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Q10.1-27E

Expert-verified

Abstract Algebra: An Introduction

Found in: Page 331

Book edition 3rd

Author(s) Thomas W Hungerford, David Leep

Pages 608 pages

ISBN 9781111569624

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Short Answer

Let $ω = \frac{(- 1 + \sqrt{- 3})}{2}$ and $ℤ [ω] = {r + s ω / r, s \in ℤ}$ . Prove that $ℤ [ω]$ is Euclidean domain with $δ (r + s ω) = (r + s ω) (r + s ω^{2}) = r^{2} - r s + s^{2}$ .

It is Proved that $ℤ [ω]$ is Euclidean domain.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Make use of Hint

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

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

For any $r + s ω$ if $δ (r + s ω) < 0$ then $r^{2} - r s + s^{2} < 0$ and $= r^{2} + s^{2} < r s$ .Then $s \leq r s^{2}$

So, $r^{2} + r s \leq r^{2} + s^{2} \leq r s$

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

Similarly if $s \geq r$ we have

$s^{2} + r s \leq r^{2} + s^{2} \leq r s$

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

Note that the above proof also shows that if $δ (r + s ω) = 0$ then $r + s ω = 0$ .

Step 2: Prove that δ(a)≤δ(ab)

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

$δ ((r + s ω) (u + v ω)) = δ (r u + (r v + s u) ω + s v ω^{2}) = δ (r u + (r v + s u) ω + s v (- 1 - ω)) = δ ((r u - s v) + (r v + s u - s v) ω) = {(r u - s v)}^{2} - (r u - s v) (r v + s u - s v) + {(r v + s u - s v)}^{2} = r^{2} u^{2} - r^{2} u v + r^{2} v^{2} - r s u^{2} + r s u v - r s v^{2} + s^{2} u^{2} - s^{2} u v + s^{2} v^{2} = (r^{2} - r s + s^{2}) (u^{2} - u v + v^{2}) = δ (r + s ω) δ (u + v ω)$

Since $u + v ω \neq 0$ by the above argument we have that $δ (u + v ω) > 0$ , and so $δ (r + s ω) \leq δ ((r + s ω) (u + v ω))$ .

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

We need to show that $r + s ω, u + v ω \in ℤ [ω]$ with $u + v ω \neq 0$ there are $q_{1} + q_{2} ω, p_{1} + p_{2} ω \in ℤ [ω]$ such that

$r + s ω = (u + v ω) (q_{1} + q_{2} ω) + p_{1} + p_{2} ω$

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

$\frac{r + s ω}{u + v ω} = \frac{(r + s ω) (u + v ω^{2})}{δ (u + v ω)} = \frac{r u + s v - r v}{δ (u + v ω)} + \frac{s u - r v}{δ (u + v ω)} ω \in ℤ [ω]$

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

Let $q_{1} q_{2} \in ℤ$ be such that $| q_{1} - \frac{r u + s v - r v}{δ (u + v ω)} | \leq \frac{1}{2}$ and $| q_{2} - \frac{s u - r v}{δ (u + v ω)} | \leq \frac{1}{2}$

Also, $p_{1} p_{2} \in ℤ$ as

$p_{1} + p_{2} ω = (u + v ω) [(\frac{r u + s v - r v}{δ (u + v ω)} - q_{1}) + (\frac{s u - r v}{δ (u + v ω)} - q_{2}) ω]$

Such that

$p_{1} + p_{2} ω = (r + s ω) - (u + v ω) (q_{1} + q_{2} ω) \in ℤ [ω]$

We then have that $(r + s ω) = (u + v ω) (q_{1} + q_{2} ω) + p_{1} + p_{2} ω$ holds that

role="math" localid="1654625799455" $δ (p_{1} + p_{2} ω) = δ ((u + v ω) [(\frac{r u + s v - r v}{δ (u + v ω)} - q_{1}) + (\frac{s u - r v}{δ (u + v ω)} - q_{2}) ω]) = δ ((u + v ω)) δ ([(\frac{r u + s v - r v}{δ (u + v ω)} - q_{1}) + (\frac{s u - r v}{δ (u + v ω)} - q_{2}) ω]) \leq δ ((u + v ω)) \frac{1}{4} < δ ((u + v ω))$

It is Proved that $ℤ [ω]$ is Euclidean domain.

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Let $R$ be an integral domain in which any two elements (not both $0_{R}$ ) have a gcd. With the notation of Exercise 25, prove that if $(b, c) ~ 1_{R}$ and $(b, d) ~ 1_{R}$ then $(b, c d) ~ 1_{R}$ .

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