I know from a theorem that:
Let $F$ be a field of characteristic $0$ and $f(x)\in F[x]$. Then $f(x)$ is solvable by radicals if and only if the Galois group of $f(x)$ is solvable.
But what if the field was not characteristic $0$?
I see the result that if $F = F_p(t)$ then this theorem is completely false. Where the Galois group is solvable but $f$ is not solvable where $f = x^p - x - t$ but why is this true?
I believe I can see why the galois group is solvable because the galois group is cyclic and hence isomorphic to $\mathbb{Z_p}$, but I cannot fully grasp why $f$ is not solvable. I tried do this by contradiction but got no result.
Can anyone help me?
Suppose $f$ is solvable by radicals. Then we should have $$F\subset F(\alpha)\subset K\subset \overline{F}$$ where $K=F(\alpha_1,\cdots,\alpha_m)$ and $\alpha_1^{n_1}\in F, \alpha_i^{n_i}\in F(\alpha_1,\cdots,\alpha_{i-1})$.
First, we may assume $p\nmid n_i$ unless $p = n_i$. For if $n_i = pr, r>1$, then we can first adjoin a r-th root and then a p-th root.
Secondly, if $p\nmid n_i, p\nmid \{F(\alpha_1,\cdots,\alpha_{i-1},\alpha_i):F(\alpha_1,\cdots,\alpha_{i-1})\}$. If $p = n_i$, then $X^p-\alpha_i^p$ is irreducible provided that $\alpha_i\notin F(\alpha_1,\cdots,\alpha_{i-1})$ and $X^p-\alpha_i^p = (X-\alpha_i)^p$. Thus $1 = \{F(\alpha_1,\cdots,\alpha_{i-1},\alpha_i):F(\alpha_1,\cdots,\alpha_{i-1})\}$.
Therefore, $\{K:F\}=\prod_{i=1}^{m}\{F(\alpha_1,\cdots,\alpha_{i-1},\alpha_i):F(\alpha_1,\cdots,\alpha_{i-1})\}$ is not divisible by $p$. But we have $p =|G(E\ /\ F)|=\{E:F\}\mid \{K:F\}$, which is a contradiction.