Lemma 91.5.1. Let $i : (X, \mathcal{O}_ X) \to (X', \mathcal{O}_{X'})$ be a first order thickening of ringed spaces. Let $\mathcal{F}'$, $\mathcal{G}'$ be $\mathcal{O}_{X'}$-modules. Set $\mathcal{F} = i^*\mathcal{F}'$ and $\mathcal{G} = i^*\mathcal{G}'$. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be an $\mathcal{O}_ X$-linear map. The set of lifts of $\varphi $ to an $\mathcal{O}_{X'}$-linear map $\varphi ' : \mathcal{F}' \to \mathcal{G}'$ is, if nonempty, a principal homogeneous space under $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{I}\mathcal{G}')$.
91.5 Infinitesimal deformations of modules on ringed spaces
Let $i : (X, \mathcal{O}_ X) \to (X', \mathcal{O}_{X'})$ be a first order thickening of ringed spaces. We freely use the notation introduced in Section 91.3. Let $\mathcal{F}'$ be an $\mathcal{O}_{X'}$-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 $\mathcal{O}_{X'}$-modules. Since $\mathcal{I}^2 = 0$ the $\mathcal{O}_{X'}$-module structure on $\mathcal{I}\mathcal{F}'$ comes from a unique $\mathcal{O}_ X$-module structure. Thus the sequence above is an extension as in (91.4.0.1). As a special case, if $\mathcal{F}' = \mathcal{O}_{X'}$ we have $i^*\mathcal{O}_{X'} = \mathcal{O}_ X$ and $\mathcal{I}\mathcal{O}_{X'} = \mathcal{I}$ and we recover the sequence of structure sheaves
\[ 0 \to \mathcal{I} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0 \]
Proof. This is a special case of Lemma 91.4.1 but we also give a direct proof. We have short exact sequences of modules
\[ 0 \to \mathcal{I} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0 \quad \text{and}\quad 0 \to \mathcal{I}\mathcal{G}' \to \mathcal{G}' \to \mathcal{G} \to 0 \]
and similarly for $\mathcal{F}'$. Since $\mathcal{I}$ has square zero the $\mathcal{O}_{X'}$-module structure on $\mathcal{I}$ and $\mathcal{I}\mathcal{G}'$ comes from a unique $\mathcal{O}_ X$-module structure. It follows that
\[ \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{I}\mathcal{G}') \quad \text{and}\quad \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{G}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) \]
The lemma now follows from the exact sequence
\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{I}\mathcal{G}') \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{G}') \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{G}) \]
see Homology, Lemma 12.5.8. $\square$
Lemma 91.5.2. Let $(f, f')$ be a morphism of first order thickenings of ringed spaces as in Situation 91.3.1. Let $\mathcal{F}'$ be an $\mathcal{O}_{X'}$-module and set $\mathcal{F} = i^*\mathcal{F}'$. Assume that $\mathcal{F}$ is flat over $S$ and that $(f, f')$ is a strict morphism of thickenings (Definition 91.3.2). Then the following are equivalent
$\mathcal{F}'$ is flat over $S'$, and
the canonical map $f^*\mathcal{J} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I}\mathcal{F}'$ is an isomorphism.
Moreover, in this case the maps
\[ f^*\mathcal{J} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I}\mathcal{F}' \]
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). We may check these conditions at stalks. Let $x \in X \subset X'$ be a point with image $s = f(x) \in S \subset S'$. Set $A' = \mathcal{O}_{S', s}$, $B' = \mathcal{O}_{X', x}$, $A = \mathcal{O}_{S, s}$, and $B = \mathcal{O}_{X, x}$. Then $A = A'/J$ and $B = B'/I$ for some square zero ideals. Since $(f, f')$ is a strict morphism of thickenings we have $I = JB'$. Let $M' = \mathcal{F}'_ x$ and $M = \mathcal{F}_ x$. Then $M'$ is a $B'$-module and $M$ is a $B$-module. Since $\mathcal{F} = i^*\mathcal{F}'$ we see that the kernel of the surjection $M' \to M$ is $IM' = JM'$. Thus we have a short exact sequence
\[ 0 \to JM' \to M' \to M \to 0 \]
Using Sheaves, Lemma 6.26.4 and Modules, Lemma 17.16.1 to identify stalks of pullbacks and tensor products we see that the stalk at $x$ of the canonical map of the lemma is the map
\[ (J \otimes _ A B) \otimes _ B M = J \otimes _ A M = J \otimes _{A'} M' \longrightarrow JM' \]
The assumption that $\mathcal{F}$ is flat over $S$ signifies that $M$ is a flat $A$-module.
Assume (1). Flatness implies $\text{Tor}_1^{A'}(M', A) = 0$ by Algebra, Lemma 10.75.8. This means $J \otimes _{A'} M' \to M'$ is injective by Algebra, Remark 10.75.9. Hence $J \otimes _ A M \to JM'$ is an isomorphism.
