Showing that $\mathbb{Z}[i]/I$ is a finite field whenever $I$ is a prime ideal, and also finding its cardinality? How to show  that $\mathbb{Z}[i]/I$ is a finite field whenever $I$ is a prime ideal? Is it possible to find the cardinality of $\mathbb{Z}[i]/I$ as well?
I know how to show that it is an integral domain, because that follows very quickly.
 A: Let $\alpha \in I$ be an element, with norm $N(\alpha)$. Any element $x \in \mathbb{Z}[i]$ can be written as $q \alpha + r$ with $N(r) < N(\alpha)$. So every element of $\mathbb{Z}[i]/I$ (viewed as an equivalence class) contains an element of norm smaller than $N(\alpha)$, and there are only finitely many such elements.
We don't have to use the assumption that $I$ is prime for the finiteness.
A: Another approach, especially if you’re only interested in counting $\Bbb Z[i]/I\,$: There’s a theorem in algebraic number theory that if $I$ is a nonzero principal ideal of the ring of integers $R$ in an algebraic number field $K$, say with $I=(\xi)$, then the cardinality of $R/I$ is $\big|\text N(\xi)\big|$, where N is the field-theoretic norm from $K$ down to $\Bbb Q$.
In the case $K=\Bbb Q(i)$, every $I$ is principal and the norm already is positive, so you can drop the absolute-value bars. In other words, if $I=(a+bi)$, then the cardinality of $\Bbb Z[i]/I$ is $a^2+b^2$. This has nothing to do with whether $I$ is prime.
A: A small variation of Arthur's argument. I wanted to do this without using the fact that the complex norm works as a Euclidean domain norm as well.

The claim is not true, if $I=\{0\}$, so let's assume that is not the case. So then there exists an element $z=a+bi\in I$ with either $a$ or $b$ non-zero. Because $I$ is an ideal, the number 
$$
z(a-bi)=(a+bi)(a-bi)=a^2+b^2\in I.
$$
This implies that the intersection $I\cap\Bbb{Z}$ is non-trivial. It is clearly a prime ideal of $\Bbb{Z}$, so the known list of prime ideals of $\Bbb{Z}$ says that there exists a prime number $p$ such that $I\cap\Bbb{Z}=p\Bbb{Z}$.
From this we can deduce two things. Namely we know that $p$ is the characteristic of the quotient ring $\Bbb{Z}[i]/I$. This implies that $\Bbb{Z}[i]/I$ is a vector space over the field $\Bbb{F}_p$. Secondly, from $p\in I\implies pi\in I$. Therefore
$$
J=p\Bbb{Z}[i]=p\Bbb{Z}\oplus pi\Bbb{Z}\subseteq I.
$$
As a subgroup of the additive group of $\Bbb{Z}[i]$ $J$ is generated by $p$ and $pi$, so it is of index $p^2$. In other words the dimension of $\Bbb{F}_p$as a vector space over
$\Bbb{F}_p$ is at most two. Therefore $\Bbb{Z}[i]/I$ has either $p$ or $p^2$ elements.
Anyway a finite integral domain $\Bbb{Z}[i]/I$ is the a (finite) field.
Both possibilities, $p$ and $p^2$ occur. The smallest examples are $\Bbb{Z}[i]/\langle 1+i\rangle$ that has only two elements, and $\Bbb{Z}[i]/\langle 3\rangle$ that has nine elements (see below for the reason why in the latter case we cannot get a field of three elements).

It is not too difficult see which case it is. Assume that $p>2$. Then we see that the cosets $i^k+I$ are distinct for $k=0,1,2,3$. In particular the multiplicative order of the coset $i+I$ is exactly four. By Lagrange this means that the order of the multiplicative group of $\Bbb{Z}[i]/I$ must be a multiple of four. So if $\Bbb{Z}[i]/I$ has only $p$ elements, then we must have $p\equiv1\pmod4$. A consequence is that if $p\equiv3\pmod4$, then the field $\Bbb{Z}[i]/I$ must have $p^2$ elements. OTOH it is known that the multiplicative group of $\Bbb{F}_p$ is cyclic of order $p-1$ (a primitive root exists!). This means that there is an integer $a, 0<a<p$ such that
$a^4\equiv1\pmod p$, $a^2\not\equiv1\pmod p$. Because in a field $\Bbb{Z}[i]/I$ the equation $x^4=1$ can have at most four solutions, the coset $a+I$ has to be equal to one of the cosets $\pm i+I$. Therefore in this case the cardinality of $\Bbb{Z}[i]/I$ is strictly less than $p^2$, hence exactly $p$.
