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Splitting field is $K=\mathbb Q (\sqrt[4]{-5}, i)$

Degree of $K$ over rationals is $8$ so the galois group $G=\text{G}(K/ \mathbb Q)$ has order $8$.

$x^4 +5$ is irreducible so there is one orbit which is the roots set $$R=\{ \sqrt[4]5 \xi, \sqrt[4]5 \xi^3, \sqrt[4]5 \xi^5, \sqrt[4]5 \xi^7 \}$$ where $\xi = e^{\frac{\pi}4 i}$.

Faithful action implies that $G \leq S_4$ and $D_8$ is the only subgroup of $S_4$ which has order $8$ so this is the isomorphism type of $G$.

But there are three copies of $D_8$, which one do we use to determine the action on the generators of $K$?

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I prefer to write the root set as $\{\sqrt[4]{-5}, i\sqrt[4]{-5}, -\sqrt[4]{-5}, -i\sqrt[4]{-5}\}$ since these are the generators you've chosen for $K$. Define automorphisms $\rho, \sigma: K \to K$ by \begin{align*} \rho: \sqrt[4]{-5} &\mapsto i\sqrt[4]{-5}\\ i &\mapsto i \end{align*} and \begin{align*} \sigma: \sqrt[4]{-5} &\mapsto \sqrt[4]{-5}\\ i &\mapsto -i \, . \end{align*} One can show that $\sigma \circ \rho = \rho^3 \circ \sigma$ by computing their action on the generators $\sqrt[4]{-5}$ and $i$. Thus the Galois group has the presentation $$ \langle \rho, \sigma \mid \rho^4 = \sigma^2 = 1, \sigma \rho = \rho^3 \sigma \rangle $$ which is a presentation for $D_8$.

For a more geometric answer, trying plotting the roots $\{\sqrt[4]{-5}, i\sqrt[4]{-5}, -\sqrt[4]{-5}, -i\sqrt[4]{-5}\}$. They form a square in the complex plane with sides parallel to the real and imaginary axes. Then $\rho$ corresponds to a rotation of the roots by $90$ degrees counterclockwise, and $\sigma$ (complex conjugation) is reflection over the real axis.

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