# Proving that this field extension is normal

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?

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Armed with one root of that polynomial, how can you express the other roots in terms of it using roots of unity? Are those roots of unity also in the field? The use of 3 as the constant term is crucial, as the result would not be Galois is your replaced $x^6 + 3$ with, say, $x^6 + 5$. –  KCd Dec 5 '12 at 2:22
$a_0^3$ is not $8i$. –  Gerry Myerson Dec 5 '12 at 23:40

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 $a_k$.

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Yes But I don't know how to generate all the roots in this case, only given 1. For example let's suppose that we have ${w\root 6 \of 3 }$ how can i generate ${w^2\root 6 \of 3 }$? –  Daniel Dec 5 '12 at 4:51
That would depend somewhat on what you mean by $\omega$. But do notice that a sixth root of unity has an expression involving a square root of $-3$, and we are talking about a field generated by a sixth root of $-3$. –  Gerry Myerson Dec 5 '12 at 5:24
I'm following the previous notation –  Daniel Dec 5 '12 at 5:34
The symbol $\omega$ does not appear in your question, neither does it appear in my answer. If you want me to understand it, you should define it. If you don't care whether I understand it or not, what are we doing here? Anyway, did you read beyond "That would depend somewhat on what you mean by $\omega$"? Because with or without $\omega$, what I wrote was intended to answer your question. –  Gerry Myerson Dec 5 '12 at 5:49
It's an element such that $w^6=-1$ –  Daniel Dec 5 '12 at 5:54

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?

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Your $\beta_0$ is not a root of the polynomial. –  JSchlather Dec 5 '12 at 3:08
The roots are with $k=1..6$. How can I generate all of these with only 1? please show me :( –  Daniel Dec 5 '12 at 3:14
Thanks @JacobSchlather. Fixed. –  DonAntonio Dec 5 '12 at 3:16