Lemma 71.8.1. 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 $(S_{k - 1})$.

## 71.8 Effective Cartier divisors on Noetherian spaces

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

**Proof.**
By our definition of the property $(S_ k)$ for algebraic spaces (Properties of Spaces, Section 66.7) and Lemma 71.6.2 this follows from the case of schemes (Divisors, Lemma 31.15.5).
$\square$

Lemma 71.8.2. 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 66.7) and Lemma 71.6.2 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 71.8.3. 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 66.12.5) is the desired effective Cartier divisor. The construction of $D$ commutes with étale localization, see proof of Properties of Spaces, Lemma 66.12.3. 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 71.8.4. 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

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 66.13.3. 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

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 71.7.8. 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 71.8.5. 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 67.30.12. 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 10.50.9). 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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