Galois group of $x^6+3$

Asked Nov 18 '17 at 12:18

Active Nov 18 '17 at 12:22

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I need your help to answer the following problem:

Let $f(x)=x^6+3\in\mathbb{Q}[x]$

Let $\alpha$ be a root of $f(x)$ and $L=\mathbb{Q}(\alpha)$.

a) Show that $L$ is a splitting field for $f(x)$ over $\mathbb{Q}$

Hint: Show that $w=\frac{1+\alpha^3}{2}$ is a primitive sixth root of unity.

b) determine $Gal(L/\mathbb{Q})$ .

I answered question a) but i don't manage to answer question b).

Thanks

edited Nov 18 '17 at 12:22

Gerry Myerson

asked Nov 18 '17 at 12:18

Start by proving $f$ is irreducible over the rationals. – Gerry Myerson Nov 18 '17 at 12:21
i proved that $f$ is irreducible over $\mathbb{Q}$ by using Einstein's cretiria with p=3 but how to use that to answer? – Nov 18 '17 at 12:27
2

Saying "I answered question a)" is a bit disingenuous, considering https://math.stackexchange.com/questions/2525878/splitting-field-over-mathbbq where you asked it and it was answered for you. – Gerry Myerson Nov 18 '17 at 12:32
Proving irreducibility (together with the rest of what you know) tells you the order of the Galois group. – Gerry Myerson Nov 18 '17 at 12:33
how that please? – Nov 18 '17 at 12:34
Well, what theorem do you know about the number of elements of the Galois group? – Gerry Myerson Nov 18 '17 at 12:36
do you mean Fundamental theorem of Galois theory? – Nov 18 '17 at 12:40
I'm not interested in what you call it. I mean, do you know a theorem that relates the order of the Galois group to the degree of the normal extension? – Gerry Myerson Nov 18 '17 at 21:04

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