The Stacks project

Lemma 115.23.8. Let $(S, \delta )$ be as in Chow Homology, Situation 42.7.1. Let $X$ be locally of finite type over $S$. Assume $X$ integral and $\dim _\delta (X) = n$. Let $D_1, D_2$ be two effective Cartier divisors in $X$. Let $Z$ be an open and closed subscheme of the scheme $D_1 \cap D_2$. Assume $\dim _\delta (D_1 \cap D_2 \setminus Z) \leq n - 2$. Then there exists a morphism $b : X' \to X$, and Cartier divisors $D_1', D_2', E$ on $X'$ with the following properties

  1. $X'$ is integral,

  2. $b$ is projective,

  3. $b$ is the blowup of $X$ in the closed subscheme $Z$,

  4. $E = b^{-1}(Z)$,

  5. $b^{-1}(D_1) = D'_1 + E$, and $b^{-1}D_2 = D_2' + E$,

  6. $\dim _\delta (D'_1 \cap D'_2) \leq n - 2$, and if $Z = D_1 \cap D_2$ then $D'_1 \cap D'_2 = \emptyset $,

  7. for every integral closed subscheme $W'$ with $\dim _\delta (W') = n - 1$ we have

    1. if $\epsilon _{W'}(D'_1, E) > 0$, then setting $W = b(W')$ we have $\dim _\delta (W) = n - 1$ and

      \[ \epsilon _{W'}(D'_1, E) < \epsilon _ W(D_1, D_2), \]
    2. if $\epsilon _{W'}(D'_2, E) > 0$, then setting $W = b(W')$ we have $\dim _\delta (W) = n - 1$ and

      \[ \epsilon _{W'}(D'_2, E) < \epsilon _ W(D_1, D_2), \]

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

\[ \text{ord}_{A_{\mathfrak p}}(a) \text{ord}_{A_{\mathfrak p}}(b) > \text{ord}_{B_{\mathfrak q}}(a) \text{ord}_{B_{\mathfrak q}}(s) \]

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 $as = b$. $\square$


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