# If $f(x)\in F[x]$ is solvable by radicals then its Galois group $Gal(E/F)$ is solvable.

In all the books that I've checked the use of roots of unity (as hypothesis) is very crucial to prove that if $f(x)\in F[x]$ is solvable by radicals then its Galois group $Gal(E/F)$ is solvable, but the professor said that it's not needed that the field must have enough roots of unity. I think this is curious, and I want to prove it. She also gave three steps to prove this and I'm struggling with them.

(1) Let $F=B_0\subseteq B_1\subseteq\dots\subseteq B_t$ a radical extension with $E\subseteq B_t$ and $B_t$ the splitting field of $f(x)\in F[x]$ with $deg(f(x))=m$. We defined $\mathcal E$ as the splitting field of $h(x)=x^m-1$ over $E$ and $\mathcal F=F(\Psi)$, where $\Psi$ are all the $m$th roots of unity. Prove that $\mathcal E$ is the splitting field of $f(x)$ over $\mathcal F$, and there for $Gal(\mathcal E/\mathcal F)$ is solvable.

I don't see why do we need that function $h(x)$ if what we are trying to do is to take the roots of unity out. Also, for $\mathcal E,\mathcal F$ we are basically adding the same elements, $\Psi$, to $E,F$, but by hypothesis we already have that $E=F(\alpha_1,\dots,\alpha_m)$ with $\alpha_i$ the roots of $f(x)$, and $\mathcal E= E(\Psi)=F(\alpha_1,\dots,\alpha_m,\xi_1,\dots,\xi_m)$ so if anything we are adding things and $f(x)$ can still get decomposed into linear factors, so what are we suppose to do?

(2) Prove that $Gal(\mathcal E/\mathcal F)\unlhd Gal(\mathcal E/ F)$ and $Gal(\mathcal E/ F)/Gal(\mathcal E/\mathcal F)\cong Gal(\mathcal F/F)$ and $Gal(\mathcal E/F)$ is solvable.

To prove that $Gal(\mathcal E/\mathcal F)\unlhd Gal(\mathcal E/ F)$, we show that $\mathcal F/F$ is a Galois extension: lets take $f(x)\in F[x]$, then by definition all the roots of $f(x)$ are in $\mathcal F$, then it is a splitting field, hence $\mathcal F/F$ is a Galois extension.
Now we can talk about the quotient $Gal(\mathcal E/ F)/Gal(\mathcal E/\mathcal F)$ and $Gal(\mathcal F/F)$, but I can't find the right isomorphism to fit this, and I'm starting to think that maybe this isn't right.
And to prove $Gal(\mathcal E/F)$ is solvable, I wanted to prove that there exists a radical extension, but to use that I would need that $F$ is of characteristic $0$, and I don't have as a hypothesis.

(3) Prove that $Gal(\mathcal E/E) \unlhd Gal(\mathcal E/F)$ and $Gal(\mathcal E/ F)/Gal(\mathcal E/ E)\cong Gal(E/F)$, and conclude that $Gal(E/F)$ is solvable.

The prove is pretty much the same as before, isn't it?

(p.s. sorry for the long post)

EDIT: I think I came up with te proof for (1). We know that $\alpha_i\in\mathcal E$ and $\mathcal F\subseteq\mathcal E$ then by the definition of $\mathcal E$ it is the splitting field of $f(x)$ over $\mathcal F$. Now lets take the tower of fields: $\mathcal F=B_0(\Psi)\subseteq B_1(\Psi)\subseteq\dots\subseteq B_t(\Psi)$ with $\mathcal E\subseteq B_t(\Psi)$ then it is a radical extension and $\mathcal E/\mathcal F$ is Galois, where $\mathcal F$ contains all the roots of unity, then $Gal(\mathcal E/\mathcal F)$ is solvable

For (2),(3) I still can't figure oit how to prove the isomorphism.

• A minor comment: it does not make sense to say that a group is solvable by radicals. Groups are just solvable, or not. – Derek Holt May 18 '14 at 12:10
• @DerekHolt, yes you're right, I've corrected it... – Ana Galois May 18 '14 at 16:47