The Stacks project

90.7 Deformations of ringed spaces and the naive cotangent complex

In this section we use the naive cotangent complex to do a little bit of deformation theory. We start with a first order thickening $t : (S, \mathcal{O}_ S) \to (S', \mathcal{O}_{S'})$ of ringed spaces. We denote $\mathcal{J} = \mathop{\mathrm{Ker}}(t^\sharp )$ and we identify the underlying topological spaces of $S$ and $S'$. Moreover we assume given a morphism of ringed spaces $f : (X, \mathcal{O}_ X) \to (S, \mathcal{O}_ S)$, an $\mathcal{O}_ X$-module $\mathcal{G}$, and an $f$-map $c : \mathcal{J} \to \mathcal{G}$ of sheaves of modules (Sheaves, Definition 6.21.7 and Section 6.26). In this section we ask ourselves whether we can find the question mark fitting into the following diagram
\begin{equation} \label{defos-equation-to-solve-ringed-spaces} \vcenter { \xymatrix{ 0 \ar[r] & \mathcal{G} \ar[r] & {?} \ar[r] & \mathcal{O}_ X \ar[r] & 0 \\ 0 \ar[r] & \mathcal{J} \ar[u]^ c \ar[r] & \mathcal{O}_{S'} \ar[u] \ar[r] & \mathcal{O}_ S \ar[u] \ar[r] & 0 } } \end{equation}

(where the vertical arrows are $f$-maps) and moreover how unique the solution is (if it exists). More precisely, we look for a first order thickening $i : (X, \mathcal{O}_ X) \to (X', \mathcal{O}_{X'})$ and a morphism of thickenings $(f, f')$ as in ( where $\mathop{\mathrm{Ker}}(i^\sharp )$ is identified with $\mathcal{G}$ such that $(f')^\sharp $ induces the given map $c$. We will say $X'$ is a solution to (

Lemma 90.7.1. Assume given a commutative diagram of morphisms of ringed spaces
\begin{equation} \label{defos-equation-huge-1} \vcenter { \xymatrix{ & (X_2, \mathcal{O}_{X_2}) \ar[r]_{i_2} \ar[d]_{f_2} \ar[ddl]_ g & (X'_2, \mathcal{O}_{X'_2}) \ar[d]^{f'_2} \\ & (S_2, \mathcal{O}_{S_2}) \ar[r]^{t_2} \ar[ddl]|\hole & (S'_2, \mathcal{O}_{S'_2}) \ar[ddl] \\ (X_1, \mathcal{O}_{X_1}) \ar[r]_{i_1} \ar[d]_{f_1} & (X'_1, \mathcal{O}_{X'_1}) \ar[d]^{f'_1} \\ (S_1, \mathcal{O}_{S_1}) \ar[r]^{t_1} & (S'_1, \mathcal{O}_{S'_1}) } } \end{equation}

whose horizontal arrows are first order thickenings. Set $\mathcal{G}_ j = \mathop{\mathrm{Ker}}(i_ j^\sharp )$ and assume given a $g$-map $\nu : \mathcal{G}_1 \to \mathcal{G}_2$ of modules giving rise to the commutative diagram
\begin{equation} \label{defos-equation-huge-2} \vcenter { \xymatrix{ & 0 \ar[r] & \mathcal{G}_2 \ar[r] & \mathcal{O}_{X'_2} \ar[r] & \mathcal{O}_{X_2} \ar[r] & 0 \\ & 0 \ar[r]|\hole & \mathcal{J}_2 \ar[u]_{c_2} \ar[r] & \mathcal{O}_{S'_2} \ar[u] \ar[r]|\hole & \mathcal{O}_{S_2} \ar[u] \ar[r] & 0 \\ 0 \ar[r] & \mathcal{G}_1 \ar[ruu] \ar[r] & \mathcal{O}_{X'_1} \ar[r] & \mathcal{O}_{X_1} \ar[ruu] \ar[r] & 0 \\ 0 \ar[r] & \mathcal{J}_1 \ar[ruu]|\hole \ar[u]^{c_1} \ar[r] & \mathcal{O}_{S'_1} \ar[ruu]|\hole \ar[u] \ar[r] & \mathcal{O}_{S_1} \ar[ruu]|\hole \ar[u] \ar[r] & 0 } } \end{equation}

with front and back solutions to (

  1. There exist a canonical element in $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_{X_2}}(Lg^*\mathop{N\! L}\nolimits _{X_1/S_1}, \mathcal{G}_2)$ whose vanishing is a necessary and sufficient condition for the existence of a morphism of ringed spaces $X'_2 \to X'_1$ fitting into ( compatibly with $\nu $.

