# Galois group of the splitting field of $x^6 - 5$

$f = x^6 - 5$ $\in \operatorname Q [x]$

I want to find a splitting field $\operatorname F$ of $f$ over $\operatorname Q$

$\sqrt{5}$ is a real root of $f$

$u$ is the 6-th root of unity

then $\operatorname F= \operatorname Q( \sqrt{5},u)$

f = x^6 - 5 is irreducible for Eisenstein's criterion, $\operatorname{char} \operatorname Q=0$ $\Rightarrow$ f is separable ⇒ F is a splitting field over $\operatorname Q$ of a separable polynomial. ⇒ $\operatorname{Gal}( \operatorname F/ \operatorname K)=(\operatorname F:\operatorname Q)$

I know that $\operatorname (Q (\sqrt{5}):\operatorname Q)=6$

How can I find the degree of $(\operatorname F : \operatorname Q (\sqrt{5}))$? I would have said $3$ but it's wrong, the solution is $2$, why?

And generally, in these excercises, how can I find the degree of the minimal polynomial of the n-th root?

• The 6th root of unity satisfies $x^6-1=0$, but that polynomial is not irreducible. Factor it into irreducibles, and see which one is the minimal polynomial for a primitive 6th root of unity. – Gerry Myerson Jul 28 '14 at 13:20
• $x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$, $x^2+x+1$ is irreducible, than the degree is $2$. Thank you, very helpful comment. – user158013 Jul 28 '14 at 13:30

$x^3 -1$ is not irreducible! That's why the degree is 2.