Proposition 5

Suppose that $\mathbf{A}$ is an integral domain and let $\mathbf{L}$ be its field of fractions. Every derivation of $\mathbf{A}$ into a vector space $\mathbf{E}$ over $\mathbf{L}$ (considered as an $\mathbf{A}$-module) can be extended uniquely to a derivation of $\mathbf{L}$ into $\mathbf{E}$.

Let $d$ be a derivation of $\mathbf{A}$ into $\mathbf{E}$ and $\bar{d}$ a derivation of $\mathbf{L}$ into $\mathbf{E}$ extending $d$; then, for $u \in \mathbf{A}, v \in \mathbf{A}, v \neq 0$, of necessity, by virtue of (19),

which proves the uniqueness of $\bar{d}$. Conversely, we show that $\bar{d}$ can be defined by formula (20); it must first be verified that if $u/v = u'/v'$ the value of the right hand side of (20) does not change when $u$ is replaced by $u'$ and $v$ by $v'$. Now, $uv' = vu'$, hence $v'(du) + u(dv') = v(du') + u'(dv)$ and therefore $v'(du - uv^{-1}dv) = v(du' - u'{v'}^{-1}dv')$, since $uv'v^{-1} = u'$ and $u'{v'}^{-1}v = u$. Thus a mapping $\bar{d}: \mathbf{L} \to \mathbf{E}$ has been defined which extends $d$; it is immediately verified that it is $\mathbf{K}$-linear and a derivation.