Splitting of cubics over $\mathbb{Q}$; radical extension

Question

Can someome help me solve this? Let $L$ be the splitting field of a cubic over $\mathbb{Q}$, and $\omega$ be a primitive cube root of unit. Prove that the extension $L(\omega)$ over $\mathbb{Q}$ is a radical extension.

My aporoach is the following: Given a polynomial $f$ and a root $\alpha$ then $L = \mathbb{Q}(\alpha,\sqrt{D})$ where $D$ is the discriminant. I got to show $L(\omega)$ is a radical extension $\implies \mathbb{Q}(\omega,\alpha,\sqrt{D})$ is a radical extension. Now I know that $\omega^3 \in \mathbb{Q}$ and $\sqrt{D}^2 \in \mathbb{Q}(\omega)$, it just remains to show $\alpha^k \in \mathbb{Q}(\omega,\sqrt{D})$, for some $k$. Can anyone help me out from here?

My guess is if I use Cardano's formulae for cubics, then cubing that might give me it, but unfortunately it does not. Am I misisng something ?

Answer 1

It is not necessarily true that $\alpha^k\in F=\Bbb Q(\omega,\sqrt D)$ for some $k$. If $\alpha\notin F$, then $F(\alpha)$ is a cyclic extension of $F$ of degree $3$ and $F$ contains a third root of unity. It is then a general fact that in this situation (see below) $F(\alpha) = F(\beta)$ with some $\beta\in F(\alpha)$ satisfying $\beta^3\in F$. This follows from Hilbert 90, or more generally from Kummer theory. This is also how you prove that an extension with solvable Galois group is solvable by radicals.

Fact: Let $F$ be a field containing the $n$-roots of unity with $\operatorname{char} F\nmid n$. If $L/F$ is a cyclic extension of degree $n$, then $L=F(\beta)$ with $\beta^n\in F$. Proof sketch: Let $\zeta$ be a primitive $n$-th root of unity. Then $N_{L/F}(\zeta)=1$, so by Hilbert 90 there is $\beta\in L^\times $ such that $\sigma(\beta)/\beta=\zeta$ where $\sigma$ is a generator of the Galois group. Then this $\beta$ works.

Answer 2

If you want to use Cardano's formula it would go like this.

W.l.o.g. we can assume that $f$ is a depressed cubic, $f(x)=x^3+px+q$. Then, by Cardano, its roots are $$ \begin{aligned} t_1&=u+v,\\ t_2&=\omega^2 u+\omega v,\\ t_3&=\omega u+\omega^2 v, \end{aligned} $$ where $u^3$ and $v^3$ are the roots of the quadratic $$(T-u^3)(T-v^3)=T^2+qT+(-p/3)^3$$ with discriminant $D=(-q/2)^2+(p/3)^3$. Furthermore, we need to choose the cube roots in such a way that $uv=-p/3$. See any book explaining Cardano, or locally here.

Anyway, $u^3$ and $v^3$ are elements of $\Bbb{Q}(\sqrt D)$, and the field $$L(\omega)=\Bbb{Q}(t_1,t_2,t_3,\omega)=\Bbb{Q}(\sqrt D, u, v, \omega)$$ is gotten as a root tower $$ \Bbb{Q}\subseteq\Bbb{Q}(\sqrt D)\subseteq\Bbb{Q}(\sqrt D,u)\subseteq\Bbb{Q}(\sqrt D,u,\omega)=L(\omega).$$ Observe that the relation $uv=-p/3$ means that $v\in\Bbb{Q}(u)$.