Quotients of $\mathbb{Z}[i]$

Let $\mathbb{Z}[i]$ be the ring of Gauss integers. For a simple representation it is all the complex numbers of the form $a+ib$ such that $a,b \in \mathbb{Z}$. It is known that $\mathbb{Z}[i]$ is a principal ideal ring.

Let $z$ be nonzero and non unit and let $(z)$ be the ideal it generates. What is the number of elements in the quotient $\mathbb{Z}[i]\big/ (z)$ ?

I guess it's $N(z)$, the classical "norm" on $\mathbb{Z}[i]$ defined by $N(z) = z\bar{z}$. This can easily be seen on a drawing (it is a reminiscence of Pythagora's proof) but can someone help me on how to prove it rigorously ?

Edit (in answer to some answer) Note that all the quotients of $\mathbb{Z}[i]$ are finite.

Edit 2. Please also note this was a question from an (advanced) undergraduate book and should only require basic skills to solve (UFD, PID, euclid domain, etc).

Solution. Marwalix provided a link to some (very) interesting document solving the issue. For the people interested, here is a sketch of this simple proof:

• the number $n(z)$ of elements in $\mathbb{Z}[i]/(z)$ and $\mathbb{Z}[i]/(\bar{z})$ is the same
• the number of elements in $\mathbb{Z}[i]/(z\bar{z})$ is $N(z)^2$ (which can be easily seen as $z\bar{z}$ is an integer).
• also, $n(z)$ is a multiplicative function of $z$, so we have

$$|\mathbb{Z}[i]/(z\bar{z}) | = |\mathbb{Z}[i]/(z)| \times |\mathbb{Z}[i]/(\bar{z})|$$

Therefore, $n(z)n(\bar{z}) = n(z\bar{z}) = N(z)^2$, which yelds the conclusion.

Solution 1:

Look at theorem 7.14 in the link below http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf