Proposition 5

*Let $K$ be a field and $L$ an extension of $K$ generated by an element $x$ such that $x \notin K,\ x^p \in K$. Let $V$ be a vector space over $L$ and $A$ a derivation of $K$ into $V$ such that $\Delta(x^p) = 0$. Then there exists a unique derivation $D$ of $L$ into $V$ extending $A$ and such that $D(x) = 0$.

By $V$, p. 24, Prop. 1, $\{1,\ x,\ \ldots,\ x^{p-1}\}$ is a basis of $L$ over $K$. For every element $u = a_0 + a_1x + \ldots + a_{p-1}x^{p-1}$ of $L$ (with $a_0,\ \ldots,\ a_{p-1}$ in $K$) we put $D(u) = A(a_0) + x\Delta(a_1) + \ldots + x^{p-1}\Delta(a_{p-1})$. It is clear that $D$ extends $A$ and satisfies (1); so it is enough to establish the relation

when $u$ is of the form $ax^i$ and $v$ of the form $bx^j$ with $a,\ b$ in $K,\ 0 \leq i < p$ and $0 \leq j < p$.

When $i + j < p$ we have $uv = x^{i+j}ab$, whence $D(uv) = x^{i+j}\Delta(ab)$. Otherwise we have $0 \leq i + j - p < p$ and so $uv = x^{i+j-p}(abx^p)$, with $abx^p \in K$, but since $A(x^p) = 0$, we have $A(abx^p) = x^pA(ab)$, whence again $D(uv) = x^{i+j}\Delta(ab)$. So we have in all cases