Finding the degree of the splitting field

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!

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 $f=(X-u)(X-h(u))(X-h^2(u))$.

• How do you know that the Galois group has order $3$? We can't even really speak of the Galois group since we don't know the extension is Galois, just the automorphism group $\text{Aut}(L|\mathbb{Q})$. Anyway, your reasoning seems to apply equally well to $x^3 - 2$ which has trivial automorphism group. – André 3000 Mar 15 '16 at 18:24
• fr.wikipedia.org/wiki/… – Tsemo Aristide Mar 15 '16 at 18:32
• What is this link supposed to tell me? $x^3 - 2$ generates a separable degree $3$ extension, too. – André 3000 Mar 15 '16 at 19:03
• This link tell you that $L$ is a separable extension of $F$, if and only if the cardinal of the Galois group $Gal(L:F)$ is the degree $[L:F]$ – Tsemo Aristide Mar 15 '16 at 19:13
• I don't see where it says that in the link. To be Galois, an extension must be separable and normal, i.e., all roots of the polynomial must also be contained in the field. I don't see where you've shown that, and $\mathbb{Q}(\sqrt[3]{2})$ is an example of an extension of $\mathbb{Q}$ that is separable but not normal: it doesn't contain the two other roots $\zeta \sqrt[3]{2}, \zeta^2 \sqrt[3]{2}$ where $\zeta$ is a primitive $3^\text{rd}$ root of $1$. – André 3000 Mar 15 '16 at 19:27

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.

• If $\alpha$ is a root there is another defined by $2\beta=\alpha+\sqrt{(\alpha)^2-\frac{4} {\alpha}}$. This makes me to be wrong about the degree. Thank you. Your answer is fine. – Piquito Mar 15 '16 at 0:41
• I gave this answer for fun, because of its connection to a famous problem. Degrees were deliberate, to give it a retro look. – André Nicolas Mar 15 '16 at 0:46
• trisection of the angle, by chance? – Piquito Mar 15 '16 at 1:03
• Yes, one works with $4t^3-3t=\frac{1}{2}$ to show that the $20^\circ$ angle is not compass and straightedge constructible. – André Nicolas Mar 15 '16 at 1:06
• Absolutely, the only constructible magnitudes are these expressed algebraically by formulas leading to extensions of $\mathbb Q$ of degree $2^n$ (and degree $3$ is not of this kind). Thank you. – Piquito Mar 15 '16 at 1:15