Proposition 14

— Let Ω be an algebraically closed extension of K.

a) The relative separable algebraic closure Ωs of K in Ω is a separable closure of K.*

b) If E and E' are two separable closures of K, there exists a K-isomorphism of E onto E'.

Let F be a separable algebraic extension of Ω_s ; since Ω is algebraically closed, there exists an 0,-homomorphism u of F into Ω (V, p. 20, Th. 1). By Prop. 9 (V, p. 42) $u(F)$ is separable over K, hence $u(F) = \Omega_s$. Therefore F is a trivial extension of $\Omega_s$ and so $\Omega_s$ is separably closed, whence a).

Let E be a separable closure of K. Since E is an algebraic extension of K, there exists a K-homomorphism v of E into Ω (V, p. 20, Th. 1). Hence $v(E)$ is separable algebraic over K, whence $v(E) \subset \Omega_s$. By V, p. 42, Prop. 9, $\Omega_s$ is separable over $v(E)$ and since the field $v(E)$ is separably closed, we have $v(E) = \Omega_s$. It follows that v is a K-isomorphism of E onto $\Omega_s$. Now b) is an immediate consequence.