Problem with a Galois extension

Question

How can i prove that Q(a,b) is a Galois extension and that its Galois group is of order 6, if a is a root of the cubic that is not 1 and b is the cubic root of 2?

Can you give me a little hint for the proof only that a finite extension (the Galois extension), i have the rest of the problem.

I think Julio is trying to say that $a$ is one of the nontrivial cube roots of $1$. (The phrase "root of the cubic" does not at all communicate this, unfortunately.) — anon, May 16 '14 at 18:32
Julio, you forgot to tell that $a$ is a cubic root of $1$ as opposed to, say a cubic root of $8$ or $216$ or $5$ or ... — Jyrki Lahtonen, May 17 '14 at 04:58
Im sorry, i forgot that. a is a cubic root of 1, that isnt the trivial one. — Julio, May 18 '14 at 01:16

score 0 · Answer 1 · answered May 16 '14 at 18:15

0

First of all argue that: since $\mathbb{Q}(a)/\mathbb{Q}$ is an extension of degree 3 and $\mathbb{Q}(b)/\mathbb{Q}(a)$ is of degree 2 we have $\mathbb{Q}(a,b)/\mathbb{Q}$ is a degree 6 extension. Moreover $\mathbb{Q}(a,b)$ is the splitting field of $f(x)=x^3-2$.

answered May 16 '14 at 18:15

Kal S.

3,781

ok, thank´s but, that´s the part that i have – Julio May 16 '14 at 18:17
@Julio So what are you missing? – Kal S. May 16 '14 at 18:18
the part of the finiteness – Julio May 16 '14 at 18:19

Problem with a Galois extension

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