on some rings whose modules have maximal submodules - American

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 50, July 1975

ON SOME RINGS WHOSEMODULES HAVE MAXIMALSUBMODULES V. P. CAMILLO

the

ABSTRACT.

It is shown

that a principal

property

every

R module

has

satisfied

by these

that

be simple.

Strong

evidence

ize

for the

an example

sional

left

full

conjecture

of Faith

[l]

ring,

domains

These

examples

is regular;

right

R module

fect.

Interest

domains,

and Bass' has

must

are deduced

giving general-

of an infinite

dimen-

of that

two questions,

ring

is never

is a max ring"

does

Now the most

a

from a theorem is to show

are,

then

to conclude as cited

to place

of Faith

that

one needs

that

ones

given

[2, p. 130],

must

in fact

only

This

to the

that

any

V-ring

P.I.D.,

seems

simple

answering

in the non-

are simple which

rings.

is an

Our first

as the examples

max ring in order

interesting

weakness

"R

on R.

rings

R is a right

result

apparent

per-

of knowledge

that

conditions

of simple

if R is a right

every

the hypothesis

these

that

there

being

thus

show

strong

that

be a product

assume

R is simple.

in the previous

very

of the fact

that,

the state

is

every

be right

nonperfect,

case)

proof

must

by the fact

and highly

whether

R over which

(a max ring)

[3] for the commutative

ring

to be an exception

a ring

prin-

module

and whether

They

not seem

simple

of Faith,

V-ring,

is enhanced

of simple

(every

in the negative.

general

in a semisimple

examples

V-rings

whether

very reasonable

(see

striking

submodule

nonregular

case

esis

socle

having

We also

two questions

question,

examples

are highly

commutative

seems

domain, submodule

K-rings.

a subring

the

noetherian

a maximal

very definitively

cited

ideal

rings

be

are right

answer

at the time,

purpose

that

[3] gave

simple,

in these

they

the last

comes

must

contains

which

a nonsemisimple,

order

they

by showing which

and Kolfmann

ideal

injective).

V-ring

that

right

a maximal

K-ring.

right

exists

conditions

linear

Cozzens cipal

right

because

of the max ring

it hypoth-

paragraph.

Received by the editors April 19, 1974. AMS (MOS) subject classifications (1970).

Primary

16A04, 16A30; Secondary

16A42, 16A46. Key

words

and phrases.

Maximal

submodule,

principal

ideal

ring,

simple

ring,

K-ring. Copyright © 1975, American Mathematical Society License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use

97

98

V. P. CAMILLO The

because able,

question

that

they

V-rings?

are

however,

erate

P.I.D.

maximal

is also

right

some

ideals,

strong

either

then

is:

seems

evidence

R/abR

Are these

to be quite for all

or R/baR

in the text

R is a V-ring

rings

simple

difficult.

for its truth.

max ring then

as is explained

P.I.D.

from this

problem

and a right

because,

a left

arises

This

to present

if R is a right interesting

naturally

We are

We show

that

a, b £ R which

is semisimple.

(before

This

Proposition

if and only if R/abR

genis

2), if R

is semisimple

for all such « and b. The

regular

second

section

ring which

only if it is regular.)

L = End Vp, generalize

is motivated

is not a V-ring. Faith's

F a field,

this

example

generated

by showing

and if R is any subring

by the example

(A commutative is the subring

by the socle

directly

given

that

by Faith

of a

ring is a V-ring of any full

if and

linear

ring

of L and the scalars.

We

if L = End VD, D a division

ring,

of L containing

socle

L, then

R is not a left

which

is a right

V-ring. I. Theorem 1. // R is a right max ring,

P.I.D.,

then

R

is simple. Proof.

Let

dR be a two-sided

ideal

in R, and let

Q be the classical

right quotient ring of R. Let U C Q be given by U = U neN^~"RU/R but

has d~n

some

a maximal

submodule

4 M. However,

r £ R.

But

M/R.

since

rd = ds

U/M

for some

is simple,

However,

the right

l + M = d~nr

r = sd~

£ M,

+ M for

. Thus,

d

~

+

on the right by d we have d~" + M =

side

is zero

since

result

suggests

d~n

£ M, so d~n

the following:

if and only if it is a max ring?

low, however, for this

tively

d~n~ d~

Then

d~n

£ M. This

U/R = 0, or that d~ I £ R, so that dR = R.

The previous

reason

is an 72 with

+M.

shows that V-ring

there

s, so thaf

M = d~n +1sd~ l + M and multiplying

d-"+1s

Now,

we present is that

straightforward

for all

determine

the

that

a proposition

a and

injective

that

generate

of a simple

seems

is quite

P.I.D.

R is a right

b which hull

If R is a P.I.D.,

question

which

if R is a two-sided

manner

is semisimple

This

suggestive.

one can show

V-ring

R a

BeThe

in a rela-

if and only if R/abR

maximal module

is

difficult.

right must

ideals.

have

(Just

a submodule

of length two, which must be of the form R/abR.) Proposition

main with classical

2. Let

right

(a) If a, b £ R and

R/abR

or R/baR

R be a right

quotient a and

max ring which

ring

is also

a right

Ore do-

Q. Then:

b generate

maximal

is semisimple.

