Gaussian integers & division theorem Q. let $\mathbb{Z}[i] = \{a + bi | a,b \in \mathbb{Z} \}$
Show that if $s,t \in \mathbb{Z}[i],$ $t\not = 0$ then there exists $r,q \in \mathbb{Z}[i]$ s.t. $ s = tq + r$ where $N(r) < N(t)$
my attempt,
if we divide s by t: $\dfrac{s}{t} = m + ni$ where $m,n \in \mathbb{Q}$
now we can define $q = q_1 + q_2i $ where $q_1,q_2$ are the largest integers s.t. $q_1,q_2 \leq m,n$ then we have $\dfrac{s}{t} = q + \alpha$ where $\alpha = \alpha_1 + \alpha_2 i \in \mathbb{C}$ where $|\alpha_1|,|\alpha_2| \leq 1/2$ so we have $s = tq + \alpha t$ now I'm having troubles defining r s.t. $N(r) < N(t)$ could someone show me how to do this?
edit: using the above definitions of $\alpha$ if we let $r = \alpha t$ and $t = x + iy$ then we get $ r = (\alpha_1 x -\alpha_2 y) + i(\alpha _2 x + \alpha _1 y)$ but the real and complex parts are not integers, if we label them $ k = (\alpha_1 x -\alpha_2 y), l = (\alpha _2 x + \alpha _1 y)$, $ r = k + il$ is there a way to choose k and l to be integers?
edit: $\alpha t = s - tq $ could I not just let $r = \alpha t$ as $s - tq $ is in $\mathbb{Z}[i]$?
any help - thanks
 A: $\newcommand{\ZZ}{\mathbb{Z}}$
It is pretty similar, but you cannot use $\le$ in $\ZZ[i]$. There's no nice ordering on the Gaussian integers. Plus, at no point do you use the well-ordering of the integers, which is critical to proving it in $\ZZ$, and necessary if you want to mimic the proof in $\ZZ[i]$.
But since you've already proven the division algorithm, why not put it to use? You may want to restate it in this form; it's a bit more useful this way.
$$ \forall a,b \in \ZZ \ \exists q, r \in \ZZ \ a = bq + r, \ |r| \le \frac{|b|}{2}$$
So if you have $\alpha = a + bi$, $\beta = c + di$, can you find $\rho$ and $\sigma$ such that $\alpha = \beta \rho + \sigma$?
Use what you know about $\mathbb{C}$ (well, $\mathbb{Q}[i]$, strictly speaking), but don't actually divide, because then you leave the integers. Just use the above division algorithm to find its components.

As motivation, we look at $\mathbb{Q}[i]$. We know $\frac{\alpha}{\beta} = \frac{\alpha\overline{\beta}}{N(\beta)}$. Since the denominator is an integer, we consider the latter expression.
We know $\alpha\overline{\beta} = m + ni$ and that $N(\beta) = p \in \mathbb{N}_0$. We use the integer division algorithm twice:
$$ m = pq_1 + r_1, \ 2|r_1| \le p $$
$$ n = pq_2 + r_2, \ 2|r_2| \le p $$
Letting $\rho = q_1 + q_2i$ and $\eta = r_1 + r_2i$:
$$\alpha\overline{\beta} = m + ni = (pq_1 + r_1) + (pq_2 + r_2)i = p(q_1 + q_2i) + (r_1 + r_2i) = N(\beta) \rho + \eta$$
Since $\eta = \alpha\overline{\beta} - N(\beta)\rho$, and $\overline{\beta}$ divides $N(\beta)$, one can see that $\overline{\beta}$ divides $\eta$. We pick $\sigma \in \ZZ[i]$ such that $\sigma \overline{\beta} = \eta$. So now we can cancel the $\overline{\beta}$ on each side to get $\alpha = \beta \rho + \sigma$, and all we need to do is verify the condition $N(\sigma) < N(\beta)$.
We know that $2|r_1| \le p$ and $2|r_2| \le p$. We look at $N(2\eta)$, for some reason:
$$N(2\eta) = N(2r_1 + 2r_2i) = (2r_1)^2 + (2r_2)^2 \le p^2 + p^2 = 2p^2$$
Messing around with the left and right sides:
$$N(2\eta) = N(2 \sigma \overline{\beta}) = 4 N(\sigma) N(\overline{\beta})$$
$$ 2p^2 = 2N(\beta)^2 = 2N(\beta)N(\overline{\beta})$$
Since $4 N(\sigma) N(\overline{\beta}) \le 2N(\beta)N(\overline{\beta})$, we can conclude $2N(\sigma) \le N(\beta)$, and since $\beta \ne 0$, we have $N(\sigma) < N(\beta)$, as desired.

Or you can do this in $\mathbb{Q}[i]$, but formalizing it is... interesting. It uses the same idea though; pick your $\rho$ such that its real part is $\le \frac{1}{2}$ away from the real part of $\frac{\alpha}{\beta}$, and same with the imaginary part. Thus, the remainder is $r_1 + r_2i$, where both components are less than or equal to one half. Taking the norm:
$$(r_1^2 + r_2^2) \le \left( \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 \right) = \left( \frac{1}{4} + \frac{1}{4} \right) = \frac{1}{2}$$
which is less than $1$.
