Determine all automorphisms of $\mathbb{Q}(\sqrt[3]{2},\omega)$

Determine all automorphisms of $\mathbb{Q}(\sqrt[3]{2},\omega)$

Here $\omega$ is a complex cube root of unity.

From an earlier part of the problem, I have that the only automorphism of $\mathbb{Q}(\sqrt[3]{2})$ is the identity map. I've found that $[\mathbb{Q}(\sqrt[3]{2},\omega):\mathbb{Q}]=6$. After that, I have a bunch of very hazy notions about how to proceed. The text I'm using has just defined Galois group and Galois extension but hasn't yet given any indication of what they're for. (It always seems to me that peeking ahead in the book to get a solution is only getting me to miss some more basic idea I'm supposed to be getting).

So I know that an automorphism, when applied to the root of a polynomial, will give me another root of the polynomial. I know that since $[\mathbb{Q}(\sqrt[3]{2},\omega):\mathbb{Q}]=6$, there is a 6th degree irreducible polynomial, a root of which will generate $\mathbb{Q}(\sqrt[3]{2},\omega)$. This polynomial, whatever it is, has 6 roots that get permuted around by the automorphisms.

So 6! automorphisms? That seems a bit much. But how do I knock it down to a reasonable number? And after that, how do I find out what they are?

Thanks.

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Not every permutation of roots is an automorphism: it must preserve the field structure to be the latter. –  anon Apr 11 '12 at 19:35
Any automorphism is determined by where it sends $\sqrt[3]{2}$ and $\omega$. $\sqrt[3]{2}$ must go to one of the three cube roots of three, while $\omega$ must go to one of the two complex cube roots of one. Thus there are at most $3 \cdot 2=6$ automorphisms. –  Chris Eagle Apr 11 '12 at 19:38
So you know that automorphisms applied to the roots of a polynomial (with rational coefficients!) give you other roots of the polynomial. This doesn't just apply to the minimal polynomial of a primitive element; it also applies to $x^3 - 2$ and to $x^2 + x + 1$... –  Qiaochu Yuan Apr 11 '12 at 19:40
If this is homework, tag it as such. –  lhf Apr 11 '12 at 19:46
@lhf: just added homework tag. Thanks –  jobrien929 Apr 11 '12 at 19:51

There is a general techinque for finding Galois groups. I shall outline them for this actual case:

1. Verify that your extension is Galois. For example, I claim that our extension is the splitting field of $x^3-2$. Indeed, the roots of $x^3-2$ are $\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2}$ and clearly $\mathbb{Q}(\sqrt[3]{2},\omega)$ is equal the field generated by these roots over $\mathbb{Q}$. Thus, our extension is the splitting field of a separable polynomial.

2. Find the degree of your extension. For us it's pretty simple since one can verify that the following tower holds

$$\begin{array}{c}\mathbb{Q}(\sqrt[3]{2},\omega)\\ \vert 2\\ \mathbb{Q}(\sqrt[3]{2})\\ \vert 3\\ \mathbb{Q}\end{array}$$

The first from the fact that we know that minimal polynomial $m_{\mathbb{Q},\sqrt[3]{2}}=x^3-2$ and the second since evidently $1+x+x^2\in\mathbb{Q}(\sqrt[3]{2})[x]$ annihilates $\omega$ and $\omega\notin\mathbb{Q}(\sqrt[3]{2})$ (since it's not real). Thus the multiplicative property of towers gives that our extension is degree six.

3.Find how many possible elements can be in the Galois group of your extension. In our case, you have noted that any $\sigma\in\text{Gal}(\mathbb{Q}(\sqrt[3]{2},\omega)/\mathbb{Q})$ must send $\sqrt[3]{2}$ to another root of its minimal polynomial, which there are three (since its minimal polynomial, as we know, is $x^3-2$), and $\omega$ must be sent to a root of its minimal polynomial $1+x+x^2$ of which there are two. Thus, the possible values that $\sigma$ sends $(\sqrt[3]{2},\omega)$ to are $(x,y)$ such that $x\in\{\sqrt[3][2],\omega\sqrt[3]{2},\omega^2\sqrt[3]{2}\}$ and $y\in\{\omega,\omega^2\}$. Since any $\sigma$ is determined by the value it takes on $(\sqrt[3]{2},\omega)$ (since they are generators) you can conclude there are at most six $\sigma$'s (not the statistical framework) and so $|\text{Gal}(\mathbb{Q}(\sqrt[3]{2},\omega)/\mathbb{Q})|\leqslant 6$.

4.Use the fact that our extension is Galois to conclude that $|\text{Gal}(\mathbb{Q}(\sqrt[3]{2},\omega)/\mathbb{Q})|$ is the degree of our extension, which is six. Aha! But we know that our Galois group has at most six elements--those $\sigma$ sending the generators to the six possible choices for $(x,y)$. But, since there MUST be six automorphisms we may conclude that all six possible automorphisms ARE automorphisms. Thus, we have found our Galois group

The obvious trick was to show that even though we can a priori only bound the number of possible automorphisms, using Galois theory we can figure out exactly how many automorphisms we SHOULD have which often forces all possible choices to be all actual choices.