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.
$f(X) = 0$ is solvable by radicals over $K$ if \deg (f)<\operatorname{char}(K)
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abstract-algebra
field-theory
galois-theory
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0Since $L:K$ is normal extension, and deg (f) < p that means $f$ is separable in $L$.. since $char(K)=p$ then $|K|= p^n$ where $n= [K:Z_p]$, $Z_p ⊆ K ⊆ L$. $[L:K]/(deg (f))!$ $⇒$ $[L:K]/p!$ $⇒$ [L:K]=(p-i_1)...(p-i_r) >= deg f for some $i_j=0,..,p-1$ – 2012-11-15
1 Answers
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My answer in finite case:
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 $K$.
And "I think" we done!