Lemma 93.14.1. Let $A' \to A$ be a surjective map of Noetherian rings with nilpotent kernel. Let $A \to B$ be a finite type flat ring map. Let $\mathfrak b \subset B$ be an ideal such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is syntomic on the complement of $V(\mathfrak b)$. Let $(B^ h, \mathfrak b^ h)$ be the henselization of the pair $(B, \mathfrak b)$. Then $B$ has a flat lift to $A'$ if and only if $B^ h$ has a flat lift to $A'$.
93.14 Deformations of henselizations
In this section we compare the deformation problem posed by an algebra and its completion. We first discuss “liftability”.
First proof. This proof is a cheat. Namely, if $B$ has a flat lift $B'$, then taking the henselization $(B')^ h$ we obtain a flat lift of $B^ h$ (compare with the proof of Lemma 93.8.8). Conversely, suppose that $C'$ is an $A'$-flat lift of $(B')^ h$. Then let $\mathfrak c' \subset C'$ be the inverse image of the ideal $\mathfrak b^ h$. Then the completion $(C')^\wedge $ of $C'$ with respect to $\mathfrak c'$ is a lift of $B^\wedge $ (details omitted). Hence we see that $B$ has a flat lift by Lemma 93.12.3. $\square$
Second 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 $P^ h$ and $(P')^ h$ the respective henselizations. We will use that taking henselizations is functorial and that the henselization of a quotient is the corresponding quotient of the henselization, see More on Algebra, Lemmas 15.11.16 and 15.12.7.
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 $C = B^ h$ and $\mathfrak c = \mathfrak bC$. Suppose that $A' \to C'$ is a flat lift of $A \to C$. Then $C'$ is henselian with respect to the inverse image $\mathfrak c'$ of $\mathfrak c$ (by More on Algebra, Lemma 15.11.9 and the fact that the kernel of $C' \to C$ is nilpotent). We choose an $A'$-algebra map $P' \to C'$ lifting the $A$-algebra map $P \to C$. These maps pass through henselizations to give surjections $P^ h \to C$ and $(P')^ h \to C'$ (for the second again using Nakayama's lemma). In particular, we find that we can endow $C'$ with an $A'$-flat $(P')^ h$-module structure lifting $C$ as an $A$-flat $P^ h$-module. Conversely, if we can lift $C$ to a $(P')^ h$-module $N'$ flat over $A'$, then $N'$ is a cyclic module $N' \cong (P')^ h/\tilde J$ (using Nakayama again) and setting $C' = (P')^ h/\tilde J$ we find a flat lift of $C$ as an algebra.
Observe that $P' \to (P')^ h$ is a flat ring map which induces an isomorphism $P'/\mathfrak p' = (P')^ h/\mathfrak p'(P')^ h$ (More on Algebra, Lemma 15.12.2). 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. $\square$
Lemma 93.14.2. Let $k$ be a field. Let $B$ be a finite type $k$-algebra. Let $J \subset B$ be an ideal such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(k)$ is smooth on the complement of $V(J)$. Let $N$ be a finite $B$-module. Then there is a canonical bijection Here $(B^ h, J^ h)$ is the henselization of $(B, J)$ and $N^ h = N \otimes _ B B^ h$.
\[ \text{Exal}_ k(B, N) \to \text{Exal}_ k(B^ h, N^ h) \]
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 henselization: given $0 \to N \to C \to B \to 0$ in $\text{Exal}_ k(B, N)$ we send it to the henselization $C^ h$ of $C$ with respect to the inverse image $J_ C \subset C$ of $J$. Compare with the proof of Lemma 93.8.8.
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
\begin{align*} \text{Ext}^1_{B^ h}(\mathop{N\! L}\nolimits _{B^ h/k}, N^ h) & = \text{Ext}^1_{B^ h}(\mathop{N\! L}\nolimits _{B/k} \otimes _ B B^ h, N \otimes _ B B^ h) \\ & = \text{Ext}^1_ B(\mathop{N\! L}\nolimits _{B/k}, N) \otimes _ B B^ h \\ & = \text{Ext}^1_ B(\mathop{N\! L}\nolimits _{B/k}, N) \end{align*}
The first equality by More on Algebra, Lemma 15.33.8 (or rather its analogue for henselizations of pairs). The second by More on Algebra, Lemma 15.99.2. The third because $\text{Ext}^1_ B(\mathop{N\! L}\nolimits _{B/k}, N)$ is $J$-power torsion, the map $B \to B^ h$ is flat and induces an isomorphism $B/J \to B^ h/JB^ h$ (More on Algebra, Lemma 15.12.2), and More on Algebra, Lemma 15.89.3. 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. $\square$
Lemma 93.14.3. In Example 93.8.1 let $P$ be a $k$-algebra. Let $J \subset P$ be an ideal. Denote $(P^ h, J^ h)$ the henselization of the pair $(P, J)$. If
$k \to P$ is of finite type, and
$\mathop{\mathrm{Spec}}(P) \to \mathop{\mathrm{Spec}}(k)$ is smooth on the complement of $V(J)$,
then the functor between deformation categories of Lemma 93.8.8
\[ \mathcal{D}\! \mathit{ef}_ P \longrightarrow \mathcal{D}\! \mathit{ef}_{P^ h} \]
is smooth and induces an isomorphism on tangent spaces.
Proof. We know that $\mathcal{D}\! \mathit{ef}_ P$ and $\mathcal{D}\! \mathit{ef}_{P^ h}$ 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 93.14.1 and the isomorphism on tangent spaces is the special case of Lemma 93.14.2 where $N = B$. $\square$
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