  2. If there exists a morphism $X'_2 \to X'_1$ fitting into ( compatibly with $\nu $ the set of all such morphisms is a principal homogeneous space under

    \[ \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X_1}}(\Omega _{X_1/S_1}, g_*\mathcal{G}_2) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X_2}}(g^*\Omega _{X_1/S_1}, \mathcal{G}_2) = \mathop{\mathrm{Ext}}\nolimits ^0_{\mathcal{O}_{X_2}}(Lg^*\mathop{N\! L}\nolimits _{X_1/S_1}, \mathcal{G}_2). \]

Proof. The naive cotangent complex $\mathop{N\! L}\nolimits _{X_1/S_1}$ is defined in Modules, Definition 17.31.6. The equalities in the last statement of the lemma follow from the fact that $g^*$ is adjoint to $g_*$, the fact that $H^0(\mathop{N\! L}\nolimits _{X_1/S_1}) = \Omega _{X_1/S_1}$ (by construction of the naive cotangent complex) and the fact that $Lg^*$ is the left derived functor of $g^*$. Thus we will work with the groups $\mathop{\mathrm{Ext}}\nolimits ^ k_{\mathcal{O}_{X_2}}(Lg^*\mathop{N\! L}\nolimits _{X_1/S_1}, \mathcal{G}_2)$, $k = 0, 1$ in the rest of the proof. We first argue that we can reduce to the case where the underlying topological spaces of all ringed spaces in the lemma is the same.

To do this, observe that $g^{-1}\mathop{N\! L}\nolimits _{X_1/S_1}$ is equal to the naive cotangent complex of the homomorphism of sheaves of rings $g^{-1}f_1^{-1}\mathcal{O}_{S_1} \to g^{-1}\mathcal{O}_{X_1}$, see Modules, Lemma 17.31.3. Moreover, the degree $0$ term of $\mathop{N\! L}\nolimits _{X_1/S_1}$ is a flat $\mathcal{O}_{X_1}$-module, hence the canonical map

\[ Lg^*\mathop{N\! L}\nolimits _{X_1/S_1} \longrightarrow g^{-1}\mathop{N\! L}\nolimits _{X_1/S_1} \otimes _{g^{-1}\mathcal{O}_{X_1}} \mathcal{O}_{X_2} \]

induces an isomorphism on cohomology sheaves in degrees $0$ and $-1$. Thus we may replace the Ext groups of the lemma with

\[ \mathop{\mathrm{Ext}}\nolimits ^ k_{g^{-1}\mathcal{O}_{X_1}}(g^{-1}\mathop{N\! L}\nolimits _{X_1/S_1}, \mathcal{G}_2) = \mathop{\mathrm{Ext}}\nolimits ^ k_{g^{-1}\mathcal{O}_{X_1}}( \mathop{N\! L}\nolimits _{g^{-1}\mathcal{O}_{X_1}/g^{-1}f_1^{-1}\mathcal{O}_{S_1}}, \mathcal{G}_2) \]

The set of morphism of ringed spaces $X'_2 \to X'_1$ fitting into ( compatibly with $\nu $ is in one-to-one bijection with the set of homomorphisms of $g^{-1}f_1^{-1}\mathcal{O}_{S'_1}$-algebras $g^{-1}\mathcal{O}_{X'_1} \to \mathcal{O}_{X'_2}$ which are compatible with $f^\sharp $ and $\nu $. In this way we see that we may assume we have a diagram ( of sheaves on $X$ and we are looking to find a homomorphism of sheaves of rings $\mathcal{O}_{X'_1} \to \mathcal{O}_{X'_2}$ fitting into it.

In the rest of the proof of the lemma we assume all underlying topological spaces are the same, i.e., we have a diagram ( of sheaves on a space $X$ and we are looking for homomorphisms of sheaves of rings $\mathcal{O}_{X'_1} \to \mathcal{O}_{X'_2}$ fitting into it. As ext groups we will use $\mathop{\mathrm{Ext}}\nolimits ^ k_{\mathcal{O}_{X_1}}( \mathop{N\! L}\nolimits _{\mathcal{O}_{X_1}/\mathcal{O}_{S_1}}, \mathcal{G}_2)$, $k = 0, 1$.

