Modulo operations over Gaussian Integers

Question

Given $a,b\in\Bbb Z[i]$, is there a definition and calculation of remainder $a\bmod b$? Could you provide examples say $35\bmod (2+3i)$, $(43+7i) \bmod (22+8i)$?

Answer 1

Yes, this is certainly something one can talk about. But first, let's look at normal modular arithmetic. If $m,n\in\Bbb Z$, you can talk about $m\pmod n$. Often, we take this to be the smallest nonnegative $d$ such that $n$ divides $m - d$. I.e., $d\in\{0,1,\dots, n-1\}$. However, mathematically, any $d'$ such that $n$ divides $m - d'$ works just as well for modular calculations. We could have taken $d\in\{1 - n,\dots, -1, 0\}$ if we wanted, everything would work out fine!

Abstract ring theory circumvents the choice of any set of valid remainders by saying "ok, rather than working with $d\in\{1-n,\dots,-1,0\}$ or $d\in\{0,1,\dots,n-1\}$, let's say that $m\pmod n$ is defined to be the set of possibilities for such a $d$ in any choice of set of remainders: $m\pmod n = \{d\in\Bbb Z\mid n\textrm{ divides } m - d\}$." This is useful because we can do the same thing in any ring (in particular, we can do this in $\Bbb Z[i]$), where there might not be an obvious choice of set of remainders.

Of course, for the sake of calculation we might want to choose a set of remainders. To find such a set, it helps to draw a picture (you might want to get out a piece of paper for this next part and actually do it to see it for yourself). In $\Bbb Z$, we draw the number line and then pick $n$ consecutive points to be our remainders. You can imagine that you identify the next integer with the first one you picked, and then the next with the second, and so on, until what you have is something like this (this is illustrating the case $n = 7$). You could have also obtained this by first marking all multiples of $n$, and then wrapping $\Bbb Z$ into a circle so that all the marked points are on top of each other (and the ones in between will be identified/smushed in the right way like in the picture). We can do the same thing in $\Bbb Z[i]$: draw $\Bbb Z[i]$ as the integer lattice in $\Bbb R^2$:

and plot whatever point you want to look "modulo." Say it's $3 + 2i$. What you want to do next is mark all multiples of $3 + 2i$ on the plane. Mark $0$, $3 + 2i$, $-2 + 3i$, $(3 + 2i)(1 + i) = 1 + 5i$, etc. You'll wind up with a lot of marked points that split your $\Bbb Z[i]$ into squares if you draw lines through the marked points parallel to the lines through $0$ and $3 + 2i$ and $0$ and $-2 + 3i$ (hopefully you're drawing this out!). If you pick a square, you've now picked a set of remainders modulo $3 + 2i$! To find out what any given $a + bi\in\Bbb Z[i]$ reduces to modulo $3 + 2i$, you want to find $a + bi$ on your grid, see where it is inside its square, then go back and find the corresponding point inside your chosen square. (Apologies for not being able to put up pictures of this process myself, if I get the chance I'll try to sketch out the process and show what's happening myself in the near future.)

Answer 2

$${35\over2+3i}={35(2-3i)\over13}={70-105i\over13}=5-9i+{5+12i\over13}$$ so $$35=(5-9i)(2+3i)+{(2+3i)(5+12i)\over13}=(5-9i)(2+3i)+{-26+39i\over13}=(5-9i)(2+3i)+(-2+3i)$$