example of a right noetherian ring that is not left noetherian

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

(

a

b

0

c

)

such that

a

is an integer and

b

,

c

are

rational

. The claim is that

R

is

right noetherian

but not left noetherian.

It is relatively straightforward to show that

R

is not left noetherian. For each natural number

n

, let

I

n

=

{

(

0

m

2

n

0

0

)

∣

m

∈

ℤ

}

.

Verify that each

I

n

is a

left ideal

in

R

and that

I

0

⊊

I

1

⊊

I

2

⊊

⋯

.

It is a bit harder to show that

R

is right noetherian. The approach given here uses
the fact that a ring is right noetherian if all of its

right ideals

are

finitely generated

.

Let

I

be a right ideal in

R

. We show that

I

is finitely generated by
checking all possible cases. In the first case, we assume that every matrix in

I

has
a zero in its upper left entry. In the second case, we assume that
there is some matrix in

I

that has a nonzero upper left entry. The second case splits
into 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

(

0

y

0

z

)

⁢

for some

⁢

y

,

z

∈

ℚ

.

Note that for any

c

∈

ℚ

and any

(

0

y

0

z

)

∈

I

, we have

(

0

c

⁢

y

0

c

⁢

z

)

∈

I

since

(

0

y

0

z

)

⁢

(

0

0

0

c

)

=

(

0

c

⁢

y

0

c

⁢

z

)

and

I

is a right ideal in

R

. So

I

looks like a rational

vector space

.

Indeed, note that

V

=

{

(

y

,

z

)

∈

ℚ

2

∣

(

0

y

0

z

)

∈

I

}

is a

subspace

of the two dimensional vector space

ℚ

2

. So in

V

there exist two
(not necessarily

linearly independent

) vectors

(

y

1

,

z

1

)

and

(

y

2

,

z

2

)

which span

V

.

Now, an arbitrary element

(

0

y

0

z

)

in

I

corresponds to the vector

(

y

,

z

)

in

V

and

(

y

,

z

)

=

(

c

1

⁢

y

1

+

c

2

⁢

y

2

,

c

1

⁢

z

1

+

c

2

⁢

z

2

)

for some

c

1

,

c

2

∈

ℚ

. Thus

(

0

y

0

z

)

=

(

0

c

1

⁢

y

1

+

c

2

⁢

y

2

0

c

1

⁢

z

1

+

c

2

⁢

z

2

)

=

(

0

y

1

0

z

1

)

⁢

(

0

0

0

c

1

)

+

(

0

y

2

0

z

2

)

⁢

(

0

0

0

c

2

)

and it follows that

I

is finitely generated
by the set

{

(

0

y

1

0

z

1

)

,

(

0

y

2

0

z

2

)

}

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 positive
integer

n

occurring as the upper left entry of a matrix in

I

. It follows that every element
of

I

can be put into the form

(

k

⁢

n

y

0

z

)

⁢

for some

⁢

k

∈

ℤ

;

y

,

z

∈

ℚ

.

By definition of

n

, there is a matrix of the form

(

n

b

0

c

)

in

I

. Since

I

is a right ideal in

R

and since

(

n

b

0

c

)

⁢

(

1

0

0

0

)

=

(

n

0

0

0

)

,

it follows that

(

n

0

0

0

)

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. Then
an arbitrary element of

I

has the form

(

k

⁢

n

y

0

0

)

⁢

for some

⁢

k

∈

ℤ

,

y

∈

ℚ

.

Note that

(

k

⁢

n

y

0

0

)

=

(

n

0

0

0

)

⁢

(

k

y

n

0

0

)

. Hence,

(

n

0

0

0

)

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

I

we have a matrix

(

m

⁢

n

y

1

0

z

1

)

⁢

for some

⁢

m

∈

ℤ

;

y

1

,

z

1

∈

ℚ

;

z

1

≠

0

.

Since

(

n

0

0

0

)

∈

I

,

it follows that

(

n

y

1

0

z

1

)

∈

I

.

Let

(

k

⁢

n

y

0

z

)

be an arbitrary element of

I

. Since

(

k

⁢

n

y

0

z

)

=

(

n

y

1

0

z

1

)

⁢

(

k

1

n

⁢

(

y

-

y

1

⁢

z

z

1

)

0

z

z

1

)

,

it follows that

(

n

y

1

0

z

1

)

generates

I

as a right ideal in

R

.

In all cases,

I

is a finitely generated.

Title

example of a right noetherian ring that is not left noetherian

Canonical name

ExampleOfARightNoetherianRingThatIsNotLeftNoetherian

Date of creation

2013-03-22 14:16:15

Last modified on

2013-03-22 14:16:15

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

18

Author

CWoo (3771)

Entry type

Example

Classification

msc 16P40

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