D+F 14.6.4

Determine the Galois group of $x^4-25$

Recall that the Galois group of a separable polynomial $f(x) \in{F[x]}$ is defined to be the Galois group of the splitting field of $f(x)$ over $F$.

$x^4-25$ partially factors over $\mathbb{Q}$ as $$x^4-25=(x^2-5)(x^2+5)$$ and the four roots are $\pm\sqrt{5} \pm\sqrt{5}i$. So, this polynomial has no repeated roots, and is thus separable. We first determine the splitting field and then use the fundamental theorem of Galois to flip the diagram of known subfields to get the Galois group lattice. We need to adjoin $\sqrt{5}$ to $\mathbb{Q}$, but notice that the two complex roots are not contained in $\mathbb{Q}[\sqrt{5}]$. So, we need to adjoin both $\sqrt{5}$ and $i$ for our splitting field to contain all 4 roots. Thus, the splitting field of $x^4-25$ should be $\mathbb{Q}[\sqrt{5},i]$. The lattice of known subfields is $\mathbb{Q}$ as the base field, then $\mathbb{Q}[\sqrt{5}]$, $\mathbb{Q}[i]$ and $\mathbb{Q}[\sqrt{5}i]$ as degree 2 extensions over $\mathbb{Q}$. Then, $\mathbb{Q}[\sqrt{5},i]$ on top as a degree 2 extension of the 3 intermediate subfields. By the fundamental theorem of Galois, the flipped image of this subfield lattice is exactly the lattice of the Galois group of $x^4-25$. This lattice is the lattice for the Klein 4-group.