Lemma 31.15.8. Let $X$ be a Noetherian scheme. Let $Z \subset X$ be a closed subscheme. Assume there exist integral effective Cartier divisors $D_ i \subset X$ and a closed subset $Z' \subset X$ of codimension $\geq 2$ such that $Z \subset Z' \cup \bigcup D_ i$ set-theoretically. Then there exists an effective Cartier divisor of the form

$D = \sum a_ i D_ i \subset Z$

such that $D \to Z$ is an isomorphism away from codimension $2$ in $X$. The existence of the $D_ i$ is guaranteed if $\mathcal{O}_{X, x}$ is a UFD for all $x \in Z$ or if $X$ is regular.

Proof. Let $\xi _ i \in D_ i$ be the generic point and let $\mathcal{O}_ i = \mathcal{O}_{X, \xi _ i}$ be the local ring which is a discrete valuation ring by Lemma 31.15.4. Let $a_ i \geq 0$ be the minimal valuation of an element of $\mathcal{I}_{Z, \xi _ i} \subset \mathcal{O}_ i$. We claim that the effective Cartier divisor $D = \sum a_ i D_ i$ works.

Namely, suppose that $x \in X$. Let $A = \mathcal{O}_{X, x}$. Let $D_1, \ldots , D_ n$ be the pairwise distinct divisors $D_ i$ such that $x \in D_ i$. For $1 \leq i \leq n$ let $f_ i \in A$ be a local equation for $D_ i$. Then $f_ i$ is a prime element of $A$ and $\mathcal{O}_ i = A_{(f_ i)}$. Let $I = \mathcal{I}_{Z, x} \subset A$ be the stalk of the ideal sheaf of $Z$. By our choice of $a_ i$ we have $I A_{(f_ i)} = f_ i^{a_ i}A_{(f_ i)}$. We claim that $I \subset (\prod _{i = 1, \ldots , n} f_ i^{a_ i})$.

Proof of the claim. The localization map $\varphi : A/(f_ i) \to A_{(f_ i)}/f_ iA_{(f_ i)}$ is injective as the prime ideal $(f_ i)$ is the inverse image of the maximal ideal $f_ iA_{(f_ i)}$. By induction on $n$ we deduce that $\varphi _ n : A/(f_ i^ n)\to A_{(f_ i)}/f_ i^ nA_{(f_ i)}$ is also injective. Since $\varphi _{a_ i}(I) = 0$, we have $I \subset (f_ i^{a_ i})$. Thus, for any $x \in I$, we may write $x = f_1^{a_1}x_1$ for some $x_1 \in A$. Since $D_1, \ldots , D_ n$ are pairwise distinct, $f_ i$ is a unit in $A_{(f_ j)}$ for $i \not= j$. Comparing $x$ and $x_1$ at $A_{(f_ i)}$ for $n \geq i > 1$, we still have $x_1 \in (f_ i^{a_ i})$. Repeating the previous process, we inductively write $x_ i = f_{i + 1}^{a_{i + 1}}x_{i + 1}$ for any $n > i \geq 1$. In conclusion, $x \in (\prod _{i = 1, \ldots n} f_ i^{a_ i})$ for any $x \in I$ as desired.

The claim shows that $\mathcal{I}_ Z \subset \mathcal{I}_ D$, i.e., that $D \subset Z$. Moreover, we also see that $D$ and $Z$ agree at the $\xi _ i$, which proves that $D \to Z$ is an isomorphism away from codimension $2$ on $X$.

To see the final statements we argue as follows. A regular local ring is a UFD (More on Algebra, Lemma 15.121.2) hence it suffices to argue in the UFD case. In that case, let $D_ i$ be the irreducible components of $Z$ which have codimension $1$ in $X$. By Lemma 31.15.7 each $D_ i$ is an effective Cartier divisor. $\square$

Comment #7190 by on

It seems too fast for me to deduce $I\subset (\prod f_i^{a_i})$ from $IA_{(f_i)}=f_i^{a_i}A_{(f_i)}$. I understand it as follows:

The localization map $\varphi:A/(f_i)\to A_{(f_i)}/f_iA_{(f_i)}$ is injective as the prime ideal $(f_i)$ is the inverse image of the maximal ideal $f_iA_{(f_i)}$. By dévissage, we deduce that $\varphi_n: A/(f_i^n)\to A_{(f_i)}/f_i^nA_{(f_i)}$ is also injective. Since $\varphi_{a_i}(I)=0$, we have $I\subset (f_i^{a_i})$. Thus, for any $x\in I$, we write $x=f_1^{a_1}x_1$ for some $x_1\in A$. We may assume that the concerning effective Cartier divisor $D_1, \dots, D_n$ are pairwise different, so that $f_i$ is a unit in $A_{(f_j)}$ for $i\neq j$. Comparing $x$ and $x_1$ at $A_{(f_i)}$ for $i>1$, we still have $x_1\in (f_i^{a_i})$ for $i>1$. Repeating the previous process, we inductively write $x_i=f_{i+1}^{a_{i+1}}x_{i+1}$ for any $i\geq 1$. In conclusion, $x\in (\prod f_i^{a_i})$ for any $x\in I$.

I will appretiate if anyone can share his or her faster ideas.

Comment #7191 by on

Very good. I will add this when I next go through all the comments. Thanks!

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