Galois Group of the polynomial $f(x)=x^4-9$ over $\mathbb{Q}$ How would one construct the Galois Group of the polynomial $f(x)=x^4-9$ over $\mathbb{Q}$?
I know first we have to find the roots, then construct the splitting field.
However what would the roots be in this case? $\{±\sqrt3,±i
\}$?
 A: As noted, the splitting field of $f$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt{3}, i)$. We know that $\mathbb{Q}(\sqrt{3}, i) \supset \mathbb{Q}$ is a Galois extension because $\mathbb{Q}(\sqrt{3}, i)$ is a splitting field of a polynomial $f$ over $\mathbb{Q}$.
Therefore $\left|\operatorname{Gal}(\mathbb{Q}(\sqrt{3}, i) / \mathbb{Q})\right| = [\mathbb{Q}(\sqrt{3}, i) : \mathbb{Q}] = 4$.
A $\mathbb{Q}$-automorphism of $\mathbb{Q}(\sqrt{3}, i)$ has to map $\sqrt{3}$ to $\pm \sqrt{3}$ and $i$ to $\pm i$ so the Galois group is given by
$$\operatorname{Gal}(\mathbb{Q}(\sqrt{3}, i) / \mathbb{Q}) =\{(\sqrt{3} \mapsto \sqrt{3}, i \mapsto i), (\sqrt{3} \mapsto -\sqrt{3}, i \mapsto i), (\sqrt{3} \mapsto \sqrt{3}, i \mapsto -i), (\sqrt{3} \mapsto -\sqrt{3}, i \mapsto -i)\}$$
because it has to have $4$ elements.
It can be easily checked that every element has order $\le 2$ so $\operatorname{Gal}(\mathbb{Q}(\sqrt{3}, i) / \mathbb{Q}) \cong \mathbb{Z}_2 \times \mathbb{Z}_2 $.
A: Note that $$x^4-9=(x^2-3)(x^2+3)=(x-\sqrt3)(x+\sqrt3)(x-i\sqrt3)(x+i\sqrt3)$$ so you have $\mathbb{Q}(\sqrt3,i).$
A: An overly (perhaps) detailed answer: You already know that the roots are $\pm\sqrt{3}$ and $\pm i \sqrt{3}$.
The base field, $\mathbb{Q}$, is over the set of rational numbers; so, the irrational number $\sqrt{3}$ is not an element of this field. You will need to adjoin $\sqrt{3}$, but this allows $-\sqrt{3}$ to be an element of your extension, too. However, this intermediate field, $\mathbb{Q}(\sqrt{3})$, is over the set of real numbers; so, the imaginary number $i \sqrt{3}$ is not an element of this field. You can adjoin it, although you would then have a field containing the nonzero elements $i\sqrt{3}$ and $\sqrt{3}$, hence also containing their ratio, $i$; conversely, you could adjoin $i$ to $\mathbb{Q}(\sqrt{3})$ and now have a field containing their product and its additive inverse, $\pm i \sqrt{3}$, as desired.
As a final note, for if/when it becomes relevant: I wrote out a detailed proof that the degree of $[\mathbb{Q}(\sqrt{2}, \sqrt{-3}):\mathbb{Q}] = 4$ in MSE 2204977. One could adapt this proof in a straightforward manner to show that the splitting field in your question also has degree $4$ as an extension of the base field $\mathbb{Q}$.
