The Stacks project

Proof. Suppose that $M'$ is as in (1). Then

The first two equalities are clear, the third holds because $M'$ is flat over $A'$, and the fourth holds by More on Algebra, Lemma 15.61.2. Thus (2) holds. Conversely, suppose $K'$ is as in (2). We may and do assume $M$ is nonzero. Let $t$ be the largest integer such that $H^ t(K')$ is nonzero (exists because $M$ is nonzero). Then $H^ t(K') \otimes _{P'} P = H^ t(K' \otimes _{P'}^\mathbf {L} P)$ is zero if $t > 0$. Since the kernel of $P' \to P$ is nilpotent this implies $H^ t(K') = 0$ by Nakayama's lemma a contradiction. Hence $t = 0$ (the case $t < 0$ is absurd as well). Then $M' = H^0(K')$ is a $P'$-module such that $M = M' \otimes _{P'} P$ and the spectral sequence for Tor gives an injective map

By the reference on derived base change above $0 = \text{Tor}_1^{P'}(M', P) = \text{Tor}_1^{A'}(M', A)$. We conclude that $M'$ is $A'$-flat by Algebra, Lemma 10.99.8. $\square$

Proof. Let $I = \mathop{\mathrm{Ker}}(A' \to A)$. By induction on the integer $n > 1$ such that $I^ n = 0$ we reduce to the case where $I$ is an ideal of square zero; details omitted. We translate the condition of liftability of $M$ into the problem of finding an object of $D^-(P')$ as in Lemma 93.12.1. The obstruction to doing this is the element

constructed in Deformation Theory, Lemma 91.15.1. The equality in the displayed formula holds as $M \otimes _ P^\mathbf {L} IP = M \otimes _ P IP$ since $M$ and $P$ are $A$-flat¹. The obstruction for lifting $M \otimes _ P Q$ is similarly the element

which is the image of $\omega (M)$ by the functoriality of the construction $\omega (-)$ of Deformation Theory, Lemma 91.15.1. By More on Algebra, Lemma 15.99.2 we have

here we use that $P$ is Noetherian and $M$ finite. Our assumption on $P' \to Q'$ guarantees that for an $P$-module $E$ the map $E \to E \otimes _ P Q$ is bijective on $J'$-power torsion, see More on Algebra, Lemma 15.89.3. Thus we conclude that it suffices to show $\omega (M)$ is $J'$-power torsion. In other words, it suffices to show that $\omega (M)$ dies in

for all $g \in J'$. Howeover, by the compatibility of formation of $\omega (M)$ with base change again, we conclude that this is true as $M_ g$ is assumed to have a lift (of course you have to use the whole string of equivalences again). $\square$

Proof. Choose an $A$-algebra surjection $P = A[x_1, \ldots , x_ n] \to B$. Let $\mathfrak p \subset P$ be the inverse image of $\mathfrak b$. Set $P' = A'[x_1, \ldots , x_ n]$ and denote $\mathfrak p' \subset P'$ the inverse image of $\mathfrak p$. (Of course $\mathfrak p$ and $\mathfrak p'$ do not designate prime ideals here.) We will denote $P^\wedge $ and $(P')^\wedge $ the respective completions.

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 = B^\wedge $ and $\mathfrak c = \mathfrak bC$. Suppose that $A' \to C'$ is a flat lift of $A \to C$. Then $C'$ is complete with respect to the inverse image $\mathfrak c'$ of $\mathfrak c$ (Algebra, Lemma 10.97.10). We choose an $A'$-algebra map $P' \to C'$ lifting the $A$-algebra map $P \to C$. These maps pass through completions to give surjections $P^\wedge \to C$ and $(P')^\wedge \to C'$ (for the second again using Nakayama's lemma). In particular, we find that we can endow $C'$ with an $A'$-flat $(P')^\wedge $-module structure lifting $C$ as an $A$-flat $P^\wedge $-module. Conversely, if we can lift $C$ to a $(P')^\wedge $-module $N'$ flat over $A'$, then $N'$ is a cyclic module $N' \cong (P')^\wedge /\tilde J$ (using Nakayama again) and setting $C' = (P')^\wedge /\tilde J$ we find a flat lift of $C$ as an algebra.

Observe that $P' \to (P')^\wedge $ is a flat ring map which induces an isomorphism $P'/\mathfrak p' = (P')^\wedge /\mathfrak p'(P')^\wedge $. 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 $g \in \mathfrak p'$. 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. The map is given by completion: given $0 \to N \to C \to B \to 0$ in $\text{Exal}_ k(B, N)$ we send it to the completion $C^\wedge $ of $C$ with respect to the inverse image of $J$. Compare with the proof of Lemma 93.8.10.

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 $N$. By More on Algebra, Lemma 15.102.1 we conclude that

is an isomorphism for any finite $B$-module $N$.

Injectivity of the map. Suppose that $0 \to N \to C \to B \to 0$ is in $\text{Exal}_ k(B, N)$ and maps to zero in $\text{Exal}_ k(B^\wedge , N^\wedge )$. Choose a splitting $C^\wedge = B^\wedge \oplus N^\wedge $. Then the induced map $C \to C^\wedge \to N^\wedge $ gives maps $C \to N/J^ nN$ for all $n$. Hence we see that our element is in the kernel of the maps

for all $n$. By the previous paragraph we conclude that our element is zero.

Surjectivity of the map. Let $0 \to N^\wedge \to C' \to B^\wedge \to 0$ be an element of $\text{Exal}_ k(B^\wedge , N^\wedge )$. Pulling back by $B \to B^\wedge $ we get an element $0 \to N^\wedge \to C'' \to B \to 0$ in $\text{Exal}_ k(B, N^\wedge )$. we have

The first equality as $N^\wedge = N \otimes _ B B^\wedge $ (Algebra, Lemma 10.97.1) and More on Algebra, Lemma 15.65.3. The second equality because $\text{Ext}^1_ B(\mathop{N\! L}\nolimits _{B/k}, N)$ is $J$-power torsion (see above), $B \to B^\wedge $ is flat and induces an isomorphism $B/J \to B^\wedge /JB^\wedge $, and More on Algebra, Lemma 15.89.3. Thus we can find a $C \in \text{Exal}_ k(B, N)$ mapping to $C''$ in $\text{Exal}_ k(B, N^\wedge )$. Thus

are two elements of $\text{Exal}_ k(B^\wedge , N^\wedge )$ mapping to the same element of $\text{Exal}_ k(B, N^\wedge )$. Taking the difference we get an element $0 \to N^\wedge \to C' \to B^\wedge \to 0$ of $\text{Exal}_ k(B^\wedge , N^\wedge )$ whose image in $\text{Exal}_ k(B, N^\wedge )$ is zero. This means there exists

Let $J' \subset C'$ be the inverse image of $JB^\wedge \subset B^\wedge $. To finish the proof it suffices to note that $\sigma $ is continuous for the $J$-adic topology on $B$ and the $J'$-adic topology on $C'$ and that $C'$ is $J'$-adically complete by Algebra, Lemma 10.97.10 (here we also use that $C'$ is Noetherian; small detail omitted). Namely, this means that $\sigma $ factors through the completion $B^\wedge $ and $C' = 0$ in $\text{Exal}_ k(B^\wedge , N^\wedge )$. $\square$

Proof. We know that $\mathcal{D}\! \mathit{ef}_ P$ and $\mathcal{D}\! \mathit{ef}_{P^\wedge }$ 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 93.12.3 and the isomorphism on tangent spaces is the special case of Lemma 93.12.4 where $N = B$. $\square$