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380 Mario Petrich Pennsylvania State...
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Lecture Notes in Mathematics Edited by A. Dold, Heidelberg and B. Eckmann, Z~Jrich
380 Mario Petrich Pennsylvania State University, University Park, PA/USA
With an Appendix by Richard Wiegandt
Rings and Semigroups
Springer-Verlag Berlin • Heidelberg • New York 19 74
AMS Subject Classifications (1970): 20-02, 20-M-20, 20-M-25, 20-M-30, 22-A-30, 16-02, 16A12, 16A20, 16A42, 16A56, 16A64, 16A80
ISBN 3-540-06730-2 Springer-Verlag Berlin • Heidelberg • New York ISBN 0-387-06730-2 Springer-Verlag New Y o r k . Heidelberg • Berlin This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under § 54 of the German Copyright Law where copies are made for other than private use, a fee is payable to the publisher, the amount of the fee to be determined by agreement with the publisher. © by Springer-Verlag Berlin • Heidelberg 1974. Library of Congress Catalog Card Number 74-2861. Printed in Germany. Offsetdruck: Julius Beltz, Hemsbach/Bergstr.
INTRODUCTION
Semigroup theory can be considered as one of the more successful offsprings of ring theory.
The relationship of these two theories has been a subject of particu-
lar attention only w i t h i n the last two decades and has generally taken the form of an investigation of the m u l t i p l i c a t i v e semigroups of rings.
The first and still
the most fundamental w o r k in this direction is due to L.M. G l u s k i n who certain dense rings of linear transformations These Lectures
represent an attempt
from the m u l t i p l i c a t i v e
studied point of view.
to put selected topics concerning both rings of
linear transformations and abstract rings, as well as their m u l t i p l i c a t i v e semigroups,
into a form suitable
for presentation
to students interested in algebra.
The Lectures are divided into three parts according to the clusters of covered topics.
part I consists of a study of certain semigroups and rings of linear transformations on an arbitrary vector space over a division ring. linear transformations two phenomena,
For dense rings of
containing a nonzero linear t r a n s f o r m a t i o n of finite rank,
from the present point of view,
are of decisive importance:
(a) its
m u l t i p l i c a t i v e semigroup is a dense extension of a completely 0-simple semigroup, and
(b) it has unique addition.
can be obtained by considering naturally
Because of (b), most
their m u l t i p l i c a t i v e
information about these rings
semigroups alone.
This leads
to a study of semigroups of linear transformations and in particular of
those satisfying
(a) above.
Hence in many instances we first estab]ish the desired
result for semigroups and then specialize
it to rings of linear transformations.
Even though the guiding idea adopted here is that first expounded by Gluskin the principal references are the books of Baer
[i] and J a c o b s o n
part II contains an investigation of various abstract rings, izations and representations.
[4],
[7]. their character-
The classes of rings under study here are semiprime
rings with minimal o n e - s i d e d ideals subject to various other restrictions. each of these classes of rings a m u l t i p l i c a t i v e b e i n g possible in view of their unique addition.
characterization
For
is provided,
this
Their m u l t i p l i c a t i v e semigroups
are either dense extensions of completely 0-simple semigroups or of their orthogonal sums.
Again some of the basic ideas here stem from G l u s k i n
reference is Jacobson
[4], but the m a i n
[7].
part III represents
a topological treatment of a left vector space, and its
ring of linear transformations, topology of modules and rings,
in duality with a right vector space.
Linear
linear compactness of vector spaces, various
IV
topologies on certsin rings of linear transformations, s topological vector space are covered here. appear here as topological rings. are due to Dieudonn~ and K~the
as well as s completion of
C e r t a i n of the rings studied earlier
Many of the basic
[2], but the chief references
ideas discussed in this part
sre the books of J a c o b s o n
[7]
[i].
The appendix contains s concise exposition of some principal achievements in the theory of linesrly compact modules and semisimple rings. consist of several characterizstions rings.
The chief original
The two main results
of linesrly compact primitive and semisimple
reference here is Leptin
12}.
Jacobson's Density Theorem
is included with a short proof.
The m e t h o d of m a x i m u m exploitstion of the m u l t i p l i c a t i v e often mskes
the use of various hypotheses more
that, st least for the rings under study here,
structure of a ring
transparent and demonstrates
the fact
the sddition is essentially extrane-
ous. The sources of these Lectures are numerous: references,
in addition to the above m e n t i o n e d
they include s vsriety of papers which sre generslly referred to in the
text.
There is slso a generous
time.
Among these are the results concerning the structure of simple rings with
minimal one-sided
sprinkling of results published here for the first
ideals in terms of Rees matrix rings,
and some related results due to Dr. E. Hotzel; author.
isomorphisms of the latter,
the remsining ones are due to the
For the sake of clarity and u n i f o r m i t y of presentstion,
have been rephrased and several new proofs hsve been provided. exercises at the end of most sections;
many known results There are several
they are designed to test the u n d e r s t a n d i n g
of the m a t e r i a l and sometimes extend the subject covered in the text. msterial
However,
the
in the msin body of the text is independent of exercises.
Part I was the subject of a one semester course in linear algebra in the Summer of 1969;
the entire Lectures
formed the content of a two semester course in topics
in ring theory in the school year 1971/72, both at the Pennsylvania State University. I sm indebted to Dr. D.E. Zitsrelli
for taking the notes
for Part I, to
Mr. JoJ. S t r e i l e i n for taking the notes for parts II and III, supplying the appendix,
and to Professor B.M. Schein
suggesting several improvements.
I am grateful
to include his u n p u b l i s h e d results, many slips,
to Dr. R. W i e g a n d t
for
for reading the m s n u s c r i p t and
to Dr. E. Hotzel
for the p e r m i s s i o n
to students in the two classes for correcting
as well as to all other persons who c o n t r i b u t e d
to the existence of these
Lectures. Statements in the text are referred to only by number: the same part,
the Arabic numerals are used,
if the statement is in
say 5.6 which is statement
6 in
V
Section 5; if the statement is in a different Part, numeral in affixed,
the number of the Part in Roman
say 1.5.6.
Since the w o r k of G l u s k i n on the subject at hand has provoked considerable interest,
it is hoped that a systematic and s e l f - c o n t a i n e d e x p o s i t i o n will propa-
gate the existing knowledge and stimulate new research in this highly promissing area of the r i n g - s e m i g r o u p cooperation.
TABLE OF CONTENTS
PART I SEMIGROUPS AND RINGS OF LINEAR TRANSFORMATIONS I.i
Definitions and notation
l
1.2
Dense rings of linear transformations
1.3
One-sided ideals of
1,4
Ideals of
1.5
Semilinear isomorphisms
34
1.6
Groups of semilinear automorphisms
47
1.7
Extensions of semigroups and rings
54
g(V)
Su(V)
7 16
and principal factors of ~ u ( V )
27
PART II SEMIPRIME RINGS WITH MINIMAL ONE-SIDED IDEALS II.l
prime rings
11,2
Simple rings
74
11.3
Maximal prime rings
85
11.4
Semiprime rings
91
65
95
ii .5
Semiprime rings essential extensions of their socles
11.6
Semiprime atomic rings
102
11.7
Isomorphisms
105 PART III
LINEARLY TOPOLOGIZED VECTOR SPACES AND RINGS III.I
A topology for a vector space
112
III.2
Topological properties of subspaces
120
III.3
Topological properties of semilinear transformations
125
111,4
Completion of a vector space
128
111.5
Linearly compact vector spaces
131
111.6
A topology for
£u(V)
136
III.7
A topology for
Su(V)
139
111.8
Another topology for
III.9
Complete primitive rings
£u(V)
143 146
VIII
APPENDIX ON LINEARLY COMPACT PRIMITIVE AND SEMISIMPLE RINGS O.
Introduction
152
i.
More about primitive rings
153
2.
Inverse limits and linearly compact modules
156
3.
Linearly compsct primitive rings
159
4.
Linearly compact semisimple rings
162
CURRENT ACTIVITY
167
BIBLIOGRAPHY
168
LIST OF SYMBOLS
175
INDEX
177
PART I
SEMIGROUPS The subject transformations dense
rings
finite mations one,
of
this part
rank h a v i n g
all o n e - s i d e d
vector
transformations
are c h a r a c t e r i z e d
of finite
ideals
are
in a vector
found,
the semigroup
taining
all
linear
duality w i t h
transformations
the given one
the u n d e r l y i n g vector
space
vector
of rank
is e x p r e s s e d
spaces.
are discussed.
in semigroups
representation
two semigroups
and rings with
part
space
of
In particular,
i with
linear
transforthe given
set of idempotents factors
an adjoint
is con-
each
are
con-
space
in
transformations
of
automorphisms
on a
of semilinear
of ideal
semigroups
is
ideals
in a vector
a discussion
to certain
all
transformations
of semilinear
of groups
of
in duality with
transformations
linear
of linear
transformation
for its principal
ends with
application
ring.
linear
ordered
linear
by means
A number This
and rings
For the ring of all
its partially
and ring of all
between
a division a nonzero
ways.
and a Rees m a t r i x
Isomorphisms
space over containing
semigroups
in several
characterized,
found.
TRANSFORMATIONS
an adjoint
structed.
For
OF L I N E A R
is a study of certain
on an a r b i t r a r y
of linear
rank
AND RINGS
extensions
and rings
of linear
transformations. The emphasis Thus many here
here
statements
for semigroups
is one
which
the m u l t i p l i c a t i v e
are usually
and are
then s p e c i a l i z e d
I.i A division all
the axioms
a commutative by
4.
The
ring
division
symbols
Throughout Denote
+,
this
the elements
case Roman
(or a skew
for a field
letters
in such a way
pair
that
is a field have
let
L
by lower
v,x,y, ....
is a function m a p p i n g The ordered
~
possibly
~X V (&,V)
under first
study. proved
AND N O T A T I O N is an algebraic
commutativi~y
system
satisfying
of m u l t i p l i c a t i o n .
and conversely.)
be a d i v i s i o n case G r e e k that
It will
V, say
postulates
ring and
letters, ~
be usually
(Thus denoted
V
be an a b e l i a n
and those of
acts on
V
V
by
group.
lower
on the left if there
(o,x) ~ ox.
is a left vector
the following
are
their usual meaning.
We say
into
of the rings
for these rings
to rings.
field or a sfield)
-, i, 0
section, of
DEFINITIONS
except
ring
structure
established
space
if
&
are fulfilled:
acts on
V
on the left
O ( x + y) = o x + o y
(oE 4, × , y E
v)
(O+~)x
(o,~ E a, x E
V)
~(~x)
= Ox+~x
= (o~)x
(x ~ V).
iX = x
The action and
those of If
~
is called
V
acts on
V
on the right,
postulates,
then
case,
using
the n o t a t i o n
Hence
the name v e c t o r
We will w r i t e
"V
Let called
the ordered
V
(left or right) the phrase
scalar m u l t i p l i c a t i o n ,
(V,4)
space
instead vector
pair
(&,V)
and
makes
of
(~,V)
(4,U)
linear
(~,V)
&
called
form
The set of all to be an abelian
linear
x(f+g)
group
V The
set
as follows. V*
V*
on the right
function
addition
V*
group
U
(&,V).
(V*,&)
forms on
for
is a right vector
Elements
of
V*
of confusion.
is to be regarded
group).
In such
A function
ss a
a case
a:V ~ U
is
V). spaces
apace
as operators
(&,A+) ~x of
of
(~,V)
where
on the right
on
into
4+
signifies
is the m u l t i p l i c a t i o n (4,V)
or functional)
into
(&,~+)
(&,U)
is easily
seen
by
(&,V)
f E V*
(i)
the usual
group
addition
can be made under
and
into
addition
o E &,
fo
of
the a b e l i a n
of homomorphisms. a right v e c t o r
(i).
We make
be defined
g
space act on
by
( x E v). space,
called
will be denoted
by
is
(~,V).
group of all h o m o m o r p h i s m s
under
×(fo) = (xf)o Then
act on the right.
danger
(x ~ V).
is an abelian
by letting,
In this
a,b,c, ....
defined
of the abelian
of linear
sp~ce.
if
transformation
= xf+xg
the abelian
First
A linear
transformations
under
it is a subgroup into
of left vector letters
as a left vector
(or a linear
group
(~,U)
and the scalar m u l t i p l i c a t i o n
o E 4, x E g+).
a linear
In fact,
itself &
V
spaces.
into
of the above
(x,y E V)
case Roman
g
group of
(here
scalars
4" is customary.
(o E 4, x E
transformations
lower
space over
be two left vector of
that
than just an abelisn vector
= xa+ys
We can consider
in
are called
a right vector
cases without
if it is clear
(rather
transformation
them by
the additive
A
that the scalars
in most
(ox)a = o(xa) We write
of
the right v e r s i o n
is called
it clear
can be used
space
(x+y)a
satisfying
(V,&)
is a (left or right)
a linear
and denote
elements
are called vectors.
the dual f,g,h,....
(or conjugate)
space of
3 A semigroup (usually
is a n o n e m p t y
mations
of
sition
(ab)x = a(bx)
written
on the right,
resulting
X
into itself,
semigroup
A linear (&,V).
written
(x C X),
on the left,
to be denoted
the c o m p o s i t i o n
will be denoted
transformation
of
g(V).
The same
the rin$ o f e n d o m o r p h i s m s
(&,V)
into
of
(~,V)
X,
binary
is a s e m i g r o u p
by
g(X).
operation
the set of all transforunder
the compo-
If the t r a n s f o r m a t i o n s
x(ab)
= (xa)b
(x E V)
and
are the
~(X). itself
(g,V) of
is called
of
~I(V);
and will be denoted
the addition
and will
an e n d o m o r p h i s m
is s s u b s e m i g r o u p
(g,V),
set endowed with of
set
is given by
by
the > e m i g r o u p o f e n d o m o r p h i s m s
or simply
an associative
For any nonempty
The set of all e n d o m o r p h i s m s
is called
£(V)
set together w i t h
called m u l t i p l i c a t i o n ) .
(i) forms
be denoted
by
by
a ring,
£(&,V)
of it
g(&,V) called
or simply
.
We have
seen that
can be given can be made made
If
U
transformations
group.
space over
4,
of
In addition, and for
(~,V)
into (~,U) + U = ~ , this set
for
U = V
the same
set can be
or a ring.
is a subgroup
is called
linear
of an abelian
into a right vector
into a semigroup
(&,U)
the set of all
the structure
of
a subspace
V
closed
of
(g,V).
under m u l t i p l i c a t i o n It is clear
that
by scalars
(~,U)
in
4,
then
is itself a vector
space. As in the case of a finite a linearly
independent
a generating and a basis
set
(every v e c t o r
of a vector
as a m i n i m a l
Any
have
two bases
denoted
to a basis
of
dim V.
every basis
To every
every vector
in
a complement
properties require proofs If mation
V
A
of a v e c t o r
further
V of
ranse of
and V
of
of
into
are U,
methods
which
then
in the form
V, and we write express
reference.
over
Va = {xa Ix E VJ
V
can be extracted.
to a basis of with
Most Zorn's
a
of the whole V
such
that
x E A, y E B;
We will
use
these
of these statements lemma).
For
is a linear
is a subspace
a, d e n o t e d
of
by rank a.
the null
V,
can be completed
the
at the end of this section. 4, and
called
of
V = A~B.
(axiom of choice,
spaces
the d i m e n s i o n
B
,
it is set
a basis
x+y
..
independent
set of vectors
a subspace
independent),
combination
has a basis;
linearly
can be completed
written
of
linear
set of vectors
one defines
is linearly
is by d e f i n i t i o n
there exists
is the rank of
is a subspace
always
independent
see the references
left vector
a; its d i m e n s i o n
N a = ix E V I xa = 0}
in
space w i t h o u t
discussion
U
A
subset
as a finite
space
generating
V,
space over a field,
finite
set or a maximal
linearly
can be uniquely
proof by transfinite and
A vector
of a subspace
subspace
B
is called
Any
V, and from every
In particular, space.
V.
generating
the same c a r d i n a l i t y
by
vector
(every
can be w r i t t e n
space
characterized
usually
dimensional
set of vectors
transfor-
U, called The set
space of
a.
the
Linear written space;
the r a n g e
space
are
a given
a function
left;
the null
vector
the dual
space
of a l i n e a r
pair
~:VX
(U,g)
U ~ ~
and
(&,V)
is c a l l e d
are d e f i n e d vector
=
(Vl,U)
+
(v2,u),
(v,ul+u2)
=
(V,Ul)
+
(v,u2) ~
~(v, u) =
of a r i g h t
and a left v e c t o r
form
u , u l , u 2 t U, o ~ &,
form
is n o n d e g e n e r a t e
and
(v,u)
stands
= 0
for all
u E
U
implies
that
v = 0,
vi)
(v,u)
= 0
for all
v 6 V
implies
that
u = 0.
In such
a case we
say
the d i v i s i o n
variously sizing
use
ring
Property (v,u)
is m o r e Let
over
(U,&,V)
LEMMA°
or
we
where
Oil
vector
should
~(v,u).
spaces
write
according
assumed
(U,&,V;~);
to the n e e d
as given).
is a left
(or a dual
Note
and
~l
that
pair)
we w i l l for emphain J a c o b s o n
is a r i g h t
vector
to: v E V
implies
for a p p l i c a t i o n ;
a vector
of dual
(U,V)
tacitly
for all
space.
a t-subspace)
If
by
t-subspace PROOF. is left
that
u = u ~,
analogously
A subspace
U
if for e v e r y
for v). of
V*
is c a l l e d
0 ~ x E V,
there
a total
exists
f ~ U
i__s_s!
pair,
vf u = (v,u) (V*,g).
t->ubspace
form then
(v,f)
the
of
function
(v E V),
(V*,g),
= vf
(v E V,
then f E
f:u ~ f u
is a l i n e a r
(U,V)
U).
(u E U),
isomorphism
is a dual
Conversely, where
of
pair
if
f u :V ~ &
(U,&)
onto
-is -
a
of
The
proof
consists
of a s i m p l e
application
of r e l e v a n t
definitions
as an e x e r c i s e .
If c o n v e n i e n t , as a t - s u b s p a c e U.
U
the b i l i n e a r
is a dual
fU = nat
is
is u s e d
convenient
with
defined
(~
(v,u ~)
be
precisely,
(U,&,V)
for
xf # 0
l.l.l &
=
is a pair
is e q u i v a l e n t
(or b r i e f l y
that
More
(~,~) vi)
(&,V)
subspace
(U,V)
4.
ring
the n o t a t i o n
vi ~)
that
the n o t a t i o n
the d i v i s i o n
space.
we m a y
space
if
(v,u)
over
vector
if it s a t i s f i e s :
v)
which
on a right
vector
(v,u)~ = (v,u~),
A bilinear
[7],
and
is a left
(ov,u) ,
V , V l , V 2 ~ V,
over
analogously
space
transformation
a bilinear
(Vl+V2,U)
where
spaces
of a right
analogously.
i)
iv)
and
of right
the
ii) iii)
such
on
and
defined
For 4,
transformations
as o p e r a t o r s
of
A dual
consider
V
we w i l l V*;
we
identify call
discussion
fU
u
fu' w h i c h
the n a t u r a l
is v a l i d
as a t - s u b s p a c e
and
of
imsse
by i n t e r c h a n g i n g U*.
amounts of
U
to c o n s i d e r i n g
in
the roles
V* of
U
and w r i t e U
and
V,
so
For a given of
a
in
U
(U,A,V)
(and
a
(v,bu) I.i.2 LEMMA. U,
if so denote
subrin$ o f
:
For a fixed b~u.
Then
b
and
=
(v,bu)
u, we o b t a i n Since
just
b~
V)
that
b
is an ad~oint
if
(2)
introduced,
a
has at most
~:a -- a*
one adjoint i n
is an isomorphism
of a
£(U,t~).
are adjoints
= (v,b'u)
(v,bu) =
u E U
in
the function
onto a subrin$ o f
If both
b
we say
(u E U, v E V).
the n o t a t i o n a*.
b E ~(U),
of
(va,u)
With
(va,u)
bu=
a E ~(V),
is an adjoint
it by
£(A~V)
PROOF.
and
(v,b~u)
is arbitrary,
of
a
in
U,
(v E V,
u E U).
for all
v ~ V
we have
then
(2) yields
which
b = b s, which
implies
proves
that
the uniqueness
of the adjoint. Suppose
that
s
has an adjoint.
Let
c~,~ E A, x,y E V;
then
for any
u E U,
we get
((ox+~y)a,u)
=
(ox+~y,a*u)
= o(x,a*u)
= o(×a,u)+T(ya,u) so that
(Ox+~y)a
Reversing the unique
= o(xa)
the roles
adjoint
Suppose
of
that
+ T(ya),
and
= a
of the adjoints,
a*
in
have
(~(×a)+T(ya),u)
is linear. we see
V, and that
a,c E 3~(V)
+ ~(y,a*u)
a*
adjoints
that
this
also
shows
that
a
is
is linear in
U.
Then
for any
v E V, u ~ U,
we obtain
(v,(a*+c*)u)
= (v,a*u+c*u) = (va,u)
which
implies
that
a*+c*
(v,(a*e*)u) which
implies
have
an adjoint
onto a subring on
U,
then
a
of
U
forms
£(U,g).
must
1.1.3 LEMMA.
For
i)
rank
(ab)<
min{ranka,
ii)
rank
(a+b)
< ranks
PROOF.
Va, then
Ab
i) Since generates
vab ~ Vab
of
By uniqueness function
a,b E £(V),
= (v(a+c),u)
Further,
=
It follows
a subring
be the zero
= (va+vc,u)
= (va,e*u)
a^c :" = (ac)*.
in
= (v,a*u) + (v,c*u)
(vc,u)
= (a + c ) * .
= (v,a*(c"u))"
that
+
((va)c,u) that
£(B,V)
= (v(ac),u)
the set of all and
~
of adjoints,
if
on
V.
Thus
~
a E $~(V)
is a h o m o m o r p h i s m a*
is the zero
which of it
function
is an isomorphism.
we have
rankb},
+ rankb. vb, we have which
rank
implies
(ac) ~ rank b.
that
If
A
is a basis of
rank
ii)
Let
generates
A
and
V(a + b )
Let then
(ab) = dim Vab < dim Va = rank a. B
and the desired
~(a,V)
~(&,V)
(U,g,V),
Then
A U B
follows.
£(&,V).
= ~a 6 ~(V) I a = ~a E ~ ( V ) f r a n k
has an adjoint
inherited
= ~(b,V)
a
in
U},
is finite],
from
£(~,V).
Note that
fL £U(&,V).
let = [a ~ ~(A,V)
n = 1,2,3,...,
the semigroup
We will omit ring.
h
Note
Ia
has an adjoint
in
U],
let
Sn,U(g,V)
division
Vb. respectively.
has finite rank];
SU(&,V)
~U(&,V)
both with
la
~U(~,V)
~U(&,V)
and for
and
let
the ring structure
Further
Va
inequality
= {a ~ ~(~,V)
is a subring of
For a given
both with
be bases of
= ta E £U(b,V) structure
I rank a < n],
inherited
from this notation that 1.3 implies
from
~(h,V).
if there is no need for stressing
that
£u(V)
is a ring,
and
~u(V)
the is a
semigroup. A nonempty
subset
b ~ I, we have and
~n,u(V)
PROOF.
is an ideal of
~u(V).
£v,(V)
It suffices
For every
Then
f E V*,
bf:V ~ &
follows easily
of a semigroup It follows
1.1.4 LEMMA.
V*.
I
ab,ba ~ I.
to show that every element let
bf
and is linear since both (f E V*)
The adjoint of
a E £(V)
in
a* a
the adjoint of
S
~u(V)
a
a
in
be the function given by
b:f ~ bf
of
is an ideal of
if for any
is an ideal of
a E S, ~u(V)
= ~(V).
that
by
S
from 1.3 that
V*
in any
a
and
f
are.
is the adjoint of
£(V)
has an adjoint
x(bf) = (xa)f Hence
bf ~ V*
a
in
V*.
is called the conjugate
of
a.
U, if it exists,
and in particular
and it
We will denote the conjugate
(which should cause no confusion). I.i.5 LEMMA. £u(V) PROOF.
Let
U
be a t-subspace o f
= is ~ ~(V) I a*U ~ U For
a E ~u(V)
let
where b
a*
V*;
then
i__~s th___~econjugate o f
be its adjoint
in
U.
in
(x ~ V).
a}.
Then for any
u E U,
(va,u)
so that
= (v,bu)
a*u = bu E U that
observe
Hence
a*IU
for
and
(U,~,V),
A
and
B
function on a set values
1
or
multiplicative with zero
if
semigroup
of
~.
references
1.2
If
(U,A,V).
Let
for each
i E IU, let
ui
in
Note
[i,0,~]
VX
For
If
The Kronecker ~
is a ring, ~
For
of
~
[i,~,k]*
[i,x,X]*
ring, ~
space)
means
the
([i], Chapter
II), Jaeobson
([1],§§7,8,9).
subspsces
in
Ui;
spaces
Ui
let
IV
of
U, and
be an index
~ ~ IV, let
v~
be a
the transformation
(v E V),
in
V for
(1)
for any
i E I U, ~ E I V •
can be uniquely written
in the form
ow
U.
We have ,u )
is the adjoint o f
is the conjugate o f
For a fixed
(v[i,v,X],u) implies
the
(or a ring or s vector
V, and for each
(2)
(u E U) [i,x,k]
in
U, and
(f E V*) [i,v,X ]
and - -
which
is to take
g.
to be the zero function
that any nonzero vector
in V*. -PROOF.
then
The identity
will denote
a division
IV) and Kothe
of
[i,v,X]*f = fuiV(vkf)
u. i
V*,
6-function
i E I U, ~ ~ 4-, ~ E IV, define
= (v,ui)vv X
li,~,%]*u = uix(v
where
= la E A I a ~ B}.
l X.
be s fixed nonzero vector
(7 6 A-, % ~ IV; similarly
1.2.1 LEMMA.
where
image in
by v[i,~,~]
for some
to
~(V).
let there be given s pair of dual vector
subspacea
and define
A\B
be an index set of all 1-dimensional
V~.
V
it suffices
DENSE RINGS OF LINEAR TRANSFORMATIONS
fixed nonzero vector on
its natural
as a subring of
is a semigroup
elements
set of all l-dimensional
[i,~,~]
S
([7], Chapter
this section IU
ring.
inclusion,
U.
for this section are Baer
I,II,V,IX);
Throughout
~u(V)
S- = ~s E S Is @ Oj.
group of nonzero
([6], Chapters
in
is identified with
to identify
in any division
The general
a
will be denoted by
O, we let
multiplicative
U
(v E V)
For the opposite
of
any sets, we write X
0
= v(a*u)
a*U c U.
is an sdjoint
1.5 gives an easier way For
= v(bu)
(2).
u E U, we have for any
= ((v,ui)xvk,u) Further,
= (v,ui)x(vk,u)
for a fixed
(3) f
is the natural
imase o f
ui v E V, = (v,uiv(vk,u))
f E V*, we have
for any
v E V,
(v[i,~,k])f
= ((v,ui)wv)f
= (v,ui)v(v~f)
(Vfu'1)~(v f) = V(fui~(vkf)) which proves
(3).
The next result
is due to Gluskin
1.2.2 THEOREM.
Every nonzero element of
the form
[i,~,k]
function
exists
Let
~ E IV
into
A-
such that a
can be uniquely written
Since
xa = (xf)v X
is linear,
in
each such
$2,u(V).
0 # a E $2,u(V).
Since
~2,u(V)
i E IU, ~ E A , X { I V , and conversely,
is a nonzero element of
PROOF.
V
for some
14].
the range of
(x E V), where
so is
f
and thus
a
is 1-dimensional,
f
is some function mapping
f E V*.
For any
there
u E U, we
get (x,a*u)
= (xa,u) = ((xf)vk,u)
Since
a ~ $2,u(V),
u E U
such that
such that
we must have
: x(f(vk,u))
a*u { U
(v,u) # O, we see that
xf = (x,ui~)
for some
a =
and all
Choosing
Then there exists x 6 V.
i 6 1U
Consequently
(x E v),
[i,v,k ] .
Suppose (x,uj) @ 0
f(vk,u) E nat U.
f 6 nat U.
~ E h-
xa = (x,ui)vv k = x[i,~,k] so that
so that
(x ~ V).
that
[i,~,X]
so that
There exists 0
(x,ui)~v I = ( x , u j ) 6 v
exists
u E U
such that
i = j.
But then also
The converse
= [j,6,D] # O.
(v ,u) # 0
so that
and thus
x E V
such that
X = ~.
ui~(vk,u ) = uj6(v
Similarly, ,u ) ~ 0
there
and thus
~ = 6.
follows
easily by (1), bilinearity
and nondegeneracy
of the
bilinear form, and (2). By direct calculation we see that multiplication [i,~,h]lj,6,~ ] = for al___!elements of
express
Gluskin
[4].
reference.
set
independent
Ul,U2, .. .,u n
corresponding
is given by (4)
~2,u(V).
1.2.3 SUPERLEMMA*. linearly
$2,u(V)
[i,~(v ,uj)6,~]
The next lemma is of fundamental without
in
Let
set
in
statement
U
importance.
It will be used repeatedly
This lemma can be found in Jacobson
(U,V)
be a pair of dual vector
Vl,V 2 .... ,v n such that
in
V
(vi,uj) = &ij
holds if we interchange
"*Name given to it by some students
there exists for
in class in admiration
spaces.
a linearly
For any independent
i -< i, j ~-- n.
the roles of
U
and
[5], and later in
and
The V.
of its many applications.
PROOF, such that
The proof
is by induction
(Vl,U) = o # 0, so
Vl,V 2,.. .,Vr,Vr+ 1
u I = uo
on -i
be s set of linearly
exist
Ul,U2, ...,u r
u E U
define
in
U
(Vk,~) = (Vk,U)
-
If
n = i, then there exists
independent
such that
r ~ = u - ~ u.(v.,u). i=l i l
n.
has the property
(vi'uj)
vectors
= 6i~i
for
(Vl,Ul) in
V
= i.
u E U
Next
for which
1 _< i, j _< r.
Then for
1 < k < r, we obtain
~ (Vk,Ui)(vi,u) i=l
= (Vk,U) - ( V k , U ) = 0
let
there
For every
(5)
and (Vr+l,U)
r ~ (v +1,u.)(v. u) i=l r i i i'
= (Vr+l,~)
+
= (Vr+l,u)
+ (
¢
"=l
(Vr+l,ui)vi,u).
(6) r
If
-- = 0 (Vr+l,U)
dicting some
linear
u; let
for every
independence
of
tr+ I = uT
It follows
that for
Now suppose and
1 < k,j < r,
(Vr+l,Uk)
Vl,V2,
..,v n
are vectors
in
Hence
(Vr+l,~)
by (5) we have
= (Vr+l,~k) + (Vr+l,Uk)
1 _< i,j _< r + i, (vi,tj)
that
Ul,U2,...,u n
Vl,V2,...,Vr,Vr+ I.
For
= O, and by (6) we obtain
~ (v r+l ,ui)vi Vr+ I = i=l
u ~ U, then (6) implies
= 6ij
where
is a linearly U
for which
independent
if
(vi,u j) = 6ij.
= O.
1 < j < r.
set of vectors
in
If j=l~u.o.j3 = 0,
i < i < n, we have
tablishes
linear
of
which
es-
Ul,U2,...,Un.
The proof of the last statement For
for
= (vk,tr+l)
so that (Vr+l,~k) t.j = ~j
then for
1.2.4 THEOREM.
= ~ # 0
(Vk,Uj)
n n 0 = (v., ~ u ~.) = ~ ( v . , u ) ~ . = o i i j=l J J j=l l 3 J
independence
contra-
of the lemma
is symmetric,
n = 1,2,...,
~n+I,u(V)\~n,u(V)
= {k~=nl[ik,~k,Xk ] lik E IU, ~ E a-, X E I V , nminimal]. n
PROOF. n minimal. We want
First
let
a =
By 2.2, each
to prove
that
with
k~=l[ik,~k,Xk]
[ik,~k,~k]
i k E IU, ~ E & , X k E IV, and
is an element
rank a = n; we will
show that
of
~u(V),
so that
V%l,...,v
n
a E £u(V).
form a basis of
Va. Suppose
that
v%
,...,v%
~iV~l = 6 2 v ~ 2 + ... + 6 ~ V X m of notation).
Then (x~
wh~re
for any
Uil)~iV~l
are linearly
=
all
dependent;
we may assume
~i # O, m ~ n, ~i ~ 0
x E V, (x'uil)~2v~2
+
..
•
+
(x,uil)6mVXm
so that [il,Yl,~l] Consequently
= [ii,62,X2] + ... + [il,~m,l m] .
that
(by a suitable
change
V
i0 n
m
a = k~=2[ik,~k,hk ] + j~=2[il,~j,hj]. For any
x E V
and
2 < j < m,
x([ij,vj,hj] + [il,6j,h j] = (x,uij)Vjvxj + (x,uil)6jvxj = (x,uljvj•. +Uil~j)vh=j = (x,u~j)~jvhj for some
%j E I U
and
~j ~ A, so that m
n
a = k~_2[£k,~k,hk ] = contradicting
the minimality of
A similar proof shows that Further,
Uil
Thus 'Uin
v
,...,v h are linearly independent. hi n are linearly independent.
n x E V, xa = k~=l(X,Uik)~kVhk
for any
subspace generated by i < k < n. -i y = ~k x; then
There exists
x E V
n -1 ya = j~__l(Yk x , u i =
which proves that each
vh
j
such that
)?jvh.
Va.
is in Va
Therefore
proof shows that each Conversely,
let
a
maps
rank a = n
Uik
is in
(x'ui.) = ~jk j
-1
j
= ?k YkVhk
and completes
k a basis of
and thus
V
into the
Vhl,...,v hn
Let Let
n.
+k=m+l~ [ik,Vk,hk]
and thus
a'U, so
a E ~n+I,u(V) \$n,u(V)
=
for
i < j < n.
vh k
the proof that
v
,...v~
is n A similar
hi a E ~n+I,u(V) \~n,u(V).
Uil'''''Uin and let
is a basis of ~iVhl ,
,OnVhn
a*U. bea basis of
Va.
Then v , . ° . , v is also a basis of Va. For any x 6 V we have n hi n xa = ~ (xf,_)v, where fl,...,fn are some functions, fk:V ~ g. By linearity of k=l K Ak a
and linear independence
of
v
,..•,v h , it follows easily that each fk is n linear, so that fk E V*. x E V, u E U, we have n n (x,a*u) = (xa,u) = ( ~ (xf~)v~ ,u) = X(k~=ifk(Vhk,U)) k=l ~ ^k n where a*u E U so that ~ f, (v, ,u) E nat U. There exists u k E U such that k=l ~ ~k n and thus fk (vh ,Uk) = 6jk for i _< j,k <__ n. Hence fk = ~ f.(v h ,Uk) E n a t U j=l j j _ J is the natural image of u ~ for some i k E I U and ~k E A . It follows that hi For any
ik k
n
n
xa = k=l~(x,uik)~kV~k = X(k=l~ Iik,Vk,hkl) not minimal,
then
a
and thus
admits also a representation
a =
n ~ Iik,~k,Xk].
k=l o f t h e form
If
n
m
a = j~l[tj,6j,~j]
is
ii with
m < n
Therefore
m
and
this implies n
minimal.
n
ii)
contradicting
Let
a =
}
rank a = n, n >
m.
then the followin$
are equivalent.
,...,u i }
are both linearly
n
independent.
rank a = n .
In such a case, V X l , . . . , V k n so
form a basis of
Va
and
U.zl,...,Uin
form a basis of
rank a = rank a* = n.
PROOF.
i) = ii).
ii) = iii). Hence
~ [ik,Nk,Xk]; k=l
~Uil
and
n
a'U,
that
is minimal.
(~Xl ..... v x
iii)
the hypothesis
is minimal.
1.2.5 COROLLARY. i)
In the first part of the proof, we have seen that
rank a = m ,
that
This was established
There exist
xka = ~kVkk E V a
xk 6 V
and thus
linear combination
of
that
is a basis of
v
,...,v k
Xl
such that
(Xk,U i ) = 8kp for 1 ~ k,p ~ n. P Since also every element of Va is a
Vkk 6 Va.
Vkl,...,Vkn
and these are linearly Va.
In particular,
independent,
it follows
rank a = n.
n
iii) ~ i).
This follows
The remaining If
in the first part of the proof of 2.4.
~
subring of
statements
is a ring and ~
immediately follow
A
A.
+ a I+ ...+ a --
group closure of
--
--
from the proof of 2.4.
is a subsemigroup
generated by
are of the form
from 2.4.
Since with
A a
n
of
~'~, by
is s semigroup, ~ A, so
C(A)
C(A)
we denote
the elements
is actually
of
the C(A)
the additive
i
A.
1.2.6 COROLLARY. ~u(V) = nU=I~n,u(V) = C(~2,u(V)) an__~d ~u(V)
is a right ideal of
PROOF.
The first equality
and the fact that
-[i,7,k]
opposite
follows
£(V). is obvious.
The second equality
follows
from 2.4
so that with each element, $2,u(V) contains n its negative. If a E ~u(V), then a is of the form a = ~ [ik,Tk,kk] , and (3) k=l implies that for each k, [ik,~k,kk]*V* ~ nat U, so the same is true for a'V*. The inclusion
=
= ~a E ~(V) I a'V* ~ nat U}
[i,-~,k]
from 1.5.
If
(ab)*V* = a*(b*v*) ~ nat U, and hence 1.2.7 DEFINITION. let of
i (A) B
For
be the idealizer
having
A
A of
as an ideal.
a ~ ~u(V)
a subsemigroup A
in
and
b 6 £(V),
(subring)
B, i.e.,
of a semigroup
the largest
for all
a 6 A}.
(ring)
subsemigroup
It is easy to see that in both cases
iB(A ) = [b E B I ba,ab E A
then
ab 6 ~u(V). B,
(subring)
12 1.2.8 PROPOSITION. i)
ig(V)(~2,u(V))
PROOF•
its natural
image in
vk E V
letting
Let
V*.
i~(V)($u(V))
is an ideal of
i) By 1.3, $2,u(V)
gu(V) ~ ig(V)(~2,u(V)).
exists
ii)
= gu(V).
gu(V)
a E i~(V)(~2,u(V)),
It suffices
such that
vku ~ 0.
~ = O(v~u) -I, we obtain
= £u(V).
to show that
Then
which
implies
that
0 # u E U, and identify
u = uio
a*u E U. for some
[i,v,k] E ~2,u(V)
[i,v,k]*u = uiv(vku ) = (uio)(vku)-l(vku)
Since
U
with
u ~ 0, there
i 6 U, ~ E ~ ;
and = uio = u
so that a u = a*([i,v,k]*u ) = (a*[i,~,%]*)u since by hypothesis we have ig(v)(~2,u(V)) ii)
= (a[i,~,k])*u E U
a[i,y,%] E $2,u(V).
Thus
a E ~u(V)
and hence
~ gu(V) •
The proof is an easy modification
of the proof of part i) and is left as
an exercise. 1.2.9 DEFINITION• Then
S
and any set
i < i < n.
2-fold
n
be a positive
is said to be a n-fold
Xl,...,x n for
Let
in
we say doubly
for every positive
1.2.10 PROPOSITION.
For
suppose
kk E I V
n ~
dim V < n, then
dim V ~
yl,...,y n
n
and let
i ~ k < n.
Further,
S
Tk ~ 0
and
i, ~n+I,u(V)
in
There exist
i < k,p < n. _
in
the subspace
a E ~n+I,U(V). ~n+I,u(V)
of
V
Then ui
The next
Let
a =
if it is n-fold
transitive independent
Yk = ~kVk k
E U ~
by default. set and
for some
such that
o k E 4,
(Xk,U i ) = ~kp~k P
[ik,~kl~k,~k];
then
So
Va
for
is contained
k=l
by
v i,
I < k < n, we obtain
~u(V)
This follows
.,v n
so that
rank a < n
and hence
xka = (Xk'Uik)~kl°kv~ k = Yk
of
is a dense rin$ of linear
immediately
theorem "locates"
linear transformation
and instead of
and
transitive.
1.2.11 COROLLARY. PROOF.
set xia = Yi
is n-fold transitive.
is n-fold
be a linearly
V.
_
generated
For
is n-fold
independent such that
is dense
P some
~(V).
[4].
~n+I,u(V)
Xl,...,x n
be any set of vectors and
a E S
we say transitive,
transitive.
is due to Gluskin
If
a subset of
integer n.
The next proposition
PROOF.
S
if for any linearly
V, there exists
Instead of 1-fold transitive,
transitive,
transitive
transitive
yl,.°.,yn
integer and
transformations.
from 2.6 and 2.10
all transitive
rank i; it appears
subrings
to be new.
of
~(V)
containing
a
13 1.2.12 THEOREM. linear
Let
transformation
~
of
be s transitive
rank i.
U = ~ui~ 6 V* I~ E ~, (where
[i,~,~]
t-subspace
of
PROOF. and
is written V*
with
for any
relative
By hypothesis
£(~,V)
contsinin$
U # 0.
for some
~ E ~-, ~ E IvJ
to the dual pair
(V*,&,V))
Su(V) = ~ N$(V) ~ Let
[i,?,~] E ~
there exists
a ~ ~
is the unique
~ £u(V).
with
~ # 0.
such that
For any
(~vx)a = ~v
.
~ E IV Hence
x E V, we have x([i,v,~]a)
so that
= (xui)v(v~s)
[i,~,B ] = [i,~,~]a ~ ~.
[i,~,D] E ~ Let
for any
0.
Hence
= (xui)~v D = x[i,~,~]
Hence
[i,~,~] E ~
with
V # 0
implies
that
~ E &, D 6 I V •
ui~,ujT ~ U
contains have
[i,~,~] 6 ~
the property
~ E ~, by transitivity
subrin$ o f
Then
and
t = u.o+u.~. If t = O, then t ~ U since U J t # 0, so t = Uk~ for some k ~ IV, , V E A-.
suppose
[i,o,~][j,~,~]
~ ~
for every
x([i,~,~] + [j,~,~])
~ E IV .
For any
We
x E V, we obtain
= (xui)ov ~ + (xuj)Tv~ = (x(ui~+uj~))v = (X(UkV))v ~ = x[~,V,~ ]
and thus U
[k,~,h]
is a subspace Let
0 ~ x6 V
[k,6,~] 6 ~ a ~ ~
= [i,o,~] + [j,T,X] ~ ~. of
V*
since
and
y ~ V.
for all
such that
t = Uk~ ~ U
closed under
Further,
There exist letting
z ~ V
y = ~v
k
proves
that
such that
such
and
which
scalar multiplication.
We know that there exists
~ E &, D ~ I V,
xa = z.
But then
it is obviously
that
zu k # 0
and
~2 = ~ ~ 52 V *(V)'
we
obtain x(a[k,(ZUk)-l~,~]) where
a[k,(ZUk)-l~,~]
particular, implies
if
that
E ~2
y ~ 0, there xu i # 0.
= (ZUk)(ZUk)-l?v
by 1.3.
Hence
exists
Since
32
= y,
is s transitive
[i,~,~] E ~2
such that
u. E U, it follows
that
U
semigroup.
In
x[i,v,~ ] = y
which
is a t-subspace
of
V*.
i
From the definition
of
U
and 2.4 it follows
32 = i [i,~,X] E $2,v,(V) Since
~2,u(V)
is an ideal of
by 2.8, part i), we have
~
By 2.6 and (7), we have and hence
~u(V) ~ ~ N ~(V).
gu(V),
l ui E U} = ~2,u(V). we have
~ ig(V)(~2,u(V)) ~u(V)
that
that = gu(V)
= C($2,u(V))
Conversely,
let
~2
(7)
is an ideal of whence
= C(~2) ~ ~
h~.
since
0 # s E ~ n ~(V).
~
is a ring
Then
n
a = k~= [ik,~k,Xk] i V%l ,...,v ~n
and we may suppose
is linearly
independent
that
n
and hence
Hence
~ ~ ~u(V).
is minimal.
By 2.5,
there exist
u j~ E U
the set with
the
14 property
(V%k,Uj%)
-i [jk,Ok ,%k] ~ ~
= 6k~O k
for
for some
1 < k < n
ok ~ 0
and
i < k,~ < n.
Consequently
and thus the element
• -i • -i a[Jk,O k ,Xk] = [ik,~k(V%k,Ujk)O k ,Xk ] = [ik,~k,kk~
is also in
~,
so by (7), we have [ik'~k'Xk]
for
1 < k < n.
But then
~ ~ ~ $2 (V) = ~ 2 = ~2,U (V)
a C C($2,u(V))
= ~u(V)
by 2.6.
Therefore
n a(V) E ~U ( v ) " Let
U~
theorem.
be a t - s u b s p a c e
Then
au(V) = au,(V )
u i o E U, t h e n quently
of
[i,~,k]
U c U~
V*
which a l s o
so
[i,o,k]
and by symmetry a l s o
The equivalence
of parts
E ~2,u,(V)
the rings in the preceding
of the
Hence i f
and t h u s
uio E U'.
U ~ C U, w h i c h p r o v e s u n i q u e n e s s o f
We are now able to characterize ways.
the requirements
a2,U(V) = a2,u,(V).
and t h u s a l s o
E a2,u(V),
satisfies
Conse-
U.
theorem in several
i), ii) and iii) is due to Jacobson
[3], part iv)
is new. 1.2.13 THEOREM.
The followin$
conditions
on a subrin$
~
of
£(A,V)
are
equivalent. i)
~u(V) c ~ c £u(V)
for some t-subspace
ii)
~
is a dense ring containing
iii)
~
is a doubly transitive
iv)
~
is a transitive
PROOF.
i) = ii).
U
a nonzero
ring containing
ring containing
This follows
of
V*•
transformation
of finite rank.
a transformation
a transformation
of
of finite rank.
rank i.
from 2.11 since a ring containing
a dense ring
is itself dense• ii) = iii). Trivially. iii) = iv). Suppose
n>
I
Let
n = min Irank a I 0 # a E ~j.
and let
basis of
Va
By double
transitivity
and let
xib = Yi
for
a E ~ Yn-i
be any vector in
there exists
i < i < n-2.
It suffices
be such that rank a = n.
b E ~
Then for any
xa = ~iXl + ... + ~nXn
V
with
such that
Let
to show that Xl,...,x n
the property
for some
~i E &,
xab = ~l(Xl b) + ... + ~ n _ 2 ( X n _ 2 b) + ~ n _ l ( X n _ l b) + ~ n ( X n b ) = ~IYl + "'' + ~n-2Yn-2 + (~n-i + ~n)Yn-i which
implies
that
rank (ab) ~ n-i Yn-i
we have
ab # O.
iv) = i).
