I proved that $\mathbb{Z}[\sqrt{-3}]$ is isomorphic to $\mathbb{Z}[x]/(x^{2}+3)$.
Because $f = (x^{2}+3)$ is irreducible by Eisenstein. Is that enough to conclude that it splits over $\mathbb{Q}$? If f spits it would mean that $\mathbb{Z}[x]/(x^{2}+3)$ is galois, doesn't it?
Now I want to find the Galois group, but for that I need the automorphisms which act on the roots. Now I'm not sure if I should look at $[\mathbb{Z}(\sqrt{-3}):\mathbb{Z}]=2$, $[\mathbb{Q}(\sqrt{-3}):\mathbb{Q}]=2$, $[\mathbb{Z}(\sqrt{3}, i):\mathbb{Z}]=4$ or $[\mathbb{Q}(\sqrt{3}, i):\mathbb{Q}]=4$.
If anyone could clear those things up for me, it would be a lot of help.
$X^2+3$ is irreducible over $\Bbb{Q}$ so $\Bbb{Q}[x]/(x^2+3)$ is a field, isomorphic to $\Bbb{Q}(\sqrt{-3})$.
$X^2+3=(X+\sqrt{-3})(X-\sqrt{-3})$ splits completely over $\Bbb{Q}(\sqrt{-3})$ so $\Bbb{Q}(\sqrt{-3})/\Bbb{Q}$ is Galois.
For each root there is an automorphism sending $\sqrt{-3}$ to that root.
Therefore, the Galois group has two elements: the identity, and $\sigma(a+b\sqrt{-3})=a-b\sqrt{-3}$.
Both restrict to automorphisms of the subring $\Bbb{Z}[\sqrt{-3}]$.
In some contexts we can say that $\Bbb{Z}[\sqrt{-3}]/\Bbb{Z}$ is Galois.
Interestingly $\Bbb{Z}[\frac{1+\sqrt{-3}}{2}]$ is a PID while $\Bbb{Z}[\sqrt{-3}]$ is not even a Dedekind domain (the ideal $(2,\sqrt{-3}-1)$ is not principal and not inversible).