Let $M$ be an algebraic closure of $\mathbb F_p$, and let $F=M(x)$. Show that $f(t)=t^p-t-x$ is not solvable by radicals over $F$, but the Galois group of the splitting field of $f$ over $F$ is cyclic.
It's an exercise from Patrick Morandi's Field and Galois Theory. I know that the Galois group of the splitting field of $f$ over $F$ is generated by $\sigma:\sigma(\alpha)=\alpha+1$, where $\alpha$ is a root of $f$. Thus, the Galois group is cyclic. However, I don't know how to prove that $f$ is not solvable by radicals.
Can anyone help me?
Best Answer
Ok, I finally got through it.
Let $K$ be the splitting field of $f$ over $F$. $K/F$ is Galois with degree $p$.
If $K$ lies in a radical extension $L$ of $F$. Then we have $$ F=F_0\subset F_1\subset F_2\ldots\subset F_r=L $$ where $F_i=F_{i-1}(\alpha_i)$ and $\alpha_i^{n_i}\in F_{i-i}$. We may assume that $\alpha_i\notin F_{i-1}$ and $n_i$ are all primes.
Let $K_i$ be $K(\alpha_1,\ldots\alpha_i)$, then $F_i \subset K_i$.
By induction, we can prove that $K_i/F_i$ is Galois with degree $p$ as follows.
First, $K_0/F_0$ is Galois with degree $p$.
We assume $K_{i-1}/F_{i-1}$ is Galois with degree $p$. $K_i=K_{i-1}(\alpha_i), F_i=F_{i-1}(\alpha_i)$. If $\alpha_i\in K_{i-1}$, then $F_i=F_{i-1}(\alpha_i)=K_i=K_{i-1}(\alpha_i)=K_{i-1}$ and $n_i=p$, since $[K_{i-1}:F_{i-1}]=p$. Because $\alpha_i\notin F_{i-1}$, $g=(t-\alpha_i)^p=t^p-\alpha_i^p$ is irreducible over $F_{i-1}$. Then the minimal polynomial of $\alpha_i$ over $F_{i-1}$ is $g$. However, $K_{i-1}/F_{i-1}$ is Galois, so $\alpha_i$ is separable, but $g=(t-\alpha_i)^p$, which shows that $\alpha_i$ is not separable.
This contradiction shows that $\alpha_i\notin K_{i-1}$. Note that $\alpha_i^{n_i}\in F_{i-1}\subset K_{i-1}$ and all $n_i$th roots of unity is in $F$. Then we have $g=t^{n_i}-\alpha_i^{n_i}$ is irreducible over $K_{i-1}$ and $[K_i:K_{i-1}]=n_i$. Then we can conclude that $K_i/F_i$ is Galois with degree $p$.
By induction, $K_i/F_i$ is Galois with degree $p$ for all $i$.
On the other hand, $K_r=F_r=L$, so $K_r/F_r$ is of degree 1, which leads to a contradiction.