Proposition 3

— Let E and F be two extension fields of a field K such that K ⊂ E ⊂ F. For F to be algebraic over K it is necessary and sufficient that E should be algebraic over K and F algebraic over E.

The condition is necessary, by V, p. 17, Cor. 2. Let us show that it is sufficient. Let x be any element of F; it is algebraic over E; let g ∈ E[X] be its minimal polynomial over E. If A is the (finite) set of coefficients of g, then g ∈ K(A)[X], hence x is algebraic over K(A) and K(A U {x}) = K(A)(x) is of finite degree over K(A). Now A ⊂ E and E is algebraic over K, hence K(A) is of finite degree over K, by Th. 2. Therefore (V, p. 10, Th. 1), K(A U {x}) is of finite degree over K, which shows x to be algebraic over K (V, p. 18, Prop. 2).