# a question about the Galois group for $x^3 - 2$

I know this question has been posed before, but there are some things that I'm still confused about.

First of all, I know that the splitting field for this particular polynomial will be the field of the rationals adjoined the cube-root of 2 and a primitive third root of unity: $\mathbb{Q}(\sqrt{2}, \gamma)$, where $\gamma^3 = 1$ and $\gamma^2 = 1 - \gamma$. I also know that since the degree of this extension is $6$, the order of my Galois group will also be $6$.

I know that my Galois group will consist of the following automorphisms:

$\sigma_i$ = identity

$\sigma_1: \sqrt{2} \mapsto \sqrt{2}\gamma$ and $\gamma$ is fixed.

$\sigma_2: \sqrt{1} \mapsto \sqrt{2}\gamma^2$ and $\gamma$ is fixed.

$\sigma_3: \gamma \mapsto \gamma^2$ and $\sqrt{2}$ is fixed.

$\sigma_4: \sqrt{2} \mapsto \sqrt{2}\gamma$ and $\gamma \mapsto \gamma^2$.

$\sigma_5: \sqrt{1} \mapsto \sqrt{2}\gamma^2$ and $\gamma \mapsto \gamma^2$.

If I didn't miss something or make some error.

My specific question is, when constructing my automorphisms, why exactly is it that I can't send $\sqrt{2}$ to something like $\sqrt{4}$ or $\gamma$ or some other random thing? Why are these automorphisms so strictly defined?

Also, my understanding is that the Galois group here would be isomorphic to some group of order 6, and there are two choices: $\mathbb{Z}_7$ or $S_3$. Is there an easier way of figuring out which it is other than finding the order of each automorphism and comparing that to the orders of elements in my two groups?

• $\gamma^2+\gamma+1=0$, not $\gamma^2+\gamma-1=0$. – egreg Jun 17 '16 at 11:00

let $j=\frac{1}{2}(-1+i\sqrt{3})$ .Clearly $\sqrt{2}\,,\sqrt{2}\,j\,$ and ${\sqrt{2}\,{j}^{2}}$ are roots of $f(x)=x^3-2$ . We claim $E=\mathbb{Q}(\sqrt{2},j)$ is splitting field $f$ over $\mathbb{Q}$ because $$E=\mathbb{Q}(\sqrt{2}\,,\sqrt{2}\,j\,,j^2\sqrt2)\subset\mathbb{Q}(\sqrt{2},j)$$ and $$j=\frac{1}{2}\sqrt{2}(\sqrt{2})^2j\implies\mathbb{Q}(\sqrt{2},j)\subset E=\mathbb{Q}(\sqrt{2}\,,\sqrt{2}\,j\,,j^2\sqrt2)$$ The polynomial $f$ is not irreducible over $\mathbb{Q}$ as a result minimal polynomial of $\sqrt2\,$ is f and $[\mathbb{Q}(\sqrt2\,):\mathbb{Q}]=3$ on the other hand $$[\mathbb{Q}(\sqrt2\,,j\,):\mathbb{Q}]=[\mathbb{Q}(\sqrt2,\,j):\mathbb{Q}(\sqrt2)][\mathbb{Q}(\sqrt2\,):\mathbb{Q}]=2\times3=6$$ in the other words $$[E:\mathbb{Q}]=6$$ so the Galois group is order 6. let
$A$: The metamorphism which fixes $j$ and maps $j\to j^2$
$B$: The metamorphism which fixes $\sqrt2$ and maps $\sqrt2\to \sqrt2\,j$
The Galois group is thus: $$\left\langle A,B\,|\,\,{{A}^{2}} \right.,\,{{B}^{3}},\,\,AB=\left. {{B}^{-1}}A \right\rangle \cong {{D}_{6}}$$
You can't send $2^{1/3}$ to $4^{1/3}$ because the image of $2^{1/3}$ has to be a root of $X^3-2$.
By the Fundamental Theorem of Galois Theorem, if the Galois group were abelian, then all subfields would be normal extensions. But this cannot be, as $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ is not a normal extension.