Calculating The Galois Group of the Splitting Field of $f=x^3-3$

If we let $f=x^3-3$ the let L be the splitting field for this polynomial I am trying to find $\Gamma(L:\mathbb{Q})$ and all intermediate field extensions.

Now as this is a splitting field and finite it is a finite normal extension and so we can use the fundemental theorem of Galois Theory so we know that:

$|\Gamma(L:\mathbb{Q})|=[L:\mathbb{Q}]$

Now we can see that if we define $\zeta=\exp{\frac{2\pi i}{3}}$ then we have that $L=\mathbb{Q}(\zeta,(3)^{\frac{1}{3}})$ and the minimal polynomial of $(3)^{\frac{1}{3}}$ over $\mathbb{Q}$ has degree 3 and the minimal polynomial of $\zeta$ has degree 2 over $\mathbb{Q}((3)^{\frac{1}{3}})$ so from the tower law we can see that:

$[L:\mathbb{Q}]=6$

Now we can see that a basis for $L$ is given by $\displaystyle{\{1,\zeta,(3)^{\frac{1}{3}},(3)^{\frac{2}{3}},(3)^{\frac{1}{3}}\zeta,(3)^{\frac{2}{3}}\zeta\}}$

So if we now consider $\mathbb{Q}$ automorphisms of $L$ then we need only consider it's action on the basis elements, so we can see that if $f$ is a $\mathbb{Q}$ automorphism then we must have that:

$f(\zeta)=\zeta,(3)^{\frac{1}{3}}\zeta$ and

$f((3)^{\frac{1}{3}})=(3)^{\frac{1}{3}},(3)^{\frac{2}{3}},-\zeta$

which will give us the 6 elements of $\Gamma(L:\mathbb{Q})$

Is what I have done so far correct? and if so how do I now find the intermediate fields? I know that they are related to the subgroups but I am unsure how?

Thanks for any help

-
In the very first line I think that + must be an = ... – DonAntonio May 10 '13 at 14:21
If you write $\,(-1)^{1/3}\,$ chances are people will think this is $\,-1\,$ – DonAntonio May 10 '13 at 14:22
@DonAntonio yeah thats what it was supposed to be, what should I write instead of $(-1)^{\frac{1}{3}}$ just $\alpha$ where $\alpha$ is a non-real third root of unity? – hmmmm May 10 '13 at 14:40
it is customary to write $\,\zeta_3=e^{2\pi i/3}\,$ to avoid confusions, or simply $\,\zeta\,$ is there's only one such root of unity. – DonAntonio May 10 '13 at 14:42
@DonAntonio cool thanks I did not know that, I will edit it now and just put $\exp{\frac{2\pi i}{3}}$ in instead I can see that this is much more clear. – hmmmm May 10 '13 at 14:44

I'll add a few things to this. You know that the Galois group has order 6, so there are only two choices: $S_3$ and $Z_3\times Z_2$. The latter group is abelian, so this means that every subgroup is normal. Thus, let $L/\mathbb{Q}$ be the splitting field and denote the Galois group by $G$.

The fundamental theorem also says that there's a bijective correspondence between subextensions $L/K/\mathbb{Q}$ such that $K/\mathbb{Q}$ is Galois and normal subgroups of $G$. Take the subextension $\mathbb{Q}(\sqrt[3]{3})/\mathbb{Q}$. This extension is not normal, so the subgroup $\textrm{Aut}(L/\mathbb{Q}(\sqrt[3]{3}))\leq G$ is not a normal subgroup. This implies that $G$ can't be abelian, so it must be $S_3$.

Now $S_3$ has the following subgroups:

• A unique subgroup of order $3$.

• $3$ subgroups of order $2$ that are all conjugate. These are the ones generated by $(1\, 2)$, $(1\, 3)$ and $(2\, 3)$.

The unique order $3$ subgroup is normal, so it corresponds to a unique quadratic Galois extension $K/\mathbb{Q}$. This must be the field extension $\mathbb{Q}(\zeta_3)/\mathbb{Q}$, which we know is Galois.

Next, we find the three non-Galois cubic subextensions of $\mathbb{Q}$. These are the field extensions corresponding to each root of $X^3-3$, so we get the extensions:

$$\mathbb{Q}(\sqrt[3]{3})/\mathbb{Q},\;\;\mathbb{Q}(\zeta_3\sqrt[3]{3})/\mathbb{Q},\;\;\mathbb{Q}(\zeta_3^2\sqrt[3]{3})/\mathbb{Q}$$

To show that these extensions are all distinct, note that if e.g. $\mathbb{Q}(\sqrt[3]{3})=\mathbb{Q}(\zeta_3\sqrt[3]{3})$, then $\zeta_3 = \zeta_3\sqrt[3]{3}/\sqrt[3]{3}\in \mathbb{Q}(\sqrt[3]{3})$. This is impossible, since $L\neq \mathbb{Q}(\sqrt[3]{3})$. The same argument applies to the other two pairs.

-
An easier way to see why the group is $S_3$ is that the Galois group of (the splitting field of) a polynomial of degree $n$ acts by permuting the roots of the polynomial, and thus is always a subgroup of $S_n$. – Hurkyl May 13 '13 at 0:25
Yes, there are much slicker ways of doing this depending on how much you know. I chose this method, since for trickier cases you have to look at the subgroup lattice. – Edvard Fagerholm May 16 '13 at 4:47