A: Let us count the number of elements in $\mathbb{Z}[i]/n\mathbb{Z}[i]$. When do two elements $a+ib$, $a_1+ib_1$ in $\mathbb{Z}[i]$ belong to the same coset of $n\mathbb{Z}[i]$?  We must have $(a-a_1)+i(b-b_1)\in n\mathbb{Z}[i]$. So, $(a-a_1)+i(b-b_1)=n(c+di)$ with $c$,$d\in \mathbb{Z}$.  This means that $a\equiv a_1\bmod{n}$ and $b\equiv b_1\bmod{n}$.  Conversely, if $a\equiv a_1\bmod{n}$ and $b\equiv b_1\bmod{n}$, then $(a+ib)-(a_1+ib_1)\in n\mathbb{Z}[i]$. If we let $S=\{a+bi\mid 0\leq a<n, 0\leq b <n\}$, then for any $u+iv$ in $\mathbb{Z}[i]$ there is a unique $a_0+ib_0\in S$ such that $(u+iv)-(a_0+ib_0)\in n\mathbb{Z}$. It follows that $\vert \mathbb{Z}[i]/n\mathbb{Z}\vert=\vert S\vert$.  Therefore, $\vert\mathbb{Z}[i]/n\mathbb{Z}[i]\vert=n^2$  since $\vert S\vert=n^2$.
Now, if $I$ is any  nonzero ideal, let $(a+ib)\in I$, $a$, $b\in \mathbb{Z}$, $a+ib\neq 0$.  Then, since $I$ is an ideal, $(a-ib)(a+ib)\in I$, so $I$ contains some nonzero integer $n$.  So, $(n)\subset I$ and we have a surjective ring homomorphism  $\mathbb{Z}[i]/n\mathbb{Z}\to \mathbb{Z}[i]/I$, therefore $\mathbb{Z}[i]/I$ is a finite ring. If $I$ is a prime ideal, $\mathbb{Z}[i]/I$ is a finite integral domain, and therefore a field.
Finding the cardinality of $\mathbb{Z}/I$ is a little more tricky.
Let $I$ be a prime ideal. The argument in the previous paragraph shows that $I$ contains a natural number and therefore a smallest natural number $n$.  This smallest natural number $n$ has to be a prime. If $n=m_1m_2$ with $m_1\neq 1$ and $m_2\neq 1$, since $I$ is a prime ideal, $m_1$ or $m_2\in I$, contradicting the minimality of $n$. So, $I$ contains a prime number $p$ and $(p)\subset I$. Then, from the surjective map $\mathbb{Z}[i]/(p)\to \mathbb{Z}[i]/I$, we see that $\vert \mathbb{Z}/I\vert$ divides $\vert\mathbb{Z}[i]/(p)\vert=p^2$. We have $I\neq \mathbb{Z}[i]$, so, $\vert \mathbb{Z}[i]/I\vert \neq 1$. Therefore, either $\vert \mathbb{Z}[i]/I\vert = p$ and
$(p)\subsetneq I$ in this case. If  $\vert \mathbb{Z}[i]/I\vert=p^2,$ $I=(p)$ in this case.  In the second case $p$ is a prime in $\mathbb{Z}[i]$.
Suppose  $(2)\subset I$. Since $2=(1+i)(1-i)\in I$, $1+i$ or $1-i$ is in $I$. Since $1+i$ and $1-i$ are associates, we may assume that $(1+i)\subset I$.  Then,  $\mathbb{Z}[i]/(1+i)$ has two elements, $0+(1+i)$ and $1+(1+i)$ since $\pm 1\equiv \pm i\pmod{1+i}$; this is because $1+i$, $-1+i$, $1-i$, $-1-i$ are associates of each other. Therefore, $\vert \mathbb{Z}[i]/(1+i)\vert=2$. Since $(1+i)\subset I\subset \mathbb{Z}[i]$ and $I\neq \mathbb{Z}[i]$, we have $I=(1+i)$ and $\vert \mathbb{Z}[i]/I\vert=2$.
Let us now assume that $p>2$. With some number theory, one can show that $\vert \mathbb{Z}[i]/I\vert=p$ if $p\equiv 1\bmod{4}$. In this case $p=a^2+b^2=(a+ib)(a-ib)\in I$, $a$, $b\in\mathbb{Z}[i]$. (See here for a beautiful proof of the fact that $p$ is the sum of two squares if $p\equiv 1\bmod{4}$.) So, $a+ib\in I$ or $a-ib\in I$, say $a+ib\in I$.  We have $(p)\subsetneq (a+ib)$; if $(p)=(a+ib)$, $p$ and   $a+ib$ will be associates. It  is easy to check this is not the case using the fact that the units in $\mathbb{Z}[i]$ are $\{\pm 1,\pm i\}$. Also, $(a+ib)\neq \mathbb{Z}[i]$ since $(a+ib)\subset I$. So $\vert\mathbb{Z}[i]/(a+ib)\vert$ is a proper divisor of $\vert \mathbb{Z}[i]/(p)\vert=p^2$. Therefore, $\vert \mathbb{Z}[i]/(a+ib)\vert=p$.  Since $(a+ib)\subset I$, $\vert \mathbb{Z}[i]/I\vert$ divides $\vert \mathbb{Z}[i]/(a+ib)\vert=p$. Since $I\neq \mathbb{Z}[i]$, $\vert \mathbb{Z}[i]/I\vert=p$ and $I=(a+ib)$.
If $p\equiv 3\bmod{4}$, then $p$ is a prime in $\mathbb{Z}[i]$. If not, let $p=\alpha\beta$. Then $p=\overline{p}=\overline{\alpha}\overline{\beta}
$. A somewhat messy argument gives $p=(a+ib)(a-ib)=a^2+b^2$. This is not possible since square of every integer is $1$ or $0\bmod{4}$, $p$ cannot be sum of two squares.
