Corollary 3

— Let E be an extension of a field K and D a derivation of E into E, mapping K into K. If L is a subextension of E which is algebraic and separable over K, then $D(L) \subset L$.

Let A be the derivation of K into L which agrees with D on K. By Prop. 4 (V, p. 129) there exists a derivation D' of L into L extending A. Now we may consider D' and the restriction D'' of D to L as derivations of $L$ into E ; since they agree on K, we have $D' = D''$ by Prop. 4, whence