Lemma 91.11.1. Let i : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}') be a first order thickening of ringed topoi. Let \mathcal{F}', \mathcal{G}' be \mathcal{O}'-modules. Set \mathcal{F} = i^*\mathcal{F}' and \mathcal{G} = i^*\mathcal{G}'. Let \varphi : \mathcal{F} \to \mathcal{G} be an \mathcal{O}-linear map. The set of lifts of \varphi to an \mathcal{O}'-linear map \varphi ' : \mathcal{F}' \to \mathcal{G}' is, if nonempty, a principal homogeneous space under \mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(\mathcal{F}, \mathcal{I}\mathcal{G}').
91.11 Infinitesimal deformations of modules on ringed topoi
Let i : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}') be a first order thickening of ringed topoi. We freely use the notation introduced in Section 91.9. Let \mathcal{F}' be an \mathcal{O}'-module and set \mathcal{F} = i^*\mathcal{F}'. In this situation we have a short exact sequence
0 \to \mathcal{I}\mathcal{F}' \to \mathcal{F}' \to \mathcal{F} \to 0
of \mathcal{O}'-modules. Since \mathcal{I}^2 = 0 the \mathcal{O}'-module structure on \mathcal{I}\mathcal{F}' comes from a unique \mathcal{O}-module structure. Thus the sequence above is an extension as in (91.10.0.1). As a special case, if \mathcal{F}' = \mathcal{O}' we have i^*\mathcal{O}' = \mathcal{O} and \mathcal{I}\mathcal{O}' = \mathcal{I} and we recover the sequence of structure sheaves
0 \to \mathcal{I} \to \mathcal{O}' \to \mathcal{O} \to 0
Proof. This is a special case of Lemma 91.10.1 but we also give a direct proof. We have short exact sequences of modules
0 \to \mathcal{I} \to \mathcal{O}' \to \mathcal{O} \to 0 \quad \text{and}\quad 0 \to \mathcal{I}\mathcal{G}' \to \mathcal{G}' \to \mathcal{G} \to 0
and similarly for \mathcal{F}'. Since \mathcal{I} has square zero the \mathcal{O}'-module structure on \mathcal{I} and \mathcal{I}\mathcal{G}' comes from a unique \mathcal{O}-module structure. It follows that
\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}'}(\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(\mathcal{F}, \mathcal{I}\mathcal{G}') \quad \text{and}\quad \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}'}(\mathcal{F}', \mathcal{G}) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(\mathcal{F}, \mathcal{G})
The lemma now follows from the exact sequence
0 \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}'}(\mathcal{F}', \mathcal{I}\mathcal{G}') \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}'}(\mathcal{F}', \mathcal{G}') \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}'}(\mathcal{F}', \mathcal{G})
see Homology, Lemma 12.5.8. \square
Lemma 91.11.2. Let (f, f') be a morphism of first order thickenings of ringed topoi as in Situation 91.9.1. Let \mathcal{F}' be an \mathcal{O}'-module and set \mathcal{F} = i^*\mathcal{F}'. Assume that \mathcal{F} is flat over \mathcal{O}_\mathcal {B} and that (f, f') is a strict morphism of thickenings (Definition 91.9.2). Then the following are equivalent
\mathcal{F}' is flat over \mathcal{O}_{\mathcal{B}'}, and
the canonical map f^*\mathcal{J} \otimes _\mathcal {O} \mathcal{F} \to \mathcal{I}\mathcal{F}' is an isomorphism.
Moreover, in this case the maps
f^*\mathcal{J} \otimes _\mathcal {O} \mathcal{F} \to \mathcal{I} \otimes _\mathcal {O} \mathcal{F} \to \mathcal{I}\mathcal{F}'
are isomorphisms.
Proof. The map f^*\mathcal{J} \to \mathcal{I} is surjective as (f, f') is a strict morphism of thickenings. Hence the final statement is a consequence of (2).
Proof of the equivalence of (1) and (2). By definition flatness over \mathcal{O}_\mathcal {B} means flatness over f^{-1}\mathcal{O}_\mathcal {B}. Similarly for flatness over f^{-1}\mathcal{O}_{\mathcal{B}'}. Note that the strictness of (f, f') and the assumption that \mathcal{F} = i^*\mathcal{F}' imply that
\mathcal{F} = \mathcal{F}'/(f^{-1}\mathcal{J})\mathcal{F}'
as sheaves on \mathcal{C}. Moreover, observe that f^*\mathcal{J} \otimes _\mathcal {O} \mathcal{F} = f^{-1}\mathcal{J} \otimes _{f^{-1}\mathcal{O}_\mathcal {B}} \mathcal{F}. Hence the equivalence of (1) and (2) follows from Modules on Sites, Lemma 18.28.15. \square
Lemma 91.11.3. Let (f, f') be a morphism of first order thickenings of ringed topoi as in Situation 91.9.1. Let \mathcal{F}' be an \mathcal{O}'-module and set \mathcal{F} = i^*\mathcal{F}'. Assume that \mathcal{F}' is flat over \mathcal{O}_{\mathcal{B}'} and that (f, f') is a strict morphism of thickenings. Then the following are equivalent
\mathcal{F}' is an \mathcal{O}'-module of finite presentation, and
\mathcal{F} is an \mathcal{O}-module of finite presentation.
