Vtyjkyuk

Problem

How many elements are there in $mathbbZ[i]/I$ where $I= langle2+2irangle$

$2+2i + I =I$ . So $2+2i$ is equivalent to $0$ . So $2+2i=0$ which implies $i=-1$ which inturn implies $2=0$.

How to move after that?

Hint:

Here is a nice way to think about what $Bbb Z[i]/I$ looks like. You are looking at elements $alpha, beta inBbb Z[i]$ and calling them equal if you can get from one to another by adding an element from $I$. For instance, $1+2iequiv_I 3+4i$ since $3+4i=(2+2i)+1+2i$.

To see what $(2+2i)$, the ideal, looks like you can start by plotting the point $alpha=2+2i$. Remember that you can add anything from $(2+2i)$ to this point $alpha$
, including:

• $2+2i$


• $(2+2i)i=-2+2i$


• $(2+2i)+(-2+2i)=4i$


• $(4i)i=-4$, and so on.

This should give you a new picture:

There are obviously and an infinite amount of red points, but I'm only drawing some. All these red points are infact the same element in the ring $Bbb Z[i]/I$, so the amount of different colours needed to fill the grid is the cardinality of $Bbb Z[i]/I$. You can continue in this way with $1+i+(2+2i)$ to get:

Another hint:

The smallest distance between two points of same colour is $sqrt2^2+2^2$.

Let $f:mathbbZ[i]to mathbbN, f(a+bi)=a^2+b^2 $, then $mathbbZ[i]$ is an ED so $mathbbZ[i]/I=a+bi+(2+2i): a^2+b^2<2^2+2^2=8$ this is because we have considered $mathbbZ[i]$ as an ED so by the division algorithm we have that every element of $dfracmathbbZ[i]I$ is of the form $(c+di)(2+2i)+x+yi$ where $x+yi=0$ or $f(a+bi)<f(2+2i)$.

We want to examine the structure of $mathbbZ[i]/I$ as a $mathbbZ-$module so since $mathbbZ$ is a PID and $mathbbZ[i]/I$ finitely generated we may use the Structure Theorem (I'm sorry but in order to give a full answer I have to use modules)

As an $mathbbZ-$module $mathbbZ[i]$ has basis $e=1,i$ and $a=(a+bi)quad a+bi,-b+ai$ and so $A=(id:a,e)=beginpmatrix a& -b\ b& aendpmatrix$. A smith canonical form of this matrix is $B=beginpmatrix d_1& 0\ 0& d_2endpmatrix$ where $d_1=gcd(a,b)$ and $d_2=dfraca^2+b^2d_1$ so there are invertible matrices $Q,RinmathbbZ^2times 2$ st

$$PAQ=BLeftrightarrow (id:e,v)(id:a,e)(id:u,a)=(id:u,v)=beginpmatrix d_1& 0\ 0& d_2endpmatrix$$
so

$$u_1=d_1v_1,u_2=d_2v_2$$
and
$$dfracmathbbZ[i]IcongdfracmathbbZu_1oplus mathbbZu_2mathbbZd_1u_1oplus mathbbZd_2u_2 cong dfrac(u_1)(d_1u_1)oplusdfrac(u_2)(d_2u_2)cong dfracmathbbZ(d_1)oplusdfracmathbbZ(d_2)Rightarrow bigg|dfracmathbbZ[i]Ibigg|=d_1d_2=a^2+b^2=8$$