The Stacks project

70.8 Effective Cartier divisors on Noetherian spaces

In the locally Noetherian setting most of the discussion of effective Cartier divisors and regular sections simplifies somewhat.

Lemma 70.8.1. Let $S$ be a scheme and let $X$ be a locally Noetherian algebraic space over $S$. Let $D \subset X$ be an effective Cartier divisor. If $X$ is $(S_ k)$, then $D$ is $(S_{k - 1})$.

Proof. By our definition of the property $(S_ k)$ for algebraic spaces (Properties of Spaces, Section 65.7) and Lemma 70.6.2 this follows from the case of schemes (Divisors, Lemma 31.15.5). $\square$

Lemma 70.8.2. Let $S$ be a scheme and let $X$ be a locally Noetherian normal algebraic space over $S$. Let $D \subset X$ be an effective Cartier divisor. Then $D$ is $(S_1)$.

Proof. By our definition of normality for algebraic spaces (Properties of Spaces, Section 65.7) and Lemma 70.6.2 this follows from the case of schemes (Divisors, Lemma 31.15.6). $\square$

The following lemma can sometimes be used to produce effective Cartier divisors.

Lemma 70.8.3. Let $S$ be a scheme. Let $X$ be a regular Noetherian separated algebraic space over $S$. Let $U \subset X$ be a dense affine open. Then there exists an effective Cartier divisor $D \subset X$ with $U = X \setminus D$.

Proof. We claim that the reduced induced algebraic space structure $D$ on $X \setminus U$ (Properties of Spaces, Definition 65.12.5) is the desired effective Cartier divisor. The construction of $D$ commutes with étale localization, see proof of Properties of Spaces, Lemma 65.12.3. Let $X' \to X$ be a surjective étale morphism with $X'$ affine. Since $X$ is separated, we see that $U' = X' \times _ X U$ is affine. Since $|X'| \to |X|$ is open, we see that $U'$ is dense in $X'$. Since $D' = X' \times _ X D$ is the reduced induced scheme structure on $X' \setminus U'$, we conclude that $D'$ is an effective Cartier divisor by Divisors, Lemma 31.16.6 and its proof. This is what we had to show. $\square$

Lemma 70.8.4. Let $S$ be a scheme. Let $X$ be a regular Noetherian separated algebraic space over $S$. Then every invertible $\mathcal{O}_ X$-module is isomorphic to

\[ \mathcal{O}_ X(D - D') = \mathcal{O}_ X(D) \otimes _{\mathcal{O}_ X} \mathcal{O}_ X(D')^{\otimes -1} \]

for some effective Cartier divisors $D, D'$ in $X$.

Proof. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Choose a dense affine open $U \subset X$ such that $\mathcal{L}|_ U$ is trivial. This is possible because $X$ has a dense open subspace which is a scheme, see Properties of Spaces, Proposition 65.13.3. Denote $s : \mathcal{O}_ U \to \mathcal{L}|_ U$ the trivialization. The complement of $U$ is an effective Cartier divisor $D$. We claim that for some $n > 0$ the map $s$ extends uniquely to a map

\[ s : \mathcal{O}_ X(-nD) \longrightarrow \mathcal{L} \]

The claim implies the lemma because it shows that $\mathcal{L} \otimes _{\mathcal{O}_ X} \mathcal{O}_ X(nD)$ has a regular global section hence is isomorphic to $\mathcal{O}_ X(D')$ for some effective Cartier divisor $D'$ by Lemma 70.7.8. To prove the claim we may work étale locally. Thus we may assume $X$ is an affine Noetherian scheme. Since $\mathcal{O}_ X(-nD) = \mathcal{I}^ n$ where $\mathcal{I} = \mathcal{O}_ X(-D)$ is the ideal sheaf of $D$ in $X$, this case follows from Cohomology of Schemes, Lemma 30.10.5. $\square$

The following lemma really belongs to a different section.

Lemma 70.8.5. Let $R$ be a valuation ring with fraction field $K$. Let $X$ be an algebraic space over $R$ such that $X \to \mathop{\mathrm{Spec}}(R)$ is smooth. For every effective Cartier divisor $D \subset X_ K$ there exists an effective Cartier divisor $D' \subset X$ with $D'_ K = D$.

Proof. Let $D' \subset X$ be the scheme theoretic image of $D \to X_ K \to X$. Since this morphism is quasi-compact, formation of $D'$ commutes with flat base change, see Morphisms of Spaces, Lemma 66.30.12. In particular we find that $D'_ K = D$. Hence, we may assume $X$ is affine. Say $X = \mathop{\mathrm{Spec}}(A)$. Then $X_ K = \mathop{\mathrm{Spec}}(A \otimes _ R K)$ and $D$ corresponds to an ideal $I \subset A \otimes _ R K$. We have to show that $J = I \cap A$ cuts out an effective Cartier divisor in $X$. First, observe that $A/J$ is flat over $R$ (as a torsion free $R$-module, see More on Algebra, Lemma 15.22.10), hence $J$ is finitely generated by More on Algebra, Lemma 15.25.6 and Algebra, Lemma 10.5.3. Thus it suffices to show that $J_\mathfrak q \subset A_\mathfrak q$ is generated by a single element for each prime $\mathfrak q \subset A$. Let $\mathfrak p = R \cap \mathfrak q$. Then $R_\mathfrak p$ is a valuation ring (Algebra, Lemma 10.50.8). Observe further that $A_\mathfrak q/\mathfrak p A_\mathfrak q$ is a regular ring by Algebra, Lemma 10.140.3. Thus we may apply More on Algebra, Lemma 15.121.3 to see that $I(A_\mathfrak q \otimes _ R K)$ is generated by a single element $f \in A_\mathfrak p \otimes _ R K$. After clearing denominators we may assume $f \in A_\mathfrak q$. Let $\mathfrak c \subset R_\mathfrak p$ be the content ideal of $f$ (see More on Algebra, Definition 15.24.1 and More on Flatness, Lemma 38.19.6). Since $R_\mathfrak p$ is a valuation ring and since $\mathfrak c$ is finitely generated (More on Algebra, Lemma 15.24.2) we see $\mathfrak c = (\pi )$ for some $\pi \in R_\mathfrak p$ (Algebra, Lemma 10.50.15). After relacing $f$ by $\pi ^{-1}f$ we see that $f \in A_\mathfrak q$ and $f \not\in \mathfrak pA_\mathfrak q$. Claim: $I_\mathfrak q = (f)$ which finishes the proof. To see the claim, observe that $f \in I_\mathfrak q$. Hence we have a surjection $A_\mathfrak q/(f) \to A_\mathfrak q/I_\mathfrak q$ which is an isomorphism after tensoring over $R$ with $K$. Thus we are done if $A_\mathfrak q/(f)$ is $R_\mathfrak p$-flat. This follows from Algebra, Lemma 10.128.5 and our choice of $f$. $\square$


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