请输入您要查询的字词:

 

单词 PropertiesOfAGcdDomain
释义

properties of a gcd domain


Let D be a gcd domain. For any aD, denote [a] the set of all elements in D that are associates of a, GCD(a,b) the set of all gcd’s of elements a and b in D, and any SD, mS:={mssS}. Then

  1. 1.

    GCD(a,b)=[a] iff ab.

  2. 2.

    mGCD(a,b)=GCD(ma,mb).

  3. 3.

    If GCD(ab,c)=[1], then GCD(a,c)=[1]

  4. 4.

    If GCD(a,b)=[1] and GCD(a,c)=[1], then GCD(a,bc)=[1].

  5. 5.

    If GCD(a,b)=[1] and abc, then ac.

Proof.

To aid in the proof of these properties, let us denote, for aD and SD, a|S to mean that every element of S is divisible by a, and S|a to mean that every element in S divides a.We take the following four steps:

  1. 1.

    One direction is obvious from the definition. So now suppose ab. Then aGCD(a,b). But bydefinition, GCD(a,b)a, so [a]=GCD(a,b).

  2. 2.

    Pick dGCD(a,b) and xGCD(ma,mb). We want to show that md and x are associates. By assumptionPlanetmathPlanetmath, da and db, so mdma and mdmb, which implies that mdx. Write x=mn for some nD. Then mnma and mnmb imply that na and nb, and therefore nd since d is a gcd of a and b. As a result, mnmd, or xmd, showing that x and md are associates. As a result, the map f:mGCD(a,b)GCD(ma,mb) given by f(d)=md is a bijection.

  3. 3.

    If da and dc, then dab and dc. So dGCD(ab,c)=[1], hence d is a unit andthe result follows.

  4. 4.

    Suppose da and dbc. Then dab and dbc and hence dGCD(ab,bc)=bGCD(a,c)=[b]. But da also, so dGCD(a,b)=[1] and d is a unit.

  5. 5.

    GCD(a,b)=[1] implies [c]=GCD(ac,bc). Now, aac and by assumption, abc. Therefore,aGCD(ac,bc)=[c].

The second property above can be generalized to arbitrary integral domain: let D be an integral domain, a,bD, with GCD(a,b)GCD(ma,mb), then dGCD(a,b) iff mdGCD(ma,mb).

随便看

 

数学辞典收录了18232条数学词条,基本涵盖了常用数学知识及数学英语单词词组的翻译及用法,是数学学习的有利工具。

 

Copyright © 2000-2023 Newdu.com.com All Rights Reserved
更新时间:2025/5/4 16:00:04