Let $J[i]$ denote the set of Gaussian integers. Show that $J[i]$ is a Euclidean ring.
The proof given is as follows:
In order to show this we must first introduce a function $d (x)$ defined for every nonzero element in $J[i]$ which satisfies:
- $d(x)$ is a nonnegative integer for every $x\neq 0\in J[ i].$
- $d(x)\leq d(xy)$ for every $y\neq 0$ in $J[i].$
- Given $u, v \in J[i]$ there exist $t, r\in J[i]$ such that $v = tu + r$ where $r = 0$ or $d(r) < d(u).$
Our candidate for this function is the following: if $x =a+ bi\in J[i],$ then $d(x) = a^2 + b^2.$ The $d(x)$ so defined certainly satisfies property $1;$ in fact, if $x\neq 0\in J[i]$ then $d(x)\neq 1.$ As is well known, for any two complex numbers (not necessarily in $J[i]$) $x, y,$ $d (xy) = d (x)d (y);$ thus if $x$ and $y$ are in addition in $J[i]$ and $y\neq 0,$ then since $d(y)\geq 1,$ $d(x) = d (x)1\leq d (x)d (y) = d(xy),$ showing that condition $2$ is satisfied. All our effort now will be to show that condition $3$ also holds for this function $d$ in $J[i].$ Thus, we merely must show that, given $x,y\in J[i]$ there exists $t, r\in J[i]$ such that $y = tx + r$ where $r = 0$ or $d(r) < d(x).$
We first establish this for a very special case, namely, where $y$ is arbitrary in $J[i]$ but where $x$ is an (ordinary) positive integer $n.$ Suppose that $y = a + bi;$ by the division algorithm for the ring of integers we can find integers $u, v$ such that $a = un + u_1$ and $b = vn + v_1$ where $u_1$ and $v_1$ are integers satisfying $|u_1|\leq \frac n2$ and $|v_1|\leq \frac n2.$ Let $t = u + vi$ and $r = u_1 + v_1i;$ then $y = a + bi = un + u_1 + (vn + v_1)i = (u + vi)n + u_1 + v_1i = tn + r.$ Since $d(r) = d(u_1 + v_1i) = u_1^2 + v_1^2 \leq n^2 /4 + n^2/4 < n^2 = d(n),$ we see that in this special case we have shown that $y = tn + r$ with $r = 0$ or $d ( r) <d(n).$
We now go to the general case; let $x\neq 0$ and $y$ be arbitrary elements in $J[i].$ Thus $x\bar x$ is a positive integer $n$ where $\bar x$ is the complex conjugate of $x.$ Applying the result of the paragraph above to the elements $y\bar x$ and $n$ we see that there are elements $t, r \in J[i]$ such that $y\bar x = tn + r$ with $r = 0$ or $d(r) < d(n).$ Putting into this relation $n = x\bar x$ we obtain $d(y\bar x- tx\bar x) < d(n) = d(x\bar x);$ applying to this the fact that $d(y\bar x- tx\bar x) = d(y- tx)d(x)$ and $d(x\bar x) = d(x)d(\bar x)$ we obtain that $d(y - tx)d(x) <d(x)d(x).$ Since $x\neq 0,$ $d(x)$ is a positive integer, so this inequality simplifies to $d(y- tx) <d(x).$ We represent $y = tx + r_0 ,$ where $r_0 = y - tx;$ thus $t$ and $r_0$ are in $J[i]$ and as we saw above, $r_0 = 0$ or $d(r_0 ) = d(y - tx) < d(x).$ This the theorem.
However, I don't understand how do they assert,
Suppose that $y = a + bi;$ by the division algorithm for the ring of integers we can find integers $u, v$ such that $a = un + u_1$ and $b = vn + v_1$ where $u_1$ and $v_1$ are integers satisfying $|u_1|\leq \frac n2$ and $|v_1|\leq \frac n2.$
I mean wouldn't the inequalities be $0\leq u_1\lt n$ and $0\leq v_1\lt n$ instead of "$|u_1|\leq \frac n2$ and $|v_1|\leq \frac n2.$"
Any clarification regarding this issue will be greatly appreciated.