Suppose 𝐹 =𝐾(𝛼) is simple and algebraic over 𝐾, so that 𝑞𝐾(𝑥) ∈𝐾[𝑥] is the minimal polynomial for 𝛼. If 𝐸 is an intermediate field, then 𝐹 =𝐸(𝛼) is also simple and algebraic over 𝐸, so that 𝑞𝐸(𝑥) ∈𝐸[𝑥] is the minimal polynomial. Moreover, 𝑞𝐸(𝑥) is a factor of 𝑞𝐾(𝑥).

𝑞𝐸(𝑥) will turn out to completely determine the intermediate field 𝐸, and since 𝑞𝐾(𝑥) only has finitely many factors in ――𝐾[𝑥], this proves the forward direction. In fact, 𝐸 will be generated over 𝐾 by the coëfficients of 𝑞𝐸(𝑥).

To show this, let 𝐸′ be the subfield of 𝐸 generated by the coëfficients and 𝐾. Then 𝑞𝐸(𝑥) ∈𝐸′[𝑥], and since 𝑞𝐸(𝑥) is irreducible in the extension field 𝐸′, so too is it irreducible in 𝐸′. Since 𝐸′(𝛼) =𝐹 =𝐸(𝛼) and 𝐹 :𝐸 :𝐸′ :𝐾, it follows (see tower of field extensions)

so [𝐸 :𝐸′] =1, i.e. 𝐸 =𝐸′, as required.

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

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

Let us consider the case 𝐾 is infinite, where we need to show that every finitely generated algebraic extension 𝐹 =𝐾(𝛼1,…,𝛼𝑟) is simple.

Arguing inductively, we may assume w.l.o.g. that 𝐹 =𝐾(𝛼,𝛽). For every 𝑐 ∈𝐾, we have the intermediate field

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

for some 𝑐′ ≠𝑐 in 𝐾, whence

so 𝐾(𝛼,𝛽) ⊆𝐾(𝑐𝛼 +𝛽). Thus 𝐾(𝑐𝛼 +𝛽) =𝐾(𝛼,𝛽), and we are done.