Assume (2). Then $J \otimes _{A'} M' \to M'$ is injective. Hence $\text{Tor}_1^{A'}(M', A) = 0$ by Algebra, Remark 10.75.9. Hence $M'$ is flat over $A'$ by Algebra, Lemma 10.99.8. $\square$
Lemma 91.5.3. Let $(f, f')$ be a morphism of first order thickenings as in Situation 91.3.1. Let $\mathcal{F}'$, $\mathcal{G}'$ be $\mathcal{O}_{X'}$-modules and set $\mathcal{F} = i^*\mathcal{F}'$ and $\mathcal{G} = i^*\mathcal{G}'$. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be an $\mathcal{O}_ X$-linear map. Assume that $\mathcal{G}'$ is flat over $S'$ and that $(f, f')$ is a strict morphism of thickenings. The set of lifts of $\varphi $ to an $\mathcal{O}_{X'}$-linear map $\varphi ' : \mathcal{F}' \to \mathcal{G}'$ is, if nonempty, a principal homogeneous space under
\[ \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G} \otimes _{\mathcal{O}_ X} f^*\mathcal{J}) \]
Lemma 91.5.4. Let $i : (X, \mathcal{O}_ X) \to (X', \mathcal{O}_{X'})$ be a first order thickening of ringed spaces. Let $\mathcal{F}'$, $\mathcal{G}'$ be $\mathcal{O}_{X'}$-modules and set $\mathcal{F} = i^*\mathcal{F}'$ and $\mathcal{G} = i^*\mathcal{G}'$. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be an $\mathcal{O}_ X$-linear map. There exists an element whose vanishing is a necessary and sufficient condition for the existence of a lift of $\varphi $ to an $\mathcal{O}_{X'}$-linear map $\varphi ' : \mathcal{F}' \to \mathcal{G}'$.
\[ o(\varphi ) \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') \]
Proof. It is clear from the proof of Lemma 91.5.1 that the vanishing of the boundary of $\varphi $ via the map
\[ \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{G}) \longrightarrow \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{I}\mathcal{G}') \]
is a necessary and sufficient condition for the existence of a lift. We conclude as
\[ \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') \]
the adjointness of $i_* = Ri_*$ and $Li^*$ on the derived category (Cohomology, Lemma 20.28.1). $\square$
Lemma 91.5.5. Let $(f, f')$ be a morphism of first order thickenings as in Situation 91.3.1. Let $\mathcal{F}'$, $\mathcal{G}'$ be $\mathcal{O}_{X'}$-modules and set $\mathcal{F} = i^*\mathcal{F}'$ and $\mathcal{G} = i^*\mathcal{G}'$. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be an $\mathcal{O}_ X$-linear map. Assume that $\mathcal{F}'$ and $\mathcal{G}'$ are flat over $S'$ and that $(f, f')$ is a strict morphism of thickenings. There exists an element whose vanishing is a necessary and sufficient condition for the existence of a lift of $\varphi $ to an $\mathcal{O}_{X'}$-linear map $\varphi ' : \mathcal{F}' \to \mathcal{G}'$.
\[ o(\varphi ) \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G} \otimes _{\mathcal{O}_ X} f^*\mathcal{J}) \]
First proof. This follows from Lemma 91.5.4 as we claim that under the assumptions of the lemma we have
\[ \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G} \otimes _{\mathcal{O}_ X} f^*\mathcal{J}) \]
Namely, we have $\mathcal{I}\mathcal{G}' = \mathcal{G} \otimes _{\mathcal{O}_ X} f^*\mathcal{J}$ by Lemma 91.5.2. On the other hand, observe that
\[ H^{-1}(Li^*\mathcal{F}') = \text{Tor}_1^{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{O}_ X) \]
(local computation omitted). Using the short exact sequence
\[ 0 \to \mathcal{I} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0 \]
we see that this $\text{Tor}_1$ is computed by the kernel of the map $\mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I}\mathcal{F}'$ which is zero by the final assertion of Lemma 91.5.2. Thus $\tau _{\geq -1}Li^*\mathcal{F}' = \mathcal{F}$. On the other hand, we have
\[ \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\tau _{\geq -1}Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') \]
by the dual of Derived Categories, Lemma 13.16.1. $\square$
Second proof. We can apply Lemma 91.4.2 as follows. Note that $\mathcal{K} = \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F}$ and $\mathcal{L} = \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{G}$ by Lemma 91.5.2, 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 91.5.6. Let $(f, f')$ be a morphism of first order thickenings as in Situation 91.3.1. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. Assume $(f, f')$ is a strict morphism of thickenings and $\mathcal{F}$ flat over $S$. If there exists a pair $(\mathcal{F}', \alpha )$ consisting of an $\mathcal{O}_{X'}$-module $\mathcal{F}'$ flat over $S'$ and an isomorphism $\alpha : i^*\mathcal{F}' \to \mathcal{F}$, then the set of isomorphism classes of such pairs is principal homogeneous under $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}( \mathcal{F}, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F})$.
Proof. If we assume there exists one such module, then the canonical map
\[ f^*\mathcal{J} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \]
is an isomorphism by Lemma 91.5.2. Apply Lemma 91.4.3 with $\mathcal{K} = \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F}$ and $c = 1$. By Lemma 91.5.2 the corresponding extensions $\mathcal{F}'$ are all flat over $S'$. $\square$
Lemma 91.5.7. Let $(f, f')$ be a morphism of first order thickenings as in Situation 91.3.1. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. Assume $(f, f')$ is a strict morphism of thickenings and $\mathcal{F}$ flat over $S$. There exists an $\mathcal{O}_{X'}$-module $\mathcal{F}'$ flat over $S'$ with $i^*\mathcal{F}' \cong \mathcal{F}$, if and only if
the canonical map $ f^*\mathcal{J} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F}$ is an isomorphism, and
the class $o(\mathcal{F}, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F}, 1) \in \mathop{\mathrm{Ext}}\nolimits ^2_{\mathcal{O}_ X}( \mathcal{F}, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F})$ of Lemma 91.4.4 is zero.
Proof. This follows immediately from the characterization of $\mathcal{O}_{X'}$-modules flat over $S'$ of Lemma 91.5.2 and Lemma 91.4.4. $\square$
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