# $f(X) = 0$ is solvable by radicals over $K$ if $\deg (f)<\operatorname{char}(K)$

Let $K$ be an infinite field s.t $\operatorname{char}(K) = p$ and let $f$ be an irreducible polynomial over $K$ of degree less than $p$. Let $L$ be a splitting field for $f$ over $K$. Show that $f(X) = 0$ is solvable by radicals.

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What have you tried so far? What do you know about the condition solvable by radicals? Do you know what this translates into in terms of Galois groups? What do you know the structure of Galois groups in the characteristic $p$ case? – KReiser Nov 15 '12 at 19:39
There is a discussion of solvability in characteristic $p$ in the text "Galois Theory" by David Cox. – KCd Nov 15 '12 at 19:41
just basic things, i mean definitions and the first theorems – Rahman Nov 15 '12 at 19:41
First question: Is $L/K$ Galois? Yes, because it is normal (as a splitting field) and separable (because $\deg f<p$). – Hagen von Eitzen Nov 15 '12 at 20:55
@ Hagen von Eitzen: and then what to do :) ? – Rahman Nov 15 '12 at 21:07

Actually any finite extension $L$ of a finite field $K$ is radical...!!!
$proof$:
If $L ≠ K$ then $L = K(a)$ for some $a ∈ L$ and $p.a = 1 ∈ K$ where $p=char(K)$.
In our question .. take $L$ to be the splitting field of $f$ over $K$.