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Abstract Algebra: An Introduction
Found in: Page 331
Abstract Algebra: An Introduction

Abstract Algebra: An Introduction

Book edition 3rd
Author(s) Thomas W Hungerford, David Leep
Pages 608 pages
ISBN 9781111569624

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Short Answer

Let ω=(-1+-3)2and [ω]={r+sω/r,s} . Prove that [ω] is Euclidean domain with δ(r+sω)=(r+sω)(r+sω2)=r2rs+s2 .

It is Proved that [ω] is Euclidean domain.

See the step by step solution

Step by Step Solution

Step 1: Make use of Hint

Now, ω3=1 and ω2+ω+1=0 can be computed directly, and means that ω=(-1+-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+sω if δ(r+sω)<0 then r2rs+s2<0 and =r2+s2<rs.Then srs2

So, r2+rsr2+s2rs

Which means r2<0 , a contradiction.

Similarly if sr we have


Which means s2<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+sω)=0 then r+sω=0 .

Step 2: Prove that δ(a)≤δ(ab)

Suppose now that r+sω,u+vω[ω] are both non-zero elements, then

δ((r+sω)(u+vω))=δ(ru+(rv+su)ω+svω2) =δ(ru+(rv+su)ω+sv(1ω)) =δ((rusv)+(rv+susv)ω) =(rusv)2(rusv)(rv+susv)+(rv+susv)2 =r2u2r2uv+r2v2rsu2+rsuvrsv2+s2u2s2uv+s2v2 =(r2rs+s2)(u2uv+v2) =δ(r+sω)δ(u+vω)

Since u+vω0 by the above argument we have that δ(u+vω)>0, and so δ(r+sω)δ((r+sω)(u+vω)).

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

We need to show that r+sω,u+vω[ω] with u+vω0 there are q1+q2ω,p1+p2ω[ω] such that


Note that r+sωu+vω is a complex number which can be written as


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

Let q1q2 be such that |q1ru+svrvδ(u+vω)|12 and |q2survδ(u+vω)|12

Also, p1p2 as


Such that


We then have that (r+sω)=(u+vω)(q1+q2ω)+p1+p2ω holds that

role="math" localid="1654625799455" δ(p1+p2ω)=δ(u+vω)ru+svrvδ(u+vω)q1+survδ(u+vω)q2ω =δ((u+vω))δru+svrvδ(u+vω)q1+survδ(u+vω)q2ω δ((u+vω))14 <δ((u+vω))

It is Proved that [ω] is Euclidean domain.

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