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

right

ideals,

either

99

RINGS WHOSE MODULES HAVE MAXIMAL SUBMODULES (b) // «R

is a maximal

right

ideal

then

R/anR

is semisimple

for all

n £ N. Proof,

(a)

Use the

same

idea

as in the proof

the chain fl~ !R C «" lb~ lR C «~ lb~ la~ lR... union.

Let

M/R

be a maximal

submodule

of Theorem

1. Consider

in g, and let U be the

of U/R.

Then

there

is an 77 such

that either:

(1) (a-h-1)"-1 £M but (a"1*.-1)"-1«-1 4M, or (2) a- \b~ la~ l)"- ! e M but a~ \b~ la~ ')"" lb~ 1 = (a~ lb~ l)n 4 M. If (1) occurs, maximal

right

let x = (a~lb~1)".

ideal

(fl-1^-1)"-1«-1

in R which

eM.

Then

contains

(x : M) = \r £ R \xr £ M \ is a

ba.

It does

not contain

b, or else

Thus

R/baR = bR/baR © (x : M)/baR. If case

(2) occurs,

an analogous

argument

shows

that

R/abR

is semi-

simple.

(b) Again let U = U_£a,«-"R of U/R. C = (a~m~

«".

:M).

Since

Then

anR/an+

R/aR

Then, induction,

C/a"

R « R/a"R, R/a"

Remarks. (B)

regular

!R © C/«*+

a"

but not

^

we may assume

2 may be proved right

1 also

works

we only require

If R isa

ideal

to be semisimple

by

under

the hypothesis

that

ring.

for a prime

that

in fact,

principal

R is a right

the right

quotient

M. Then

ring

the ideal

Q/M is injective

II. Writing

functions

of Faith's

example,

socle

of a full linear

finite

rank,

a fact

right

ideal

Goldie

ring

in question

domain

max ring if and only

Q oí R is injective

found in [3, Proposition

zation

containing

which

is right

be generated

prin-

by a

element.

(C)

then,

principal

Theorem

since

which

ideal

we have

R is.

(A) Proposition

R is a semiprime

right

is simple

= anR/an+

so that

submodule

^ M, but a~k + 1 £ M. Look at

C is a maximal

R

R/an+!R

cipal

and let M/R be a maximal

Then there is a k such that «

since

a unique

simple

if R is a right

and has

R is hereditary.

V-ring,

a maximal

The

idea

module

for

submodule

for this

is

l.l]. opposite

scalars

as explained

ring is a collection

we use

with

without

explicit

we have

the following

in the introduction.

of all linear mention

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

Recall

transformations

below.

generali-

the

of

'

V. P. CAMILLO

100 Theorem

3. // R is a subring

L = End WD, containing Proof.

Let

socle

of an infinite

L, then

X be a basis

dimensional

R is not a left

and pick

out countable

each i £ N let f. e R be defined by fix)

x 4 x.. Then XR/. is direct, for it p .f

=x

full

linear

ring

V-ring. subset

ix.|x.

£ N\.

For

and /.(x) = 0 for x e X;

+• • • + p f =0 then, for any k we

have

PlW and, since

+ — + Pkf¿xk) + ~' + Pjnixh)-0

p .f .(*k) = 0 for i 4 k, we have p,/,(ï.)

0 for all x £ X so that p J, Further, In fact,

=

= 0.

if 1(g) denotes

if Im L.

= 0 and thus, p.fAx)

the left

annihilator

D Im L _ for any two linear

of g £ L, then

transformations

/(/, ) C /(/,).

then

l(L,)

C

l(L2), for if pL. = 0, then for any x e W, L Ax) = L Ay) so that pL Ax) = pLj(y) = 0 and pL 2 = 0. Thus, the maps W. defined by (pf )H . = pf. ER/, is direct,

this yields

are well defined.

a map H: SR/y ~*R/,

Since

given by

(Z?/;> =Z(V¿K=(Z?>r Now, R/j is simple. linear

transformation

since

R contains

For if pf. 4 0 then pf Ax A 4 0 so there is a

g with socle

L.

gpf Ax A - x But clearly,

and

Im g Cx.D.

Then

from the construction

g £R

of /,,

gpf.

=

f. so that Rpf, = Rf, and thus Rf, is simple. We now show for a simple

Xld1+...

that

the map

noninjective

+xydn.

H cannot

module.

Then,

For,

be extended let

(1) H = tf,

/j =(/n + 1)W = /„ + 1t/,so

and'that

Rf.

and let

that

is there-

tf Ax/)

=

/j =fn + 1tf1 or

a contradiction.

REFERENCES 1. J. Cozzens,

Homological

properties

Bull. Amer. Math. Soc. 76 (1970), 75-79. 2. C. C. Faith,

Notes

Lectures

on injective

in Math., no. 49, Springer-Verlag,

of the ring of differential

polynomials,

MR 41 #3531. modules

Berlin

and quotient

rings,

and New York, 1967.

Lecture

MR 37

#2791. 3. L. A. Kolfmann,

Rings

over

which

each

module

has

a maximal

Mat. Zametki 7 (1970), 359-367 = Math. Notes 7 (1970), 215-219.

submodule,

MR 41 #6913-

DEPARTMENT OF MATHEMATICS, UNIVERSITY OF IOWA, IOWA CITY, IOWA 52242 License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use