0
$\begingroup$

Let's consider the polynomial $f=x^6+3 \in \Bbb Q[x]$. I have to prove that for some root $\beta $ of $f$. The extension $ \Bbb Q (\beta) $ is galois.

In other words if $ {\root 6 \of { - 3} } $ denotes some root of $f$. Then the extension $ {\Bbb Q}\left( {\root 6 \of { - 3} } \right)/{\Bbb Q} $ Is Galois

What I tried

Let's denote $ w = e^{\frac{i\pi}{3}}$ i.e the primitive 6-root of unity $w^6=1$. Let's denote $a_0 = \root 6 \of 3 e^{\frac{{i\pi }} {6}} $. The roots of $f$ are $ a_k = \root 6 \of { - 3} w^k \,\,\,k = 0..5 $ Given $a_0$ I want to generate all the roots, If I generate $a_1$ I'm done, but I don't know how. Here are some of my computations: $ \eqalign{ & a_0 = \frac{{\root 6 \of 3 }} {2}\left( {\sqrt 3 + i} \right) \cr & a_0 ^2 = \frac{{\root 3 \of 3 }} {2}\left( {1 + i\sqrt 3 } \right) \cr & a_0 ^3 = 8i\, \cr} $ I want to generate $ a_1 = a_0 w = \frac{{\root 6 \of 3 }} {2}\left( {\sqrt 3 + i} \right)\frac{1} {2}e^{\frac{{i\pi }} {3}} = i\root 6 \of 3 $ . How can I do it?

  • 0
    $a_0^3$ is not $8i$.2012-12-05

2 Answers 2

0

Can you find all the roots of that $f$?

Can you show that if any one of them is in an extension $K$ of $\bf Q$, then all of them are?

Do you know that that's enough to imply that the extension is Galois?

EDIT: Adopting the notation in the body of the current version of the question, we have $\omega={1\over2}+{\sqrt{-3}\over2}$ and $a_0^3=\sqrt{-3}$ from which it follows that $\omega=(a_0^3+1)/2$ and $a_k=2^{-k}a_0(a_0^3+1)^k$ Thus, any extension of the rationals containing $a_0$ contains all of the $a_k$.

  • 0
    I did not put , that number2012-12-06
0

The roots of $\,f(x)=x^6+3=0\,$ are

$\beta_k:=\sqrt[6] 3\;w^k\,\,,\,w:=e^{\pi i/6}\,\,,\,\,k=1,2,...,6$

So what root $\,\beta\,$ do you like the most to check?

  • 0
    Thanks @JacobSchlather. Fixed.2012-12-05