Degree of splitting field of $x^6-2$ over $\mathbb{Q}$

by user zero   Last Updated September 11, 2019 19:20 PM

Let $f=x^6-2$, find the degree of splitting field of $f$ over $\mathbb{Q}$.

I calculated the roots of $f$ are $\pm \sqrt[6]{2},\pm e^{i \pi/3}\sqrt[6]{2},\pm e^{2i \pi/3}\sqrt[6]{2}$. I suspect that the splitting field of $f$ is $\mathbb{Q}(e^{i \pi/3}\sqrt[6]{2})$ but I don't know how to prove this. Any idea?

Answers 2

The splitting field of $f$ is $\mathbb{Q}(e^{i \pi/3},\sqrt[6]{2})$, which is not the same as $\mathbb{Q}(e^{i \pi/3}\sqrt[6]{2})$.

It has degree $12$ since $e^{i \pi/3}$ has degree $2$ and $\sqrt[6]{2}$ is real.

April 26, 2016 00:32 AM

The splitting field of $f$ is the field $\mathbf Q(\zeta,\sqrt[6]2)$, where $\zeta$ is a primitive $6$th root of unity. The minimal polynomial of $\zeta$ is the cyclotomic polynomial $\;\Phi_6(x)=x^2-x+1$. so that $$[\mathbf Q(\zeta:\mathbf Q]=2, \enspace[\mathbf Q(\sqrt[6]2:\mathbf Q]=6.$$ Furthermore, these extensions are linearly independent and thus $$\mathbf Q(\zeta\sqrt[6]2):\mathbf Q]=[\mathbf Q(\zeta:\mathbf Q][\mathbf Q(\sqrt[6]2:\mathbf Q]=12.$$

April 26, 2016 00:50 AM

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