Proof. The implication (1) \Rightarrow (2) follows from Modules on Sites, Lemma 18.23.4. For the converse, assume \mathcal{F} of finite presentation. We may and do assume that \mathcal{C} = \mathcal{C}'. By Lemma 91.11.2 we have a short exact sequence
0 \to \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{F}' \to \mathcal{F} \to 0
Let U be an object of \mathcal{C} such that \mathcal{F}|_ U has a presentation
\mathcal{O}_ U^{\oplus m} \to \mathcal{O}_ U^{\oplus n} \to \mathcal{F}|_ U \to 0
After replacing U by the members of a covering we may assume the map \mathcal{O}_ U^{\oplus n} \to \mathcal{F}|_ U lifts to a map (\mathcal{O}'_ U)^{\oplus n} \to \mathcal{F}'|_ U. The induced map \mathcal{I}^{\oplus n} \to \mathcal{I} \otimes \mathcal{F} is surjective by right exactness of \otimes . Thus after replacing U by the members of a covering we can find a lift (\mathcal{O}'|_ U)^{\oplus m} \to (\mathcal{O}'|_ U)^{\oplus n} of the given map \mathcal{O}_ U^{\oplus m} \to \mathcal{O}_ U^{\oplus n} such that
(\mathcal{O}'_ U)^{\oplus m} \to (\mathcal{O}'_ U)^{\oplus n} \to \mathcal{F}'|_ U \to 0
is a complex. Using right exactness of \otimes once more it is seen that this complex is exact. \square
Lemma 91.11.4. Let (f, f') be a morphism of first order thickenings as in Situation 91.9.1. Let \mathcal{F}', \mathcal{G}' be \mathcal{O}'-modules and set \mathcal{F} = i^*\mathcal{F}' and \mathcal{G} = i^*\mathcal{G}'. Let \varphi : \mathcal{F} \to \mathcal{G} be an \mathcal{O}-linear map. Assume that \mathcal{G}' is flat over \mathcal{O}_{\mathcal{B}'} and that (f, f') is a strict morphism of thickenings. The set of lifts of \varphi to an \mathcal{O}'-linear map \varphi ' : \mathcal{F}' \to \mathcal{G}' is, if nonempty, a principal homogeneous space under
\mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(\mathcal{F}, \mathcal{G} \otimes _\mathcal {O} f^*\mathcal{J})
Lemma 91.11.5. Let i : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}') be a first order thickening of ringed topoi. Let \mathcal{F}', \mathcal{G}' be \mathcal{O}'-modules and set \mathcal{F} = i^*\mathcal{F}' and \mathcal{G} = i^*\mathcal{G}'. Let \varphi : \mathcal{F} \to \mathcal{G} be an \mathcal{O}-linear map. There exists an element
o(\varphi ) \in \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}')
whose vanishing is a necessary and sufficient condition for the existence of a lift of \varphi to an \mathcal{O}'-linear map \varphi ' : \mathcal{F}' \to \mathcal{G}'.
Proof. It is clear from the proof of Lemma 91.11.1 that the vanishing of the boundary of \varphi via the map
\mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(\mathcal{F}, \mathcal{G}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}'}(\mathcal{F}', \mathcal{G}) \longrightarrow \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}'}(\mathcal{F}', \mathcal{I}\mathcal{G}')
is a necessary and sufficient condition for the existence of a lift. We conclude as
\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}'}(\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}')
the adjointness of i_* = Ri_* and Li^* on the derived category (Cohomology on Sites, Lemma 21.19.1). \square
Lemma 91.11.6. Let (f, f') be a morphism of first order thickenings as in Situation 91.9.1. Let \mathcal{F}', \mathcal{G}' be \mathcal{O}'-modules and set \mathcal{F} = i^*\mathcal{F}' and \mathcal{G} = i^*\mathcal{G}'. Let \varphi : \mathcal{F} \to \mathcal{G} be an \mathcal{O}-linear map. Assume that \mathcal{F}' and \mathcal{G}' are flat over \mathcal{O}_{\mathcal{B}'} and that (f, f') is a strict morphism of thickenings. There exists an element
o(\varphi ) \in \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(\mathcal{F}, \mathcal{G} \otimes _\mathcal {O} f^*\mathcal{J})
whose vanishing is a necessary and sufficient condition for the existence of a lift of \varphi to an \mathcal{O}'-linear map \varphi ' : \mathcal{F}' \to \mathcal{G}'.
