Suppose \(F = K[\alpha_1, \cdots, \alpha_n]\) where \(\alpha_1^{n_1} \in K\) for some \(n_1\) and for each \(i\) we have \(\alpha_i^{n_i} \in K[\alpha_1, \cdots \alpha_{i-1}]\) for some powers \(n_i\). We want to show that \(F = E[\beta_1, \cdots \beta_m]\) where each \(\beta_i\) satisfies a similar condition.
Solution
Theorem.
Let \(F=K(\alpha_1,\dots,\alpha_n)\) be
radical over \(K\), and let \(K\subseteq E\subseteq F\). Then there exist
\(\beta_1,\dots,\beta_m\in F\) such
that \[
F=E(\beta_1,\dots,\beta_m),\qquad \beta_r^{N_r}\in
E(\beta_1,\dots,\beta_{r-1})\text{ for some }N_r>0.
\]
Proof. 1. Define \[ S:=\{\alpha_i:\alpha_i\notin E\}. \] Then \(F=E(S)\) since \(K\subseteq E\) and adjoining exactly the generators outside \(E\) produces \(F\).
List \(S\) in increasing index order: \[ \alpha_{i_1},\alpha_{i_2},\dots,\alpha_{i_m},\qquad i_1<\cdots<i_m. \] Then \[ F=E(\alpha_{i_1},\dots,\alpha_{i_m}). \]
Let \[ E_0:=E,\qquad E_r:=E_{r-1}(\alpha_{i_r}),\ r=1,\dots,m. \] 3.1 For each \(r\), there is \(n_{i_r}>0\) with \[ \alpha_{i_r}^{n_{i_r}}\in K[\alpha_1,\dots,\alpha_{i_r-1}] \] by the radical hypothesis. Each factor on the right has index \(<i_r\), hence is in \(E_{r-1}\). So \[ \alpha_{i_r}^{n_{i_r}}\in E_{r-1}. \] Set \(N_r:=n_{i_r}\).
By definition, \[ E_m=E(\alpha_{i_1},\dots,\alpha_{i_m})=E(S)=F. \] Thus taking \(\beta_r:=\alpha_{i_r}\) gives \[ F=E(\beta_1,\dots,\beta_m), \] and each step satisfies \[ \beta_r^{N_r}\in E_{r-1}. \] Hence the tower is radical over \(E\). ∎