Gaussian Integers

Let $i=\sqrt{-1}$ and $\Z[i]$ is a set of ${a+bi|a,b\in \Z}$. For addition: $a + bi + c + di = (a+c) + (b+d)i \in \Z[i]$. For multiplication: $(a+bi)(b+ci)=(ac-bd)+(ad+bc)i \in \Z[i]$. So $\Z[i]$ is a ring on integral domain.

Proposition 1

$\Z[i]$ is a Euclidean Domain

Proof:

Revise the definition of Euclidean Domain: if $\exist \R \to \Z^+$

1. If $0\neq a,b \in \R$, then $\lambda[a] < \lambda[ab]$.
2. if $a,b\in \R$, $b\neq 0$, there exist $q,r\in\R$ such that $a=qb+r$ wither $r=0$ or $\lambda[r]<\lambda[b]$.

Here the $\lambda$ is like a "norm" function.

So, first, define a "norm" function first: $\alpha = a+bi \in \Z$, so $\lambda(\alpha) = |\alpha| = a^2+b^2$

For Gaussian Integer: $\lambda(\alpha\beta) = \lambda(\alpha)\lambda(\beta) > \lambda(\alpha)$.

Consider $\alpha=a+bi$ and $\beta=c+di$. $$\frac{\alpha}{\beta} = \frac{(a+bi)(c-di)}{c^2+d^2} = \frac{ac+bd}{c^2+d^2} + \frac{bc-ad}{c^2+d^2}i$$ Assume $q_1,q_2 \in \Z$ and $-\frac{1}{2} \leq r_1,r_2 \leq \frac{1}{2}$ such that: $$q_1 + r_1 = \frac{ac+bd}{c^2+d^2}$$

$$q_2 + r_2 = \frac{bc-ad}{c^2+d^2}$$

So we have : $$\alpha\beta^{-1} = (q_1 + q_2i) + (r_1 + r_2i)$$

$$\alpha = \beta(q_1 + q_2i) + \beta(r_1 + r_2i)$$

Since $\alpha,\beta,(q_1 + q_2i) \in \Z[i]$, so $\beta(r_1 + r_2i) \in \Z[i]$

Now we only need to prove: $$\lambda(\beta(r_1 + r_2i)) = \lambda(\beta)(r_1^2+r_2^2) \leq \frac{1}{2} \lambda(\beta) \leq \lambda(\beta)$$ So $\Z[i]$ is a Euclidean Domain.

Another "Integer"

Let $\omega = \frac{-1+\sqrt{-3}}{2}$ and $\Z[\omega]$ is a set of ${a+b\omega|a,b\in \Z}$. For addition: $a + b\omega + c + d\omega = (a+c) + (b+d)\omega \in \Z[\omega]$. For multiplication $(a+b\omega)(c+d\omega)=(ac-bd)+(ad+bc-bd)\omega \in \Z[\omega]$. So $\Z[\omega]$ is a ring on integral domain.

Proposition 2

$\Z[\omega]$ is a Euclidean Domain

Proof:

First, obviously, $\omega$ is one of the root of $x^2+x+1=0$. So we have $\omega^2=-1-\omega$

Define our "norm" function: $\alpha = a+bi \in \Z$, so $\lambda(\alpha) = \alpha\bar\alpha = a^2+b^2 - ab$​

We have $\lambda(\alpha\beta) = \lambda(\alpha)\lambda(\beta) > \lambda(\alpha) = a^2+b^2-ab$.

Still consider $\alpha=a+bi$ and $\beta=c+di$. $$\frac{\alpha}{\beta} = \frac{\alpha\bar\beta}{\lambda({\beta})} = \frac{(a+b\omega)(c-d\omega)}{c^2+d^2 - cd} = \frac{ac+bd}{c^2+d^2 -cd} + \frac{bc-ad}{c^2+d^2-cd}\omega$$ Same operation, we take $q_1,q_2 \in \Z$ and $-\frac{1}{2} \leq r_1,r_2 \leq \frac{1}{2}$ such that: $$q_1 + r_1 = \frac{ac+bd}{c^2+d^2-cd}$$

$$q_2 + r_2 = \frac{bc-ad}{c^2+d^2-cd}$$

We have: $$\alpha\beta^{-1} = (q_1 + q_2\omega) + (r_1 + r_2\omega) = r+s\omega$$ Think another way, since $\alpha,\bar\beta\in\Z[\omega]$ and $\lambda[\beta]\in\Z$. We can be sure that $\alpha\beta^{-1} = r+s\omega$ where $p,q$ are rational numbers.

So things are clear: $$\alpha = \beta(q_1 + q_2\omega) + \beta(r_1 + r_2\omega)$$

$$\lambda(\beta(r_1 + r_2\omega)) = \lambda(\beta)(r_1^2+r_2^2-r_1r_2) \leq \frac{3}{4}\lambda{\beta} \leq \lambda(\beta)$$