The Stacks project

Proof. This proof is the same as the proof of Lemma 93.12.3 but easier. We suggest the reader to skip the proof. Choose an $A$-algebra surjection $P = A[x_1, \ldots , x_ n] \to B$. Let $S_ P \subset P$ be the inverse image of $S$. Set $P' = A'[x_1, \ldots , x_ n]$ and denote $S_{P'} \subset P'$ the inverse image of $S_ P$.

Suppose $A' \to B'$ is a flat lift of $A \to B$, in other words, $A' \to B'$ is flat and there is an $A$-algebra isomorphism $B = B' \otimes _{A'} A$. Then we can choose an $A'$-algebra map $P' \to B'$ lifting the given surjection $P \to B$. By Nakayama's lemma (Algebra, Lemma 10.20.1) we find that $B'$ is a quotient of $P'$. In particular, we find that we can endow $B'$ with an $A'$-flat $P'$-module structure lifting $B$ as an $A$-flat $P$-module. Conversely, if we can lift $B$ to a $P'$-module $M'$ flat over $A'$, then $M'$ is a cyclic module $M' \cong P'/J'$ (using Nakayama again) and setting $B' = P'/J'$ we find a flat lift of $B$ as an algebra.

Set $C = S^{-1}B$. Suppose that $A' \to C'$ is a flat lift of $A \to C$. Elements of $C'$ which map to invertible elements of $C$ are invertible. We choose an $A'$-algebra map $P' \to C'$ lifting the $A$-algebra map $P \to C$. By the remark above these maps pass through localizations to give surjections $S_ P^{-1}P \to C$ and $S_{P'}^{-1}P' \to C'$ (for the second use Nakayama's lemma). In particular, we find that we can endow $C'$ with an $A'$-flat $S_{P'}^{-1}P'$-module structure lifting $C$ as an $A$-flat $S_ P^{-1}P$-module. Conversely, if we can lift $C$ to a $S_{P'}^{-1}P'$-module $N'$ flat over $A'$, then $N'$ is a cyclic module $N' \cong S_{P'}^{-1}P'/\tilde J$ (using Nakayama again) and setting $C' = S_{P'}^{-1}P'/\tilde J$ we find a flat lift of $C$ as an algebra.

The syntomic locus of a morphism of schemes is open by definition. Let $J_ B \subset B$ be an ideal cutting out the set of points in $\mathop{\mathrm{Spec}}(B)$ where $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is not syntomic. Denote $J_ P \subset P$ and $J_{P'} \subset P'$ the corresponding ideals. Observe that $P' \to S_{P'}^{-1}P'$ is a flat ring map which induces an isomorphism $P'/J_{P'} = S_{P'}^{-1}P'/J_{P'}S_{P'}^{-1}P'$ by our assumption on $S$ in the lemma, namely, the assumption in the lemma is exactly that $B/J_ B = S^{-1}(B/J_ B)$. We conclude that our lemma is a consequence of Lemma 93.12.2 provided we can show that $B_ g$ lifts to an $A'$-flat $P'_ g$-module for $g \in J_ B$. However, the ring map $A \to B_ g$ is syntomic and hence lifts to an $A'$-flat algebra $B'$ by Smoothing Ring Maps, Proposition 16.3.2. Since $A' \to P'_ g$ is smooth, we can lift $P_ g \to B_ g$ to a surjective map $P'_ g \to B'$ as before and we get what we want. $\square$

Proof. This proof is the same as the proof of Lemma 93.12.4 but easier. We suggest the reader to skip the proof. The map is given by localization: given $0 \to N \to C \to B \to 0$ in $\text{Exal}_ k(B, N)$ we send it to the localization $S_ C^{-1}C$ of $C$ with respect to the inverse image $S_ C \subset C$ of $S$. Compare with the proof of Lemma 93.8.7.

The smooth locus of a morphism of schemes is open by definition. Let $J \subset B$ be an ideal cutting out the set of points in $\mathop{\mathrm{Spec}}(B)$ where $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is not smooth. Since $k \to B$ is of finite presentation the complex $\mathop{N\! L}\nolimits _{B/k}$ can be represented by a complex $N^{-1} \to N^0$ where $N^ i$ is a finite $B$-module, see Algebra, Section 10.134 and in particular Algebra, Lemma 10.134.2. As $B$ is Noetherian, this means that $\mathop{N\! L}\nolimits _{B/k}$ is pseudo-coherent. For $g \in J$ the $k$-algebra $B_ g$ is smooth and hence $(\mathop{N\! L}\nolimits _{B/k})_ g = \mathop{N\! L}\nolimits _{B_ g/k}$ is quasi-isomorphic to a finite projective $B$-module sitting in degree $0$. Thus $\text{Ext}^ i_ B(\mathop{N\! L}\nolimits _{B/k}, N)_ g = 0$ for $i \geq 1$ and any $B$-module $N$. Finally, we have

The first equality by More on Algebra, Lemma 15.99.2 and Algebra, Lemma 10.134.13. The second because $\text{Ext}^1_ B(\mathop{N\! L}\nolimits _{B/k}, N)$ is $J$-power torsion and elements of $S$ act invertibly on $J$-power torsion modules. This concludes the proof by the description of $\text{Exal}_ A(B, N)$ as $\text{Ext}^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$ given just above Lemma 93.12.4. $\square$

Proof. We know that $\mathcal{D}\! \mathit{ef}_ P$ and $\mathcal{D}\! \mathit{ef}_{S^{-1}P}$ are deformation categories by Lemma 93.8.2. Thus it suffices to check our functor identifies tangent spaces and a correspondence between liftability, see Formal Deformation Theory, Lemma 90.20.3. The property on liftability is proven in Lemma 93.13.1 and the isomorphism on tangent spaces is the special case of Lemma 93.13.2 where $N = B$. $\square$