n, i, i'} \cap \pi _ n^{-1}(V) \not= \emptyset $ for some $\{ i, i'\} \in P(I_ n)$. Hence there exists an irreducible component $W \subset D_{n, i} \cap D_{n, i'}$ with $\dim _\delta (W) = d - 1$. In particular we see that $\epsilon _ W(D_{n, i}, D_{n, i'}) > 0$. Applying Lemma 115.23.8 repeatedly we see that

with $\epsilon (V, \{ D_ i|_ V\} )$ as in (115.23.11.1). Since $V$ is quasi-compact, we have $\epsilon (V, \{ D_ i|_ V\} ) < \infty $ and taking $n > \epsilon (V, \{ D_ i|_ V\} )$ we see the result.

Note that by construction the difference $X_ n \setminus U_ n$ has $\dim _\delta (X_ n \setminus U_ n) \leq d - 3$. Let $T_ n = \pi _ n(X_ n \setminus U_ n)$ be its image in $X$. Traversing in the diagram of maps above using each $b_ n$ is closed it follows that $T_0 \cup \ldots \cup T_ n$ is a closed subset of $X$ for each $n$. Any $t \in T_ n$ satisfies $\delta (t) \leq d - 3$ by construction. Hence $\overline{T_ n} \subset X$ is a closed subset with $\dim _\delta (T_ n) \leq d - 3$. By the claim above we see that for any quasi-compact open $V \subset X$ we have $T_ n \cap V \not= \emptyset $ for at most finitely many $n$. Hence $\{ \overline{T_ n}\} _{n \geq 0}$ is a locally finite collection of closed subsets, and we may set $U = X \setminus \bigcup \overline{T_ n}$. This will be $U$ as in the lemma.

Note that $U_ n \cap \pi _ n^{-1}(U) = \pi _ n^{-1}(U)$ by construction of $U$. Hence all the morphisms

are proper. Moreover, by the claim they eventually become isomorphisms over each quasi-compact open of $X$. Hence we can define

The induced morphism $b : U' \to U$ is proper since this is local on $U$, and over each compact open the limit stabilizes. Similarly we set $J = \bigcup _{n \geq 0} I_ n$ using the inclusions $I_ n \to I_{n + 1}$ from the construction. For $j \in J$ choose an $n_0$ such that $j$ corresponds to $i \in I_{n_0}$ and define $D'_ j = \mathop{\mathrm{lim}}\nolimits _{n \geq n_0} D_{n, i}$. Again this makes sense as locally over $X$ the morphisms stabilize. The other claims of the lemma are verified as in the case of a quasi-compact $X$. $\square$