Find the Galois group of $\mathbb{Q}(\sqrt[3]{2}, \sqrt{2}): \mathbb{Q}$.
My Attempt: Clearly the identity on $\mathbb{Q}(\sqrt[3]{2}, \sqrt{2})$, i.e., $\sigma_1=id_{\mathbb{Q}(\sqrt[3]{2}, \sqrt{3})}$, is in $\Gamma(\mathbb{Q}(\sqrt[3]{2}, \sqrt{2})/\mathbb{Q})$. My gut tells me that there is more.
By monomorphism, for any $\sigma \in \Gamma(\mathbb{Q}(\sqrt[3]{2}, \sqrt{2})$, we must have $\sigma(\sqrt{2})=\pm\sqrt{2}$ and $\sigma(\sqrt[3]{2})=\sqrt[3]{2}$. And it seems that $\mathbb{Q}(\sqrt[3]{2}, \sqrt{2})=\{a+p2^{\frac{1}{6}}+q\sqrt[3]{2}+r\sqrt{2}+s2^{\frac{2}{3}}+t2^{\frac{5}{6}}: a, p, q, r, s, t,\in \mathbb{Q}\}$. But to write and test out all candidates for $\mathbb{Q}$-automorphisms on $\mathbb{Q}(\sqrt[3]{2}, \sqrt{2})$, i.e., $$\sigma_i: a+p2^{\frac{1}{6}}+q\sqrt[3]{2}+r\sqrt{2}+s2^{\frac{2}{3}}+t2^{\frac{5}{6}} \mapsto a\pm p2^{\frac{1}{6}}+q\sqrt[3]{2}\pm r\sqrt{2}\pm s2^{\frac{2}{3}}\pm t2^{\frac{5}{6}}$$, seems beyond painful. Is there a more intelligent way to do it?
Any hint would be greatly appreciated.
Let $k = \mathbb{Q}(\sqrt[3]{2}, \sqrt{2})$.
First, let us note that as a ring, $k$ is generated by $\sqrt[6]{2}$. Thus, to describe an automorphism $f : k \to k$, it suffices to determine $f(\sqrt[6]{2})$.
Now since $f$ is an automorphism, we see that $f(\sqrt[6]{2})$ must be a root of $X^6 - 2$. Since $k$ is a subfield of $\mathbb{R}$ and the only roots of this polynomial in $\mathbb{R}$ are $\pm \sqrt[6]{2}$, we see that those are also the only roots in $k$. So there are only 2 candidate ring homomorphisms $k \to k$ which could be automorphisms, and it is easy to verify that both of the maps are indeed automorphisms.
Therefore, the automorphism group has two elements and is thus isomorphic to $\mathbb{Z} / 2 \mathbb{Z}$.