(ab)
and
Since
= Xn-i b = Yn-i ~ 0
But this contradicts This follows
ab E 2.
from 2.12.
minimality
Yn_l a = Xn_ I,
Xnb = Xn_l b = Yn-l'
x E V, we obtain
of the rank of
a.
n = i.
be a
let
15 1.2.14 Exercises. i)
Let
~
be
finite field having
a
space of finite dimension n. both
IV ii)
and Let
IV,
mn_l ~
have
[ik,~k,~k] £ k=l be nonzero elements of ~U(~,V).
a)
(Uil
[l
b)
vXI
= B
and
~i
and
V*
(~,V) have
be a vector
mn
elements,
and
t,6t,~t] , with n and m minimal, t~l[J = Show that a = b if and only if m = n and
n × n matrices
• .. u i n ) = (Ujl
elements and V
elements.
a. =
there exist invertible
m
Show that both
b =
A
and
B
over
&
such that
"'" U j n ) A ,
,
LVXnJ c) ~i ~2
0 B =
A
. . 0
?n
The entries of the row and column matrices above are vectors, with square matrices over
A
and the entries of the resulting matrices iii) ef = 0
A set
idempotents minimal,
E
for any
are again vectors.
of nonzero idempotents of a ring e,f E E, e ~ f.
~
is said to be orthogonal if
Show that the sum of a finite number of orthogonal
is again an idempotent.
£ [ik,~k,)~ k] E ~u(V) with n k=l find necessary and sufficient conditions in order that a be an idempotent.
Show that for any
n >
there exists a set e = el+e2+...+e iv)
the m u l t i p l i c a t i o n
is performed according to the usual row by column rule
0, e E ~u(V)
el,...,e n
For
a =
is an idempotent of
rank n
of o r t h o g o n a l idempotents in
if and only if
~2,u(V)
such that
n
Prove that every nonzero ideal of a transitive subsemigroup of
g(V)
is
itself transitive. v)
Prove that every nonzero ideal of a dense subring of
~(V)
is itself
dense. vi) Va
Show that an element
s
of
~(V)
is idempotent if and only if
elementwise fixed. vii)
For
a,b E g(V),
prove that
frank a - r a n k b l <-- rank (s + b ) .
a
leaves
16 viii)
Let
(U,~,V)
be a dual pair, I
be a nonempty subset of
IU
and
S = {nit I i E I, T E A], T = [ li,~,~] ~ 3u(V ) I i E I]. Show that ix) a ring
S
is a subspace of
U
if and only if
The least positive integer
n
C(T) n 3u(V) = T.
for which
nr = 0
~, if such exists, is the characteristic of
have characteristic zero.
Show that
g
and
for all elements
~; otherwise
3u(A,V )
~
r
of
is said to
have the same characteristic.
For further results on dense rings of linear transformations see Dieudonn~
[2],
[3], Gluskin [4], Jacobson [3], [4],([6], Chapter IX, Section 10),([7], Chapter IV, Section 15), Ribenboim ([i], Chapter III, Section 2). found in Erdos
1.3
ONE-SIDED IDEALS OF
We fix a pair of dual vector spaces V~
Some related material can be
[i], Rosenberg [i], Rosenberg and Zelinsky [i], Zelinsky
a subspace of
[3].
3u(V)
(U,A,V), and for
UI
a subspace of
U,
V, we use the notation = ~a E 3u(V ) IVa c- V ' ,
~,(V')
a*U ~ U'}.
It is clear that
,%u,(v') = 1%,(v) n gu(v'). We will sometimes write Clearly
~u,(V)
also of
3.
3
instead of
is a right ideal and
3u(V ) ~U (V ~)
(1) and
32
instead of
32,u(V ).
is a left ideal of
We will consider only right ideals and subspaces of
$1u(V) U.
lation and proof of dual statements for left ideals and subspaces of left as an exercise; for this two different proofs are possible:
and thus
The formuV
will be
direct by re-
taining the notation already used, or using 1.2 one may consider the isomorphic copy of
~
in
£(U)
1.3.1 LEMMA.
and then dualize the proofs to be presented here.
We have n
~,(V') PROOF.
Let
= [k~=l[ik,Wk,Xk ] =
By symmetry and using (i), it suffices to show that n ~ , ( V ) = ~k~__l[ik,Vk,Xk ] lUik E U ~, ~k E A, vXk E VJ.
0 # a E ~u,(V).
Then
a*U c U I --
so that that
l Uik ~ U j, ~ E A, VXk E V'}.
and
a =
i
[ik,~k,k k]
where
(2) n
is minimal
k=l
v
,...,v X are linearly independent by 2.5. There exist u. E U such kk n Jp (Vkk,Ujp) = 6kpO k for some ~k ~ 0 and i <_ k,p _< n. Hence for I ___ k _< n,
we have -i [ik,Vk,% k] = a[jk,o k ,X k] ~ ~ , ( V ) since
~UI(V)
is s right ideal of
~.
But then
[ik,~k,~k]*U ~ U ~
which implies
17 that u. ~ U ~ and thus ~k inclusion is obvious.
a
is in the right hand side of
(2).
The opposite
The next result is new. 1.3.2 PROPOSITIONthen
form,
(Ut,~,V ~)
Suppose that
form to
(UI,g,V t)
V~X U ~
From the definition of in
UI
alv,
If
~ l ( V ~)
is equal to
b ~ ~u~(V~),
sentation of
a*Iul,
b, we extend
b
(x E V). (u E U)
and
~
maps
have
a =
suppose 'that
~ k=l [ik'~k'~k ]
that
Uil
, ...
n
c
a~ # O.
Consequently
Fixing this repreV
by defining
defined by
the adjoint of
a
in
0 # a E ~ut(V~).
(U~,V ~)
for
U.
Consequently
Then by 3.1, we
1 < k < n, and we may
is a dual pair, and by 2.5 we have there exists
v E V~
such that
Hence
n va = ~ (v,u i )~kVx = k =I k k (v'uil)~iV~l and
is a homomorphism.
on all of
is zero, let
Since
i < k < n.
by
~uI(VI ).
are linearly independent
for
a
u. E U j, v E V~ ik Xk
is minimal.
,Uin
(v'ui k) = ~kl
with
~
~
n b = k~_l[ik,?k,Xk ] . =
Then the function
onto
To show that the kernel of
~
of the
a~ E ~ul(V~), where the adjoint of
it follows that
and it is clear that
is evidently
~ I ( V I)
the restriction
(a E ~ u , ( V ' ) ) .
to a function
n cu = k~=lUik~k(V~k,U )
i.e.,
Define the function
then by 2.6 we have
xa = k=l~(x,uik)~kV k
a~ = b
is a dual pair,
is nondegenerate.
~:a --
a~
is a dual pair with the induced bilinear
~uI(V I) ~ ~ut(VI).
PROOF. bilinear
If
the kernel of
~
~ 0
is zero and
~
is an isomorphism.
The next lemma will be quite useful. 1.3.3 LEMMA. PROOF. for any
If
i E IU
For any
a ~ £u(V),
a # 0, then
xa @ 0
and
Since for every
(xa,uj) ~ O.
for some
= (Y,Ui)?(vxa)
i E IU, there exists that
so that
y E V
[i,~,X]a ~ O.
Hence for any
= (xa,uj)~vx # 0
~2 a = 0
or
a~ 2 = O, then
0 # x E V.
Hence
a = 0.
x = ~v X
and thus
y E V, we obtain
y([i,~,X]a)
xa # O, it follows
if
= (Y,Ui)(xa). such that
Further,
~ E ~-, ~ 6 IV, we obtain a[j,~,X]
# O.
(Y,Ui) # 0
there exists
j E IU
x(a[j,~,X])
and we have such that
18 Recall
that a lattice
one-to-one
correspondence
isomorphism
of a lattice
which preserves
preserves
either meet or join or the partial
lattices,
consult
Sz~sz
1.3.4 THEOREM.
[i].
a l l risht
U'
and
isomorphism
(U I
is
of the lattice
B
is a
(it suffices
For matters
is due to Jacobson
that it
concerning [5].
subspace
a
o__ff U)
of all subspaces
of
U
onto the lattice
of
3u(V ).
We have already
Un
onto a lattice
The function
ideals of
PROOF.
order).
The next result
)<:U I ~ % l ( V ) is a lattice
A
both meet and join
are distinct
observed
subspaces
Hence
for any
that
[i,~,~] E % , ( V ) \ % s , ( V ) .
that
of
U'
[i,~,~]*U ~ U'
Consequently
and
UH
is a right
U, we may suppose
v k E V, ~ E A-, we have
For any subspaees
%,(V)
of
~<
ideal of
3.
If
that
u.l E U ' \ U '~. [i,7,k]*U ~_ U n, so
and
is one-to-one.
V, we have
% , (v) n ~,, (v) = 3u,n u" (v) which
implies
suffices form
that
)~ is a meet homomorphism.
to show that
%,(V) Let
){
for some
31
is onto,
U'.
be a nonzero
Note
right
i.e., that
If
0 # a E ~, ~i~.
then
ideal of
a3 2 ~_ ~ N 3 2
By 3.3, we also have
zero right
ideal of
32.
[i,6,p] Hence
[i,?,k] E ~I2
p ~ I V,
u.j E U
since
= [i,v,k ] [j,
with
V # 0
As in the proof of 2.12, ~2 = ~
for some '~
that
such
the proof,
ideal of
3
it
is of the
$, and let ,{ ~ A-, k ~ IV].
is a right
a32 # 0, which
It follows
P E IV, then there exists
to complete
0 = ~o(V).
U' = [ui(~ ~ U Io E A , [i,~{,k] ~ 31
of
Hence
that every right
implies
U' # 0.
that
ideal
that
If
and
32
is an ideal
312 = ~ ~ 32
is a non-
[i,~,k] 6 312' 6 E A,
(v ,uj) = o # 0, so that
o-1 -1 6 V ,p] 6 ~2"
implies
that
it follows
N 3 2 = ~ [i,v,k]
that
[i,6,p] E ~2 U~
for any
is a snbspace
of
6 ~ fl, U
and
l ui E U ~} n
%,(V) = C('~2)
so that minimal.
There exist
i <__ k,p _< n.
c_~. u.JP 6 U
t~
Therefore
0
# a = k~__l[ik,Vk,kk ] _
such that
(Vkk,Ujp)
be in
= ~kp(~k
31
with
for some
n dk # 0
and
Hence [ik,Vk,kk]
since
Let
is a right ideal. 31 = % 1 ( V ) .
= a[Jk,okl~k,kk ] ~ Thus
u. ~ U ~ ik
which by 3.1 implies
that
a E "~u'(V)"
19 1.3.5 DEFINITION.
For
~
a ring,
let
~2 = { i a.b. I ai,bi ~ ~i}. i=l z l It is clear that a,b E ~t.
j~2
is an ideal,
and that
A ring
~ a
is called simple of a semigroup
if
0
is said to be regular
(or a ring)
S
is regular
regular.
This concept was introduced
1.3.6 LEMMA. PROOF.
A semigroup
Let
.,Uin } v
for some
(regular)
by von Neuman
~VXl,...,VXn}
and
are linearly
I
n
a = aba are
rings are also
minimal.
independent
n
Then both
by 2.5
E V and u. E U such that (v , u ) = ~kpOk, ~k ]p ~k ip o k ~ O, ~k ~ O, and 1 < k,p < n. Hence
a
if
if all its elements
[i]; regular
is a regular simple ring. n 0 @ a k ~ l [ i k ' ~ k ' % k ] ~ ~u(V) with
n
and thus
~ 2 ~ O.
rings.
There exist
(v k,Ujp ) = 6kpG k
n
(k~=l[ik,Yk,kk])(p~=l[Jp,';l~;l°;1 If
for all
gu(V)
, . .
nonzero
ab = 0
are its only ideals and S
b E S.
known as von Neumann
and
(or a ring)
for some
vectors
if and only if
Such a ring is called a zero ring.
An element
~Uil
,~2 = 0
n
Dp])(k~=l[ik,~k,kk ]) = k~__l[ik,~k,kk ]
is regular. is a nonzero
ideal of
gu(V),
then by 3.4 we have
subspace
Ul
of
U, and by the dual of 3°4, also
nonzero
subspace
Vl
of
V.
V~ = V
and thus
I = ~u(V).
Consequently,
I = ~I(V)
Since
is regular,
~u(V)
1 = ~uI(V)
I = ~u(V I)
= ~ ( V ~)
for some
for some
so that
U ~ = U,
it is not a zero ring,
and
hence must be simple. 1.3.7 DEFINITION.
~£(~)
For
any ring,
I
the set
= 0
for all
a E ~j
~r(~) = [r E ~ I ar = 0
for all
a ~ ~]
=
~r E ~
is the left annihilator
is the right annihilator
of
of
It is clear that both appears
~
to be new.
ra
~,
~. ~(~)
and
~r(~)
are ideals of
~.
The next result
20 The followin$ statements
1.3.8 THEOREM,
concerning ~ subspace
U~
of
U
are equivalent.
i)
(U~,V)
is a dual pair.
ii)
~u~(V)
is a simple ring.
iii)
~%(~,(V))
= 0.
iv)
~,(V)
is transitive.
v)
~,(V)
is a regular ring.
If any of these conditions holds, PROOF. ~I(V)
i) = ii).
~ ~uI(V).
that
~(V)
Since
is simple. By simplicity of
iii) = iv).
Let
x,y E V
is the zero function.
that
For any
[i,o,X]s ~ 0
for some
~uI(V), we must have
with
So suppose
i E IUI , we have
[j,~,~] E ~uJ(V).
a ~ ~uI(V).
iv) = v).
Since
E ~u~(V). ~uI(V)
transformation of rank i. such that ~u~(V)
~(V)
= ~u~(V)
regularity,
Let
Hence
~uI(V )
a E ~uI(V)
that
is a dual p a i r
~e consider next principal 1.3.9 THEOREM.
[i,o,~] [j,~,~] ~ 0
(v ,uj) # 0.
~u~(V)
For any
0
o # 0,
implies for some
Thus
is transitive.
since all elements of
and thus there exists
where
where
and the hypothesis
is a nonzero right ideal,
[i,~,~]a # 0 (U~,V)
xO = y
x = ovk, y = ~v
and
= 0.
= ~(v x,uj)(v x,uj)-IO-I~v~ = y
u.l E U r ' ~ E ~
there exists
y = 0, then
But then
j E IU~
~%(~l(V))
it contains a linear
Thus by 2.12, there exists a t-subspace
is regular by 3.6, so
v) = i).
If
Hence
[i,o,l] E ~u~(V)
It follows that
[j,(v~,uj)o-l~,D]
x # 0.
y ~ 0.
x[j'(vX'u')-Io-I~'~]j where
~u~(V) ~ ~u~(V ). is a dual pair, 3.2 implies that
This proves the last statement of the theorem and by 3.6 implies
ii) = iii).
# 0.
then
(UI,V)
~(V)
U~
of
V*
are of finite rank.
is regular. v~ ~ V.
Then
such that u.j 6 U ~
[i,~,~] E ~ ( V ) ,
and by
[i,~,~] = [i,~,~]a[i,~,~].
such that
Hence
(v%,uj) ~ 0, which implies
ideals.
s E 5, th___~e~rincipal
risht ideal senerated b_~y a
equals a~ = ~a,u(V). PROOF. with
If
n minimal.
a = 0, the assertion
But
is trivisl,
Then s typical element of
aS
so let
0 ~ a =
is of the form
~ [ik,~k,Xk] k=l
21 m
n
m
at~=l[Jt,~t,P t] = (k~='l[ik,~k,Xk])(~l[Jt,6t,Pt ])
(3)
= k= ~ I t~=ml[ik,~k(VXk,Ujt ) ~t'~t ] with
Jt E IU, ~t
E ~, Pt ~ IV"
P q=~l[%q'°q'~q]
is of the form U~q = a*(Usq~q)
E a*Uo Hence for q Usq E U, Tq 6 A so that
for some n
U~q
On the other hand, a typical element of
with
u~
.
*
(a*u sq)~q = k~=l[lk,~k,~k ] _
Consequently,
for any
%a*u (v)
I < q < p, we have
n u sq~q = k=l~U.lk~k(V~&k~USq)~q.
x E V, we have
n P P P x ( ~ [%q,Oq,~q]) = ~ ( x , u ) 6 v = ~ ~ (x Ulk)Vk(Vkk,USq)TqOqV q=l q=l ~q q Vq q=l k=l ' " p n = x(~ ~ [ik,Vk(VXk,U )~ O ,~q]) q=l k=l Sq q q
q
so that P q~=l[~q,Oq,~q] E IU, "rq E ~ - , q a~ = ~a.u(V).
with
s
The set
s~
P = k=l ~ ~ [ik ,Vk(VXk, u s q )~ qOq, Vq] q=l "0q E I V .
A c o m p a r i s o n o f (3) and (4) q u i c k l y shows t h a t
is indeed the principal right ideal generated by
is clearly a right ideal and
a E aS
since
The next corollary is a supplement 1.3.10 COROLLARY.
U
$
a
since
a~
is regular by 3.6.
to 3.4.
The isomorphism
finite dimensional subspaces of of
(4)
x:U' ~ ~u~(V)
maps .the lattice of all
onto the lattice of all principal risht ideals
~u(V). PROOF.
Let
U'
be an n-dimensional
has a basis of the form for some a*U ~ U ~.
ok # 0
and
Uil , "''Uin
subspace of
Let
i < k,p ! n, and let
On the other hand, if
V~k E V
U
where
be such that
n . -i a = k~__l[ik,~k ,hk ].
u E U ~, then
n u = k~=lUik~k
n -1 n a*u = k=l ~ u.ik o k (v~k ,u.ik ) ~K = k~=lUikVk = u.
n > 0.
Then
U'
(V~k,Uip) = ~kpOk It is clear that
and thus
22
Hence
U' c a*U
and thus
u'x Conversely, a E ~.
if
~u,(v)
=
~
a*U = U'.
1.3.11 REMARK.
~a.u(V)
=
is a principal
Thus by 3.9, we obtain
correspond minimal
By 3.4 and 3.9, we have =
a~.
right ideal of
(a*U)x = ~a,u(V)
It follows
5,
then
~ = a~
from 3.10 that to l-dimensional
risht ideals,
for some
= ~.
and from its proof,
subspaces
of
U
that they are all the sets
of the form R i = {[i,~,X] Similarly
all minimal
IV ~ 4, X ~ Iv3
left ideals are ~iven by
Lk = [ [ i ' v , k ] Further,
L k = ~[i,o-l,k]
(k ~ IV). where
and
(vk,ui) = o # 0
is sn idempotent.
1.3.12 PROPOSITION. has no proper PROOF. vectors
I i ~ IU, ~ ~ ~j
R i = [i,o-l,k]~,
[i,o-l,k]
(i E IU).
V
is a finite dimensional
vector
space,
then
V*
t-subspaces. Let
in
If
dim V = n, let
V, and let
U
be a linearly
Xl,...,x n
be a t-subspace
of
V*.
independent
set of
There exist
f. E U J independent.
such that
x.f. = 6.. for i < i,j < n and fl'''" f are linearly Hence i j lj --' n dim U ~ n and since U is a subspace of V* and dim V* = n, it follows that U = V*. 1.3.13 COROLLARY. finite dimension
If
(U,V)
is a dual pair and either
U
or
V
is of
n, then the other is also and each can be identified with the dual
of the other. For n xn
~
any ring and
matrices
over
~;
1.3.14 COROLLARY. satisfies
d.c.c,
n
a positive
integer,
also write d.c.c, If
~
denote the ring of all n instead of "descending chain condition".
is a subrin$ of
PROOF.
Let
~
~U(&,V)
for either left or risht ideals,
the same finite dimension, ~ = ~(~,V) ~ ~ d i m V ' and risht ideals,
by
~
containin$
then both
satisfies
U
d.c.c,
~U(~,V) and
V
and
are of
for both
left
and all its left and risht ideals are principal. U1 ~ U2 ~
...
be a descending
chain of subspaces
of
U.
Then
by 3.4 we have that of
~ i (V) ~ ~U (V) ~ ... is s descending chain of right ideals 2 Since each ~ui(V ) is also a right ideal of ~ and ~ satisfies
Su(V).
d.c.c,
for right ideals,
there exists
i > n
and thus by 3.4,
U. = U
-either
1
U
or
dimensional,
V
satisfies
n
for all n
d.c.c,
such that i > n.
~ui(V ) = ~ n ( V ) Consequently
for all
by 3.4, or its dual,
-
V
is finite
and hence by 3.13 both are of the same finite dimension,
for subspaces,
so either
U
or
and we may
23 write
V* = U.
Using 1.4 we obtain
Again by 3.4 and its d u a l
~
£u(V) = ~(V)
satisfies d.c.c,
and hence
~ = ~(V) = ~(V).
for both left and right ideals.
The
last statement follows from 3.10. 1.3.15 COROLLARY. risht ideals; V
(and thus also PROOF.
~u(V)
satisfies d.c.c, for principal
it satisfies d.c.c, for risht ideals
follows from 3.10 since a vector space evidently
for finite dimensional subspaces.
statement follows from 3.14, and the converse Here,
as before,
PROOF.
U.
Let
such that On
A
Fo___rr a,b E ~(U), A
be a basis of
b~ = u
define
a
if
function
d
aU ~ bU,
aU.
(possible since
The direct part of the second
from 3.10 and the first statement.
the "left" part of the statemenls
1.3.16 LEMMA.
U
if and only if
U) is finite dimensional.
The first statement
satisfies d.c.c,
left and principal
(or left ideals)
follows from dual statements. then
For every
u 6 A
A c aU c bU).
by letting
du = ~,
and extend d to a linear transformation on U. n ax = k =~lukOk= for some u k E A, o k E g, and thus n
a = bc
for some
choose an element
Extend
A
let
map
d
For any
n
c E ~(U).
to a basis B\A
~ B
of of
onto zero,
x E U, we have
n
= bd(~k=lUkOk) = b[~l(dUk)°k]= = b(~l~k°k)=
bd(ax)
n
Letting
c = da, we get
For
S
a semigroup
a = bc
with
(or ring)
c ~ ~(U).
let
ES
be the set of all idempotents of
S
partially ordered by e < f Let V'
~(U,V)
if and only if
be the set of all dual pairs
is a subspace of
including the pair
V, dim V ~
If
(A,~) onto
a ! a~
(U~,V~), where
is finite,
under
U~
is a subspace of
U,
the induced bilinear form,
(O,0), partially ordered by
(U~, V~) ! (U#, V~j)
(A,!)
e = ef = fe.
and
(B,~)
(B,!)
if and only if
are partially ordered sets,
is a one-to-one
if and only if
function
~
a~ ! a ~ "
The next result is new. 1.3.17 THEOREM.
The function
~ : e -- (e*U,Ve) is an order isomorphism of
U ~ ~ U ~, V ~ ~ V'.
E~u(V )
~
defined by (e E ESu(V ))
onto
~(U,V).
mapping
then an order isomorphism of A
onto
B
such that
24
PROOF.
Let
k=l There exist
[ik,Wk,hk]
xk ~ V
-i = 6kp~k .
(Vkk,Uip) u
. i I ' "''Uin Let
V'.
=
2
There exist
Since by 2.5,
be an element u i P E U' Uil
'Uin
3.13 also
: n
and hence
Letting dently
dimnU'
e = k=l~ [ik,~[l,kk],k Ve ~ V'.
If
it follows
of
~(U,V)
the set
so that
equation
Similarly
e,f E E~
e*U = U ~
and assume
Hence implies
Let
c 6 ~(U),
lent to:
e~U ~
which
~k ~ 0
dim V' = n, by
Since
forms a basis of
since
(Vkk,Uip)
for some
and
Uj
= 6kpTk.
ok 6 A
Evi-
and thus
kk
e = ef
e~p = f~. and
f'U, Ve ~ Vf
1.3.18 PROPOSITION.
fe
the first gives
means
that
that
to:
e*U = f*U
a ~ £(V) = f'e*.
3.2,
The last
e = f. which can be
e* = f*c
the last statement
e = df is equiva-
e~e ~ &rank e"
3.13,
and 1.4 we obtain
= ~e,u(Ve) = £(Ve) ~ ~rank e"
which
and
eq0 < f~p.
(i), 3.17,
= ~(Ve),(Ve)
and
e = ef = fe
0 # e E E~, we have
= ~e,u(V)~u(Ve)
~e*u(Ve)
Ve = Vf for some
is in turn equivalent
which means
For ~
e~ = (U~,VJ).
e* = f*b = f*(f*b)
By 3.16 and its dual,
Using 3.9 and its dual, e~e = e $ ~ e
Then
e* = f*b
and
e < f
which
d E £(V).
and therefore
together with
The relation
e* = f'e*
for some
PROOF.
that
f = ae
f = ae = (ae) e = e = fe
e,f E E~°
written as
be a basis of
n -i n n ~ (x,u i )~k v = ~ (~ ~ v ,Uik)~kl = ~ OkV = x k=l k ~k k=l p=l P kp Vkk k=l kk
by 3.16 and its dual yields b 6 £(U).
,Uin
and
is a dual pair.
for some
independent
e2 = e
Ve
v kl ,. ..,Vkn
= 6kpT k
n ~ okv
x =
so we must have
(e*U,Ve)
and let
k=l
Ve = V ~.
Let
that
Uil
we see that then
so that
form a basis of
(Vkk,Uip)
--
xe =
Then
1 < k,p ! n
independent
Vkl,...,Vkn
are linearly
x 6 V',
minimal
= p~__l~k(V>k,Ulp)~pVkp
such that
i < k,p < n, and
for
are linearly
e'U,
n
~n
xke = xke
form a basis of
(U',V')
(Xk,U i ) = 6k p P
v 1 ,...,v I n
the vectors
with
2 = k=l ~ p=l i [ik,~k
= e = e
such that
~kVkk By 2.5,
n ~ [i,,~k,~k] ~ E~ k= 1 K
0 # e =
25 1.3.19 DEFINITION. every finite subset isomorphic to
~
F
Uj
~
is a locally matrix rin$ over a ring
there exists a subring
n
Su(A,V)
be the subspace of
U
e E 3
of the proof of 3.10 is an idempotent in
quently
a i E ~U j(V)
aa i = a(abi) = ab i = a i n ( O Vs.) U Vs. i= I l
generated by
an idempotent Letting
b E 3
e = a+b
~
3
and
a
Un a~fU. i=l l
U'
Then
is
constructed in the first part
a i = sb i
1 < i < n.
al,s2,...,s n
a I, ...,a n E eSe.
a3 = ~ , ( V ) .
for some
Next let
V~
b.l E 3.
It Conse-
be the subspace of
By a dual argument, we conclude that there exists
such that
3b = ~u(V ~ '), a i = aib for i < i < n 2 - bs, we obtain e = e, ea = a, be = s, and hence
a i = aaib = (ea)ai(be) 6 e~e
for
([7], Chapter IX, S e c t i o n 15, Theorem 3).
theorem can be found in Faith and Utumi
and
s = ab
1 < i < n.
The above theorem in c o n j u n c t i o n with 11.2.8 implies Litoff's Jacobson
F
A.
with the property
and hence for
if for
containing
such that
generated by the set
finite dimensional and the linear transformation
V
of
to show that for every finite set
3, there exists an idempotent
follows from 3.9 that
~
~
n.
is a locally matrix rin$ over
By 3.18, it suffices
of elements of Let
of
~
for some positive integer
1.3.20 THEOREM. PROOF.
A ring
theorem,
see
Different proofs of Litoff's
[i] and Steinfeld
[i]. We will prove a
g e n e r a l i z a t i o n in 11.2.19 due to Hotzel. 1.3.21EXERCISES. i) ii)
iii)
Find all zero rings without proper ideals. Show that for any semigroup
(or ring)
S,
the following are equivalent.
a)
S
b)
every principal right ideal is generated by idempotent,
c)
every principsl
be a ring and
~
be s right ideal of
~
~
is a right ideal of
~.
v) of
~
left ideal is generated by an idempotent.
Show that every right ideal of iv)
Let
is regular,
Prove that
g(V)
which is a regular ring.
is s regular semigroup.
Show that dense subrings of
3u(V )
are precisely the simple right ideals
3u(V ) . vi)
Let
S
be a semigroup and
e
be an idempotent of
S.
Show that
G e = {x E S Ix = ex = xe, e E xS n S x ] is a subgroup of Ge
S
containing every subgroup of
is called a maximal subsroup of
S.
S
having
e
ss its identity;
26 vii)
Let
e E E£(V).
Prove
that
1 - e
is an idempotent, N e = V ( I - e),
V = N e G V¢, G e = {a E £(V) INs = Ne, Va = Ve} having
e
as its identity and
transformations on viii)
For a)
ix) if = xf
b)
(a*U,Va)
c)
V = N a ~ Va.
linear
(f E V*).
~u(V),
is a dual pair,
(&,V)
and
x E V, let
Prove that the m a p p i n g
~:x ~ x
is called the natural isomorphism of
x)
g(V)
show that the following are equivalent.
is contained in a subgroup of
For any vector space
V**;
is the maximal subgroup of
is isomorphic to the group of invertible
Ve.
a E ~u(V), a
G
V
x:V* ~ g
be defined by
is an isomorphism of
into
V
into
V**.
Prove that the following statements concerning a vector space
V
are
equivalent.
(Hint:
a)
V
b)
V*
is finite dimensional.
c)
V*
has no proper t-subspaces.
d)
~:x - i
e)
~(V)
If
xi)
is finite dimensional.
V
maps
V
onto
V**.
is a simple ring.
is infinite dimensional,
Let
show that c), d), e) fail.)
be a d i v i s i o n ring and
n
be a positive integer•
For the ring
An
a)
find all left (right) ideals,
b)
w h i c h of them are minimal, maximal,
c)
show that
£
is a simple,
n for both left and right ideals.
principal,
simple?
regular ring satisfying d.c.c,
and a.c.c.
Use the results in the text but also give a direct proof w h e n e v e r you can. xii)
Let
e
f = e+g
if
xiii)
If
and
f
for some idempotent •
g
in
is a dense subring of
finite rank, prove that ideal of
be idempotents in a ring
~ ~ $(V)
~
£(V)
~.
orthogonal
to
e < f
if and only
e•
containing a nonzero transformation of
is an ideal of
~
contained in every nonzero
~.
For more i n f o r m a t i o n on this subject see Behrens •
Show that
i
Dzeudonne [2], G l u s k i n Chapter IV, Sections
[4], J a c o b s o n
([i], Chapter II, S e c t i o n 3),
[5],([6], Chapter VIII, Sections 3,4),([7],
15,16), R i b e n b o i m
([i], Chapter III, Sections
2,3), Rosenberg
[i].
27
1.4
IDEALS OF
1.4.1 DEFINITION. ideal if
g(V)
A nonempty
x E S, y E I
implies
AND PRINCIPAL FACTORS OF subset
xy E
I
I
of a semigroup
(yx E I).
If
I
is an ideal of
S
S\I,
~nd
with
I @ S, then the set
if
xy~
I
if
xyE
I
I
is a left (risht) is an ideal of
S
if
S.)
the multiplication:
=fxy
S
(Hence
and only if it is both a left and s right ideal of
is not an element of
[~u(V)
Q = (S\I)
for
(J 0, where
0
x,y E S \ I ,
x*y
x*0 is a semigroup denoted by Let S
called
= 0-0
the Rees Nuotient
= 0 of
S
relative
to the ideal
I
and is
S/I. J(a)
(i.e.,
= 0*x
denote
the principal
the smallest
ideal of
S
ideal generated containing
by an element
a).
a
of a semigroup
It is easy to see that
J(a) = a U a S U S a U SaS, and if
S
Let
has an identity,
J(a) = SaS.
I(a) = {b E J(a) I J(b) # J(a)j.
empty or is an ideal of as an ideal of I(a)
then
= ~,
let
For
C
J(a),
S.
One verifies
The Rees quotient
is called the principsl
J(a)/I(a)
easily that
J(a)/l(s), factor of
where
S
l(a)
l(a)
containing
is either
is regarded a
(if
= J(a)).
+ than
a cardinal number,
let
¢, and for a c-dimensional
C
denote
vector
space
the least cardinal number (g,V),
greater
let
(a,V) = ~b E g(~,V) I r a n k b < ~} together with
the multiplication inherited from g(~,V), where Q is a cardinal + 0 < ~ < ¢ . As usual, we will omit & if there is no need for stressing
number,
the division
ring.
The next result appears
1.4.2 THEOREM. --°f g(V)
PROOF.
g (V)
since
generated g(V)
gb+(v) •
vector
is s principal
That
find principal J(b)
The set {gQ(V) I0 < Q < ¢+j
for a =-dimensional
addition,
g (V)
ideals. by
to be new.
b
space
b 6 g(V)
consists
has an identity.
g(V)
and let
of elements
the set of all idesls
g~(V) ~ gb(V)
ideal if and only i f
is sn ideal of Let
constitutes
V, and
follows
~
from 1.3.
rank b = b.
of the form
__if Q # b.
__In
is not a limit cardinal.
cbd
Again by 1.3, such an element
We will
first
The principal
ideal
where
c,d E g(V)
is contsined
in
28
Conversely, extend Vb by
A
let
to a basis
such
that
(possible
For each
a function
on
c
B\A, of
for any
d
8
choose
C\Aa,
x E V, we have
xa =
n
(xa)dbc Consequently,
° (
i(Yid))bc
a = adbc E J(b)
~ E V
xd = x8
for each
c
and extend ~ aiy i i=l
such
that
on
g(V),
and
into
~b = x
V.
it to a linear
d
Extend if
A8
x E AS,
transformation
~. E A, Yi E A, and I
n
J(b)
and
A
x E A, define
on
thus
n
° ° thus
Va
rank a ~ rank b
xc = x8 -I
for some
of
mapping
since
transformation
by letting
°
and
function
(possible
n
1
be a basis
an element
function
the set
A
be a o n e - t o - o n e
it to a linear
a
Let
independent
by letting
Define on
rank a < b-
Let
and extend
V
arbitrarily
Then
V.
x E AS,
define
C
then
is linearly
A8 ~ Vb).
to a basis
V.
of
A8
since
arbitrarily
define
B
the set
hypothesis).
Now
a E gb+(V),
i°l °iY
= g +(V). b
Next larger
let
than
for some
J
rank b
is a limit
assertions
Let from
But
~(V).
Jacobson
and
then
denote
The next
constitutes PROOF. C ( ~ 2 v,(V)) ,
be the least
with
proves
J = g
Q
(V).
g
then
number + b = Q
by the above.
c E J
that
cardinal
cardinal,
rank b = b
there exists
which that
the semigroup
g
can be
such
If
that
(V) c_ j.
Since
the
The proof of the r e m a i n i n g
(V)
found
provided in Baer
9),([7], Chapter vector
with
the a d d i t i o n
([i], Chapter
IV, S e c t i o n space
V,
inherited
V, A p p e n d i x
i),
17).
the set
+ or
~ <
c
the set of all ideals By 4.2 and
So by 4.2, £(V)
b
For a c - d i m e n s i o n a l
I~ = ¢
= ~(V),
Q
is left as an exercise.
+ ~£Q(V)
let
is not a limit
then
J
we have
IX, S e c t i o n
1.4.3 COROLLARY.
a
for any
rank b < a,
corollary
([6], Chapter
If
b ~ J(c) c
is trivial,
of the theorem
£a(V)
b E J.
J = J(b)
cardinal
inclusion
ideal of
for all
b, and hence
rank b < rank c. opposite
be any
no
1.3, g
(V)
has no other
and of
~
is an infinite
cardinal]
£(V).
these
are ideals of £(V). Since by 2.6, + for ~ ~ ¢ and Q finite is closed under
ideals
than
those
listed
in the statement
addition.
of the
corollary.
We have
seen in 2.2 that
the nonzero
written
in the form
[i,~,X]
where
written
in the form
[i,0,k]
for any
elements
of
This
is a special
case of
=
[i,~(vk,uj)6,~].
the following
can be uniquely
I V , the zero
function
i E IU, X E IV, and the m u l t i p l i c a t i o n
given by [i,~,k] [j,~,~]
$2,u(V)
i E IU, ~ E ~ , h E
construction.
is is
29 1.4.4 DEFINITION. element not in with
Phi
exists that
Let
the value at
% E A
I
G, and let
By
denote
(i,0,~)
We will prove next that for all finite
group
Ln(&)
of
positive
~ ~ A
Then
Let
i E I
such
is a semigroup
is isomorphic
a division
ring and
matrices
over
n
as a matrix
and
to a Rees matrix
&
a positive
integer,
(under multiplication)
linear group of degree
n
over
the is
g.
is new. (&,V)
be a vector space , U
n < dim V.
! t-subspace o f
is an n-dimensional
subspace of
B
B
Vl, ... ,v n
fi__~x! basis
V*, n
Let
n U = [A I A
A.
i ~ I, there
can be considered
subspace
of __
be an
with multiplication
S = ~°(I,G,A;P)
P = (pxi)
is an n-dimensional
B E nv
0
P:A × I ~ G °,
there exists
nV = {BIB
A A .. u I , .,u n
basis
g
inteser such that
For every
Let
S.
invertible
result
1.4.6 THEOREM.
be a group,
that for every
(IX G X A) U 0
~n+I,u(V)/~n,u(V)
(or full)
The following
sets, G
acting as zero.
n < dimV. For
n Xn
called the seneral
set
O.
Here
is called the sandwich matrix of
1.4.5 DEFINITION.
0
= (i,gp~jh,~)
instead of
called a Rees matrix semisroup.
semigroup
be nonempty with
PXi ~ 0, and for every
~°(I,G,A;P) (i,g,h)(j,h,D)
where we write
A
(~,i), be any function such
such that
P~i ~ 0.
and
G° = G U 0
_of _
of
V}, U} .
B, and for every
A E nU
fix a
Let
Gn,U(&,V ) = ~°(nu,Ln(~),nv;P ) where
P = (PBA)
with v1
=
.
PBA
A
is invertible, Qn,u(a,V)
let of
e
B A2 VlU
• ..
B nA I VlU
B A VnUl
B A VnU2
"'"
vBuA I n nJ
A
[ U l " ' " Un]
=
vB n
if this matrix
B AI vlu
and
PBA = 0
Finally,
let
= ~n+l,U(a,V)/$n,U(g,V),
be the function defined b_.x:
Gn,U(&,V )
otherwise.
e
maps the zero of
and
e:h ~ (b*u,[~ij],vb)
(h ~ ~,U(a,V) \ 0 )
Qn,u(g,V)
onto the zero
30
where
b*
is the conjugate
xb = Then
of
is an isomorphism i.
Let
and
~ b*U~ f. = ~ u. ~.. J i=l I ll
~ (xfj)v b j=l "''
PROOF.
b
of
~,U(~,V)
G = Gn,U(~,V),
for
i < j < n
onto
(i)
(x E V).
Gn,U(~,V ).
Q = ~,U(~,V),
and write
v1 BA [ViUj] =
.
A A [u I ... Un].
vB n In order
to prove that
have that
b*u E nU
have
b*U C_ U
Define
and
and
the
@
maps
Q
into
[~ij ]
is
invertible.
dim(b'u)
G, we must
= rank b = n
show that for
By 1.5
so that
and 2 . 5 ,
b C Q\0,
we
respectively,
we
b*U E n u.
functions
f. by t h e f i r s t formula in (1). ~4e w i l l show t h a t ] Vb Vb fl,...,fn form a basis of b*U. In fact, since v I ,...,v n is a basis of vb, vb vb f o r any x E V, xb i s a u n i q u e l i n e a r c o m b i n a t i o n o f t h e v e c t o r s v 1 ,...,v n Then
f.:V ~ A, and from the linesrity of b, it follows easily that each f. ] J linear. Thus f . E V* f o r 1 < j < n. F u r t h e r f o r any x E V, f E V , u s i n g l notation i n ( 1 ) , we o b t a i n
is
the
n x(b*f)
generated
as we have seen above,
follows
that
formula
in (i) implies
2.
= (~ (xf.)v\(b)f j=l J J
n ~ f (v~bf) j=l j by f l ' " ' ' ' f n '
b*f =
so that
V*
= (xb)f
that
let
have For
(the
b,c E Q \ O
and
by the
n
where T
in
Q
is
in
g(h,V))
= n.
Since
given if
T
of ~
is
where
vectors
form a basis Therefore
bf)),
dim(b'u)
b*U.
for
rank (bc)
It
Now t h e
Q
into
of
= n,
fl'''''fn"
maps
by:
the subspace
second
G.
b,c E Q'\0, = n
otherwise and
rank(bc)
[~ij ]
Vbc c Vc
and
dimVbc
b*c*U = b*U.
as in (i), and
n Vc xc = j=l >~, (xgj )vj , we then obtain
T
b*U ~ b ' V * c
fl'''''fn
product
=
b'V* c
{~ij] E Ln(A ).
the multiplication
Vbc = Vc; similarly f.j
that
is generated
and
that
0 Hence
T
b*U = b * v = T
Recall
follows
2 (xfj)(v j=l
Consequently
and
bc = { b c
we
It
=
n c*U gj = i~=lUi ~ij
for
i _< j _< n
(x 6 v),
= dimVc
= n,
31 n
c*U ]vVC (xb)c = J~--I[ (xb) ~ n ~ i~__lUi Yij ) j n n n b*u ]vVb~ n c*U Vc ~ [X(m~=~lum ~mk ) k j (if~=lUi Yij)vj j=l [ k=l ~
~ub,U
n n ~,. [ ~ . ( Z v , Vb u.c~U k=l mK i= 1 K i
x
j=l
j=l
m=l m
'
)~..]~ lj
vVC j
x( ~ u b*c*U , Vbc x(bc) m= I m °mj)Vj =
(2)
where we have set • Vb
c'U,
[oij] = [~ij ] IV i nj By hypothesis
rank(bc)
= n, so that
] [~ij ] .
(2) together
with
(i) yields
e:bc ~ (b*c*U,[oij],Vbc) since
the linear
forms f. and the matrix [~ij ] are unique. J ~ Vb c*U~ and thus also Iv i uj j and we obtain
is invertihle
(b~)(c@)
, Vb
c*U~
] [Yij ],Vc)
(b*c*U,[oij],Vbc) Suppose
next
that
1 ~ i < n, let
the hypothesis
rank(bc)
dependent. Hence, n XlbC = i~=2oi(xibc) with
rank(bc)
x.1
< n
Then
implies
bc = 0
in
V
with
for some
o i.
in
For
(bc)@ = 0.
xib = vVbi "
XlbC,...,XnbC
we may suppose
i _< j _< n, let
Then for
Q, so that
the property
that the vectors
vb c*U v i uj = (xib)(c*hj) and
= (bc)~.
loss of generality,
c*U c*h. = u . J J
the property
< n.
be any vectors
without
[~ij ]
= (b*U,[~ij],Vb)(c*U,[Yij],Vc) = (b'U, [~ij ] IV i uj
For
Consequently
h.j
Then
are linearly
that
be any vectors
in
U
i < i,j < n, we obtain -_ = (xibc)h j
hence
Vb c*U
v I uj which
implies
n n = (XlbC)h j = [i=2 ~ o.(x.bc)]h. i i J = i~_2°i(xibc)hj
of the remaining (b@)(ce) 3.
= 0
rows and thus
which
completes
To show that
(b*u,[~ij],Vb)
Vb c*U~ [v i uj j
that the first row of the matrix
~
r Vb
c'U,
iv i uj
j
that
is one-to-one,
let
= (c*U,[Yij],Vc)
with
is a linear combination
is not invertible.
the proof
[~ij] = [Yij], Vb = Vc, which
n ~ Vb c*U~ = i=2 ~ o.~v. i i u.J )
0
b,c E Q \ O
the notation
by (i) evidently
Consequently
is a homomorphism. and suppose
as above.
implies
that
Then b = c.
that
b*U = c'U,
32 4.
To show that
e
~n A f.j = =If~lu~iji
Then for Since
is onto,
for
let
[~ij ]
is invertible,
are linearly
[~ij] E Ln(~),
n B xb = J=~l(xfj)vj
i < j < n;
i <-- j _< n, we have that
fl'''''fn
A ~ nu,
f.j E A, which
it follows
independent.
then implies
the
first formula
there exist
in (3) and the calculation already
proved,
parison of (i) and (3), we deduce onto
that
of
b
is linear.
in (3) that
x. ~ V
in part i
the uniqueness
Define
(3)
that
from the first formula
Hence
B E n V.
(x E V).
x.f. = 6.. for 1 <__ j,i <__ n. But then by (3), x .b = v B. j i jl j j clearly vb ~ B, we obtain B = Vb and thus r a n k b = n.
From the statements
and
such that
Jso that
v B. E vb. Since j It further follows from
of the proof that [~ij ]
b8 = (A,[~ij],B).
b*U = A.
in (i), and a com-
Therefore
~
maps
Q
G. 5.
It remains
zero entry. a = aba
ab
to show that every row and every column of
a E Q \ 0, we may consider
for some
= rank(aba) both
For
b E ~u(V)
< rank(ab)
and
ba
by 3.6.
so that
rank(ab)
it follows
= n
of
Since
a
contains Su(V).
a non-
Hence
rank(ab) j r a n k a
and similarly
Q.
p
as an element of
By 1.3, we obtain
are nonzero elements
a(ba) = (ab)a = a E Q \ 0 ,
a
rank(ba)
= n.
is arbitrary
Thus
and
that the sandwich matrix has the required
property. 1.4.7 LEMMA.
If
S, then there exist PROOF.
a
and
c,d E S
b
are nonzero elements
such that
of a Rees matrix
Exercise.
We are now able to find all ideals and all principal Recall
semisroup
a = cbd.
that an ideal
I
of a semigroup
(or ring)
S
factors
of
is proper if
~u(V). I ~ 0, I # S.
The next result is new. 1.4.8 THEOREM. i) ~,v,(V)
If
dim V = m
I i < n < m}
and all principal ii)
Let
If
V
be a pair of dual vector spaces.
is finite,
then
ar___~erespectively
factors of
~u(V)
and all principal
in 4.6.
12 < n < m]
and proper ideals
{~n,u(V)
In = 2,3 .... }
the sets of all (distinct)
and proper
ideals
~u(V).
factor different
constructed
then
ar__~erespectively
factors of
Each principal
~n,v,(V)
the sets of all (distinct)
= gu(V) = g(V).
is infinite dimensional,
i Qn,u(V) In = 1,2 .... }
semigroup
(U,V)
from
QI,u(V)
is isomorphic
to a Rees matrix
33 PROOF. fications ~n,u(V) if
of the arguments is an ideal of
m ~ n Let
I
Qn,u(V)
be a nonzero a
and
b
to be presented
ideal of
F.
