The Stacks project

Proof. We construct a quasi-inverse to the forgetful functor of the lemma. Namely, suppose that $(A, U)$ is an object of $\mathcal{D}\! \mathit{ef}_ X$. The given map $X \to U$ is a finite order thickening and we can use it to identify the underlying topological spaces of $U$ and $X$, see More on Morphisms, Section 37.2. Thus we may and do think of $\mathcal{O}_ U$ as a sheaf of $A$-algebras on $X$; moreover the fact that $U \to \mathop{\mathrm{Spec}}(A)$ is flat, means that $\mathcal{O}_ U$ is flat as a sheaf of $A$-modules. In particular, we have a filtration

with subquotients equal to $\mathcal{O}_ X \otimes _ k \mathfrak m_ A^ i/\mathfrak m_ A^{i + 1}$ by flatness, see More on Morphisms, Lemma 37.10.1 or the more general Deformation Theory, Lemma 91.5.2. Set

viewed as sheaf of $A$-algebras on $Y$. Since $R^1f_*\mathcal{O}_ X = 0$ we find by the description above that $R^1f_*(\mathfrak m_ A^ i\mathcal{O}_ U/\mathfrak m_ A^{i + 1}\mathcal{O}_ U) = 0$ for all $i$. This implies that the sequences

are exact for all $i$. Reading the references given above backwards (and using induction) we find that $\mathcal{O}_ V$ is a flat sheaf of $A$-algebras with $\mathcal{O}_ V/\mathfrak m_ A\mathcal{O}_ V = \mathcal{O}_ Y$. Using More on Morphisms, Lemma 37.2.2 we find that $(Y, \mathcal{O}_ V)$ is a scheme, call it $V$. The equality $\mathcal{O}_ V = f_*\mathcal{O}_ U$ defines a morphism of ringed spaces $U \to V$ which is easily seen to be a morphism of schemes. This finishes the proof by the flatness already established. $\square$