Show $f(x)$ is not solvable by radicals and $Gal(E/K) \cong Z_p$.
I was just reading Galois theory from the textbook "Galois Theory - Rotman" and I stumbled acrossed an exercise that looked interesting. Because the next chapter it says that $f(x)$ is solvable iff the Galois group of $f$ is solvable.
So I have two questions, if we have a field of characteristic $0$, this means that the statement $f(x)$ is solvable iff the Galois group of $f$ is solvable is true correct?
Second question is, how do I approach this question? Because I know that this statement is saying that the Galois group is solvable but yet $f$ is not solvable by radicals. Any sort of approach would be helpful!
Showing $f$ is irreducible:
Suppose that $x$ was a root of $f$ in $F_p[x]$. Then $x+1$ is a root as well since $f(x+1) = (x+1)^p - (x-1) - t = x^p -x +1^p - 1 - t = f(x) = 0.$ As this equation holds for any root of $f$, we see that: $\{ x, x+1, ..., x+(p-1) \}$ are all distinct roots of $f$ which implies that $f$ is separable and hence splits over $F_p[x]$.
Now we see that each monic irreducible factor of $f$ is of the form $(x + k)$ for some $k$, $0 \leq k \leq p-1$. Since $F_p[x] = F_p[x + k]$ for all $k$, we have that each irreducible factor of $f$ has the same degree, call this $n$. So $p$ is a multiple of $n$. So $n = 1$ or $p$. $n = 1$ is impossible since $f$ has no roots in $F_p$ so it forces $n = p$ so $f$ is irreducible.