PlanetMath: quadratic imaginary norm-Euclidean number fields

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quadratic imaginary norm-Euclidean number fields

(Theorem)

Theorem 1. The imaginary quadratic fields $\mathbb{Q}(\sqrt{d})$ with $d = -1,\,-2,\,-3,\,-7,\,-11$ are norm-Euclidean number fields.

Proof. $1^\circ.$ $d \not\equiv 1\;($ mod $\;4)$ , i.e. or . Any element $\gamma$ of the field $\mathbb{Q}(\sqrt{d})$ has the canonic form $\gamma = c_0+c_1\sqrt{d}$ where $c_0,\,c_1\in\mathbb{Q}$ . We may write $\gamma = (p+r)+(q+s)\sqrt{d}$ , where is the rational integer nearest to and the one nearest to . So $\vert r\vert \le \frac{1}{2}$ , $\vert s\vert \le \frac{1}{2}$ . Thus we may write

$\displaystyle \gamma = \underbrace{(p+q\sqrt{d})}_\varkappa+ \underbrace{(r+s\sqrt{d})}_\delta,$

where $\varkappa$ is an integer of the field. We then can estimate

$\displaystyle 0 \le$ N $\displaystyle (\delta) = (r+s\sqrt{d})(r-s\sqrt{d}) = r^2-s^2d = r^2+s^2\vert d... ...\le \left(\frac{1}{2}\right)^2\!+2\left(\frac{1}{2}\right)^2 = \frac{3}{4} < 1,$

and therefore $\vert$ N $(\delta)\vert < 1$ . According to the theorem 1 in the parent entry, $\mathbb{Q}(\sqrt{-1})$ and $\mathbb{Q}(\sqrt{-2})$ are norm-Euclidean number fields.

$2^\circ.$ $d \equiv 1\;($ mod $\;4)$ , i.e. $d\in\{-3,\,-7,\,-11\}$ . The algebraic integers of $\mathbb{Q}(\sqrt{d})$ have now the canonic form $\frac{a+b\sqrt{d}}{2}$ with $2\,\vert\,a\!-\!b$ . Let $\gamma = c_0+c_1\sqrt{d}$ where $c_0,\,c_1\in\mathbb{Q}$ be an arbitrary element of the field. Choose the rational integer such that $\frac{q}{2}$ is as close to as possible, i.e. $c_1 = \frac{q}{2}+s$ with $\vert s\vert \le \frac{1}{4}$ , and the rational integer such that $\frac{q}{2}+t$ is as close to as possible; then $c_0 = \frac{q+2t}{2}+r = \frac{p}{2}+r$ with $\vert r\vert \le \frac{1}{2}$ . Then we can write

$\displaystyle \gamma = \frac{p}{2}+r+(\frac{q}{2}+s)\sqrt{d} = \underbrace{\frac{p+q\sqrt{d}}{2}}_\varkappa+ \underbrace{(r+s\sqrt{d})}_\delta.$

The number $\varkappa$ is an integer of the field, since $p-q = 2t \equiv 0\,\,($ mod $\,2)$ . We get the estimation

$\displaystyle 0 \le$ N $\displaystyle (\delta) = r^2+s^2\vert d\vert \le \left(\frac{1}{2}\right)^2\!+11\left(\frac{1}{4}\right)^2 = \frac{15}{16} < 1,$

so $\vert$ N $(\delta)\vert < 1$ . Thus the fields in question are norm-Euclidean number fields.

Theorem 2. The only quadratic imaginary norm-Euclidean number fields $\mathbb{Q}(\sqrt{d})$ are the ones in which $d = -1,\,-2,\,-3,\,-7,\,-11$ .

Proof. Let be any other negative (square-free) rational integer than the above mentioned ones.

$1^\circ.$ $d \not\equiv 1\;($ mod $\;4)$ . The integers of $\mathbb{Q}(\sqrt{d})$ are $a+b\sqrt{d}$ where $a,\,b\in\mathbb{Z}$ . We show that there is a number $\gamma$ that can not be expressed in the form $\gamma = \varkappa+\delta$ with $\varkappa$ an integer of the field and $\vert$ N $(\delta)\vert < 1$ . Assume that $\gamma := \frac{1}{2}\sqrt{d} = \varkappa+\delta$ where $\varkappa = a+b\sqrt{d}$ is an integer of the field ( $a,\,b\in\mathbb{Z}$ ). Then $\delta = \gamma-\varkappa = -a+(\frac{1}{2}-b)\sqrt{d}$ and N $(\delta) = \vert a\vert^2+\vert d\vert\cdot\vert\frac{1}{2}-b\vert^2$ . Because cannot be 0, we have $\vert\frac{1}{2}-b\vert \ge \frac{1}{2}$ and thus

$\displaystyle \vert$ N $\displaystyle (\delta)\vert \ge 0+\vert d\vert\left(\frac{1}{2}\right)^2 = \frac{\vert d\vert}{4}\ge \frac{5}{4} > 1.$

Therefore $\mathbb{Q}(\sqrt{d})$ can not be a norm-Euclidean number field ( $d = -5,\,-6,\,-10$ and so on).

$2^\circ.$ $d \equiv 1\;($ mod $\;4)$ . Now $\vert d\vert \ge 15$ . The integers of $\mathbb{Q}(\sqrt{d})$ have the form $\varkappa = \frac{a+b\sqrt{d}}{2}$ with $2\,\vert\,a-b$ . Suppose that $\gamma = \frac{1}{4}+\frac{1}{4}\sqrt{d} = \varkappa+\delta$ . Then $\delta = \gamma-\varkappa = (\frac{1}{4}-\frac{a}{2})+ (\frac{1}{4}-\frac{b}{2})\sqrt{d}$ and

$\displaystyle \vert$ N $\displaystyle (\delta)\vert \ge \left\vert\frac{1}{4}-\frac{a}{2}\right\vert^2+... ...right\vert^2 \ge \left(\frac{1}{4}\right)^2\!+15\left(\frac{1}{4}\right)^2 = 1.$

So also these fields $\mathbb{Q}(\sqrt{d})$ are not norm-Euclidean number fields.

"quadratic imaginary norm-Euclidean number fields" is owned by pahio.

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