Solvability of Artin-Schreier Polynomial

Question

I'm having a hard time trying to prove that the polynomial

f(x) = x^p - x - 1 in Z_p[x]

is not solvable by radicals even though its Galois Group is solvable.

So far, I have shown that the polynomial is irreducible in the base field and that its Galois group is isomorphic to Z_p. Since the Galois group is cyclic, it is solvable. I have also shown that if n is a root of f then the pth root of n is not a root of f by the definition of solvable by radicals given to us, which is definition 2 in the link below:

http://www.math.brown.edu/~abrmovic/MA/f1314/251/Zijian-notes.pdf

Answer 1

Show that the polynomial is irreducible. Show that the decomposition field of that polynomial is generated by any one of its roots; in particular, the degree of the decomposition field is then $p$. Show that thay field is not generated by the pth roots of any element of the prime field —or that if an element of that field is a pth root of the prime field, it is already in the prime field.

Answer 2

The OP specified that the characteristic zero definition from Zijian Yao's lecture notes is to be taken literally. Given that I want to put this question to rest by declaring that

The claim is false as stated.

The polynomial $x^p-x-1$ is irreducible over $\Bbb{F}_p$ (many proofs for this fact can be found by searching our site). Therefore its zeros generate the field $L=\Bbb{F}_{p^p}$ that is the unique degree $p$ extension of the prime field (and also the splitting field of $f(x)$).

But $L=\Bbb{F}_p(\zeta)$, where $\zeta$ satisfies the equation $\zeta^{p^p-1}=1$. Therefore all the zeros of $f(x)$ are "expressible by radicals" and, according to Definition 2, $f(x)$ is thus deemed solvable by radicals.

---

A very concrete counterexample: when $p=2$ the zeros of the polynomial $x^2-x-1=x^2+x+1$ are cubic roots of unity.