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I am trying to prove that $K=\mathbb{Q}(2^{1/3}, i\sin{2\pi/3})$ is Galois extension over $\mathbb{Q}$. It is easy to see that $K=\mathbb{Q}(2^{1/3},i\sqrt{3})$. I know it is Galois since $K$ is a splitting field of the separable polynomial $f(x)=x^3-2$.

Now I am trying to show this using the other method, that is by explicitly computing the automorphisms. I found that $|Aut(E/F)|=6$. However I am having a hard time proving $[K:\mathbb{Q}]=6$. In particular I am having a hard time finding an irreducible degree 2 polynomial which will say that $[K:\mathbb{Q}(2^{1/3})]=2$.


marked as duplicate by Dietrich Burde, Davide Giraudo, kjetil b halvorsen, Did, Daniel W. Farlow Apr 6 '15 at 10:57

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  • 2
    $\begingroup$ Something’s wrong. The sine of $2\pi/3$ is $\sqrt3$. You want $i\sin(2\pi/3)$. But if you notice that $K$ is also equal to $\Bbb Q(2^{1/3},\omega)$, you’ll be in good shape. $\omega=\exp(2i\pi/3)=(i\sqrt3-1)/2$, a root of $X^2+X+1$, and a primitive cube root of unity. $\endgroup$ – Lubin Apr 6 '15 at 1:33
  • $\begingroup$ @Lubin: Sorry for the typo! Thanks for your help $\endgroup$ – user104221 Apr 6 '15 at 1:46
  • $\begingroup$ For the degree see here. $\endgroup$ – Dietrich Burde Apr 6 '15 at 7:55

Once you have proven that $Aut(K/\mathbb{Q})$ is of cardinal $6$, you just need to know that $[K:\mathbb{Q}]$ is below $6$, which is an easy thing to do because :

$$[K:\mathbb{Q}]=[K:\mathbb{Q}(2^{\frac{1}{3}})][\mathbb{Q}(2^{\frac{1}{3}}):\mathbb{Q}] $$

We easily get that :


And :

$$[K:\mathbb{Q}(2^{\frac{1}{3}})]=1\text{ or } 2$$

because $x^2+3$ is a polynomial with coefficients in $\mathbb{Q}(2^{\frac{1}{3}})$ and $i\sqrt{3}$ is one of its root (for the inequality you do not need to show it is irreducible). Finally :

$$[K:\mathbb{Q}]\leq 6 $$

Now : $6=|Aut(K/\mathbb{Q})|\leq [K:\mathbb{Q}]\leq 6$ which allows you to say that $[K:\mathbb{Q}]= 6$.

  • $\begingroup$ No, splitting field is the right terminology (decomposition field is only in french ?). A field extension with at least one root is called a rupture field. $\endgroup$ – mercio Apr 6 '15 at 9:35
  • $\begingroup$ @mercio Thank you for correcting my english I'll delete this part immediately. $\endgroup$ – Clément Guérin Apr 6 '15 at 9:36

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