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We know the following theorem from Galois theory:

Let $F$ be a field of characteristic $0$ and $f(x) \in F[x]$. Then $f(x)$ is solvable by radicals if and only if the Galois group of $f(x)$ is solvable.

For fields of characteristic $p > 0$ one direction of this theorem is not true. We could take $F = \mathbb{F}_p(t)$ and $f(x) = x^p - x - t \in F[x]$. Then the Galois group of $f(x)$ is $\mathbb{F}_p$ which is solvable, but $f(x)$ is not solvable by radicals.

What about the other direction? If $f(x) \in F[x]$ is solvable by radicals, is the Galois group of $f(x)$ solvable even when $F$ has characteristic $p > 0$?

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There is a possibility the the polynomial is not separable, and hence has no Galois group. The standard example is $F=\mathbb{F}_p(t)$, $f(x)=x^p-t$. Here $f(x)$ is solvable by radicals, as $z=t^{1/p}$ is its only root, of multiplicity $p$. But the splitting field of $f(x)$ is $F(z)$, and that is not a Galois extension of $F$.

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Thanks. I'm aware of this example, I probably should have specified that I'm more interested in the case where the Galois group exists (ie. $f(x)$ is separable). Do you know if there is a counterexample for that case? –  spin Jun 24 '12 at 21:37
    
@spin: I think that if the splitting field of a polynomial is contained in a root tower extension that does not involve joining of $p^{th}$ roots, then the splitting field is separable, hence Galois, and also solvable. I also think that is the polynomial is separable, and its Galois group is both solvable and $\gcd(p,|G|)$, then the splitting field is contained in a root tower extension provided that the base field has the necessary roots of unity. I'm afraid I don't have a reference. Anyway, I think that that kind of counter examples don't exist, but I won't bet the family fortune on it. –  Jyrki Lahtonen Jun 24 '12 at 22:02
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