Show that the field of quotients of $\mathbb{Z}[2i]$ is $\mathbb{Q}[i]$

Question

Let $D = \mathbb{Z}[2i]$. I need to obtain the field of quotients of $D$. To do that, I take any to elements $a+2bi$ and $c+2di\neq 0$ in $D$ an take the quotient of them as $$\frac{a + 2bi}{c + 2di} = \frac{ac + 4bd}{c^{2}+4d^{2}} + \frac{-2ad + 2bc}{c^{2}+4d^{2}}i.$$ Then, we obtain that $F\subset \mathbb{Q}[i]$, where $F$ is the field of quotients of $\mathbb{Z}[2i]$. In fact, I have to show that $F=\mathbb{Q}[i]$, but I don't know how to show the inclusion $\mathbb{Q}[i]\subset F$. Thanks for your suggestions.

Answer 1

It is clear that $D := \mathbb{Z}[2i] \subseteq \mathbb{Q}[i]$. Since $\mathbb{Q}[i]$ is a field, that means that the quotient field $F$ of $D$ can be seen as a subfield of $\mathbb{Q}[i]$.

$F$ is the smallest subfield of $\mathbb{Q}[i]$ containing $D$. Now :

1. Since $\mathbb{Z} \subseteq F$, we have $\mathbb{Q} \subseteq F$.
2. Since $2i \in F$ and $\frac{1}{2} \in F$, we get $i \in F$

Hence, $F = \mathbb{Q}[i]$.