the ring $\mathbb{Z}[i]/<2+2i>$

Question

prove that characteristic of $\mathbb{Z}[i]/<2+2i>$ is $4$ and it has exactly $8$ elements.

I need to know, is this below approach works for characteristic?

My attempt

I saw $c+di\in <2+2i>$ if $c+di=(2+2i)(a+bi)$ for some $a+bi\in\mathbb{Z}[i]$.

i.e. if $c+di=(2a-2b)+ (2a+2b)i$ for some $a, b\in\mathbb{Z}$

i.e. if $2a-2b=c$ and $2a+2b=d$

i.e if $c+d=4a$ and $d-c=4b$

i.e if $\frac{c+d}{4}, \frac{d-c}{4}\in\mathbb{Z}$

So that from above, if we consider $4(p+qi)$ for any $p+qi\in\mathbb{Z}[i]$ we saw $c=4p$ and $d=4q$ and hence $\frac{c+d}{4}=p+q\in\mathbb{Z}, \frac{d-c}{4}=q-p\in\mathbb{Z}$. So that,

$4(p+qi)\in <2+2i>$ for any $p+qi\in\mathbb{Z}[i]$.

Hence for every $a+bi+ <2+2i>\in\mathbb{Z}[i]/<2+2i>$ we have,

$4(a+bi+ <2+2i>)= 4(a+bi)+<2+2i>=<2+2i>=\text{ zero element of ring }\mathbb{Z}[i]/<2+2i>$

Hence characteristic of given ring is $4$.

Further, i dont know how to show given ring has exactly $8$ elements. I saw various post on MSE for the question of above type but unable to understand the answer( I understand the case for $\mathbb{Z}[i]/<a+bi>$ when $a$ and $b$ are relatively prime. But here $a$, $b$ are not prime)

Please help.

Answer 1

Note that $$(2+2i)(1-i)=4$$ so $4\in \langle 2+2i\rangle$. You also need to know that this is the smallest, meaning the characteristic is not $2$ or $3$. So suppose $$2 = (a+bi)(2+2i)$$ Then $$2=(2a-2b)+(2a+2b)i$$ This means that $b=-a$, so $$2=4a$$ which is impossible. This also rules out $3=(a+bi)(2+2i)$.

Answer 2

$\newcommand{\r}{\langle 2+2i\rangle}$ You got $\frac{c+d}{4}$ and $\frac{d-c}{4}$ both belong to $\mathbb Z$. From $2a-2b = c$ and $2a+2b = d$ you get that $c,d$ are multiples of $2$. But they need not be multiples of $4$ i.e. the step then deducing $c = 4p, d = 4q$ is incorrect. For example, $c=2,d=6$ work but don't satisfy this property.

However, what is true, is that if $c= 4p,d=4q$ then $c+di$ belongs to $2+2i$. What this shows is that the characteristic is less than four. We need to show that it is not equal to $1,2,3$.

That is made clear from the fact that the elements $1+\r,2+\r,3+\r$ are not equal in $\r$. I leave you to see this. Thus, $1+\r$ has to be multiplied by at least $4$ to get $0 + \r$. This proves the characteristic result.

Tip : Suppose you find a number $N$ satisfying the property that $N1 = 0$ where $1,0$ are the unit and zero of the ring respectively, and you can't reduce $N$. Then, $N$ is the characteristic of the ring.

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If you want to write down all the elements of the ring, you do : $c+di + \r = a+bi + \r$ means that $(c-a)+(b-d)i \in \r$, so there exist $x,y$ so that $2(x-y) = c-a$ and $2(x+y) = b-d$. In particular, if we consider the elements $$ \{a+bi + \r: a \in \{0,1,2,3\}, b \in \{0,1\}\} $$

Then these are all distinct. To see this, note that if two elements from this set have different or same $i$ coefficient. I leave you to argue from here.

Now, suppose you have any other $x+iy$. First use the characteristic to bring it to $a,b \in \{0,1,2,3,4\}$. Then, note that $-2-2i$ or $2-2i \in \r$ can be added to bring it to the given set. Hence we are done.