Suppose is simple and algebraic over , so that is the minimal polynomial for . If is an intermediate field, then is also simple and algebraic over , so that is the minimal polynomial. Moreover, is a factor of .

will turn out to completely determine the intermediate field , and since only has finitely many factors in , this proves the forward direction. In fact, will be generated over by the coëfficients of .

To show this, let be the subfield of generated by the coëfficients and . Then , and since is irreducible in the extension field , so too is it irreducible in . Since and , it follows (see tower of field extensions)

so , i.e. , as required.

For the converse, assume that there are only finitely many intermediate fields . The extension must be finitely generated, for otherwise the infinite sequence of subextensions

would give infinitely many intermediate fields. If is a Galois field so is , and then by Finite extension of a Galois field is simple.

Let us consider the case is infinite, where we need to show that every finitely generated algebraic extension is simple.

Arguing inductively, we may assume w.l.o.g. that . For every , we have the intermediate field

But there are only finitely many such intermediate extensions. Since is infinite, we must have

for some in , whence

so . Thus , and we are done.