# constructing the Galois group for $\mathbb{Q}(\sqrt2, \sqrt[3]2)$

Suppose we have the extension field $\mathbb{Q}(\sqrt2, \sqrt[3]2)$.

For simplicity, let $\alpha = \sqrt2$ and $\beta = \sqrt[3]2$

First, I can show that $\mathbb{Q}(\alpha, \beta) = \mathbb{Q}(\alpha + \beta)$ by subset arguments, so to find an irreducible polynomial (which would be irreducible by the rational zeros theorem) I can do the following:

$x = \alpha + \beta$

$x - \alpha = \beta$

$(x - \alpha)^3 = 2$

$x^3 + 4x - 2\alpha(x^2 + 1) = 2$

$x^3 + 4x - 2 = 2\alpha(x^2 + 1)$

So..

$f(x) = (x^3 + 4x -2)^2 - 8(x^2 + 1)^2$

Which is of degree $6$, and so my extension is a degree $6$ extension.

I can also verify this using the tower law for extensions, since:

$[\mathbb{Q}(\alpha, \beta) : \mathbb{Q}(\alpha)] \cdot [\mathbb{Q}(\alpha) : \mathbb{Q}] = 3 \cdot 2 = 6$

So the Galois group for this extension will be isomorphic to a subgroup of $S_6$. That much is clear. Where I need help is in constructing the automorphisms. Besides the identity automorphism, I can have the following:

$\pi_1: \alpha \rightarrow -\alpha$

$\pi_2: \beta \rightarrow \beta^2$

$\pi_3: \alpha \rightarrow -\alpha, \beta \rightarrow \beta^2$

But if $\pi_2(\beta) = \beta^2$, then $\pi_2(\beta^2) = \beta^4 = 2\beta$ which seems weird. Something is wrong with what I am doing.

And what about something like this?

$\pi_4: \beta \rightarrow -\beta$

$\pi_5: \beta \rightarrow -\beta^2$

• That is not a Galois extension. – Lord Shark the Unknown Jul 18 '17 at 11:28
• Okay, I figured there was something very wrong. In order for it to be one, we'd need to also adjoin the 3rd root of unity, correct? – Yabbadule Jul 18 '17 at 14:37

Just keep in mind that an element of the Galois group over the rationals must take a root of a rational polynomial to another root. So how many choices do you get for $\sqrt{2}$ (which is a root of $x^{2} - 2$), and how many for $\sqrt[3]{2}$ (which is a root of $x^{3} - 2$)?
You may want to check what are the other roots of $x^{3} - 2$.