I want to show that each fractional ideal of $\mathbb Z[\sqrt{-3}]$ is of the form $\mathbb Z[\sqrt{-3}]a$ or $\mathbb Z[1/2+1/2\sqrt{-3}]a$ with $a\in\mathbb Q(\sqrt{-3})$.
Initial considerations:
Write $R=\mathbb Z[\sqrt{-3}]$ and $S=\mathbb Z[1/2+1/2\sqrt{-3}]$. Given a fractional ideal $I$ of $R$, there exists a $q\in\mathbb Q(\sqrt{-3})$ such that $q\cdot I\subset R$. Note that $q\cdot I$ is an $R$-ideal. Since $S$ is a PID, we know that if $q\cdot I$ is also an $S$-ideal, then $q\cdot I=Sa$ for some $a\in S$, so we are done. If $q\cdot I$ is not an $S$-ideal, then we need to show that $q\cdot I$ is of the form $R\cdot a$ for some $a\in\mathbb Q(\sqrt{-3})$. Since $q\cdot I\subset R$ and $1\in R$, we would need that $a\in R$, so we would need to show that in that case $q\cdot I$ is principal. Or alternatively: any $R$-ideal $J$ that is not principal, is of the form $S\cdot a$ with $a\in \mathbb Q(\sqrt 3)$. Since $1\in S$, we would have $a\in R$, so any non-principal $R$-ideal is a principal $S$-ideal of the form $S\cdot r$ with $r\in R$. How to show that?
My attempt:
Let $J$ be a non-principal $R$-ideal. Since $R$ is Noetherian, there exists finitely many $a_i\in R$ such that $I=\sum_i Ra_i$. Since $S$ is a PID, there exists $a\in S$ such that $\sum_i S_i a_i=Sa$. Thus we can write $a_i=x_i a$ with $x_i\in S_i$. Note that $a$ is the greatest common divisor of the $a_i$ in $S$, so $\sum S x_i=1$. We now have that $$ \sum_i Ra_i=\sum_i R(x_ia)=Ra\cdot \sum_i Rx_i. $$ Is this the right direction? How to continue?
An example for intuition:
I know that $I=(2,1+\sqrt{-3})$ is not principal, so we should have $I=Sr$ for some $r\in R$. Since $I^2=R2\cdot I$, we get $$ I^2=Sr\cdot Sr=Sr^2=R2\cdot Sr=S2\cdot Sr=S\cdot 2r, $$ so $r=2$ should work. It is indeed true that $(2,1+\sqrt{-3})=\mathbb Z[1/2+1/2\sqrt{-3}]\cdot 2$.
Recall $R=\mathbb Z[\sqrt{-3}]$ and $S=\mathbb Z[1/2+1/2\sqrt{-3}]$. Let $I$ be a fractional $R$-ideal. Then $\exists q\in\mathbb Q(\sqrt{-3})^\times$ such that $q\cdot I\subset R$. So WLOG we may assume that we work with an $R$-ideal $I\subset R$.
Let $y_0\neq 0\in I$ be such that $\vert y_0\vert$ is minimal, where $\vert\cdot\vert$ is the Euclidean norm (this is possible, since $R$ is discrete). I will show that $I=Ry_0$ or $I=Sy_0$.
Let $y\in J$. Choose $q\in R$ such that $\vert y/y_0-q\vert$ is minimal. Then $\vert y/y_0-q\vert\leq 1$ (see here if it's not clear why this is true). Hence $\vert y-qy_0\vert\leq \vert y_0\vert$.
We consider two cases:
If $\vert y/y_0-q\vert<1$, then $y=qy_0$ by minimality of $\vert y_0\vert$, hence $y\in Ry_0$.
If $\vert y/y_0-q\vert=1$, then $y/y_0\in S\setminus R$ (again from the same source as before), thus $y\in\mathbb Z[1/2+1/2\sqrt{-3}]y_0$.
We conclude that if for all $y$ with corresponding $q$ we have that $\vert y/y_0-q\vert<1$, then $J=Ry_0$. If there exists an $y\in I$ such that with the corresponding $q$ we have $\vert y/y_0-q\vert=1$, then $y=(m+n(1/2+1/2\sqrt{-3}))y_0$ for some $m,n\in\mathbb Z$ with $n$ odd, and therefore we also have that $(1/2+1/2\sqrt{-3})y_0\in I$, which shows that $I=Sy_0$.