Lemma 31.15.8 (0AGB)—The Stacks project

Lemma 31.15.8. Let $X$ be a Noetherian scheme. Let $Z \subset X$ be a closed subscheme. Assume there exist

a collection of integral effective Cartier divisors $D_ i \subset X$, $i \in I$
a closed subset $Z' \subset X$ all of whose irreducible components have codimension $\geq 2$ in $X$ (Topology, Definition 5.11.1)

such that $Z \subset Z' \cup \bigcup _{i \in I} D_ i$ set-theoretically. Then there exist integers $a_ i \geq 0$ with $a_ i = 0$ for almost all $i \in I$ such that the effective Cartier divisor

\[ D = \sum a_ i D_ i \]

is contained in $Z$ and such that the inclusion morphism $D \to Z$ is an isomorphism away from codimension $2$ in $X$ (in the sense that there exists an open $U \subset Z$ such that $D \cap U \to Z \cap U$ is an isomorphism and such that every irreducible component of $Z \setminus U$ has codimension $\geq 2$ in $X$). When $Z$ is nowhere dense in $X$ existence of the $D_ i$, $i \in I$ and $Z'$ 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 $\mathcal{I}_{Z, \xi _ i} \subset \mathcal{O}_ i$. Of course $a_ i > 0$ only if $D_ i$ is an irreducible component of $Z$ and hence $a_ i > 0$ only for a finite number of $i \in I$. 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 $x \in D_ i$ and $a_ i > 0$. 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 15.122.2) 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. $\square$

Comments (6)

Comment #7190 by Tongmu He on April 15, 2022 at 04:59

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 Johan on April 15, 2022 at 09:58

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

Comment #7318 by Johan on May 04, 2022 at 15:56

Again thanks very much. The change is here.

Comment #9555 by Branislav Sobot on August 13, 2024 at 04:03

Are you assuming in the statement that there are finitely many divisors $D_i$ ? If so, I suggest you add it. If not, then I don't understand what would the sum of these divisors be since we only defined finite sums of (effective Cartier) divisors. It also doesn't hurt to add to the statement that $a_i$ are nonnegative integers.

Comment #9556 by Branislav Sobot on August 13, 2024 at 04:21

Also, for the proof of the final statement, what happens with the irreducible components of $Z$ which have codimension zero in $X$ ? For instance, what if $X$ is irreducible and $Z=X$ , how do you find $D_i$ ?

Comment #10297 by Stacks Project on June 30, 2025 at 09:19

Okay, I am now explicitly allowing an infinite number of $D_i$ , I have spelled out thr codimension $2$ condition more precisely and I have added the requirement that $Z$ is nowhere dense in the last statement of the lemma. See this commit.

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