Commutative Ring Theory/Derivations

Proposition (alternative construction of the universal derivation):

Let $S$ be a unital $R$-algebra. Note that $S\otimes _{R}S$ becomes an $S$-module via the linear extension of the operation $t(s\otimes u):=(ts)\otimes u$. We then have a morphism of $S$-modules

$\phi :S\otimes _{R}S\to S,s\otimes t\mapsto s\cdot t$,

where the dot indicates the algebra multiplication of $S$. Set $I:=\ker \phi$ and $\Omega '_{S/R}:=I/I^{2}$. Then

$d':S\to \Omega '_{S/R},~~t\mapsto t\otimes 1-1\otimes t+I^{2}$

is a derivation, and we have an isomorphism $\Theta :\Omega _{S/R}\to \Omega '_{S/R}$ inducing a commutative diagram

Proof: Note first that $d'$ is a derivation. This takes some explaining. First, note that for arbitrary $\alpha ,\beta$ the element $\alpha \otimes \beta -\beta \otimes \alpha$ is in $I$. Moreover, from this follows that the element

${\displaystyle }$

is in $I^{2}$ for $\alpha ,\beta ,\gamma ,\delta \in S$ arbitrary.

Hence, from the universal property of $d:S\to \Omega _{S/R}$, we obtain a unique morphism of $S$-modules $\Theta :\Omega _{S/R}\to \Omega '_{S/R}$ that makes the diagram

commutative. We construct an inverse map to $\Theta$. Namely, on $B\times B$ we can define the map

$(\alpha ,\beta )\mapsto \alpha d\beta$ $\Box$