If
from 1.3, and that
a E F, b E I, and
elements
of
c,d E Qn,u(V) \ 0 ,
a = cbd E I, which implies ~n+I,u(V) \ ~ n , u ( V ) .
and is left as an exercise.
follows
Qn,u(V).
to the Rees matrix semigroup
for some
simple modiThat
~n,u(V)
# ~m,u(V)
of 2.4.
are nonzero
is isomorphic
a = cbd
the proof of part i) requires
F = ~D~u(V )
is a consequence
then both
have
We will prove part ii);
that
Next let
I
and thus
rank a = r s n k b
= n>0,
By 4.6 we know that
Gn,u(V ).
Hence by 4.7, we
c,d E F.
Consequently
is the union of some of the sets
b 6 1
and
rank b = n > m >
O.
Then
n b = __k~=l[ik'~k'~k ] with n minimal by 2.4, and v ,...,v k are linearly indeXl n pendent by 2.5. There exist u. E U such that _(vx ,u~ ) = Ok6kp for some 3p k Jp m 1 O k ~ 0 and i < k,p < n. Letting a = p~__l[jp,~p= ,Xp], we obtain n
m
.
-i
ba = (k~=l[ik,~k,~k])(p~=l[jp,Op
2 k=l p=l where both sets
[ik,~k(V
U.ll,...,Uim
2.5, we have that
rank(ha)
the above,
it follows
that
must have
Sn+l,u(V) ~
I.
unbounded,
then we have
Now let property:
J(a) = FaF). property
is
and = m
vhi,...,VXm and hence
Consequently,
O.
~n+I,u(V)
and hence
if and only if
of
The last statement
isomorphisms
with zero removed except
to a Rees matrix semigroup as a modified
for
constructed
replica
for some
It follows
Consequently F
Thus by
rankm.
of
I
n ~
2.
F
are
with
F
we
the
having
(since
this
that for any
l(a) = Sn,u(V)
containing
a
is given by
= Qn,u(V).
now follows [~u(V)
from 4.6.
as a tower of principal
factors
n = I.
Each of these factors
in 4.6,
so we can think of each of the
of the preceding
By
is arbitrary,
of higher rank
the only ideal of
factor of
of
m < n
is an ideal of
= ~n+I,u(V)/~n,u(V)
about
independent.
and no element
rank b = n.
I
an element
Since
J(a) = ~n+I,u(V).
In light of 2.6 and 4.8, we can view
"layers"
J(a)
rank n
that the principal
J(a)/l(a)
I.
I = ~n,u(V)
By the part already established,
in turn implies
Qn,u(V)
contains ~
E
if the ranks of elements
Then
an element
are linearly
I
~m+I,u(V) \ ~ m , u ( V )
I = F, otherwise
contains
b E F, J(a) = J(b) which
. ,u )o ,X ] = z.~ [ ' ,Xp] ~k Jp P P p=l :P'~P
a E F, rank a = n >
J(a)
,~p])
one.
is isomorphic
34 1.4.6 EXERCISES. i)
Find all ideals of
ii)
Characterize
iii)
g~(V)
and of
all one-sided
For a dual pair
£ (V).
ideals of a Rees matrix semigroup.
(U,V), show that the sets of all minimal right ideals of
the following semigroups and rings coincide:
b) ~.~u(V), iv)
c) ~u(V),
d) gu(V),
For any subspace
a ~ £(V),
show that
V'
a) Sn,u(V)
of
V, let
~(V)a = Xva(V ). ~:V ~ ~ Ev~(V )
left ideals of
is isomorphic
£(V).
~v,(V) = ~b E ~(V) IVb ~ V'}.
(V l
is a subspace of
to the lattice of all principal
Throughout
this section,
l,
For
V
V) onto the lattice
of all
Deduce that the lattice of all left ideals of
1.5
spaces.
n >
Prove that the m a p p i n g
is an isomorphism of the lattice of all subspaces of principal
for any
e) ~u(V).
left ideals of
$(V)
~(V).
SEMILINEAR ISOMORPHISMS
let
(U,g,V)
and
(U',h',V ~)
be pairs of dual vector
Our purpose now is to express every isomorphism of a ring
~
onto a ring
~', where
~u(~,v) ~
~u,(&',v') ~ ' ~£u,(a',v'),
C£u(~,v).
in terms of mappings of respective pairs of dual vector spaces. 1.5.1 DEFINITION. ordered pair
(w,a)
A semilinear
where
additive h o m o m o r p h i s m
w
(written on the right) of
(ov)a = (ow)(va) If
a
is one-to-one and maps
are said to be isomorphic, In the case that
instead of
(w,a)
V
(~,a)
V
(h,V)
into
onto
into
At
V'
(AI,V j)
and
a
onto
V',
and
(~,a)
such that
is called a semilinear isomorphism
w
(&,V) ~
(g,V)
and
(&',V')
(&',Vt).
is the identity automorphism,
is a linear transformation.
is an
is an
(i)
g = 4', V = V'), and then
to he denoted by
& = &~
h
(o E 4, v E V).
(semilinear a u t o m o r p h i s m if also
transformation
transformation of
is an isomorphism of
the semilinear
We will usually write
w h e n there is no danger of confusion.
a
This is also justified by
the following statement. 1.5.2 LEMMA.
Let
a
be a nonzero additive h o m o m o r p h i s m o f
there exists at most one function exists,
it is a ring homomorphism.
w:h ~ ~' If
a
satisfying
V
into
(I), and if such an
is one-to-one,
then
~
is also.
V~
Then
35
PROOF. then
for
= w'.
Let
x E V
that
any
o E 4, we h a v e o,~ E 4, =
(o7)w =
and h e n c e
o~ = ~w,
that
then
It f o l l o w s function trivial
of
can also
right
vector
Let
[(o+~)x]a
=
(o~)(xa)
that
spaces
(U,&) b(u~)
=
or
b
and
into
Let
a function
semilinear
this
is
LEMMA.
If
b = c
transformation
Let
(v,bu ~) = and
o~ = ow'
and
(i),
thus
(o~)(~w)(xa)
(dx)a +
(Tx)a
homomorphism.
=
(~x)a
of
is
a
so
If
a
ox = ~x
is o n e - t o - o n e which
w
implies
we w a n t
U,
the
be a o n e - t o - o n e
transformation
is a
the o n e - t o - o n e n e s s
to show
of a s e m i l i n e a r
is dual,
(u E on
that
the zero
is o n e - t o - o n e , when
(Ut,& I)
that
a given
transformation
in p a r t i c u l a r
of
of
(i) b e c o m e s
~ E 4)
(2)
left).
be o u r
dual
pairs
and w r i t e
instead
(v,u)
~(v~,u').
(m,a)
be a s e m i l i n e s r
transformation
is an a d j o i n t
of
(w,a)
= (va,u~) '
(v ~ V, u ' ~ U ' ) .
a generalization
of the definition
(~,a)
c
of
if
b
has
=
(v,cu ~)
the a d j o i n t
an a d j o i n t
(U~,& ~) is o n t o
be a n o t h e r
(v,bul)w so that
satisfy
of
(g,V)
into
if
(3) o f an a d j o i n t
given in
i.
is on___~e-to-on___~e, and PROOF.
a
definition
b:U ~ ~ U
(v,bu')~
b
w ~
(o~+~w)(xa)
to r e q u i r e
( U I , & ~ , V ~ ; ~ ~)
DEFINITION.
1.5.4
(Tw)(xa)
that
(written
1.5.3
(2) of S e c t i o n
=
(bu)(~w)
instead
Observe that
and
that
=
=
=
is a r i n g
is u s e f u l
The
(vl,u~) '
Then
(ox+Tx)a
(~)(xa)
case w h e r e
lemma
~(v,u),
(&~,V~).
=
it s u f f i c e s
in the c a s e
This
(U,&,V;g)
(ow)[(~x)a]
w
(ow)(xa)
(the
and
(b,w)
so
=
+
Thus
is s e m i l i n e a r .
and we w r i t e
of
[o(~x)la
=
=
~
be o m i t t e d .
transformation
(owJ)(xa)
~
is o n e - t o - o n e .
onto
exception),
=
If b o t h
and
(ox)a ~
from 5.2
B
=
= o~ + ~w.
~ = T; h e n c e
xa # O.
(ow)(xa)
[(o~)x]a
(Ow)(7~),
(o+T)w
that
then
[(o+~)~](xa)
and
such
Let
[(o~)®](xa) so
be
into then
adjoint
(va,ul) I = for all
is u n i q u e .
b,
of
it is u n i q u e
(U,&). a
if
(b,w a
-1)
is a
is o n t o
then
is o n e - t o - o n e . (~,a).
(v,cnl)w
Then
for any
u ~ E U ~, we have
(v E V)
v E V, w h i c h Further,
Moreover,
and
implies
that
bu ~ = cu ~.
Hence
36
(v,b(u'+s'))~
= (va,(u'+s'))' =
t (v,bu)w
+
= (va,u')'
(v,bs')~
+ (va,s')'
s [(v,hu ~ ) + ( v , b s ) ] ~
=
= (v,bu'+bs')w which implies b(u ~ +s')
that
(v,b(u ~ +s'))
= b u ' + b s ~.
= ( v , h u ~ + b s j)
= (va,u%)'
= (va,u')'o
= [(v,bu')(om-1)]m implies
b(u'o)
=
that
If
a
maps
A similar
This
=
onto
V'
(h,w -I) and
and t h u s
= bs', (va,se) j
(va,u')'
proof
shows
that
if
lemma
shows
that
by d u a l i z i n g
b
(b,m-l),
If
a
for
all
v E V
is a s e m i l i n e a r
=
that
of
= [(v,bu')~][ow-l~]
bu'
so
an a d j o i n t
v E V
= (v,(bu')(om-1))~
(v,(bu')(om-l))
Therefore
V
(v,bs')
is also
(v,b(u'~))
(bu')(ow-l).
(v,bu ~) =
all
Also
(v,b(u'o))~
which
for
is onto,
then
for any
which
then
a
the c o n c e p t
or s i m p l y
that
a
implies
that
u ~ = s ~.
be o n e - t o - o n e .
of a d j o i n t , and
thus
v E V, we h a v e
then
must
and
transformation.
h
are
we m a y
say
adjoints
that
a
of e a c h
other. 1.5.5 one-to-one
LEMMA. function
of
is an a d d i t i v e ~
onto
h',
b
homomorphism
of
is a f u n c t i o n
V
onto
mapping
U'
V', ~
is a
onto
U,
such
that
then
lemma
(~,a)
is
(o E 4,
(v,bu')~ = (va,u')'
(v E V, u ' E U ' ) ,
a semilinear
PROOF.
This
In this
lemma
is s s p e c i a l
Clifford
(ov)a = (O~)(va)
follows
isomorphism of
from 5.2
the h y p o t h e s e s case
and P r e s t o n
of
(&,V)
on
a
and
the h o m o m o r p h i s m
([i], C h a p t e r
(5)
(&',V')
b
may
theorem
c:I I ~ G, w i t h
3, S e c t i o n
that
and o n t o
matrix
adjoint
b.
isomorphism
d:A ~ G ~, w i t h
d:X - dh, and
and
onto
The
following
semigroups,
see
4). the R e e s
matrix
(written
on
the
left),
the
right)
O~ = 0 ~,
(written
semisroup
S' = ~ ° ( I I , G J , A I ; P I ) .
c:i I -- ci, ,
w : G -- G ~, o n t o
~ : A ~ A I, o n e - t o - o n e
with
be i n t e r c h a n g e d . for R e e s
functions: ~ : I I ~ I, o n e - t o - o n e
such
onto
(4)
and 5.4.
1.5.6 LEMMA. Let 0 be an i s o m o r p h i s m of o S = ~ (I,G,A;P) o n t o the R e e s m a t r i x s e m i s r o u p exist
v C V),
on
Then
there
37
(i,g,X)~ = ( -li,(c ~lig)~d%,X~)-I
((i,g,~) E S),
(6)
d ~ p ~r, i ~ = (p~,~iJci~)~
(X E A, i
(7)
PROOF. form
First note that the nonzero
(i,p$~,~)A~ where
to be denoted
Phi ~ O.
i, such that
(l,pi~,l)6 = (k,PSkl,~)
Pll ~ 0.
for some
idempotents
of
Fix an element in Then
S
A
are the elements of the
and an element in
(l,Pl~,l)
k E I l, ~ E A~.
E I~).
is an idempotent
Any element
(i,g,X)
I, both and hence of
S
can be written in the form -l -i (i,g,~) = (i,e,l)(l,Pllg, l)(l,Pll,X) where
e
is the identity of
G.
There exists
D 6 A
(8) such that
P~i # 0, so
(k,P~kl,') = (1,P;~,l)~ = [(1,Pi~g-lp~,~)(i,g,1)]~ = (l,p;~g-lp~,.)~ (i,g, 1)~ which shows that form
(k, , ).
(i,g,l)~
is of the form
( , ,~); similarly
(l,g,X)~
Using this and motivated by (8), we define the functions
q:i ~ qi,o), d:% -- dx, ~
is of the ~,
by
(i,e,l)~ =
(~i,qi,'r)
(i E I),
(l,Pllg,l)~ = (k,P~k (gw),~)
(g E G°),
-1 (l,Pll,X)~
(~ E A).
= (k,P~kldx,~)
Then by (8), we obtain (i,g,X)e = (i,e,l)~(l,pi~g,l)@(l,pi~,X)~ = (~i,qi,~)(k,PSkl(gw),~)(k,PSkldx,X~)
(9)
= (~i,qi(gw)dx,X~). Further, by the definition of -
(k,P~kl(gh)m,~)
~, we have
~:G ~ G ~
and
0w = Or; also
-i -i -i = (l,Pllgh,l)0 = [(l,Pllg,l)(l,Pllh,l)]0
= (l,Pl~g, l)e (i, Pl~h, I) e = (k,PSkl(g~),~)(k,P~kl(h~),~) = (k,P~kl(g~)(h~),~) which implies
(gh)w = (gw)(hw).
definition of
~, we have
If
g~ = e ~, the identity of
-i ~-i ~-i -i (l,Pllg,l)0 = (k,P~k (g~),~) = (k,P~k ,~) = (l,Pll,l)~ so that
g = e, and
•
is one-to-one.
G j, then by the
38 Using
(9), we see that for any
(k,qladl,~) maps
G
= (l,g,l)~.
onto
For
Hence
(j,e,~) E S ~, let
implies If
that both
= (k,ql(gW)dl,~)
such that
so
a = g~
and
(i,h,~)
= (j,e,~)~ -I.
= (i,h,X)@
~
and
~
We obtain by (9),
= (~i,qi(h~)d%,X~)
are onto.
~i = ~j, then again by (9), we have (i,e,l)~
which
(k,qladl,~)
g E G
G j.
(j,e,B) which
a E G ~, there exists
implies Further,
= (~i,qi,~)
that
i = j.
using
(9) we obtain
(i,e,~)20
-i -I -i = (j,(qj qidl )w ,l)e
= (~j,qj(q]lqi),T)
Consequently
= (i,P~i,~)0
~
is one-to-one,
= (~i,qi(P~iw)d
property, =
p%i w Finally, functions
~
letting and
~ =
c
d
~
is also.
,~) ,
[(i,e,X)~] 2 = (~i,qidx,XTl) 2 = (~i,qidxP i so by the homomorphism
similarly
~,~ iqid~,X~)
we have
xpx~, ~iqi •
g-l
and
(i0)
-1
ci~ = q ~ i ~
as in the statement
-i
for all
of the lemms,
i ~ E I ~, we obtain
the
and (6) and (7) follow
from (9) and (i0), respectively. 1.5.7 DEFINITION. a semilinear b
which maps
U~
to be isomorphic, Strictly
A semilinear
isomorphism onto
U.
The dual pairs
and we write
speaking we should write
not necessary
by 5.4.
We will
fusion.
Again by 5.4, we know
(U~,& ~)
onto
Then
(~,a)
(~-l,a-l)
the function
~(~,a) c~(~,a)
is an isomorphism PROOF.
o__f £U(&,V)
Exercise.
a
instead
-- (&~,V ~) (~,a)
(U I ,~ I ,V ! )
snd
is called
has an adjoint are then said
(UI,&~,V~).
(b,m,a)
of
that then
and hence
Let
(U,£,V)
if
instead
of
since all we need is the existence
also write
(U,£),
1.5.8 LEMMA.
(U,&,V),
onto
(U,&,V) ~
(w,a):(A,V)
l l (U I ,g ,V )
(U,&,V)
actually
(UI,~I,VI).
isomorphism
of
(m,a) (b,m -I)
isomorphism
be a semilinear
b
but this is
which
is then unique
if there is no danger of conis a semilinear
of dual pairs
is ~ semilinear
(~,a),
of
isomorphism i>omorphism
isomorphism
is an equivalence of of
(U,&,V) (UI,&I,V ~)
onto onto
defined by = a
-i
ca
(c E £U(~,V))
ont_____o £U~(&
,V ), and
of
relation.
rank c = rank(c~(~,a) ).
39 We wish
to show that every isomorphism
of a semigroup
S
onto a semigroup
S ~,
for which $2,U(h,V) with
dim V >
isomorphism
several
important
(~,a)
of
dim V = I.
Let
of Section
O
8
/ P%~i'
/ = (v%,
define
,Iv;P ) i )'
functions
Then
(w,a)
a
and
b
i.
then
Since
of the general
S = $2,U(&,V)
case;
onto
E V, u i E U ~, v ki fi V I
and
S~
as Rees matrix
be as i n 5 . 6 .
as
at
~h__!e b__e_e-
semigroups
O
t-
Suppose
that
S ~ = ~ (Iu~,~&
and
.
,Iv~,p
J
)
with
dim V >
1, a n d
b_~
= u ici(Tw - i)
i C U'). (ui~
isomorphism
of
E V),
(U,~,V)
(ii) (12) onto
(U',~/,V ~)
with adjoint
= ~, of
the
first
the representation
Ow = 0 1
~
for
proves
2.
S
of
b(u~T)
implies
= (~w)d v ~
statement
consists
of
of nonzero
elements
a verification
of
the
= ~
b
0a = 0 I, a
# 0 I.
so that
It follows
~ = 6.
of
V
in the form
is single valued. that
We thus have
~
If
= B~
~v
=
and hence
~v
and
?v X
that
a
k~ _l]a maps
is defined
one sees that For any
a
k = ~.
But
is one-to-one.
= [(~d ~ -I- l)w-lw]d V
onto
i v~k ~ -~~ =
~v~~
(13)
V'.
in a way completely b
~-
is a one-to-one
analogous function
to that of of
U~
onto
a, so by a s~milar U.
o E ~, ~v k E V, we obtain by (ii),
[o(~vk)]a = [(~)vk]a = (o~)~dkv'k~ = (°~)[(v~)d~vk~ ] = (o~)[(~v)a] which verifies For any
is
(~vx)a = ( 6 v ) a # 0,
0 ~ ~ Wv~ E V ~, we obtain
/(~d k~--I l)w-lv
Now
false
V)
(vw)dxv~
argument,
for some semi-
then establish
in 5.5.
then also
which
for the treatment
(~v
The p r o o f
and
Further
{(w,a)
We will
and show that it is in general
= (~)dkv~
I$2, U (~,
unique,
(U',A',V').
(~v)a
b and ¢(w,a)
hypotheses
u i E U, v both
~,c,~,d,~
is a semilinear
PROOF,
is crucial
Phi = (v~ ,ui)
with
and l e t
U I
of an isomorphism
be an isomorphism
2, consider
-
S = ~ (Iu,~&
~ S' ~ gU,(h',V')
[4].
= $2,U ~(~ ' ,V ' ), fix vectors
$innin$
S
onto
of this result
The next result
1.5.9 THEOREM.
to
(U,h,V)
consequences
it is due to Gluskin
S'
$2,U,(£',V')
i, is the restriction
linear
for
c S c gU(&,V),
(4). ~vx E V
and
u~l E U ~, we obtain
by (7),
(ii),
and (12),
40
(7vk,b(u~T))w = (~vk,u ici(T~-l))w d
= (~) which verifies
3.
Let
(xia,b
-i
Xl,X2,X 3
a
is additive
be linearly
such that
and will do this in several
independent
(xi,gj) = 6ij
gj)1 = (xi,gj) ~ = 6ijw = ~ij
for for
vectors
in
1 ! i,j < 3. I < i,j < 3. -
for
a
/
,uiT)'
(5).
We will now show that
gl,g2,g3 E U
= (3~0)[(vk,u i)ci]wT
~ J ~T I u I x(vX~,ui ) = ((~w)dxvx~ , iT) ` = ((~vx)
V.
steps.
There exist
By (5), we have 3 If ~ o.(x.a) = 0, then
--
i=
1
z
i
i < k < 3, we have 3
3
-1 o k
and hence 4.
= Ok(Xka,b
xla,x2a,x3a Now let
x
-
gk) =
~ o.(xia,b i=l I
are linearly
and
y
-
lgk) = (i~__lOi(xia),b
igk) = 0
independent.
be linearly
independent
vectors
in
V.
The vectors
xa,ya,xa+ya are linearly dependent so by part 3 we have that the vectors -i -i -i xsa ,ysa ,(xa + y a ) a are linearly dependent. Since x and y are linearly -i independent, we must have (xa + y a ) a = ox+Ty for some o,T E 4, whence xa+ya Using
= (ox+Ty)a.
(14) and (5), we obtain for any
(14) u I E U ~,
(x,bul)w + (y,bul)~ = (xa,ul) I + (ya,u')' = (xa + y a , u l ) ' = ((Ox+~y)a,ul) I = (Ox+~y,bul)~ There exists yields
u E U
lw = o~.
such that
=
[O(x,bu I) + T ( y , b u l ) ] ~ .
(x,u) = i, (y,u) = 0; letting
Consequently
o = i
and analogously
(15)
u ~ = b-lu,
T = i.
Formula
(15)
(14) then
becomes (x+y)a for the case in xhich 5.
Again
let
x
z
are linearly
It follows easily
so that
xa + y a
vectors
x,y.
y
(x,y E V) are linearly
If
x = 0
but both nonzero,
independent
that if
linearly independent. xa+ya+
and
x,y E V.
are linearly dependent and
= xa+ya
x+y
Using
= (x+y)a,
y = 0, (16) is trivial. z
be a vector
(here we are using
(16) several
which a
let
independent.
~ 0, then both sets
za = x a + ( y + z ) a
Therefore
or
(16)
implies
V
the hypothesis x,y+ z
and
If
x
such that that x+y,z
and
y
y
dim V >
i).
are
times, we obtain
= [x+ (y+z)]a
is additive.
in
= [(x+y)+
that (16) is valid
sin = ( x + y ) a + z a for linearly dependent
41 By 5.5, (m,a) adjoint 6.
is a semilinear isomorphism of
(U,~,V)
onto
It remains to verify the last statement of the theorem.
[i,~,~] ~ 32,U(~,V )
and
with
= ~(od-i l)~-iv = od-l-i [(v
= [(ov~)a-l} [i,~,X]a
_i } [i,v,k]a = [(od -I_I)~-I( v
-l'Ui)~}wdkvk~ = od-l-i (p
_l,Ui)~vx}a
-i .~)(~)dkvk~
-
= od -I i d
~-
~
_i p'
-i
_li(C !li~)wdkv~
~,~
~(~,a)I~2,U(~,V)
i
= o(v',u' , )'(c _li~)~dkvkl
~
= (ov~)[~-li,(c~!liV)~dk,k~] which shows that
~-~i
= (~v~)@
= @
and completes the proof.
1.5.10 DEFINITION. Let S be a subsemigroup of gU(g,V) U ~ { i ,V i). An isomorphism ~ of S onto S i
and
semigroup of
by a semilinear isomorphism =
4'
For
ov I~ E V', using (13), (ii), (7), (6), we obtain
(ov~)(a-l[i,?,X]a)
~(~,a) Sl
(UI,~',V I)
b.
(~,a)
of
(U,g,V)
onto
S'
be s sub-
is said to be induced
(U',&',V ~)
if
This same definition is to be used for subrings of
£U(~,V)
and
£U,(8',V'). In particular 5.9 says that every isomorphism of is induced by a semilinear isomorphism.
$2,U(h,V)
onto
$2,U,(& ,V ~)
We will show that the conclusion is valid
under much less restrictive circumstances. 1.5.11LEMMA. 32,u(V )
If
is a subsemisroup of
gu(V)
containing
~2,u(V),
is the unique nonzero ideal contained in every nonzero ideal of
PROOF.
Write
52 = $2,u(V).
52 ~ S ~ ig(v)(32) S, and let
and thus
a C I\0.
ba E 32 N I. But then
S
For any
c E I
52
By 2.8, part i), the hypothesis is an ideal of
S.
By 3.3, there exists
b ~ 32
c C 32 , there exist
g,h C 32
and hence
Let such
I
S.
implies that
be a nonzero ideal of
ha # O; also
such that
c = g(ba)h
For semigroups of matrices,
1.5.12 THEOREM. Sl
Let
be a subsemigroup of
dim V > i, and let
~
by 4.7.
32 ~ I.
We are now able to prove the main result of this section, due to Gluskin
guise, and then by Gluskin
then
this was first proved by Halezov
[4].
[2] in a different
[3]. S
be a subsemigroup of
gUI(&I,V ')
containing
be an isomorphism of
S
gU(&,V)
containing
32,U,(~',V'), onto
S ~.
Then
$2,U(g,V),
suppose that ~
is induced by a
42 semilinear isomorphism
(o0,a) of
unique extension of
to s homomorphism of
PROOF. ideal of
~
~2,U~(f~,V~),
Letting
hypotheses of 5.9, so
~
i ~ IU
$2,u(A,V) S.
gu(A,V)
V~
into
is the unique nonzero
must map e
gU~(A~,V ~) (~)dxv~
V ~, and this representation is unique since such that
~
we see that
in the form
(vX,u i) = o ~ 0
so that
is the
Since the same kind of
a
~2,u(A,V)
satisfies the
is induced by a semilinear isomorphism
q0 be a homomorphism of
onto
and ~(u~,a) guI(A i ,V J).
into
it is clear that
e = ~I~2,u(A,V )
write an arbitrary element of V
(U',AI,V l)
gu(A,V)
contained in every nonzero ideal of
~2,uI(A~,VI).
Let
onto
First note that by 5.11, we have that
S
statement is valid for onto
(U,A,V)
(w,a).
which extends
~.
as we may since is one-to-one.
~v X = (~vx)[i,o-l,kl.
a
We maps
There exists
Hence for any
c ~ gU(A,V), we obtain [ ( ~ ) d k v ~ ]a-lca = (~vk>ca = [ (~v k) [i,(~-l,k]}ca
(~vx)a[a-~([i,o-~,X]c)a} = (Vv~)a{ [i,o-~,X]c}~
= t (Vv~) [i,c-~,~l}a(c~) = (~v~)a(c~) = { (w)d~v~J (c~) and thus
cop = a
-I
ca.
Consequently
same calculation with Since
cp
is the unique extension of
~
to a homomorphism of
it is also the unique extension of
Defining
£(&1,Vl).
contained in
~(w,a)
~(~,a)
~U(&,V), we see that
~(~,a)
~UI(&I,V I)
if
subsemigroup of
U~
~
gu(A,V)
containing
into
to such a homomorphism.
~
~U(&,V)
onto
to an isomorphism of
is contained in
~(~,a)
to s, the ~(0~,a)IS = ~'
by the same formula as on
SU(~,V),
preserves rank, and Ii~u(&,V )
are t-subspaces of
~2,U,(&,V)
then
Sl
~(&,V) is
~(~,a)I~U(~,V) into
is
[~UI(&~,VI).
V*, then every isomorphism of a
~2,UI(&,V )
onto a subsemigroup of
can be extended to an (additive and multiplicative)
~(~,V).
1.5.13 COROLLARY. be a sabring of
£(&,V)
to a homomorphism of U~
8UI(g,V)
containing
automorphism of
since ~
and
~
is an extension of S
e
c ~ S, shows that
is in fact an isomorphism of
on all of
For example, if
the unique extension of Furthermore,
let
extends
and for
~e know by 5.8 that
gU,(~,V)
~
~0
£UI(~J,VJ).
onto
q0 = ~(0~,a)" Since
instead of
gu~(AI,V~),
~
Let i
i
~Ut(~ ,V )
be an isomorphism of
~
be a subrin$ of i
containing ~
onto
5.12 and is thus a ring isomorphism of
£U(g,V) i
containing
~UI(~ ,V ), suppose that rD.~~. ~
Then
onto
~.
~
~U(g,V),
~
dim V > i, and
satisfies the conclusions of
43 The ring version restriction
of 5.12, with
on the dimension of
the rings
~l = ~u,(V) ' the result was established
isomorphism
every multiplicative cannot be dropped. ~(V') ~ 4'.
~'
as in 5.13 and no
[3]; for
the remarkable
~ = Su(V),
[2].
property
that every
of a ring in that class is automatically additive,
isomorphism For if
is a ring isomorphism.
The hypothesis
dim V = i, it is easy to see that
But there exist division
multiplicative
and
earlier by Dieudonn~
In 5.13 we have a class of rings with multiplieative
~
V, is due to Jacobson
isomorphisms
which
rings
(even fields)
are not additive.
i.e.,
dim V >
i
~(V) ~ 4; similarly
for which
For example,
there exist
let
4 = 4~
be
the field of real numbers and n be any odd integer, n # i; then the mapping n is a multiplicative automorphism of ~ 4 but is not additive. It follows
x ~ x
that the corresponding isomorphism which
automorphism
in turn implies
of
£(V)
that
is not induced by a semilinear
dim V >
1
cannot be dropped in 5.12 or
5.13. For
0 < ~ _< (dim V) +, let
1.5.14 COROLLARY. that
PROOF.
Then
If
~
Conversely and
~
i) ii) iii) iv)
v)
4
and
if
of
4'
g ,U(g,V)
is induced by a semilinear
(w,a)
¢(w,a)
preserves
is a semilinear
~ = 4 ' , then
be cardinal numbers
statements
onto
such
__if --and onl~ __if S = 4 ~
g4,,U,(4~,V~),
isomorphism
rank
(~,a)
we must have
isomorphism of
and
(4~,V ~)
then by
of
(U,4,V)
~ =
(U,4,V)
is the required
~(~,s),l~
4, U(g, V) For vector spaces (A,V)
1.5.15 COROLLARY. the following
and
g4,U (4'V) ~ ~s',U ;(g~'V~)
is an isomorphism
(U ,& ' ,V ' ), and since
(U ,4 ' ,V ' )
i
fl ~U(4,V).
(UJ,4J,V').
5.12 we know that
onto
isomorphism. with
dim V >
i,
are equivalent.
& ~ 4', dim V = dim V' (4,V) ~
(4~,V').
g (4,V) ~ g
4
~(4',V ~)
for some cardinal
numbers
Q,4 ~ >
i.
g(~,V) ~ g(a',V').
£(4,v) ~ £(a',v').
PROOF. one-to-one a
dim V >
i < 4,s' < (dim V) +.
an__~d (U,a,V) ~
onto
Let
g4,U (&'V) = ~4(&'V)
i) = ii).
Let
function mapping
to all of
V
w
be an isomorphism
a basis
B
of
V
of
4
onto
onto a bssis
by defining n
va = k~=l(~k~)(VkS)
n if
V = k~=l~kVk
(v k E B).
~ B~
and of
V ~.
s
be a Extend
44 Verification
that
a
is a semilinear isomorphism of
(~,V)
onto
( ~ , V ~)
is left
as an exercise. ii) = iii).
This follows
iii) = iv).
This follows from 5.12 with
iv) = v) and v) = i).
from 5.14. U = V*, U ~ = V ~*.
This follows from 5.13 with
This corollary says that under the hypothesis determines
(~,V)
the semigroups
and the ring
We now investigate when two semilinear (or ring) isomorphism. of
G
induced by
If
G
g, i.e.,
1.5.16 DEFINITION. (x E V)
(~,V).
V
Let
m
(&,V)
U
of
V*
then
a
PROOF.
Let
a
defined by
xm
= ?x
is a semilinear auto-
but most of the proof is old.
be a vector space.
If
is s multiplication.
i E IU
the hypothesis
dim V,
?, or simply a multiplication.
a
is any function $2,U(&,V)
Conversely,
for some
every multi-
£(V).
be as in the statement of the theorem.
and there exists
~2,u(V),
and
the inner automorphism
into itself which commutes with all elements of
t-subspace
g
g
~ ~ O, m~ = (~ _l,m )
~ l i c a t i o n commutes with all elements of
x = ~vl_
i, any of the following
induce the same semigroup
~ ~ &, the transformation
The next result is new,
1.5.17 THEOREM. mapping
in
isomorphisms
c
It is easy to verify that for
U ~ = V ~* .
the division ring
is a group, we denote by -I xc = g xg (x E G). g
For
and
~(~,V).
is called the m u l t i p l i c a t i o n induced by
m o r p h i s m of
dim V >
up to a semilinear isomorphism:
g (~,V), g(~,V),
U = V*
such that
Let
(vX,ui) = o # O.
0 ~ x E V. Since
Then
[i,o-l,k]
is
implies
[(xa,ui)o-l~-l]x = (xa,ui)o-lvk = x(a[i,o-l,~]) = x([i,o-l,k]a) = (~(vk,ui)o-lvk)a = (~vx)a = xa. It follows that
a
some scalar
we have
~x
there exists
maps
b E ~2,u(V)
x
into the subspace of
xa = ?x x.
If
such that
x
and
xb = y.
V y
generated by
x, and thus for
are nonzero vectors,
by 2.10
Using the hypothesis, we obtain
~yy = ya = (xb)a = x(ba) = x(ab) = (xa)b = (~xX)b = ~x(Xb) = VxY and hence x # 0. quently
~x
Let
=
o
a = m
Conversely,
~y.
But then
=
~x
be the zero function on
is a constant and we have V.
Then
if
b ~ £(V),
m b = bm
.
then for every
x E V, we have
= (~x)b = ~(xb) = ( x b ) m
xa
=
~x
0 = (0a)o = (Oo)a = 0a.
•
x(m b) = ( x m ) b and thus
~
= x(bm~)
for all Conse-
45 If C(S)
S
is a semigroup
[a ~ S I xa = ax 1.5.18 LEMMA. PROOF.
(or ring), by
for all
For
C(S)
we denote
the center of
S, i.e.,
x E S} .
~ ~ &, m
is linear if and only if
~ E C(&).
Exercise.
1.5.19 COROLLARY. PROOF.
If
a E £U(g,V),
C(£U(g,V))
=[m
IV E C(~)} .
a ~ C(£U(~,V)) , then
by 5.18 we also have
by 5.18, we have
m~ E £(g,V),
£(~,V).
for any
Further,
a = m
that
for some
~ E C(~).
and by 5.17, m
? E ~
by 5.17,
Conversely,
let
and since
~ E C(A).
commutes with all elements
Then
of
x E V, f E V*, we obtain
x(m~f) = (xmv)f = (vx)f = v(xf) = (xf)v = x(fv) which proves
that
m*f = f~
again a multiplication). and therefore
m
( ~ , V ~)
If
and of
a semilinesr
Consequently
(~,a)
onto
~ E & . PROOF.
( ~ , a ~)
into
of
Let
into
Then
that
m v E £U(&,V)
transformations
respectivel~,
then
of
(~0o0 ,as )
(&,V) is
and
( ~ , a ~)
The equation
a~a -I
-i
~(w,a) = ~(w~,a ~)
ca = a'-ica '
to
isomorphisms
if and only if
of
aI = m a
for
to
to
as
a~a -I = m
a ~ = m a.
is equivalent
(c E ~U(&,V))
commutes with all elements
is equivalent
be semilinear
I_~n suc____~h~ case, w ~ = ¢ _i w.
in turn can be w r i t t e n
equivalent
implies
is
(g~,VH).
~(~,a) = ~(~i,al)
(a'a-l)c = c(a' a -1) i.e.,
which
are semilinear
(&~,V~),
(A,V)
(~,a)
(UI,&I,VI).
a which
and
( ~ , V ~)
transformation
1.5.21 COROLLARY.
some
m*U c-- U
of a multiplication
Exercise.
PROOF.
(U,&,V)
the conjugate
E C(£U(&,V)).
1.5.20 LEMMA. into
(in particular,
for some
(c E £U(&,V)), of
~U(&,V).
~ E A-;
The last statement
(17)
Hence by 5.17,
the last expression
of the corollary
formula
(17)
is evidently
follows
from 5.20.
1.5.22 EXERCISES. A ring
(~,+,.)
is said to have unique addition
if for any ring
defined on the same set and under the same multiplication, This concept was introduced i)
by R.E. Johnson
Show that a ring
every isomorphism
of
~t
that
(~,-) + = ~.
[i].
has unique addition onto
we have
is additive.
if and only if for any ring
~,
46 ii)
A ring
~
all of whose elements are idempotent is called a Boolean rin$.
Show that a Boolean ring has characteristic
2, is commutative~
and has unique
addition. iii)
Show that the ring
Z
of integers has only one automorphism,
has an infinite number of automorphisms.
Hence
Z
wheras
~Z
is not a ring with unique ad-
dition. For any ring iv)
Let
~
~, by
3+
denote the additive group of
be a ring and
F
3.
be a prime field (i.e., F
is either the field
of rational numbers or the ring of residue classes of integers Prove that if ~ 2 = 0.
not additive F+
[F0~~ ~F,
then
~ ~ F, and if
Give an example of a prime field (and hence
F
mod a
~ + ~ F +, then either F
prime).
~ ~ F
w i t h an a u t o m o r p h i s m of
does not have unique addition)
or
~F
which is
and an automorphism of
which is not multiplicative. v)
Prove that if
~
is a ring for which
3 + ~ Q+/Z +
rational numbers over the additive group of integers), example of a semigroup vi) g(V)
S
Show that for a vector space
and any nonzero ideal of An alsebra
over a field (A,+,.)
4
then
with zero and with the property
(g,*,A,+,.)
V
~(V)
with
dim V >
(the additive group of ~ 2 = 0. S ~ ~0~
with scalar m u l t i p l i c a t i o n +
i, any nonzero ideal of
*
and addition
and m u l t i p l i c a t i o n
~(ab) = (~a)b = a(~b)
etc.
+, and a ring
(~ E 4, a,b E A)
dim V >
Let i.
(U,4,V) Show that
A.
be a pair of dual vector spaces where ~U(4,V)
x(~a) = ~(xa)
algebra i s o m o r p h i s m of inner a u t o m o r p h i s m of
(U,4,V) ~U(4,V) £(4,V),
4
can be made into an algebra over
is a field and 4
by defining
(x E V, ~ ~ 4, a E £U(4)).
prove that every algebra isomorphism of linear isomorphism of
The concepts of
for an algebra refer to tbe same notion for both the
vector space and the ring structure of vii)
(4,*,A,+)
., and satisfying
where we have denoted both "multiplications ~' by juxtaposition. homomorphism,
~.
have isomorphic a u t o m o r p h i s m groups.
is a system consisting of a vector space
with the same addition
subalgebra,
Give an for no ring
onto onto
~U(g,V)
(UI,4,V~). £U~(4,V )
onto
~UI(4,V~ )
Deduce that for
is induced by a V = V l, every
can be extended to an algebra
and that every algebra a u t o m o r p h i s m of
£(&,V)
is
inner. viii)
Prove that if
4
is a field,
leaves the elements of the center of
then every a u t o m o r p h i s m of ~(&,V)
£(g,V)
which
fixed is an inner automorphism.
47 ix)
Show by an example
is0
have a unique the form
addition.
that e subring
Such an example over a 2-element
a21 0
0:
all 0
0
~
a32 a32 0 is an automorphism
In fact,
need not
of all matrices
of
show that the mapping
~
a21
0
a31+alla21
a32
but is not additive.
Find another
addition
for
~I~
which
it a ring. x)
State
striction
and prove
xi)
Show
that if
and
Chapter
[4], Jacobson
[1],[2],
2, Sections
([7], Chapter
Mihalev
Gluskin
We will discuss
and Lam
[I], R~dei
ii),
/i], [2],
[1], (~2],
IX, Section [l], Johnson
If], Mihalev [i], Rickart
12), and
[i], [1],[2],
AUTOMORPHISMS groups
in the preceding
we write
all functions dim V >
auto-
Throughout
of functions
on a set
on the right. i,
auEomorphisms
automorphisms
of semilinear
section.
Multiplication
space with
is the group of all linear
Johnson
Martindale
of certain
is the group of all semilinear A
Jacobson
and Steinfeld
notation.
is a fixed vector
/I],[2],
IV, Section
/2],
[5],[6].
only a few properties
is taken to be their composition;
of
see
6), Dieudonn~
[I], Fajans
12),([6], Chapter
OF SEMILINEAR
encountered
gidelheit
[i], R.E.
[1],[2],
[i], Wolfson
we fix the following
(&,V)
Halezov
transformations,
II, Section
ii), ([7], Chepter
III,IV),
3, Section
GROUPS
we have essentially
this section,
IX, Section
[3], Mackey
/i], Morita
1.6
the re-
with all idempotents
and semilinear
([i], Chapters
12), Jodeit
[i], Stephenson
and commutes
([i], Chapter
[2],[6],[7],
[i], Kaplansky
and Satalova
morphisms
on isomorphisms 4), Behrens
Baer
5.14 and 5.15 without
must be a multiplication.
7,8 and Chapter
Y
[3], Rjabuhin
is additive a
[3], ([6], Chapter
IV, Section
Kiokemeister
in 5.17
V, Section
and for related material
of 5.12,
V.
i, then
information
([1], Chapter
Gewirtzman
of
a
dim V >
For further
Gluskin
the ring version
on the dimension
~2,U(&,V),
Baer
of
field.
addition
~
Jail001 Lall 0!]
a31 a32
makes
of a ring with unique
is given by the ring
of
of V
V, (invertible
linear
transformations), M
I
is the group of all multiplications is the group of all semilinear
(m
= (e
,m ) with ? inner automorphisms (i.e.,
W E ~ ),
48 £ = £(A,v). Furthermore,
for
S
a semigroup
(or ring),
let
~(S)
be the group of all automorphisms
~(S)
be the group of all inner automorphisms
where an inner automorphism
of
S, of
S,
of a semigroup or a ring is defined by g -I (x ~ S), as in the case of groups, with the provision that g exists,
must have an identity If set
H
e
element and
g
is a subset of a group
{g E G I gh = hg
for all
must have an inverse
relative
G, then the centralizer
h E H}.
For subsets
H
of
and
H
K
-i = g xg g i.e., S
x¢
to it. in
of
G
is the
G, let
HK = [hk lh E H, k E K]. PROPOSITION.
1.6.1
Further,
A
a normal
subsrou~
is the centralizer
PROOF.
o_f~ ~ ,
~
onto
subgroup
6(£)
o_~f ~
PROOF. For any
with kernel
and of
By 5.8,
c E ~
implies
5.21, M
and
thus
that
~
[
for any
M
and is a normal
M ~ A = im
maps
and maps
~
into
of
~
subgroup
I ~ ~ C(a-)}
onto
of
~,
M.
I
i__~s
= C(M).
~.
~
maps
and by 5.13,
= ~(w,a)(~,b) M
Hence
¢
M
maps
is a normal
~
onto
6(£).
= (ab) -i c (ab) = c¢(w~,ab)
and hence
~
is a homomorphism.
is s normal subgroup of
m
I
$(~).
we have
Hence
7 E g , we have
to show that
onto
= b -I (a-lca)b
(a-lca){(w,b)
~(w,a)~(~,b)
((~,a) E >~), is a homomorphism
I/M ~ $(£).
6(£),
(~,a),(w,b) E ~ , =
I
~ / M ~ ~(~),
is also a normal subgroup of
It remains
i_nn ~
is an isomorphism
~:(w,a) ~ {(w,a)
M,
and
is the kernel of
Since
M
I = MA, a n d
Th__.~emappin $
c~(~,a){(~,b) which
of
~ ~ m -i
Exercise.
1.6.2 THEOREM. of
The mapping
= (¢ _l,m ), it follows I
onto
and the kernel of $(~).
Hence
let
~
~
and
that
M c
restricted
(¢ ,a) E I.
By
~ / M ~ 6(£). I
and to
I.
For any
o E g, x E V, we obtain (ox)(m a) = (7ox)a = ~-l(~g)~(xa) = c[(~x)a] and hence
m a
is linear.
= o~(xa)
= o[[(~e
_l)x]a}
= o[(xmv)a ] = o[x(mva)]
This
together with 5.20 yields
C(~v,a) = ~m a = e m a E a(~). 7 Finally, ~Z~,a~<)
let
~(w,a~,,E
= ~(5~,b)" =
J(~).
Then there exists
But then 5.21 implies
= e _15~ c -i £ J(~) so that V inverse image of ~(£) under ~.
that
(~,a) E I. In
a linear automorphism a = m~b
for some
It follows
particular,
that
I/M ~ ~(£).
~ £ ~ I
b
such
that
and
is the complete
49 1.6.3 COROLLARY. are equivalent:
The following conditions on a semilinear isomorphism
i) ~ E $(&),
ii) ~(~,a) E ~(£),
iii) m a
(~,a)
is linear for some
rE&. PROOF.
Exercise.
1.6.4 COROLLARY.
The identity m a p p i n g i~s ~h___eeonly a u t o m o r ~ h i s m of
leaves fixed every element of PROOF.
Let
automorphism
~
(~,a)
Hence
(w,a) E M
S
PROOF. (~,a)
and
£(g,V)
Sl
S
an___~d S ~
~
and
~(g',V~),
so t h a t
~(w,a)~(i',a ')
these are extensions
of
b
of
Now
~2,U(&,V),
If then
PROOF.
®
0
onto
~
~
~(,
and
,
~
satisfies
Iva = v
for all
is a subgroup of
B
~(w',a~) '
of
say
are isomorphisms of
the c o n d i t i o n s
we m u s t have
of 6.4,
and s i n c e
~ = ~. V
and let
v ~ B].
~,
the m a p p i n g
and
® n A = ~V'
((~,a) E ®)
~ ~
onto
~(&)
are iso-
~ = ~l.
respectively,
6(&), ~
V e r i f i c s t i o n that
m o r p h i s m of
an___dd ~
~
~:(~,a) is on isomorphism of
~
~(w,a) =
~
for some
~ = ~(w,a) = ~£"
For the remainder of this section, we fix a basis
1.6.6 PROPOSITION.
for some semilinear
a = m
Consequently
® = [(w,a) E ~
which
~2,u(V), we have
= ~(g,V)" ~',
£
V*.
are induced by semilinear isomorphisms,
~(~,a) ~
and
o_L
0 = ~(w,a)
be as in 5.12.
~
• (w , ,a l ), respectlvely.
