# Galois group isomorphic to $S_3$

I am trying to find the proper subfields between $\mathbb Q(2^{1/3},\omega)$ and $\mathbb Q$ . where $\omega$ is a cube root of unity . I have found that finding the galois groups and the proper subfields one has to be very careful , and its very easy to conclude something wrong .

Here taking this particular case , finding the possible automorphism which keep $\mathbb Q$ fixed we observe the following things ,

The vector space of $\mathbb Q(2^{1/3},\omega)$ over $\mathbb Q$ will have elements of the form ,

$a+b 2^{1/3}+c 2^{2/3} +d\omega +e 2^{1/3} \omega +f 2^{2/3} \omega$ .

I find it quite problematic to find whether mapping one "root " to the other gives an automorphism or not .

Is it true that first we need to find the set of linearly independent "basis" of the extension before we find the set of automorphism .

May be my question is very vague but my question is what are the very basic facts one needs to take care while finding galois group and the proper subfields.

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Do you mean $\omega$ is a primitive cube root of unity ? –  Belgi Dec 28 '12 at 13:41
I don't understand the relation between the post and the title - Do you think (or know ?) that the Galois group is $S_3$ and you just need help to prove it ? –  Belgi Dec 28 '12 at 13:42
@Belgi yes . i know the fact , but i was to investigate more on this problem and understand completely –  Theorem Dec 28 '12 at 13:44
Did you figure what is the degree of the extension ? –  Belgi Dec 28 '12 at 13:45
@Belgi : degree of extension is 6 . –  Theorem Dec 28 '12 at 13:56

Hints:

1. $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=?$ because $x^{3}-2\in\mathbb{Q}[x]$ is irreducible (why $?$ )

2. $[\mathbb{Q}(\omega):\mathbb{Q}]=2$ because $\frac{x^{3}-1}{x-1}=x^{2}+x+1$ is the minimal polynomial of $\omega$ over $\mathbb{Q}$(If you know more theory than you know it is $\phi(3)=2)$

3. Since the degrees are co-prine, what is $[\mathbb{Q}(\sqrt[3]{2},\omega):\mathbb{Q}]$ ?

4. In general, where can an automorphism (over $\mathbb{Q}$) map some (algebraic) $\alpha$ ?

5. You may want to use the fact that there are only two groups of order $6$, one is abelian (even cyclic) and the other is not ($S_3$), you can show the Galois group is $S_3$ in other ways ofcourse

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Very good hints. –  Bombyx mori Dec 28 '12 at 14:21
@user32240 - thanks! –  Belgi Dec 28 '12 at 14:26

$Q(2^1/3,ω)$ over $Q$ is normal being the splitting field of $x^3-2$ so you can map any root of x^3-2 to another

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reason for downvote???? –  Koushik Dec 28 '12 at 13:52
I did not downvote, but you should really say root of what –  Belgi Dec 28 '12 at 13:56
Also, there is still a lot of work to do to get to the answer from here... –  Belgi Dec 28 '12 at 14:10

I suppose this is not an exercise problem, so let me try to explain more on the matter. Firstly, an automorphism leaving $Q$ fixed must leave all polynomials with rational coefficients fixed, hence maps every root of one polynomial to another root of the same polynomial. This could be regarded as a precise way of saying that automorphisms preserve the algebraic structures on a field. Conversely, for every conjugate of $\alpha$, there is a unique automorphism of $F( \alpha )$ mapping $\alpha$ to that conjugate. Hence, the number of automorphisms of $F( \alpha )$ is equal to the degree of $\alpha$ over $F$, when $F( \alpha )$ is a separable splitting field of $F$. Notice, however, that $2^{1/3}$ is not a generator of your field, so you have to find a single generator of that extension to directly apply the above. Of course you could employ the above to the case of two generators and conclude from here the structure of the group of the extension.
In summary, to compute the Galois group of an extension it is key to choose some convenient basis, and to find the conjugates of the generators.

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