Consider the two rings $R_1=\{\mathbb{Z}[\sqrt{-3}]=a+b\sqrt{-3}\mid a, b \in \mathbb{Z}\}$ and $R_2=\{\frac{a+b\sqrt{-3}}{2}\mid a,b \in \mathbb{Z},a\,\,\textrm{and}\,\,b\,\,\textrm{both even or both odd}\}$.
Let $F_1$ and $F_2$ be the fields of fractions induced by $R_1$ and $R_2$ respectively. I want to describe the elements in these fields, and in particular determine whether they are equal, or if one is contained in the other. My suspicion at this point is that $F_1=F_2$, but the fields are throwing me for a loop:
Here is my attempt at showing equality:
I'm pretty confident that $F_1=\mathbb{Q}[\sqrt{-3}]$. If: $$\alpha = \frac{a+bi\sqrt{3}}{c+di\sqrt{3}} $$ is an arbitrary element of $F_1$, then by multiplying top and bottom by $(c-di\sqrt{3})$, we can express $\alpha$ in the form $p+qi\sqrt{3}$, where $p,q \in \mathbb{Q}$. Hence $F_1 \subseteq\mathbb{Q}[\sqrt{-3}]$. Conversely, any element of $\mathbb{Q}[\sqrt{-3}]$ can be written as the quotient of two elements of $R_1$, so $\mathbb{Q}[\sqrt{-3}]\subseteq F_1$. We conclude that $F_1=\mathbb{Q}[\sqrt{-3}]$.
Now, I would like to show that $F_2=\mathbb{Q}[\sqrt{-3}]$. Inclusion in one direction is straightforward: any quotient of two elements in $R_2$ will take the form $(a+bi\sqrt{3})/(c+di\sqrt{3})$ and hence belong to $\mathbb{Q}[\sqrt{-3}]$ by an argument similar to above. Now let $\beta \in \mathbb{Q}[\sqrt{-3}]$ be arbitrary. Then we may write $\beta$ in the form: $$\beta = \frac{v}{w}+\frac{x}{y}i\sqrt{3}, $$ with $v,w,x,y\in \mathbb{Z}$ and $w\neq 0$ and $y\neq 0$. Simple algebraic manipulation shows that: $$\beta = \frac{(2v)(2y)+(2x)(2w)i\sqrt{3}}{(2w)(2y)+0\cdot i \sqrt{3}}, $$ which shows that I have written $\beta$ as the quotient of two elements of $R_2$: namely, $\frac{1}{2}(4vy+4xwi\sqrt{3})$, and $\frac{1}{2}(4wy+0\cdot i \sqrt{3})$ -- note that the coefficients in each element are both even. So it appears that $\beta \in F_2$ and because $\beta\in \mathbb{Q}[\sqrt{-3}]$ was arbitrary, we have that $\mathbb{Q}[\sqrt{-3}] \subseteq F_2$. Hence $F_2 = \mathbb{Q}[\sqrt{-3}] = F_1$.
Is this legitimate? I feel a little bit strange about the previous paragraph (in particular, given that I haven't cancelled the 2's, but I suppose in the definition of $R_2$, it is not stipulated that $a/2$ and $b/2$ be written in reduced form).
Thank you for your help!
Instead of specific verification, there is a simplier method to arrive $F_1=F_2$.
On the one hand, $$\forall a,b\in\mathbb Z,a+b\sqrt{-3}=\frac{2a+2b\sqrt{-3}}{2}\Rightarrow R_1\subset R_2\Rightarrow F_1\subset F_2$$ and on the other hand, $$\forall a,b\in\mathbb Z,\frac{a+b\sqrt{-3}}{2}\in F_1\Rightarrow R_2\subset F_1$$ Since $F_1$ is closed to those operations of field, $F_2\subset F_1$.