Finding the degree of the splitting field

Question

Find the degree of the splitting field of $f(x)=x^3-3x-1$ over $\mathbb{Q}$. I know that this polynomial is irreducible by using Eisenstein's criteria(by letting first $x=y+1$), and for every cubic polynomial $f∈\mathbb{Q}[x]$, the splitting field of $f$ over $\mathbb{Q}$ is a radical extension. But how can I start my work? Thanks!

Answer 1

The degree is 3. Since $f$ is irreducible, $L=Q[x]/(f)$ is a separable field of degree $3$ since the characteristic of $Q$ is zero. $Gal(L:Q)$ has order 3 and is generated by $h$. Let $u$ be a root of $f$ in $L$ we can take $u$ to be the class of $x$ in $Q[x]/(f)$, $h(u)\neq u$ since $u$ is not in $Q$. $h(u),h^2(u)$ are also roots of $f$ and remark that $u,h(u), h^2(u)$ are distinct (example, $h(u)=h^2(u)$ implies $h(h(u))=h^3(u)=u, h^2(u)=u, h(h^2(u))=h(u)=u$ contradiction). This implies that $L$ is a splitting field and $f=(X-u)(X-h(u))(X-h^2(u))$.

Answer 2

Our cubic is irreducible over the rationals by the Rational Roots Theorem.

Let $x=2t$. Our equation becomes $4t^3-3t=\frac{1}{2}$. Let $t=\cos\theta$. The roots $t$ are the $\cos(\theta)$, where $\cos(3\theta)=\frac{1}{2}$.

The roots $t$ are therefore the cosines of $20^\circ$, $100^\circ$, and $140^\circ$. Note that $\cos(140^\circ)=-\cos(40^\circ)=-(2\cos^2(20^\circ)-1)$. So the roots of our equation are all contained in $\mathbb{Q}(\cos(20^\circ))$, and therefore the splitting field has degree $3$ over the rationals.

Answer 3

Considering the $\DeclareMathOperator{\Gal}{Gal} \Gal(f)$ as a permutation group on the roots of $f$, since $f$ is irreducible, then $\Gal(f)$ must be a transitive subgroup of $S_3$. There are only two such transitive subgroups: $A_3 \cong \mathbb{Z}/3\mathbb{Z}$ and $S_3$ itself.

Recall that the Galois group of an irreducible polynomial $h$ of degree $n$ is a subgroup of $A_n$ iff the discriminant of $h$ is a square. (See Dummit and Foote, $\S14.6$, Proposition $34$ or this post, for example.) Since the discriminant of $f$ is $$ -4 (-3)^3 - 27 (-1)^2 = -27(-4 + 1) = 81 = 9^2 \, , $$ then $\Gal(f) \cong A_3 \cong \mathbb{Z}/3\mathbb{Z}$.