I am trying to calculate the Galois group of $f = X^4+20$. Here is my attempt so far.
Firstly we determine the roots of $f$ in al algebraically closed field, in this case in $\mathbb{C}$. Write $x = re^{i\phi}$ for a root of $f$. Then we have \begin{align*} (re^{i \phi})^4 = -20 \ & \Longleftrightarrow \ r^4e^{4\phi i} = 20e^{k\pi i}, \ k \in \{1,3,5,\ldots\}, \\ & \Longleftrightarrow \ r^4 = 20 \textrm{ en } 4\phi = k\pi, \ k \in \{1,3,5,\ldots\}, \\ & \Longleftrightarrow \ r = \sqrt[4]{20} \textrm{ en } \phi = \frac{\pi}{4} \cdot k, k \in \{1,3,5,\ldots\}. \end{align*} So for $\phi \in [0,2\pi]$ we find that the roots are given by $$ x = \sqrt[4]{20} e^{\pi/4}, \sqrt[4]{20} e^{3\pi/4}, \sqrt[4]{20} e^{5\pi/4}, \sqrt[4]{20} e^{7\pi/4}. $$ Notice that $e^{\pi/4} = e^{2\pi i /8} = \zeta_8$, an eight primitive root of unity. Hence, the roots are given by $$ x = \sqrt[4]{20} \cdot \zeta_8, \sqrt[4]{20} \cdot \zeta_8^3, \sqrt[4]{20} \cdot \zeta_8^5, \sqrt[4]{20} \cdot \zeta_8^7. $$ So we see that the splitting field of $f$ is given by $\Omega_f = \mathbb{Q}(\sqrt[4]{20} \cdot \zeta_8, \zeta_8^2)$. Notice that $\zeta_8^2 = \zeta_4 = i$, and that we can $\sqrt[4]{20} \cdot \zeta_8$ denote as $\sqrt[4]{-20}$, since $(\sqrt[4]{20} \cdot \zeta_8)^4 = 20 \zeta_8^4 = -20$. So from this we see that $\Omega_f = \mathbb{Q}(\sqrt[4]{-20},i)$.
Now the field extension $\mathbb{Q} \subset \mathbb{Q}(\sqrt[4]{-20})$ is of degree $4$, since the polynomial $f = X^4+20$ is irreducible, namely Eisenstein for $p = 5$.
Now I am trying to show that $\mathbb{Q}(\sqrt[4]{-20}) \subset \mathbb{Q}(\sqrt[4]{-20},i)$ has degree $2$. Then we could conclude that $\# Gal(f) = 8$, and as a subgroup of $S_4$ it has to be $D_4$, the Dihedral group on 8 elements. However, I am stuck at this last step, how do I show this extension has degree 2? I could not show that $i \notin \mathbb{Q}(\sqrt[4]{-20})$.
It is enough to show that $f$ remains irreducible over $\Bbb{Q}(i)$ because that would show that the splitting field has degree $4$ over $\Bbb{Q}(i)$. Since $[\Bbb{Q}(i):\Bbb{Q}] = 2$ and $f$ is irreducible of degree 4 over $\Bbb{Q}$, $f$ can have no root in $\Bbb{Q}(i)$. So if $f$ is reducible over $\Bbb{Q}(i)$, $f$ would factor as a product of two quadratic polynomials. The constant terms of these polynomials would be the product of two roots of $f$. Those products are $\pm\sqrt{20}$ and $\pm i\sqrt{20}$ which are not in $\Bbb{Q}(i)$. Hence $f$ is irreducible over $\Bbb{Q}(i)$.