Compute the following Galois group. Then determine whether or not it is a normal field extension or not. If the extension is not normal, find a normal extension of $\mathbb{Q}$ in which the extension field is contained.

$G(\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}, i) / \mathbb{Q})$.

Attempt : This does not have a normal field extension since $\sqrt[3]{2}$ has roots in $\mathbb{C}$ and we need a normal field extension over $\mathbb{Q}$. I am unsure how to find one that is normal though. I believe there are four automorphisms then. The $\sqrt{2}$ goes to $\sqrt{2}$ and $-\sqrt{2}$, and $i$ goes to $i$ and $-i$.

  • $\begingroup$ What do you mean by $2^{\frac{1}{3}}$ has no solution in $\mathbb{C}$? $\endgroup$
    – LASV
    Nov 13, 2013 at 22:52
  • $\begingroup$ It has complex roots which are not in the rationals. $\endgroup$
    – user105998
    Nov 13, 2013 at 22:53
  • $\begingroup$ Yes, but in your field extension you are adjoining $i$. How do you know it is not the splitting field of a family of polynomials? It is true that $\mathbb{Q}(2^{\frac{1}{3}})$ is not a normal extension of $\mathbb{Q}$, but that is not the field extension we are looking at $\endgroup$
    – LASV
    Nov 13, 2013 at 22:55

1 Answer 1


A Galois extension wich contains $\sqrt[3]2$ must contains the 3-rd primitiva root of 1 $\zeta_3=\frac{1}{2}+\frac{i\sqrt2}{2}$.

A Galois extension wich contains $\zeta_3$ and $i$ must then contain $\sqrt3$.

Follows that $\mathbb{Q}(\sqrt2,\sqrt[3]2, i )$ is not normal over $\mathbb{Q}$.

Consider now $\mathbb{Q}(\sqrt2,\sqrt[3]2, i, \sqrt3 )$. It is normal because is the splitting field over $\mathbb{Q}$ of the polynomial $(x^3-2)(x^2-2)(x^2+1)$ and contains your extension.

It's also interesting to observe that $\mathbb{Q}(\sqrt2,\sqrt[3]2, i ,\sqrt3)$ is the smallest Galois extension of $\mathbb{Q}$ wich contains $\mathbb{Q}(\sqrt2,\sqrt[3]2, i )$, beacause $[\mathbb{Q}(\sqrt2,\sqrt[3]2, i \sqrt3 ):\mathbb{Q}(\sqrt2,\sqrt[3]2, i )]=2$.

We want now calculate the Galois group of $\mathbb{Q}(\sqrt2,\sqrt[3]2, i, \sqrt3 )$. This group must have order 24.

Now observe that $\mathbb{Q}(\sqrt2,\sqrt[3]2, i, \sqrt3 )$ has two subestension , $\mathbb{Q}(\sqrt[3]2, \zeta_3)$ and $\mathbb{Q}(\sqrt2, i)$, wich are Galois subestension with Galois groups, respectively, $S_3$ and $\mathbb{Z}_2 \times \mathbb{Z}_2$.

But $\mathbb{Q}(\sqrt[3]2, \zeta_3) \cap \mathbb{Q}(\sqrt2, i)=\mathbb{Q}$ and $\mathbb{Q}(\sqrt[3]2, \zeta_3)\mathbb{Q}(\sqrt2, i)=\mathbb{Q}(\sqrt2,\sqrt[3]2, i, \sqrt3 )$. Then we obtain that the Galois Group of $\mathbb{Q}(\sqrt2,\sqrt[3]2, i, \sqrt3 )$ is $S_3 \times \mathbb{Z}_2\times \mathbb{Z}_2$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .