How many elements are in Z[i]/<3+i>. Explain your answer.
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 $\displaystyle \frac{\mathbb{Z}[i]}{<a+bi>}$ has $\displaystyle a^2 + b^2$ elements. to see this start with the fact that $\displaystyle \mathbb{Z}[i]$ is a euclidean domain. so for any $\displaystyle x \in \mathbb{Z}[i],$ there exist $\displaystyle s, r \in \mathbb{Z}[i]$ such that $\displaystyle x=s(a+bi) + r$
and either $\displaystyle r=0$ or $\displaystyle |r|^2 < a^2 +b^2.$
(First $\displaystyle a+bi \not = 0$ but that is a trivial thing).
What you wrote is correct, however, I do not see how it follows from $\displaystyle 0\leq |r|^2 < a^2+b^2$.
Because if you take $\displaystyle a+bi=1+i$, for example, then there is non-uniqueness for the division algrorithm.
Notice if $\displaystyle r=n+mi$ then we want $\displaystyle 0\leq n^2+m^2 < 2$. There are $\displaystyle 5$ choices for $\displaystyle n,m\in \mathbb{Z}$ but there are only $\displaystyle |1+i|^2 = 2$ congruence classes.