First proof. This follows from Lemma 91.11.5 as we claim that under the assumptions of the lemma we have
\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(\mathcal{F}, \mathcal{G} \otimes _\mathcal {O} f^*\mathcal{J})
Namely, we have \mathcal{I}\mathcal{G}' = \mathcal{G} \otimes _\mathcal {O} f^*\mathcal{J} by Lemma 91.11.2. On the other hand, observe that
H^{-1}(Li^*\mathcal{F}') = \text{Tor}_1^{\mathcal{O}'}(\mathcal{F}', \mathcal{O})
(local computation omitted). Using the short exact sequence
0 \to \mathcal{I} \to \mathcal{O}' \to \mathcal{O} \to 0
we see that this \text{Tor}_1 is computed by the kernel of the map \mathcal{I} \otimes _\mathcal {O} \mathcal{F} \to \mathcal{I}\mathcal{F}' which is zero by the final assertion of Lemma 91.11.2. Thus \tau _{\geq -1}Li^*\mathcal{F}' = \mathcal{F}. On the other hand, we have
\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(\tau _{\geq -1}Li^*\mathcal{F}', \mathcal{I}\mathcal{G}')
by the dual of Derived Categories, Lemma 13.16.1. \square
Second proof. We can apply Lemma 91.10.2 as follows. Note that \mathcal{K} = \mathcal{I} \otimes _\mathcal {O} \mathcal{F} and \mathcal{L} = \mathcal{I} \otimes _\mathcal {O} \mathcal{G} by Lemma 91.11.2, that c_{\mathcal{F}'} = 1 \otimes 1 and c_{\mathcal{G}'} = 1 \otimes 1 and taking \psi = 1 \otimes \varphi the diagram of the lemma commutes. Thus o(\varphi ) = o(\varphi , 1 \otimes \varphi ) works. \square
Lemma 91.11.7. Let (f, f') be a morphism of first order thickenings as in Situation 91.9.1. Let \mathcal{F} be an \mathcal{O}-module. Assume (f, f') is a strict morphism of thickenings and \mathcal{F} flat over \mathcal{O}_\mathcal {B}. If there exists a pair (\mathcal{F}', \alpha ) consisting of an \mathcal{O}'-module \mathcal{F}' flat over \mathcal{O}_{\mathcal{B}'} and an isomorphism \alpha : i^*\mathcal{F}' \to \mathcal{F}, then the set of isomorphism classes of such pairs is principal homogeneous under \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( \mathcal{F}, \mathcal{I} \otimes _\mathcal {O} \mathcal{F}).
Proof. If we assume there exists one such module, then the canonical map
f^*\mathcal{J} \otimes _\mathcal {O} \mathcal{F} \to \mathcal{I} \otimes _\mathcal {O} \mathcal{F}
is an isomorphism by Lemma 91.11.2. Apply Lemma 91.10.3 with \mathcal{K} = \mathcal{I} \otimes _\mathcal {O} \mathcal{F} and c = 1. By Lemma 91.11.2 the corresponding extensions \mathcal{F}' are all flat over \mathcal{O}_{\mathcal{B}'}. \square
Lemma 91.11.8. Let (f, f') be a morphism of first order thickenings as in Situation 91.9.1. Let \mathcal{F} be an \mathcal{O}-module. Assume (f, f') is a strict morphism of thickenings and \mathcal{F} flat over \mathcal{O}_\mathcal {B}. There exists an \mathcal{O}'-module \mathcal{F}' flat over \mathcal{O}_{\mathcal{B}'} with i^*\mathcal{F}' \cong \mathcal{F}, if and only if
the canonical map f^*\mathcal{J} \otimes _\mathcal {O} \mathcal{F} \to \mathcal{I} \otimes _\mathcal {O} \mathcal{F} is an isomorphism, and
the class o(\mathcal{F}, \mathcal{I} \otimes _\mathcal {O} \mathcal{F}, 1) \in \mathop{\mathrm{Ext}}\nolimits ^2_\mathcal {O}( \mathcal{F}, \mathcal{I} \otimes _\mathcal {O} \mathcal{F}) of Lemma 91.10.4 is zero.
Proof. This follows immediately from the characterization of \mathcal{O}'-modules flat over \mathcal{O}_{\mathcal{B}'} of Lemma 91.11.2 and Lemma 91.10.4. \square
Comments (2)
Comment #2955 by Ko Aoki on
Comment #3082 by Johan on