Lemma 29.32.4. Let $R \to A$ be a ring map. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules on $X = \mathop{\mathrm{Spec}}(A)$. Set $S = \mathop{\mathrm{Spec}}(R)$. The rule which associates to an $S$-derivation on $\mathcal{F}$ its action on global sections defines a bijection between the set of $S$-derivations of $\mathcal{F}$ and the set of $R$-derivations on $M = \Gamma (X, \mathcal{F})$.
Proof. Let $D : A \to M$ be an $R$-derivation. We have to show there exists a unique $S$-derivation on $\mathcal{F}$ which gives rise to $D$ on global sections. Let $U = D(f) \subset X$ be a standard affine open. Any element of $\Gamma (U, \mathcal{O}_ X)$ is of the form $a/f^ n$ for some $a \in A$ and $n \geq 0$. By the Leibniz rule we have
\[ D(a)|_ U = a/f^ n D(f^ n)|_ U + f^ n D(a/f^ n) \]
in $\Gamma (U, \mathcal{F})$. Since $f$ acts invertibly on $\Gamma (U, \mathcal{F})$ this completely determines the value of $D(a/f^ n) \in \Gamma (U, \mathcal{F})$. This proves uniqueness. Existence follows by simply defining
\[ D(a/f^ n) := (1/f^ n) D(a)|_ U - a/f^{2n} D(f^ n)|_ U \]
and proving this has all the desired properties (on the basis of standard opens of $X$). Details omitted. $\square$
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