If $f$ is an irreducible cubic over $\mathbb{Q}$ and $L$ is its splitting field, then $L$ is radical (condition on discriminant). Let $f$ be an irreducible cubic polynomial in $\mathbb{Q}[x]$ and let $L/Q$ be its splitting field. I want to show that $L/\mathbb{Q}$ is radical if and only if the discriminant is of the form $D = -3m^2.$ If $L$ is a radical extension, then I know that $L = \mathbb{Q}(\alpha)$ for some $\alpha^3 \in \mathbb{Q},$ and that the Galois group of $L/\mathbb{Q}$ is $S_3$ or $A_3$ depending on whether $D$ is a square. How do I proceed from here? There were previous questions similar to this, and they all seem to say something about extending $L$ by $\sqrt{-3}$ but I have no idea where that came from, hence the new question.
 A: If the discriminant is $-3m^2$ then the splitting field $L$ contains
$\sqrt{-3}$ and so also $\omega=\frac12(-1+\sqrt{-3})$, a primitive cube
root of unity. Therefore $L$ is a Kummer extension of $\Bbb Q(\sqrt{-3})$,
so $L=\Bbb Q(\sqrt{-3},\alpha)$ where $\alpha^3=\beta\in\Bbb Q(\sqrt{-3})$.
As $L$ is Galois over $\Bbb Q$, $L$ also contains a cube root of $\overline \beta$, the complex conjugate of $\beta$, which we can take to be $\overline\alpha$. Therefore $L=\Bbb Q(\sqrt{-3},\overline\alpha)$.
By Kummer theory, either $\overline\alpha/\alpha$ or $\overline\alpha/\alpha^2$ is a cube in $\Bbb Q(\sqrt{-3})$. The latter case leads to
Gal$(L/\Bbb Q)$ being cyclic, which is false. Therefore $\overline\alpha=\gamma^3\alpha=(\gamma\overline\gamma)^3\overline\alpha$ for
some $\gamma\in\Bbb Q(\sqrt{-3})$. Therefore $\gamma\overline\gamma=1$,
so $\gamma=\delta/\overline\delta$ by Hilbert 90, with $\delta\in\Bbb Q(\sqrt{-3})$. Then $s=\delta^3\alpha=\overline{\delta^3\alpha}\in\Bbb Q$.
Also $L$ contains $\Bbb Q(s^{1/3})$ which must be the field generated
by the real root of the original polynomial.
A: We have a cubic irreducible polynomial $$f(x)=\sum_{n=0}^3 c_n x^n = \prod_{j=1}^3 (x-\alpha_j) \qquad \in \mathbb{Q}[x]$$
Using $\alpha_1+\alpha_2+\alpha_3=-c_2 \in \mathbb{Q}, \alpha_1\alpha_2\alpha_3 = -c_0$ and $D = \prod_{i \ne j} (\alpha_i - \alpha_j) \in \mathbb{Q}$,
with $\sqrt{D} = (\alpha_1-\alpha_2)(\alpha_1-\alpha_3)(\alpha_2-\alpha_3)$  we find that  $L = \mathbb{Q}(\alpha_1,\sqrt{D})$ is the splitting field of $f$, with $[L:K] = 3$ and $[L:\mathbb{Q}]=3$ or $6$ depending on $D$ being a square.

If $D= -3 m^2$ then $L/\mathbb{Q}$ is a radical extension 
$L = K(\alpha_1)$ where $K = \mathbb{Q}(\sqrt{D}) = \mathbb{Q}(\zeta_3)$. 
Take any $\beta \in L, \beta \not \in K, \beta^3 \in K$. We know it exists from the Cardano's formula 
So we have the tower $L/K(\beta)/K$, but $[K(\beta):K] \ | \ [L:K]$ means $[K(\beta):K] = 3$ and $K(\beta) = L$ ie. $$L = \mathbb{Q}(\sqrt{-3}, \sqrt[3]{\gamma})$$
where $\gamma =\beta^3= a+b \sqrt{-3}$.

If $L/\mathbb{Q}$ is a radical extension then $D= -3 m^2$ 
$L = K(\sqrt[3]{\gamma})$ for some $\gamma \in K = \mathbb{Q}(\sqrt{D})$ and $[L:K] = 3$. But $L/K$ is Galois so that $Gal(L/K)$ contains $3$ elements of the form $\sigma(\sqrt[3]{\gamma}) = \lambda$ where $\lambda$ is another root of $x^3-\gamma$, ie. $\lambda = \zeta_3^k\sqrt[3]{\gamma}$ and $\zeta_3 \in L$.
But $L/K$ doesn't have any quadratic subfield so that $\zeta_3 \in K$ and hence $K = \mathbb{Q}(\zeta_3)$ and $D = -3 m^2$.