U
By 5.17, we must have
which agree on
By 5.12, both
onto
and t h u s
ba = ab.
and by 6.2, we obtain
Let
onto
Then
For every element
so that
1.6.5 COROLLARY. morphisms of
for some t-subspace
be such an automorphism. by 5.13.
b = b~(~,a) = a-lba E & •
~2,u(V)
®
= ~A,
is a subgroup of
w i t h trivial kernel
~
and that
~
is a homo-
is left as an exercise
(for ~'onto"
part use the function a c o n s t r u c t e d in the proof of 5.15)o For
(~,a) E ~
and extend of
~
define
7
on
V
V, we have that
~
maps
V
onto itself.
~ v =
(vi
E B)
"
°l = 02 = "'" = ~n = 0
Hence
Let Since
Suppose
v ~ = va
for every
s
B
that
maps
v ~ B,
onto a basis
v ~ = 0, where
Then
0 = v~ =
~ ~.(v.a) i= I i l
since the vectors
w h i c h implies that
vla,v2a,...,Vna
are linesrly independent.
v = 0 and 7 is one-to-one. Therefore ~ E A and (w,a) = (~,a~-l)(~&,~). ---l as maps V onto V and is one-to-one; for o E g, x E V, we have (ox)a74
and for
V.
n
~.v. i= l i i
Hence
as follows.
to a linear transformation on
= [(ox)a]7-1 = [(om)(xa)]7-1 = (o~)(xa~-l)
v E B, v ~ = va
(~&,7) E A
implies that
w h i c h proves that
~
= ®A.
v a ~ -I = v.
Thus
(w,a~ -I) E ®
and
50 If
(~,a) E ® N A, then for every
implies that
a = lV
since
v E B, va = v
(w,a) E A
and
B
since
(~,a) E ~, w h i c h
is a basis of
V.
We now discuss a group theoretic construction which will clarify some of the concepts we have studied, 1.6.7 DEFINITION. we w r i t e
~:h ~ ~h'
see Hall
Let
H
The set
and
We say that a group
G
of
The mappings
H
and
identify
and
with
1.6.8 THEOREM. then if and
G G
If
is a group, A
A (determined by
and
PROOF.
Let
G
H
with
H
it is convenient
to
and we do so henceforth. H
with a group
G, G = HA, H ~ A = i.
and a normal subgroup product o f
be a semidirect product of
(more precisely
are easily seen to be isomorphisms
is a semidirect product of a group
is a semidirect
A
to a group c o n s t r u c t e d as
into the semidirect product;
is a normal subgroup o f
G
~,).
is isomorphic
a ~ (l,a)
is a group having ! subgroup H N A = i, then
G
their respective images, G
be a homomorphism,
(hh~,(a~h~)al)
~") if
h ~ (h,l)
A
~:H ~ ~(A)
is a semidirect product of
A, respectively, H
be groups and
together with the m u l t i p l i c a t i o n
H with
"determined by a h o m o m o r p h i s m above.
A
HX A
(h,a)(h~,a ~) =
is a s e m i d i r e c t product of
([i], Chapter 6, Section 5).
H
H
with
with
A.
A
A,
Conversely,
such that
G = HA
A. Then
I(h,a)(h',a')](h',a" ) = (hh~(a~h,)a')(h~,a") I
= (hh'h",[(a,~h,)a
ii
t
l~h,,a ) = (hh'hH,(a~h,~h,,)(a ~hH)a ~)
= (hh'h",(a~h,h,,)(a'~h,,)a")
= (h,a)(h'h~,(a'~hH)S")
= (h,a)[(h',a')(hH,a")] and
the multiplication
is associative. (h,a)(1,1)
Further
= (hl,(a~l)l)
= (h,a),
(l,l)(h,a) = (lh,(l~h)a) = (h,a), so that
(i,i)
is the identity of
(h-l,a-l~
_l)(h,a)
G.
Finally,
= (hh-l,(a-l~
_l)~h a) = (l,a-la)
h w h i c h shows that
= (i,i)
h
(h -1, a -13h_l )
is a left inverse of
(h,a).
It follows that
G
is a group. Now we note that (h'a)-l(l'b)(h'a) = (h~l'a -i ~ -i )(h,(b~h)a h =
so
that
A
is a normal
subgroup
(l,(a-l~h_l~h)(b~h)a) of
G.
It
is clear
) = (l,a-l(b~h)a), that
G = HA
and
H ~ A = i.
51 Conversely, such that
let
G = HA
the form
g = ha
implies that g~ = h~a~;
G
and
be a group with a subgroup H N A = i.
with
Since
h E H, a E A.
H
and a normal subgroup
G = HA, any
g E G
can be w r i t t e n in
This r e p r e s e n t a t i o n is unique since
h~-lh = a~a -I ~ H N A = i, so
h = h ~, a = a j.
A
Next let
ha = h~a ~
g = ha,
then gg~ = (ha)(h~a ~) = hh~(h~-lah~)a~,
where
h~-lah ~ E A
a~h = h~lah
since
A
is a normal subgroup of
(a E A), we see that
~h E G(A).
(a~h)~ht = h ~ - l ( h - l a h ) h ~ so the m a p p i n g then can write
~ : H ~ ~(A)
(i)
For
G.
For
h E H, w r i t i n g
h,h ~ E H, we have
(hh~)-la(hh ') = a~hh~
defined by
~:h ~ ~h
(h E H)
is a homomorphism.
We
(i) as gg~ = (hh~)[(a~h~)a~ ]
or, w r i t i n g the elements of (2),
G
(2)
as ordered pairs in
H X A,
(i) becomes by virtue of
(h,a)(hJ,a I) = (hh~,(a~h~)a~). Therefore
G
is a semidirect product of
With the n o t a t i o n of 6.8, are equivalent
let
to saying that
follows easily that
B ~ H.
H; in fact, G
H
G/A = B.
H
(one also says that the extension
Let
G
ii)
~
maps
Every element of
splits over
H
with
A).
A
by
For a discussion of group
H
with
A
determined
A.
G(A).
commutes with every element of
~ $(h)
is a semidirect product o f with
A.
~
®~ = ~ G I. ~' N I
C(&)
with
A
is a normal subgroup of ®
with
A.
is also a semidirect product of
and
I
is
is a normal subgroup of
~,
so that by
A g a i n by 6.6 we have ~(h)
with
It is left as an exercise to show that
= iV, A
A
A.
is a semidirect product of
and hence
and
H
We have seen in 6.1 that
6.6 and 6.8, ~
I = ~JA,
It
is an extension of
Exercise.
a semidirect product o f
Now let
G
onto the identity o f
1.6.10 PROPOSITION.
® ~ 6(A)
exactly once.
Then the followin$ statements are equivalent.
is the direct product o f
PROOF.
A
be a semidirect product o f
~:H - 6(A).
G
PROOF.
G = HA, H n A = i
([i], § 50).
i)
iii)
The conditions
is a very special extension called a split extension of
1.6.9 PROPOSITION.
H
A.
intersects each coset of
by
by the h o m o m o r p h i s m
with
In terms of (group) extensions, G
A
extensions see R~dei
H
A.
®' ~ $(h),
I, so that the proof of the
second statement of the proposition follows along the same lines as the proof of the first.
52 1.6.11 REMARK. product o f
G(&)
We can make 6.10 more precise b y w r i t i n $ the semidirect
with
A
(and thus also of
~(&)
with
pairs a n d exhibiting the determinin$ h o m o m o r p h i s m account th___~eisomorphism in 6.6 between (~,a) E ~
as the product
of the h o m o m o r p h i s m Let
~
~ E G(&)
(m,b) E ~
For any
and
(~,a~-l)(s~,~)
~.
To this end, we take into
G(~),
th__~erepresentation o f
in the proof o f 6.6, and the construction
in the proof of 6.8.
and define
b
vb = i=l(~i~)vi ~ Then
~
A) in terms of ordered
on
V
by
n v = i~=l~iVi
if
and we obtain the function
v E B, we have
va =
o.v. i=l 1 1
(v E B).
M :a ~ b-lab
for some
oi E &
n
(a E A). and
Let
vi E B
a E A.
whence
n
v(b-lab) = (va)b = (~lOiVi)bi = = (oiw)v i •= i=l since
b
leaves the elements of
B
fixed.
=
~ :a ~ a~
(a E A)
where
~
n
va
Hence
~ (~.~)v. i=l z l
if
va =
O.v. i=izl
(v E V, v i ~ B).
(3)
Thus the semidirect product m u l t i p l i c a t i o n is given by i
(~,a)(~',a') where
aw
= (~x~',a w • a ' )
(~,~'
then for any
v E B
we have
can be represented as a
for
a , a ' ~ A),
i s g i v e n by ( 3 ) .
We are now able to express automorphisms of
c
e G(A),
i < i < n
vc = B ×B
and every
~ o.v. i=l z i
0
in matrix form.
for some
m a t r i x over
v E B, and
~
A
c~i E f~
having
otherwise.
and
oi
If
c E £,
v. E B. z
in its (v,vi)-entry
This m a t r i x
C = [~ij]
row finite (i.e., every row has at most a finite number of nonzero entries). versely,
it is clear that to every row finite
uniquely an e n d o m o r p h i s m of
(A,V).
B× B
matrix
C
is an isomorphism of
£
A, where
are added and m u l t i p l i e d as usual
Now let
c E ~
c~(®,a) where
onto the set of all row finite
B ×B ("row by
is possible because of row finiteness).
and
(0~,a) E ~ .
Then, as we have seen above,
= a -ica = 7 -l[(a~ -l)-Ic(a~-l)j~
~ = (~A,-a) E A, a~ -I = (~,a~ -I ) E ®.
for some
Con-
It is easy to show that the c o r r e s p o n d e n c e
c ~ C = [~ij]
column" m u l t i p l i c a t i o n
is
one can associate
matrices over
the matrices
Hence
o i E A, v i E B, and thus
For
v E B, we have
(4)
vc =
~ q.v. i=l z t
53
v(a~-l)-ic(a~-~
---l aa E ~
using the fact that 7
to
A = [~ij]
n = ( -~=l oli v ii) a ~
= (vc)a~-i
in matrix
twice.
Then if
c
=
n ~ (a,~)v. i= I ~ l
corresponds
to
C =
[Wij]
and
form, we obtain by (4), -1
[Wij]~(w,a)
= A
We have thus proved,
using 5.13,
the following
if the corresponding
endomorphism
For the finite dimensional some time,
see Jacobson
to Halezov
[I] and Gluskin
1.6.12 THEOREM. elements in
of
in
£
vector
([2], Chapter
Let
~ = £(A,V)
£, and for
[Wij~]A. Call a matrix
C
invertible
spaces 2,
the next result has been around
Section 5);
the semigroup version
for
is due
[31. (g,V)
be a vector
in matrix
[~ij] E £,
result.
is invertible.
form.
space with
Let
m E 6(&),
dim V > A
i, and write
be an invertible
the
matrix
let
[~ij]~ = [Vij~] , [Vij]¢ A = A-I[wij]A. Then
~(~,A) = S e A
is an automorphism
is of this form.
Moreover,
of
~, and conversely,
~(w,a) 6 ~(~(~,V))
We see that every automorphism
of
~
every automorphism
i f and only if
form
~
and an inner automorphism.
From the above discussion
~
is a semidirect
C(A) with
matrices
over
A
product of
w C ~(A).
is the product of an automorphism
that
of
of the
and 6.10, we see
the group of invertible
BX B
with multiplication i (w,A)(~',A')
where
for
=
(tlJoJ/,A w
"A')
A = [~ij], A ~ = [~ij~].
1.6.13 EXERCISES. i)
Define a semidirect
a homomorphism nition
of
H
product of a semigroup
into the endomorphism
is to be the same as for groups.
semigroup of semilinear is a subspace
of
transformations
V; let
semigroup
Let on
(&,V) V.
rank (w,a) = dimVa.
Ca(V ) = [(~,a) E ~(V) I r a n k a Find all ideals of ga(V)
~(V),
show that
and that each maximal
with a maximal
subgroup of
subgroup g(V)
P (V) Q
of
H
with a semigroup of
A, otherwise
(as groups).
by using
the defi-
he a vector space, ~(V)
be the
(~,a) E ~(V),
Va
For
show that
Define < Q}.
is a semidirect ~(V)
A
product of
is a semidirect
~(~)
product of
with G(A)
54
ii)
Show that
~2(V ) ~ ~ O ( I v , , G ,IV;P),
where
G
is the semidirect product of
_
~(&)
with
_
~
(as groups) determined by
Pgi = (~&'(vD'ui))" iii)
~(s,z) for
Let
(Hint: Consider
(g,V)
defined
by
x~(a,z ) = xa +z
translation.
Let
+ T ~ V
a)
For
s E f~
( x E V)
is
is called a homothetic
A, H, and
homothetic transformations,
T
and
~
~lli~&_ and
called
z 6 V, the function
an sffine
transformation;
transformation;
a(ml,z)
is called a
be the groups of all affine transformations,
and translations,
(the additive group of
A N T = iV, so that
with
xli,(w,~),k ) = (x,ui)w~vx.)
be a vector space.
~ E g , ~(m ,z)
=
~:C(A) ~ 6([i~& )
respectively.
Show that
is s normal subgroup of A, A = AT, + is a semidirect product of A with V ; express this
A
V), T
explicitly. M ~ ~&
b)
, T
is s normal subgroup of
a semidirect product of iv)
~-
H
is
(automorphism) groups related to vector spaces,
see
with
Find the center of the following groups;
v)
Use this to strengthen the assertion of 6.5.
For information on various Baer ([i], Chapter VI), Dieudonn~ 1.7 1.7.1 DEFINITION. is an ideal of ii)
the property
o
[4], Gluskin
[2],14],[6],
~lotkin
([i], Chapter IV).
EXTENSIONS OF SEMIGROUPS AND RINGS
i) A semigroup
K
is an extension of a semigroup
S
if
S
K.
A congruence
~:S ~ S I
if
~,A,I,®,A,H.
Show that "element ~' in the statement of 6.4 can be substituted by
"idempotent".
only if
H, H = MT, M fJ T = ~V' so that
V+; express this explicitly.
that
~
on a semigroup
a ~ b
implies
is a homomorphism, ~p = y~
is an equivalence relation on
ac o bc, ca o cb
then the relation
is a congruence,
is a congruence on
S
o
for all defined on
a,b,c ~ S. S
by:
called the congruence induced by
S, then the set
S/~
of all classes of
with
If
x o Y ~.
o
S
if and
Conversely, can be given
s m u l t i p l i c a t i o n in a natural way. iii)
An extension
only congruence on
K
dense extension
of
containing iv) for all
K
K
K
of
S
is dense if the equality relation on
whose restriction to S
is maximal
S
K
is the
is the equality relation on
if there exists no dense extension
KI
S.
A
of
as a proper subsemigroup.
A semigroup
S
is weakly reductive
x E S, implies that
a = b.
if for any
a,b E S, ax = bx, xa = xb
S
55 The m a i n result of this section is the statement that dense extension of each of its nonzero ideals £u(V)).
1.7.2 DEFINITION.
w r i t t e n on the right,
and
Let
S
be a semigroup.
x(Xy) = (x0)y
%
The set I
(X,O)(%~,O)
~(S)
I
= (~h ,PO )
ha
and
pa
defined by:
sometimes write
~ = (~,O) E ~(S)
1.7.3 LEM~iA.
I
and
~(S)
PROOF.
Let
= ~wa'
S,
The pair
S.
For
XPa = xa
under the I
) = (xp)01
a E S, the
(x E S)
a, respectively;
xw = xp, wx = ~x
(x E S).
are called m s = (Xa,Pa) We will
(a E S, w E ~ ( g ) )
~(S).
Exercise. Let
K
be an e x t e n s i o n of a semigroup
and
p
k
are functions on xkx = kx,
T
S
I
I a ~ S}.
= T(K:S):k ~ ( k , p k )
Then
(x,y E S).
~(S) = ~ a
~a w = ~aw
is an ideal of
1.7.4 LEMMA.
k
on
is a right translation,
= ~(~ x), x(pp
of
p
With the n o t a t i o n as in 7.2, we have
~a so that
with
P
(X~)x ~(S)
%a x = ax
a.
S, w r i t t e n on
of b i t r a n s l a t i o n s of
where
inner left and inner right translations induced by is the inner b i t r a n s l a t i o n induced by
on
(xy)p = x(yp)
(x E S), is a s e m i g r o u p , t h e translational hull functions
~
(x,y E S); a function
is a left translation,
(x,y ~ S). I
multiplication
A function
~(xy) = (~x)y
is a right translation if
is s b i t r a n s l a t i o n if
where
is a m a x i m a l
For this we need some preparation.
the left, is a left translation if
(X,p)
gu(V)
(and the c o r r e s p o n d i n g statement for
K
and
~
(k,pk)
E ~(S)
(k E K)
(x E S).
into
~:a ~ ~a
Let
defined by
xp k = x k
is a h o m o m o r p h i s m o f
= TIS , we have
S
S.
~(S)
a n d maps
S
onto
~(S).
is one-to-one if and only i f
S
Letting
is weakly
reductive. PROOF.
That
from the a s s o c i a t i v i t y in xa = xb
(x E S)
K.
and that
For
a,b E S,
which implies that
~
f
is a h o m o m o r p h i s m follows immediately
~a
=
~h
is one-to-one
if and only if
ax
=
bx~
if and only if
S
is weakly
S
The next
reductive. Hence for a w e a k l y reductive
S, we may identify
result is due to Grillet and Petrieh 1.7.5 THEOREM. let
~ = T(K:S).
Let
Then
K K
with
be an extension of a weakly reductive semigroup is a dense extension of
S
if and only i f
one-to-on____ee, and is a maximal dense extension if and onl~ i f ~(S).
If we identify
S
w i t h all subsemigroups o f ~(S)
•
~(S).
[i].
with ~(S)
~(S),
~
then dense extensions o f
containing
~(S)
T
also maps S
S
and
is K
onto
may be identified
and maximal dense extensions with
56 PROOF.
Let
congruence then
for
on any
T.
o
~a
Let ~(S).
if
K
already K
K
so
that
dense
proved
that
K\~(S).
and
that
of
p
of
ax = bx,
of
~(S)
o
be a
a ~ b,
×a = xb. in the c o n g r u e n c e
that
K
is a d e n s e
containing
w o i,
then
the h y p o t h e s i s , aw = aw I
for
is the e q u a l i t y
S
K
is not
that
of
S,
~(S)
~(S) of
for w h i c h
see
that
for
~(S) any
~
whose E ~(S),
a
we o b t a i n
all
a ~ S
relation,
K~
~(S)
since
so
then
is a p r o p e r
is a d e n s e
a maximal •
dense must
is a m a x i m a l
is a d e n s e ~(S)
T
S
is
is a d e n s e
containing
extension
be onto.
dense
extension
of
~(S)
subsemigroup
extension of
S
S.
Hence
To p r o v e
extension
of
of
of
the ~(S).
~(S).
as a p r o p e r
subsemigroup,
a a~x'
on
S
~xP
extension
= ~x a
~ k x = a~x'
by
and w r i E i n g
T = T(K:~(S))
a dense
has
of
~
(x E S). ~Xp
=
= ~xa
(k,p)
the p r o p e r t y
~(S).
(x E S).
that
ka
aT = ~T
Therefore
~(S)
It is e a s y
k ~,
=
with
p
a
P
a ~ w
is a m a x i m a l
w
to
•
so that
dense
K
extension
~(S).
For
S
a semigroup,
in P e t r i c h
[3],
1.7.6 maximal its
dense
ideals
containing Let
extension
of
S,
of
T
since
a,b E T
S
the e q u a l i t y K
T
S
2
appears S
= {ab I a,b ~ S}. here
for
the
be a s e m i g r o u p
of
S.
Then
K
The n e x t
first
such
it f o l l o w s
D.
relation,
is a d e n s e
is m e n t i o n e d
time.
that
2
S
is a m a x i m a l
= S
dense
of
that
K
K
containing
is a d e n s e
S.
and
let
extension
of
D
If then
so g
bd E
T
and
is a c o n g r u e n c e o
extension
restricted of
S.
But
S
then
D
K
K
of
K
be a
of e a c h
of
Let
then
obtain
whose
is a d e n s e
T.
s 6 S,
is an ideal
on
to
Since
extension
containing K as a s u b s e m i g r o u p . If 2 since S = S. H e n c e for any d ~ D, we
of
theorem
S. be an ideal
is an ideal
is an ideal
since
Let
extension
extension some
let
its p r o o f
THEOREM.
PROOF.
S
and
we
=
(~,p) ~ ~ ( S ) ,
Consequently is not
let If
Then ~x
check
and
is c o n t a i n e d
If
7.3 and
~
extension
to show
~a
k
and
KT,
be an e x t e n s i o n
a E
Define
theorem relation.
we c o n c l u d e
T
relation.
wa = w l a
with
it s u f f i c e s
let
o
is o n e - t o - o n e ,
so by
extension
K
is a m a x i m a l
Let and
that
on a s u b s e m i g r o u p
w = i
be a d e n s e
containing
have
the
so by h y p o t h e s i s proves
the e q u a l i t y
Thus
of
is the e q u a l i t y
is o n e - t o - o n e .
Hence
i.
S
~(S).
K
converse, We
is
maw
=
Identifying
properly
T
TIS
' ~a ~ o ~a ~l,
reductive. of
that
if
~(S)
~ W~a ~aw
extension
to
xa o xb,
a congruence
to
~a
= ~a'
weakly
Recalling
be
restriction we h a v e
be as in the s t a t e m e n t
ax ~ bx,
if and o n l y
Let
K
restriction
h a = ~b , p a = p b , w h i c h
by
extension
and
whose
x E S,
Consequently induced
S
K
of
T.
to
for ~ S
Consequently
restriction
restricted
be a d e n s e
sd = a(bd)
is the e q u a l i t y g
D
s = ab
to
relation T
is
S on the
is K
57 equality
relation on
T, and since
D
is a dense extension of
is the equality relation on all of which by maximality extension
of
of
K
that
Thus D = K.
For a weakly
reductive
1.7.8 LEMMA.
K
is a maximal
dense
semisroup
S
for which
S 2 = S, ~(S)
N(S).
Let
0 ~ (k,p) E ~(S).
S = ~°(I,G,A;P)
[i].
be s Rees matrix
Then there exist a subset
~:A ~ G
(written on the left) and a subset
~:F ~ G
(written on the risht) ~(i,a,~)
A F
o_~f I of
A
semisroup
and let
and functions and functions
~:A ~ I, ~:F ~ A,
satisfying
(~i,(~i)a,~)
if
i E A
= 0
otherwise
(i,a(~),D~)
if
D E r
(i,a,B)p = 0
iii)
Therefore
that S
Exercise.
The next lemma can be found in Petrich
ii)
T, it follows
is a dense extension of
dense extension of each of its ideals containin$
PROOF.
i)
D
T.
1.7.7 COROLLARY. is a maximal
implies
D.
otherwise
i E A, PD(~i) # 0
if and only if
~ E [~, p(D~) i # 0
an_d i_~f s_~o, then
PD(~i)(q°i) = ( ~ ) p ( D ~ ) i , for all
(i,a,~) E S \ 0 .
PROOF. k, Q
The condition
is nonzero.
there exists # 0
implies If
means
But if, e.g., Xx # 0
y E S
such that
then since
y(kx) # 0
p # 0, and conversely,
that at least one of the functions
so that
S
is a Rees matrix semigroup,
(yp)x # 0
# O, then there exists
j 6 1
k(i,a,D)
-l = k[(i,a,D)(j,pDj,D) ] =
-1 [k(i,a,D)](j,pDj,D)
that
X(i,s,~)
Suppose next that is the identity
of
is of the form
l(i,l,D) G.
= (j,b,D) # 0
Then for some
(j,b,~) = ( j , b , ~ ) ( m , p ~ , ~ )
=
= k[(i,l,D)(m,p~,~)] Consequently
j = k
of
depends
k(i,l,~)
and
b = c
only upon
and
and hence both functions
k(i,a,D)
which implies
i
(l,p) # 0
such that
yp # 0.
are nonzero.
p~j # 0
and
# 0
( , ,~). and
k(i,l,~)
m ~ I, we have
= (k,c,~) # O, where
p~nn # 0
and thus
[X(i,l,~)](m,p~,~) = k(i,l,v)
= (k,c,~) # O.
which shows that the value of the first i.
Thus
two indices
58 This makes
it possible
~:A - I, ~ : A ~ G If now
to define
by the formula
k(i,a,D)
A = [i ~ I I k(i,l,D)
k(i,l,z)
= (~i,~i,v)
# 0, then for some
j E I, pDj # 0
= (~i,~i,p)(j,p~ja,p)= with
i ~ A.
The same calculation
h(i,l,D)
= 0
so that
argument
establishes
For any
i ~ A.
and the functions
~ 0]
(i E A). so that
(~i,(~i)a,D)
shows
that if
This proves
k(i,a,H ) = 0, then we must have
part i) of the lemma;
a symmetric
part ii).
i,j E I
and
D,v 6 A, we obtain
(~i
1 (j,I,D) [%(i,i,~)]
if
i E A
= ~(J'
L (j,l,~)0
otherwise
= {(j,pg(~i)(~i),~ )
if
i ~ A, P~(~i)
# 0
(1) 0
otherwise
and analogously [(j,l,H)p](i,l,m)
Since by hypothesis,
For the remainder
1.7.9 THEOREM.
For
as in 7.8 and define
I
Then
a E ~u(V),
b
{
let
is essentially S = ~2,u(V) a
~(B~)vH~
(2) are equal,
the assertions
taken
be a pair of dual vector
from petrich
[2].
an___~d 0 # (k,p) ~ ~(S),
an___dd b if
(U,h,V)
let
A,~,~,F,~,~
b_!e
by
~ E F
( T v E V),
[0
=
(2)
follow.
functions
(VvH)a =
~ 0
otherwise
of this section,
The next result
b(uiT)
~ E F, P ( ~ ) i
the left hand sides of (I) and
in part iii) of the lemma
spaces.
,(D~)p(D~)i,m)
= {(J 0
otherwise
u i(~i)T
if
0
otherwise
is the ad~oint kr = sr,
i E A
of
rp = ra
a
(uiT E u). in
U, and
(r ~ ~2,u(V)).
(3)
59 PROOF.
For any
~vp ~ V
and
u.rm ~ U, using 7.8 we obtain
((~vg)a'ui~) = { (~(p,)v ~,ui~ )
if
(O,ui,~)
g ~ F
otherwise
=~(~)(v ~,ui)~ if p E T 0
otherwise
=S~(v
[
,u i)(~i)m
if
i E A
0
otherwise
(~v ,u i(~i)T )
if
(~v ,0)
otherwise
i C A
= (~v , b ( u i T ) ) hence
b
is an adjoint of
unique sdjoint of
s
For any
E V
?v
in and
a
so that
a E £u(V)
and 1.2 implies that
U. [j,~,~] E ~2,u(V), using 7.8 we obtain
~vg(~[j,o,v]) =I~vp[~J'(~J)°'~]
if
j E A
0
otherwise
~(v ,u .)(~j)~v
if
0
otherwise
S~(~)(vD~,uj)~v v 0
if
j E A
D ~ F
otherwise
=I ~(D~)v ~[j,~,v] 0
if ~ E F otherwise
= VvD(a[j,o,~]) which proves the first part of (3).
~vB([j,~,~]p ) =
~
Finally
v [j,o(~,),v~]
L 0
~
if
~ E F
otherwise (v ,uj)o(v~)v
=L 0
if
v ~ F
otherwise
= ~(vp,uj)ov a = VVD([j,~,~]a ) which proves the second part of (3).
60 The following result is new; 1.7.10 THEOREM.
gu(V)
for
g(V)
it is due to Gluskin
[5].
is a maximal dense extension of each of its nonzero
ideals. PROOF. of
By 5.11 we know that
K = gu(V).
Since
weakly reductive and
S
S = ~2,u(V)
S 2 = S.
it suffices to show that
and thus T
for all
r E S
(k,p) E ~(S) If
a,b E K
S
is
according to 7.5 and 7.6,
is one-to-one and maps
with the n o t a t i o n of 7.9.
(a - b ) r = 0
it follows easily that
Thus to prove the theorem,
T = ~(K:S)
onto part follows from 7.9 since for a given (~a,pa) = (~,p)
is contained in every nonzero ideal
is a Rees matrix semigroup,
K
~(S).
The
we obtain
and
which by 3.3 implies
onto
a~ = bT, that
then
a = b
ar = br
and thus
is one-to-one. 1.7.11 COROLLARY. PROOF.
For every nonzero ideal
I
gu(V), we have
~(I) ~ gu(V).
This follows directly from 7.10 and 7.5.
We will now obtain the corresponding results £u(V).
of
for nonzero
For this we need the concepts and statements
of those we have seen for semigroups.
(ring) ideals of
for rings which are analogues
To this end we substitute semigroups with
rings w i t h the following modifications. In 7.1:
i) the definition remains
speak of ideals; only ideal of
iii) an e x t e n s i o n
K
K
the same; of a ring
whose intersection with
S
is
ii) instead of congruences, 0
we
S
is essential if
is the
0
(such an ideal is called larse);
iv) is expressed by saying that the annihilator ~(S) = ~a E S l as = sa = 0 is
for all
s C Sj
0. In 7.2 we only add that every left and every right translation also be an
additive homomorphism. In 7.6 the condition (i.e.,
S2 = S
can be w e a k e n e d to
the same kind of condition with
With these modifications,
S
2
defined as in ring theory).
all statements
of this assertion is left as an exercise
n [ ~_isiti~_ I si,ti ~ S} = S
in 7.3-7.7 remain valid;
the proof
(work with ideals instead of congruences
as is usual in ring theory). 1.7.12 LEMMA.
~u(V)
PROOF.
is a nonzero ideal of
~u(V),
If
I
is contained in every nonzero ideal of
so that by 5.11 we must have
£u(V),
~2,u(V) ~ I.
The following result appears to be new.
then
[[iI
£u(V).
is a nonzero ideal of
But then by 2.6 we obtain
61 1.7.13 THEOREM.
£u(V)
is a maximal
essential
extension of each of its nonzero
ideals. PROOF.
We know by 3.6 that
(ring and semigroup) ideal of theorem 3
~ = £u(V). it suffices
is regular,
that
condition
~
we have
essential
follows
p
= p132 .
that
32
maps
32
(~,pl)
essential
~
of
~
of 7.5,
onto
is a homomorphism.
to prove the
extension
5, it suffices
isomorphism
the
in every nonzero
of 7.6, in order
is a maximal
extension of
as in 7.4 is a (ring)
of
3.
in order
Since
to prove
to show that
~(3).
By the ring
That the kernel of
T
is
from 3.3.
To prove that I
£
is contained
~(~) = 0, so by the ring analogue
analogue of 7.4, we know that trivial
is regular and hence satisfies
Hence by the ring analogue to show that
is a maximal
• :~ ~ ~(3)
3 = 3u(V)
32 = 3; by 7.12, 3
T
maps
Then for any
into itself;
onto
r E 32,
is a Rees matrix
E ~(32).
~
semigroup).
Thus
p~
maps
(X,p) E ~(3). e 6 32
rp ~ = ra
Next let
such that
32
into itself.
X~ = X132,
r = re
hlr = h(re) = (hr)e E 32
a E ~
there exists
~Ir = ar,
let
there exists
similarly
By 7.9,
~($),
It follows
(recall
so that
~
that
such that
(r E 32).
n
If
c E 5, then
c = k~=irk
for some
n
n
and similarly
Cp = ca. that
T
1.7.14 COROLLARY. PROOF.
Since
maps
then implies
n
Xc = h ( ~k=l rl) ~Xrk ~ = k=l
which proves
r k E ~2' which
a E £,
£
n ac
= k=l ~ X'rk = ~ =lark = ak~=irk =
onto
it follows
For every nonzero
This follows directly
that
T
maps
a
onto
(X,p),
~(3). ideal
1
of
~u(V),
we have
from 7.13 and the ring analogue
~(I) ~ ~u(V).
of 7.5.
1.7.15 EXERCISES. i)
Show that every regular
semigroup
is weakly
reductive.
Deduce
that a
regular ring has trivial a n n i h i l a t o r ii) that
~
reductive
Let
~
be an isomorphism
induces
an isomorphism
and we identify
to a homomorphism
of
(~,p)~ = (~,p), where iii)
Let
respectively. Now let
S
S
with
~(S) ~x =
into
(X,p) ~ ~(S)
Show that the function
of
~
be invertible
onto a semigroup ~(S~). ~
If
S
S ~.
Prove
is weakly
is the unique extension
Do the same for rings.
of
(Hint: Let
xp = [(x~-l)p]~.)
~, p p
S
onto
show that
~(S~)o
and and
~(S)
~(S),
[~(x~-l)]~,
be a semigroup We say that
of a semigroup
~
be a left and right
are permutable and suppose
that
if X
translation
(Xx)p = k(xp) and
p
of
S,
(x ~ S).
are permutable.
62
6(X,p) is an automorphism
of
S.
(k,p),(~t,0 I) E ~(S), S
with
~(S),
morphism
of
iv)
that
Let
~
and
p
v)
Let
~
x(Xy) = (xp)y
Deduce
6(X,p) = e(x,p)l S
induced by
(X,p).
show that for any
In such a case~
where
s(X,p)
if we identify
is the inner auto-
Do the same for rings.
We call
be any ring and
(k,p),(k',p ~) E ~(~).
Deduce
that if
~(~) = 0
6(k,p )
Show
that for every
and
(X,p),(X~,pt)
a
E ~(~),
are permutable. be any ring, ~
(x,y E ~),
and
Prove
p
be functions
that for any
on
~
such that
r,s E ~,
k(r+s)
- (kr+ks)
E ~r(~),
b)
(rs)p - r(sp) E Z~(~),
(r+s)p
- (rp+sp)
~ ~(~).
that if
~r(~)
= ~£(~)
it suffices
e.g., when
= 0, for a pair
that for all
~
is a subring
~
define
(k,O)
of functions
x,y E ~, x(ky) = (xp)y. of
£u(V)
containing
on
~
Show
Su(V)
to be a
that this is
for some dual
(U,V). vi)
In any ring
xoy that under
vii) Define
Let
the circle
a
and
a mapping
a~
~(a,at )
Find an invertible
and
morphisms, viii)
Q(~)
be elements
~
~ a
of a ring
by
~
with
identity.
such that
a o a
~
= a' o a
= 0.
(X ~ ~).
oxoa (X,p) ~ ~(~)
the automorphism
prove
for which
~(a,a~)
is called
with
X
and
p
are permutable
quasi-inner.
and all generalized
Denoting
inner auto-
~(~) c ~(~) ~ Q(~) ~ ~(~).
~ = ~U(&,V)
a)
~(~) = ~ b l ~
b = ~V+a,
b)
Q(~) = is bl~
b ~ ~U(&,V)
c)
If
dim V >
i, prove
a ~ ~, b and
is finite dimensional,
d)
If
V
is infinite
e)
If
g
is a field with more
than one element.
is a semigroup
the sets of all quasi-inner
respectively,
o
by
For
V
composition (x,y E ~).
composition
bitzanslation
~(a,a I) = ~(k,p); ~(~)
the circle
= x+y+xy
~(a,a,):x
and
reductive,
are permutable.
X(rs) - (kr)s E ~r(~),
the case,
by
is weakly
pl
a)
bitranslation
Show
S
(x E S)
inner automorphism.
~
pair
If and
(Xr)pl -X(r0 I) E ~(~).
r E R,
then
prove
~(S)
seneralized
k
:x - (x-lx)p
b
statements.
is invertible].
then
dimensional,
the following
invertible].
then
~ @ $(~) = ~(~) = Q(~). ~ = ~(~) C ~(~) C Q($).
than two elements,
then
~(~)
has more
63 f) ix) onto
Q($) = ~(~)
Let
B ~.
B
and
if and only if
B~
Prove that
~
O(&) = 6(g).
be B o o l e a n rings and
~
be a one-to-one
is a ring isomorphism if and only if
of the corresponding semigroups of
B
and
B~
function of
~
B
is an isomorphism
under the circle composition.
Give
an example of an isomorphism of the additive groups of two Boolean rings which is not multiplicative. x)
Prove that if
~
is a Boolean ring,
Boolean ring has trivial annihilator,
so
then
~(~)
is also.
~ ~ ~(~) ~ ~(~)
of a Boolean ring into a Boolean ring with identity.
Show that a
provides an embedding
Do the same for a ring without
proper zero divisors. xi)
Prove that every extension of a Boolean ring by a Boolean ring is again
Boolean. xii)
Let
~
be a commutative ring without proper zero divisors.
the field of quotients of
3,
embedding of
Show that
xiii)
~
into
Let a ring
zero divisors
~. ~
and
~
be the image of
~
in
~
Let
~
be
in the usual
~(~) ~ i~(~l).
be an extension of a ring
if and only if
~
~.
Show that
~
has no proper
has no proper zero divisors and the extension is
essential. xiv)
Let
Prove that
~ ~
be a ring with
~(~) = 0
and
~
has unique addition if and only if
be an essential extension of ~
has unique addition,
the identity a u t o m o r p h i s m is the only endomorphism of
~
~.
and that
leaving every element of
fixed. xv)
A semigroup
(essential)
(or ring)
S
is said to be uniform if
extension of each of its nonzero ideals
S
is a maximal dense
(cf. 7.10 and 7.13).
Show
that the following semigroups and rings are uniform:
xvi)
a)
every simple semigroup or ring wlth identity,
b)
~x I0 < x < i}
under multiplication,
c)
ix I0 < x < i}
under multiplication,
d)
nonnegative
integers under addition,
e)
the semigroup of all transformations on a nonempty set,
f)
the ring of integers,
g)
the ring of polynomials
Let
R
be the field and topological space of real numbers, R[x]
ring of polynomials over functions m a p p i n g
R
set of all fractions zeros, and
in one indeterminate over a field.
R
in the indeterminate
into itself. p(x)/q(x)
deg q(x) - deg p(x) ~
maximal essential extension of
A . n
p(x),q(x) E R[x],
Show that
be the
be the ring of continuous
For every nonnegative integer n, let
such that n.
x, C
ic(An) = A °
q(x)
An
be the
has no real
and that
A°
is a
64
For the material
concerning
see Grillet
and Petrich
in Petrich
([5]j Chapter
Petrich
[i],[2],
B.E. Johnson
R~dei
ideal extensions
[i], Petrich IIl);
transformations
Lambek
([i], Chapter
related
article
a systematic
see Everett
([1],§§52,53,54,85);
[i] and the survey
linear
[i],[4],
in rings,
Petrich
12].
13].
subject,
in semigroups,
discussion
[i], MacLane
(c) in both semigroups
to the present
IV), Petrich
and translations
and
and rings~
see
For semigroups consult
can be found
[i], M~ller
and rings
Gluskin
[4],[5],
of
65
PART II
SEMIPRIM_E RINGS W I T H M I N I M A L This
is a study
of the class
restrictions
such as simplicity,
properties.
Prime
rings with
one of these being containing one-sided provided
that
a nonzero ideals
of finite
transformation
results
rank on an a r b i t r a r y
concordantly.
prime
rings w h i c h
are essential
in several
ways.
of semigroup
socles
Isomorphisms
of Rees m a t r i x
Each properties class.
of the classes
of rings
of the m u l t i p l i c a t i v e
This
is possible
are
all
with
a nonzero
the rings linear
of this
socle.
isomorphic
transformation
in several ways terpreted
in 1.2.
After
of n e e d e d
specified
II.l.l NOTATION. the set of all elements a ~ A, b E B.
If
of the entire
rings.
have
belonging
unique
by the
to the given
addition.
several
characterizations says
that
transformations
latter
itself
characterization
rings
they are p r e c i s e l y
containing
class was
of the class
of prime
a nonzero
characterized
can thus be in-
of concrete
rings
appearing
characterization.
an i n t r o d u c t i o n
ring unless
This
characterization
all the c h a r a c t e r i z a t i o n s
arbitrary
charac-
isomorphisms
PRIME RINGS
of linear
rank.
as semi-
are also
of rings,
characterized
of the rings
these rings
The first m e n t i o n e d
as an abstract
is also
characterizations
rings
of finite
to isomorphisms
the semi-
as well
socles
class
The nonzero and
found.
is to give
One of these to dense
in the second m e n t i o n e d
collect
section
rings
is
of the
transformations
space.
components
of their nonzero
semigroups
because
representation
characterizations
vector
atomic
of the latter
studied here
II.i The purpose
Semiprime
uniquely
rings
several
ways,
transformations
rings with minimal
and a m a t r i x
dimensional
extensions
are e x t e n d e d
Simple
into h o m o g e n e o u s
For a subclass
in several
as the ring of all linear
or a finite
is d e c o m p o s e d
socle
ways
additional
to certain
rings of linear
rank.
to give
as well
relative
are c h a r a c t e r i z e d
in several
is d e c o m p o s e d
title w i t h various
to dense
of finite
ring
group
socle
are used
transformations
socle of a semiprime
terized
a nonzero
IDEALS
and m a x i m a l i t y
they are isomorphic
These
linear
in the
atomicity
are c h a r a c t e r i z e d
for them.
ring of all
of rings
ONE-SIDED
If of
A = ~a},
A
and p r e l i m i n a r y theorem.
results,
We denote
by
we will ~
an
otherwise.
and
~
concepts
in a single
B
which
we write
are n o n e m p t y
subsets
are sums of elements aB
instead
of
of
let
AB
denote
of the form
ab
with
~a}B;
Ab
~,
has
a similar
66
meaning.
By induction, we define
that
is the set of all sums of elements of the form
An
si~
A n = (An-I)A
for
n >
i, w h e r e
A 1 = A, and see
sla2 .°. s
where
n
A11.1.2 DEFINITION.
if
In = 0
ideals.
for some
An (left, right,
n.
A ring
For s nonempty subset
~ B
two-sided)
is semiprime of
~,
b E B}
~r,~(B) = [r E ~ Ibr = 0
for all
b ~ BJ
Then
A
~] (~)
and
A semiprime ring L
is a two-sided ideal of
~
B
~/r(~)
~
be a left ideal of
N
is nilpotent
if it has no nonzero nilpotent
for all
It is clear that both
Let
of
the sets
are the left and the right annihilators of
II.l.3 LEMMA.
I
= [r E ~ I rb = 0
~,~(B)
PROOF.
ideal
in
3, respectively.
are nilpotent
ideals of
~.
has no one-sided nilpotent ideals. ~
such that
Ln = 0
and let
and a typical element of
An
A = I~.
is a sum of
elements of the form Pl
P2
Pn
(k~_-l~k_,irk,l ) (~k=l~k,2rk,2)
'''
(k~_-l~k,nrk,n)
'
But (~irl)(£2r2) ... (~nrn) = ~ i ( r i £ 2 ) . . . (rn_l%n)r n = 0 where
~i 6 L, r i E ~,
by hypothesis.
Thus
since
ri%i+ 1 ~ L
L ~ ~%(~)
where
and ~(~)
L n = 0. = 0
Hence
An = 0
by hypothesis,
so
A = 0
so that
L = 0.
The case of right ideals is snslogous. It follows from the lemma that the absence of nilpotent (2) right ideals,
(3) ideals are equivalent statements.
(i) left ideals,
We will be dealing mainly
with such rings.
if
11.1.4 DEFINITION. 2 A = A. Note that if
an
A
A (left, right,
two-sided)
ideal
is a minimal one-sided ideal of
ideal on the same side as
A; so by m i n i m a l i t y of
~, A
A
of
~
is idempotent
then
A2 ~ A
either
A 2= 0
and or
A2
is
A 2 = A.
Hence a minimal one-sided ideal is either nilpotent or idempotent° 11.1.5 LEMMA.
Every idempotent minimal one-sided ideal of
~
i_~s generated by
a__nnidempotent. PROOF.
Let 2
a E L, then La = L L.
L
since
Define
L
L
be an idempotent minimal left ideal of
= O, a contradiction, is a minimal
~ : x ~ xs
(x C L).
hence
La # 0
for some
3.
If a E L
La = 0
for all
and thus
left ideal. Consequently ea = a for some e in -I Then 0~ is a left ideal of ~ contained in L,
67
and is d i f f e r e n t Hence in
from
L
e 2 - e ~ 0~ -I
L, so
since
Le = L.
Hence
The following
~ e = L.
statement
11.1.6 COROLLARY.
e ~ O~ -I.
e 2 = e.
implies
will
By m i n i m a l i t y
Thus
Again
prove
In a semiprime
Le
the case
quite
of
L, we have
is a nonzero of right
0~ -I = O.
left ideal
ideals
contained
is analogous.
useful.
ring every m i n i m a l
one-sided
ideal
is gener-
ated by a n idempotent. II.l.7 NOTATION. denotes
by induction, with
If
A
and
the set of all elements we
a i ~ A. 11.1.8
nilpotent
that
An
left,
right or two-sided of
ab
subsets
with
of w r i t i n g
An .
S
ideal
remains
Further,
The a n n i h i l a t o r s
with
S
~z(S)
of a semigroup
a E A, b E B°
is the set of all elements
For a semigroup
definition
ideals.
are n o n e m p t y
the same convention
DEFINITION.
the semigroup nilpotent
see
Again
B
of the form
aB
of the form if
zero,
S, AB
Defining
An
ala 2 ... a n
A = ~a}
will
the d e f i n i t i o n
be used.
of a
the same as in the ring case w i t h
is a semiprime and
~r(S)
semigroup
are defined
if it has no
as in the case
of a ring.
Note group,
that in spite of the different
for a subset
A
11.1.9 LEMMA.
of
Exercise.
II.i.i0
DEFINITION.
ideal
L
of
S
different
of
S
from
L Let
Sa = L
if
S
~
semigroup
S
S
is semiprime.
with
such
in a ring or a semiAn = 0
in
nilpotent
a nonzero
left
(right,
two-sided)
are called
corresponding
subset
left ideal;
statement
twoideal
0-minimal).
Then a nonempty
is s m i n i m a l
~.
ideals.
left
(right,
ideals
or a ring.
0 # a E L The
An
zero,
no nonzero
theory
be a semigroup for every
of
if and only if
has no o n e - s i d e d
if it contains
(in semigroup
satisfying holds
definitions
in
In a semigroup
is m i n i m a l
II.i.ii LEMMA.
converse
An = 0
A semiprime
PROOF.
sided)
~,
is valid
L
the for right
ideals.
Let
PROOF.
The
L
the property
have
is a left any
ideal.
Let
L~
0 ~ a E L t, we have Conversely,
S, and
let
Sa = L
or
and
following
~(S)
suppose
0 ~ s E L. Sa = 0.
arguments
are valid
in the s t a t e m e n t be a left L = Sa ~ that
Then
S Sa
The latter
would be a nonzero
L~
for both
of the lemma.
ideal of so that
is semiprime,
S
that
L~ = L
and
let
nilpotent
cannot
ideal.
