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I gonna show that if $F\subset L$ is solvable and Galois the theorem above holds.

Suppose $F\subset L$ is Galois and radical.

Since L is radical over F, we have $F\subset F_1\subset F_2 ...\subset F_n=L$ where $F_{i+1} =F_i(\gamma_i)$ and $\gamma_i^{m_i}\in F_i$ for some $m_i >0$

Each $c_i$ is the $m_i^{th}$ root of unity.

We can adjoin the $m_i^{th}$ roots of unity to each subfield above to show that $F(c_1,c_2,...,c_n)\subset L(c_1,c_2,...,c_n)$ is radical.

i.e. $F(c_1,c_2,...,c_n)\subset F_1(c_1,c_2,...,c_n)\subset F_2(c_1,c_2,...,c_n) ...\subset F_n(c_1,c_2,...,c_n)=L(c_1,c_2,...,c_n)$

where $F_{i+1}(c_1,c_2,...,c_n) =F_i(c_1,c_2,...,c_n)(\gamma_i)$ and $\gamma_i^{m_i}\in F_i(c_1,c_2,...,c_n)$ for some $m_i >0$

Note that $F_n(c_1,c_2,...,c_n)$ is Galois over $F(c_1,c_2,...,c_n)$

Let $K_i= F_i(c_1,c_2,...,c_n)$

a) We need to show that $K_{i-1}\subset K_i$ is Galois and its Galois group is cyclic.

Since $\gamma_i^{m_i}\in F_{i-1}\subset K_{i-1}$ and $c_i\in K_{i-1}$ we have $x^{m_i}-\gamma_i^{m_i}\in K_{i-1}[x] $ with distinct roots $\gamma_i,\gamma_i c_i,...,\gamma_i c_i^{m_i-1} $

Hence the splitting field of $x^{m_i}-\gamma_i^{m_i}$ is $K_{i-1}(\gamma_i)=K_i$. Hence, the Galois assertion is proved.

$\forall \sigma, \tau \in Gal(K_i/K_{I-1})$ $\sigma(\gamma_i)=\gamma_i c_i^a$ and $\tau(\gamma_i)=\gamma_i c_i^b$ where $1\leq a,b\leq m_i$.

Hence, $\sigma\tau(\gamma_i)=\gamma_i c_i^{b+a}=\tau\sigma(\gamma_i)$

Therefore, abelian assertion is proved.

b) We now have the following chain:

$Gal(K_n/k_n)\subset .... \subset Gal(K_n/K_0)$

Let us consider $K_{i-1}\subset K_i \subset K_n$. Note that $K_n$ is Galois over $K_{i-1}$.

From (a) we have that $Gal(K_i/K_{i-1})$ is cyclic hence, $Gal(K_n/K_{i-1})/Gal(K_n/K_i)$ is cyclic and we have $Gal(K_n/K_{i-1})$ is normal in $Gal(K_n/K_i)$

Therefore $Gal(K_n/K_0)$ is radical which implies that Gal(L/F) is radical.

Are there any flaws or loopholes?

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