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?

  • $\begingroup$ GAP tells me that the Galois group is $S_3 \times S_3$ so your splitting field should have degree $36$ . $\endgroup$ – Marc Bogaerts Feb 12 '17 at 2:55
  • $\begingroup$ @MarcBogaerts What is GAP? $\endgroup$ – bellcircle Feb 12 '17 at 15:48
  • $\begingroup$ It is a (marvellousà compupter assisted algebra system. Gap stands for Groups, Algorithms and Programs. There is even a tag about it. More information here $\endgroup$ – Marc Bogaerts Feb 12 '17 at 18:37

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) $.

  • $\begingroup$ Is there any other solutions without Hansel's lemma? I'm not familiar with fields such as $\mathbb Q_{19}$. $\endgroup$ – bellcircle Feb 10 '17 at 3:57
  • 1
    $\begingroup$ Well, perhaps a quicker way is to note that $ \mathbf Q(2^{1/3}) $ is totally ramified at $ 3 $, while $ \mathbf Q(\alpha) $ is unramified at $ 3 $. $\endgroup$ – Starfall Feb 10 '17 at 9:36

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.

  • $\begingroup$ It's easier to look at the behavior of $ 3 $ instead of $ 2 $. $\endgroup$ – Starfall Feb 10 '17 at 22:59

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.


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.