Prove that $\Bbb{Z}[i]/I$ is finite where I is an ideal of $\Bbb{Z}[i]$

Question

Show that for any nontrivial ideal $I$ of $\Bbb{Z}[i]$, $\Bbb{Z}[i]/I$ is finite.

$\Bbb{Z}[i]$ is a PID, so $I=\langle{a+ib\rangle}$. Now $\Bbb{Z}[i]/I$ has elements of the form $c+id+\langle{a+ib}\rangle$. Now I have a vague idea how it'll be finite: $a+ib=0$ gives $a^2=-b^2$. So..?

Think that both $a^2+b^2$ and $i(a^2+b^2)$ will be in $I$. Does that make you see the light? — Jyrki Lahtonen, Apr 19 '15 at 17:30
@JyrkiLahtonen yes. So we'll view the $c$ and $d$ modulo $a^2+b^2$ — zed111, Apr 19 '15 at 17:33
http://math.stackexchange.com/questions/23358/quotient-ring-of-gaussian-integers — user26857, Apr 20 '15 at 08:43

Martin Brandenburg · Answer 1 · 2015-04-19T17:39:28.433

2

Let $a+ib \in I \setminus \{0\}$. Then $n:=a^2+b^2$ is a positive integer contained in $I$. It follows that $\mathbb{Z}[i]/I$ is a quotient of the ring $\mathbb{Z}/n\,[x]/(x^2+1)$, which is of course finite.

edited Apr 19 '15 at 17:39

answered Apr 19 '15 at 17:35

Martin Brandenburg

181,922

I don't know torsion groups, will look them up. – zed111 Apr 19 '15 at 17:36
I've replaced it with a direct argument. – Martin Brandenburg Apr 19 '15 at 17:39

Prove that $\Bbb{Z}[i]/I$ is finite where I is an ideal of $\Bbb{Z}[i]$

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