I’m reading David Cox’s “Primes of the form $x^2+ny^2$”. Right after Corollary 10.20 he says:
First, consider all lattices which have complex multiplication by $\sqrt{-3}$. This means that we are dealing with an order $\mathcal{O}$ containing $\sqrt{-3}$ in the field $K=\mathbb{Q}(\sqrt{-3})$. Then $\mathcal{O}$ must be either $\mathbb{Z}[\sqrt{-3}]$ or $\mathbb{Z}[\omega]$ with $\omega=e^{2\pi i/3}$, and since both of these have class number $1$, the only lattices are $[1,\sqrt{-3}]$ and $[1,\omega]$. Thus, up to homothety, there are only two lattices with complex multiplication by $\sqrt{-3}$.
Here I don’t understand why $\mathcal{O}$ is one of that two options. I know that $\mathcal{O}=\mathbb{Z}+\mathbb{Z}\left(f\cdot\frac{1+\sqrt{-3}}{2}\right)$ for some $f\in\mathbb{Z}_{+}$, but why we get only those two?
Given an order $O$ and a complex lattice $L$ sent to itself by $O$ then $L= r J$ for some ideal $J\subset O$ and $r\in \Bbb{C}$.
Let $K=Frac(O)$ then $\{ zJ,z\in \Bbb{C}^*\}$ corresponds to $\{ a J,a\in K^*\}$ which is an element of the ideal class monoid $$ClM(O)=M(O)/Frac(O)^*$$ where $M(O)$ is the monoid of all the non-zero $O$-modules $\subset K$
(not the same as the ideal class group $Cl(O)=I(O)/Frac(O)^*$ obtained by restricting to the invertible fractional ideals)
We are told that $Cl(O)$ is trivial. Thus every maximal of $O$ not containing $[O_K:O]=2$ is invertible and principal. If $J$ is contained in some such principal invertible maximal ideal $aO$ replace it by $a^{-1}J$. Doing so iteratively we can assume that either $J=O$ or the only maximal ideal containing $J$ is $2O_K$. In the last case, if $J\supset 2O$, replacing $J$ by $2^{-1}J$, we can assume that $J$ is not contained in $2O$. From there the only choice is $J= O$ or $J=2O_K$, ie. $$ClM(O)=\{O,2O_K\}$$