subrings of noetherian rings - American Mathematical Society

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 46, Number 2, November

1974

SUBRINGSOF NOETHERIAN RINGS EDWARD FORMANEK AND ARUN VINAYAK JATEGAONKAR ABSTRACT. Let generated

is

R.

right

Generalizing

Noetherian that

S be a subring

S-module.

if

and

Rç

Su. for

has

a result

S is commutative a finite

a right

Let

S be a subring S-module.

R-module

of a ring

that trivial)

Noetherian

Eakin

M. Nagata

[10])

to be a commutative

ring.

[1] have

extended

S-module.

then

that

show

u -S =

Artinian

This

so

if R is right We also

! such

is Noetherian,

(and

by taking

later

u

R such

It is well known

eral as can be seen

[3] (and

{u ,,•••,

ring

that

or semi-

yields

a re

algebras.

therian ring then R is a right

is assumed

we show

S is Noetherian.

set

so as a right

on group

R such that R is a finitely

S is a right Noetherian

then

generating

1 < í < m, then

of Clifford

a right

if

of P. M. Eakin,

simple iff it is respectively suit

of a ring

Clearly,

Eakin's

ring.

R = (q q)

theorem

R is finitely that

The converse

and

S = ( j ^).

showed

that

is false

as

in gen-

However,

the converse

D. Eisenbud

to some

generated

if S is a right Noe-

P. M.

holds

if R

[4] and J. E. Björk

mildly

noncommutative

situa-

tions.

In this theorem.

note,

The

The second theorem

we provide

first

version

version

and,

when

two mildly answers

improves applied

noncommutative

a question

upon

raised

Eisenbud's

to group

versions

by J. E. Björk

generalization

algebras,

yields

of Eakin's [8J.

of Eakin's

a theorem

of Clifford

[2, p. 343]. As usual, Recall

that

infinite

a module

direct

Mr, is finite that

all rings,

sum of nonzero

integer

there

if it does It is known

exists

of M contains

the uniform

dimension

to be unitary. not contain

any

[7, p. 216] that

a nonnegative

submodules

is called

are assumed

dimensional

submodules.

then

sum of nonzero

such

and modules

MR is finite

dimensional

any direct

The least

subrings

if

integer

n such

at most

22 terms.

of M and is denoted

as d(M) or dAM). _

K

Received

by the editors

AMS(MOS) subject

September

17, 1973.

classifications(l970).

Primary 16A46, 13E05; Secondary

16A26, 16A38. Key words therian

and phrases.

rings, P. I. rings,

Commutative

Eakin's

theorem,

Noetherian Clifford's

rings, theorem

noncommutative on group

Noe-

algebras.

Copyright © 1974, American Mathematical Society License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use

181

182

E. FORMANEK AND A. V. JATEGAONKAR Our proofs

each

case,

related

of our main results

we start

object

sents

with a Noetherian

is Noetherian.

no problem

have

in lots

a certain object

strategy

and want to show

Due to the Noetherian

of situations.

out, we show that the following

in common.

induction,

Once the dubious

that

some

this

pre-

cases

lemma or the idea underlying

In

are singled

its proof is

applicable. 1. Lemma. therian

Proof. Since

M be a finite

essential

Let

dimensional

submodule

above

sequence

a submodule

K of Al such

that

It follows

that

K Ci M = (0).

\Mn : n — > 1!.

lemma

than rings. 2. Lemma.

Then

any

Let

Let

that

every

We proceed

S be a subring

proper

S, it suffices

that

T is a domain.

M itself

Consider

ated

Lx

For,

the

of L.

and faithful

ian ring

T-torsion

generated

so

over the domain

Noetherian

to treat

and torsion-free

so

T.

over

when

are finitely

It follows over

AsS.

generated

is finitely

Note that generated

of L.

It is in fact since

S.

is Noetherian

of T such

If N 4 (0) then

N = (0).

M/L

T = 5/ann^L.

N of L.

over

over

over

our choice

T-module.

over

elements

submodule

R.

S.

Since

of M which

L is Noetherian

the case

generated

over

annRL.

R-submodule

ring

of M is Noetherian

L is Noetherian

submodule

Noetherian

of M which

possible

for modules

above.

S is Noetherian

is finitely

image

stated

indicated

over

is Noetherian

it is a proper

[9, p. 53], and

It remains

which

by y, contradicting

Further,

free and so) faithful

generated

if x, y are nonzero

is a nonzero

theorem

of a commutative

R-submodules

to show that

S and annihilated

module

Eakin's

is finitely

L with largest

over

over

to

and essen-

\K © M : n > 1\ stops,

the strategy

R-homomorphic

to show that

xy = 0 then

lemma)

is direct

m

Since the chain

M be an R-module

S, choose

K + M

is

) for

) ® Mm — CM n and diMn ) = diM m ). n

is essentially

which

Among the nonzero

over

of M (or Zorn's

the sum

Its proof illustrates

R-module

Proof. sume

which

diM ) = diM

n

The following

rather

of M.

integers

say,

dimensionality

is Noe-

of submodules

constant;

tial in M. For n — > m, we have (KDM does

chain

If M/N

M is Noetherian.

of nonnegative

by d{M), it is eventually

n > 772. We can use the finite

obtain

module.

N of M then

\M : n > 1\ be an ascending

diM ) is a nondecreasing

bounded all

Let

for every

an R-sub-

L is finitely

gener-

L/N

is a (torsion-

that

T is a Noether-

T. In this

T; so, it is finite

License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use

case,

dimensional

L is finitely

over

T,

If

S.