Step 1. Construction of the obstruction class. Consider the sheaf of sets

\[ \mathcal{E} = \mathcal{O}_{X'_1} \times _{\mathcal{O}_{X_2}} \mathcal{O}_{X'_2} \]

This comes with a surjective map $\alpha : \mathcal{E} \to \mathcal{O}_{X_1}$ and hence we can use $\mathop{N\! L}\nolimits (\alpha )$ instead of $\mathop{N\! L}\nolimits _{\mathcal{O}_{X_1}/\mathcal{O}_{S_1}}$, see Modules, Lemma 17.31.2. Set

\[ \mathcal{I}' = \mathop{\mathrm{Ker}}(\mathcal{O}_{S'_1}[\mathcal{E}] \to \mathcal{O}_{X_1}) \quad \text{and}\quad \mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_{S_1}[\mathcal{E}] \to \mathcal{O}_{X_1}) \]

There is a surjection $\mathcal{I}' \to \mathcal{I}$ whose kernel is $\mathcal{J}_1\mathcal{O}_{S'_1}[\mathcal{E}]$. We obtain two homomorphisms of $\mathcal{O}_{S'_2}$-algebras

\[ a : \mathcal{O}_{S'_1}[\mathcal{E}] \to \mathcal{O}_{X'_1} \quad \text{and}\quad b : \mathcal{O}_{S'_1}[\mathcal{E}] \to \mathcal{O}_{X'_2} \]

which induce maps $a|_{\mathcal{I}'} : \mathcal{I}' \to \mathcal{G}_1$ and $b|_{\mathcal{I}'} : \mathcal{I}' \to \mathcal{G}_2$. Both $a$ and $b$ annihilate $(\mathcal{I}')^2$. Moreover $a$ and $b$ agree on $\mathcal{J}_1\mathcal{O}_{S'_1}[\mathcal{E}]$ as maps into $\mathcal{G}_2$ because the left hand square of ( is commutative. Thus the difference $b|_{\mathcal{I}'} - \nu \circ a|_{\mathcal{I}'}$ induces a well defined $\mathcal{O}_{X_1}$-linear map

\[ \xi : \mathcal{I}/\mathcal{I}^2 \longrightarrow \mathcal{G}_2 \]

which sends the class of a local section $f$ of $\mathcal{I}$ to $a(f') - \nu (b(f'))$ where $f'$ is a lift of $f$ to a local section of $\mathcal{I}'$. We let $[\xi ] \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_{X_1}}(\mathop{N\! L}\nolimits (\alpha ), \mathcal{G}_2)$ be the image (see below).

Step 2. Vanishing of $[\xi ]$ is necessary. Let us write $\Omega = \Omega _{\mathcal{O}_{S_1}[\mathcal{E}]/\mathcal{O}_{S_1}} \otimes _{\mathcal{O}_{S_1}[\mathcal{E}]} \mathcal{O}_{X_1}$. Observe that $\mathop{N\! L}\nolimits (\alpha ) = (\mathcal{I}/\mathcal{I}^2 \to \Omega )$ fits into a distinguished triangle

\[ \Omega [0] \to \mathop{N\! L}\nolimits (\alpha ) \to \mathcal{I}/\mathcal{I}^2[1] \to \Omega [1] \]

Thus we see that $[\xi ]$ is zero if and only if $\xi $ is a composition $\mathcal{I}/\mathcal{I}^2 \to \Omega \to \mathcal{G}_2$ for some map $\Omega \to \mathcal{G}_2$. Suppose there exists a homomorphisms of sheaves of rings $\varphi : \mathcal{O}_{X'_1} \to \mathcal{O}_{X'_2}$ fitting into ( In this case consider the map $\mathcal{O}_{S'_1}[\mathcal{E}] \to \mathcal{G}_2$, $f' \mapsto b(f') - \varphi (a(f'))$. A calculation shows this annihilates $\mathcal{J}_1\mathcal{O}_{S'_1}[\mathcal{E}]$ and induces a derivation $\mathcal{O}_{S_1}[\mathcal{E}] \to \mathcal{G}_2$. The resulting linear map $\Omega \to \mathcal{G}_2$ witnesses the fact that $[\xi ] = 0$ in this case.

