FInd $\Bbb{Z}[i]/(i)$ and $\Bbb{Z}[i]/(1+i)$ and find GCD of $85$ and $1+13i$ in $\Bbb{Z}[i]$. So for finding $\Bbb{Z}[i]/(i)$ I have that an ideal is the whole ring if it contains a unit. So ($i$) contains a unit,  namely $i$ as
$$(-i)(i)=1$$
thus $i$ is a unit forcing $(i)$ to be all of $\Bbb{Z}[i]$ thus $\Bbb{Z}[i]$ is the trivial ring.
For $\Bbb{Z}[i]/(1+i)$ I know $1+i=0$ so $i= -1$ and $-i= 1$ so we only have $1$ and $-1$ which is iso to $\Bbb{Z}_2$ is this correct?
And for dividing $85$ by $1+13i$ I know
$$\frac{85}{1+13i}=-6i$$
but I wanna know why we round the fractions $\frac{1}{2}$ and $\frac{13}{2}$ down to $0$ and $6$ Then when I plug this for $q$ into
$$85=q(1+13i)+r$$
and solve for $r$ I get $7+6i$. where do I go from there do I divide $1+13i$ by $7+6i$ to show the next remainder is $0$
Cause that equals 1+i which means $r=0$ in the next step. So is the GCD just $$7+6i$$?
Thanks in advance!
EDIT: this post has been up for over 12 hours and no one has helped can someone comment if I’m on the right track or not

You solved everything you asked but it is necessary to talk about the prime factorization of the Gaussian integers for a better solution for your last question.
Here, $85=(2+i)(2-i)(4+i)(4-i)$ and $1+13i=(2+i)(4+i)(1+i)$ are prime factorization of these numbers up to units, so upto units, $GCD(85,1+13i)=(2+i)(4+i)=7+6i$.
I learned these things today from : https://en.wikipedia.org/wiki/Gaussian_integer#:~:text=Basic%20definitions,-The%20Gaussian%20integers&text=In%20other%20words%2C%20a%20Gaussian,is%20thus%20an%20integral%20domain.
It also says: Unfortunately, except in simple cases, the prime factorization is difficult to compute, and Euclidean algorithm leads to a much easier (and faster) computation. I think this is a simple case.