L
of
and a ring
It is easy
such
is a left ideal alternative
a semigroup
occur
contained since
then for
is minimal.
be s minimal S
to see that
0 # L ~ ~ L; L
S.
then
in
left L
ideal of and
a E ~(S),
thus
L
68
11.1.12
LEMMA.
A ring
[R
i_~s semiprime
if and only
if
~
is a semi~rime
semigroup. PROOF. an ideal I = 0
Necessity.
of
[R
snd
and hence
Let
~
be an ideal of
h~
with
by hypothesis
I n = 0.
implies
Then
C(I)
= 0.
But
ideals
of
C(I)
is then
is semiprime.
Sufficiency.
Obvious.
II.l.13 COROLLARY. and
I
[C(1)] n = 0, w h i c h
In a semiprime
ring
[R, minimal
one-sided
concerning
a n idempotent
[D0~ coincide.
II,i.14 LEMMA. semiprime i)
ring ~e
ii)
[R
following
left
is a d i v i s i o n
e~
is a m i n i m a l
PROOF.
i) =
group.
Let
e = ra
for some
statements
e~e
ii).
ring.
right
Since
ideal. ~e0{e = [Re ~ ~e~, we have
0 # s 6 ~e~e, then r E ~.
ere E e~e,
0 ~ a % ~e
ii) =
i).
(ae)ba # 0 y E e~e which
and since
that that
e
0 # ebs E
of ii) and iii)
II.l.15 Then
introduced
DEFINITION.
the subring
of
[R
[Ai}iEl,
and is d e n o t e d
be equal
to zero.
below Let
is an additive
that
[Ra = [Re.
Hence
[Ai}iE I
so
*ab~ ~ 0
e~e
is a d i v i s i o n
implies
which
of
~e
also
so that
ring,
there
exists
c = ce = cyeba E [Ra Ks c ~e
is a minimal
so that left
ideal.
by symmetry.
multiplicative, "division
hence
the same
ring" by ~'group with
to D i e u d o n n ~
statements zero".
The
[2].
be a n o n e m p t y @ A. iEl l
b E ~
implies
that
elements
and thus
for some
now follows
is due
the nonzero ring.
aba # 0
a E ~
entirely
by
e~e,
c E [Re, we obtain
I.ii
by r e p l a c i n g
generated
by
Since
Conversely, and
of
a = ae E a~
For any
0 # a E ~e
in a semigroup
notion of socle
implies
is a d i v i s i o n
Hence
e~e.
The proof of 1.14 is almost remain v a l i d
~e~e
Then
= e.
~ e c [Ra.
The e q u i v a l e n c e
so
is semiprime.
y(eba)
for every
e~e
= (ere)s
is the identity
group,
0 ~ a 6 ~e.
[R
and thus
such
proves
~a = ~ e
Let
since
and i.ii
that
Consequently
form a m u l t i p l i c a t i v e
([Ra) 2 # 0
of a
ideal.
e = ra = (er)(ea) where
e
are equivalent.
is a m i n i m a l
e~e
iii)
The
is called
family
of subrings
of
~.
a sum of the subrings
~ A i. A sum of an empty set of subrings is defined to iE I A sum of all m i n i m a l right ideals of [R is the socle ~ of [R.
69 Note
that
all the elements
r I + r 2 + ... + r n a right
where
ideal of
sum of all minimal
every minimal O # a E L.
that
by
inclusion = a(ex)
from
1.5,
i.ii,
where
~e~e
and
Let
of
~.
thus 0 } b E
proves
ideal
and hence
that
11.1.17
a 6 ~b~
from right
II.l.18
and
II.l.19
ideals
PROOF.
e
of
~
left
coincides
such
~
is
with
~
Thus
the
next
ideal and
that
that
so that exists
if
contains
~ e = L.
x E ~, we have
there
~
Further
is semiprime,
We prove
y E ~,
if and only
that
left
that
= 0, it follows
for some
ideal
to prove
since
a~.
for some
~(~)
proving
the opposite
b = ax = exD~ # 0
(ae)x and
thus
0 # exye E e~e, c E e~e
such
that
1.14,
~, we easily a form
= ae = a ~b~ = s~.
By i.ii,
a~
is a m i n i m a l
right
that
the set
theoretic
union
of all minimal
left
and
coincide.
left
ideals
socle.
is called
Because
interested
of 1.16,
in semiprime
rin$
~
is an ideal
the left socle,
whereas
we need not make
rings.
The
of
the
following
~. the socle
left-right
concept
is
[4].
DEFINITION.
A ring
LEMMA.
If
~
First note
~ = ~e~e &
act on
see
that
~ ~ 0 right
is a simple
spaces that
that every m i n i m a l
Letting
~
be a m i n i m a l
The socle of a semiprime
the risht
of dual v e c t o r
Define
that
therefore
is a sum of its minimal
~.
L
b~ c
: s(exyec)
the proof
w h e n we are
to J a c o b s o n
By
R i, and that
~ ~.
COROLLARY.
is called
e.
ideals
has a minimal it suffices
let
ring by 1.14.
The sum of all m i n i m a l
implies
~
Hence
It follows
a E ~
It follows of all minimal
(i.e., ~
right
the form
Consequently
which
(U,V)
that
'~R, then
b(yec)
distinction
are of
the socle
an idempotent 2 ~ ( *a~)a = (~a) = ~a
is a division
(exye)c = e.
~,
By symmetry,
and hence ex # 0. Since 2 } 0. Hence exye # 0
(ex~)
due
rin$
there exists
also holds.
also
above
1.14
ideal.
left ideal
aS # 0.
~
left ideals.
right
By
for some m i n i m a l
In a semiprime
It follows
it has m i n i m a l
~a = L
ri E Ri
socle
~.
II.i.16 LEMMA.
PROOF.
of a n o n z e r o
~2 ~ 0
right
implies
ideal
by
with
its socle
then
that
~
U = ~e
is a left and
U (U,&)
= e(rs)e
~ ~ Su(V)
fo__~rsome
pair
g.
is of the form
ring and
(er,se)
if it coincides
rin$,
rin$
on the left and on
(g,V)
~:V xU ~ &
atomic
over a d i v i s i o n
is a d i v i s i o n V
is atomic ideals).
is semiprime. V = e~
Hence
1.6
for some idempotent
is a minimal
left ideal of
on the right by m u l t i p l i c a t i o n is a right v e c t o r (r,s E ~).
Then
~
space over
in ~.
is e v i d e n t l y
70 bilinear.
If
and hence
erse = e # 0
(er,se) # O
er # 0, then
for some
(U,A,V;~) = (U~V) For every
e~
= ~
= V
r E ~
and hence
ers = e
(er,se) # 0. ~
for some
Dually
is nondegenerate.
se # 0
s E implies
Consequently
is a pair of dual vector spaces.
r E [~, let
r
be defined by
(es)r = e(sr) then it follows inm~ediately that
(s 6 ~); r ~ £(~,V).
~(se) = (rs)e then similarly
by 1.11, so
which shows that
r ~ £(U,g).
Define
r
by
(s E ~);
Since
((es)r, te) = (e(sr),te) = esrte = (es,r(te)) = (es,~(te)), it follows that
9
The m a p p i n g e ~ O, we have
)<:r ~ r
hence
(r E ~)
ker X # 0{
it follows that For any
is the adjoint of
Thus
r E b~, we have
rank ~ = i.
Since
in
U, and hence
X
I~
is simple and
ker X
is an isomorphism of
(er)e = ere = (ere)e
Su(V)
r E £u(V).
is easily seen to be a homomorphism.
and since
ker X = 0.
r
is an ideal of
where ~u(V),
Since
is an ideal of
~
into
~,
~u(V).
ere ~ &, e E V, and the set
T = {r E '~ I ~ E au(V)] must be a nonzero ideal of For
~.
Hence
[i,y,k] 6 $2~u(V), we have
T = ~
and thus
X
maps
u.l = te, y = epe, vl = er
~
into
Su(V).
for some
t,p,r E ~{, and thus (es)[i,y,~] = ( e s ) ( t e ) ( e p e ) ( e r ) = which shows that
(teper))¢ =
that
onto
X
maps
~
[i,~{,h].
~u(V).
e(steper) = (es)teper
Since
Therefore
~2,u(V)
generates
~u(V),
it follows
~ ~ Su(V).
The following concepts will play an important role. II.i.20 DEFINITION. a ring.
If
~
acts on
Let M
M
be an additively w r i t t e n abelian group and
on the right satisfying:
(x + y ) a
= xa + y a ,
x(a+b)
= xa+xb,
for any
be
x,y E M, a,b E ~,
(×a)b = x(ab), then
M
for w h i c h
is a .risht ~ - m o d u l e or simply an ~-module. xa ~ N
~ B ~_ ~, let the form
ab
AB with
for all
x E N, a E ~
is an ~ - s u b m o d u l e of
be the set of all elements of a ~ A, b E B.
An additive subgroup
M
M.
If
N
of
M
~ ~ A c_ M,
which are sums of elements of
71 An ~ - m o d u l e (2) M and
M
~
are denoted
r E 5,
there exists
11,1.21 module. AB ~
An c
ideals
a ring
A
B ~
I
or of
that
~
m E M M
and
since
II.i.23 semisroup ideal
of
K
and
Let
I
~
A
irreducible
and
is a prime
B
of
ring
if for any ideal
0
ring
is a semiprime
A
2,
if the of
~,
ideal
of
~.
ring and can be taken
if and only
and
B
M
as
if for any nonzero
mA = M
then
be nonzero
is faithful since
mA
0 = mAB = MB
Consequently
nonzero
ideals.
on
to
S
with
AB # 0 zero
ideals
and
then
K
S
that
K ~
of
~,
A ~ 0,
and
there
is an ~ - s u b m o d u l e which
and
of
is impossible
~
and a dense
ideals,
On
or K
since
is a prime extension
is c o n t a i n e d
ring.
of a
in every nonzero
such S
x = y.
define
It is easy
is the equality S
is a dense
A semigroup
the
I.
to see
relation.
extension
is completely
Thus
following
of
0 - ~
K P I = 0
relation:
since
x T y
if
that
T
is a congruence
Thus
T
must be
K, but
then
the
1 = 0.
if it is isomorphic
to
semigroup.
is not a customary
and Preston result
the Jscobson
Density
in G l u s k i n
but
definition
it turns out
of a completely to be equivalent
([i], Chapter 3, S e c t i o n
of this
are due to 3 a c o b s o n proved
A
is a s e m i g r o u p
x,y E I
theory of semigroups,
principal
if
Since
AB = 0,
be an ideal of
11.1.24 DEFINITION.
Clifford
If
S
restriction
relation
s Rees m a t r i x
This
~-module.
proper
nonzero
if either
whose
equality
element
tins is prime.
m A # 0, so that
If
without
Further,
of a semiprime
ring,
is faithful.
LEMMA.
has no proper
S
M
M
S.
PROOF.
and only
that
is irreducible.
B ~ 0
and
AB # O.
primitive
irreducible
such
I.
is a prime
~, we have
be a primitive
be a faithful
if it has a faithful
if for any ideals
if and only
Every
~
if for any nonzero
is .semiprime
the d e f i n i t i o n
II.i.22 LEMMA. Let
~
r E ~
the zero of both
mr # 0.
is primitive
A ~
is semiprime
of
that
m E M,
(caution:
is faithful such
ideal
I
for some 0
I.
with
note
B
and
is a prime
PROOF.
exists
on
A ~
~
Also
and
M
K
that
~
M
M
m E M
An ideal
is in conformity
a definition.
~
that either
implies
(i) mr @ 0
from
Further,
A ring
of
is prime.
Hence
0).
an element
I
if
different
DEFINITION.
implies 0
I
This
by
A n ideal
I
ideal
M
is irreducible
has no ~ - s u b m o d u l e
section.
Theorem, [4].
[4].
The
2).
0-simple
semigroup
to the usual
We are now ready
The i m p l i c a t i o n
ii) = v)
one,
to prove
is a special
[3].
implications
i) = 1.2.13 i) and vii) = 1 . 2 . 1 3 i )
implications
appeared
ii) ~ v) ~
in Petrich
the
case of
see J a c o b s o n
The remaining
The equivalences
in the see
[4].
1.2.13 i) are
72 11.1.25 THEOREM. i)
~
The followin$ conditions on a rin$
~
is a semiprime ring with a minimal right ideal
are equivalent. A
such that
~r,~(A) = 0. ii)
~
is a primitive rin$ with a nonzero socle.
iii)
~
is ~ prime ring with a nonzero socle.
iv)
~
is an essential extension of its simple socle.
v)
~
is isomorphic to a dense ring of linear transformations on a vector
space containing a nonzero t r a n s f o r m a t i o n of finite rank. vi) vii)
~O~
is
~
has a completely 0-simple
PROOF.
dense extension of a completely O~simple semigroup.
a
i) = ii).
We consider
ideal contained in every nonzero ideal of
A
as an ~-module.
Then its m i n i m a l i t y as a
right ideal implies that it is irreducible and the condition
~r,~(A) = 0
that it
is faithful. ii) = iii).
This is a special case of 1.22.
iii) = iv).
Let
E
be the socle of
be a minimal right ideal of
~.
If
I ~ ~r(~) = 0, a contradiction. is a prime ring.
Hence
exists an idempotent where
re E E
that
es E ~
Similarly
bra ~ 0
and
for some
ce # 0.
s E ~.
again by 1.17.
0 ~ A 2 = (c~)(~),
A = e~.
Further,
that
contained in
d E I
1
so that
that
dE c
idempotent, we have If
J
is an ideal of
is prime we must have iv) = v).
Letting
and thus ~
E
~
of
~u(V)
Su(V).
Thus
containing
Su(V).
since
By 1.6, there b(re)a # 0
c = bre, we see
(~c 'e~)2 ~ 0
that
so that
0 ~ d ~ A N I
since
the hypothesis yields u E ~.
It follows that
d ~ = A.
This shows that I = ~.
Since
dE
is
On the other hand, I
~
J ~ ~ = O, then also ~
contains every contains a nonzero
~ ~ ~u(V).
Finally,
J ~ = 0, and since
is an essential extension of
~.
~, by 1.19 there exists a pair By 1.7.13, £u(V)
By the ring v e r s i o n of 1.7.5, ~
of linear transformations.
Letting
A
is simple.
be the socle of
of dual v e c t o r spaces such that extension of
~
a ~ A. and thus
implies that
and thus
that
such that
J = O.
A
A ~ I.
~, which implies
~2 # 0
b(re) E I.
for some
~
implies
a = ea
~, and
and thus
(N I ~ ) ( ~ A ) # 0
b E I, r E ~,
i.ii together with
d(ud) # 0
ideal of
P~ ~ ~£(~) = 0
d = esc, we deduce
a nonzero right ideal of
minimal right ideal of
be a nonzero
Hence
The hypothesis
Letting
and hence
I
~ A # 0, and thus
for some
such that
by 1.17, which implies
0 ~ c E I
esc # 0
e
~,
~ I ~ = 0, then
(U,V)
is a maximal essential
is isomorphic to a subring
1.2.13 asserts that
~
is a dense ring
73 v) = vi). (U,V)
By 1.2.13, we have
of dual vector spaces.
dense extension of
~2,u(V).
0-simple semigroup
~2,u(V).
vi) = vii). calculation),
~ ~ ~,
where
Su(V) ~ ~J ~ ~u(V)
By 1.7.10, we have that Hence
~J
for some pair
gu(V) = ~£u(V)
is a maximal
is a dense extension of the completely
Since a Rees matrix semigroup has no nonzero proper ideals
the same holds for a completely 0-simple semigroup.
(simple
The desired
conclusion now follows from 1.23. vii) = i).
The hypothesis
question is unique.
Letting
in every nonzero ideal of implies that
~
see that for every
a right ideal of
since
~
A
diction.
~.
a E A
M
contains nonzero idempotents,
such that
since
er E R.
~I~.
If
B
~
is a right ideal of
~
Further,
J ~ O, then
~,
A
since
A
and thus
A
be a
B = A
A
is also
A, then or ~
B
is
B = 0. by 1.13
is a right ideal of ~, J = ~r,~(A) is 2 0 ~ A ~ AI C AJ = O, a contra-
~r,~(A) = 0. M
is defined in a dual manner
on the left.
to a (right) ~ - m o d u l e by making
The d e f i n i t i o n of a "left ~ - s u b m o d u l e " ,
"left primitive ring" should be clear.
left and right can be interchanged.
with
in
Hence for any
snd thus also of
I c j
right socles coincide in this case. ideal
a = ae.
contained in
"irreducible",
Now note that conditions vi)
and vii) contain no left or right modifier, which implies conditions
~
contained
the hypothesis Let
Consequently
which by m i n i m a l i t y implies that
is a minimal right ideal of
If
~
By a simple c a l c u l a t i o n in a Rees matrix semigroup, we e ~ ~
A left ~ - m o d u l e act on
~
is an ideal of
there exists
Therefore
"faithful",
Since
I
ar = (ae)r = a(er) E A
is semiprime.
an ideal of
~.
R.
also a right ideal of Consequently
that the completely 0-simple ideal
cannot have nilpotent ideals and is thus semiprime.
minimal right ideal of
r E ~, we have
implies
I = C(~), we see that
that in the remaining
Indeed 1.16 says that left and
In condition i) we may take a minimal
~%,~(A) = O; in ii) a left primitive
left
ring; in v) the linear trans-
formations can be taken on a left or right vector space. II.i.26 DEFINITION. the semigroup socle Exercise
g
The set theoretic union of minimal right ideals of of
~
is
~.
1.28,v) below implies that
after 1.16 that
~
~
is an ideal of
~IR.
We have remarked
agrees with the set theoretic union of sll minimal
left ideals
in a semiprime ring. II.i.27 COROLLARY.
A rin$
~
of dual vector spaces if and only if maximal relative to ii) or iii) semigroup socle.
i__ssisomorphic t___0_o£u(V) ~
for some pair
(U,V)
satisfies the conditions of 1.25 and is
or iv) for a fixed socle or vi) o__~rvii) for a fixed
PROOF.
This follows from 1.25,
1.2.13,
1.7.10 and 1.7.13.
II.i.28 EXERCISES. i)
Let
~
be a ring and
I
be a nonzero ideal of
~.
Prove the following
statements. (a) ~(I)
= 0
~(I) = 0
where (b)
~(I)
and
~
is an essential extension of
I
if and only if
= lr E ~ I rI = Ir = 0].
Mr(I ) = 0
and
~
is an essential extension of
I
if and only if
~r,~(I) = 0. ii)
Show that a ring
~
for which
~
is completely 0-simple must be a
division ring. iii)
Show that a prime ring with a nonzero socle and without nonzero nilpotent
elements must be a division ring (an element
r
is nilpotent
if
rn = 0
for some
positive integer n). iv)
Let
the ring
v) that
~
and
~
in 1.25.
denote
the semigroup socle and the socle, respectively,
~ = C(~)
(b)
The sets of minimal
(c)
Every
Let
A
Let
0 # a E A.
A
and is contained in every nonzero ideal of
be a minimal 0
~
right ideal and
~. ~
coincide.
is also a (left) right ideal of a
be an element of a ring
or a minimal right ideal of
13],[4],15].
~.
Show
~. ~
and let
for this section are Dieudonn~
[2],[3], G l u s k i n
Further material concerning primitive rings and rings
of linear transformations can be found in Behrens
([i], Chapter II, Section
i),
(17], Chapter II, Section 2 and Chapter IV, Sections 9, 15), Kaplsnsky
Part II, Section 6), McCoy Section 2), Wolfson
~.
aA = a~ = A.
The main original references
Jacobson
R, ~, and
be an idempotent minimal right ideal of a ring
Show that
[4} and J a c o b s o n
(left) right ideals of
(left) right ideal of
is either
of
Prove the following statements.
(a)
aA vi)
~
([I], Sections
20,21,23-27), R i b e n b o i m
([4],
([i], Chapter III,
[i],[3],[4].
II.2 After some preliminaries, sided ideals in several ways.
SIMPLE RINGS
we will characterize simple rings with minimal oneWe will supplement these results by establishing
further properties of dual pairs of vector spaces in connection with these rings.
75
II.2.1 product a k x
DEFINITION.
~ o p and
If
x p b
for some
II.2.2 DEFINITION. i)
a~b
ii)
Then
~
£, R, ~
are b i n a r y on
X
relations
given by:
on a set
a k o p b
X,
their
if and only
if
x E X. S, define
the following
relations.
= b U Sb.
a U aS = b U bs.
are called Green's
Clearly
£
left
generate
then
p
~ = ~ o ~.
principal b
and
relation
On a semigroup
a USa
aR b ~
iii)
k
is the b i n a r y
L(a)
and
~
are equivalence
ideal of
S
generated
the same principal = Sa.
relations.
left
If
a
statements
In a Rees m a t r i x
The set
a, so that
ideal.
The c o r r e s p o n d i n g
II.2.3 LEMMA.
relations.
by
hold
semisroup,
a~b
L(a)
= s USa
if and only
has a left identity, for right
is the
if
a
say
and a = ea,
ideals.
any two nonzero
elements
are
~-related. PROOF.
For
the v e r i f i c a t i o n
S = ~ that
o
(I,G,M;P)
(i,a,p) £
and
(i,a,p),(j,b,~)
(j,a,D)
and
E S
where
(j,a,~) ~% (j,b,v)
a # 0, b # 0, is left as an
exercise. II.2.4 a function of abelian
DEFINITION. ~
groups
a E A, r E ~. and
B
(respectively
A
and
If
are said
II.2.5
If
mapping
~
A
and
into
B
B
are ~ - m o d u l e s
(respectively
is an ~ - h o m o m o r p h i s m
(~a)r = ~(ar) is one-to-one
(respectively and onto
to be ~ - i s o m o r p h i c ,
r(a~)
it is called
to be denoted
if
~
left ~ - m o d u l e s ) ,
is a h o m o m o r p h i s m
= (ra)~)
for all
an N - i s o m o r p h i s m
by "A ~ B
and
A
as ~ - m o d u l e s "
as left ~ - m o d u l e s ) .
LEMMA.
The
followin$
conditions
on idempotents
e
and
f
of a ring
N
are equivalent. i) ii)
e~ ~
f~
N e --~ N f
iii)
There
iv)
e g f.
PROOF. Then
as left N - m o d u l e s .
exist
ii) =
a E ~f,
as N - m o d u l e s .
a,b E ~
iii).
b E Ne
such
Let
~:Ne ~ ~f
f =
since
ab = e, ba = f.
be an N - i s o m o r p h i s m ,
and
-i -i e = eqxp = a%0 =
-i %0
that
f~-~
=
~
=
(af)~ -I = a(f~ -I) = ab,
(be)~
is also an N - i s o m o r p h i s m .
= b(~)
=
ba,
and
a = e~, b = f q - i
76 iii) =
iv).
Letting
c = fbe, we obtain
e = ab = a(ba)b = a(fbe) which
implies iv) =
and also
that
ii).
e £
Let
c.
One
shows
e £ b, b ~
b = fb = be.
f.
Letting
and
= ae E ~ c
similarly Then
that
e = xb
c ~
and
f
so that
f = by
e ~ f.
for some
x,y E
a = xf, we obtain
ab = (xf)b = x(fb) ba = b(xf)
c ~ ~e
= xb = e,
= bx(by)
= b(xb)y
= bey = by = f,
a = af. Then of
the m a p p i n g s ~e
r~
into
~ : r ~ ra
~f
and
= rab = re = r and
~
(r E ~e), into
~e,
and similarly
~:r ~ rb
(r E ~f),
respectively.
r@~ = r
For any
for any
are ~ - h o m o m o r p h i s m s r E ~e
r ~ ~f,
which
we have proves
that both
are ~ - i s o m o r p h i s m s .
The equivalence II.2.6 function
~f
of i) and iii)
DEFINITION.
P =
(phi)
row-independent
now
follows
For a d i v i s i o n
mapping
A x I
if for any finite
ring
into set
by symmetry.
&
g
and nonempty
sets
I
is a A × I-matrix
over
g; F
kl,k2,...,k n E A
and
A, any is left
and scalars
n
O 1 , O 2 , . . . , O n ~ g, o I = 02 =
... = o n = 0; right
II.2.7 division
the relation
DEFINITION
ring,
p =
(cf.
(p%i) &; by
over
having
a finite number
(aik) + (bik) =
(aik + b i k ) on the right
is possible
exercise
to show
over
with
£
Caution. We are
ready
precisely
and iv)
in the next
by a m p l i f i e d version
of the Rees
Section
2).
a ring
for
the m a i n
the rings
on
gu(V)
closely
theorem
The r e m a i n i n g
here
to this,
by Hotzel
in semigroups,
the
of m a t r i c e s
it the Rees m a t r i x
ring
rings".
pair [4].
is due
It follows
easily
ideal
up to an
(U,V).
are,
This was
The e q u i v a l e n c e to J a c o b s o n
[l] and its proof
see C l i f f o r d
are new.
£
ring.
one-sided
time;
over
by entries
It is left as an
of this section.
for the first
equivalences
addition
B).
by G l u s k i n
be a
A * B = APB, where
"Rees
for some dual
&
I ×A-matrices
we call
a minimal
related
established
appears
and
is called
[2] and r e d i s c o v e r e d
theorem,
A
sets,
column-independent
multiplication
a Rees m a t r i x
result
analogously.
with
given by
in this way:
rings w i t h
implies
be nonempty
together *
I
and right
row-by-column
class of rings
of i) and v) was somewhat)
A
the set of all
entries,
p, or simply
that simple
by D i e u d o n n ~
denote
i 6
is defined
and
of the hypothesis
matrix
established
I
is the usual
An u n r e l a t e d
1.2.13
Let
of nonzero
that we obtain
sandwich
isomorphism,
equivalence
1.4.4).
and m u l t i p l i c a t i o n
because
finally
from 1.25 and
column-independence
~(I,g,A;P)
multiplication (which
for every
be a left r o w - i n d e p e n d e n t
A × I-matrix only
~ o.p. i = 0 j=l ] ~.j
The
(due to Hotzel
it represents and preston
[3].
of i)
the ring
([i], C h a p t e r 3,
77 II.2.8 THEOREM.
i)
The followin$ conditions on a rin$
i__s_s~ simple ring w i t h !
ii)
is a prime
iii)
(respectively ~rimitive)
is isomorphic
v)
~
vi)
~
i) = v).
Since
v) = iv). A = e~
satisfies
~
and
First let
right ideal of
But then
A
since
A = e~
let
I
follows that
C(~)
~
e E ~.
If
A = e~
A
~,
and thus
I~ = ~ I = O.
contains its negative since
right ideals.
A
is a
and thus the equality ~, minimal
I [~ R = 0.
It
We have just seen that every
is a minimal right ideal of ~
~.
Hence with every element
is the union of its minimal
It follows that the elements of ~ are sums of elements in R. n t = ~ t~ with ti ~ ~ and n is minimal. i=l
Let
have the property that
hypothesis exists
then
w h i c h implies that
[E)l and suppose that
a
0 ~ t ~ I
and thus
We know by 1.3.11
r ~ ~,
is a right ideal of
~
~
D~
R.
be a nonzero ideal of
the ideal
~.
and
~ e(~L~) c A
which implies that
I ~ = RI = 0
a C A
minimal right ideal of ~,
has a
is a nonzero ideal of
is an ideal of
~0~. C o n s e q u e n t l y
Hence
condition iv) of 1.25, [E~
be a minimal right ideal of
in view of its m i n i m a l i t y in
of
~ = C(~).
~.
for some idempotent
ar = (ea)r = e(ar) E A
prevails.
~
~.
by simplicity of
Next
for principal risht ideals.
is isomorphic to a Rees matrix ring.
completely 0-simple ideal = C(~)
atomic ring.
to a dense ring of linear transformations of finite rank.
has a completely 0-simj~le ideal
PROOF.
that
minimal__ right ideal.
is a simple ring satisfying d.c.c,
iv)
are equivalent.
I ~ R = 0
e.~ ~ ~
implies
such that
that
n >
t i = tie i
i.
for
Since
~
I < i < n.
0 = (tl+t2+...+tn)[el+e2+...+en-
is completely 0-simple,
The there
Thus
el(e2+ ...+en) ]
n
n
= t + i=2 ~ t.(e~ • i + " . ' + e i -i + e i + l + ' "" + e n ) - i~=2tiel (e 2 + " ' ' + en)' so that n t = i~=2[i[el(e2 + ... + e n ) - (e I + ... + e i _ 1 + ei+ 1 + ... + e n ) ] where all the summands are in I N ~ # 0, and since ~
I.
I n ~
Consequently
R
and 1.2.13 implies that (U,V)
of vector spaces.
~,
c o n t r a d i c t i n g the m i n i m a l i t y of
is an ideal of
But
where
Hence
~
Su(V) ~ ~' ~ £u(V)
Su(V) = C(~2,u(V)) ~,
n.
Therefore
I n ~ = ~
is contained in every nonzero ideal of ~ ~ ~i
unique completely 0-simple ideal of ~' = ~u(V).
~, we must have
and thus
[E~, which by 1.25
for some dual pair
by 1.2.6 and
~2,u(V)
which by hypothesis implies
is the
that
is a dense ring of linear transformations of finite rank.
78 iv) = iii). Su(V)
Agsin by 1.2.13,
satisfies
d.c.c,
iii) = ii). ideal
A.
If
The hypothesis
B
0 ~ (b) ~ B ~ A
primitive)
implies
each implies
(b)
that
By 1.2.13,
we hsve
ring
of
zxs = onto
~(V)
~ o~ z .
the ring
~
g.
z
and
~
let
in
U.
o~
For sny
oX~ # 0
W = [wi}i~ I
right
hss the property
As above,
we conclude
thst
(respectively
in an atomic ring,
~
is also simple.
X E A
of dusl vector
Z = ~z }%EA
of
~:a ~ ( ~ )
A × A-matrices
s E £(V),
(U,V)
by letting that
~
V
snd represent (aE~(V))
is an isomorphism
over
&
the subspace is equsl
and
under the usual
of
to
of
V
spanned by the
Va, so that
identi-
be a basis of
of
U, let
A
have nonzero
a ~ ~u(V)
and
elements}. b
(i)
be the adjoint
We set
be as above
where and
P~i = (z~'wi) bw. = i
(i E I, ~ E A).
~ w ~.. it follows j jl'
that
(z a,wi) = (zk,bwi)
jEI (~O%
z ,wi) = (z~,~wjTij)j
~ox~P~i= ~ i P ~ j T j i
tation, we obtain that for the linear ~u(V) = [ A E ~ I
AF = PB.
"
We let
there exists
A = (o%),
Conversely,
transformation
s
(i E I, ~ E A)
that
~u(V) = £u(V)
B = (~ji) , so in the matrix no-
the last formula,
we must have
a column finite
going backwards,
a E ~u(V).
I × 1-matrix
such that
~u(V).
b
hence
in 1.6.12
for some
implies
Recslling
by
for a pair
&
a finite number of columns
P = (p~i)
so that
principsl
we have
~(V) = ~ A E ~ l o n l y
Letting
a minimal
are equivalent
over
of all row-finite
for which
~(V)
Now
A.
We fix a basis
We have remarked
addition and multiplication.
a
1.3.15 ssys that
0 ~ B c A, then for any
is also a prime
~ ~ ~u(V)
ss A × A - m a t r i c e s
of
of
~
that the socle of the ring is simple,
the elements
fying
contains
generated
the minimality
1.25 implies
spaces over a division
vectors
~
such that
By 1.25, prime and primitive
iv) = vi).
£(V)
Further,
ring~
ii) = i).
where
that
~
right ideal
contradicting Further,
~ ~ ~u(V).
right ideals.
is a right ideal of
0 # b ~ B, the principal
is atomic.
we have
for principal
B
AP = PB}.
implies
Consequently
over (2)
~ ~(V), we have by (i) and (2) the form of matrices
in
79
n
Suppose all
i E I
that
n
j=l~~'pX j nji = 0
for every
i E I.
Then
j~==l(~jzXj,wi)
= 0
for
( ~ o.z.,u) = 0 for all u E U. By the nondegeneraey of the j=l J J n form, we must have ~ o.zx = 0, which by linear independence of the j=l 3 j
bilinear
and hence
z%. implies Ol = o2 = ''" = On = 0. Thus p is left row-independent; J one verifies analogously that P is also right column-independent.
vectors
For u
a E ~u(V),
= ~w.T . ik j I Jl k'
v
z~a =
by 1.2.6, we have
a = k=l[ik,~k,Xk]~ .
Letting
= ~Jo, z , we obtain k D Ak~ D
k=l (zx' Uik)~kVXk
=
~ (z~,~w.T. i )~k(~OX z ) = ~ ( p ~ j T J i k ~ k O X k ) z D. k=l J J J k D k ~ k,j,B
(3)
As above we let A = ( o X ) where z a = ~ z , and now form a matrix n M A = ( ~ T.i ~k°X )' Then M is a I x A - m a t r i x over & having only a finite k=l J k k~ A number of nonzero entries; in matrix notation (3) becomes PM A = A. Suppose that C = (ci~)
is an I X A - m a t r i x
for which
PC = A
and let
over
A
with only a finite number of nonzero
M A = (mi~).
Then
P(M A - C) = 0
which
entries
implies
that
• Phi (miB- ciD ) = 0 for all ~,D E A. Since at most a finite number of mid ci~ ~i may be nonzero, this is a finite sum, and thus the right column-independence of P implies
that
equation
mi~ = ciB
PM A = A
for all
uniquely
Define a function
i,D.
Consequently
determines
~:~u(V) ~
C = MA
which
shows that the
M A.
[~(I,A,A;P)
by
~:A ~ M A.
For
A,A ~ E ~U(V),
we then have AA ~ = (PMA)(PMA~)
= P(MAPMAI ) = P(M A * M A ~ ) ,
AA ~ = PMAA ~ , which
implies
that
ring homomorphism. finite,
MAAI = M A * M A ~ . Let
M E
~(I,~,A;P),
only a finite number of columns
AP = (PM)P = PB
where
B = MP
from (i) and (2) above that onto.
Similarly
If
M A = 0, then
is an isomorphism
of
then for A
A @ ~u(V)
onto
= M A+MAt
and also
entries,
IX I-matrix over
4.
M A = M, which proves
and the kernel of ~(I,~,A;P).
so that
A = PM, we see that
contain nonzero
is a column-finite
A = PM A = 0 ~u(V)
of
MA+At
~
is trivial.
~ A
is a is row-
and It follows that
~
is
Therefore
80 vi) = i). of
~
and
Let
A
N =
~(I,A,A;P)
be a nonzero
way that the nonzero
rows of
labeled
We fix
1,2,...,n.
be a Rees matrix
element of A
J.
are labeled
k E I
and
ring.
We may choose 1,2,...,m
~ E A
Let
J
m n i=l~P~i (X=I~aiXP'Ak ) = 0
i < i < m 1 < X < n buk ~ 0
since
P
since
P
c
b lk =
is left row-independent,
C*A*D
i E I, % E A, we denote by entries
above and letting
only in the (i,h)-position,
form
(g)ih'
A
'~
the latter
minimal
(a C A)
whose
(i,h)-entry
the notation
~ c p a p d p,t,6,s jp pt t@ 6s so
to
that
J
contains
is nonzero Now
all matrices
c of the
~.
imply that there exists
that
(pki)i h
is a nonzero
With the same notation,
Phi # 0.
Then
~ If
~
~
that
For
is a completely
g
Conse-
A
onto
that
-i (pxi)i
-l (pxi)i k. *~
is a
is semiprime. O-simple
is a simple atomic ring and an ideal of a cardinal
the ring of all ~ X ~ - m a t r i c e s
of
-1 Aik = (pxi)i k * ~ *
ring which by 1.14 implies
since
idempotent.
the mapping
is easily seen to be an isomorphism
is a division
II.2.10 NOTATION. denote
with
c # 0, D = (1)kl, we obtain
It is also easy to verify
right ideal of
C(~)
P
showing
ring.
II.2.9 COROLLARY. then
Consequently
eix = c t,6 ~ PDta @p@k = CbDk.
which shows
must be equal
= (Pxi ih
AiX = { (a)ik I a E A} . Hence
i <_ i _< m,
-1
is a simple
-i )<:a ~ (Phia)i~
If
for
for
A # 0.
the l x A - m a t r i x
ej(~
-i)
(Phi)ikP(Phi)ik
ai% = 0
Continuing
where
since
on the matrix
-i
quently
J
(c)ix
in which case
arbitrarily
But then
The conditions
-1
but then
are
x~=nlPBiaikPhk •
n h=l~a.lhpkk = 0
that
contradicting
are zero.
C = (c)i D
= CPAPD = (ejo) = (cb k)i h
can be" chosen in
columns
k E I, ~ E A.
and all the remaining
introduced
implies
is right column-independent;
for every
For any is
which
in such a
and the nonzero
and let
i=l b k = 0, then
be an ideal
the notation
over
number and A
ideal of
~
for a rin$
~,
~. A
a division
ring,
let
A
>t
with only a finite number of nonzero
entries. This is an extension over
A.
Further,
A~
identity gxx-matrix. identify
U
with
V*
~
where
An
for the ring of all n × n-matrices I
has cardinality
We have seen in 1.3.12-3.15 and have
know for which dual pairs For convenience,
of the notation ~(I,A,I;Q)
that for
~(V) = ~u(V) ~ Adi m V .
(U,A,V)
we introduce
is
SU(A,V)
and
Q
is the
dim V < ~, we can
It is then of interest
isomorphic
a new concept.
~
to
to a ring of the form
81 11.2.11 DEFINITION. subsets if
lu i l i E l j
(vi,uj) = 6i j
Let
and for all
11.2.12 NOTATION. ~, respectively
form
G of n i =~Irigi
M
If
senerates
M
The usual definition
i, and let
ii)
~ ~ r ~
iii)
and
V
of a completely
M
can be written
in the
r.z E ~, gi E G. with zero, an idempotent
is
to the ordering
0-simple
(U,~,V)
semigroup
idempotent,
S
is that
cf. 1.24.
S
has a
The next
12].
be a pair of dual vector
then the following
statements
and a cardinal
have a common set
G
of ~enerators
have biorthogonal
primitive
generates
then
& ~ F
spaces with
are equivalent.
number
3.
consistin$
o f primitive
~.
ii) = iii).
A.
biorthogonal
A symmetric Suppose
that
dim U = dim V = 3. ~ = F
3.
E = (E i I i E I}
consists
A ~ F shows
that
that
generates
and
The hypothesis G.
with
For every of
be the non-
n ~ AE. which proves k=l ~k generates ~.
A = E
both
~
and
We will
~,
that
and let
show that
Z
and
W
are
and let
x
be
U, respectively.
be a nonzero vector
[ik,~k,X k ] E
• " "n Zl,Z2,...,z
and let
W = [u i I [i,~,X] ~ G}. V
&
has cardinality
proof shows G
over
I
Let
where
are I X 1-matrices
that the set
A simple argument
(x,ui) = o.
a k E ~,
of
entries,
It is clear
idempotents.
bases of
v = ov
bases. and
The elements
Z = [vx I [i,~,k] ~ G],
some
relative
of
then a
e = ef = fe.
Let
E i = (1)i i.
zero columns of
Let
m
right) ~-module,
~
i) = ii).
i E I, let
such that
right, multiplication).
(respectively
if every element n i =~Igiri) for some
only a finite number of nonzero
E
the
the left and the right ~-module
ring
i f i) occurs,
PROOF.
orthogonal
Then
are biorthogonal
idempotents.
U
Moreover,
denote
is a left
~ = ~U(g,V),
~
V, respectively,
is left, respectively
for some division
and
orthogonal
and
[i], see also Dieudonn@
11.2.15 PROPOSITION.
i)
U
proper ideals and a primitive
is due to Hotzel
dim V >
~
and minimal
e < f ~
result
of
In a ring or a semigroup
if it is nonzero
zero, no nonzero
and
M
(respectively
11.2.14 DEFINITION. primitive
~
the action
11.2.13 DEFINITION. subset
be a dual pair of vector spaces.
i,j E I.
Let
(i.e.,
(U,V)
[v i l i E I]
Hence
in
V.
implies
Choose any that
ui E ~
[i,l,~] =
~ ak[ik,~k,k k] k=l
for
82
v = uvD = (x,ui)v D = x[i,l,~] n
n
= X(k~__laklik,~k,Xk]) proving that Let
Z
generates
= k~=i(xak,Uik)~kVhk ,
V.
[i,v,X],[j,6,D] E G
be distinct.
[i,~,h] [j,~,D] = so that
(vX,uj) = (vD,ui) = O.
similarly
X # ~.
and w r i t i n g
all
VXk.
[j,~,D] [i,~,X] = 0
Since
(v%,ui) # O, it follows that
that
vx =
There exists
i
and
W = ~w k I k ~ K}, we also have
n ~ ~,v , where k=l ~ h k such that
v
v E Z h' X k
and
that
v~
and
i # j
This establishes a one-to-one c o r r e s p o n d e n c e b e t w e e n
Z = ~Zk I k E K}
Now s u p p o s e
Then
is
Z
and
W,
(wi,zj) = 6i j.
different
from
[i,(vX,ui)-l,x] E G; thus
n 0 ~ (v~,ui) = k~=lOk(Vhk,Ui) = 0 by the above,
a contradiction.
This shows that
proJing that
Z
V.
Hence
Z
and
iii) = i).
is a basis of W
Z
is linearly independent,
A similar proof shows that
W
thus
is a basis of
U.
are b i o r t h o g o n a l bases. In the proof of "iv) = vi)" in 2.8, we take the given biorthogonal
bases and immediately obtain that
P = (phi) = ((z~,wi))
and the rest of the proof shows that
is the identity matrix
~U (g'V) ~ gdim V'
We finally prove the last statement of the proposition. Hence assume that -i Let e = (pxi)i h where Phi ~ 0. We have seen in the proof of
~U(g,V) ~ F .
"vi) = i)" in 2.8 that F F
&~ = e * F
is a simple atomic ring. ~ ~U l(A ,V ~) , w h e r e
Thus
U~ = F
SU(~,V) ~ ~ U ~ ( g ' , V I)
so that
& ~ F.
* e
is a d i v i s i o n ring isomorphic
Now 1.14, *e
and
to
r
and that
together w i t h the proof of 1.19, shows that V I = e*F
w i t h the induced bilinear form.
w h i c h by 1.5.15 implies that
We have seen above that both
biorthogonal bases w h i c h evidently implies that
(U,V) dimU
and
g ~ &t (U~,V j)
= dimV
and
and
dimV
= dimV I
must have d i m U ' = d i m V '.
Hence all four vector spaces have the same dimension. It is easy to see that
consists of all matrices in F all of whose nonth zero entries are situated in the i row. Let I be an index set of rows and columns of matrices entries
in
in the columns
F , and let
cjP~i = 0
for V~.
then
m ~i~=l(cjPli)i~j
i < j < m; but then It is clear that
dim V = dim V ~ = ~.
B = [(1)i D I ~ E I}.
DI,N2,...,~n,
m i~=l(Cj)i~ * (1)iBj = O, then
basis of
Vt
B
If
A E V~
has nonzero
n 1 A = k=l~(aiDkP~i)iXA * (1)i~ k" = 0
If
which evidently implies that
c. = 0 for 1 < j < m. Consequently B is a J is of cardinality ~, which finally yields
83 11.2.16 COROLLARY. and only if
~
A rin$
~
i~s isomorphic to a rin$ of the form
is a simple atomic ring for which
~
and
~
&
if
have a common set of
generators consisting of orthogonal primitive idempotents. 11.2.17 COROLLARY.
If
Note that in the case and
V
FQ ~ &b' then dim V < ~
that
U
that
~u(V) = ~u(V) = £(V) ~ gn
and
F ~ &
and
Q = b.
(U,V)
is a dual pair, 1.2.3 implies
admit biorthogonal bases, confirming what we already know, viz., for
n = dim V
snd
g
a division ring.
can be carried one step further as the next proposition, due to Msckey II.2.18 PROPOSITION. dim U = di_~mV PROOF. V and
and
Let
is denumerable, Let
(U,V) then
be a dual pair of vector spaces such that U
and
Z = ~zi I i = 1,2,3 .... }
U, respectively.
Jo = min~j I (zj'w I) # Oj
V and
admit biortho$onal bases. W = ~wi I i = 1,2 .... }
We will construct biorthogonal bases
Y = [Yi I i = 1,2,...]
of
V
and
and set
This
[2], shows.
U, respectively, by induction.
Xl = (Zjo'Wl)-iZjo '
be bases of
X = ~x i li = 1,2,...} Let
Y l = Wl' hence
(xl,Yl) = i. Assume that (xi,Yj) = 6ij
Xl,...,Xk, and
for
Jk = min ij I zj and set
Xk+ I = Zjk
i < i,j ~ k.
yl,...,y k
have been
Suppose first that
chosen in such a way that k
is odd.
Let
is not contained in the subspace generated by x I ..... Xn} k i~=l(Zjk,Yi)X i.
Then for
1 < t < k, we obtain
k (Xk+l,Yt) = (Zjk,Yt) - i=1~(Z.jk,Yi)(xi,Yt) = (Zjk,Yt) - (Zjk,Yt) = O. Next let
Jk+l = min~j I (Xk+l,Wj) ~ 0}
and set
k Yk+l = (W.jk+l - i~=lYi(Xi'Wjk+l))(xK~i .... W.jk+l)-l. For
i ~ t < k, we have k
-i
(xt'Yk+l) = [(xt'wjk+l ) - i~='l(xt,Yi)(xi ,w-Jk+l )](Xk+l'Wjk+l ) -i = [(xt'wjk+l) - (xt'Yt)(xt'w'jk+l)](Xk+l'Wjk+l)
= 0
and also k (Xk+l'Yk+l) = [(Xk+l'Wjk+l ) - i~__l(Xk+l,Yi) (x.l ,w.Jk+l )] (Xk+l,Wjk+l)-i
=
i.
84 If Yk+l
k
is even, we interchange
This process defines
that the sets X
X
generates
of
V
V
and
Y
the roles of
the sequences
spanned by some
xi's. V
Similarly
and
Y
II.2.19 PROPOSITION. (U,V)
that
SU(h,V)
Let
We construct a sequence
K =
is a subring of Let
I
en~en c_ I But then
Since
That
Hence
X
and
Y
are
to
h.
K.
[i].
~U(h,V)
for a dual
to
h
where
~ < ~
.
be a set consisting of nonzero elements of of idempotents of eI
~
such that en+ l
as follows. b I E e~el;
be an idempotent
By the suppose
for which
K
for
it follows
that
K
is the desired subring.