Step 3. Vanishing of $[\xi ]$ is sufficient. Let $\theta : \Omega \to \mathcal{G}_2$ be a $\mathcal{O}_{X_1}$-linear map such that $\xi $ is equal to $\theta \circ (\mathcal{I}/\mathcal{I}^2 \to \Omega )$. Then a calculation shows that

\[ b + \theta \circ d : \mathcal{O}_{S'_1}[\mathcal{E}] \to \mathcal{O}_{X'_2} \]

annihilates $\mathcal{I}'$ and hence defines a map $\mathcal{O}_{X'_1} \to \mathcal{O}_{X'_2}$ fitting into (

Proof of (2) in the special case above. Omitted. Hint: This is exactly the same as the proof of (2) of Lemma 90.2.1. $\square$

Lemma 90.7.2. Let $X$ be a topological space. Let $\mathcal{A} \to \mathcal{B}$ be a homomorphism of sheaves of rings. Let $\mathcal{G}$ be a $\mathcal{B}$-module. Let $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {B}(\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}}, \mathcal{G})$. There exists a map of sheaves of sets $\alpha : \mathcal{E} \to \mathcal{B}$ such that $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {B}(\mathop{N\! L}\nolimits (\alpha ), \mathcal{G})$ is the class of a map $\mathcal{I}/\mathcal{I}^2 \to \mathcal{G}$ (see proof for notation).

Proof. Recall that given $\alpha : \mathcal{E} \to \mathcal{B}$ such that $\mathcal{A}[\mathcal{E}] \to \mathcal{B}$ is surjective with kernel $\mathcal{I}$ the complex $\mathop{N\! L}\nolimits (\alpha ) = (\mathcal{I}/\mathcal{I}^2 \to \Omega _{\mathcal{A}[\mathcal{E}]/\mathcal{A}} \otimes _{\mathcal{A}[\mathcal{E}]} \mathcal{B})$ is canonically isomorphic to $\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}}$, see Modules, Lemma 17.31.2. Observe moreover, that $\Omega = \Omega _{\mathcal{A}[\mathcal{E}]/\mathcal{A}} \otimes _{\mathcal{A}[\mathcal{E}]} \mathcal{B}$ is the sheaf associated to the presheaf $U \mapsto \bigoplus _{e \in \mathcal{E}(U)} \mathcal{B}(U)$. In other words, $\Omega $ is the free $\mathcal{B}$-module on the sheaf of sets $\mathcal{E}$ and in particular there is a canonical map $\mathcal{E} \to \Omega $.

Having said this, pick some $\mathcal{E}$ (for example $\mathcal{E} = \mathcal{B}$ as in the definition of the naive cotangent complex). The obstruction to writing $\xi $ as the class of a map $\mathcal{I}/\mathcal{I}^2 \to \mathcal{G}$ is an element in $\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {B}(\Omega , \mathcal{G})$. Say this is represented by the extension $0 \to \mathcal{G} \to \mathcal{H} \to \Omega \to 0$ of $\mathcal{B}$-modules. Consider the sheaf of sets $\mathcal{E}' = \mathcal{E} \times _\Omega \mathcal{H}$ which comes with an induced map $\alpha ' : \mathcal{E}' \to \mathcal{B}$. Let $\mathcal{I}' = \mathop{\mathrm{Ker}}(\mathcal{A}[\mathcal{E}'] \to \mathcal{B})$ and $\Omega ' = \Omega _{\mathcal{A}[\mathcal{E}']/\mathcal{A}} \otimes _{\mathcal{A}[\mathcal{E}']} \mathcal{B}$. The pullback of $\xi $ under the quasi-isomorphism $\mathop{N\! L}\nolimits (\alpha ') \to \mathop{N\! L}\nolimits (\alpha )$ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {B}(\Omega ', \mathcal{G})$ because the pullback of the extension $\mathcal{H}$ by the map $\Omega ' \to \Omega $ is split as $\Omega '$ is the free $\mathcal{B}$-module on the sheaf of sets $\mathcal{E}'$ and since by construction there is a commutative diagram

\[ \xymatrix{ \mathcal{E}' \ar[r] \ar[d] & \mathcal{E} \ar[d] \\ \mathcal{H} \ar[r] & \Omega } \]

This finishes the proof. $\square$

Lemma 90.7.3. If there exists a solution to (, then the set of isomorphism classes of solutions is principal homogeneous under $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S}, \mathcal{G})$.

Proof. We observe right away that given two solutions $X'_1$ and $X'_2$ to ( we obtain by Lemma 90.7.1 an obstruction element $o(X'_1, X'_2) \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S}, \mathcal{G})$ to the existence of a map $X'_1 \to X'_2$. Clearly, this element is the obstruction to the existence of an isomorphism, hence separates the isomorphism classes. To finish the proof it therefore suffices to show that given a solution $X'$ and an element $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S}, \mathcal{G})$ we can find a second solution $X'_\xi $ such that $o(X', X'_\xi ) = \xi $.

