I am atttempting to solve problem 16.12.6 from Artin's algebra book.
Prove that if the Galois group of a polynomial $f$ is a nonabelian simple group, then the roots are not solvable.
My attempt
Say $f\in F[x]$ where $F$ is a subfield of the complex numbers. Let $K_f=$ the splitting field for $f$ over $F$.
If $f$ is constant, $K_f=F$ and $Gal(K_f/F)=Gal(F/F)=\{id\}$, which is abelian.
Suppose $Gal(K_f/F)$ is a nonabelian simple group. Then $deg(f)>1$. Say $K_f=F(α_1,...,α_n)$. I want to show that $α_1$ isn't solvable.
Assume $α_1$ is solvable over $F$. Then there is a chain of subfields $F=F_0 \subset F_1 \subset ... \subset F_s \subset \mathbb C$ such that $α_1\in F_s$ and each $F_{j+1}/F_j$ is a Galois extension of prime degree.
It would be nice if $F_1$ were a subfield of $K_f$, but sadly I don't think that can be shown.
Here is what I would do, though, if $F_1$ were a subfield of $K_f$
Since $F_1/F$ is Galois, this would imply that $F_1=K_f^H$ for some $H \lhd Gal(K_f/F)$, hence $F_1=K_f^{\{id\}}$ or $F_1=K_f^{Gal(K_f/F)}$, hence $F_1=K_f$ or $F_1=F$. Since $[F_1:F]$ is prime, we must have $F_1=K_f$. So $|Gal(K_f/F)|$ is prime, so $Gal(K_f/F)$ is cyclic and thus abelian, a contradiction.