The Stacks project

Proof. Taking base change, this immediately reduces to the following problem: given a valuation ring $R$ with fraction field $K$ and an algebraic space $X$ proper over $R$ and a coherent $\mathcal{O}_{X_ K}$-module $\mathcal{F}_ K$, show there exists a finitely presented $\mathcal{O}_ X$-module $\mathcal{F}$ flat over $R$ whose generic fibre is $\mathcal{F}_ K$. Observe that by Flatness on Spaces, Theorem 77.4.5 any finite type quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ flat over $R$ is of finite presentation. Denote $j : X_ K \to X$ the embedding of the generic fibre. As a base change of the affine morphism $\mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(R)$ the morphism $j$ is affine. Thus $j_*\mathcal{F}_ K$ is quasi-coherent. Write

as a filtered colimit of its finite type quasi-coherent $\mathcal{O}_ X$-submodules, see Limits of Spaces, Lemma 70.9.2. Since $j_*\mathcal{F}_ K$ is a sheaf of $K$-vector spaces over $X$, it is flat over $\mathop{\mathrm{Spec}}(R)$. Thus each $\mathcal{F}_ i$ is flat over $R$ as flatness over a valuation ring is the same as being torsion free (More on Algebra, Lemma 15.22.10) and torsion freeness is inherited by submodules. Finally, we have to show that the map $j^*\mathcal{F}_ i \to \mathcal{F}_ K$ is an isomorphism for some $i$. Since $j^*j_*\mathcal{F}_ K = \mathcal{F}_ K$ (small detail omitted) and since $j^*$ is exact, we see that $j^*\mathcal{F}_ i \to \mathcal{F}_ K$ is injective for all $i$. Since $j^*$ commutes with colimits, we have $\mathcal{F}_ K = j^*j_*\mathcal{F}_ K = \mathop{\mathrm{colim}}\nolimits j^*\mathcal{F}_ i$. Since $\mathcal{F}_ K$ is coherent (i.e., finitely presented), there is an $i$ such that $j^*\mathcal{F}_ i$ contains all the (finitely many) generators over an affine étale cover of $X$. Thus we get surjectivity of $j^*\mathcal{F}_ i \to \mathcal{F}_ K$ for $i$ large enough. $\square$