Solvability of $X^p-X-a$ In remark 3.28 of J. S. Milne's online note, Fields and Galois Theory, it claims that if $F$ is a field with characteristic $p$, then $X^p-X-a=0$ is not solvable by radicals even though it is separable with abelian Galois group. 
It is obvious that $X^p-X-a$ is separable, but I do not know how to show its Galois group is abelian and why it is not solvable by radicals! 
 A: The claim is false: the splitting field of $ X^2 - X - 1 $ over $ \mathbb F_2 $ is $ \mathbb F_4 $, which is obtained by adjoining a cubic root of unity to $ \mathbb F_2 $, therefore it is a radical, and solvable, extension.
If that does not satisfy you, the splitting field of $ X^3 - X - 1 $ over $ \mathbb F_3 $ is $ \mathbb F_{27} $, which is obtained by adjoining a $ 26 $-th root of unity to $ \mathbb F_3 $, so it is a solvable extension. Indeed, any finite extension of $ \mathbb F_p $ is solvable, since any such extension is obtained by adjoining a root of unity of sufficiently high order to $ \mathbb F_p $.
A: What Milne wants to say is that when characteristic is not 0, the fact that Galois group is solvable is not sufficient to guarantee the solvability by radicals of a polynomial.
A typical example is to consider $F = \mathbb{F}_p(t)$, the rational function over $\mathbb{F}_p$ where $p$ is a prime. Let $f = x^p-x-t$, then one can check that the Galois group of $f$ is cyclic $p$-group, however, $f$ is not solvable by radicals in the sense that we cannot get all roots of $f$ by successively adjoining n-th roots.
A: Milne's remark is contradicted by the so called theory of Artin-Schreier: for any field $F$ of prime characteristic $p$, the polynomial $f(X) = X^p - X - a $, with $a \in F$, either has all its roots in $F$, or is irreducible over $F$ and its splitting field is a Galois extension of $F$, with Galois group cyclic of order $p$  (this is easy to show; the converse is a theorem of Artin-Shreier: any Galois extension of $F$ of degree $p$ is of the form just described). Perhaps we should recall that a separable polynomial $g \in k[X]$ is solvable by radicals over $k$ iff the Galois group of its splitting field over $k$ is solvable (in the sense of group theory). Since a cyclic group is solvable, Milne's claim is  wrong.
Edit.  Perhaps the terminology "solvable by radicals" is a bit misleading. In Lang's "Algebra", chap. VIII, § 7, a finite extension $K/k$ is said to be solvable by radicals iff it is separable and is contained in a finite extension $E/k$ admitting a tower decomposition $(E_i)$ s.t. each step $E_{i+1}/E_i$ is one of the following types: 1) It is obtained by adjoining a root of unity   2) It is obtained by adjoining a n-th root of an element of $E_i$, $n$ prime to  the characteristic   3) It is obtained by adjoining a root of a polynomial $ X^p - X - a$  if the characteristic is $p$. 
But even if one does not accept condition 3), Milne's claim is wrong for instance over finite fields. 
