The Stacks project

Proof. Immediate from Chow Homology, Lemma 42.26.3 and the fact that $D$ is the zero scheme of the canonical section $1_ D$ of $\mathcal{O}_ X(D)$. $\square$

Proof. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the quasi-coherent ideal sheaf of denominators of $s$, see Divisors, Definition 31.23.10. By Divisors, Lemma 31.34.6 we get (2), (3), and (4). By Divisors, Lemma 31.32.9 we get (1). By Divisors, Lemma 31.32.13 the morphism $\pi $ is projective. We still have to prove (5). By Chow Homology, Lemma 42.26.3 we have

Hence it suffices to show that $\text{div}_{\mathcal{L}'}(s') = [D]_{n - 1} - [E]_{n - 1}$. This follows from the equality $s' = 1_ D \otimes 1_ E^{-1}$ and additivity, see Divisors, Lemma 31.27.5. $\square$

Proof. Note that the quasi-coherent ideal sheaf $\mathcal{I} = \mathcal{I}_{D_1} + \mathcal{I}_{D_2}$ defines the scheme theoretic intersection $D_1 \cap D_2 \subset X$. Since $Z$ is a union of connected components of $D_1 \cap D_2$ we see that for every $z \in Z$ the kernel of $\mathcal{O}_{X, z} \to \mathcal{O}_{Z, z}$ is equal to $\mathcal{I}_ z$. Let $b : X' \to X$ be the blowup of $X$ in $Z$. (So Zariski locally around $Z$ it is the blowup of $X$ in $\mathcal{I}$.) Denote $E = b^{-1}(Z)$ the corresponding effective Cartier divisor, see Divisors, Lemma 31.32.4. Since $Z \subset D_1$ we have $E \subset f^{-1}(D_1)$ and hence $D_1 = D_1' + E$ for some effective Cartier divisor $D'_1 \subset X'$, see Divisors, Lemma 31.13.8. Similarly $D_2 = D_2' + E$. This takes care of assertions (1) – (5).

Note that if $W'$ is as in (7) (a) or (7) (b), then the image $W$ of $W'$ is contained in $D_1 \cap D_2$. If $W$ is not contained in $Z$, then $b$ is an isomorphism at the generic point of $W$ and we see that $\dim _\delta (W) = \dim _\delta (W') = n - 1$ which contradicts the assumption that $\dim _\delta (D_1 \cap D_2 \setminus Z) \leq n - 2$. Hence $W \subset Z$. This means that to prove (6) and (7) we may work locally around $Z$ on $X$.

Thus we may assume that $X = \mathop{\mathrm{Spec}}(A)$ with $A$ a Noetherian domain, and $D_1 = \mathop{\mathrm{Spec}}(A/a)$, $D_2 = \mathop{\mathrm{Spec}}(A/b)$ and $Z = D_1 \cap D_2$. Set $I = (a, b)$. Since $A$ is a domain and $a, b \not= 0$ we can cover the blowup by two patches, namely $U = \mathop{\mathrm{Spec}}(A[s]/(as - b))$ and $V = \mathop{\mathrm{Spec}}(A[t]/(bt -a))$. These patches are glued using the isomorphism $A[s, s^{-1}]/(as - b) \cong A[t, t^{-1}]/(bt - a)$ which maps $s$ to $t^{-1}$. The effective Cartier divisor $E$ is described by $\mathop{\mathrm{Spec}}(A[s]/(as - b, a)) \subset U$ and $\mathop{\mathrm{Spec}}(A[t]/(bt - a, b)) \subset V$. The closed subscheme $D'_1$ corresponds to $\mathop{\mathrm{Spec}}(A[t]/(bt - a, t)) \subset U$. The closed subscheme $D'_2$ corresponds to $\mathop{\mathrm{Spec}}(A[s]/(as -b, s)) \subset V$. Since “$ts = 1$” we see that $D'_1 \cap D'_2 = \emptyset $.

Suppose we have a prime $\mathfrak q \subset A[s]/(as - b)$ of height one with $s, a \in \mathfrak q$. Let $\mathfrak p \subset A$ be the corresponding prime of $A$. Observe that $a, b \in \mathfrak p$. By the dimension formula we see that $\dim (A_{\mathfrak p}) = 1$ as well. The final assertion to be shown is that

where $B = A[s]/(as - b)$. By Algebra, Lemma 10.124.1 we have $\text{ord}_{A_{\mathfrak p}}(x) \geq \text{ord}_{B_{\mathfrak q}}(x)$ for $x = a, b$. Since $\text{ord}_{B_{\mathfrak q}}(s) > 0$ we win by additivity of the $\text{ord}$ function and the fact that $as = b$. $\square$

Proof. Since we are proving an equality of cycles we may work locally on $X$. Hence this reduces to a finite sum, and by induction to a sum of two effective Cartier divisors $D = D_1 + D_2$. By Chow Homology, Lemma 42.24.2 we see that $D_1 = \text{div}_{\mathcal{O}_ X(D_1)}(1_{D_1})$ where $1_{D_1}$ denotes the canonical section of $\mathcal{O}_ X(D_1)$. Of course we have the same statement for $D_2$ and $D$. Since $1_ D = 1_{D_1} \otimes 1_{D_2}$ via the identification $\mathcal{O}_ X(D) = \mathcal{O}_ X(D_1) \otimes \mathcal{O}_ X(D_2)$ we win by Divisors, Lemma 31.27.5. $\square$

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 $X$. $\square$