# Gaussian Integers ring

Let $\mathbb Z[i] =\{a+bi \mid a, b \in \mathbb Z\}$ be a Gaussian Integer set.

1) Show that ideal $I = (2+2i)$ is not a prime ideal. Find all elements of quotient ring $\mathbb Z[i]/I$ and find its characteristic.

2) Is quotient ring $\mathbb Z[i]/(1-i)$ a field?

So far I have this:

If I understand definition of prime ideal correctly, it should be enough to find such an $a$ and $b$ which gives me an element of the ideal $I$ but neither of them is element of $I$. Formally written: $a \notin I, b \notin I, a b \in I$

which gives me for example this solution:

$a = 2 + 0i \\ b = 1 + i$

is it correct?

I have already found the answer for elements of this ring here: Find all elements of quotient ring

so remaining thing is characteristic. Is it $0$?

And for 2nd task: To get a ring, I would need $1 - i$ to be irreducible on $\mathbb Z[i]$, but it is not, since I can factorize it to $i (-1 -i)$ - so this quotient ring is not a field.

Are my conclusions correct? Please correct me if I'm wrong. Thank you very much.

The first part is correct, provided you show that $2\notin(2+2i)$ and $1+i\notin(2+2i)$.
However, from $2=(x+yi)(2+2i)$ we get $|2|^2=|x+yi|^2\,|2+2i|^2$, that is $$4=8(x^2+y^2)$$ which is impossible. Do similarly for $1+i$.
The element $1-i$ is indeed irreducible. Indeed, if $1-i=z_1z_2$, we get $$|1-i|^2=|z_1|^2|z_2|^2$$ or $$2=|z_1|^2|z_2|^2$$ and so either $|z_1|=1$ or $|z_2|=1$. The elements $z\in\mathbb{Z}[i]$ such that $|z|=1$ are $1,-1,i,-i$ and all of them are invertible.
It's not a contradiction with $1-i=i(-1-i)$, exactly because $i$ is invertible.
Since $\mathbb{Z}[i]$ is a principal ideal domain (as you should know), if an element $z$ is irreducible, the ideal $(z)$ is maximal, so $\mathbb{Z}[i]/(z)$ is a field.