finite factor ring with $4$ elements

Question

How to show that $\mathbb Z[i]/(2)$ is a finite factor ring with $4$ elements?

I know that $(2)=(1+i)(1-i)$, and the factor group is of a form $\{a+bi+(2)\}$. But how to show that this factor group has only $4$ elements?

Answer 1

Apply Euclidean division in $\mathbb{Z}[i]$ : Define $d: \mathbb{Z}[i]\setminus\{0\} \to \mathbb{N}$ by $$ d(a+bi) = a^2+b^2 $$ Then, $(\mathbb{Z}[i], d)$ is a Euclidean domain, and hence, for any $\alpha \in \mathbb{Z}[i]$, there are $\beta, \gamma \in \mathbb{Z}[i]$ such that $$ \alpha = 2\beta + \gamma, \text{ and } d(\gamma) < d(2) = 4 $$ whence $\overline{\alpha} = \overline{\gamma}$ in $\mathbb{Z}[i]/(2)$.

Now, $d(\gamma) < 4$, then consider the cases :

a) $d(\gamma) = 3$ : There is no such $\gamma$ (why?)

b) $d(\gamma) = 2$ : Then $\gamma = 1\pm i$.

c) $d(\gamma) = 1$ : Then $\gamma \in \{\pm 1, \pm i\}$

Now note that $\overline{1+i} = \overline{1-i}$ since $(1+i) - (1-i) \in (2)$. Similarly eliminate all the other overlaps, to obtain $$ \mathbb{Z}[i]/(2) = \{\overline{0}, \overline{1}, \overline{i}, \overline{1+i}\} $$