Quotient ring and gaussian integers

Question

In a certain exercise I am asked to identify the ring $R = \mathbb{Z}[i]/\langle2i\rangle$. What I did is the following: $R = \{a+bi, c+di \in \mathbb{Z}[i]: a = c, b - d \equiv 0$ $($mod $2)$ (this is because the equivalence relation is defined such that the substraction of two elements must be in the ideal). However, I have a question to solve:
Since each $a$ defines a class (supposing that the second condition is satisfied) and there are infinite ones, aren't there infinite equivalence classes? As a consequence of this, isn't the ring infinite? I saw here another case similar to this one, but in this case the ideal was generated by $2+i$ ending up with the result that the quotient ring was equivalent to the ring $\mathbb{Z}/5\mathbb{Z}$. Could somebody help me?

Answer 1

You are modding by the ideal generated by $2i$, which is not just the set of integer multiples of $2i$, but the set of all multiples, so your equivalence relation isn't the right one. For example, $2$ is in the ideal generated by $2i$, since it's $(2i)\cdot(-i)$, so $2$ should also be equivalent to $0$. The right definition is that $a - c$ is even and $b - d$ is even. (In fact the ideal generated by $(2i)$ is the same as the ideal generated by $2$, since $i$ is a unit in this ring.)