Pick $\alpha : \mathcal{E} \to \mathcal{O}_ X$ as in Lemma 90.7.2 for the class $\xi $. Consider the surjection $f^{-1}\mathcal{O}_ S[\mathcal{E}] \to \mathcal{O}_ X$ with kernel $\mathcal{I}$ and corresponding naive cotangent complex $\mathop{N\! L}\nolimits (\alpha ) = (\mathcal{I}/\mathcal{I}^2 \to \Omega _{f^{-1}\mathcal{O}_ S[\mathcal{E}]/f^{-1}\mathcal{O}_ S} \otimes _{f^{-1}\mathcal{O}_ S[\mathcal{E}]} \mathcal{O}_ X)$. By the lemma $\xi $ is the class of a morphism $\delta : \mathcal{I}/\mathcal{I}^2 \to \mathcal{G}$. After replacing $\mathcal{E}$ by $\mathcal{E} \times _{\mathcal{O}_ X} \mathcal{O}_{X'}$ we may also assume that $\alpha $ factors through a map $\alpha ' : \mathcal{E} \to \mathcal{O}_{X'}$.

These choices determine an $f^{-1}\mathcal{O}_{S'}$-algebra map $\varphi : \mathcal{O}_{S'}[\mathcal{E}] \to \mathcal{O}_{X'}$. Let $\mathcal{I}' = \mathop{\mathrm{Ker}}(\varphi )$. Observe that $\varphi $ induces a map $\varphi |_{\mathcal{I}'} : \mathcal{I}' \to \mathcal{G}$ and that $\mathcal{O}_{X'}$ is the pushout, as in the following diagram

\[ \xymatrix{ 0 \ar[r] & \mathcal{G} \ar[r] & \mathcal{O}_{X'} \ar[r] & \mathcal{O}_ X \ar[r] & 0 \\ 0 \ar[r] & \mathcal{I}' \ar[u]^{\varphi |_{\mathcal{I}'}} \ar[r] & f^{-1}\mathcal{O}_{S'}[\mathcal{E}] \ar[u] \ar[r] & \mathcal{O}_ X \ar[u]_{=} \ar[r] & 0 } \]

Let $\psi : \mathcal{I}' \to \mathcal{G}$ be the sum of the map $\varphi |_{\mathcal{I}'}$ and the composition

\[ \mathcal{I}' \to \mathcal{I}'/(\mathcal{I}')^2 \to \mathcal{I}/\mathcal{I}^2 \xrightarrow {\delta } \mathcal{G}. \]

Then the pushout along $\psi $ is an other ring extension $\mathcal{O}_{X'_\xi }$ fitting into a diagram as above. A calculation (omitted) shows that $o(X', X'_\xi ) = \xi $ as desired. $\square$

Lemma 90.7.4. Let $f : (X, \mathcal{O}_ X) \to (S, \mathcal{O}_ S)$ be a morphism of ringed spaces. Let $\mathcal{G}$ be a $\mathcal{O}_ X$-module. The set of isomorphism classes of extensions of $f^{-1}\mathcal{O}_ S$-algebras

\[ 0 \to \mathcal{G} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0 \]

where $\mathcal{G}$ is an ideal of square zero1 is canonically bijective to $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S}, \mathcal{G})$.

Proof. To prove this we apply the previous results to the case where ( is given by the diagram

\[ \xymatrix{ 0 \ar[r] & \mathcal{G} \ar[r] & {?} \ar[r] & \mathcal{O}_ X \ar[r] & 0 \\ 0 \ar[r] & 0 \ar[u] \ar[r] & \mathcal{O}_ S \ar[u] \ar[r]^{\text{id}} & \mathcal{O}_ S \ar[u] \ar[r] & 0 } \]

Thus our lemma follows from Lemma 90.7.3 and the fact that there exists a solution, namely $\mathcal{G} \oplus \mathcal{O}_ X$. (See remark below for a direct construction of the bijection.) $\square$

