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Answers without the blur. Sign up and see all textbooks for free! Q10.1-27E

Expert-verified Found in: Page 331 ### Abstract Algebra: An Introduction

Book edition 3rd
Author(s) Thomas W Hungerford, David Leep
Pages 608 pages
ISBN 9781111569624 # Let ${\mathbit{\omega }}{\mathbf{=}}\frac{\mathbf{\left(}\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{-}\mathbf{3}}\mathbf{\right)}}{\mathbf{2}}$and ${\mathbit{ℤ}}\mathbf{\left[}\mathbf{\omega }\mathbf{\right]}{\mathbf{=}}\mathbf{\left\{}\mathbf{r}\mathbf{+}\mathbf{s}\mathbf{\omega }\mathbf{/}\mathbf{r}\mathbf{,}\mathbf{s}\mathbf{\in }\mathbf{ℤ}\mathbf{\right\}}$ . Prove that ${\mathbit{ℤ}}\mathbf{\left[}\mathbf{\omega }\mathbf{\right]}$ is Euclidean domain with ${\mathbit{\delta }}\mathbf{\left(}\mathbf{r}\mathbf{+}\mathbf{s}\mathbf{\omega }\mathbf{\right)}{\mathbf{=}}\mathbf{\left(}\mathbf{r}\mathbf{+}\mathbf{s}\mathbf{\omega }\mathbf{\right)}\mathbf{\left(}\mathbf{r}\mathbf{+}\mathbf{s}{\mathbf{\omega }}^{\mathbf{2}}\mathbf{\right)}{\mathbf{=}}{{\mathbit{r}}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbit{r}}{\mathbit{s}}{\mathbf{+}}{{\mathbit{s}}}^{{\mathbf{2}}}$ .

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

See the step by step solution

## Step 1: Make use of Hint

Now, ${\omega }^{3}=1$ and ${\omega }^{2}+\omega +1=0$ can be computed directly, and means that $\omega =\frac{\left(-1+\sqrt{-3}\right)}{2}$ 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 $\delta \left(r+s\omega \right)<0$ then ${r}^{2}-rs+{s}^{2}<0$ and $={r}^{2}+{s}^{2}.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 $\delta \left(r+s\omega \right)=0$ then $r+s\omega =0$ .

## Step 2: Prove that δ(a)≤δ(ab)

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

$\delta \left(\left(r+s\omega \right)\left(u+v\omega \right)\right)=\delta \left(ru+\left(rv+su\right)\omega +sv{\omega }^{2}\right)\phantom{\rule{0ex}{0ex}}=\delta \left(ru+\left(rv+su\right)\omega +sv\left(-1-\omega \right)\right)\phantom{\rule{0ex}{0ex}}=\delta \left(\left(ru-sv\right)+\left(rv+su-sv\right)\omega \right)\phantom{\rule{0ex}{0ex}}={\left(ru-sv\right)}^{2}-\left(ru-sv\right)\left(rv+su-sv\right)+{\left(rv+su-sv\right)}^{2}\phantom{\rule{0ex}{0ex}}={r}^{2}{u}^{2}-{r}^{2}uv+{r}^{2}{v}^{2}-rs{u}^{2}+rsuv-rs{v}^{2}+{s}^{2}{u}^{2}-{s}^{2}uv+{s}^{2}{v}^{2}\phantom{\rule{0ex}{0ex}}=\left({r}^{2}-rs+{s}^{2}\right)\left({u}^{2}-uv+{v}^{2}\right)\phantom{\rule{0ex}{0ex}}=\delta \left(r+s\omega \right)\delta \left(u+v\omega \right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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

## Step 3: Prove that a=bq+rand either  r=0R  or  δ(r)≤δ(b)

We need to show that $r+s\omega ,u+v\omega \in ℤ\left[\omega \right]$ with $u+v\omega \ne 0$ there are ${q}_{1}+{q}_{2}\omega ,{p}_{1}+{p}_{2}\omega \in ℤ\left[\omega \right]$ such that

$r+s\omega =\left(u+v\omega \right)\left({q}_{1}+{q}_{2}\omega \right)+{p}_{1}+{p}_{2}\omega$

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

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

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

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

Also, ${p}_{1}{p}_{2}\in ℤ$ as

${p}_{1}+{p}_{2}\omega =\left(u+v\omega \right)\left[\left(\frac{ru+sv-rv}{\delta \left(u+v\omega \right)}-{q}_{1}\right)+\left(\frac{su-rv}{\delta \left(u+v\omega \right)}-{q}_{2}\right)\omega \right]$

Such that

${p}_{1}+{p}_{2}\omega =\left(r+s\omega \right)-\left(u+v\omega \right)\left({q}_{1}+{q}_{2}\omega \right)\in ℤ\left[\omega \right]$

We then have that $\left(r+s\omega \right)=\left(u+v\omega \right)\left({q}_{1}+{q}_{2}\omega \right)+{p}_{1}+{p}_{2}\omega$ holds that

role="math" localid="1654625799455" $\delta \left({p}_{1}+{p}_{2}\omega \right)=\delta \left(\left(u+v\omega \right)\left[\left(\frac{ru+sv-rv}{\delta \left(u+v\omega \right)}-{q}_{1}\right)+\left(\frac{su-rv}{\delta \left(u+v\omega \right)}-{q}_{2}\right)\omega \right]\right)\phantom{\rule{0ex}{0ex}}=\delta \left(\left(u+v\omega \right)\right)\delta \left(\left[\left(\frac{ru+sv-rv}{\delta \left(u+v\omega \right)}-{q}_{1}\right)+\left(\frac{su-rv}{\delta \left(u+v\omega \right)}-{q}_{2}\right)\omega \right]\right)\phantom{\rule{0ex}{0ex}}\le \delta \left(\left(u+v\omega \right)\right)\frac{1}{4}\phantom{\rule{0ex}{0ex}}<\delta \left(\left(u+v\omega \right)\right)$

It is Proved that $ℤ\left[\omega \right]$ is Euclidean domain. ### Want to see more solutions like these? 