The Stacks project

This is discussed in [Section 5.3, Ravi-Murphys-Law] and [Theorem 3.3, Ran-deformations].

Lemma 92.10.6. In Example 92.9.1 let $f : X \to Y$ be a morphism of schemes over $k$. If $f_*\mathcal{O}_ X = \mathcal{O}_ Y$ and $R^1f_*\mathcal{O}_ X = 0$, then the morphism of deformation categories

\[ \mathcal{D}\! \mathit{ef}_{X \to Y} \to \mathcal{D}\! \mathit{ef}_ X \]

is an equivalence.

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

\[ 0 = \mathfrak m_ A^ n\mathcal{O}_ U \subset \mathfrak m_ A^{n - 1}\mathcal{O}_ U \subset \ldots \subset \mathfrak m_ A^2\mathcal{O}_ U \subset \mathfrak m_ A\mathcal{O}_ U \subset \mathcal{O}_ U \]

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 90.5.2. Set

\[ \mathcal{O}_ V = f_*\mathcal{O}_ U \]

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

\[ 0 \to (f_*\mathcal{O}_ X) \otimes _ k \mathfrak m_ A^ i/\mathfrak m_ A^{i + 1} \to f_*(\mathcal{O}_ U/\mathfrak m_ A^{i + 1}\mathcal{O}_ U) \to f_*(\mathcal{O}_ U/\mathfrak m_ A^ i\mathcal{O}_ U) \to 0 \]

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. $\square$


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