What is a generator for an ideal such that $I=\{a+bi|a+b \text{ is even}\}$?

Question

I had this problem where i had the application $\varphi: \mathbb Z[i] \Rightarrow \mathbb Z/(2)$ where $\varphi(a+bi)=\bar{a}+\bar{b}$. I had to find the kernel and prove that is a factor ideal. I proofed that the kernel is formed by all the complex numbers such that $a+b$ is even but I need to find the generator to the ideal.

Answer 1

A generator of the ideal is $1+i$.

Let $a+bi$ be a multiple of $1+i.$

Then $a+bi= (x+yi)(1+i)=(x-y)+(x+y)i,$ so $a+b=2x$ is even.

On the other hand, if $a+b$ is even, then so is $b-a$,

so $\dfrac{a+bi}{1+i}=\dfrac{(a+bi)(1-i)}2=\dfrac{a+b}2+\dfrac{(b-a)i}2\in \mathbb Z[i]$.

Answer 2

Since $\mathbb{Z}[i]$ is a Euclidean domain and this implies that it is a PID we have that any ideal $I \subset \mathbb{Z}[i]$ is generated by any element in $\mathbb{Z}[i]$ with the smallest norm; where the norm is defined by $a+bi \mapsto a^2 + b^2$. Therefore we need to find a non-zero $a,b$ that satisfies $a+b \equiv 0 \ (\text{mod } 2)$ that has the smallest possible $a^2+b^2$.

Therefore $1+i$ works; $1-i$, $-1+i$, and $1-i$ also work.