Let F/K have a finite number of intermediate fields. Choose u in F so that the dimension of K(u) over K is maximal. Suppose this remains a proper subfield of F, and draw v from F-K(u). Consider all intermediate fields K(u+av) where a is taken from K. Since K is infinite, there must be a distinct pair a and b such that K(u+av) = K(u+bv). It follows that (u+av)-(u+bv) = (a-b)v is in K(u+av). Divide by a-b, multiply by a, and subtract from u+av to get u. And since we have u, we also have v. Thus K(u+av) properly contains K(u), and pulls in v, which contradicts the selection of u. There must be some u with K(u) = F, and F is a simple extension.
Conversely, assume F = K(u). Let u satisfy the monic irreducible polynomial p(x). If E is an intermediate extension, factor p(x) in E, and find the irreducible factor q(x) such that q(u) = 0. Adjoin the coefficients of q to K to obtain a subfield of E. Call this subfield L, and suppose L is properly contained in E. Of course q remains irreducible over L, and q(u) is still 0. The dimension of L(u) over L is the degree of q. Yet, the dimension of E(u) over E is also the degree of q, and L(u) and E(u) are both F. Therefore E and L have the same dimension over K, and since one contains the other, they are equal. Our subfield E is generated by the coefficients of the irreducible polynomial q(x).
The polynomial p(x) factors uniquely in the ring F[x]. These prime factors clump together in finitely many ways to build intermediate polynomials q(x). Therefore there are finitely many intermediate fields.
As a corollary, separable finite extensions are simple. Let F/K be such an extension and let E/K be its normal closure. Now E is galois and finite, with galois group G. Intermediate extensions correspond to the subgroups of G, and there are finitely many of these. Therefore there are finitely many extensions between F and K, and F/K is a simple extension K(u) for some primitive element u.
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