Question

Let $\mathit{\omega }\mathbf{=}\frac{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{-}\mathbf{3}}\mathbf{)}}{\mathbf{2}}$and$\mathit{\mathbb{Z}}\mathbf{\left[}\mathbf{\omega }\mathbf{\right]}\mathbf{=}\mathbf{\{}\mathbf{r}\mathbf{+}\mathbf{s}\mathbf{\omega }\mathbf{/}\mathbf{r}\mathbf{,}\mathbf{s}\mathbf{\in }\mathbf{\mathbb{Z}}\mathbf{\}}$ . Prove that$\mathit{\mathbb{Z}}\mathbf{\left[}\mathbf{\omega }\mathbf{\right]}$ is Euclidean domain with$\mathit{\delta }\mathbf{(}\mathbf{r}\mathbf{+}\mathbf{s}\mathbf{\omega }\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{r}\mathbf{+}\mathbf{s}\mathbf{\omega }\mathbf{)}\mathbf{(}\mathbf{r}\mathbf{+}\mathbf{s}{\mathbf{\omega }}^{\mathbf{2}}\mathbf{)}\mathbf{=}{\mathit{r}}^{\mathbf{2}}\mathbf{-}\mathit{r}\mathit{s}\mathbf{+}{\mathit{s}}^{\mathbf{2}}$.

Short answer

It is Proved that$\mathbb{Z}\left[\omega \right]$ is Euclidean domain.

Step-by-step solution

Make use of Hint

Now, ${\omega }^3=1$and ${\omega }^2+\omega +1=0$can be computed directly, and means that $\omega =\frac{(-1+\sqrt{-3})}{2}$is a primitive cube root of unity.

First we must show that the image of$\delta$is contained in the non-negative integers.

For any$r+s\omega$if$\delta (r+s\omega )<0$then$r^2-rs+s^2<0$and $=r^2+s^2<rs$.Then $s\le r{s}^2$

So, $r^2+rs\le r^2+s^2\le rs$

Which means $r^2<0$, a contradiction.

Similarly if $s\ge r$we have

$s^2+rs\le r^2+s^2\le rs$

Which means $s^2<0$, also a contradiction. Therefore the mage of$\delta$ is contained in the non-negative integers.

Note that the above proof also shows that if$\delta (r+s\omega )=0$ then$r+s\omega =0$ .

Prove that δ(a)≤δ(ab)

Suppose now that $r+s\omega ,u+v\omega \in \mathbb{Z}\left[\omega \right]$are both non-zero elements, then

$$\begin{gathered} \delta \left((r+s\omega )(u+v\omega )\right)=\delta (ru+(rv+su)\omega +sv{\omega }^2) \\ =\delta (ru+(rv+su)\omega +sv(-1-\omega )) \\ =\delta ((ru-sv)+(rv+su-sv)\omega ) \\ ={(ru-sv)}^2-(ru-sv)(rv+su-sv)+{(rv+su-sv)}^2 \\ =r^2{u}^2-r^{2}uv+r^2{v}^2-rs{u}^2+rsuv-rs{v}^2+s^2{u}^2-s^{2}uv+s^2{v}^2 \\ =(r^2-rs+s^2)(u^2-uv+v^2) \\ =\delta (r+s\omega )\delta (u+v\omega ) \end{gathered}$$

Since $u+v\omega \ne 0$by the above argument we have that $\delta (u+v\omega )>0$, and so$\delta (r+s\omega )\le \delta \left((r+s\omega )(u+v\omega )\right)$.

Prove that a=bq+rand either  r=0R  or  δ(r)≤δ(b)

We need to show that $r+s\omega ,u+v\omega \in \mathbb{Z}\left[\omega \right]$with $u+v\omega \ne 0$there are $q_1+q_2\omega ,p_1+p_2\omega \in \mathbb{Z}\left[\omega \right]$such that

$r+s\omega =(u+v\omega )(q_1+q_2\omega )+p_1+p_2\omega$

Note that $\frac{r+s\omega }{u+v\omega }$is a complex number which can be written as

$\frac{r+s\omega }{u+v\omega }=\frac{(r+s\omega )(u+v{\omega }^2)}{\delta (u+v\omega )}=\frac{ru+sv-rv}{\delta (u+v\omega )}+\frac{su-rv}{\delta (u+v\omega )}\omega \in \mathbb{Z}\left[\omega \right]$

We may not apply the same stratergy as in example 7.

Let $q_1{q}_2\in \mathbb{Z}$be such that $|q_1-\frac{ru+sv-rv}{\delta (u+v\omega )}|\le \frac{1}{2}$ and $|q_2-\frac{su-rv}{\delta (u+v\omega )}|\le \frac{1}{2}$

Also, $p_1{p}_2\in \mathbb{Z}$as

$p_1+p_2\omega =(u+v\omega )[(\frac{ru+sv-rv}{\delta (u+v\omega )}-q_1)+(\frac{su-rv}{\delta (u+v\omega )}-q_2)\omega ]$

Such that

$p_1+p_2\omega =(r+s\omega )-(u+v\omega )(q_1+q_2\omega )\in \mathbb{Z}\left[\omega \right]$

We then have that $(r+s\omega )=(u+v\omega )(q_1+q_2\omega )+p_1+p_2\omega$holds that

role="math" localid="1654625799455" $$\begin{gathered} \delta (p_1+p_2\omega )=\delta \left((u+v\omega )\left[\left(\frac{ru+sv-rv}{\delta (u+v\omega )}-q_1\right)+\left(\frac{su-rv}{\delta (u+v\omega )}-q_2\right)\omega \right]\right) \\ =\delta \left((u+v\omega )\right)\delta \left(\left[\left(\frac{ru+sv-rv}{\delta (u+v\omega )}-q_1\right)+\left(\frac{su-rv}{\delta (u+v\omega )}-q_2\right)\omega \right]\right) \\ \le \delta \left((u+v\omega )\right)\frac{1}{4} \\ <\delta \left((u+v\omega )\right) \end{gathered}$$

It is Proved that$\mathbb{Z}\left[\omega \right]$ is Euclidean domain.