I am trying to determine the Galois group $G(\mathbb{Q}[\sqrt{2}, \sqrt[3]{2}]/\mathbb{Q})$. I am fairly confident I have the correct answer, but I need someone to confirm my work since I have just taught myself this material today.

First, note that $K = \mathbb{Q}[\sqrt{2}, \sqrt[3]{2}]$ is a field extension of degree $6$. However, it is not a splitting field of some polynomial with coefficients in $\mathbb{Q}$. Thus, the order of the Galois group will be strictly less than $6$ since it is not a Galois extension.

Consider the polynomial $f(x) = (x^2 - 2)(x^3 - 2)$. This polynomial has $3$ roots in $K$, namely $\sqrt{2}$, $-\sqrt{2}$, and $\sqrt[3]{2}$. Any $\mathbb{Q}$-automorphism of $K$ will permute these three roots. Note that for any such automorphism, $0 = \phi(0) = \phi(\sqrt{2} - \sqrt{2}) = \phi(\sqrt{2})+\phi(-\sqrt{2})$. Hence, the only two possible automorphisms are the identity and that which swaps $\sqrt{2}$ with $-\sqrt{2}$.

We conclude that $G(K/\mathbb{Q}) \cong \mathbb{Z}_2$.

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    $\begingroup$ I thought the Galois group of $E/F$ is defined to be the Galois group of the Galois closure of $E$ over $F$. The Galois closure of $E/F$ is the smallest extension of $E$ which is Galois over $F$. You seem to be defining it to be the Galois group of a Galois extension of $F$ which is maximal among those contained in $E$. I don't think that is even well-defined in general (there could be multiple such extensions with different Galois groups). $\endgroup$
    – anon
    Apr 13 '14 at 21:21
  • $\begingroup$ The textbook I am using defines the Galois group, denoted G(K/F), as the group of automorphisms of K that fix all elements of F. Here's a link from Harvard that backs up what I am saying (see bottom of page 3): math.harvard.edu/~nate/teaching/UPenn/2007/fall/math_371/… $\endgroup$
    – Kaj Hansen
    Apr 13 '14 at 22:03
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    $\begingroup$ Although "defining" the Galois group of $E/F$ to be the group of automorphisms of $E$ that fix $F$ is within one's rights, approximately half the population will misconstrue what is being asked, just as did @anon. $\endgroup$ Apr 13 '14 at 22:20
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    $\begingroup$ Very interesting, Paul Garrett. So it's similar to the confusion surrounding the dihedral group notation $D_n$? I was previously unaware of this discrepancy. Are there any names that are less ambiguous? $\endgroup$
    – Kaj Hansen
    Apr 13 '14 at 22:23
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    $\begingroup$ The automorphism group of $E$ over $F$: $$\operatorname{Aut}(E/F)$$ $\endgroup$
    – user14972
    Apr 13 '14 at 22:27

Well, $G(\mathbb{Q}[\sqrt{2}, \sqrt[3]{2}]/\mathbb{Q})$ is not Galois. What you have computed here is the automorphism group $Aut(\mathbb{Q}(\sqrt[6]{2})/\mathbb{Q})$ rather than the Galois group (i.e., automorphism group of $(\mathbb{Q}(\sqrt[6]{2}),\sqrt{-3})$, the closure $\mathbb{Q}[\sqrt{2},\sqrt[2]{3}]$ over $\mathbb{Q}$.


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