Universal algebraic differentials and A-derivations

Let $A\stackrel{\phi }{\to }B$ be a ring homomorphism of commutative rings (Commutative ring), so that $B$ is an $A$-algebra. Let $F$ be the free $B$-module on the set underlying $B$ (Unital left and right modules over a ring; unqualified module means left module), with basis symbol $[b]$ for $b\in B$, and let $R\subseteq F$ be the $B$-submodule generated by all elements

$[b+b^{\mathrm{\prime }}]-[b]-[b^{\mathrm{\prime }}],[b{b}^{\mathrm{\prime }}]-b[b^{\mathrm{\prime }}]-b^{\mathrm{\prime }}[b],[\phi (a)]$

with $b,b^{\mathrm{\prime }}\in B$ and $a\in A$. The module of algebraic differentials of $B$ over $A$, also called the module of Kähler differentials, is the quotient $B$-module

${\mathrm{\Omega }}_{B\mathrm{/}A}:=F\mathrm{/}R,$

and the map $\mathrm{d}={\mathrm{d}}_{B\mathrm{/}A}\text{\hspace{0.17em}⁣}:B\to {\mathrm{\Omega }}_{B\mathrm{/}A}$, $\mathrm{d}b:=[b]+R$, is the universal $A$-derivation of $B$ over $A$. By construction $\mathrm{d}$ is additive, satisfies the Leibniz rule

$\mathrm{d}(b{b}^{\mathrm{\prime }})=b\mathrm{d}{b}^{\mathrm{\prime }}+b^{\mathrm{\prime }}\mathrm{d}b(b,b^{\mathrm{\prime }}\in B),$

and is $A$-constant, meaning $\mathrm{d}(\phi (a))=0$ for every $a\in A$; these are exactly the three families of relators above.

Now let $M$ be a $B$-module. An $A$-derivation of $B$ into $M$ is a map $D\text{\hspace{0.17em}⁣}:B\to M$ that is additive, is $A$-constant in the sense that $D(\phi (a))=0$ for all $a\in A$, and satisfies the Leibniz rule $D(b{b}^{\mathrm{\prime }})=bD(b^{\mathrm{\prime }})+b^{\mathrm{\prime }}D(b)$. The set of such maps is written ${\mathrm{Der}}_A(B,M)$; it is a $B$-module under the pointwise operations, since $M$ is a $B$-module.

Three conventions are part of the definition. First, ${\mathrm{\Omega }}_{B\mathrm{/}A}$ is presented by the whole set $B$ of generators, so no finiteness of $B$ over $A$ and no finite generation, finite presentation or Noetherian hypothesis is assumed anywhere in this definition. Second, $\mathrm{d}1=0$, because $1_B=\phi (1_A)$ makes $[1_B]$ one of the relators; consequently $\mathrm{d}(-b)=-\mathrm{d}b$, $\mathrm{d}(b^n)=n{b}^{n-1}\mathrm{d}b$ for $n\ge 1$ by induction on the Leibniz rule, and $\mathrm{d}(ab)=a\mathrm{d}b$ for $a\in A$, $b\in B$. Third, the extreme case $B=A$ (with $\phi$ the identity) gives ${\mathrm{\Omega }}_{A\mathrm{/}A}=0$: every $[b]$ is the relator $[\phi (b)]$, so the quotient is the zero module and the only $A$-derivation of $A$ into any $A$-module is the zero map.