# Gaining insight in the definition of derivations over schemes

This question refers to Definition 34.1 of the chapter "morphisms of schemes" of the Stacks Project. In particular, let $f: X \rightarrow S$ be a morphism of schemes and $F$ an $O_X$-module. An $S$-derivation of $F$ is a map $D:O_X \rightarrow F$ such that it is additive, it satisfies the Leibniz rule and it annihilates the image of the morphism $f^{-1} O_S \rightarrow O_X$. Suppose in particular that $S$ is the spectrum of field $k$. Then what is the meaning/interpretation of $D$ annihilating the image of $f^{-1} O_{Spec(k)} \rightarrow O_X$? I know that $O_{Spec(k)} \cong k$ but still i can not connect this definition to the algebraic definition in the context of derivations over a ring and modules of differentials, which i understand.

## 1 Answer

It seems to me that you already understand the meaning here, but hopefully this will be helpful.

The image of the map $f^{-1}O_S \rightarrow O_X$ identifies the "constants" that a derivation will vanish on; combined with the Leibniz rule this ensures that the derivation is $S$-linear. When $S$ is the spectrum of a field $k$ this image is indeed $k$, and when $X$ is an affine variety (so we can think about derivations as "partial differentiation followed by evaluation at a point") we get the familiar fact that the derivative of a constant is 0.

• Thanks Thom, yes this confirms my intuition. Dec 5, 2012 at 15:22