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 $Q$.
[Math] $f$ is solvable by radicals, but the splitting field $L:Q$ not radical extension.
field-theorygalois-theory
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Note that you are not asked to prove that there is a chain of simple irreducible radical extensions $\Bbb Q \subset E_1 \subset E_2 \subset \ldots \subset \Bbb Q(\zeta_n)$, but that there is a chain $\Bbb Q \subset \ldots \subset L$ such that $\Bbb Q(\zeta_n) \subset L$.
If a normal extension $K \subset L$ has Galois Group isomorphic to $\Bbb Z/p \Bbb Z$, then $K(\zeta_p) \subset L(\zeta_p)$ is a normal simple irreducible radical extension :
$K \subset K(\zeta_p)$ is a normal extension, thus $Gal_{K(\zeta_p)}(L(\zeta_p))$ and $Gal_L(L(\zeta_p))$ are normal subgroups of $Gal_K(L(\zeta_p))$ and there is a well-defined restriction map $Gal_K(L(\zeta_p)) \to Gal_K(L) \times Gal_K(K(\zeta_p))$.
This map is injective because $L$ and $\zeta_p$ generate $L(\zeta_p)$. Furthermore, the composition of this map with either projection has to be surjective.
Since $|Gal_K(L)|=p $ and $|Gal_K(K(\zeta_p))|$ are coprime, the only subgroups of their product are products of subgroups, so the map has to be surjective. Thus it is an isomorphism.
let $x \in L \setminus K$. Then $K(x) = L$. Let $\sigma$ be a generator for $Gal_K(L) = Gal_{K(\zeta_p)}(L(\zeta_p))$, and look at $y = \sum_{k=0}^{p-1} \zeta_p^{-k} \sigma^k(x) \in L(\zeta_p)$. Then, $\sigma(y) = \sum_{k=0}^{p-1} \zeta_p^{-k} \sigma^{k+1}(x) = \zeta_p y$.
Thus $\sigma(y^p) = \sigma(y)^p = \zeta_p^p y^p = y^p$. Since $y^p$ is fixed by $\sigma$, $y^p \in K(\zeta_p)$.
Since $X^p - y^p = \prod(X - \zeta_p^k y) = \prod(X - \sigma^k(y))$, this polynomial must be irreducible over $K(\zeta_p)$ (there is exactly one orbit of the action $Gal_{K(\zeta_p)}(L(\zeta_p))$ on its roots)
Finally, $K(\zeta_p,y) = L(\zeta_p)$ because there is an obvious inclusion and their degree over $K(\zeta_p)$ are the same.
Next, if $K \subset L$ is a normal simple irreducible radical extension of degree $p$, and $K \subset K'$ is a normal extension, then $K' \subset K'L$ is still a normal simple irreducible radical extension, of degree $1$ or $p$. The polynomial $X^p - y^p$ can't suddenly become reducible unless all the $y$ are in $K'$, because $\Bbb Z/p\Bbb Z$ has no nontrivial subgroup.
With this, you can show that if $K \subset L$ and $L \subset M$ are solvable by simple irreducible radical extensions, then so is $K \subset R$ (just add all the roots one after the other).
Finally, with an induction argument, you can finally show that since $K \subset K(\zeta_p)$ is abelian of degree dividing $p-1$, it is solvable, and since $K(\zeta_p) \subset L(\zeta_p)$ is solvable, $K \subset L(\zeta_p)$ is solvable, which implies that $K \subset L$ is too. And by decomposing any abelian extension into cyclic prime extensions, you obtain that any abelian extension is solvable.
Now we apply this to $\Bbb Q \subset \Bbb Q(\zeta_{47})$. Since its Galois group is isomorphic to $\Bbb Z / 2 \Bbb Z \times \Bbb Z / 23 \Bbb Z$, we need to add $\zeta_2$ (which is already there, it's $-1$) and $\zeta_{23}$. For this one, we need $\zeta_{11}$, which needs $\zeta_5$, which needs $\zeta_4$.
The resulting chain (showing the degrees of the extensions) is : $\Bbb Q \subset^2 \Bbb Q(\sqrt{-1}) \subset^2 \Bbb Q(\sqrt{-1},\sqrt 5) \subset^2 \Bbb Q(\zeta_{20}) \subset^2 \Bbb Q(\zeta_{20}, \sqrt{-11}) \subset^5 \Bbb Q(\zeta_{220}) \subset^2 \Bbb Q(\zeta_{220}, \sqrt{-23}) \\ \subset^{11} \Bbb Q(\zeta_{5060}) \subset^2 \Bbb Q(\zeta_{5060}, \sqrt{-47}) \subset^{23} \Bbb Q(\zeta_{237820}) \supset \Bbb Q(\zeta_{47})$.
Theoretically, you can compute explicitly at each step what are the things you are taking $n$th roots of and express everyone in terms of radicals, though it gets messy really fast. (I only explicited the quadratic subfields of the $\mathbb Q(\zeta_p)$)
Let $f$ be an irreducible cubic, whose roots generate non-normal cubic extensions, and let $L$ be its splitting field, which has degree $6$ and Galois group $S_3$. Let $K$ denote a cubic subfield of $L$. It is not difficult to show that $K$ is radical if and only if the discriminant of $f$ (or $K$) has the form $D = -3m^2$. If $L$ is radical in your sense, then either $K$ must be radical, or $L$ must be radical over its quadratic subfield $k$. Since $L/k$ is cyclic, $L$ can be normal only if $k = {\mathbb Q}(\sqrt{-3})$ and $L/k$ is Kummer. But since $L = K(\sqrt{{\rm disc}\ K})$, we have $k = {\mathbb Q}(\sqrt{-3})$ if and only if the discriminant of $f$ has the form $-3m^2$. Thus $L$ is radical in your sense if and only if $K$ is radical, which is the case if and only if the discriminant of $f$ has the form $-3m^2$.
Best Answer
$f(x)= x^3 -3x+1$ 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!