Corollary

— Let E be a separably closed extension of K and F a separable algebraic extension of K ; then there exists a K-homomorphism of F into E.

Let Ω be an algebraic closure of E ; we have $\Omega_s \subset E$ and it suffices to treat the case where $E = \Omega_s$. Since F is an algebraic extension of K, there exists a K-homomorphism u of F into Ω (V, p. 20, Th. 1). Since the field $u(F)$ is separable over K, we have $u(F) \subset \Omega_s$ and u defines a K-homomorphism of F into $\Omega_s = E$.