Lemma 10.131.8. Let $\varphi : A \to B$ be a ring map.
If $S \subset A$ is a multiplicative subset mapping to invertible elements of $B$, then $\Omega _{B/A} = \Omega _{B/S^{-1}A}$.
If $S \subset B$ is a multiplicative subset then $S^{-1}\Omega _{B/A} = \Omega _{S^{-1}B/A}$.
Proof. To show the equality of (1) it is enough to show that any $A$-derivation $D : B \to M$ annihilates the elements $\varphi (s)^{-1}$. This is clear from the Leibniz rule applied to $1 = \varphi (s) \varphi (s)^{-1}$. To show (2) note that there is an obvious map $S^{-1}\Omega _{B/A} \to \Omega _{S^{-1}B/A}$. To show it is an isomorphism it is enough to show that there is a $A$-derivation $\text{d}'$ of $S^{-1}B$ into $S^{-1}\Omega _{B/A}$. To define it we simply set $\text{d}'(b/s) = (1/s)\text{d}b - (1/s^2)b\text{d}s$. Details omitted. $\square$
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