# Is the splitting field of $g=x^3-3x-1$ over $\Bbb Q$ a radical extension?

Since $g=x^3-3x-1$ is irreducible over $\Bbb Q$, and has square discriminant.

If $L$ is the splitting field of $g$ over $\Bbb Q$, since $g$ has square discriminant we have $\text{Gal}(L/\Bbb Q) = A_3$

So $[L:K]=3$.

My question is, is $L/\Bbb Q$ a radical extension? Is $g$ solvable by radicals over $\Bbb Q$?

Some of my attempt:

Now $g$ is solvable by radicals if $L$ is a radical extension and $g$ splits over $L$, but the converse isn't in general true.

So first we will try to check if it's a radical extension.

Since $[L:K]=3$, we know that there cannot be an intermediate field between $L$ and $K$, since $[L:M][M:K]=3 \implies$ $[L:M]=3 \implies M=K$ or $[M:K]=3 \implies M=L$.

So $L=K(\alpha)$. For $L$ to be radical, we need $\alpha^3\in \Bbb Q$, which has minimal polynomial $f(x)=x^3-q$ where $q=\alpha^3$.

Since $g(x) = x^3-3x-1$ is cubic and $\alpha$ is a root of $g(x)$, and $f(x)$ is cubic, we use the fact that the minimal polynomial is unique to imply that $g(x)|f(x)$.

not fully sure if the following is true

$g(x)$ splits as $(x-\alpha)(x-\beta)(x-\gamma)|(x-\alpha\omega)(x-\alpha\omega^2)(x-\alpha)$

Where $\omega$ is a primitive third root of unity, So $\beta$ and $\gamma$ are $\alpha\times$primitive root of unity.

Now I am confused and cannot progress.

A radical extension for me is defined as a tower of simple radical extensions.

I have that $g$ is solvable if $L$ lies in a radical extension. I.e. $M$ is a radical extension of $g$ such that $g$ is solvable by radicals, then I want to see if $L= M$ to see if $L$ is a radical extension, or if $L\subsetneq M$

• For every cubic polynomial $f\in\Bbb Q[x]$, the splitting field of $f$ over $\Bbb Q$ is a radical extension. – Lubin Dec 16 '15 at 19:24