Given an irreducible cubic $f = x^3 + px + q$ over $\mathbb{Q}$ I'm trying to show that, for $\alpha$ a root of $f$, $\mathbb{Q}(\alpha)$ is a radical extension of $\mathbb{Q}$ if and only if $-\Delta/3$ is a square.
My ideas so far are that if it is a radical extension then $\alpha^n \in \mathbb{Q}$ for some $n$ and we must have $n \ge 3$ because $f$ is the minimal polynomial of $\alpha$. I also know that the Galois group of $f$ is either $A_3$ or $S_3$ depending on whether $\Delta$ is a square or not.
I'd also like to use the fact that $K(\alpha)$ is a radical extension iff it's Galois group is solvable but how does the condition on the discriminant tell us that this is a Galois extension?
Thanks
Best Answer
If $\Bbb Q(\alpha)$ is a radical extension then it is $\Bbb Q(\sqrt[3]a)$ for some rational $a$, and its Galois closure is obtained by adjoining a third root of unity, so it has to be $\Bbb Q(\sqrt[3]a, \sqrt{-3})$.
Since in an $S_3$ extension the only quadratic subfield is $\Bbb Q(\sqrt \Delta)$, we must have $\Bbb Q(\sqrt \Delta) = \Bbb Q(\sqrt {-3})$ and so $-3\Delta$ has to be a square.
Conversely, suppose $-3\Delta$ is a square. I could just tell you to apply Cardan's formula but this is no fun.
Let $\beta,\gamma$ be the other two roots.
Then $(\alpha + \zeta_3 \beta + \zeta_3^2 \gamma)^3 = ((\alpha^3+\beta^3+\gamma^3)-\frac 32(\alpha^2\beta + \alpha\beta^2+\alpha^2\gamma+\alpha\gamma^2+\beta^2\gamma+\beta\gamma^2) + 6\alpha\beta\gamma) + \frac32 \sqrt{-3}(\beta-\alpha)(\gamma-\beta)(\alpha-\gamma)$
The first half is symmetric so is expressible in terms of the coefficients and is rational. The second half is $\frac 32 \sqrt{-3 \Delta}$ which is also rational, so $(\alpha + \zeta_3 \beta + \zeta_3^2 \gamma)^3 = b \in \Bbb Q$.
Moreover $(\alpha + \zeta_3 \beta + \zeta_3^2 \gamma)(\alpha + \zeta_3^2 \beta + \zeta_3 \gamma) = \alpha^2+\beta^2+\gamma^2 - \alpha\beta-\beta\gamma-\gamma\alpha \in \Bbb Q$, and $\alpha+\beta+\gamma \in \Bbb Q$, and so $\Bbb Q(\alpha) = \Bbb Q(\alpha+\beta+\gamma,\alpha+\zeta_3\beta+\zeta_3^2\gamma,\alpha+\zeta_3^2\beta+\zeta_3\gamma) = \Bbb Q(\alpha+\zeta_3\beta+\zeta_3^2\gamma) = \Bbb Q(\sqrt[3]b)$ for the right choice of cube root.