Minimal field extension of $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$ and Galois group

Find a minimal field extension $L$ of $\mathbb{Q}(\sqrt{2})$ such that $L$ is normal over $\mathbb{Q}$

$L$ is normal over $\mathbb{Q}$ which means that it is the splitting field of polynomials in $\mathbb{Q}[X]$. Also $\mathbb{Q}(\sqrt{2}) \subset L$.

So $L$ is a splitting field of the polynomial $X^6-2$. I am unsure what the minimal field extension would be. Would it be $\mathbb{Q}(\sqrt{2}, i)$? I do not think so since $i^6 -2 \neq 0$. Is it just $\mathbb{Q}(\sqrt{2})$ ?

Describe the structure of $G=Gal(L/\mathbb{Q})$

If $L=\mathbb{Q}(\sqrt{2})$ does this mean that $Gal(L/\mathbb{Q})$ is generated by just one automorphism $\tau : \tau(\sqrt{2})=\sqrt({2})$, and is thus trivial?

The minimal polynomial of $\;\sqrt2\;$ over the rationals is indeed $\;x^6-2\;$. Let us observe that

$$x^6-2=\left(x^3-2^{1/2}\right)\left(x^3+2^{1/2}\right)=$$

$$=\left(x-2^{1/6}\right)\left(x+2^{1/6}\right)\left(x^2+2^{1/3}x+2^{1/6}\right)\left(x^2-2^{1/3}x+2^{1/6}\right)$$

If $\;\xi:=e^{\pi i/3}\;$ , then this is a primitive root of unity of order $\;6\;$, whereas $\;\xi^2\;$ is a primitive root of unity of order $\;3\;$ , and thus we can write down the roots of $\;p(x)=x^6-2\;$ as

$$2^{1/6},\,2^{1/6}\xi,\,2^{1/6}\xi^2,\,-2^{1/6},\,-2^{1/6}\xi^2,\,-2^{1/6}\xi^2\;\;(\text{observe that}\;\xi^3=-1\,)$$

Since it must be $\;Q(\sqrt2)\subset L\;$ and $\;L/\Bbb Q\;$ normal, it must be all the roots of the above polynomial must be in $\;L\;$ , and thus the minimal $\;L\;$ as wanted indeed is the splitting field of the above polynomial over the rationals, and we have:

$$L=\Bbb Q(\sqrt2\,,\,\xi)\;,\;\;\text{and}\;\;[L:\Bbb Q]=[L:\Bbb Q(\sqrt2)][\Bbb Q(\sqrt2):\Bbb Q]=2\cdot6=12$$

since the minimal polynomial of $\;\xi\;$ over $\;\Bbb Q(\sqrt2)\;$ is the same as over the rationals: $\;x^2-x+1\;$ (Can you see why?).

We could then form the fields tower

$$\Bbb Q\stackrel 6\subset\Bbb Q(\sqrt2)\stackrel 2\subset\Bbb Q(\sqrt2,\,\xi)=:L\,,\,\,\text{and we can write:}$$$${}$$

$$\begin{cases}&\Bbb Q(\sqrt2)=&\text{Span}_{\Bbb Q}\left\{\,1,2^{1/6},2^{1/3},\,2^{1/2},2^{2/3},2^{5/6}\,\right\}\\{}\\&L=&\text{Span}_{\Bbb Q(\sqrt2)}\left\{\,1,\xi\,\right\}\end{cases}\implies$$$${}$$

$$\implies L=\text{Span}_{\Bbb Q}\left\{\,1,2^{1/6},...,2^{5/6},\xi,2^{1/6}\xi,...,2^{5/6}\xi\,\right\}$$

Observe now that any element in $\;G:=Gal(L/\Bbb Q)\;$ must map $\;w\;$ to either $\;w\;$ or $\;w^5\;$ (the roots of $\;x^2-x+1)$, and $\;2^{1/6}\;$ to any of the roots of $\;x^6-2\;$ . We have the next two automorphisms:

$$\sigma:\;\begin{cases}2^{1/6}\mapsto2^{1/6}w^2\\w\mapsto w^5\end{cases}\;\;,\;\;\tau:\;\begin{cases}2^{1/6}\to-2^{1/6}w\\w\mapsto w\end{cases}$$

and it's easy to check that $\;\sigma^4=\tau^3=1(=Id_{\Bbb Q})\;$. We also have that

$$\tau\sigma\tau\;:\;\begin{cases}2^{1/6}\mapsto2^{1/6}w^2 \\w\mapsto w^5\end{cases}\implies\tau\sigma\tau=\sigma$$

With the above relations we get already $\;|\langle\sigma,\,\tau\rangle|=12\;$ , so this is the whole Galois Group and it is isomporhic with $\;T:=$ the dicyclic group of order $\;12\;$