Ideal of Gaussian Integers Ring. Congruence Classes

Question

Let $I = \{a+bi \in \mathbb{Z}[i] \; / \; [a]_2 = [b]_2\}$ be an ideal of $\mathbb{Z}[i]$ where $[a]_2 = \{a+2n \;/\; n \in \mathbb{Z}\}$

(a) Prove that I is generated by $1 + i$

I solved it this way:

|$\subseteq$ $$[1]_2=[1]_2\Rightarrow 1+i \in I$$

|$\supseteq$ $$ z \in I \Rightarrow z = x+yi \; / \; [x]_2=[y]_2 \Rightarrow y-x=2n\;/\;x,y,n \in \mathbb{Z}\Rightarrow y=x+2n\; \land \; x=y-2n \;/\;x,y,n \in \mathbb{Z} \Rightarrow z \in (1+i) $$

But I’m not sure if it is correct.

(b) How many elements does $\mathbb{Z}[i]/(1+i) $have?

I think $\mathbb{Z}[i]/(1+i)=\{[1+i]\}$ but I don’t know how to prove it.

I really appreciate your help.

Answer 1

(a) It is definitely correct that $I=(1+i)$, although I have some comments that I hope will be helpful.

First, in your proof that $(1+i)\subseteq I$, you've only proved that $1+i\in I$. Note that this is enough to show that $(1+i)\subseteq I$ if you know that $I$ is an ideal. I think you should show that $I$ is an ideal of $\Bbb{Z}[1+i]$.

Second, I'm not sure that I understand your proof that $I\subseteq(1+i)$. Below, I've written a few steps that you can add to make it more clear. Note: the main observation is that $2=(1+i)(1-i)$.

$$\text{Let }z\in I$$ $$z=x+iy\text{ for some }x,y\in\Bbb{Z}\text{ with }x\equiv y\text{ mod }2$$ $$y=x+2n\text{ for some }n\in\Bbb{Z}$$ $$z=x+iy=x+i(x+2n)=x(1+i)+2in=(1+i)\cdot\left[x+i(1-i)n\right]$$ $$z\in(1+i)$$

(b) $\left\vert\Bbb{Z}[i]/(1+i)\right\vert=2$.

Note that for every $x,y\in\Bbb{Z}$ there are two possibilities: either $x\equiv y$ modulo $2$, or $x\not\equiv y$ modulo $2$. You've already shown that

$$(1+i)=\{x+iy\,\vert\,x\equiv y\text{ mod }2\}.$$

It is enough to show that

$$1+(1+i)=\{x+iy\,\vert\,x\not\equiv y\text{ mod }2\}.$$

Answer 2

$I$ is precisely the kernel of the ring epimorphism $\psi:a+bi\mapsto [a]_2+[b]_2$ from $\Bbb Z[i]$ to $\Bbb F_2$ where $[x]_2$ denotes the residue class $x+2\Bbb Z$. Therefore $\frac{\Bbb Z[i]}{I}\approx\Bbb F_2$ and $|\frac{\Bbb Z[i]}{I}\vert=2$. $$\forall z_1,z_2\in\Bbb Z[i]\;\;1+i=z_1z_2\implies 2=|z_1|^2|z_2|^2\implies z_1=1\lor z_2=1$$ Therefore $\langle 1+i\rangle$ is maximal because $1+i$ is irreducible and thus prime as $\Bbb Z[i]$ is a principal ideal domain. Therefore we must have $I=\langle 1+i\rangle$ because $\langle 1+i\rangle\subseteq I$; i.e. $1\equiv 1\;\text{mod}\;2$ and so $1+i\in I$.