Exercise 8, § 16

Let $k$ be a field and $L$ an extension of $k$. Show that the following conditions are equivalent:

(i) $L$ is a separable algebraic extension of a finitely generated pure transcendental extension.

(ii) $L$ is a separable extension of $k$ and

(To prove (ii) ⇒ (i) we may proceed as follows: let x_1, ..., x_n in L be such that dx_1, ..., dx_n form a basis of $\Omega_k(L)$ and put K = k(x_1, ..., x_n). One first shows that tr. deg_k(K) = n and hence that K is a pure transcendental extension of k. Let L' be a finite algebraic extension of K contained in L. By comparing the canonical mapping $u : \Omega_k(K) \otimes_K L' \to \Omega_k(L')$ with the canonical mapping $\Omega_k(K) \otimes_K L \to \Omega_k(L)$, show that u is injective. Calculate [$\Omega_k(L') : L'$] (V, p. 134, Cor. 1). Deduce that u is bijective and hence that $\Omega_k(L') = 0$. Show that L' is separable over K; deduce that L is separable over K.)