Remark 90.7.5. Let $f : (X, \mathcal{O}_ X) \to (S, \mathcal{O}_ S)$ and $\mathcal{G}$ be as in Lemma 90.7.4. Consider an extension $0 \to \mathcal{G} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0$ as in the lemma. We can choose a sheaf of sets $\mathcal{E}$ and a commutative diagram

\[ \xymatrix{ \mathcal{E} \ar[d]_{\alpha '} \ar[rd]^\alpha \\ \mathcal{O}_{X'} \ar[r] & \mathcal{O}_ X } \]

such that $f^{-1}\mathcal{O}_ S[\mathcal{E}] \to \mathcal{O}_ X$ is surjective with kernel $\mathcal{J}$. (For example you can take any sheaf of sets surjecting onto $\mathcal{O}_{X'}$.) Then

\[ \mathop{N\! L}\nolimits _{X/S} \cong \mathop{N\! L}\nolimits (\alpha ) = \left( \mathcal{J}/\mathcal{J}^2 \longrightarrow \Omega _{f^{-1}\mathcal{O}_ S[\mathcal{E}]/f^{-1}\mathcal{O}_ S} \otimes _{f^{-1}\mathcal{O}_ S[\mathcal{E}]} \mathcal{O}_ X\right) \]

See Modules, Section 17.31 and in particular Lemma 17.31.2. Of course $\alpha '$ determines a map $f^{-1}\mathcal{O}_ S[\mathcal{E}] \to \mathcal{O}_{X'}$ which in turn determines a map

\[ \mathcal{J}/\mathcal{J}^2 \longrightarrow \mathcal{G} \]

which in turn determines the element of $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits (\alpha ), \mathcal{G}) = \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S}, \mathcal{G})$ corresponding to $\mathcal{O}_{X'}$ by the bijection of the lemma.

Lemma 90.7.6. Let $f : (X, \mathcal{O}_ X) \to (S, \mathcal{O}_ S)$ and $g : (Y, \mathcal{O}_ Y) \to (X, \mathcal{O}_ X)$ be morphisms of ringed spaces. Let $\mathcal{F}$ be a $\mathcal{O}_ X$-module. Let $\mathcal{G}$ be a $\mathcal{O}_ Y$-module. Let $c : \mathcal{F} \to \mathcal{G}$ be a $g$-map. Finally, consider

  1. $0 \to \mathcal{F} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0$ an extension of $f^{-1}\mathcal{O}_ S$-algebras corresponding to $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S}, \mathcal{F})$, and

  2. $0 \to \mathcal{G} \to \mathcal{O}_{Y'} \to \mathcal{O}_ Y \to 0$ an extension of $g^{-1}f^{-1}\mathcal{O}_ S$-algebras corresponding to $\zeta \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ Y}(\mathop{N\! L}\nolimits _{Y/S}, \mathcal{G})$.

See Lemma 90.7.4. Then there is an $S$-morphism $g' : Y' \to X'$ compatible with $g$ and $c$ if and only if $\xi $ and $\zeta $ map to the same element of $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ Y}(Lg^*\mathop{N\! L}\nolimits _{X/S}, \mathcal{G})$.

Proof. The stament makes sense as we have the maps

\[ \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S}, \mathcal{F}) \to \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ Y}(Lg^*\mathop{N\! L}\nolimits _{X/S}, Lg^*\mathcal{F}) \to \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ Y}(Lg^*\mathop{N\! L}\nolimits _{X/S}, \mathcal{G}) \]

using the map $Lg^*\mathcal{F} \to g^*\mathcal{F} \xrightarrow {c} \mathcal{G}$ and

\[ \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ Y}(\mathop{N\! L}\nolimits _{Y/S}, \mathcal{G}) \to \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ Y}(Lg^*\mathop{N\! L}\nolimits _{X/S}, \mathcal{G}) \]

using the map $Lg^*\mathop{N\! L}\nolimits _{X/S} \to \mathop{N\! L}\nolimits _{Y/S}$. The statement of the lemma can be deduced from Lemma 90.7.1 applied to the diagram

\[ \xymatrix{ & 0 \ar[r] & \mathcal{G} \ar[r] & \mathcal{O}_{Y'} \ar[r] & \mathcal{O}_ Y \ar[r] & 0 \\ & 0 \ar[r]|\hole & 0 \ar[u] \ar[r] & \mathcal{O}_ S \ar[u] \ar[r]|\hole & \mathcal{O}_ S \ar[u] \ar[r] & 0 \\ 0 \ar[r] & \mathcal{F} \ar[ruu] \ar[r] & \mathcal{O}_{X'} \ar[r] & \mathcal{O}_ X \ar[ruu] \ar[r] & 0 \\ 0 \ar[r] & 0 \ar[ruu]|\hole \ar[u] \ar[r] & \mathcal{O}_ S \ar[ruu]|\hole \ar[u] \ar[r] & \mathcal{O}_ S \ar[ruu]|\hole \ar[u] \ar[r] & 0 } \]

and a compatibility between the constructions in the proofs of Lemmas 90.7.4 and 90.7.1 whose statement and proof we omit. (See remark below for a direct argument.) $\square$

