# Commutative Ring Theory/Derivations

Proposition (alternative construction of the universal derivation):

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

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

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

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

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

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

${\displaystyle }$

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

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

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

${\displaystyle (\alpha ,\beta )\mapsto \alpha d\beta }$ ${\displaystyle \Box }$