The ring $\mathbb{Z}[i]/(2)$

Question

Let $\mathbb{Z}[i]/(2)$ be the ring of Gaussian Integers and $I=(2)$ be the ideal generated by $2$. Then $I=\{2(x+yi) | x,y \in \mathbb{Z} \}$. Therefore the elements of $\mathbb{Z}[i]/(2)$ are the cosets $a+bi +I$.

Here we have $4$ cases depending on whether $a, b$ are even or odd integers and therefore we have $4$ cosets $I, 1+I, i+I$ and $1+i+I$. Now we have that $ord(1+I)=2$ since $1+I + 1+I =2+I =I$. Similarly we have that $ord(i+I)=2$ and $ord(1+i+I)=2$ and therefore $(\mathbb{Z}[i]/(2),+)\simeq(\mathbb{Z}_{2}\times\mathbb{Z}_{2},+)$.

Now what about $(\mathbb{Z}[i]/(2),\cdot)$? We know that it not a field because it contains zero divisors. Indeed $(1+i+I)(1+i+I)=I$. So we know it is commutative ring with identity. But can we find if it is isomorphic to a familiar ring?

Answer 1

Using the isomorphism $\mathbb{Z}[i] \cong \mathbb{Z}[X]/\langle X^2+1 \rangle$ we get $$\mathbb{Z}[i]/\langle 2 \rangle \cong \mathbb{Z}[X]/\langle X^2+1,2 \rangle = \mathbb{F}_2[X]/\langle X^2+1 \rangle = \mathbb{F}_2[X]/\langle (X+1)^2 \rangle \cong \mathbb{F}_2[Y]/\langle Y^2 \rangle.$$ This is the ring of dual numbers over $\mathbb{F}_2$. It is one of the $4$ rings with $4$ elements (see here).