Proposition 10

— *Let E and K' be two subextensions of the same extension of K and let $E' = K'(E)$. Suppose that E is algebraic over K, hence E' algebraic over K' (V, p. 18, Cor. 2).

a) If E is separable over K, then E' is separable over K'.

b) Conversely if E' is separable over K' and E and K' are linearly disjoint over K, then E is separable over K.

Assertion a) follows directly from Prop. 6 (V, p. 39).

Under the hypothesis b), let F be a subextension of E of finite degree over K. Then F and K' are linearly disjoint over K, hence the K'-algebra $F_{(K')} = K' \otimes_K F$ is isomorphic to $K'(F)$. Since $K'(F)$ is a subextension of E' of finite degree over K' and E' is algebraic and separable over K', the K'-algebra $K'(F)$ is etale. In other words, the K'-algebra $F_{(K')}$ is etale, and now Cor. 2 of Prop. 4 (V, p. 32) shows that F is etale over K. Thus we have shown E to be separable over K.