SUBRINGS OF NOETHERIAN RINGS K is any essential torsion K.

so unfaithful

Since

over

T-submodule

Lt

T.

1 shows

induction

An alternate sult of [5J.

T,

This

is a R-submodule

Lemma

Noetherian

over

of L then

proof

of Lemma

We now prove

that

Lt C

L/K

is Noetherian

T.

Now a routine

over

and

o

2 can be obtained

a version

generated

t e T such

and so

L is Noetherian

the lemma,

is finitely

a nonzero

of L, L/Lt

that

proves

yields

L/K

183

of Eakin's

by using

theorem

the main re-

which

answers

a

question of J. E. Björk [8, p. 376]. 3. Theorem.

Let

ring R such

that

a Noetherian

ring.

Proof.

Clearly,

S-module. loss

S be a commutative

R is finitely

it suffices

two-sided

of a right

as a right

to show that

In view of the Noetherian

that for every

subring

generated

Then

R is Noetherian

induction,

ideal

Noetherian

S-module.

as a right

we may assume

/ of R, R/I

S is

is Noetherian

without as a right

S-module. Suppose

R is not a prime

prime ideals radical

Pj,

P(R)

tive integer morphic

• • -, P,

of R.

ring.

Then there

image

k > 1 since

By Levitski's

of a finite

theorem,

direct

We now treat

which

Since

ring.

A theorem

There

a nonzero

d e Z(R)

R-module

a positive

satisfies

R/(K

such

that

that

integer

h such

in Rz,Ry

A result

exists

222. Since

the centre Z(R)-

chain of T-submod-

that

dz,Ry(Mn)

=

K of R such that the K O M^ = (0) for all

) < o». Using the Z(R)-embedding over

Z(R).

Rd C /< © M . It is easily

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P. I.

a Z(R)-

dimensional

Then

due iden-

R is a prime there

integer

R is a finite

© /VLj is unfaithful

that

Thus,

a polynomial

of End Rs,

[6] now shows

Choose a Z(R)-submodule and essential

R/P(R).

R is a prime ring.

End Rc

Let ¡M : n > l! be an ascending

exists

we see that

when

that

' fot some positive

since dZ(R.(M ) = dZ(R.(M

Z(R)(m),

= (0) for some posi-

Noetherian.

with a subring

it follows

for all n>h.

sum K + M, is direct

n>h

of

Then R embeds in I1JR/P¿: 1 < t < k\

case

shows

R Z(R)

of R is a domain,

of R.

number

If P(R) = (0) then we must have

Rs

due to Formanek

module. Set T = SZ(R), ules

[ll]

R is isomorphic

monomorphism

dz,R)(M.)

makes

the (dubious)

and Small

tity.

Z(R)

{P(R)\

sum of the right

R is not a prime ring. R-module

to Procesi

a finite

I. It is clear that {P(R)Y~ 1/{P(R)Y, 1 < i < I, is an R-homo-

if P(R) 4 (0) then Rs is Noetherian. as a right

exists

of R such that (\{P ^ 1 < i < k\ is the prime

This

seen

R •*

provides

that

M Ci

184

E. FORMANEK AND A. V. JATEGAONKAR

(Rd + Al, ) = Al, for ail n > h. So, Alrz/Al,n ^ iRd + Mn )/iRd + AlnJ as Tno — modules. that

Since

R/Rd

the chain

ring.

By Lemma

The following

theorem

Let

subset

1

Thus

S and so over

R™. is Noetherian

2, R„ is a Noetherian

improves

T, it follows which

module.

upon Eisenbud's

makes

T

G

generalization

of

theorem.

4. Theorem. a finite

over

{Al : 72 > 1 ¡ stops.

a Noetherian

of Eakin's

is Noetherian

R = %\u.S:

S be a subring

\u,,

■ ■ ., u

Im1

1 < z < m\. series)

tion series)

as a right

dimensional

and semisimple

Proof.

After

! of R such

Then

composition

a right

if and only

R.

that

Assume

u .S = Su. i

R-module

Every

as a right

adjustment

that

for

(resp.

simple

right

there

exists

I < i < m and

1 '

—

is Noetherian

if it is Noetherian

S-module.

a trivial

of a ring

—

(resp.

has a

has a composi-

R-module

is finite

S-module.

if necessary,

we may assume

that

«1 = 1. Let

Al be a Noetherian

right

R-module

momorphic

image

of Al is Noetherian

an infinite

direct

sum

Zorn's

lemma,

®\w

n

over

in AL.

w,d

u. e W + Sjtzz S: 1 < 72 11. Since n

Inductively

assume

, = 0 then it suffices

—

exists

s eS

u .S = Su . for i

i

that

---,/%.

Al contain Using

the sum that,

R-ho-

W+

for each

'

4 0 and

The claim is trivial

that the claim

is valid

for

for k, 1
1\ of nonzero

we get a 5-submodule

2,\w S: n > 1! is direct

such

—

\s € ^ +

s , s, , • • • , s, , in '

1

*'

S such that su,

Set d,k+l , = d.s'. k Clearly,

w^,

k+i

, = u,

k+l

,s,

s u . — u .s\

Then 0 4 w ,d, 1 k u,k+l ,s = w ,d, 1 k+i ,u,k+l , £ W + ~ï\wn S:t2>1!. — . 4 0.

Further,

we can choose

w,1d,k+l,u, k + ,l £ W + 1.\w n S: 1 — n,

such

that

Then, ' for l — < i— < k,' we have

w 1,d,k + ,u . = w 1,d,uk 1s 1. elf + *-^ /\w l 1 This

for i = 1, • • •, k.

111

n

on k and proves

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