Thread: Splitting field for X^3-5

Hi,

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

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


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


$\displaystyle [\mathbb{Q}(\alpha,\omega):\mathbb{Q}(\alpha)] >1 $ because $\displaystyle \omega \in \mathbb{C} \setminus \mathbb{R} $ and yet $\displaystyle \mathbb{Q}(\alpha) $ is a subfield of $\displaystyle \mathbb{R} $. Also the degree is $\displaystyle \leq 3 $ because $\displaystyle X^3 -1 $ has $\displaystyle \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 $\displaystyle \omega^2 + \omega = \sqrt{2} $

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

I can't see how do to this?

Thanks for any help!

The roots of $\displaystyle X^3-1$ are $\displaystyle X_1=1, X_2=e^{\frac{2 \pi i}{3}}, X_3 = e^{\frac{4 \pi i}{3}}$ whereby $\displaystyle X_2$ and $\displaystyle X_3$ have minimal polynomial $\displaystyle T^2+T+1$

Of course, I made a stupid mistake:

$\displaystyle \omega^2 + \omega = -1 $!

thanks