Corollary 2

— Let E be a subextension of L and U the vector sub-L-space of $\Omega_K(L)$ generated by the differentials of elements of E. For an element x to be algebraic over E it is necessary and sufficient that $dx \in U$.

For each $y \in L$ let D(y) be the residue class of $dy$ mod U. Then D is an E-derivation of L into $\Omega_K(L)/U$. Since K is of characteristic 0, every algebraic extension of E is separable (V, p. 37, Cor.) ; if $x \in L$ is algebraic over E, we have $Dx = 0$ by Prop. 4 (V, p. 129), that is, $dx \in U$.

If $x \in L$ is transcendental over $E$, there exists an E-derivation $A$ of $E(x)$ into $L$ such that $\Delta(x) = 1$ ($V$, p. 128, Prop. 3); by Th. 1, $A$ extends to an E-derivation D of $L$ into L. Let $\varphi$ be the linear form on $\Omega_K(L)$ such that $D = \varphi \circ d$; we have $\varphi(dy) = 0$ for $y \in E$ and $\varphi(x) = 1$, whence $dx \notin U$.