## 91.6 Application to flat modules on flat thickenings of ringed spaces

Consider a commutative diagram

$\xymatrix{ (X, \mathcal{O}_ X) \ar[r]_ i \ar[d]_ f & (X', \mathcal{O}_{X'}) \ar[d]^{f'} \\ (S, \mathcal{O}_ S) \ar[r]^ t & (S', \mathcal{O}_{S'}) }$

of ringed spaces whose horizontal arrows are first order thickenings as in Situation 91.3.1. Set $\mathcal{I} = \mathop{\mathrm{Ker}}(i^\sharp ) \subset \mathcal{O}_{X'}$ and $\mathcal{J} = \mathop{\mathrm{Ker}}(t^\sharp ) \subset \mathcal{O}_{S'}$. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. Assume that

1. $(f, f')$ is a strict morphism of thickenings,

2. $f'$ is flat, and

3. $\mathcal{F}$ is flat over $S$.

Note that (1) $+$ (2) imply that $\mathcal{I} = f^*\mathcal{J}$ (apply Lemma 91.5.2 to $\mathcal{O}_{X'}$). The theory of the preceding section is especially nice under these assumptions. We summarize the results already obtained in the following lemma.

Lemma 91.6.1. In the situation above.

1. There exists an $\mathcal{O}_{X'}$-module $\mathcal{F}'$ flat over $S'$ with $i^*\mathcal{F}' \cong \mathcal{F}$, if and only if the class $o(\mathcal{F}, f^*\mathcal{J} \otimes _{\mathcal{O}_ X} \mathcal{F}, 1) \in \mathop{\mathrm{Ext}}\nolimits ^2_{\mathcal{O}_ X}( \mathcal{F}, f^*\mathcal{J} \otimes _{\mathcal{O}_ X} \mathcal{F})$ of Lemma 91.4.4 is zero.

2. If such a module exists, then the set of isomorphism classes of lifts is principal homogeneous under $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}( \mathcal{F}, f^*\mathcal{J} \otimes _{\mathcal{O}_ X} \mathcal{F})$.

3. Given a lift $\mathcal{F}'$, the set of automorphisms of $\mathcal{F}'$ which pull back to $\text{id}_\mathcal {F}$ is canonically isomorphic to $\mathop{\mathrm{Ext}}\nolimits ^0_{\mathcal{O}_ X}( \mathcal{F}, f^*\mathcal{J} \otimes _{\mathcal{O}_ X} \mathcal{F})$.

Proof. Part (1) follows from Lemma 91.5.7 as we have seen above that $\mathcal{I} = f^*\mathcal{J}$. Part (2) follows from Lemma 91.5.6. Part (3) follows from Lemma 91.5.3. $\square$

Situation 91.6.2. Let $f : (X, \mathcal{O}_ X) \to (S, \mathcal{O}_ S)$ be a morphism of ringed spaces. Consider a commutative diagram

$\xymatrix{ (X'_1, \mathcal{O}'_1) \ar[r]_ h \ar[d]_{f'_1} & (X'_2, \mathcal{O}'_2) \ar[r] \ar[d]_{f'_2} & (X'_3, \mathcal{O}'_3) \ar[d]_{f'_3} \\ (S'_1, \mathcal{O}_{S'_1}) \ar[r] & (S'_2, \mathcal{O}_{S'_2}) \ar[r] & (S'_3, \mathcal{O}_{S'_3}) }$

where (a) the top row is a short exact sequence of first order thickenings of $X$, (b) the lower row is a short exact sequence of first order thickenings of $S$, (c) each $f'_ i$ restricts to $f$, (d) each pair $(f, f_ i')$ is a strict morphism of thickenings, and (e) each $f'_ i$ is flat. Finally, let $\mathcal{F}'_2$ be an $\mathcal{O}'_2$-module flat over $S'_2$ and set $\mathcal{F} = \mathcal{F}'_2|_ X$. Let $\pi : X'_1 \to X$ be the canonical splitting (Remark 91.4.9).

Lemma 91.6.3. In Situation 91.6.2 the modules $\pi ^*\mathcal{F}$ and $h^*\mathcal{F}'_2$ are $\mathcal{O}'_1$-modules flat over $S'_1$ restricting to $\mathcal{F}$ on $X$. Their difference (Lemma 91.6.1) is an element $\theta$ of $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}( \mathcal{F}, f^*\mathcal{J}_1 \otimes _{\mathcal{O}_ X} \mathcal{F})$ whose boundary in $\mathop{\mathrm{Ext}}\nolimits ^2_{\mathcal{O}_ X}( \mathcal{F}, f^*\mathcal{J}_3 \otimes _{\mathcal{O}_ X} \mathcal{F})$ equals the obstruction (Lemma 91.6.1) to lifting $\mathcal{F}$ to an $\mathcal{O}'_3$-module flat over $S'_3$.

Proof. Note that both $\pi ^*\mathcal{F}$ and $h^*\mathcal{F}'_2$ restrict to $\mathcal{F}$ on $X$ and that the kernels of $\pi ^*\mathcal{F} \to \mathcal{F}$ and $h^*\mathcal{F}'_2 \to \mathcal{F}$ are given by $f^*\mathcal{J}_1 \otimes _{\mathcal{O}_ X} \mathcal{F}$. Hence flatness by Lemma 91.5.2. Taking the boundary makes sense as the sequence of modules

$0 \to f^*\mathcal{J}_3 \otimes _{\mathcal{O}_ X} \mathcal{F} \to f^*\mathcal{J}_2 \otimes _{\mathcal{O}_ X} \mathcal{F} \to f^*\mathcal{J}_1 \otimes _{\mathcal{O}_ X} \mathcal{F} \to 0$

is short exact due to the assumptions in Situation 91.6.2 and the fact that $\mathcal{F}$ is flat over $S$. The statement on the obstruction class is a direct translation of the result of Remark 91.4.10 to this particular situation. $\square$

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