Proposition 6

— Let E be an extension of K. a) If E is algebraic and separable over K, every element of E is algebraic and separable over K. b) Conversely, let A be a set of elements of E, algebraic and separable over K and such that $E = K(A)$; then E is algebraic and separable over K. If E is algebraic and separable over K, the same is true of the extension K(x) of K for every $x \in E$, whence a). Under the hypothesis b), the extension E is algebraic over K (V, p. 18, Cor. 1).

Let F be a subextension of E of finite degree over K. By V, p. 11, Cor., there exist elements $x_1, \ldots, x_m$ of A such that $F \subset K(x_1, \ldots, x_r)$ and we have

By the hypothesis on A, the algebras $K[x_1], \ldots, K[x_m]$ are etale over K; hence the same is true of $K[x_1] \otimes \cdots \otimes K[x_m]$ (V, p. 32, Cor. 1). Now F is isomorphic to a subalgebra of a quotient algebra of $K[x_1] \otimes \cdots \otimes K[x_m]$, and so is etale (V, p. 30, Prop. 3).