请输入您要查询的字词:

 

单词 ExampleOfARightNoetherianRingThatIsNotLeftNoetherian
释义

example of a right noetherian ring that is not left noetherian


This example, due to Lance Small, is briefly described in Noncommutative Rings, by I. N. Herstein, published by the Mathematical Association of America, 1968.

Let R be the ring of all 2×2 matrices(ab0c)such that a is an integer and b,c are rational. The claim is that R is right noetherianPlanetmathPlanetmath but not left noetherian.

It is relatively straightforward to show that R is not left noetherian. For each natural number n, let

In={(0m2n00)m}.

Verify that each In is a left idealMathworldPlanetmathPlanetmath in R and that I0I1I2.

It is a bit harder to show that R is right noetherian. The approach given here usesthe fact that a ring is right noetherian if all of its right ideals are finitely generatedMathworldPlanetmathPlanetmathPlanetmath.

Let I be a right ideal in R. We show that I is finitely generated bychecking all possible cases. In the first case, we assume that every matrix in I hasa zero in its upper left entry. In the second case, we assume thatthere is some matrix in I that has a nonzero upper left entry. The second case splitsinto two subcases: either every matrix in I has a zero in its lower right entry or some matrix in I has a nonzero lower right entry.

CASE 1: Suppose that for all matrices in I, the upper left entry is zero. Then every element of I has the form

(0y0z) for some y,z.

Note that for any c and any(0y0z)I, we have(0cy0cz)I since

(0y0z)(000c)=(0cy0cz)

and I is a right ideal in R. So I looks like a rational vector spaceMathworldPlanetmath.

Indeed, note thatV={(y,z)2(0y0z)I}is a subspacePlanetmathPlanetmath of the two dimensional vector space 2. So in V there exist two(not necessarily linearly independentMathworldPlanetmath) vectors (y1,z1) and (y2,z2)which span V.

Now, an arbitrary element(0y0z)in I corresponds to the vector (y,z) in V and (y,z)=(c1y1+c2y2,c1z1+c2z2) for some c1,c2. Thus

(0y0z)=(0c1y1+c2y20c1z1+c2z2)=(0y10z1)(000c1)+(0y20z2)(000c2)

and it follows that I is finitely generatedby the set{(0y10z1),(0y20z2)}as a right ideal in R.

CASE 2: Suppose that some matrix in I has a nonzero upper left entry. Then there is a least positiveinteger n occurring as the upper left entry of a matrix in I. It follows that every elementof I can be put into the form

(kny0z) for some k;y,z.

By definition of n, there is a matrix of the form(nb0c)in I. Since I is a right ideal in R and since(nb0c)(1000)=(n000),it follows that(n000)is in I. Now break off into two subcases.

case 2.1: Suppose that every matrix in I has a zero in its lower right entry. Thenan arbitrary element of I has the form

(kny00) for some k,y.

Note that(kny00)=(n000)(kyn00). Hence,(n000) generates I as a right ideal in R.

case 2.2: Suppose that some matrix in I has a nonzero lower right entry. That is, in Iwe have a matrix

(mny10z1) for some m;y1,z1;z10.

Since(n000)I, it follows that(ny10z1)I.Let(kny0z)be an arbitrary element of I. Since(kny0z)=(ny10z1)(k1n(y-y1zz1)0zz1), it follows that(ny10z1)generates I as a right ideal in R.

In all cases, I is a finitely generated.

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 23:40:35