Use the the division algorithm and the fact that $I= (\alpha)$

Question

Prove that the quotient ring $Z[i]/I$ is finite for any non zero ideal $I$ of $Z[i]$.

Answer 1

$ℤ[i]$ is euclidean, and especially a principal ideal domain, so any ideal $I$ can be written as $I = ℤ[i]x$ for some $x ∈ ℤ[i]$. It’s also an integral domain, so the multiplication with $x$ yields an isomorphism $ℤ[i] \cong I$, so $I$ is a free $ℤ$-module of rank $2$.

By the structure theorem of finitely generated abelian groups, there is a $ℤ$-basis $x_1, x_2$ of $ℤ[i]$ and divisors $d_1 \mid d_2$, such that $d_1x_1, d_2x_2$ is a $ℤ$-basis of $I$. By the above, deduce $d_1 ≠ 0$ and $d_2 ≠ 0$. What does this imply about $ℤ[i]/I$?

Answer 2

Here is a geometric way to think of it. I is nonzero so has some nonzero element $a+bi$. I is an ideal, so $i(a+bi)=ai-b \in I$. Think of I as a lattice in $\mathbb{C}$, so $a+bi$ is some nonzero vector and multiplication by $i$ acts as rotation by $\frac{\pi}{2}$, so that $v:=a+bi$, and $w:=ai-b$ are $\mathbb{Z}$-linearly independent elements of I. Then the 2-dimensional lattice $\mathbb{Z}v + \mathbb{Z}w$ is a subset of $I$ and the quotient $\mathbb{Z}[i]/I$ has at most the number of elements in a fundamental domain for that lattice, which is some square with a vertex at the origin and rotated by the angle of $a+bi$. In particular, it is a bounded region with finitely many points.

Answer 3

$ I\neq 0\,$ so there is a $\,0\neq \alpha\in I,\,$ so an integer $\,0\neq n=\alpha\bar\alpha\in I.\ $ Thus, $ $ mod $\,I\!:\ n\equiv 0\ $ so $\ a+bi\,\equiv\, \bar a + \bar b i,\,$ for $\,\bar x = x\ {\rm mod}\ n.\,$ But there are at most $\,n^2\,$ such $\,\bar a + \bar b i$.