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Give an example of a polynomial $f$ over $Q$ which is is solvable by radicals, but the splitting field $L$ for $f$ is not radical extension over $Q$.

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up vote 4 down vote accepted

$f(x)= x^3 -3x+1$ has three real roots.. so $L$ in $ R$.

$gal(L:Q)$ is $A_3$ hence soluble group hence there exist $M$ containing $L$ with $M:Q$ is radical extension. assume that $L:Q$ is radical then [L:Q]= 3 then $L=Q(a)$ with $a^n$ in $Q$

$f/x^n - a^n$

let $r$ be another root of $f$ then $r^n=a^n$ and $r/a$ is the $n^{th}$ real root of the unity and hence $r=+- a$ then we have at most two zeros for $f$ in L. which is impossible!

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