Maps out of the module of differentials are the same as derivations.

Lemma 10.131.3. The module of differentials of $S$ over $R$ has the following universal property. The map

$\mathop{\mathrm{Hom}}\nolimits _ S(\Omega _{S/R}, M) \longrightarrow \text{Der}_ R(S, M), \quad \alpha \longmapsto \alpha \circ \text{d}$

is an isomorphism of functors.

Proof. By definition an $R$-derivation is a rule which associates to each $a \in S$ an element $D(a) \in M$. Thus $D$ gives rise to a map $[D] : \bigoplus S[a] \to M$. However, the conditions of being an $R$-derivation exactly mean that $[D]$ annihilates the image of the map in the displayed presentation of $\Omega _{S/R}$ above. $\square$

Comment #1219 by David Corwin on

Suggested slogan: Maps out of the module of differentials are the same as derivations

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