Consider $f(x)=x^5-6x+2\in\Bbb Q[x]$ not solvable. Show that $f(x)$ doesn't have a root in a radical extension of $\Bbb Q$.
I want to prove it by contradiction:
Suppose there is a root of $f$, $u\in E=E_0$ and $E$ is a radical extension such that we have the field chain $E=E_0\supseteq E_1\supseteq...\supseteq E_n=\Bbb Q$ and $\forall i=0,...,n-1 $, $E_i=E_{i+1}(u_i)$ for some $u_i\in E_i$ and $u_i^{n_1}\in E_{i+1}$ for some $n_i\in\Bbb N$
First I showed that $[E_0:\Bbb Q]=[\Bbb Q(u_{n-1},...u_1,u_0):\Bbb Q]=...\le\prod\limits_{i=0}^{n-1}n_i<\infty$ so the extention is algebraic, but I feel like that information is useless in this exercise.
I need to arrive to the contradiction that "$f(x)$ is solvable by radicals". By definition this is equivalent to "There exists a radical extension of $\Bbb Q$ that contains a splitting field of $f(x)$".
A natural thing would be if $E_0$ was that extension... We are allowed to use the following fact:
Let $E$ be an intermediate field $\Bbb C\supseteq E\supseteq \Bbb Q$ and $p(x)\in E[x]$ of degree at most $4$.
Any help would be appreciated
Unfortunately $E$ need not be the splitting field of $f$, but there is the next best thing --- a solvable extension of $E$ is.
Indeed, suppose $u$ is a root of $f(x)\in\mathbb{Q}[x]$ in some solvable extension $E/\mathbb{Q}$. Splitting the linear factor $(x-u)$ from $f(x)$ then gives a degree 4 polynomial $g(x)\in E[x]$, which may or may not be irreducible. But you know every polynomial of degree $\leq 4$ is solvable by radicals so there is a radical extension $F$ of $E$ that contains all roots of $g$, hence all roots of $f$. This contradicts the assumption $f$ is not solvable by radicals.