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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