Why the ring of integers of $ \mathbb Q(\sqrt{-19}) $ is $ \mathbb Z\left[\frac{1+\sqrt{-19}}{2}\right] $?
I came up with this question when I was trying to understand this: https://math.stackexchange.com/a/3086739/549397. And this result seemed to be mentioned in an algebraic number theory book when I was looking for references.
When $d\equiv1\pmod 4$ is squarefree, then the ring of integers of $K=\Bbb Q(\sqrt d)$ is ${\cal O}_K=\Bbb Z[\frac12(1+\sqrt d)]$. Pretty much every textbook on algebraic number theory proves this.
For $$r+s\sqrt d\in\Bbb Q(\sqrt d)$$ to be an algebraic integer, it is necessary and sufficient for its trace $$(r+\sqrt d)+(r-\sqrt d)=2r$$ and norm $$(r+\sqrt d)(r-\sqrt d)=r^2-ds^2$$ to be integers. If $r$ is an integer, then from the norm being an integer it follows that $s$ is. But if $2r$ is an odd integer, then $(2r)^2-d(2s)^2\equiv0\pmod 4$ from which $2s$ must be an odd integer too.
