# Thread: order of an ideal ring

1. ## order of an ideal ring

Can someone please show me how to find the order of an ideal complex ring?

For this problem, I guess there are 10 elements in the ring.

Thank you.

2. Originally Posted by john_n82

Can someone please show me how to find the order of an ideal complex ring?

For this problem, I guess there are 10 elements in the ring.

Thank you.
in general the ring $\frac{\mathbb{Z}[i]}{}$ has $a^2 + b^2$ elements. to see this start with the fact that $\mathbb{Z}[i]$ is a euclidean domain. so for any $x \in \mathbb{Z}[i],$ there exist $s, r \in \mathbb{Z}[i]$ such that $x=s(a+bi) + r$

and either $r=0$ or $|r|^2 < a^2 +b^2.$

3. Originally Posted by NonCommAlg
in general the ring $\frac{\mathbb{Z}[i]}{}$ has $a^2 + b^2$ elements. to see this start with the fact that $\mathbb{Z}[i]$ is a euclidean domain. so for any $x \in \mathbb{Z}[i],$ there exist $s, r \in \mathbb{Z}[i]$ such that $x=s(a+bi) + r$

and either $r=0$ or $|r|^2 < a^2 +b^2.$
(First $a+bi \not = 0$ but that is a trivial thing).

What you wrote is correct, however, I do not see how it follows from $0\leq |r|^2 < a^2+b^2$.

Because if you take $a+bi=1+i$, for example, then there is non-uniqueness for the division algrorithm.
Notice if $r=n+mi$ then we want $0\leq n^2+m^2 < 2$. There are $5$ choices for $n,m\in \mathbb{Z}$ but there are only $|1+i|^2 = 2$ congruence classes.