The Stacks project

Lemma 114.20.3. In the situation of Remark 114.20.2 assume that $\mathcal{F}$ is flat over $U$. Then the vanishing of the class $\xi _{U'}$ is a necessary and sufficient condition for the existence of a $\mathcal{O}_{X \times _ B U'}$-module $\mathcal{F}'$ flat over $U'$ with $i^*\mathcal{F}' \cong \mathcal{F}$.

Proof (sketch). We will use the criterion of Deformation Theory, Lemma 90.11.8. We will abbreviate $\mathcal{O} = \mathcal{O}_{X \times _ B U}$ and $\mathcal{O}' = \mathcal{O}_{X \times _ B U'}$. Consider the short exact sequence

\[ 0 \to \mathcal{I} \to \mathcal{O}_{U'} \to \mathcal{O}_ U \to 0. \]

Let $\mathcal{J} \subset \mathcal{O}'$ be the quasi-coherent sheaf of ideals cutting out $X \times _ B U$. By the above we obtain an exact sequence

\[ \text{Tor}_1^{\mathcal{O}_ B}(\mathcal{O}_ X, \mathcal{O}_ U) \to q^*\mathcal{I} \to \mathcal{J} \to 0 \]

where the $\text{Tor}_1^{\mathcal{O}_ B}(\mathcal{O}_ X, \mathcal{O}_ U)$ is an abbreviation for

\[ \text{Tor}_1^{h^{-1}\mathcal{O}_ B}(p^{-1}\mathcal{O}_ X, q^{-1}\mathcal{O}_ U) \otimes _{(p^{-1}\mathcal{O}_ X\otimes _{h^{-1}\mathcal{O}_ B}q^{-1}\mathcal{O}_ U)} \mathcal{O}. \]

Tensoring with $\mathcal{F}$ we obtain the exact sequence

\[ \mathcal{F} \otimes _\mathcal {O} \text{Tor}_1^{\mathcal{O}_ B}(\mathcal{O}_ X, \mathcal{O}_ U) \to \mathcal{F} \otimes _\mathcal {O} q^*\mathcal{I} \to \mathcal{F} \otimes _\mathcal {O} \mathcal{J} \to 0 \]

(Note that the roles of the letters $\mathcal{I}$ and $\mathcal{J}$ are reversed relative to the notation in Deformation Theory, Lemma 90.11.8.) Condition (1) of the lemma is that the last map above is an isomorphism, i.e., that the first map is zero. The vanishing of this map may be checked on stalks at geometric points $\overline{z} = (\overline{x}, \overline{u}) : \mathop{\mathrm{Spec}}(k) \to X \times _ B U$. Set $R = \mathcal{O}_{B, \overline{b}}$, $A = \mathcal{O}_{X, \overline{x}}$, $B = \mathcal{O}_{U, \overline{u}}$, and $C = \mathcal{O}_{\overline{z}}$. By Cotangent, Lemma 91.28.2 and the defining triangle for $E(\mathcal{F})$ we see that

\[ H^{-2}(E(\mathcal{F}))_{\overline{z}} = \mathcal{F}_{\overline{z}} \otimes \text{Tor}_1^ R(A, B) \]

The map $\xi _{U'}$ therefore induces a map

\[ \mathcal{F}_{\overline{z}} \otimes \text{Tor}_1^ R(A, B) \longrightarrow \mathcal{F}_{\overline{z}} \otimes _ B \mathcal{I}_{\overline{u}} \]

We claim this map is the same as the stalk of the map described above (proof omitted; this is a purely ring theoretic statement). Thus we see that condition (1) of Deformation Theory, Lemma 90.11.8 is equivalent to the vanishing $H^{-2}(\xi _{U'}) : H^{-2}(E(\mathcal{F})) \to \mathcal{F} \otimes \mathcal{I}$.

To finish the proof we show that, assuming that condition (1) is satisfied, condition (2) is equivalent to the vanishing of $\xi _{U'}$. In the rest of the proof we write $\mathcal{F} \otimes \mathcal{I}$ to denote $\mathcal{F} \otimes _\mathcal {O} q^*\mathcal{I} = \mathcal{F} \otimes _\mathcal {O} \mathcal{J}$. A consideration of the spectral sequence

\[ \mathop{\mathrm{Ext}}\nolimits ^ i(H^{-j}(E(\mathcal{F})), \mathcal{F} \otimes \mathcal{I}) \Rightarrow \mathop{\mathrm{Ext}}\nolimits ^{i + j}(E(\mathcal{F}), \mathcal{F} \otimes \mathcal{I}) \]

using that $H^0(E(\mathcal{F})) = \mathcal{F}$ and $H^{-1}(E(\mathcal{F})) = 0$ shows that there is an exact sequence

\[ 0 \to \mathop{\mathrm{Ext}}\nolimits ^2(\mathcal{F}, \mathcal{F} \otimes \mathcal{I}) \to \mathop{\mathrm{Ext}}\nolimits ^2(E(\mathcal{F}), \mathcal{F} \otimes \mathcal{I}) \to \mathop{\mathrm{Hom}}\nolimits (H^{-2}(E(\mathcal{F})), \mathcal{F} \otimes \mathcal{I}) \]

Thus our element $\xi _{U'}$ is an element of $\mathop{\mathrm{Ext}}\nolimits ^2(\mathcal{F}, \mathcal{F} \otimes \mathcal{I})$. The proof is finished by showing this element agrees with the element of Deformation Theory, Lemma 90.11.8 a verification we omit. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 09DQ. Beware of the difference between the letter 'O' and the digit '0'.