Splitting field for X^3-5

**Ant** · May 18th 2013, 04:52 AM

Hi,

Let $f=X^3 -5$ , the $\Sigma_f = \mathbb{Q}[\alpha,\omega]$ with $\alpha$ real cube root of 5 and $\omega$ primitive root of 1.

$[\mathbb{Q}(\alpha):\mathbb{Q}]=3$ because $X^3-5$ is irreducible.

But what is $[\mathbb{Q}(\alpha,\omega):\mathbb{Q}(\alpha)]$ ?

$[\mathbb{Q}(\alpha,\omega):\mathbb{Q}(\alpha)] >1$ because $\omega \in \mathbb{C} \setminus \mathbb{R}$ and yet $\mathbb{Q}(\alpha)$ is a subfield of $\mathbb{R}$ . Also the degree is $\leq 3$ because $X^3 -1$ has $\omega$ as a root. So the possibilities are 2 or 3.

If it's 2. Then because the min poly has to be monic and $\omega^2 + \omega = \sqrt{2}$

I think that we have to express $\sqrt{2}$ as a linear combination (with rational coefficients) of $\alpha$ and $\alpha^2$ .

I can't see how do to this?

Thanks for any help!