Proposition 9

— Let F be an algebraic extension of K and E a subextension of F. For F to be separable over K it is necessary and sufficient that F should be separable over E and E separable over K.

Suppose first that F is separable over K; then E is separable over K by Prop. 1 (V, p. 36). Moreover, every element of F is separable over K (V, p. 39, Prop. 6) hence over E (V, p. 39, Cor. 2) and so F is separable over E (V, p. 39, Prop. 6).

Conversely, suppose that F is separable over E and E separable over K. Denote by x an element of F and by $f \in E[X]$ the minimal polynomial of x over E. Since E is algebraic over K, Th. 2 (V, p. 18) shows that there exists a subextension E' of E of finite degree over K such that $f \in E'[X]$; then f is at the same time the minimal polynomial of x over E and over E', and since x is separable over E (V, p. 39, Prop. 6) it is so also over E' (V, p. 39, Prop. 5). Write $F' = E'(x)$; then F is separable and of finite degree over E', and since E is separable over K, E' is separable and of finite degree over K (V, p. 36, Prop. 1). Hence F is separable and of finite degree over K, by V, p. 32, Cor. 2. Therefore (V, p. 39, Prop. 6) x is separable over K. We have now shown that every element of F is separable over K, hence F is separable over K (V, p. 39, Prop. 6).