Then
I ~ e~e n # 0
n I N ek~e k @ 0
and hence
n = 1,2,...,
Next
and thus n by 1.3.18 and the last ring is simple by 1.3.6.
en3en ~ hrank e I
We
h, then every denumerable subset
isomorphic
e ~e c en+l~en+ n n 1
We will show that
argument implies that idempotents,
~
el,e2,..,
be a nonzero ideal of
el3e I ~
and
this is possible again in view of the proof of 1.3.20.
3.
since
U.
This is due to Hotzel
have been defined and let
U e ~e . n= 1 n n
Xk+ 1
One shows
is contained in the subspace
be any ring isomorphic
there exists an idempotent
bn+l,e n E e n + ~ e n + I, let
~
B = [b i li = 1,2,...]
el,e2,...,e n
to define yl,Y2, ....
is a locally matrix ring over
of vector spaces over a division ring
proof of 1.3.20, that
Let
is contained in a subring of PROOF.
3.
zk
generates
are now in a position to extend this result.
3
V
and
U, respectively.
Recall that 1.3.20 asserts
pair
and
are linearly independent as in the proof of 1.2.3.
follows from the fact that every
b i o r t h o g o n a l bases of
of
U
xl,x2,..,
ek3e k ~ I, so that
for
I = K.
for some
e
k = 1,2,..., which by the same Since
K
contains nonzero
it must be simple.
Next let
A
be a minimal right ideal of
an idempotent
e
so that
el~e I.
A = e(el~el) = e~e I.
division ring, and it is clear that
By [.6,
A
Hence by 1.14,
e(el~el)e = e~e ~ A.
is generated by e(el~el)e
is a
Furthermore,
e~e = e(el~el)e c_ eKe c_ e~e so that
eKe = e~e ~ h.
K, proving that is isomorphic to over
eKe
K
SKe(eKe,eK)
where
eK
is a minimal right ideal of
The proof of 1.19 now shows that
(Ke,eK)
K
is a dual pair of vector spaces
with the b i l i n e a r form induced by ring multiplication.
Note that
eK = e(n=iO en3en ) = nU=l(e~en ) ' =
left vector space over en3en ~ hrank en of
But then 1.14 implies that
is a simple atomic ring.
e~e
for
where
n = 1,2, ....
according to 1.3.18.
e~el, we may extend it to a basis
e~en c_ e~en+ 1
and
e~en
is a
Further,
dim e~e = rank e since n n S t a r t i n g with a basis V l , V 2 , . . . , V r a n k el
Vl'V2''''' v rank el'''''Vrank e 2
and with an obvious inductive process construct a denumerable basis of
of eK.
e~e2, The
85 same c o n s t r u c t i o n applies
to the right vector space
= dim Ke = dim eK ~ ~o" and
eK
If
~
have biorthogonal bases,
from 2.18.
In either case,
Ke, which shows that
is a finite cardinal, and if
~
then 1.2.3 says that
is infinite
2.15 implies that
Ke
the same conclusion follows
SKe(eKe,eK) ~ &
with
~ ~ ~o"
II.2.21 EXERCISES. i)
Show that in any semigroup Green's relations
~
and
~
commute and that
is an equivalence relation. ii)
Let
~ = ~(I,g,A;P)
be a Rees matrix ring.
a)
Find a Rees m a t r i x r e p r e s e n t a t i o n of the semigroup socle of
b)
Characterize
the right ideals of
c)
Characterize
the elements in the semigroup socle of
d)
Simplify the results in a), b), c) in the case that
matrix to obtain as simple characterizations iii) if
Show that in a regular ring
~e~e
is a division ring.
iv)
Let
£(V)/~(V) ideals.
V
~,
~.
~. ~. p
is the identity
as possible.
an idempotent
e
is primitive if and only
be a vector space of countably infinite dimension.
Show that
is s simple regular ring with identity and without minimal o n e - s i d e d Give
an example of a simple ring without an identity and without minimal
one sided ideals. For further results on this subject, Dieudonn~
[2], Hotzel
Rosenberg
[i], W o l f s o n
[i], Jacobson
see Behrens
([i], Chapter II, Section 5),
[3],([7], Chapter IV, Section 15), Pless
[i],
[1],[3],
II.3
M A X I M A L PRIME RINGS
The theorem of this section gives a number of abstract characterizations of the ring of all linear transformations of finite rank on a vector space.
The remaining
results deal with further special cases of primitive rings with a nonzero socle. It will be convenient to introduce several new concepts. 11.3.1 DEFINITION.
A semigroup
isomorphic to an ideal of ~
for some ring
N~).
II.3.2 LEMMA.
If
simple atomic rin$ PROOF.
Let
K
K ~
~
S
is an r i - s e m i s r o u p
for some ring
~
so that
be an ideal of
K
(ring-ideal)
(equivalently,
S
is a completely 0-simple ri-semigroup, such that
K
is the semigroup socle of ~b~ ~.
Then
by 2.8, and in view of 1.25 it is clear that ideal of
~
K
~ = C(K)
if
S
is
is an ideal of
then there exists ~
and
~ = C(K).
is a simple atomic ring
is the unique completely 0-simple
must be the semigroup socle of
2.
86 II.3.3 DEFINITION. to a proper
A simple atomic ring is r-maximal
if it is not isomorphic
right ideal of a simple ring.
II.3.4 DEFINITION. not isomorphic
A completely
0-simple
ri-semigroup
to a proper right ideal of a completely
is r-maximal
0-simple
We are now ready for the first main result of this section.
The equivalence
of i) and ii) is due to Gluskin
[4], that of ii) and iv) to Wolfson
remaining part is new.
characterization
Jacobson
Another
([7], Chapter
II.3.5 THEOREM.
IV, Section
~
is an r-maximal
ii)
~
i__{sisomorphic
of these rings
[i]; the
csn be found in
16).
The following
i)
if it is
ri-semigroup.
conditions
on a ring
~
are equivalent.
simple atomic ring.
to the ring of all linear
transformations
of finite rank
on a left vector space. iii)
[P~
iv)
has a completely
~
implies
r-maximal
is a simple atomic ring in which
ideal
~r,~(L)
~
and
= 0
~ = C(~).
for a left ideal
L
L = ~.
PROOF. vector
0-simple
i) = ii).
spaces.
By 2.8, we have
~ ~ ~u(V)
for some pair
(U,V)
of dual
By 1.2.6, we have ~u(V) = is ~ ~(V) I a'V* ~ nat U]
which
implies
r-maximality
that of
ii) = iii). assume that 0-simple
gu(V)
is a right ideal of the simple ring
~, we must have We may take
R = $2,v,(V)
ri-semigroup
simple atomic ring
Hence
there exists
which
implies
~' ~ gUl( V' )
where
is a right ideal of
It is easy to verify S = ~°(I,G,A;P) is a nonempty
subset of
the columns of o (I , ,G,A;P I)
P
I, p1
indexed by
even though
V
that
S
K
KI =
(U~,V ')
S ~ $2,UI( V I ).
Thus
socle of a
of vector
spaces
~ ~ K ~ KI
right ideals of a Rees matrix
the semigroups
of the form
is the A x l l - m a t r i x
~°(I',G,A;p'),
obtained
from
I \ 1 I, where by abuse of notation,
~°(I1,~-,Iv1;p') I
and
$2,uJ(VI).
there may exist
$2,U,( V ) =
Hence by
of a completely
is the semigroup
k E A
such that
i' 6 I'. Hence
$(V).
N ~ ~(V).
is a left vector space,
a dual pair
that
that the nonzero
are precisely
Consequently
to a right ideal
By 3.2, we know
such that K'
~ = $(V), where
is isomorphic
S.
~'.
gu(V) = $(V).
~0
where J
11 ~
IV,
(I V ,~h-,Iv,;P).
since
P
semigroup where
11
by eliminating
we have written
pxil = 0
for all
87 If
dim V = i, then
which
forces
IV~
~2,uI(V~ )
K~ = ~2, uI(V t) = ~2,V ~,(V ~ ) ' isomorphism of ~ onto K ~. i j E I ~, then for any [i,?,X] ~ 0. 1.3.3.
U~ = ~u~
(~,a)
U~
a
iii) = iv). a right ideal of = ~2,v,(V). Let
L
L = ~v,(S)
completely
of
~2,v,(V),
where
be a left ideal of
transformation
a
onto
[i,~,~]K ~ = 0
Then
V
contradicting Hence
of 1.5,9 can be defined.
(&~,V ~) and
We conclude
and
that
K = S
that
~(~,a)l~ = @.
~2,V~,(V ~)
which
proving
of
~ = ~2,u(V)
and since
of the former implies
so that
for all
on
with
semigroup.
is not used.
K~
It then follows
implies
that
~2,u(V)
that
is
~2,u(V)
~ = ~(V).
Then the dual of 1.3.4 implies
that
V, and it is easy to see that =
= ~r,N(~v,(S))
s E S}.
It follows
(1)
(V)
~S~
such that
s(a*f) = (ss)f = Of = 0.
be an for all
that in the proof of 1.5.9 the im-
is both
~(V). S
0
pxi~ = 0
semigroup.
= ~(V)
S ~ = ~f E V* I sf = 0
linear
(&,V)
~ = Su(V).
for some subspace
and let
in the statement
the r-maximslity
~u(V)
~r,~(L)
b
@
0-simple
1
we have
ring
so that
is indeed a Rees matrix
and observe
under
We may take
But then
dim V >
is closed under addition
~
is also s division
dim V ~ = i
must have the same property,
and
K ~ = ~2,U ~(V ~) = ~2,V~,(V ~).
is an r-maximal
have
~, ~
isomorphism
the image of
Kl
~ E g-, we obtain
l i ~ I ~, ~ E ~] that
so
% E IV~
~°(I~,~I~-,Iv~;P I)
is a semilinear
that
and
and the functions
plicit hypothesis
Consequently
If for some
By the isomorphism
1.5.6 applies
ring,
But then
Suppose next that
j E It
Consequently
We let
is a division
to have only one element.
If
S # V, then there exists
Ss = 0. that
Then for any
a'V* ~ S ~
a nonzero
s E S, f ~ V*, we
and thus
0 # a E ~
(V). S~
Consequently, that
if
L = ~v,(V) iv) = i).
Further e E ~ which
let
A
by 1.6. is minimal
~r,~(L)
(U,V)
S = V
which
implies
= ~(V) = ~. Let
~
satisfying
be a minimal Hence
~I
iv) be a right ideal of a simple ring
right ideal of
= (e~)~ ~ c ~
that
~ = ~(V)
~
~.
= A
in view of its minimality
ring, and we may suppose 1.3.4, we have
= 0, then by (i) above we obtain
in
= ~u~(V )
for some subspace
Then
A = e~
so that ~.
Thus
where U
A~
U~ of
for some idempotent
is a right ideal of
~
~,
is a simple atomic
is a t-subspace
U~
~I.
of
V*.
By
and by 1.3.8, we have that
is a dual pair. The following argument
and notation
dual of 1.3.4 and (i) above, we obtain
is relative
to the dual pair
that if s subspace
S
of
V
(U,V). has
(V) = 0, then S = V. Let 0 # f E V*. Then S = Iv E V I vf = 0} S• subspace of V, and hence by the preceding statement, we must have ~ Consequently
By the
the property
is a proper (V) # 0. S±
88
s~ = [ u ~
ols~ = 0
Let
0 ~ g E S ~, then for every
and
g
sE
for all
v E V
are linearly independent,
s] ¢ O.
we have:
if
vg # 0
according to 1.2.3 applied to the dual pair
f = g~
for some
r ~ g-
which implies that
vf = O, then
then there exists
so that
f E U.
v ~ V
vg = 0.
such that
If
vf = 0
(V*,V), a contradiction.
Therefore
U = V*
and thus
f and Thus
U = U~
~ = ~.
The equivalence of ii) and iv) in the next corollary is due to W o l f s o n [i]; the rest is new. II.3.6 COROLLARY.
The following conditions on a ring
i)
~
is a primitive ring with an r-maximal socle.
ii)
~
is isomorphic
~
are equivalent.
to a ring of linear transformations on a left vector space
containing all transformations of finite rank. iii)
~I~
iv)
~
is a dense extension of an r-maximal completely 0-simple semigroup. is a primitive ring with a nonzero socle
a left ideal PROOF.
L
implies
i) = ii).
~u(V ) ~ ~i ~ £ u ( V ) But then
We know by 1.25 that
for some dual pair
~u(V) = ~(V)
ii) = iii).
and thus
We may take
is a dense extension of "ii) = iii)" in 3.5 that iii) = iv). L
a ~ ~r,N(L) that
~
and hence
~
ba # 0
and thus
for some ring
Since
~
~i
for
satisfying
is r-maximal, so is
for a left vector space
by 1.25.
~u(V).
V.
Hence
It was shown in the proof of
is a primitive ring with a nonzero socle ~r,N(L) = 0.
for some
Further let
b ~ L.
xa ¢ 0
~r,~(LA~)
and = 0.
0 ~ a E ~,
~.
Let
then
It follows from 1.25 and 1.3.6
and thus there exists
x = bach, we have
a ~ ~r,~(Ln~),
~r,~(L) = 0
is an r-maximal completely O-simple semigroup.
such that
is a regular ring,
Hence for
in which
~ £(V).
$(V) c ~ c ~(V)
$2,v,(V)
~
N ~ ~
(U,V).
~(V) ~ ~
~2,v,(V)
As before,
be a left ideal of
~
L ~ ~.
c E ~
x E L n ~
such that
ba = bacba.
which shows that
By 3.5 we must have
L N ~ = ~
so that
L ~ ~. iv) = i). ~r,~(L) = 0. 0 ~ a E ~, so
Let
L
As above
then
ab # 0
be a left ideal of the socle ~
is regular~ for some
Lab ~ 0, w h i c h implies that
hypothesis implies
that
L ~ ~.
and thus
b E ~ La # 0.
L
~
of
~
and suppose
is also a left ideal of
by 1.25 and 1.3.3. Consequently
Now 3.5 implies that
But then
~r,~(L) = 0 ~
that ~.
If
0 ~ ab ~
and the
is an r-maximal simple
ring.
The next result characterizes rings isomorphic space
V; other c h a r a c t e r i z a t i o n s
to
can be found in Leptin
£(V)
for some left vector
([2], Part I) and W o l f s o n [i].
89 II.3.7 COROLLARY.
The followin$ conditions on a ring
~
are equivalent.
i)
~
is a maximal primitive ring with an r-maximal socle.
ii)
~
is isomorphic
to the ring of all linear transformations on a left vector
space. iii)
~
is a maximal dense extension of an r-maximal completely O-simple semi-
group. is a semiprime ring having a minimal
iv)
right ideal
and is m a x i m a l among the semiprime rings
~r,~(A) = 0 ideal with
~'
A
such that
having
A
as a right
~ r , ~ ( A ) = 0. The equivalence of i), ii) and iii) is an immediate consequence of 1.27
PROOF. and 3.6.
i) = iv). such that
By 1.25,
~
~r,~(A) = O.
and having ideal of
A
is a semiprime ring and has a m i n i m a l right ideal
Let
be a semiprime ring containing
as a (automatically minimal)
~.
By 1.6, we have
e,e ~ E A, f E B.
Then
right ideal.
A = e~ = et~ I
A = ~
and 2.8, by 2.5 there exist y E ~,
~i
~ e~ ~ = e 1 ~ a,b E ~
and
= A, so that
such that
Let
B = ~
~
B
A
as a subring
be a minimal right
for some idempotents A = e~ j.
e = ab, f = be.
In view of 1.25 If
x E B
and
then xy = (fx)y = (baba)xy = b(ab)axy = fbe(axy) E fb ~e/~~ = fbA c f~ = B
which proves that right ideal of ~
of
~.
B
~.
is a right ideal of Consequently
But then
the socle
~
of
Since each minimal right ideal of
it follows that their sum r-maximality of
~
~, we have
Now the m a x i m a l i t y of iv) = ii).
~
Again
~u(V)
~
~ = ~J
since
~
~
where
are precisely
Ri = ~ [ i , ~ , X ]
must also be a minimal
is contained in the socle
~
and thus also of
satisfies
as a primitive ring with socle ~ ~ ~l
B
is a minimal right ideal of
is a right ideal of
~u(V) ~ ~' ~ £u(V)
dual vector spaces by 1.25 and 1.2.8. ideals of
~
~.
the conditions in 1.25. ~ = ~
implies
for some pair
and
~(V)
coincide. = 0
semiprime,
vxa = 0
Since
then
for all
the m a x i m a l i t y of
X E IV • ~,
£(V)
Ri
[i,~,~]a = 0
(vD,ui) J O, so that
of
I V ~ ~, X E I v ] .
and we have seen that
a ~ ~r,£(V)(Ri), such that
R i.
(U,V)
the sets
Thus the hypothesis on
for some
~ = ~.
We have seen in 1.3.11 that the minimal right
It follows easily that the sets of all minimal right ideals of the rings
~r,~(Ri)
~, By the
~
implies
satisfies
But then
~'
for all
a = 0 that
~ ~ g, ~ ~ I V . which
for
which shows that
3' = ~(V)
is semiprime and
the conditions
is a minimal right ideal of
(v ,ui)w(v a ) = O
we conclude
that
~u(V), ~
in 1.25, ~(V).
There exists ~ 9 0
it is
If ~ ~ Iv
yields
~r,~(v)(Ri) = O.
w h i c h finally yields
By
~ ~ ~(V).
90 The implication Wedderburn
"i) = ii)" in the next corollary
(or Wedderburn-Artiu)
theorem,
these rings was given by Gluskin [4],([7], Chapter
IV, Section
11.3.8 COROLLARY.
[i].
is known as the Second
A multiplicative
characterization
For further characterizations,
16) and Wolfson
The following
[ii.
conditions
~
are equivalent.
~
is a simple ring satisfying
ii)
~
is isomorphic
to a full matrix
iii)
~
is isomorphic
to the ring of all linear transformations
iv)
vector
d.c.c,
on a ring
i)
dimensional
of
see Jacobson
for right ideals.
ring
&n
over a division
ring
&.
on a finite
space.
~
is a prime
(respectively
~
is a simple
primitive)
ring satisfying
d.c.c,
for right
ideals. v) 0-simple
PROOF. minimal
ring a n d
~
is a maximal
dense extension
of a completely
semigroup. i) = v).
The second part of the hypothesis
right ideal, which
for some dual pair by 1.3.14 implies
that
But then 1.7.10 yields which is a completely extension
But then
dim V < ~. that
By 1.27,
which implies
dim V < ~.
that
d.c.c,
Since
ii) ~
We deduce
vector iii).
space
that
~ ~ ~u(V)
and thus
dense extension ~D~
has a
for right ideals which
= ~(V)
Consequently
~
~ ~ £(V). of
is a maximal
implies
that
is also simple, we must have
Consequently
for right ideals by 1.3.4.
iv) = iii).
gu(V)
is a maximal
~
~2,v,(V) dense
semigroup.
ideals and is both prime and primitive
dimensional
d.c.c,
the second part of the hypothesis
(U,V).
that
by 2.8 implies
satisfies
Consequently
semigroup.
0-simple
for some dual pair
satisfies
~u(V)
[k~(V) = g(V)
O-simple
of a completely
v) = iv).
together with simplicity
(U,V).
implies
nat U = V* Hence
~
~ ~ £u(V)
~u(V) = £u(V)
by 1o3.13 so that
£(V)
satisfies
d.c.c,
for right
~ ~ ~(V)
for a finite
by 1.25,
from 1.25 and 1.3.14
that
V.
This is a well-known
fact proved in elementary
texts on linear
algebra. iii) = i). Further
~(V)
fies d.c.c, Again
First note that
= gv,(V)
where
~(V)
is simple by 1.3.6 since
dim V < ~
so that 1.3.4 implies
~(V) = ~(V). that
~(V)
satis-
for right ideals. the hypothesis
corresponding
hypothesis
on right
ideals in i) and iv) can be substituted
on left ideals,
be taken on a left or a right vector
and in iii) the linear
space.
by the
transformations
can
91 11.3.9 EXERCISES. i)
Let
ideal of
~
~
be a primitive ring with a nonzero socle
~.
Show that a right
which contains s m i n i m a l left ideal must contain
A nonempty subset
A
for some n o n e m p t y subset
of a ring B
of
~;
~
~.
is called s left annihilator if
A = ~%,~(B)
a risht annihilator is defined dually.
It is
clear that a left (right) annihilator is a left (right) ideal. ii) ~(V)
Let
V
he a vector space.
coincide with right
Show that the principal right
(left) annihilators of
dimensional if and only if every right
£(V).
(left) ideal of
(left) ideals of
Deduce that £(V)
V
is a right
is finite (left)
annihilator. iii)
Let
~
be a regular ring with identity.
right (left) ideals of ~(V)
~
a sum of two right
is a principal
right
(left) annihilators
For the first statement, if A and B 2 e = e and b, respectively, consider
Show that s sum of two principsl
(left) ideal of
is s right
~.
Deduce that in
(left) annihilator.
(Hint:
are principal right ideals generated by e + (l-e)bx(l-e)
where
(l-e)h
= (l-e)bx(l-e)b.) iv)
Show that in any ring the intersection of any set of right
annihilators is a right
(left) annihilator.
of any set of principal right v)
Let
= ~(V).
V
Deduce
that in
(left) ideals is a principal
be a finite dimensional vector space, U
~(V)
(left) the intersection
right (left)
ideal.
be a subspace of
V*
and
Show that ~u(V) = ~r,$( [c ~ ~ I c*U = 0}).
vi)
Give an example of a ring
~
and two principal right ideals of
~
whose
~
whose
intersection is not a principal right ideal. vii)
Give an example of a ring
and two principal right ideals of
sum is not a principal right ideal.
II~4
SEMIPRIME
RINGS
We are interested here in semiprime rings with a nonzero socle. us a general result,
This will give
several special cases of which will be treated in greater
detail in the two succeeding sections.
Some new notions will come in handy
(recall
1.15). 11.4.1 DEFINITION. every
i ~ I, we have
Ai, i E I, if =
~ Ai . iEl
~ =
A family
~AiJiE I
A. • ~ A. = 0. I j#i j
~ A. iEl l
and the family
of subrings of
A ring
~
~Ai}i61
~
is independent
if for
is a direct sum of its subrings is independent,
to be denoted by
92
11.4.2
DEFINITION.
subsemigroups denoted
by
Such
A semigroup
S , ~ ~ A,
S =
~ ~ o@A
a "sum"
S
if
S =
as follows.
zeroes
0~.
Let
Let
S =
with
zero
and
S S~ = S
is an orthogonal ~ S~ = 0
sum of its
if
~ # ~,
to be
•
of semigroups
proceed
S
U S
may be termed
~S~c~A_
[ U
(S
'internal".
Externally,
be a family of pairwise
\03]
U 0, where
0
disjoint
is an extra
we may
semigroups
symbol,
with
with multi-
~A plication
a*b
in all other groups
= ab
a,b ~ S for some ~ ~ A and ab # 0 , and ct c~ In this case also we say that S is an ortho$onal
cases.
S c~, ~ 6
theorem,
if
A, with
see D i e u d o n n @
the same notation. [3] and J a c o b s o n
For
the origin
([7], C h a p t e r
a*b
= 0
su.__mmof semi-
of a part of the next
IV, Section
3),
the rest is
new. II.4.3 minimal
THEOREM.
risht
Let
ideals
~
be a semiprime
assumed R
=
U
nonempty. Ri,
~
=
be the socles
of
~lR
i T j <= R. --~ R. z j and
Then each of all all
~
~
.
%.
minimal
=
~
~
= C(~
yield T
that
~.
ring by
elE
each of these
left
and
steps,
~
If implies
which ~
that
~ ~
¢ 0, then
and
Hence
this
follows 1.16
R
~
by:
Let
J = I/~
is an ortho$onal ~
sum
is a direct
sum of
definitions
with
~
immediately
that
ideals
.
Consequently since
~.
these
are
contains
is ~ - i s o m o r p h i c
~
R
.
definitions ~
agree ~
ideals of
as
e, e ~.
and by 2.5
each
that
a minimal
isomorphic
in some
Consequently it follows ~
the
idempotents
of
are contained
~
and
ideals
the c o r r e s p o n d i n g
~
from
the left socle of left
for some
right
they
(and of
intersection in
~
relation.
and
L ~ = ~e ~
that
is an ideal of
ideal
~,
are minimal
~, R, % ,
that both
a relation
the analo$ous
are minimal
~X~
~R
of a sum of subrings,
then implies
every right
and
L~
we conclude
ideals of
and hence
of all
, ~.
L = ~e,
the same
J).
of
relation
and
e~
define
(~ E
of
ideal
We know by
L
as ~ - m o d u l e s .
By the very d e f i n i t i o n
= C(R),
ideal
~, ~, ~
1.6 we have
left ideals yield
1.13),
If
I
R
), ~ = C(£),
isomorphism.
then by
a union of m i n i m a l
~
atomic
the same
the socle of
On
is an equivalence
=
is an equivalence
they are isomorphic
reversing
~
0-simple
is a simple
of a module
By 1.14, we have
minimal
R.
is a completely
left ~ - m o d u l e s ,
[~.
=
Each
That
agrees w i t h
respectively. Then
Furthermore,
definition
~,
U
~Ri}i~ I
i~l
as ~ - m o d u l e s .
left ideals
PROOF.
that
and
the set
Ri
~
i~l
rin$ with
Let
By using
is also
in a semiprime is an ideal of
~
= C(~
),
~.
right
to any minimal
ideal which right
ideal
in
93 R~,
so we must
since both R
have
~
R
and
= 9~.
R~
It follows
are ideals
of
that
R(~
~ 9
[F~i, so that
~ ~
R
= 0
if
~ #
is an orthogonal
sum of all
•
We prove next ideal of
~
a~ = R i
that
contained
and thus
Let
I
~
a~
that
idempotent
minimal
left
for every
t 6 R
there exists
iv)"
conclude
I Q ~
that
Let
y ~ Rj
J
are N - i s o m o r p h i c , =
that
quently
~_ I.
~
~
shown above
nonzero
R i = a~ ideal of
.
~
is also
in
right
again by
.
in 2.8, we
such
let
the union of
by idempotents.
~
through w i t h obvious
Hence ~
of "v) = iv)"
by [.6 are g e n e r a t e d
and
let
ideals e.
that
Hence
t = te.
modifications,
Ri
x~ = e.~l
that
I Q ~
idempotents,
Rj•
in
~
of •
implies
that
~ 0
The
and we
x £ R.
By 1 . 6 ,
~.
of
a,b ~ ~
Ri = e i ~ R.
and
such
that
for some
J, since
and
and R.
z ~ ~
b,zay ~ ~ .
It
Conse-
ideals.
which
, it follows
Then
e i = xz
= bx(zay) £ nonzero
.
By hypothesis
the existence
= beiaY
~
and e.
has no proper
generates
0 # x 6 J, y ~ R
and
2.5 implies
Further,
~
that
that
e
be a m i n i m a l
right
In the proof we have
an idempotent
idempotents
and
Since
contains
which
y = ejy = b ( a b ) a y
J = ~
It was ~
of
which by
ab, e.j = ha.
follows
.
Ri
# 0.
be an ideal
for some
~
Let
By the dual of i.ii, we have
implies
is a m i n i m a l
above,
in 2.8 now goes
f o r some m i n i m a l r i g h t
R. = e.~
ei
ideals,
which R.
ideal of
ring.
0 # a ~ R i.
a~ = R
the first p a r a g r a p h
proof of "v) =
atomic
and let ~
we conclude
be a nonzero
From
is a simple
in
Ri ~
the dual of i.ii,
= ~ .
~
now yields
that
it must be simple.
I ~ ~
I = ~.
= ~
Finally,
Therefore
~
and
thus
since
~
is a simple
atomic
ring. We know a minima[ then
A = e~
and
A
~
socle of
.
%(
=
~
~(%)
we deduce ~ ~j
consider
seen above %)
= 0
is equal ~
A
~
atomic
~u(V)). for
in v i e w
to zero
since
~
~
is,
~
~
But
right
of
%,
~
socle
= 0 ~ a = 0
is a simple
atomic
[~c~A
sum of all
~ of
in
that
is completely
then
in
ideal
A~ = ~ ~ c
of its m i n i m a l i t y ideals
R
the family
contained
then
the semigroup
~
is a direct
of
is a m i n i m a l
~ ~ 8, we have ~
that
right
ring,
a E ~
that
ideal
A
by 1.6.
Therefore
and if
= 0,
if
e E ~
~, m i n i m a l
that
right
Conversely,
idempotent
ideal of
, it follows ~
.
But in a simple
(e.g.
We have Hence
for some
~
is the union of all m i n i m a l ~
0-simple
that every m i n i m a l
ideal of
is a right
quently
But
from above
right
~
is,
.
.
,
= A
Conse-
the semigroup
is completely O-simple. ~~ ~
so that so that ring.
=
0.
a t ~£(%).
Thus
is independent. ~
e~
is ~
a = 0 Since
and
94 11.4.4 COROLLARY.
A semiprime ~
i_~f [i~
has a completely 0-simple
if
is a minimal right ideal of
Ri
product, and PROOF. ideal of Also
~
= (~)Ri,
~
contained in
~
has a minimal
right ideal if and only
In fac___~t, in the n o t a t i o n of th__~etheorem, contained in
~ , then
~
= ~Ri,
the ring
the semigroup ~roduct.
The first statement
(fi~)Ri
~
ideal.
~
is an ideal of
follows
from 4.3 and 1.13.
and thus must coincide with ~[~
contained
in
R
Further, ~R. i ~ since ~
is an is simple.
so it must coincide with
since a completely 0-simple semigroup has no proper nonzero ideals.
II.4.5 EXERCISES. i) that A
Let
A
AB = 0
into
B
and
if
B
be R - i s o m o r p h i c minimal right ideals of a ring
B2 = 0
and
AB = A
A
of
A
into
A 2 # 0, find
~,
a sum of all minimal right ideals
of
~
~; a homogeneous component of the left socle is defined analogously.
of
3. iii)
A
If
of
ii)
to
is called a homogeneous
component of the socle
Show that a homogeneous component of the socle of a ring
Show that the nonzero socle
homogeneous components.
~
of a ring
~
~
is an ideal
is a direct sum of its
Also prove that each homogeneous component of
direct sum of pairwise ~ - i s o m o r p h i c minimal second statement,
consider independent
Show
B.
is a minimal right ideal of a ring
which are ~ - i s o m o r p h i c
~.
B 2 # 0, and that an ~ - h o m o m o r p h i s m of
is either a zero ~ - h o m o m o r p h i s m or an ~ - i s o m o r p h i s m .
the form of all ~ - h o m o m o r p h i s m s If
if
right ideals of
3.
~
(Hint:
is a For the
families of minimal right ideals and use
Zorn~s lemma.) iv)
Let
F
be a homogeneous component of the socle of a ring
be a sum of all nilpotent minimal right ideals of N
is a nilpotent ideal of v)
~
and that
~
contained in
~, F.
and let
N
Show that
N = F N ~r,~(F).
Give an example of a ring which coincides with a homogeneous
component of
its socle and contains both idempotent and nilpotent minimal right ideals. vi)
Prove that if a homogeneous component
semiprime, socle of
then
F
F
of the socle of a ring
is a simple atomic ring and a h o m o g e n e o u s
~
is
component of the left
~.
For general references on this subject, 2), Jacobson ([7], Chapter IV, Sections
see Behrens
1,2,3).
([i], Chapter IV, Sections
i,
95
II.5
SEMIPRIME RINGS E S S E N T I A L EXTENSIONS OF THEIR SOCLES
The principal result here is a collection of c h a r a c t e r i z a t i o n s of semiprime rings which are essential extensions of their nonzero socles.
As a preparation,
we first study the relationship of prime and semiprime ideals in any ring. element
a
of s ring
11.5.1 LEMMA. i)
I
ii)
I
If
A
and
B~
PROOF.
I, then either B
i) = ii).
a E I
and
or
B
there exists we obtain
If
n ~ p.aq. In i= I i 1
such that
of
generated by
a.
~
are equivalent.
b ~ I.
~
such that
(a) c
ABc
I, then either
I
or
(b) ~ I
so that
I
a E ~,
is an integer,
~
such that
a ~ I.
(a)(b) c AB + ~ A B ~
For any
and thus
r,s Pi qi E ~}. ' '
ABc
I.
(i)
Suppose that
A ~
I; then
b E B, using (i) and the hypothesis,
b E I
since
a ~ l.
Hence
B G I.
Trivial. Every semiprime ideal of s rin$
~
is the intersection o f prime
~.
PROOF.
Let
I; then
I
which implies
r ~ I, then We let
be a semiprime
(~r~) 2 c ~(r!Rr)~ ~
(r) 3 c ~riR ~ I if
or
(a)(b) ~ I, then either
be right ideals of a E A
11.5.2 LEMMA.
riRr ~
1
~
b 6 I, First note that for any
iii) = i).
ideals of
a ~ I
are right ideals of
(a) = I n a + r a + a s + A
ideal of
I.
ii) = iii).
Let
the principal
The followin$ conditions on an ideal
(a)(b) ~
or
either
let (a) denote
is a prime ideal.
If
iii) Ac
~,
For an
rxr ~ I
r E ~\ I
I
that
for some
ideal of where r E
I.
2. ~r~
Let
such that
r 2 = rlxr I ~ I.
Xl,X2,...
such that
Starting with
p
such that
We can inductively define sequences
rn+ I = rnXnr n ~ I.
Let
M = irl,r2,...}.
r.j[xj(rj_iXj_l) ... (ri+iXi+l)(rixi)]r i = rj+ I E M. c,d E M,
then
~r~ ~
I.
cxd E M
for some
r ~ P
r I = r, there exists
r i[(xiri)(xi+Iri+l) ... (xj_irj_l)xj]rj = r j+ I E M,
if
so
But then
We will use this in the following form:
easily o b t a i n
Consequently,
and suppose that
x C ~.
and wish to find a prime ideal
P ~ I, w h i c h will prove the lemma.
r 6 ~
is an ideal,
x E ~.
rl,r2,.., For
and
xI E and
i ~ j, we
96 Now
let
~
for w h i c h standard
be
Zorn's
lemma
to show
that
P
P + (a) D
P
which
there exists that
then
the p a r t i a l l y
J fl M = ~
and
argument
is in fact
r ~ P
and
~
P
x E ~.
If
d 6
P
ideals I E ~.
P.
In order
a,b ~ P.
and hence
We have
Consequently
is a prime
ideal.
(a)(b) ~ Since
and is also denoted its ~-th
Any subring sum
consisting ponents
Let
[~#o~A_
(also
of
be a family
~ ~
of rings.
On the C a r t e s i a n
PROOF.
consists then
Thus
subdirect
element
homomorphism,
projection
of the rings
defined
earlier
~
.
of
~
to be d e n o t e d
homomorphisms
are onto
The subdirect
by
is a
sum of
the same way.
is sometimes
and external
An ideal ideals.
I
direct
called
sums
"internal
is v e r y
direct
close
and the
of
~
i_~s semiprime
if and only i f
I
i__~s
A ring
~
is semiprime
if and only
~
is a
of the first
~ I
statement
for some prime
It follows
that
is the content
verification ideals
the m a p p i n g
I r ~
of 5.2;
rings
~
the proof of
and is omitted. of
~
by
(r + l ) a 6 A
, ~ ~ A, we can identify
~
with
If
~
is semiprime.
is
statement
is an i s o m o r p h i s m are evidently if
~
a subring
~ • If now I is a n i l p o t e n t ideal of ~, then In is a n i l p o t e n t o~A ~ ~ , which implies that I ~ = 0 . Since ~ C A is arbitrary, we obtain Consequently
~
the first
~ (~/I) and the p r o j e c t i o n h o m o m o r p h i s m s of the image o~A ~ is a subdirect sum of prime rings ~ / I • Conversely,
sum of prime
if
rings.
of a s t r a i g h t f o r w a r d
0 =
of the proposition.
onto.
product)
of internal
of ~rime
Necessity
sulficiency
into
all of whose
to every
of the rings
is rarely made.
sum of prime
~
projection
"sums" will be labeled
sum" of rings
the i n t e r s e c t i o n
of
The m a p p i n g which
The resulting
~ ~ with only a finite number of nonzero como~A ~ (also discrete) direct sum of the rings ~ . Any ring
II.5.4 PROPOSITION.
semiprime,
we have
of
to any of these
distinction
•
is the ~-th
subdirect
The r e l a t i o n s h i p
subdirect
~ ~
component
is an external
A "direct sum".
~
by
of all elements
isomorphic
which
P E ~
~ ~ define the a d d i t i o n and m u l t i p l i c a t i o n coordinatewise. o6A ~ structure is a ring called a complete direct sum (also direct product)
subdirect
P
I.
11.5.3 DEFINITION.
.
seen above
then
product
associates
of A
Then
( P + (a)) n M # ~
p,
J
so
p
P N M = @.
that
element
that
( P + (b)) ~ M.
(a)(b) c
( P + ( a ) ) ( p + (b)) c
implies
that
of all
I N M = ~
has a maximal
implies
contradicting
and 5.1
since
ideal, we suppose
and s i m i l a r l y
for some
P P, M
P ~
that
(under inclusion)
~ # ~
(p + (a)) N M
c ~
by c o n t r s p o s i t i v e
shows
a prime
set
Then
of
cxd E M
exd E
J.
by m a x i m a l i t y
cxd E and thus
ordered
I ~
is a of
ideal of I = 0.
97 II.5.5 DEFINITION. product
II S
~A
Let
~S }¢~ A
be a family of semigroups.
The C a r t e s i a n
together w i t h c o o r d i n a t e w i s e m u l t i p l i c a t i o n is a direct product of
(7
semigroups
S , ~ ~ A, and is also denoted by
~ S •
o~A ~ II.5 • 6 LEMMA.
Let
~ ~ S , where for each 0~A ~
S =
~:(k'P) -- (~'lS '~IS )c~:A is a n
isomorphism of
Then the m a p p i n g
((k,p) ~ fl(S))
~I ~(S(~). o~A PROOF. Let X be a left translation of S, a E S , and suppose that ha ~ 0. 2 Then a = xy since S = S so that Xa = X(xy) = (Xx)y # 0 which implies that ~y Xx ~ S • Hence ~ maps S~ into itself. Consequently ~ maps [~(S) into II ~(S(7).
fl(S)
(7, S 2 = S • ~
onto
V e r i f i c a t i o n of the remaining properties of
~
is left as an exercise.
~A II.5.7 DEFINITION. 0-simple semigroups
A semigroup which is an orthogonal sum of completely
is a primitive r e g u l a r semigroup.
It is clear that a primitive regular semigroup is indeed regular, more that each of its nonzero idempotents of s primitive regular semigroup.
is primitive;
and further-
this is the usual d e f i n i t i o n
The two definitions are equivalent,
see c l i f f o r d
and preston ([i], Chapter 6, Section 5). II.5.8 LEMMA.
--Let ~R =
~ ~ c~A
minimal right ideals. if for some
only
A subset
(7 E A
Necessity.
same as in 5.4. some
(7 E A.
0
is a semi~rime ring having ct
R
of
~
E R,
a~
First note that
Now let
Let
~
is a minimal right ideal of
and some minimal right ideal
R = ~a E ~ l a~ PROOF.
, w h e r e each ct
R
= 0 ~
if
R
of
~
~
if and
, we have
~ # ~}.
is a semiprime ring; the proof is the
be a minimal right ideal of
3,
then
R~
~ 0
for
E R~ ; then there exists 0 ~ a E R such that an = a . (7 cy ~ c~ (7 By the dual of i.ii, we have s~ = R so that s ~ = R~ w h i c h again by the dual of i.ii implies that
@ a
R~
is a minimal
D = [r E R I r ~ If
r E D
and
s E R, then
a right ideal of (recall that a ~ R, we have we obtain in
R
~.
~
r
0 # ab ~ D.
r~
~ a
and
.
Let
~ # (7}.
= (r~)(s~) E R~
~
= 0 b
E ~
and
rs E R, so that
be such that
a b
is semiprime so it has trivial right annihilator).
Let
b E ~
Consequently
w h i c h implies that
=
for all
(rs)~ 0
an~ = a ; letting
Sufficiency. and
Let
= 0
right ideal of
D
be such that
b~
D
is
# 0
Then for some
= b , b~
= 0~
is a nonzero right ideal of
~
if
~ # ~,
contained
R = D.
s,r E R
and suppose that
a # 0.
~ R(7, which by the dual of i.ii implies that
Then
0
# s
r~ = a~x~
= an
E R
for some
98 x~ ~ ~.
Now letting
obtain
ax = r.
x E ~
be such that
Consequently
dual of i.ii implies
that
We are now ready
R
a~ = R
x~
= x~
for every
and
x~ 6 = 06
0 ~ a ~ R
is a minimal right ideal of
if
6 # ~, we
w h i c h again by the
~.
to prove the m a i n result of this section.
This result is new.
Another c h a r a c t e r i z a t i o n of the rings in the next theorem was given by Jaffard 11.5.9 THEOREM. i)
~
ii)
~
~
~
are e@uivalent.
i~s ~ semiprime ring essential extension of its nonzero socle.
Ever~ nonzero ideal of
iii) each
The following conditions on a ring
[ii.
i_~s isomorphic
~
contains an idempotent minimal right ideal.
to a subrin$ o f
~ ~ containing the socle thereof, where ~A ~ -is - _ a prime ring with minimal right ideals.
iv)
~
is a dense extension of a primitive regular semigroup.
v)
~
has a primitive regular ideal having nonzero intersection with every
nonzero ideal of PROOF.
~.
i) = ii).
Since
right ideal is idempotent. we have then
~ N I ~ 0
since
~
a = a l + a 2 + ... + a
components of b I E %1
~
is semiprime,
Let
~
such that
1.6 implies
the socle of
ai E ~
, where
n I
alb I @ 0; on the other hand
and thus
For a nonzero ideal
~i
~
~
~.
Let
0 # a E ~ A I,
are some homogeneous
aib I = 0
I.
But then
I
for
1 < i < n
Hence
since
0 ~ ab l
contains an idempotent minimal
right ideal. ii) = iii). hence of
~
~.
The hypothesis
is semiprime. If
A
for every
By 5.2,
implies that
~
the ideal
is the intersection of prime ideals
0
cannot have nilpotent
is an index set of homogeneous components of the socle
~ E A
we can choose a prime ideal
I
such that
I
ideals and
~
N ~
of
= 0
~,
since
is simple. Assume that
N I , then (a) is a nonzero ideal and hence by o~A ~ hypothesis contains a m i n i m a l right ideal R of ~. Hence R ~ (a) c N I
0 ~ a ~
--
thus 66
R c
I
for all
and hence
But
R
Consequently
N I ~ A
-
is a prime ring and the m a p p i n g
onto a subdirect sum of rings
-
~
.
o~A
and ~
is contained in some homogeneous component
R c 6 6 n 16 = 0, a contradiction. -
~ = ~/I~
~ E A.
I,
oi ~i is a simple atomic ring and thus there exists
~
distinct homogeneous components annihilate each other. = alb I E ~ i
that every minimal
~.
is an essential extension of
for some
n By 4.3, each
6.
denote
~:r ~ (r + I )o~ A
= 0.
Hence
~
is an isomorphism of
99 The canonical ~
~ I~ = 0~.
an ideal of
homomorphism
Hence ~
.
~
contained
~e
ideal,
~e"
Next let @ A
M
Since
) ~
since
~e
and thus
that every ~e
is
is a prime ring, its socle is contained that
~e
right ideal of
right ideal
M = [(a
@e
is the socle of
~.
Hence
~
right ideals.
be a minimal
and a minimal
right ideal of
~
and we conclude
is a prime ring with minimal
on
of a simple atomic ring, it follows
is a minimal
in the socle of
is one-to-one
so that the latter is a simple atomic ring and also
From the properties
minimal right ideal of
in every nonzero
~e :N ~ ~e = ~/I~
~ ~e~ e
M~
~ ~
of
N ~ . Then by 5.8, o~A ~ such that
~
I a e E Me,
a
= 0
if
there exists
~ # e}.
(i)
o6A ~ Furthermore,
we must have
M~ ~ ~e~ e
and thus
~ ~ . o~A ~ iii) = iv). We may suppose that
M ~ ~.
Consequently
~
contains
the socle of
~ ~
R
with property
-- c ~ A
are pairwise disjoint. group socle
~
of
We know by 1.25 that
~
.
a -f'r(:y
of
.el~
U ~',
~ =
~
M
the hypothesis Consequently ! b E ~e with ab = O.
~
®
, .
c~
R ei
into
~e'
given in (I) above is a minimal implies that
~' ~ ~
M c-- ~.
and thus
~ ~ ~, then
It follows
is a primitive
that
(ab)~ ~'
regular
is an isomorphism
(a~)~A
of
~
It follows
~ =
(a~7)(b~)
semigroup.
that
G(~
).
= k a I~ ' PI~
a' E ~ ,
we obtain
~np
maps
O-simple.
If
for all
[1 ~ by 5.8 and c~ o~A 9'~ onto ~ . a E ~
and
W E A, and thus ~
and therefore
t h e mapping
(a ~ ~ ) Hence by 5.6, we have that
~ (k a I~ ' Pa I~ ) ~ ( k , p )
kl~
is a minimal right ideal
sum of semigroups
By 1 . 7 . 5 ,
)
Me
right ideal of
= 0
is an orthogonal
into
If
is completely
a ~ (kal ~ , pal~
where
~
c~A
is an isomorphism of
IRe, then
is a dense extension of the semi-
C~' a'r[ ~,
o~A ~ Then
BIR
'01 c~
Define
(Y ~' =
iii), and that
~
((a)~A
= O a I~ , are isomorphisms
E ~),
mapping
~A For
I
a' = (a~)~A
~ (Xa, la , Pa':~
) ~ (X:'~a)
E :(~)
(2>
i00 where
a = aJ~
a E ~, then istence of
and
~(~)
for some
a~ E ~
for which
mapped onto
(Xa,Pa).
also maps
is the inner part of
a ~ ~
~
onto
~ E A. a'~
It follows ~(~)
If
I
a
and
ar~
= 0~
which by 1.7.5 implies
that
~
~I~
by:
is an ideal of
a o b
that
[PS~ such tLat
if and only if
is the equality relation.
be the equality relation which v) = i).
Let
semigroup socle of ~.
If
R
regular, Hence
= C(R)
J ~ ~' = 0 Thus
~
~
But
~ ~
Let:
I
~.
R
implies
that
~
must
J = I U ~R
that
that I
Then
I
where of
~
R~
is
rxa E ~' N R = 0.
~
J = 0.
~ = ~.