Remark 90.7.7. Let $f : (X, \mathcal{O}_ X) \to (S, \mathcal{O}_ S)$, $g : (Y, \mathcal{O}_ Y) \to (X, \mathcal{O}_ X)$, $\mathcal{F}$, $\mathcal{G}$, $c : \mathcal{F} \to \mathcal{G}$, $0 \to \mathcal{F} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0$, $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S}, \mathcal{F})$, $0 \to \mathcal{G} \to \mathcal{O}_{Y'} \to \mathcal{O}_ Y \to 0$, and $\zeta \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ Y}(\mathop{N\! L}\nolimits _{Y/S}, \mathcal{G})$ be as in Lemma 90.7.6. Using pushout along $c : g^{-1}\mathcal{F} \to \mathcal{G}$ we can construct an extension

\[ \xymatrix{ 0 \ar[r] & \mathcal{G} \ar[r] & \mathcal{O}'_1 \ar[r] & g^{-1}\mathcal{O}_ X \ar[r] & 0 \\ 0 \ar[r] & g^{-1}\mathcal{F} \ar[u]^ c \ar[r] & g^{-1}\mathcal{O}_{X'} \ar[u] \ar[r] & g^{-1}\mathcal{O}_ X \ar@{=}[u] \ar[r] & 0 } \]

Using pullback along $g^\sharp : g^{-1}\mathcal{O}_ X \to \mathcal{O}_ Y$ we can construct an extension

\[ \xymatrix{ 0 \ar[r] & \mathcal{G} \ar[r] & \mathcal{O}_{Y'} \ar[r] & \mathcal{O}_ Y \ar[r] & 0 \\ 0 \ar[r] & \mathcal{G} \ar@{=}[u] \ar[r] & \mathcal{O}'_2 \ar[u] \ar[r] & g^{-1}\mathcal{O}_ X \ar[u] \ar[r] & 0 } \]

A diagram chase tells us that there exists an $S$-morphism $Y' \to X'$ compatible with $g$ and $c$ if and only if $\mathcal{O}'_1$ is isomorphic to $\mathcal{O}'_2$ as $g^{-1}f^{-1}\mathcal{O}_ S$-algebra extensions of $g^{-1}\mathcal{O}_ X$ by $\mathcal{G}$. By Lemma 90.7.4 these extensions are classified by the LHS of

\[ \mathop{\mathrm{Ext}}\nolimits ^1_{g^{-1}\mathcal{O}_ X}( \mathop{N\! L}\nolimits _{g^{-1}\mathcal{O}_ X/g^{-1}f^{-1}\mathcal{O}_ S}, \mathcal{G}) = \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ Y}(Lg^*\mathop{N\! L}\nolimits _{X/S}, \mathcal{G}) \]

Here the equality comes from tensor-hom adjunction and the equalities

\[ \mathop{N\! L}\nolimits _{g^{-1}\mathcal{O}_ X/g^{-1}f^{-1}\mathcal{O}_ S} = g^{-1}\mathop{N\! L}\nolimits _{X/S} \quad \text{and}\quad Lg^*\mathop{N\! L}\nolimits _{X/S} = g^{-1}\mathop{N\! L}\nolimits _{X/S} \otimes _{g^{-1}\mathcal{O}_ X}^\mathbf {L} \mathcal{O}_ Y \]

For the first of these see Modules, Lemma 17.31.3; the second follows from the definition of derived pullback. Thus, in order to see that Lemma 90.7.6 is true, it suffices to show that $\mathcal{O}'_1$ corresponds to the image of $\xi $ and that $\mathcal{O}'_2$ correspond to the image of $\zeta $. The correspondence between $\xi $ and $\mathcal{O}'_1$ is immediate from the construction of the class $\xi $ in Remark 90.7.5. For the correspondence between $\zeta $ and $\mathcal{O}'_2$, we first choose a commutative diagram

\[ \xymatrix{ \mathcal{E} \ar[d]_{\beta '} \ar[rd]^\beta \\ \mathcal{O}_{Y'} \ar[r] & \mathcal{O}_ Y } \]

such that $g^{-1}f^{-1}\mathcal{O}_ S[\mathcal{E}] \to \mathcal{O}_ Y$ is surjective with kernel $\mathcal{K}$. Next choose a commutative diagram

\[ \xymatrix{ \mathcal{E} \ar[d]_{\beta '} & \mathcal{E}' \ar[l]^\varphi \ar[d]_{\alpha '} \ar[rd]^\alpha \\ \mathcal{O}_{Y'} & \mathcal{O}'_2 \ar[l] \ar[r] & g^{-1}\mathcal{O}_ X } \]

