# Galois group is solvable but $f$ is not solvable

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?

• I found a similar question math.stackexchange.com/questions/1054548/…, but I don't understand the answer in that post. Why when $p=n_i$, $1 = [F(\alpha_1,\cdots,\alpha_{i-1},\alpha_i):F(\alpha_1,\cdots,\alpha_{i-1})]$. Shouldn't it be $p$? Dec 6, 2020 at 14:59
• He shows that when $n_i=p$ the splitting polynomial for $\alpha_i$ is actually $X-\alpha_i$. Dec 6, 2020 at 15:22
• @ancientmathematician but $\alpha_i \notin F(\alpha_1,\cdots,\alpha_{i-1})$? Dec 6, 2020 at 16:11
• It boils down to $X^p-\alpha$ never being separable in characteristic $p$. A solution of $t-t^p=x$ is the formal power series $$t=x+x^p+x^{p^2}+\cdots.$$ Dec 6, 2020 at 17:08

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.