Galois group of $\mathbb Q(\sqrt{4+\sqrt 7})/\mathbb Q$

Question

Let $\alpha =\sqrt{4+\sqrt 7}$ and $\beta =\sqrt{4-\sqrt 7 }$. I have to compute $Gal(\mathbb Q(\alpha )/\mathbb Q)$. So, I found that $Gal(\mathbb Q(\alpha )/\mathbb Q)=\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$, that $\mathbb Q(\beta^2)= \mathbb Q(\alpha^2)$. I show that the roots of the minimal polynomial of $\alpha $ are $\{\alpha ,-\alpha ,\beta ,-\beta \}$. But now that ask me to enumerate the element of the galois group. I did it's $\{id, \sigma ,\tau,\sigma \tau\}$ where

\begin{align*} \sigma : \alpha &\longmapsto -\alpha \\ \beta &\longmapsto -\beta \end{align*} and \begin{align*} \tau : \alpha &\longmapsto \beta \\ \beta &\longmapsto \alpha \end{align*}

But in my correction they wrote

\begin{align*} \sigma : \alpha &\longmapsto -\alpha \\ \beta &\longmapsto \beta \end{align*}

Question : Who is right ? me or the correction ?

The problem it's that this group is abelian, so if they are right, $$\sigma \tau(\alpha )=\beta \neq -\beta =\tau\sigma (\alpha )$$ so I think I'm right, but in the doubt I prefer to ask you.

Do you want the Galois group of $\mathbb{Q}(\alpha)$ (is this Galois?), or of $\mathbb{Q}(\alpha,\beta)$? — Martin Brandenburg, Jan 31 '16 at 15:23
Since the extension is Galois and $\alpha $ and $\beta $ are conjuguate, $\mathbb Q(\alpha )=\mathbb Q(\alpha ,\beta )$ @MartinBrandenburg — Surb, Jan 31 '16 at 15:53

score 2 · Accepted Answer · answered Jan 31 '16 at 15:18

Notice that $$\alpha\beta =\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=\sqrt{16-7}=3,$$ so $\beta=\frac{3}{\alpha}$. As a consequence, any homomorphism $\varphi$ such that $\varphi(\alpha)=-\alpha$ is also such that$$\varphi(\beta)=\varphi\left(\frac{3}{\alpha}\right)=\frac{3}{\varphi(\alpha)}=\frac{3}{-\alpha}=-\beta.$$ So the error is in your correction. Your answer is right.

Galois group of $\mathbb Q(\sqrt{4+\sqrt 7})/\mathbb Q$

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