Number of elements in the ring $\mathbb Z [i]/\langle 2+2i\rangle$ The question is : 
Show that $I=\langle 2+2 i\rangle$ is not a prime ideal of $\mathbb Z[i]$. Also find the number of elements in $\mathbb Z[i]/I$ and its characteristic.
My try:
I started with elements $2, 1+i \in \mathbb Z[i]$. Clearly, the product is $0$ in the coset representation.
The part where i am struck at is the number of elements and the characteristic.
Well, $2+2i+\langle 2+2i\rangle=0+\langle 2+2i\rangle$
Can we directly write $i^2=-1$ in the coset representation?
 A: The following is not the most elementary approach, but it puts the question in a nice perspective (in my humble opinion):
There is an isomorphism $\Bbb{Z}[i]\cong\Bbb{Z}[X]/(X^2+1)$, given by $i\ \mapsto\ X$, yielding an isomorphism
$$\Bbb{Z}[i]/(2+2i)\cong\Bbb{Z}[X]/(X^2+1,2+2X).$$
The characteristic of the latter is the least positive integer in the ideal $(X^2+1,2+2X)\cap\Bbb{Z}$ of $\Bbb{Z}$. Its number of elements is precisely the index of the ideal $(X^2+1,2+2X)$ in $\Bbb{Z}[X]$, which is the same as the resultant of $X^2+1$ and $2+2X$, which is in itself a nice fact to prove.
A: One can show that if $a+bi$ is nonzero, $$\Bbb Z[i]/(a+bi)$$ has exactly $a^2+b^2$ elements. Consider the free $\Bbb Z$-module, $M=\Bbb Z^n$. Let $e_1,\ldots,e_n$ be the canonical basis elements. Suppose $f_i=\sum a_{ij}e_j$, and $N=\langle f_1,\ldots,f_n\rangle$. Suppose moreover that $\det (a_{ij})=d\neq 0$. Then $M/N$ has $d$ elements: using Smith's Normal Form, we can obtain $Q,P$ invertible matrices over $\Bbb Z$ such that $QAP={\rm diag}(d_1,\ldots,d_n)$, i.e. make an appropriate change of basis so that we get generators $f_1',\ldots,f_n'$ with $f_i'=d_ie_i'$. Since $d\neq 0$; no $d_i$ is zero, so we get $d_1\ldots d_n=d$ choices for a linear combination in $M/N$ which gives our result. Now consider $\Bbb Z[i]$. This is isomorphic to $\Bbb Z^2$ as a $\Bbb Z$-module with basis $\{1,i\}$. The ideal $(a+bi)$ is generated by $a+bi$ and $-b+ai$, i.e. we have generators $f_1=(a,b)$ and $f_2=(-b,a)$ of the submodule $(a+bi)$. Then $A=\begin{pmatrix}a&b\\-b&a\end{pmatrix}$ has determinant $a^2+b^2$.

Note that $2=(1-i)(1+i)$, so $2(1+i)=(1-i)(1+i)^2$. The elements $1+i,1-i$ are irreducible in $\Bbb Z[i]$, since they have prime norm, so they are prime, and they are associates, since $i(1-i)=1+i$, and $i$ is a unit. Thus $(2(i+1))=((1+i)^3)$. It follows you're looking at  $$\Bbb Z[i]/(p^3)$$ where $p=1+i$. Let $\mathfrak a=(p)$. From $(1+i)^2=1+2i-1=2i$, we get $\mathfrak a^2=(2)$, so $\mathfrak a^4=(4)$. We may look then at $(\Bbb Z[i]/\mathfrak a^4)(\mathfrak a^3/\mathfrak a^4)$.
The denominator is $(\Bbb Z/4\Bbb Z)[i]$. This has $16$ elements. Thus your ring, being a nontrivial quotient of this, has at most $8$ elements. Using Paul's comment you may show your ring has exactly $8$ elements, and has characteristic $4$.
A: I usually solve it by drawing the square generated by $2+2i$.  
....|....
..xbbbx..
..aoooa..
--aoooa--
..aoooa..
..xbbbx..
....|....

I count 16 points:


*

*the 9 points marked o are distinct

*the 3 points marked a on the left and the right are congruent mod $(2+i)$

*the 3 points marked b on the top and bottom are congruent mod $(2+i)

*the 1 points marked x are the same

