enter image description here

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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.