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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.

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  • $\begingroup$ Not every positive integer can be the norm of a Gauss integer. $4k+3$ primes, for example, aren't. $\endgroup$ – ajotatxe Apr 30 '15 at 15:51
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    $\begingroup$ what's the link with my question ? $\endgroup$ – Tlön Uqbar Orbis Tertius Apr 30 '15 at 15:51
  • $\begingroup$ I thought that you thought that there is an element in $\Bbb Z[i]/(z)$ for each integer between $0$ and $|z|^2-1$. If you didn't, please ignore my previous comment. $\endgroup$ – ajotatxe Apr 30 '15 at 15:53
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    $\begingroup$ You can use areas to calculate this. In $\Bbb{Z}[i]$ there is a square of area $1$ attached to each element in such a way that you cover the entire plane without overlap. If you look at the ideal generated by $z=a+ib$ it is generated as an abelian group by $z$ and $iz=-b+ia$. In cartesian coordinates you can then attach a parallelogram with side vectors $(a,b)$ and $(-b,a)$ to each element of $I$ - also in such a way that they cover the entire plane without overlap. You should then observe that these parallelograms are actually squares with area $a^2+b^2$. $\endgroup$ – Jyrki Lahtonen Apr 30 '15 at 17:18
  • $\begingroup$ Thank you Jyrki, but that's exactly what I meant when I said "this can easily be seen on a drawing". In fact, the point is to prove that the number of points with interger coordinates in a square does not depend on the position of the square, but only on its size. This is intuitive, but not so easy to prove. $\endgroup$ – Tlön Uqbar Orbis Tertius Apr 30 '15 at 18:08
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Look at theorem 7.14 in the link below http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf

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  • $\begingroup$ This proves the thing. I've added the solution to my original post as an edit. Thank you ! $\endgroup$ – Tlön Uqbar Orbis Tertius Apr 30 '15 at 18:25

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