Proposition 4

— Let $K$ be a field, $L$ a separable algebraic extension of $K$ and $V$ a vector space over $L$.

a) Every $K$-derivation of $L$ into $V$ is zero.

b) If $A$ is a derivation of $K$ into $V$, there exists a unique derivation $D$ of $L$ into $V$ which extends $A$.

Let $D$ be a $K$-derivation of $L$ into $V$. If $E$ is a subextension of $L$ of finite degree over $K$, then the $K$-algebra $E$ is etale, and so $\Omega_K(E) = 0$ ($V$, p. 33, Th. 3), whence $D|_E = 0$ by the universal property of $\Omega_K(E)$ (III, p. 569). Since $L$ is the union of a family of subextensions of finite degree over $K$, we have $D = 0$, whence a).

Let $A$ be a derivation of $K$ into $V$. If $D'$ and $D''$ are two extensions of $A$ to a derivation of $L$ into $V$, then the difference $D' - D''$ is a $K$-derivation of $L$ into $V$; hence it is zero by a) and so $D' = D''$.

It remains to prove the existence of an extension of $A$. Zorn's Lemma (E, III, p. 20) implies the existence of a maximal extension $D_0$ of $A$ to a derivation defined on a subfield $L_0$ of $L$ containing $K$.

Let $x$ be in $L$ and $g$ the minimal polynomial of $x$ over $L_0$. Since $L$ is algebraic and separable over $K$, $x$ is algebraic and separable over $L_0$ ($V$, p. 39, Prop. 6 and p. 39, Cor. 2); therefore $x$ is a simple root of $g$ ($V$, p. 39, Prop. 5), whence $g'(x) \neq 0$ (IV, p. 17, Prop. 7). If we define $g^{D_0}(x)$ as in $V$, p. 127, there exists thus an element $u$ of $V$ such that $g^{D_0}(x) + g'(x) \cdot u = 0$; by Prop. 1 (V, p. 127) there exists a derivation $D$ of $L_0(x)$ into $V$ extending $D_0$ and such that $D(x) = u$. In view of the maximal character of $(L_0, D_0)$ we thus have $L_0(x) = L_0$, whence $x \in L_0$. Since $x$ was arbitrary we conclude that $L_0 = L$.