I = 0.
be the
is an ideal of
s E ~, r E R, and since
is an ideal of
If
~
be a minimal right ideal of
implies that
which proves If
o
a = b, has the property
a = axa = s(rxa)
The ideal
and the hypothesis implies
is a nonzero ideal of .
or
I = (~k~)(~R).
for some
w h i c h by hypothesis
extension of its nonzero socle
Therefore
by 4.3.
such that
~.
is a
R, then the congruence
a,b ~ I
consider
a = sr
R c F~
~
1 = 0.
~ ~ ~,
~ I = 0.
is the socle of
~I ~ R = 0
I N ~ = 0, where
Then
x E ~J
is
~(~)
is a dense extension of
~.
and thus
contradiction.
into
be the primitive regular ~deal in v), and let
there exists
property
~
~
~
a E ~' N I, then
a = 0
~ # ~; in (2), a ~
But then the hypothesis
forces
is not contained in
and if
if
if
implies the ex-
~.
dense extension of the primitive regular semigroup defined on
On the other hand,
the hypothesis
that the isomorphism w h i c h maps
the primitive regular semigroup iv) ~ v).
~(~).
As above,
then has the
But then
R = 0, a
By 4.3, we have that such that
Consequently
is a nonzero ideal of
~ [~ I = 0, then B ~,
is an essential then
~I ~
~, and thus must contain a completely 0-simple component
always contains a nonzero idempotent
so that
I
cannot be nilpotent.
is a semiprime ring.
Note that the rings minimal right ideals. according to 5.8,
for each
a prime ring by 1.25.
~
in the theorem are subdirect sums of prime rings with
For if
~
is a subring of
~ ~ containing its socle, then (~A ~ must contain the socle of ~ and is thus
~ ~ A, ~
It is clear that the rings in the theorem do not represent
the general case of subdirect sums of prime rings with minimal right ideals.
In
order to deduce a consequence of the theorem we first need a ring analogue of 5.6. II.5.10 LEMMA.
Let
~ = ¢~A~'
where for each
~ E A, ~ 2 = ~
mapping ~:(k'P) ~ (~I~ ' PI~ )o~A is an isomorphism of
~(~)
onto
~ ~(~ QEA
).
((k,0) E Q ( ~ ) )
.
Then the
i01
PROOF.
Let )~ be a left translation of n a = i~__ixiYi for some xiY i E [Rc~ since
Then
3, a ( ~c~' and suppose that 3 2~ = ~
ka # 0.
so that
n 0 ~ ~a = ~(i~__ixiYi) = i=l~(Xxi)Y i. Since
~c~
ha ~ ~
.
is an ideal of
3
and
This shows that
~
maps
valid for r i g h t I~ ~(~ ).
translations
That
c~A ~ of a direct
~
-lY- ~ 51(~, we must have 3
which
into itself.
implies
that
indeed
maps
f~(~)
into
sum a n d
is
left
as
an exercise.
The following conditions on a ring
~
are equivalent.
i)
~
i__s_s~ semiprime ring maximal essential extension of its nonzero socle.
ii)
3
is a complete direct sum of maximal primitive rings.
iii)
F~
is a maximal dense extension of a primitive regular semigroup.
PROOF.
i) = ii).
Since
~
is semiprime, it has a trivial annihilator and
hence the ring analogue of 1.7.5 implies that
~u
and thus
has the remaining properties follows easily from the definition
II.5.11 COROLLARY.
3.
~
_(kxi)Yi ~ ~c~
Similarly the same result is
By 4.3, we have (A ,Vc~)
~ ~ ~(~), where
~
is the socle of
~ =
+G ~ , where each ~ is isomorphic to a ring of the form c~A c~ c~ by 2.8 so that ~2c~ = ~c~" Hence 5.10 yields [~(~) ~ II ~(~c~ )" By
c~A 1.7.13, we have
~(~c~ ) ~ £U (Ac~'Vc~)' where
ii) = iii). pairs
£U (4~'Vc~)
By 1.27 and 1.7.13, we have
(U ,4c,Vc~).
F~ ~
But by 1.7.10, gU (4(,V(~)
completely O-simple semigroup
is a maximal primitive ring.
~ gU (~c~'Vc~) for some dual ~fA is a maximal dense extension of the
~2,U (Ac~'V~)" Now using 1.7.5 and 5.6, we obtain
-~ ~ g,, (4 ,v ) ~ E P(~o , (4 ,v )) ~ Q ( ~ ® ~2,u c~A Uc~ c~ c~ o~A ~,u ~ a c~A where
¢~A~ ~) 52'U (A ,V )
(4 ,v ))
is a primitive regular semigroup which again by 1.7.5
implies that iii) holds. iii) = i). socle.
containing that
By 5.9, ~
is a semiprime ring essential extension of its nonzero
By the ring analogue of 1.7.5, we may identify
~
~(~), where
~
is the socle of
~.
is a dense extension of the semigroup socle
regular semigroup.
On the other hand
~ =
~ %'
%
~ ~2,U (4 ,V )
as
before.
$2,U(A,V)
~(~)
~
of
3
which is a primitive
~ O ~ , a~..A In the proof of 1.7.13, we have seen that the
correspondence which to each bitranslation of to
with a subring of
~ ~U (4 ,V ), ~ =
c~A
~
~
It follows from the proof of 5.9
is an isomorphism of
~
~U(4,V)
~I'~(~U(4,V)) onto
associates its restriction ~(~2,U(4,V))
so the same
102 must hold for
~
and
~ .
5.6, by taking direct dense extension
of
Hence
products,
~(~),
identified
extension of
~.
by hypothesis
of maximality
ring analogue
of 1.7.5 yields
zero socle
~f~(~) ~ ( ~
yields
We have already
above with
implies
that ~
which
~ ~(~).
seen that
that
)
~(~)
~, so
~
~
together with 5.10 and
By 1.7.5, ~(~) ~(~)
is a maximal
is a maximal
is a dense extension
= ~(~)
is a maximal
and thus essential
of
dense
~
~ = ~(~).
which The
extension of its non-
~.
II.5.12 EXERCISES. i)
Establish
ii)
the relationship
Show that in 5.9,
iii)
this for semiprime
implies
1
and on subdirect
~
ideals,
([i], Chapter
SEMIPRIME
several
(a)r
the principal
the last two references right ideals.
II.6.1 THEOREM.
thereof.
contain
further
The followin$ conditions
is a semiprime
rin$ satisfying
i) = ii).
rings.
Let
of
Then
~.
(al) r ~
aI = Xll+Xl2+
from 2.8 and 1.3.6 that each such that
Xli = xliei,
It follows
that
resular
ideal
By 4.3, we have
(a2) r ~
...
generated
~
~ =
a by
of a ring a.
~,
The equivs-
[i] and also to F. Sz~sz
information
on rings satisfying
is new. ~
ar___~eequivalent.
for principal
such that G ~
for some
right ideals.
~ = C(~).
, where
~
are simple atomic
chain of principal
right ideals
Xli E % i , 1 < i < n.
is a regular ring.
and thus letting
atomic rings
rinss.
be a descending
... + X l n ~
~
on a rin$
d.c.c
is a direct sum of simple atomic has a primitive
([i], Sections
2).
For an element
The rest of the theorem
ii)
see McCoy
of semiprime
is due to Faith
atomic ring.
PROOF.
an analogue of
ATOMIC RINGS
right ideal of
is a semiprime
iv)
Establish
ill, Section
characterizations
i)
iii)
~".
if and only if for sny
on prime and semiprime sums, Behreus
fence of ii) and iii) in the next theorem [1],[2];
is prime
b E I.
first establish
d.c.c f o r p r i n c i p a l
direct sums.
[g~" may be replaced by "ideal of
a ~ I
and then will deduce some consequences we denote by
snd internal
or
II.6 We will
of
of a ring
that either
external
ideals.
For more information 18-20),
part v), "ideal
Show that an ideal
a,b ~ ~, ~a~b ~ I
between
Hence
there exists
e = e l + e 2 + ... + e n ,
we obtain
We know e. E s I = ale.
103
(al) r = al~ = ( X l l + X l 2 + ... + X l n ) ~ +...+ = Xll%l Xln~ n Xll~+''" =
+Xln~
= (Xll) r + ... + (Xln) r. Since
for any
i
we have
(ai) r ~
(al)r,
this discussion
is valid
for any
a i = Xil + ... + X i n , and we obtain (Xll) r + (x12) r + ... + (Xln) r ~ Since
n ~ ~ i=l ~i
the sum
is direct
each of these
d.c.c,
(xji)r ~ ~ ,
(x2i)r ~
is a principal
for principals
right
...
right
ideals.
for
that
(am) r = (am+l) r = ...
i < i < n,
ideal of
~i
By 2.8
Hence we can find
(Xmi)r = (Xm+l,i) r = ... But then
we conclude
....
i
(Xli)r ~ where
and
(x21) r + (x22) r + ... + (X2n) r ~
and thus
for ~
m
satisfies
each
such that
1 < i < n.
satisfies
d.c.c,
for principal
right
ideals. ii) = iii). minimal
We have
principal
right
seen in the proof of "iii) = ii)" in 2.8 that every
ideal
is a minimal
right
ideal.
Since
by 1.6, every minimal
right
ideal is of the form
e/E
Hence
right
ideal of
a minimal
every principal
e~.
Let
a E ~
(a)r ~ e ~
and consider
(a)r.
for some idempotent
eI
~
contains
If
(s) r
~
for some idempotent right
is not a minimal
such that
el~
is semiprime, e.
ideal of the form right
is a minimal
ideal,
right
then
ideal.
Since a = e l a + (a - els) 6 e l ~ + (a - ela)r, we obtain
(a)r ~ e l ~ + (a- ela)r.
Conversely,
for any integer
n ( a - ela ) = n a - e l ( n s ) ~_ ( a ) r + e l ~ and for
s 6 ~, we have
el~+ (a-ela)r If
(al) r
(a)r.
e~
Hence
letting
right
contained
a 2 = a I -e2al,
in
we have
principal
(al)r, where
(al) r = e ~ +
ideals.
as a sum of elements
Thus
e2
right
ideals
that
~
is a direct
ideal of We deduce
r ~
that
(a2) r.
of steps because
e.~.
+en~
of the d.c.c,
so that
Therefore
~
a
sum of simple
atomic
for
can be written
is an atomic
i
which by 4.3 implies
implies
Hence
(a)r = e l N + e ~ + . . .
of minimal
right
is an idempotent.
(a2) r.
~ e~+(a2)
which
(a)r = e l ~ + ( a l ) r.
we can find a minimal
must end after a finite number
right
(a)r
a I = a - e l a , we obtain
ideal,
(a)r = e l ~ + e ~ + ( a 2 ) r This process
(a)r
(a- ela)s = a s - elas E (a)r+ el~ ~
is not a minimal
of the form for
~
c
n, we have
rings.
ring
104 iii) = iv). we have obtain ~(y
Let
O ~ , where each ~ ¢6A ~ , where ~ is [he semigroup
= C (~)
~ ~
~ ~
c ~ n ~ -- ~
~ =
Q ~
= 0
= 0
and
~
[J ~ a6A o~
is a regular semigroup,
each element of ~
= C(~ )
some
~ .
Consequently Let
Since for each
it follows that
nilpotent. and
~
~ = C(~), where 3.
Consequently
coincide w h i c h
Then
~
~
~
~
Hence for
there exists
Let
bi 6 % i
~
[RC(~)
Hence
~ (y ~-- I
so that
I
for some cannot be
~
be a semiprime ~
implies
~ha~
atomic ring and and
~
~
~
is atomic,
be its semigroup
is a regular ring.
~.
~.
By 6.1, we have
Hence by 5.9, we conclude that
such that
[~
is
~ =
~ ~ , where each ~ is simple and c~A c~ and we know that the ~ annihilate each ai ~ ~ i
a i = aibia i
and
~i
~ ~j
if
i ~ j,
and thus
= alblal + a 2 b 2 a 2 + ' ' ' + a n b n a n ... + a n ) ( b l + b 2 +
... + b n ) ( a l + a 2
+ ... + a n )
is a regular ring.
II.6.3 COROLLARY.
PROOF.
[R~. Let
By 1.13, the minimal right ideals of
together with the hypothesis
= (al+a2+
then
,
is s sum of elements of
I ~ ~ # 0.
a = a l + a 2 + ... + a n, where
al+a2+'''+an
and hence
~ ~ A,
~
Further,
c~
is semiprime.
atomic hence regular by 2.8 and 1.3.6,
=
, and for each
is completely 0-simple,
is a dense extension of
s dense extension of
a
[~.
is a primitive regular ideal of
has a nonzero intersection with
other.
is an ideal of
is an ideal of
~
Thus
We have seen in the proof of "iv) ~ i)" in 6.1 that every nonzero ideal
PROOF.
of
~
~
~ . c~
A part of the proof of "v) = iv)" in 2.8 can be
~ ~ ~ and :~A c~
II.6.2 COROLLARY. socle.
~ # ~, we
Hence in particular,
~ E A, ~ ~
that each element of
applied to the present situation showing that ~ =
For
By 2.8,
N = O(~).
be a nonzero ideal of
~ A, where
~
is actually an orthogonal sum of
is a sum of elements of some
by 2.8; it follows
iv) = i). I
~
socle of
since the sum is direct.
~ =
is a primitive regular semigroup. and
is a simple atomic ring.
If
~
is a primitive regular ideal of
is a dense extension of
~L~
for a ring
3,
~.
This follows immediately from 6.1 and 6.2.
The next corollary is known as the First W e d d e r b u r n
(or W e d d e r b u r n - A r t i n )
Theorem w i t h "semisimple" instead of "semiprime" w h i c h are in this case equivalent. For other c h a r a c t e r i z a t i o n s of these rings, Jans ([i], Chapter 2). II.6.4 COROLLARY.
Also consult Behrens A ring
~
see Fuchs and Szele
[i], G o l d m a n
[i],
([i], Chapter III, Section 3).
is semiprime and satisfies d.c.c,
for right ideals
if and only if it is a direct sum of a finite number of matrix rings over d i v i s i o n rings.
105 PROOF.
~ 6 , where each ~ is simple c~A~ atomic. Since the sum is direct, every right ideal of ~ is also a right ideal (y of 3. Hence ~ satisfies d.c.c, for right ideals, which by 2.8 and 1.3.14
implies
Necessity.
that
infinite,
By 6.1, we have
% ~ ~ffn ' n
X n
~ =
matrices
over the division
then for any infinite sequence
obtain an infinite d e s c e n d i n g chain Therefore
A
ring
~cl.
~i,c~2,... ~ A, letting
Jl ~ J2 ~
If
A
is
Ji =
~ ~ we k>i ~k' "''' c o n t r a d i c t i n g the hypothesis.
must be finite
The proof of the s u f f i c i e n c y is left as an exercise. II.6.5 EXERCISES. i) for
Prove that each of the following conditions on a ring
~
~
is sufficient
to be a division ring. a)
[F~
b)
~
is a primitive regular semigroup.
c)
For every
is a regular ring with only one nonzero idempotent. 0 # e ~ ~,
there exists a unique
x ~ ~
such that
= axa.
(These conditions are trivially necessary.) ii)
Let
ideal of
~ =
~ ~ , w h e r e each ~ ~A ~ is a direct sum of some ~
~ a)
A
b)
each
is a simple atomic ring. and conversely.
is finite if and only if ~
is isomorphic
only if every minimal
~
Show that every
Deduce that
satisfies d.c,c,
for two-sided ideals,
to a matrix ring over a division ring if and
two-sided ideal of
II.7
~
satisfies d.c.c,
for right ideals.
ISOMORPHISMS
We consider first isomorphisms of some of the rings we have encountered in the two preceding sections.
After that we give a construction of all isomorphisms of
two Rees m a t r i x rings. 11.7.1 LEMMA. >emigroup socles of
Let
~
and
~
and
~
w h i c h is a division ring.
~
and = of
~
~
G
~ ~
~
and
~
of
~/ =
~
~
onto
onto
AI
and
~
onto
~ ~/t
onto
~.
= ~I~ ' we have an i s o m o r p h i s m
~
~ ~ = ~~ of
~i
and
~
b e the
has no ideal can be uniquely
By 6.1, we have
is a simple atomic ring.
~ and ~ ~ ~ It is easy to see that
such that
~
~.
@ ~ /, w h e r e
~IEA~ respectively. A
~
be an isomorphism of
~//, where each
aEA ~ % and ~11,
function ~
Let
and assume that
Then every isomorphism o f
extended to a ring isomorphism o f PROOF.
b~e semiprime atomic rings, ~
respectively,
~
onto
~ aEA ~
Hence
are the semigronp socles ~
for all ~i~ .
~ =
induces a one-to-one ~ E A.
Letting
106 i ,V )I , where ~--~U (h(~,V) and r~~ ~--:---~U,(£<~ dim V(~> i c~ (y in view of the hypothesis on ~. We know from 1.5.12 that every isomorphism @
By 2.8, we have
$2~U(£,V)
onto
isomorphism
@
restriction
of
SU(&,V)
onto
~'
.
~2,UJ(h',V'), of
~U(h,V)
~
to
where
onto
SU(L,V)
that
Since
'~a
admits
~ =
~ ~
r = r I + . . . + r n,
hence
of
~
extended
Using 1.5.12 again,
is the unique extension
a uniqu~ ~xtension
the function +
morphism
l, can be uniquely
•'u'(A''V')"
..
+
,~:r ~ rl~a I where
dim V >
of
@
of
to an
we see that the
to an isomorphism
of
SU j(A',v'),
We deduce onto
~
~
onto
~
to an isomorphism
defined
~'
Let
rn~ n
At each step,
~
c~
-
.
extension
of
by
r.i % '~'-i' is a unique extension
is the unique
Ii.7.2 LEMMA.
~
of
~
the extension
of each
isomorphigms
to an isomorphism
be an isomorphism
~c
of a rin$
of
[~
~
to an isoare unique,
onto
~'
onto a rin~
~'
Then
the function
e-:(~,o) ~x = [ k ( x e - l ) ] G ,
~vhere
onto
~(.~) PROOF.
-
(k,~)
((>..,0)
x0 = [(x6-l)p]G
w ~ = .~, r r6 ()-,0) ( ~ ( ~ ) and
(x ~ ~ ' ) ~
such t h a t
(r t ~).
For
x,y Q ~',
[(xy)
= {.),l(xy)e-£]}e
= [~[(x+y)6-1]]6 =
so that
~
x(ly) = x[k(ye-t)]6 =
which proves
that (~$)x
and similarly one-to-one From
=
},r x
=
and
x ( ~',
Ikr(X@-l)}@
= kx+Xy,
p-
is a right
translation;
further
= [ (x6-l) [;, ( y e -i ) ] ; e
= [(×e-1)p]ey
If also
that
= (x-6)y ,
(~2,~) ~ ~(~),
= {),[
x(~,7) = x%0-~, showing easily
= (fx)y,
= [>(x6-1+y6-1)]e
analogously
lL[~(xe-1)]e]
and onto follows r ( ~
= lX(xe-1)]ey
{.[(xe-i)oj(ye-1)]e
(~,p) ~ fZ(~').
o__[f N(~)
we o b t a i n
[k(xe-1)J6+[),(ye-i)]e
is a left translation,
is an i s o m o r p h i s m
= {k[(xe-l)(y6-£)]]6
= [[x(,,,:e-l)](ye-l).~e ~(x+y)
~ ~(~)),
~
then
= X-6x
is a homomorphism.
from the fact
that
@
we obtain
= [r(xs-l)j6 = (r@)x = krsX
shares
That
@
is also
these properties.
i07
so that
--Xr= Xr~, and analogously
II.7.3 LEMMA.
Let
~
Pr = PrG --
which proves
be an essential extension of a ring
be a maximal essential extension of a ring onto
It
more~ X
I ~.
~
onto
~
if and onl~ if
~
~r ~ =
X
~r~"
I, ~(I) = 0, 3 ~
Then ever~ isomorphism
ca_nn.be uniquely . . extended . to an isomorphism maps
the formula
__°f ~
into
~
of
~.
1
Further-
is a maximal essential extension of
I. PROOF.
Let
8
be an isomorphism of
the ring analogue of 1.7.5 implies that ~(I), that all
and ~
T' = 7(3t:I l)
Hence
onto
X = T~Tt-I
~(I)
It
m = T(3:I)
is an isomorphism of
is an isomorphism of
x E I.
I
onto
~(I ~) = 0
and hence
is an isomorphism of
3'
~(I')
Then
onto
~(I').
with the property
is an isomorphism of
~
into
3
into
Now 7.2 asserts
~t
~x ~ = nx8
for
such that for all
x ~ I, we have x~ so that
X
extends
Let also for any
~
r ~ 3
x0
~x@ T ~.
be an extension of and
6
to an isomorphism of
~
into
3 ~.
Then
x ~ I, we obtain
(r~)(x@) = (r~)(xw) = (rx)~ = (rx)x = (rx)(x~) = (rx)(x~) and similarly
(xS)(r~) = (x~)(rA).
r ~ - rk { ~ t ( I t ) .
Bun
~,(I')
~,Cl
t) e ~' = ~C~'} = o
~ t ( I t) = 0.
x ~ I
is an ideal of
which together with the hypothesis implies
Since
that
Consequently
3'
is arbitrary,
~'
it follows that
satisfying
is an essential extension of
r~ = rx
and thus
~ = X
since
It
r ~ ~
is
arbitrary. The necessity in the last statement of the lemma is obvious; from the fact that
~X
is an essential extension if
sufficiency follows
I'.
The next result is new. II.7.4 THEOREM. socle
~
Let
~
be a semiprime tins essential extension of its nonzero
and assume that
3
has no ideal which is a division ring.
ring which morphism
is a maximal essential extension of its socle ~
of the semigroup socle
~
of
~
onto
can b e uniquely extended to a rin$ i s o m o r p h i s m X
~
~
onto
PROOF.
Let
3 ~ ~
if and only if
Since
According ~.
~t = C ( ~ ) ,
to 7.1, ~
By 7.3, ~
R
Hence
the semigroup socle o__[f 3
onto Rt
3 t
admits a unique extension
Rt
be a
into
3 ~.
~
of
~t
Furthermore, ~.
It follows from 5.9 that
is a primitive regular ideal of
by 6.1 we conclude that
can be uniquely extended
Let
Then every iso-
is a maximal essential extension of
be an isomorphism of
is a primitive regular semigroup. ~t
3
X
~.
~t
is a semiprime atomic ring.
to a ring isomorphism
to an isomorphism
X
of
3
0
of
into
~ ~t
onto
108
Consequently If
~
~=
X
provides an extension of
is another such extension of ~'.
But then
=
~I~
The last statement
8
~ = C(R)
• = X
and
if every m u l t i p l i c a t i v e
~/ = ~(~')
exercise
i), a ring
~
A semiprime rin$
~
~
~F~
onto
~I~ j
for some ring
~
that
~'
nonzero socle.
By the ring analogue of 1.7.5, we may suppose that
in a maximal essential
extension
the proof of uniqueness
~
of the socle ~,
~"
6'
satisfies
in 7.3, it follows
X
of
~ = X
of
~
such that
£
and
BA
~'
Hence
is embedded ~ = ~I~'
~J~.
Similarly as in
(x~"~ R, r E ~).
~', we have that
which proves that
is said to be invertible
AB
3'
into
R' ~
T r~ = T rX, where
by 5.9 which by 1.7.5 yields is additive.
We now consider isomorphisms of Rees matrix rings. division ring
ac-
that
is a dense extension of
Consequently
Since
the conditions of 7.4 and ~ e n c e
(xg)(r~) = (xg)(rk)
be the semigroup socle of But
[4].
is a semiprime ring essential extension of its
to an isomorphism
(r~p)(xg) = (rk)(x¢),
~'.
has can be characterized m u l t i p l i c a t i v e l y
is the semigroup socle of
admits a unique extension
£
An I X I'-matrix
B
if there exists an I' X l-matrix
over a A
over
are the identity matrices of the c o r r e s p o n d i n g sizes.
The following p r o p o s i t i o n is due to Hotzel
[l], its proof appears here for the first
time. 11.7.6 PROPOSITION. matrix
rings.
X = (xij)
Let
over
h
~
Let
~ = ~(I,h,A;P)
be an i s o m o r p h i s m o__ff g
we write
finite A × A ~ - m a t r i x over &', an___ddsuppose that
X~ = (xij~).
h', B
P$ = AP~B.
i s an i s o m o r p h i s m o f
~
onto
and
~' =
onto
~(I',hJ,A';P ')
Next let
A
be an invertible row
~J,
Then the function
X
defined by
(X E ~) Conversely ever~ isomorphism of
can be so constructed. To prove the direct part, we let (X*Y)X
= B(X*Y)$A
be Rees
g', and fo____rran____yymatrix
be an invertible column finite I'X I-matrix over
x:X ~ B(X$)A
PROOF.
X"
extension of its nonzero socle, has unique addition.
be an isomorphism of that
r~0 = r k.
and
which has no ideal which is s division
it follows
Rt
e
(snd is thus a
cording to 5.9,
T = T(~":R').
imply that
[i], see also Gluskin
a ring with the properties
Letting
~¢.
has unique addition if and only
isomorphism onto another ring is additive
rin$, a n d is an essential
~
into
by the uniqueness of both
The next corollsry is due to Rickart
II.7.5 COROLLARY.
where
~
follows immedistely from 7.3.
ring isomorphism).
Let
to an isomorphism of
~, then
and hence
Recall that by 1.5.22,
PROOF.
9
X,Y 6 •
and obtain
= B(XPY)$A
= B(X~)(P$)(Y$)A = B(X$)(AP'B)(Y$)A
(XX)P'(Y X) :
(X~) * (Yx) ,
~
onto
~
109 Since B
X
is obviously
easily
implies
additive,
that
For the converse, division assume
ring,
that
k
maps
we let
the statement
~
it is a homomorphism. ~
k
onto
~'
be an isomorphism
in the proposition
is not a division
The imvertibility
of
A
and
and is one-to-one. of
~
onto
is trivially
ring and consider
~'.
If
satisfied.
the commutative
~
is a
We thus
diagram
X
au(a,v) . where
the vertical
2.8, and of
{(~,a)
(U,h,V)
the isomorphisms
is the isomorphism
onto
of describing exploiting
arrows denote
(U',h',V')
in the proof of "iv) = vi)" in
induced by the semilinear
provided by 1.5.13.
the isomorphism
the commutativity
, au, (k' ,v')
X
isomorphism
(~,a)
The idea of the proof consists
as in the statement
of the proposition
by
of the above diagram.
Let
z=iz~IxeA], z' = { z ' t be bases of
V, U, V'
I X ' E Ar}
I
and
w'={w~,li'E
U ~, r e s p e c t i v e l y .
(m,a)
linear isomorphism
w=iw iliE f], Let
I' } b
be t h e a d j o i n t
of
the
semi-
and l e t
A = (ceXX,)kEA,X,EA,,
B = (~i,i)i,Ei,,i~i
where =
~ z' k'6h '~kk' ~''
z a Then
A
is row finite,
size and are invertible
B
is column
since both
b-lw i =
finite and both a
and
b -I
(1)
~ w' i'£f' i'll'i" A
and
B
are one-to-one
have the required and onto.
Further,
we obtain Pki w = (Zk,Wi)tO = (zk,b(b-lwi))~
= ( ~
~,z'
X'EA'
r
= (z a,b-lwi )I
w'
i'' i'~l'
'
i'~i~i)
Wj I k'' i ') ~i'l
Z /
=
= which
~ X'EA'
~ ~k'( i'El I
E
in terms of matrices We also have
E
zi,s -I =
(2)
'Px
]~'6A' i'{l '(~xk
''i'~i'i
has the form ~ %E A °k 'Xz k
P~ = AP'B
as required.
so that by (1)
we obtain
110
z!
z' a -I ~EA
= ~-~ ( ° k ' x '~) lEA which
XEA
£~ , ~ k ~ ' z "
~
~' i
•
~
(~
~'EA'
,)z',
(o, , . ~ ) ~
)EA
~ ~
~
implies
where
(o%,X~)~k , = 61, ~, (t',~' E i'), (3) X£A i s the Kronecker d e l t a f u n c t i o n . I t follows from (2) and (3) that
6k~p/
¢
XEA ~'6A' i
XEA
I' !
~'EA'
i'61' ~EA
A
(4)
= i'EI' ~J P'' i i ~ i 'i" Finally
let
X = (xix) E ~{. X
~
PX
=
Following
( Y], p . . x .
).
iEl ~i ~ and hence we must express
(z{,a-l)ea
a
-i
ca
the above diagram,
A,D6 i
in matrix
~
e
we have
a -I ~
form.
(5)
ca Using
(i) and (4), we obtain
= (~ol,k(zkc))~ lEA = ( ~
= ~
~ (a.,.p..×.
) ~ ( z a)
= iEl ~ ( )~6A ~ (<~k,kPki)tv) (xiotu) (zPa)
which
=
~ ( i ~t iEl
=
L X'6A'
in matrix notation becomes
isomorphisms
E i'EI'
Z
C~ . , z J , ) A
).'EA' PA
<" (px,i,~i,i ' , ' z. (x ipw)C~pl,)z l, iEl
P'B(Xco)A.
Hence
continuing
the sequence
of
in (5), we have
a proving
~PiJ ,)i / ~(i , i )E( x i1
that
-i
ca ~ P'B(Xo%)A - B(X&)A
X X = B(X~)A.
II.7.7 COROLLARY.
If
g ~ g', card i = card A'.
~(I,h,A;P) Further,
e~ ~(I',A',A';P'),
hl ~-~ h{,
then
if and only if
card I = card I', h ~ g~,
card I
= card I'. II.7.8
i)
EXERCISES,
Show
that an essential
extension
of a semiprime
that every semiprime
ring can be embedded
the same for "prime"
instead of "semiprime".
ring is semiprime.
into a semiprime
ring with
Deduce
identity.
Do
iii ii)
Let
A
has no nonzero
iii) each
Let
~
be a m i n i m a l nilpotent
~c~A
ideal then
t~ ,}0/~ A,
and
is a simple
right
elements,
atomic
ring.
of a semiprime A
must
be two
ring
~.
be a d i v i s i o n
families
of rings
For an i s o m o r p h i s m
each (r)
that
there exists
~ ~ A, an i s o m o r p h i s m E
II ~ c~A a)
, we have substitute
b)
assume
c)
combine
([5], Chapter
of
a complete
items
L~
~
function
onto
~ = r ~ .
that each
For a d i s c u s s i o n petrich
~
(r),~
a one-to-one
Also
direct
~
of
~ ~ c~6.A o~
~
of
A
such
consider
that
is a m a x i m a l
primitive
if
A of
that
onto
onto
A~
and for
for every
the following
sum by a direct
that
and assume
~
c~ II ~ j, show ~ 6 A ~ o~
Show
ring and an ideal
variants:
sum, ring,
a) and b).
of isomorphisms V, S e c t i o n
2),
of
and
the translational for a survey
hull of a semigroup,
consult
petrich
[3].
see
112
PART III
L I N E A R L Y T O P O L O G I Z E D V E C T O R SPACES AND RINGS This is essentially a topological
treatment of some of the rings studied
previously provided with a topology induced either by the vector space or by multiplication.
For a pair of dual vector spaces
U-topology on
V
(U,V)
over a division ring
is discussed here in considerable detail.
Several properties of
subspaces and of semilinear
transformations
It is proved that
the finite topology is a completion of
U-topology.
U*
with
Linearly
relative
topologies
to this topology are e s t a b l i s h e ~
compact modules are shown to be complete,
vector spaces are c h a r a c t e r i z e d in several ways. for the ring of all continuous
the ring of all continuous
4, the
V
with the
and linearly compact
The finite and the uniform
linear transformations are discussed.
For
linear transformations of finite rank various properties
of one-sided ideals relative
to the relativized finite topology are established.
A
completion of a semiprime ring which is an essential extension of its nonzero socle, with a topology and thus a uniformity III.l
induced by its socle,
A TOPOLOGY FOR A V E C T O R SPACE
For a pair of dual vector spaces induced in a natural way by
U
(U,V)
we will introduce a topology on
and the bilinear
general topology will be assumed; follow Kelley
is constructed.
form.
V
A rudimentary knowledge of
for the terminology and basic results we will
[i].
The C a r t e s i a n product of a family of sets the particular case when
for all
is denoted by XA
In
instead of
X. We will denote a topological
T
~ E A, we write
~ X . o6A ~
~ X , ~A ~ since in this case we are actually dealing w i t h the set of all functions from A into
X = X
~X } ~ A ~
is a topology on
simply write
X
X
space by
(X,T)
where
X
is a nonempty set and
given as a collection of (open) sets;
sometimes we will
if there is no danger of confusion.
For a nonempty family
IX } ~ A
of topological spaces,
their ~roduct is the
topological space on the set the sets
U = ~ (x)
fixed
and
~
U~
Ix
~ X with the topology whose subbase consists of c~A ~ E U~} where ~ E A and Up is an open set in X~ for
(it suffices
to take
U~
ranging over either a base or a subbase
113 of
X~).
This topology
topological
is called
the product
(or Tychonoff)
We now take each has a subbase
~ X . o~A ~ to have the discrete
topology and the
space is denoted by X
consisting
topology.
Then the product
~ X aEA ~
of the sets
S(~;y) = i(x ) I x~ = y], where
~ E A
product E A
and
y E X~.
If
X~ = X
for all
topology by taking all sets of the form and
y E X.
of all functions
Specializing
from
X
~ E A, we obtain s subbase
for the
S(~;y) = if E X A I ~f = Y}
where
further by letting
into itself,
A = X, we obtain
and a subbase
X X,
the set
for it is given by the sets
S(x;y ) = [f E X X I xf = y}. The following
concepts
Ill.l.l DEFINITION. X A, where
A
are due to Jscobson
For
A
has the discrete
Hence s base
and
X
nonempty
topology,
for the finite
[5]. sets,
is the finite
topology
consists
the product
topology of
topology on X A,
of the sets
= ~f ~ x A I ~i f = x i'
i = 1,2 ..... n}
B(~I, ...,~n;Xl,...,XnJ n = N B i=l ((~i;xi )" 111.1.2 DEFINITION. topology.
Let
Then the finite
topology of
(A,V)
topology of
be a vector space and give ~V
relstivized
spaces
(U,g,V)
to
V*
A
the discrete
is the finite
V*.
We now fix s pair of dual vector the image of
U
in
V*
under
the isomorphism
and recall
f:u ~ fu' where
that
nat U
Vfu = (v,u)
is (vE V),
see I.l.l. 111.1.3 DEFINITION. topology on U
into
U
V*, where
Equivalently, where
For
for which
the latter
as open sets in
U
V*).
symmetrically
the finite
If needed
of
U
is the
f
(or
be a homeomorphism finite
topology of
K = L N nat U
where
the U-topology
of
L V
of V*.
U
onto
nat U,
This amounts
Kf -I, where
K
of
to
ranges over
runs over all open sets in by considering
nat V
in
U*
topology.
111.1.4 NOTATION. V.
the V-topology
just the sets of the form
nat U
with
that
the relativized
all open sets in One defines
a dual pair,
f:u ~ f u (u E U) is a homeomorphism is endowed with the finite topology.
we may require
the latter is given
declaring
(U,&,V)
the isomorphism
For
(U,&,V)
for emphasis,
we write
a dual pair, TU(~,V )
Tu(V )
instead of
denotes ~u(V).
the U-topolosy o f
i14 111.1.5 LEMMA.
A base for the finite
topology
of
V*
consists
of all sets of
the form B(Vl,V2 ..... Vn;61,62 .... ,6n) = i f E V* I vif = ~i' where each set PROOF.
~Vl,V2,...,Vn}
From above,
linear independence. set
B(vl,v2,...
is linearly
i = 1,2 ..... n}
indel~endent.
a base is given by all such sets without
The proof
,Vn~.$i,~2 '
that for
..,6n}
B(zl,z2,...,Zk;~l,V2,...,~k )
iVl,V2,...,Vn}
is either empty or equal
with
~Zl,Z2,...,Zk]
the requirement
linearly dependent,
of
the
to some
linearly
independent,
is left as
an exercise. III.i.6 COROLLARY.
A base
for
Tu(V )
C
consists
of a!l sets
= iv ~ V I (v'u i) = ~i'
i = 1,2
(Ul,U2,...,Un;61,62,...,6n) where
{Ul,U2,...,Un} PROOF.
is a linearly
If W For
A
of
For a dual pair
= 0
for all
v ~ S},
T~=
Iv~
Vl(v,u)
~ 0
for all
u~
and
subset of a vector
A; we also write u E U, we write
on
binary operation -i x ~ x
making
W,
r}. then
[Wl,W2,...,Wn]
[A]
denotes
instead of
Tu(V )
U, di___m_mT < ~}
the subspace o f
[[Wl,W2,...,Wn}].
of
0
ha___~s
as an open base,
(x,y) - x • y
of
G.
(G,',T)
it a group,
is continuous.
consisting
and a topology
is jointly
continuous
In such a ease,
We write
(G,T)
T
in a topological
to give the neighborhood
G.
the preceding
G
G, a binary for which
and the operation
G
Since
the
of in-
if there is no need for
for any two elements
of
in order
system
on
is said to be compatible with
translation
group these are homeomorphisms,
it suffices of
a left and a right
of a set
T
or simply
the group operation or the topology.
x,y 6 G, there exists
reason,
T
v m = iv] z, u m = [u] m
group is a triple
the group structure
the identity
and
as an open subbase.
G
version
group,
V
Exercise.
A topological
emphasizing
space
The neighborhood ~
subspace o f
[u ~ I u ~ U]
operation
U.
! subspace o f
u l(v,u)
{ T~ I r
PROOF.
S
~u~
111.1.8 COROLLARY.
ii)
(U,V),
s ~=
generated by
i)
subset of
U, let
is a nonempty
v E V
independent
Exercise.
Iii. I.7 NOTATION. subspace
n} '''''
G
sending
x
onto
to give a topology on a
(or a base or subbase
We will use these facts without corollary will be particularly
y, and
express mention. useful.
thereof)
For this
of
i15 III.l.9 COROLLARY.
Th_~e neighborhood
system of
v E V
Tu(V)
for the topology
has i)
{ v + T z IT
ii)
~V+U~
subspace
lu E U}
III.l.10 LEMMA.
U, di___mmT < ~}
as an o~en base,
as an open subbase.
In a topological
group
(G,T)
the following
statements
are
equivalent. i)
T
ii)
is a Hausdorff
topology.
The intersection
o f a l l neighborhoods
The identity o f
G
of the identity of
G
equals
the
identity. iii)
has a neighborhood
base
(subbase)
whose intersection
is the identity. PROOF.
Exercise
(see the remarks
IIl.l.ll LEMMA. G
Every
is a homeomorphism o f PROOF.
If
H
xH
each of which
vector space
is a topological V
making
viz.
A
i) ii) iii)
of
H
in
G
is
where
(with the obvious
(6,V)
is a vector
definition),
T
is a
and such that the
is jointly continuous.
A subset
special
case, which will be assumed
topology. H
subspsce
The following
(L,V,T)
group under addition
only in the following
a maximal
III.i.14 LEMMA. V
ring
has the discrete
III.i.13 DEFINITION. is under inclusion
division
group is closed.
G, then the complement
is a triple
(6,v) ~ 6v
We are interested
space
group
is open by i.ii.
it a topological
scalar multiplication
throughout,
in a topological
Every open subgroup o f ~ topological
is an open subgroup of
A topological
topology on
translation
onto itself.
Exercise.
the union of cosets
space, &
left and every right
G
III.I.12 COROLLARY. PROOF.
above).
of a vector space of
V
is a hyperplane
if
H
V.
statements
concerning a subspace
S
of a vector
are equivalent. S
is a hyperplane.
dim(V/S) S = fm
= 1. for some
0 # f ~ V*, where
fZ
is taken relative
to the dual pair
(V*,V). PROOF. between
The equivalence
subspaces
of
V
of i) and ii) follows
containing
S
from the well-known
and subspaces
of
V/S.
correspondence
116 i) = iii). where
Let
v
be a vector in
V
the sum is direct in the group sense,
x E V, we can write easily that
f
x = x+
is linear,
iii) = i).
Let
v
(xf)v so
not contained in since
for unique
f E V*
and
be a vector in
V
S
S.
Then
V = S ~
is a hyperplane.
x E S
and
xf E A.
Iv],
Hence for any
It follows
f~ = S. such that
vf = o # O.
Then for every
x E V, we can write x = (x- (xf)o-lv) + (xf)o-lv ~ f.L + Thus
v ~ f±
generated by
implies
f~" +
f~
v.
Hence
fm
For
(U,V)
a dual pair and
and
111.1.15 LEMMA. of
V, we have
Iv] = V, where
Iv]
is the subspace of
S
a finite d i m e n s i o n a l
is a basis of
S
S ~ S zz
and
for any subspace
v @ S, then the set
and
(vi,u) = 0
which implies that
THEOREM.
IIl.l.16
S mz ~ S
(A,V,Tu(V))
fu
na__!t U = [ f ~ V* I f
~) c
If
Consequently
x,y E V
(x+y)+T
If is
such that
(v,u) # 0
and
u E S~
topolo$ical vector space,
(fu:V ~ (v,u)
(v ~ V)).
and finite dimensional subspace
±,
6 E ~, then
T
- ( x + T ~) = ~ x + T ±, proving that V
V
Tu(V )
and in
Moreover,
of
Tu(V )
6 ( x + T ±) = 6 x + T z, where
6
[I u ~ = 0 uEU
is com-
so
is its own neighborhood. (A,V,Tu(V))
Since the bilinear form associated with
it follows that
U, we have
(i.e., addition and inversion are
the scalar m u l t i p l i c a t i o n is continuous,
topological vector space. degenerate,
But then
u E U
the additive $rou~) structure of
patible with the additive group structure of continuous).
V.
and the equality holds.
i_~ss_ H a u s d o r f f
are continuous
of
i__~ continuous}.
For any
(x+T'~)+(y+T
i = 1,2,...,n. Hence
topology compatible with
which all functions
PROOF.
for
v ~ S ram.
S
~V,Vl,V 2 ..... Vn}
linearly independent which by 1.2.3 implies the existence of
is the weakest
subspace
S ~ Z = S.
[Vl,V 2 ..... Vn}
(v,u) # 0
V
is a hyperplane.
It is easy to see that
PROOF.
f~ +
[v].
which by i.i0 implies that
is a
(U,V)
is non-
~u(V)
is
Hausdorff. In order to prove that open for any
6 E a.
fu
is continuous,
First note that
and
x E u ~, we obtain
z+x
E ~fu I
and thus
(z,u) = ~ z+ut
8fu I = ~v E V I (v,u) = 6}.
and
~-- ~ful
(x,u) = 0
proving that
'
-i 6f u
it suffices to show that
so that 6fu I
For
(z+x,u)
is
z E 6fu I
= ~.
Hence
is open and hence that
fu
is continuous. Next let of
V
be a topology on
and in which all
T-open since
group.
•
fu
is
fu
V
compatible with
are continuous.
continuous.
But t h e n
For
the additive group structure
u E U, the set
7u(V ) ~ T
since
(V,~)
u z = Of u-I
is
is a topological
117 If
f E nat U, then
f = f u for some u E U and we have seen above that fu To prove the converse, we first let f be a nonzero element of V*
is continuous. for w h i c h v E V
fZ = Of -I
such that
subspace so that
T
is a closed subspace of
v ~ f~.
of
U
such that
T ± + fm # V.
T ~ ~ f~. implies
Since
V \ fZ
the dual pair
[f] = [f]m& c
(nat T) ~
that
f
f±
= nat T
(nat T) ~ ~
Consequently
implies that
cannot be continuous.
v ~ T ± + fz
is a hyperplane so we must have
by 1.15.
f ~ nat U
f # O, there exists
It follows easily that
(V*,V), we obtain
By contrapositive, we deduce that which implies
Since
there exists a finite dimensional
( v + T ±) N f~ = ~.
By 1.14 we know that
Considering
V.
is open,
This proves
[f]~
which
f 6 nat T c nat U.
0f -I
is not closed,
the last assertion of the
theorem.
We consider next a converse of the foregoing theorem. III.l.17 LEMMA.
l__nn~ topological vector space
ation of continuous PROOF.
It suffices
E h, both we let
6 E g
Since N ~
f +g
f
and
and
M
of
Consequently
Then
which are continuous and
Assuming that
vf+vg
and thus
such that
vf+Mg
= (xf+vg)+
N
Nf + v g = 6.
(vf+xg)-
f,g E V*
= 6, so that
there exists a n e i g h b o r h o o d Nf = 6 - vg
v
x(f+g)
f,g E V*
are continuous.
v E 6(f+g)-I
Hence
a neighborhood
f+g
to show that for fq
is continuous,
(6 - vg) f-l-
(A,V,T), every linear combin-
linear forms is continuous.
of
= $.
are continuous,
vf = ~ - v g v
such that
Similarly,
For any
E h.
there exists
x E N ~ M, we now have
6 = 26 - 6 = 6.
v E N n M ~ ~ ( f + g)-l, proving that
6(f+g)-i
is open and thus
is continuous. Next let
tinuous.
If
open since
q E h.
If
q # O, then for any f
is continuous,
III.i.18 THEOREM. ha___~san open subbase satisfying
~ = O, then
~
N H = 0
Let
fq
is the zero function and hence con-
6 E 4, we have
proving that (h,V,T)
fo
~(fo) -I = (6o-l)f -I
which is
is continuous.
be a Lopological vector space.
fo_~rthe n e i g h b o r h o o d > y s t e m of
0
Assume that
consistin$ o f hyperplanes
and let
HEY
u ={f~ Then
U
is a t-subspace o f
v*l f
is continuous]
V*, T = Tu(V )
n U = { f E V* I fZ ~ A H. -i=l l PROOF. V = f m @ Iv]
First let
for any vector
6f -I= [ x E Since
H
H 6 ~.
Vlxf
and
for some
H i E ~].
By 1.14, we have
v ~ fm
Then for any
= ~] = { y + o v l y E
is open so is its translate
H = f~
H+qv
for some
f ~ V*
and thus
6 E g, we obtain
H, (~v)f = 6] = for any
o E h.
U (H+~v). But the union of open
118
sets is again open w h i c h implies that In view of 1.17, U
so that
-i
is a subspace of
implies the existence of f E U
6f
vf ~ 0
H t ~
ix open. V*.
such that
which proves
Let
v ~ H.
that
U
Hence
f
is continuous.
0 # v E V.
The hypothesis
By the above, H = f~-
is a t-subspace of
for some
V*.
-i For any H
H E ]~, we also have
is Tu(V)-open.
topology
Since
~
But the sets
"ru(V )
topology If
um
so t h a t
f E U, then
for some
f ~ U
which shows that
is a subbase of the neighborhood system of
T, it follows that
• -open.
