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 $K$?


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.