I factored $x^4-5$ into $(x-\sqrt[4]{5})(x+\sqrt[4]{5})(x-i\sqrt[4]{5})(x+i\sqrt[4]{5})$, determining the roots: $x=\pm\sqrt[4]{5}$ and $x=\pm i\sqrt[4]{5}$. The $\mathbb{Q}$-automorphisms of $x^4-5$ are the identity automorphism, the one exchanging $\sqrt[4]{5}$ and $-\sqrt[4]{5}$ and the one exchanging $i\sqrt[4]{5}$ to $-i\sqrt[4]{5}$.
Now, how do I go on to find the Galois group? I found an explanation about the Galois group being isomorphic to $\mathbb{Z}_2\times \mathbb{Z}_2$, but I would love to have a more profound explanation of why.
(I am a high school student seeking some insight on Galois theory for my high school thesis on the roots of polynomials.)
Best Answer
It cannot be $\mathbb{Z}_2 \times \mathbb{Z}_2$ for the following reason:
The splitting field for $f(x) = x^4-5$ is $K = \mathbb{Q}[\sqrt[4]{5}, i]$. We have the tower of fields $\mathbb{Q} \subset \mathbb{Q}[\sqrt[4]{5}] \subset K$. We have $[\mathbb{Q}[\sqrt[4]{5}]:\mathbb{Q}] = 4$, and since $i \notin \mathbb{Q}[\sqrt[4]{5}]$, then $x^2+1$ is irreducible in this field, so we must have $[K:\mathbb{Q}[\sqrt[4]{5}]] = 2$. (See footnote)
Therefore, $[K:\mathbb{Q}] = [K:\mathbb{Q}[\sqrt[4]{5}]] \cdot [\mathbb{Q}[\sqrt[4]{5}]:\mathbb{Q}] = 8$. Recall that the order of $\operatorname{Gal}(f/\mathbb{Q})$ is equal to the degree of the splitting field of $f$ over $\mathbb{Q}$, so $|\operatorname{Gal}(f/\mathbb{Q})| = 8$. Further, since $f$ has $4$ roots, we have $\operatorname{Gal}(f/\mathbb{Q}) \subset S_4$. Note that subgroups of order $8$ inside of $S_4$ are Sylow $2$-subgroups, and all Sylow $p$-subgroups are isomorphic to each other for a given $p$, so it will suffice to simply find a single subgroup of order $8$ inside of $S_4$ and classify it.
Alternatively, you can list the automorphisms by hand, and you'll know your list is exhaustive once you've written down $8$ of them, then compare to groups you're already familiar with (e.g. $\mathbb{Z}_8, \ \mathbb{Z}_4 \times \mathbb{Z}_2, \ D_4, \ \mathcal{Q}_8$, etc).
Footnote: The degree of an extension $F[\alpha]$ over $F$ is equal to the degree of an irreducible polynomial in $F[x]$ with $\alpha$ as a root. So for example, $\mathbb{Q}[\sqrt[4]{5}]$ is a degree $4$ extension over $\mathbb{Q}$ since $\sqrt[4]{5}$ is a root of $g(x) = x^4-5$, which is an irreducible polynomial.