Degree of splitting field of $x^6+3x^3-2$ over $\mathbb Q$

Question

This is the sequel of the question Is $x^6 + 3x^3 -2$ irreducible over $\mathbb Q$?

Thanks to the answer, I finally found out the answer and I also figured out that the splitting field of $x^6 + 3x^3 -2$ over $\mathbb{Q}$ is $\mathbb Q(\sqrt[3]{2},\alpha,\zeta) $, where $\alpha= \sqrt[3]{\frac{ \sqrt{17}-3}{2}}$ and $\zeta$ is a primitive 3rd root of unity.

To compute the degree of the splitting field, I attempted as follows:

$[\mathbb Q(\sqrt[3]{2},\alpha,\zeta):\mathbb Q]=[\mathbb Q(\sqrt[3]{2},\alpha,\zeta):\mathbb Q(\sqrt[3]{2},\alpha)][\mathbb Q(\sqrt[3]{2}, \alpha):\mathbb Q(\alpha)][\mathbb Q(\alpha):\mathbb Q]$

Since $\zeta$ is imaginary, I easily found out that $[\mathbb Q(\sqrt[3]{2},\alpha,\zeta):\mathbb Q(\sqrt[3]{2},\alpha)]=2$.

Also, as I found out at the previous question, $[\mathbb Q(\alpha):\mathbb Q]=6$.

However, I have trouble figuring out $[\mathbb Q(\sqrt[3]{2}, \alpha):\mathbb Q(\alpha)]$. I guess "there is no $\beta \in \mathbb Q(\alpha)$ satisfying $\beta^3=2$", so the polynomial $x^3 -2$ does not have linear factors in $\mathbb Q(\alpha)$, concluding that $[\mathbb Q(\sqrt[3]{2}, \alpha):\mathbb Q(\alpha)]=3$

I want to prove "". Does anyone have any idea?

Answer 1

We only need to show that $ 2 $ has no cube root in $ \mathbf Q(\alpha) $. $ X^6 + 3X^3 - 2 $ has the single root $ X = 5 $ in $ \mathbb F_{19} $, which lifts to a root in $ \mathbf Q_{19} $ by Hensel's lemma to give an embedding $ \mathbf Q(\alpha) \to \mathbf Q_{19} $. However, $ X^3 - 2 $ is irreducible modulo $ 19 $, thus irreducible in $ \mathbf Z_{19}[X] $, and since $ \mathbf Z_p $ is integrally closed, in $ \mathbf Q_{19}[X] $. It follows that, in particular, $ X^3 - 2 $ is irreducible in $ \mathbf Q(\alpha)[X] $, and thus $ 2^{1/3} $ has degree $ 3 $ over $ \mathbf Q(\alpha) $.

Answer 2

If you can prove that 2 does not divide $[O_K:Z[\alpha]]$ , then by Dedekind-Kummer the ideal (2) should factorise analogously to $x^6+3x^3-2=x^3.(x+1).(x^2-x+1)(mod 2)$.But if $\sqrt[3]2$ was in K, then $(2)=(\sqrt[3]2)^3$, but that would contradict the factorisation mentioned before.

Answer 3

A simpler, but more tedious approach goes like this:Obviously if $x^3-2$ isn't irreducible in $K=Q(\alpha)$, then the cube root of 2 is in $Q(\alpha)$. Let $\sqrt[3]2 = b_0+\alpha.b_1+ \alpha^2.b_2+ \alpha^3.b_3+\alpha^4.b_4+\alpha^5.b_5$. Therefore, taking the trace from K down to Q on both sides we get $0=6b_0$, so $b_0=0$. Since $N(\alpha)=-2$, then the ideal $(\alpha)$ must be prime, lying over 2.The same applies to the ideal $(\alpha-1)$. From before we know that $\alpha$ divides $\sqrt[3]2$.Also, the ideal $(\sqrt[3]2)$ is prime in $Q(\sqrt[3]2)$, and $(\alpha)$ is prime lying over it in $Q(\alpha)$.Since $[Q(\alpha):Q(\sqrt[3]2)]=2$, then the ideal $(\sqrt[3]2)$ must factorise in one of the following ways:

$\textbf1) (\sqrt[3]2)=P$

Since $(\alpha)$ is prime lying over $(\sqrt[3]2)$, then it must be P.So $\sqrt[3]2=\alpha.u,$ where u is a unit.But by taking norms this is seen to be impossible.

$\textbf2)(\sqrt[3]2)=P^2$

Again $P=(\alpha)$.Now $\sqrt[3]2=\alpha^2.u,$ where u is a unit with norm 1.Cubing both sides and using $\alpha= \sqrt[3]{\frac{ \sqrt{17}-3}{2}}$ we get $4=(13-3\sqrt{17}).u^3$. Taking norms we get $4^6=23^3$, which is obviously wrong.

$\textbf3)(\sqrt[3]2)=PQ$

From this we get that $(2)=P^3Q^3$, but we know that $|N(\alpha)|=|N(\alpha-1)|=2$, so they are prime and lie over 2, so P and Q must be $(\alpha)$ and $(\alpha-1)$. Therefore $2=\alpha^3(\alpha-1)^3.u^3$. Since $2=\alpha^3(\alpha^3+3)$ by definition of $\alpha$, we would have that $(\alpha-1)^3|\alpha^3+3=-(\alpha-1)(1+\alpha+\alpha^2)(7+2\alpha^3)$.But this is impossible.

None of the cases hold, so our assumption was wrong.