In the context of Galois theory the splitting field of $x^4-3$ is isomorphic to $D_8$. Therefore one of the elements of this group is $\sigma$, which maps $i\rightarrow i$ and $\sqrt[4]{3}\rightarrow w\sqrt[4]{3}.$
Now the field extension is the set $\Bbb Q(\sqrt[4]{3},i)=\{a_0+a_1\sqrt[4]{3}+a_2\sqrt[4]{9}+a_3\sqrt[4]{27}+a_4i+a_5i\sqrt[4]{3}+a_6i\sqrt[4]{9}+a_7i\sqrt[4]{27}\}$
Applying $\sigma^2$ to this we see it fixes elements of the form $a_0+a_1\sqrt[4]{9}+a_4i+a_6i\sqrt[4]{9}$.
Now my question is, considering the Galois correspondence , What field does the subgroup $\{e,\sigma^2\}$, correspond to ?
Is it $\Bbb Q(i,\sqrt[4]{9})$?
My confusion lies in understanding what splitting field has elements $a_0+a_1\sqrt[4]{9}+a_4i+a_6i\sqrt[4]{9}$.
Note that $(\sqrt[4]{3})^2=(3^{1/4})^2=3^{1/2}=\sqrt{3}$, so the intermediate field corresponds (by the Galois correspondence) to the intermediate field $\color{red}{\mathbf{Q}(\sqrt{3},i)}$. (Quick check: $[\mathbf{Q}(\sqrt{3},i):\mathbf{Q}]=4$ and the subgroup $\langle \sigma_2\rangle$ has order 2 thus index $4$ in $D_8$, since it represents a reflection of the square.)
Alternatively, you could write out where the generators $\sqrt[4]{3}$ and $i$ get sent to under all the elements of $D_4$, and work from there.