such that $g^{-1}f^{-1}\mathcal{O}_ S[\mathcal{E}'] \to g^{-1}\mathcal{O}_ X$ is surjective with kernel $\mathcal{J}$. (For example just take $\mathcal{E}' = \mathcal{E} \amalg \mathcal{O}'_2$ as a sheaf of sets.) The map $\varphi $ induces a map of complexes $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\beta )$ (notation as in Modules, Section 17.31) and in particular $\bar\varphi : \mathcal{J}/\mathcal{J}^2 \to \mathcal{K}/\mathcal{K}^2$. Then $\mathop{N\! L}\nolimits (\alpha ) \cong \mathop{N\! L}\nolimits _{Y/S}$ and $\mathop{N\! L}\nolimits (\beta ) \cong \mathop{N\! L}\nolimits _{g^{-1}\mathcal{O}_ X/g^{-1}f^{-1}\mathcal{O}_ S}$ and the map of complexes $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\beta )$ represents the map $Lg^*\mathop{N\! L}\nolimits _{X/S} \to \mathop{N\! L}\nolimits _{Y/S}$ used in the statement of Lemma 90.7.6 (see first part of its proof). Now $\zeta $ corresponds to the class of the map $\mathcal{K}/\mathcal{K}^2 \to \mathcal{G}$ induced by $\beta '$, see Remark 90.7.5. Similarly, the extension $\mathcal{O}'_2$ corresponds to the map $\mathcal{J}/\mathcal{J}^2 \to \mathcal{G}$ induced by $\alpha '$. The commutative diagram above shows that this map is the composition of the map $\mathcal{K}/\mathcal{K}^2 \to \mathcal{G}$ induced by $\beta '$ with the map $\bar\varphi : \mathcal{J}/\mathcal{J}^2 \to \mathcal{K}/\mathcal{K}^2$. This proves the compatibility we were looking for.

Lemma 90.7.8. Let $t : (S, \mathcal{O}_ S) \to (S', \mathcal{O}_{S'})$, $\mathcal{J} = \mathop{\mathrm{Ker}}(t^\sharp )$, $f : (X, \mathcal{O}_ X) \to (S, \mathcal{O}_ S)$, $\mathcal{G}$, and $c : \mathcal{J} \to \mathcal{G}$ be as in ( Denote $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ S}(\mathop{N\! L}\nolimits _{S/S'}, \mathcal{J})$ the element corresponding to the extension $\mathcal{O}_{S'}$ of $\mathcal{O}_ S$ by $\mathcal{J}$ via Lemma 90.7.4. The set of isomorphism classes of solutions is canonically bijective to the fibre of

\[ \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S'}, \mathcal{G}) \to \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Lf^*\mathop{N\! L}\nolimits _{S/S'}, \mathcal{G}) \]

over the image of $\xi $.

Proof. By Lemma 90.7.4 applied to $X \to S'$ and the $\mathcal{O}_ X$-module $\mathcal{G}$ we see that elements $\zeta $ of $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S'}, \mathcal{G})$ parametrize extensions $0 \to \mathcal{G} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0$ of $f^{-1}\mathcal{O}_{S'}$-algebras. By Lemma 90.7.6 applied to $X \to S \to S'$ and $c : \mathcal{J} \to \mathcal{G}$ we see that there is an $S'$-morphism $X' \to S'$ compatible with $c$ and $f : X \to S$ if and only if $\zeta $ maps to $\xi $. Of course this is the same thing as saying $\mathcal{O}_{X'}$ is a solution of ( $\square$

Remark 90.7.9. In the situation of Lemma 90.7.8 we have maps of complexes

\[ Lf^*\mathop{N\! L}\nolimits _{S'/S} \to \mathop{N\! L}\nolimits _{X/S'} \to \mathop{N\! L}\nolimits _{X/S} \]

These maps are closed to forming a distinguished triangle, see Modules, Lemma 17.31.7. If it were a distinguished triangle we would conclude that the image of $\xi $ in $\mathop{\mathrm{Ext}}\nolimits ^2_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S}, \mathcal{G})$ would be the obstruction to the existence of a solution to (

[1] In other words, the set of isomorphism classes of first order thickenings $i : X \to X'$ over $S$ endowed with an isomorphism $\mathcal{G} \to \mathop{\mathrm{Ker}}(i^\sharp )$ of $\mathcal{O}_ X$-modules.

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