The Stacks project

Typo.In the statement of this lemma, it should be "Given any section $s : \mathcal{O}_X \to \mathcal{A}$".

Thanks for pointing out the typos once more! They are fixed in this commit.

Instead of this lemma, one could prove the following more general result and obtain Lemma 17.28.11 as a Corollary:

Lemma. Let $X$ be a topological space, let $\mathcal{O}$ be a sheaf of rings over $X$ and let $\mathcal{A},\mathcal{B}$ be sheaves of $\mathcal{O}$-algebras. Suppose $\mathcal{I}\subset\mathcal{B}$ is an ideal sheaf with $\mathcal{I}^2=0$. Then $\mathcal{I}$ is naturally a $\mathcal{B}/\mathcal{I}$-module. Let $\varphi:\mathcal{A}\to\mathcal{B}/\mathcal{I}$ be a morphism of $\mathcal{O}$-algebras ($\mathcal{I}$ becomes an $\mathcal{A}$-module via $\varphi$). Denote $\operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})$ to the set of $\mathcal{O}$-algebra homomorphisms $\mathcal{A}\to\mathcal{B}$ that lift $\varphi$ to $\mathcal{B}$. Then $\operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})$ is a $\operatorname{Der}_\mathcal{O}(\mathcal{A},\mathcal{I})$-torsor via $$\begin{align*} \operatorname{Der}_\mathcal{O}(\mathcal{A},\mathcal{I})\times\operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})&\to\operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})\\ (D,\psi)&\mapsto D+\psi. \end{align*}$$

Proof. The map is well-defined: It is clear that $D+\psi$ is $\mathcal{O}$-linear, that it preserves the multiplicative unit and that as a map of sheaves of abelian groups, it's a lifting of $\varphi$ to $\mathcal{B}$. It is left to show that $D+\psi$ is multiplicative. For this, we need first to prove the following: Let $\psi\in\operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})$, and let $h,\tau$ be local sections respectively of $\mathcal{A}$ and of $\mathcal{I}$, over the same open subset $U\subset X$. Then we claim that $h\tau=\psi(h)\tau$, i.e., that $\varphi(h)\tau=\psi(h)\tau$. It suffices to verify the equality on germs at $x\in U$, and we leave this as an exercise to the reader (use the natural $\mathcal{B}_x/\mathcal{I}_x$-module structure on $\mathcal{I}_x$). Now, one has $$\begin{align*} (Df+\psi f)(Dg+\psi g) &=\psi(fg)+(\psi f)Dg+(\psi g)Df,&\text{for }\mathcal{I}^2=0,\\ &=\psi(fg)+fDg+gDf,&\text{by the claim,}\\ &=\psi(fg)+D(fg),&\text{by Leibniz's rule.} \end{align*}$$ Thus, $D+\psi\in \operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})$ and the map from the statement constitutes a group action of $\operatorname{Der}_\mathcal{O}(\mathcal{A},\mathcal{I})$ on $\operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})$. It is clear that the action is free. To show transitivity, let $\psi,\psi'\in\operatorname{Lift}_\mathcal{O}(\varphi,\mathcal{B})$. We have to show that $D=\psi'-\psi\in\operatorname{Der}_\mathcal{O}(\mathcal{A},\mathcal{I})$. On the one hand, it is clear that $D$ is $\mathcal{O}$-linear. On the other hand, using the same facts as in the last computation, we get $$\begin{align*} \psi'(fg)=\psi'(f)\psi'(g)&=(\psi(f)+D(f))(\psi(g)+D(g))\\ &=\psi(fg)+fD(g)+gD(f), \end{align*}$$ which shows Leibniz's rule for $D$. $\square$