H = fz = Of
~ ~ Tu(V ) .
Conversely,
if
u E U, then
0
u ± = 0u -I
form a subbasc of the n e i g h b o r h o o d system of
"ru(V ) ~ "r.
f~ = 0f -I
Consequently
for the
0
is
in the
T = "ru(V ) .
is an open neighborhood of
0
and hence must
contain a basic open set ~ H. with H i ~ ~. Conversely, let f E V* and assume i= 1 l n fi ~ U as above. that fz ~ N H i for some H i 6 ]~. Then H i = f+i for some i=l n Hence f~ ~ ~i f.~ = Ill,f2,... fn ]m But then -- i=l i ' . f E
[f]~" = f~C__ Ill,f2,...,fn ] ~ -
= [fl,f2,... ,fn] c_ U
in view of 1.15. 111.1.19 LEMMA.
Let
S
be a proper subspace of a vector space
and extend
A
to a basis
a basis of
S
a basis of
V/S.
Conversely,
if
A
B
o_~f V.
is a basis of
is s_ system of r.epresentatives of the cosets of is a basis of PROOF. then
v =
S
Then S, C
C = ib+S
V.
is basis of
makin~ u_£
Let
A
I b ¢ B\A} V/S
C, then
and
b_!e is D
B = A U D
V. Let
A, B
>-~ <~ibi biEB
v+S
showing that
C
and
C
be as in the first part of the lemma.
If
v ¢ V,
so that
=
generates
~ (Tibi+s = ~ o i ( b i + S ) , bi~B b .~B i V/S.
Assume that
~
oi(bi+S)
= S.
Then
b .EB\A i
~.b.
E S
and hence
biEB\A i i A. C
Since
B
~ o.b hi E B\A i i
is linearly independent,
must be a linear c o m b i n a t i o n of elements of this is possible only if all
is linearly independent and thus a basis of Now let
we obtain di E D that
v+S
and B
A, B, C =
D
be as in the second part of the lemma.
~ o i ( d i + s ) = ~ <sidi+S d.EDI dieD
s E S.
generates
and
Hence V.
s =
o i = O.
~ ria i ai~A
Suppose that
Hence
V/S.
so that
and thus
v =
v =
~ ~idi+s dieD
For
for some
~ Old i + ~ Tia i died aiEA
~ o.d + ~_~ Tia i = 0. dieD t i siVA
Then
v ~ V,
proving
119
oi(d i + S )
=
~ o.d i + S diED i
dieD where
d. + S l
E C
and
independence
of
C
independence
of
A.
111.1.20 of
S
III.l.21
by:
T = Tu(V )
ii)
T
iii)
has
The
ii) =
an open
S
all
oi = 0
that all
independent of a vector
conditions
i) =
ii),
iii).
This
by linear
~i = 0
and hence space
V,
on a topological
In the former = V.
case,
Then
by linear
a basis
of
V.
the c o d i m e n s i o n
vector
space
of
~.
thus
v + B
open.
Let
system of 1.19
Hence
by
Let
~
system
of
system of
0
the c o d i m e n s i o n
The statement
= H
or
0
consisting
from
so that
[I H)), 1.19
v E H.
on
0.
T
Hence
T~ c C
of
from 1.19
of V If
B
B E ~
B ~ ~ Then B
and
generated
by
exists
H
v
codim
~
implying B 6 ~
B = n
C = A U [v,x2,...,Xn] A U ~x2,...,XnJ
which
H
which
that = i.
T ~.
at least one
for
B E ~,
to a basis
Therefore
be a basis
is a h y p e r p l a n e
since of
V.
and
that
H
is
of the neighbor-
then
it follows
D =CU
the subspace
v ~ B
B c H
shows
~ X l , X 2 , . . . , x n} H.
Since
of a finite number
that
suppose
and
contain
is a hyperplane.
T c
Thus
dim V/G < ~
in
i < i < n,
such
and
as an open subbase
can be extended that for
codim G < ~
is a hyperplane.
B E ~, we have
contained
having
of a Let
dim V/(G ~[H) = dim V / G + i < ~.
Then for some
is the i n t e r s e c t i o n
is r~-open
there
of
n = i.
dim ( G / ( G n H)) = d i m V / H
be the set of all hyperplanes and
that
since
where
that
C D iXl,X2,...,Xi_l,Xi+l,...,Xn}
each
= V
for
codim (G ~ H) = codim G < ~.
follows
consisting
of the intersection
is trivial
suppose
G+H
~ G/(G ~h H)
be the topology
follows
that
(V/(G N H ) ) / ( G / ( G
H E ~
that each
f; H. HEY be a basis of V
V.
are hyperplanes,
G+H
so that
(G+H)/H
that any basis
It also
0 ~ v E
Hi
is an open n e i g h b o r h o o d ~
from
it follows
G ~ H
It then
Let
hood
generated
o_~
(~ B = 0. BEB 1.18 and 1.14.
from
to show
Then
V/G ~
iii) = ii).
for which
is finite.
where
V/H =
we have
dim G/(G A H) = i.
V.
U
for the ne.ishborhood
follows
It suffices
be a hyperplane.
Further,
~
codimension
... fl H n
H
t-subspace
subbase
G = H1 ~ H2 N
of
is linearly
'
for w h i c h
of h y p e r p l a n e s
H
B
following
finite number
~.
yields
For a subspace
for some
of finite
PROOF.
of
Hence
which
~ H = 0. H@~ has an o p e n base ~ fo_~r th_~e n e i s h h o r h o o d
~
subspaces
member
i ~ j.
~ Tia i = 0 siVA
= $
codim S = dim V/S.
THEOREM.
of hyperplanes
G+H
if
~ Tia + $ ai~A i
are equivalent.
i)
let
~ d. + S j
Therefore
DEFINITION.
is defined
(~,V,T)
d. + S I
so that
= -
of
V
B = ir~iHi ,
of members
T = T ~.
Assume
of that
N B = 0, Let A B~ Then the subspace
containing
B
so that
120 H E ~.
But
v ~ H, contradicting
ii) = i).
the assumption.
Therefore
N H = O. HE~
This is a part of 1.18.
III.i.22 EXERCISES. i)
Show that
Tu(V )
is the discrete
topology
if and only if
V
is finite
dimensional. ii)
Show
tivized
to
iii)
that for every finite dimensional
S
that for any
fl,f2,...,fn,f
vf. = 0 for i = 1,2,...,n, then i combination of fl,f2,...,fn.
is to be found in Dieudonn~
mation on this subject, (I6], Chapter
Section
see Behrens IX, Section
~ V*
satisfying:
topology.
We will
given relative
introduced
tacitly
to which
topolosy,
PROOF.
f
is a linear
and statements
proved
([i], Chapter
2), Bourbaki
6),(I7], Chapter
PROPERTIES
assume
II, Section II, Section
in this
For more infor([i], § I),
3 and Chapter
IV,
Conversely,
that
S
vigj = ~ij
is dense
vf = ~ ~ 0.
let
S
Since
vg = 6 ~ 0
in
of
S ~, S ~ ±
V*, where
V*
S
V*
and let
is dense we have
of
that V*.
S
is
the
Further
be a basic neighborhood
There exists
S ~ B(v;~ ) ~ ~.
of
let
f E V*
0, where
of
Letting V*.
and
Vl,V2,...,v n
By 1.2.3 there exist gl,g2,...,g n E S such that n i ~ i,j ! n. Set g = j~__igj(vjf+6j); then for any I ! i < n, we
vig = j~_Ll(vigj)(vjf+ 6j) = v i f + 6 i g - f E B(vl,v2,...,Vn;61,62,...,6n ).
g E S
is endowed with
is a t-subspace
have
that
(U,V)
etc. are defined.
0 ~ v C V.
independent. for
to the
that a dual pair
as well as
showing
be a t-subspace
B(vl,v2,...,Vn;61,62,...,6n ) are linearly
S
section
in relation
is dense if and only if it is a t-subspace.
Assume
such that
A subspace
V
OF SUBSPACES
of subspaces
in the whole
the topology on
g E S riB(v;6), we obtain
proving
rela-
v ~ V, if
[5].
TOPOLOGICAL
III.2.1 PROPOSITION.
so that
for any
[i],[2],13 ] and Jacobson
We consider here some simple properties
f 6 V*
V, Tu(V )
18). 111.2
finite
of
vf = O, we must have that
The origin of most of the concepts
Jacobson
S
is discrete.
Show
section
subspace
Consequently
(f+B(vl,v2,...,Vn;61,62,...,6n))
is dense in
V*.
N S
121 The following lemma is due to Dieudonn~ 111.2.2 LEMMA. PROOF.
For
S
! subspace o f
We denote the closure of
with
V, S ~
S
= {v E V I ( v + T z) n S ~ ¢
[i].
by
is the closure of
S.
S.
By 1.19, we have
for every subspace
T
of
U
(i)
dim T < ~},
also note that S aa
= {v ~ V I if
Now let s E S. where
v E S
Then
(s,u) = 0
from (2) that
s E S, then
and suppose that for some
( v + u m) N S # ~
(x,u) = 0
for all
and thus
by (1), say
u E U, we have
s E
(v,u) = (s,u) = 0
(v + u a) N S.
C
v ~ S + T ±.
of
V.
U
(V*,V). f
Then
f E V*
xf = 0
111.2.3 COROLLARY. PROOF.
for all
s = v+x u.
It follows
with
nat U
T~ ~
T
f
on
C
U
f E fame
~
T aa
x E S
v E V\S.
for which
S +T a
f
Hence by
( v + T a) ~ S
and extend
by letting
Also denote by
fa, where
for all
of
be a basis of
and let
xf = 0
=
B U {v} for all
its linear extension
to
is taken relative to the dual pair
= T
since
dim T < m.
which by (2) shows that
Every finite dimensional subspace o f
V
But then v ~ S~± .
is closed.
Exercise.
111.2.4 LEMMA. PROOF.
For an___~ysubspace
T
of
U, we have
T ~a~
= T ±,
Exercise.
III.2.5 COROLLARY. some subspace PROOF. Conversely,
B
arbitrary.
and
Using 1.15, we obtain
~ U, vf # 0, and
T If
of S
if
S
is closed,
S = T~
Let
Xl,X2,...,x n E V ~ ff.x. E T a. i=l I I
S
of
V
is closed if and only i f
S = T~
for
then
(S£) ~ = S
for some subspace =
T ~az
=
T~
=
T
by 2.2 with of
S~
s subspsce of
U.
U, then by 2 . 4 , w e have
S
is closed by 2.2.
III.2.6 LEMMA. PROOF.
A subspace
U.
S ±~
and thus
Let
Define a function
x E B, vf # 0, and otherwise V.
(s,u) = 0 Hence
by the assumption on
(I) there exists a finite dimensional subspace which implies that
all of
(2)
v E S~ m .
To prove the converse, we identify
to a basis
(v,u) = 0}.
If
T
is an n - d i m e n s i o n a l
{tl,t2,...,tn] such that
be a basis of
(xi,t j) = 6ij.
If
subspace o f
In particular, we have n
U, then
codim T ~ = n.
T. By 1.2.3, there exist n i~=l~i(xi+T a) = T a, then
n
0 = ( ~ ~kXk,ti) = ~ ffb(x' ~ti) = ffi k=l k=l ~
122
ix[ + Tj.} n =l
so that fixed
v ~ V,
let
is s l i n e a r l y
~i =
(v~ ti)
n (v - ~ ~ x., i= I ~ i Lj) = and
thus
v-
£ ~.x. E T z. i=l I 1
the v e c t o r s
x i +T ±
independent
for
(v,
i <
of v e c t o r s
V/'f ~
v+T
and
in
V / T m.
For
a
Then
n - 71 (v t.)(x. i= l ' 1 l'tj ) =
tj)
Consequently
generate
set
i < n.
(v,
tj)-
(v,tj)
0
=
~ =
~ @i(xi+T~) w h i c h p r o v e s that i=l form a basis. Hence codim T ±
thus
= dim V / T ~ = n. III.2.7
PROPOSITION.
The
followin$
conditions
o__nn_a s u b s p s c e
S
of
V
ar___~e
equivalent. i)
S
i__~sopen.
ii)
S
is c l o s e d
iii)
S = Tm
PROOF. Since
S
for
i) =
codim
some
finite
ii).
is open,
d i m T < oo
and
in any
it c o n t a i n s
Hence
T z c_ S
S < ~. dimensions[
topological a basic
which
subspace group,
open
implies
T
every
o__[f U. open
neighborhood
T = T m z - -~
Sm
of by
subgroup 0,
say
1.15,
is closed. T ~-, w h e r e
so
d i m S ~ < co.
By 2.2 and 2.6, we h a v e codim
ii) =
iii).
It
The b e g i n n i n g
of
V / T "L
which
is a l s o iii) =
i).
III.2.8
the
S = d i m V/S = dim V/S mm = d i m S z < ~.
follows proof by
from
2.5
of 2.6
that
shows
contrapositive
In s u c h
a case,
S
S = Tm
that
if
implies
The
basic
open
COROLLARY.
The
followin$
some
subspsce
is i n f i n i t e
that
is a b a s i c
COROLLARY.
for
T
T
open
is f i n i t e
neighborhood
neishborhoods
o__[f 0
T
of
dimensional, dimensional. of
are
U. then
0.
the o n l y
open
sub-
spaces. 111.2.9
conditions
on a s u b s p a c e
S
of
V
are
equivalent. i)
S
is a c l o s e d
ii)
S
is an o p e n
iii)
Also subspace
S = um
note
PROOF.
then
that
containing
III.2.10
a closed
for
hyperplane. hyperplsne.
some
a hyperplane S
and
COROLLARY.
f = fu E
nat U
so
S
is e i t h e r
thus m u s t
Every
Necessity.
hyperplane
0 # u E U.
be e i t h e r
subspace
In p a r t i c u l a r ,
of
for any
that by 2.9 we h a v e and
thus
V
nat U = V*.
that
closed S
or d e n s e
or
is c l o s e d
if and o n l y
0 ~ f E V*, F = u~
since
S z±
is a
V. if
nat
U = V*.
F = [v C V I vf = Oj
for some
0 # u ~ U.
But
is
123 Sufficiency. basis if
B
of
x E B\{v].
and obtain
Let
V.
f
Then
S c uz
hyperplanes
S
Let
u~
be a subspace of
f ~ V* = nat U
and
v ~ u m,
S
Let
S
uz
with
T
of
v
to a
and
u E U, we have
xf = 0 f = f
u
v~ >+s v { u~+S
for some
u~+S
V
and let
v E V \ S.
( v + T z) N S = @
be a basis of
=
is the intersection of (closed
0 # u E U.
U, we have
n
and thus
vf = i
is the intersection of some closed
be a closed subspace of
[Ul,U2,...,Un}
it follows that
Extend
for which
Every closed subspace o_~f V
finite dimensional subspace Let
v ~ V\S.
V
and thus must be closed.
III.2.11 LEMMA.
v ~ Tm+S.
and
and hence for some
Hence
and open) hyperplanes of the form PROOF.
V
be the linear form on
T.
Then for some
so that
Then
n
~ u++s =
=~ (u~+s)
i=l l
i 1
i,
u~-+S ~ V
Hence
and since
: u~,. i.e., S ~_ u~'i Consequently
u.a
is a hyperplane,
v ~ u+1 while
S c_ u~, as
required. The following p r o p o s i t i o n is due to Dieudonn6 111.2.12 PROPOSITION. topology of
V/S
If
S
[i].
is s closed subspace o f
coincides with
• .(V/S), where
V, then the quotient
(S~,V/S)
is a dual pair with
S the bilinear form PROOF. u ~ S~
If
(v+S,u) v+S
so that
(v+S,u)
If
v+S
(v,u) ~ 0
and thus
(v,u) # 0
(v,u),
} S, then (v + S , u )
v-v'
t S
and thus
(v-v',u)
= 0
for all
whicll shows that the function which to each pair
the value
so that
(v ~ V, u t Sin).
then
(v,u) = (v',u)
associates
bilinear.
that
= (v,u)
= v' + S ,
v ~ S ~ O.
is single valued.
For every
(v + S , u )
This function is clearly
so by 2.11 there exists
} 0.
u ~ S~
u ~ S ~, there exists
Consequently
such that v ( V
such
the new bilinear form is non-
degenerate. Let
m:V ~ V/S
topology of
V/S.
be the canonical h o m o m o r p h i s m and let A subbase for
~
~
denote the quotient
is given by
g = { A ~ V/S IA~ -l= u< u~ u], and a subbase for the topology
a
~
6 = Ts~(V/S ) ,
is given by
{u ° luE s~j,
where
u° = { v + s Then for
l(v+s,u) = 0] = { v + s l ( v , u )
= 0j = { v + s l v ~
u E S "~, we have
u~°-l={v~
vlv+s~
u °] = i v ~
vlv<
u~] = u<
u-).
124
so
u
o
E $
which
for some hence
u ~ U.
proves
that
Since
0 E u ~, it follows
u E u ± ± ~ S m.
g c g.
S C [.
The next III.2.13 space o f
V.
PROOF. that v {
if
Using
v ~ A,
= i.
is due
A+B
Let
A +
implies
that
For every
u E Am
A
(V,Uo)
= (v,u)-
( A + [v]) ~.
i)
(U,V)
ii)
Let
Now
of
(U,V)
B
be a finite
V
dimensional
suffices
v @ A.
sub-
(v,~)
( A + Iv]) ~ ±
to show
Then
u
= (v,u)-
x 6
suppose
that
= 0 Then
= (x- (X,Uo)V,U) But
then
[v]
the opposite
pair
that
E A ± and we may o u - Uo(V,U ) E A m and also
(X,Uo)(V,U)
inclusion
Prove
if and only
be a dual
B, it clearly
for some
let
~ A+
be a dual pair.
V'
of
x - (X,Uo)V E A m ~ = A.
A + [v];
EXERCISES.
Let
and
We suppose
(V,Uo)(V,U)
= (x,u)-
so that
( A + Iv]) m ± c
subspace
# 0
u E A m , we have
III.2.14
proper
be a closed
is closed.
x E A+(X,Uo)V and thus
[2].
on the d i m e n s i o n
[v]
0 = (x,U-Uo(V,U)) for all
u~] : u ° E
to M a c k e y
induction
then
u - u (v,u) E o
0 ~ A, so
is closed.
(v,U-Uo(V,U)) so that
-i Then At1 : u j" -i S c A~ : u ~" and
A E ~.
~ = 6.
PROPOSITION. Then
(Am) ~ = A
(V,Uo)
Therefore
result
that
let
Consequently
A = {v~Iv~ so that
Conversely,
that
if
always
(U,V')
holds.
is a dual
pair
for no
nat U = V*.
and write
$ = Su(V).
Prove
the following
statements. a)
The m a p p i n g S - S~
is an i n c l u s i o n
(S
inverting
function m a p p i n g
the lattice of all subspaces S~±
is a subspace
b)
S ~
c)
~%,$(~T(V))
d)
~r,$(~u(S))
e)
~T(V)
For
the dual pair
of
for every
of
the
V) lattice of all subspaces
of
into
U. subspace
= ~ ( T ~)
S
for every
of
V.
subspace
T
of
U.
= ~ ±(V) for every subspace S of V. S = is E $ IT A ~ Na} for every finite d i m e n s i o n a l
subspace
T
of
U. iii) V*
if and only
if
V
(V*,V),
is finite
prove
that
dimensional.
Tm± = T
for every
subspace
T
of
125 iv)
Let
(U,V)
complete lattice
be a dual pair.
(a lattice
l.u.b, and a g.l.b, v) (U/S~,S)
Let
in
(U,V)
Show that closed subspaces of
~u(V)
relativized
S
be a subspace of
to
L
has a
V.
Show that
~U/sZ(S)
is weaker
S.
for this section include Bourbaki
([6], Chapter IX, Sections
Sections 5,6), K~the
form a
L).
be a dual pair and
The references
V
is complete if every nonempty subset of
can be made into a dual pair in such a way that
than the topology
Jacobson
L
([i], § I), D~eudonne " ~ [i] , [2],
7,8),(/7], Chapter II, Section 4 and Chapter IV,
([i], ~ i0), Mackey
•2], O r n s t e i n
/i], R i b e n b o i m
([i],
Chapter Ill, Section 2).
III.3
T O P O L O G I C A L PROPERTIES OF S E M I L I N E A R T R A N S F O R M A T I O N S
For dual pairs
(U,V)
and
(UI,V ~)
and a semilinear t r a n s f o r m a t i o n
we examine here the conditions under which also one-to-one or onto. Dieudonn~
III.3.1 THEOREM.
The next theorem is due to J a c o b s o n
PROOF.
Let
(U,&,V)
(V,Tu(V))
into
Suppose that
i
and
transformation o__ff (&,V)
function from
into
a:V ~ V j,
is continuous and if so, w h e n it is [5], see also
[i].
a semilinear
function
a
by:
a
and
Then
s
if and only if
is continuous,
o,~ E &
be dual pairs and
(~,VI).
(V~,~u~(V~))
v ~ ~ = ( v a , u ~ ) ~ -I
&, and for any
( U I , ~ , V I)
into
and for every
(v E V).
Then
~
(~,a)
be
is a continuous a
has an adjoint.
u~ E U~
define a
is a function from
V
x,y E V, we obtain
(ox+Ty)~ ~ = ((~x+Ty)a,u')~w
-l
= ((o~)(xa) + (T~)(ya),ul)IW -I = (OW~-I)[(Xa,Ul)I~ -I] + (TW~-I)[(ya,ul)I~ -I]
= o(x~') +T(y~') proving that a w a
~
E V*.
Further, u~
is a composition of the following functions:
w h i c h is continuous by hypothesis, -i
which is continuous since both
continuous
function
b
which is T u ~ ( V ~ ) - c o n t i n u o u s by 1.16, and
and
g~
are discrete.
linear form which by 1.16 implies that
Consequently
u ~ E nat U.
Now define a
by bu ~ = u
For any
fu ~ g
if
u ~ = fu
--
v ~ V, u I E U ~, letting
(v,bu')w showing that
b
= (v,u)~
bu ~ = u, we have
= (Vfu)~
is the adjoint of
(u E U).
a.
= (v2')~
u ~ = fu
--
= (va,u')'~-l~
so that
= (va,u')
u~
is
126
Conversely, a
at
0,
b
be
the a d j o i n t
since
by
1.8
the
family
neighborhood
system
of
0
in
V',
open
For
in
and
let
V.
so that
(bu') m
v.
Let
x E
v E
u '~
v+x
E
a -i
is a n e i g h b o r h o o d
(bu')a;
then
implies
u'¢a -1
III.3.2
that
showing
[ u ~ m I u~ %
show
of
0
A function if
Cf
Dieudonn6
III.3.3
= 0
and
Let
spaces
Va = V ~ i)
only
note
Consequently,
va = 0 ~
using
a
of
2.2,
A
into A.
On
= 0.
completes
bu' ~ U
the o t h e r But
of hand,
then
the proof.
a topological
The
next
result
space
B
is due
to
is
that
if
bU'
is d e n s e
is o n e - t o - o n e any
u ~
pairs
adjoint
in
and
(~,a)
b.
b__ee
Relative
statements
to
hold.
U
and o_o_o~.
U, we h a v e
for all u' ~
u' E U'
U' ~
,v~}
for w h i c h ~
we
= v~,
u = bu'
for
some for
for uI ~
i =
in
1.5.4
first
be a b a s i s
via
(bU')-
is d e n s e Then
is open,
[v l'~v~,.
Let vi 6 V
with
( U ' , T v , ( U ' ) ) , th_~e f o l l o w i n $
0a -I = 0 ~
Va = V'.
b
be dual
(g~,V I)
v %
(1)
(bU')-.
we o b t a i n
bU j
that
b
(U',A',V')
into
for all
~
u = bu'
and
Uj
asserts
let
SI
of
SI
let
some
= 0
U.
and
S = u' {
such
that
be
b
is "one-to-one.
a finite for
dimensional
each
In subspace
1 ~ i < n, c h o o s e
[vl,v2,...,Vn].
a
Then
S 'm
that
(v~,ul) ' = 0
1,2,...,n,
since (v~,u l ) = 0 ~
we o b t a i n Sm
the
is
is a n e i g h b o r h o o d
= 0.
(v + x a , u ' ) '
for
u~ma -I
we h a v e
in
(va,u') t = 0
is o n e - t o - o n e
u ~ b(S '~)
and
C
and
for
= 0
(bU~) - ~ = U ~ Suppose
to s h o w
V~
vector
space
and
if
that
(v,bu')
ii)
a subbase
that
of
is ~ontinuous].
(£,V)
if and only
if and First
v = 0a -l ~
of
(U,TV(U))
is o n e - t o - o n e
PROOF.
order
continuity
u~ma -I ; h e r e
(bu') ±
which
(U,V),
set
(U~h,V)
transformation
a
ii)
open
v+
(xa,u')'
thus
u'~a -l
pair
a topological
for e v e r y
THEOREM.
topological i)
from
v + (bu') ~ ~_
(va,u')' a dual
constitutes to s h o w i n g
thus
and h e n c e
For
to p r o v e
[i].
a semilinear the
f
is o p e n
that
= 0
v + (bu') m c
U']
reduces
and
£u(V) : [a E £(v) la
open
It s u f f i c e s
(x,bu')
that
COROLLARY.
a.
the p r o o f
u,~ a -l , we w i l l
v ~
of
b(S ~ )
is an o p e n
= S m.
subspace
(via,u')1 Here of
U.
S
: 0
~
(vi,bul)
is a finite Hence
b(S 'm)
= 0 ~
dimensional is open,
(vi,u)
: 0
subspace and
thus
of
V
for any
so u
i
that ~
U I,
127
b(u~ + S ~ )
= bu ~ + b ( s 'm)
neighborhood system of Suppose roles of
that
a
b
and
is also open. u I in
v~ 6
in (i), we duduce
that
(b-10) ~.
b(v ~z)
Let
v £
v~
S
of
V
with
This
entails
But
the
together
for all
then
s'
E v 'a
v ~ E (Sa) ± ±
U.
(b-lo) ~ ~ Va
which
b
Va
Va
s'
Hence
E v ~a
together with
Va
,s
)'
since
111.3.4 COROLLARY. onto onto
PROOF.
With
(V~,Tu~(VI))
implies
there exists a finite di-
v ~ S.
Then
(v,bu ~) = 0
w h i c h implies
bu ~ = bs'.
= 0. Sa
(2) shows that
V ~.
and the hypothesis
(bul) m
v ~ E (b-10) m
Va
implies Now (2)
Hence
b(u ~ -s~)=0
(v~,u 'in
s~) ~ = 0.
2.2 shows
is finite dimensional. Va = (Va) ma.
By 2.2,
that Therefore
(Va) ± ±
is
is closed.
then interchanging
is dense in
the
b(v') a.
= (v
that
U~
for all
S ~
for which
that
(v',u')'
which proves
Interchanging
so that
S m ~ b(vlm).
the roles of
Finally,
if
must be both dense and closed and thus
(V,Tu(V)) (U,&,V)
of
implies
is closed.
Consequently
Sa~
hypothesis
is one-to-one,
we duduce that then
the
form a base of the
(2)
(va,u~) ~ = 0
v 6 S.
and hence by 1.15, v' E Sa
the closure of If
existence
with
Va
(Va) m = b-10
the property
and assume that
(bu~) ±
that
u ' + S ~±
is open.
is an open subspace of
bu' E ( b u ' ) m a ~
which
h
(b-10) z.
is an open subspace of
uI E U~
and thus
(Va) ~ z =
Then
mensional subspace
the sets
is open; we will show that
b
Va c Let
Since
U~, it follows that
b
and
b
in part i),
is both one-to-one and open,
Va = V ~.
the n o t a t i o n of 3.1, a if and only if
a
a
is a h o m e o m o r p h i s m o f
is a semilinear isomorphism o f
(U~,A',V~). Exercise.
111.3.5 EXERCISES. i) space and
Let
a
V ~.
be a semilinear transformation of a vector space Construct
(V~*,V~). ii)
Let
formation of
Then (U,V) V
U ~-) topology. For more
into
b
the adjoint
b
of
relative
is called the conjugate of
and
(U~,V ~)
V ~, and let
Show that
a
be dual pairs, V
is continuous
Chapter IV, Section 7), R i b e n b o i m
let
(respectively
information on this subject,
transformations,
a
a
V
into a vector
to the dual pairs and is denoted by
a
(V*,V) s*.
be a semilinear trans-
V ~) have the U - ( r e s p e c t i v e l y
if and only if see Dieudonn~
a*(nat U ~)
C
nat U.
[i],[2], J a c o b s o n
[5],([7],
([i], Chapter III, Section 2); on continuous
see Rosenberg and Zelinsky
[i].
linear
128 III.4 Before aboarding notions
related
Let
X
OF A VECTOR SPACE
the main subject of this secLion,
to completeness
be a nonempty
Cartesian product
COMPLETION
X × X.
of uniform
spaces,
set; a (binary)
we briefly
see Kelley
relation on
The set of all binary
X
summarize
([i], Chapter
the 6).
is a subset of the
relations
on
X
forms a semigroup
under the operation hop Further, and
= {(x,z) E X X X
X-I
I (x,y) ~ h
and
(y,z) E 9
for some
y E X].
is defined by X-I = [(x,y) 6 X × X I (y,x) E X} so that ( -i)-i = ~-i o The identity relation or the dia$onal is defined by
(~ o p)-i = p
~(X) = i(x,x) Ix ~ X}, and is usually denoted by
~.
For
~ ~ X XX
and
A ~ X,
we let k[A] = [y E X I (x,y) E k and for
A = ix}, we simply write
A uniformity satisfying
~
for a set
the following
%[x]. X
if
k E ~,
then
~ ~ &;
(~)
if
% E ~,
then
p o p ~ X
if
~ E ~,
(6)
if
~,0 E ~,
(~)
if
X E ~
The pair
completely
%
then
and
family
~
if
~ E ~, then
(6 ~)
the intersection
~,
space.
that
family of binary relations
on
X
p 6 ~;
then
p 6 ~.
A subfamily
a member of
~.
Note
One easily verifies
satisfies ~-i
(~),
contains
of
the following
on a set
~
is a base
for
X
assertion.
is a base for some
(~) and
a member of
of any two members
(X,~),
~
that in view of (¢), a base of
of
the topology
g
~
~, contains
a member of
of the uniformity
~.
~, briefly
the
is the family
{T ~ X I for every One shows easily
for some
of binary relations
if and only if
topolosY of
(k o p)[A] = k[p[A]].
X N p ~ ~;
~. ~
space
that
E ~;
contains
(V')
uniform
-i
is a uniform space.
~
For a uniform
It follows
is a nonempty
X ~ p ~ X xX,
determines
A nonempty uniformity
then
(X,~)
if each member of
x E AJ,
requirements:
(~)
(?)
for some
g
x E T
there exists
X E ~
is indeed a topology,
such that
and thus
(X,g)
%Ix] c T] . is a topological
129 A function mapping a uniform space uniformly continuous
if for each
(X,~)
into a uniform space
(Y,~)
is
p E ~, we have
{ (x,y) E X × X I (f(x),f(y)) E p] E ~. Further,
in the case that
formly continuous, (Y,~)
then
f f
is one-to-one,
onto and both
f
and
f-i
is a uniform isomorphism and the spaces
are uni-
(X,~)
and
are uniformly equivalent. As in the case of topological spaces,
(X,~)
is a uniform space,
of
~
to
of
(x,~)
if
Y
is reflexive, p ~ m
and
where
S
to be denoted by
transitive,
p ~
n.
A net
eventually in
A
D
[S n I n ~ D}
directed by
m E D
(X,~)
[Sn I n ~ D}
converges
p ~ D
D
(S,~),
Sn ~ A
for all
Sn E A
to a point
n ~ D; it is
whenever x
n > m.
relative to
~
(X,~)
is complete if every Cauchy net in the space converges to a point in the
(Sm,S n) 6 k
A uniform subspace of a uniform space
(X*,~*)
(i.e.,
(X,~)
f
m,n ~
k.
Further,
is dense if it is dense in is a pair
(f,(X*,~*)), where
is a uniform isomorphism onto a dense
X*; a completion is H a u s d o r f f if
the uniform topology is Hausdorff).
whenever
(X,~)
A completion of
is a complete uniform space and
subspsce of
if it
is a Cauchy net if for each
(X,~)
~.
A net
x.
there exists
the uniform topology of
if
such that
A net is a pair
~. t ~
space.
such that
D, directs
and is also denoted by
if
such that
in a uniform space
k ~ D
~,
A
and
is a uniform subspace
m,n E D, there exists
is eventually in every 3 - n e i g h b o r h o o d of i net
(Y,~)
is a d i r e c t e d set.
X
is the r e l a t i v i z a t i o n
on a nonempty set
is ij!n a set
if there exists
in a topological space
D
Y, and
>,
and if for any
In such a case
is a function on a set
[S n I n ~ D].
is a nonempty subset of
~ = [~ n (Y x Y) Ik E %0
Y, or the relative uniformity for
A binary relation,
(S,>)
the set
(X*,~*)
is a Hausdorff uniform space
A H a u s d o r f f completion of a H a u s d o r f f
uniform space is unique up to a uniform isomorphism and can be referred
to as the
completion. If identity
(G,g) 0
of
is an abelian topological group, G, let [N
N = [(x,y) t G x G I x - y
IN
is a neighborhood of
is a base for a uniformity with
~
(verify~).
can be omitted)
~
It follows
on
G
t N].
of the
0j
and the uniform topology for
in addition,
N
Then the family
that the topology of any topological
is a uniform topology.
refers to this uniformity;
for every neighborhood
~
coincides group
(abelian
The concept of a complete topological group specializing to a complete module, vector
space or a ring, we have in mind the uniformity of their additive structure. 111.4.1LEMMA. topology.
If
V
is a vector space,
then
V*
is complete in the finite
130
PROOF. gV
Since
is complete
([i]~ p. 192),
~
is discrete,
being
the Cartesian
in order
is a closed subspace convergent a net in
net in V*
relativized
topology,
by Kelley
to
f
and
with
Yfn = yf
V*.
Since
Hence
to the convergence
m E D
whenever
of
n such that
Xfn = xf
n ~ m 3.
V*
= vf
to the point it follows
whenever
(ox+Ty)f n
whenever
Finally
be
is the
of
is discrete,
vf
V*
~fn In ~ DJ
~vf n I n ~ DJ ~
that
mI
such that
let
the convergence
Since
such
to show that
to show that every
the topology of
([i], p. 217),
convergence).
m2
it suffices
=
n > m. (ox+Ty)f
n >_ m2;
[here
there exists
= ( o x + T y ) f n = ~(Xfn) + T ( y f n )
But then
f
= o(xf) + T ( y f )
is linear so that
f { V*.
Hence
V*
is closed and
complete.
thus
III.4.2 THEOREM. isomorphism where
~
of
V
Let
into
is the finite
PROOF.
nat V. V
By definition,
be a dual Eair and
Then
is a dense subspace
The function ~u(V)
f
subspace
x -y E T~ ~ (fx-fy)t [ (x,y) I x - y
T
U*, and
on
V
of
U
and
finite
is complete
f
of
V
in the onto
a homeomorphism
topology of
U*.
For
x,y E V, we obtain for all
t E T
is a uniform neighborhood
[(fx'fy ) I (fx - fy)t = 0
U*
isomorphism
which makes
the relativized
f t = (x -y,t) = 0 x-y = 0 for all t ~ T.
E T ~]
of
is an algebraic
is the topology
nat V, the latter having
any finite dimensional
f:v ~ f be the natural v is the completion of (V,Tu(V)) ,
(f,(U*,~))
topolosy.
By 2.1, nat V
onto
The set
(U,V)
U*.
finite topology by 4.1.
of
in
In view of Kelley
it suffices
([l] , p. 66),
f ~ 6 V,
spaces.
([i], p. 194),
the property
(ox+Ty)f n > m.
of complete
x,y C V, there exists
there exists
such that
m > ml,m2,m 3
whenever
(pointwise
v E V, there exists
n >_ ml; m3
is equivalent
v E V
Hence by Kelley
is complete,
to some point
product
o,T E &
exists
converges
to some
that for every For any
By Kelley
converges
for every
whenever
&V.
V*
which
{ fn I n h D} vf
product
to show that
of
V*
it is complete.
for all
in
V
while
t E T]
is a uniform neighborhood in U*. Since f is one-to-one, we conclude that both -i f and f carry uniform neighborhoods onto uniform neighborhoods, proving that both are uniformly
continuous.
dense subspace
of
U*, where
completion
of
(V,Tu(V)).
completion
is unique
For
nat V = U*.
f
is a uniform isomorphism
the latter is complete.
Since both
up to a uniform
III.4.3 COROLLARY. if and only if
But then
(U,V)
(V,~u(V)) isomorphism a dual ~ ,
and
Thus
by Kelley V
(f,(U*,~))
(U*,~)
of
V
onto a
is a
are Hausdorff,
the
([I], p. 197).
is complete
in the U-topology
131
111.5
L I N E A R L Y C O M P A C T VECTOR SPACES
We consider here linearly and weakly topologized and linearly compact vector spaces and establish relationships III.5.1 DEFINITION. M
If
M
is a topological module if
of these notions with those already encountered.
is s module
~
(left or right) over a ring
is a topological ring, M
under addition and the action of
~
upon
M
is jointly continuous.
is a linearly topologized module if the topology of neighborhood system of
0
M
consisting of submodules.
~,
then
is a topological group Further, M
admits an open base for the A topological vector space is
weakly topologized if it has an open base for the neighborhood system of
0
con-
sisting of subspaces of finite codimension. We are interested here in Hausdorff
topological vector spaces over discrete
division rings. As a supplement
to and partly a consequence of 1.21, we have the following
result. III.5.2 PROPOSITION. (V,T)
The following conditions on a topological vector space
are equivalent. i)
ii) iii)
V
is weakly
T = Tu(V ) V
topologized.
f o r some t-subspace of
V*.
is linearly topolo$ized and every open subspace has finite codimension.
PROOF.
Items i) and ii) are equivalent by 1.21.
i) = iii).
Obviously a weak topology is linear;
the second part of iii) follows
immediately from 2.7. iii) = i).
Each subspace in an open base of the n e i g h b o r h o o d system of
hypothesis must have finite codimension, The next result further elucidates weakly topologized spaces
If
linear forms on
PROOF. V*.
By 1.17, U
Let
~
subspaces of
V
Since
0 ~ x ~ V, there exists containing function extend
x f
f
is linearly
•
D
linearly
by
V*.
is a subspace of
T
V*.
We show next that
is Hausdorff, we infer that
B t ~
such that
by letting: vf = i to all of
V.
then the set
Furthermore,
U
of
Tu(V ) ~ T
is a weak topology.
and whose intersection with on
topologized,
is a t-subspace o f
be an open base of the neighborhood system of
V.
0
is indeed weakly topologized.
the relationship between linearly and
(&,V,7)
and the equality holds if and only i f
of
V
(cf. 1.18).
III.5.3 PROPOSITION. all continuous
so
if
For any
x ~ B. B
Let
U
0
[i B = 0. D
is a t-subspace
consisting of Hence for any
be a basis of
forms a basis of
v ~ D\B, vf = 0 o ~ &, we obtain
if
B.
V
Define a
v E D [I B, and
132
of -I = iv E V I vf = o] : [y + z I Y E B, z ~
[D\B],
zf = o]
~ [D\B] I zf = O} =
= B+[z
(B + z) U z E [D\B] zf=o
which is a union of open sets and is thus open. and thus so
fu
that
U
is a t-subspace of
V*.
is continuous and 1.16 yields T
is a w e a k topology.
and 1.18, we have linear forms on
V.
T = 7uI(V) = Tu(V )
Since
Tu(V ) ~
Conversely,
T = Tu~(V )
where
U
let
U~
V/S
M,
then
PROOF.
for every
M/N
Let
# N.
If
[K~c~A
N
if
~:M - M / N
V
is weakly
0
it follows that
K s~ + N.
+N
topologized and
S
is
The next result
for linearly topologized modules.
N
in
M
III.5.5 DEFINITION. m+N
But then
0
in -I
and thus K~ ~ F~ -i P~ and hence K ~
If
N
M/N.
is closed,
m ~ K~+N.
0 Then
is a union of open sets
an open base for the neighborhood system of
$
and thus
be the canonical homomorphism.
~ an open submodule of
and since
so that
n e i g h b o r h o o d of
A family
1.21
is a closed submodule of a linearly topolo$ized
fi K ~ = 0 in M/N. ckA ~ Let P be an open neighborhood of
then the set
U = U~
is also weakly topologized.
K ~ = K
K m
m { N
N N = ~
we have
be an open base for the neighborhood system of
c~ E A, the set
Then
By 5.2,
i__sslinearly topologized in the quotient topolosY.
thus open w h i c h makes
M,
be a weak topology.
that the corresponding statement also holds
consisting of submodules and let
(m+K~)
T
xf = i u E U,
T = Tu(V ), then 5.2 implies
also has this property,
V, then
111.5.4 PROPOSITION.
m+N
If
with
for any
is the vector space of all continuous
Note that 2.12 can be rephrased thus:
module
T.
f E U
fu = u
aS required.
a closed subspsce of asserts
Consequently
With this notation,
m~ ~ K ~
M/N.
in
K~
Then
P~
-i
~ ~ A.
which proves M/N m
such that
which proves
for some
is a submodule and
is a linear variety,
Let
there exists
P 0
U (K + n ) and is hEN m E M be such that
that
is an open Since that
N ~ P~
[K TI]o~A__
-i
,
is
consisting of submodules. is an element of a module
or simply s subvariety,
of
M.
of sets has the finite intersection property if the intersection
of any finite number of members concept, due to Lefschetz
in
$
is nonempty.
We now come to a fundamental
[i]; it represents a w e a k e n i n g of the n o t i o n of com-
pactness. 111.5.6 DEFINITION.
A linearly topologized module
M
is linearly compact if
every family of closed subvsrieties w h i c h has the finite intersection property has nonempty intersection.
133 111.5.7 PROPOSITION. M,
then
M/N
PROOF.
By 5.4, M / N
closed subvarieties of ~ : M ~ M/N
If
N
intersection,
M
Let
iC }0~ A
be e family of
having the finite intersection property and let Then
tC~ ~-I~jo~A
is a family of closed
having the finite intersection property and thus has a nonempty
say
C.
~ #
Thus
c~
( fi C
=
-i)~ ~ fi c ~ -i u = fi c -- o~A ~ c~:A ~
o~A M/N
topology.
is linearly topologized. M/N
be the canonical homomorphism.
subvsrieties of
and hence
is a closed submodule of a linearly compact module
i__~slinearly compact in the quotient
is linearly compact.
III.5.8 PROPOSITION.
A discrete linearly compact vector space is finite di-
mensional. PROOF.
Let
be s basis of
V
V
be a discrete and infinite dimensional vector space.
and for every
x E B, let
S
be the subspace of
V
Let
B
generated
X
by
B\[x] .
For . any .
.Xl,X2,
,x n ~ B, we obtain
(x I + x 2 + ... + x i _ I) + x i + (xi+ 1 + ... + x n )
E x i+S
xi
n
so
i-l~(xi + S x ' )
# ~"
Since
V
is discrete,
the family
IX+Sx]x~ B
consists of
i
closed subvsrieties and has the finite intersection property. where
xi,x E B.
n m ~ ~ixi = x+j~__l~jZj i= I
Then
so that
Let
y =
~ ~ixi ~ x+S x i=l
n m ~ ~.x - ~ ~.zj i= I i i j=l j
x =
where
z
E B and x ~ z for 1 < j < m. By linear independence we must have x = x i J J -_ , for some i. It follows that y ~ x + S x for all x # x i for i < i < n and thus ~L ( X + S x )
= @.
Consequently
V
is not linearly compsct.
x6B A family if for each
~
of subsets of a uniform space
~ C ~
there exists
next result is due to Dieudonn~ III.5.9 THEOREM. PROOF.
F E ~
(X,~)
such that
is said to contain small sets
F c X[x]
for some
x E X.
The
[5].
Every linearly compact module is complete.
In view of Kelley
([i], p. 193), it suffices to show that every family
of closed sets w h i c h has the finite intersection property and contains small sets has nonempty intersection. compact module
M
neighborhood system of exists
F E ~
It follows that x E
Hence let
~
he s family of subsets of a linearly
satisfying these requirements and let
such that
0
consisting of submodules of F ~ K[x]
F ~ iY E M I x - y
for some e K]
~ M.
x ~ M~ where
which, we claim,
be an open base for the For every
K E E, there
~ = ~(y,Z) l y - z ~ K } .
is equivalent
to
N (f+K). For if F ~ [y E M I x - y % K}, then for any f ~ F, x - f 6 K so fE F x E f+K and thus x E A ( f + K ) . Conversely, if x E ~I ( f + K ) , then x ~ f + K f6F fEF
134 for every
f E
F, so
x - f 6 K
and
thus
f t ty ~ M I x - y L K]
for every
f ~ F.
Let
We have
just
Let
F+K
x+K
~
seen
~ Q
that
and
f~ F K ~ ~{, there exists
for every
x C
N
(f+K).
Then
for any
F ~ ~
such
f 6 F, we have
that x E
F+K
f+K
E ~. and thus
fEF F+K.
k' E K.
If
f C F,
For any
k 6 K, we
f+k so that
F+K~
=
If
F+K
$
has
n
closed, H 6 $
there such
x = h+k z ~ M
that
that
which
quently
Then
~ q,
that
K
F+K
is open and since = x+K
is a
then
such
that
(x+K)
For
Now
Since
N F = ~,
contradicting x ~
this
K
iet
g t H.
F
there
x 6 H+K,
implies
= h + k+ K = x+K
Therefore
6 ~
since
of closed
compactness
x %
;~ F = ¢.
H+K
m ~ H.
let that
seen above.
is a family
by linear
such
(z - h) - (z - g) 6 K
g % h +K
property.
F ~ ~
Q
which
int{,rsection;
Further
for all
and thus
Therefore
property
exists
as we have
z -m E K
H 0, Fc:_ ( x + K )
intersection
there
h ~ H, k 6 K.
implies
property.
intersection
has nonempty
K ~ }{
H+K
z - h,z - g E K
g -h E K
q
N F.
for some such
have
intersection
the finite
exists
for some
i=l
the finite
x ~
follow
i = 1,2,...,n,
~ Fi ~ ¢
the family
Suppose
for
n
implies
x = f+k'
By hypothesis
It
fi (Fi + K i) E
having
that
= x+K.
also he closed,
Fi+K i E Q
subvarieties
that
= x + ( k - k ') t x + K
i=l since
implies
then have
Therefore
it must
subvariety.
the hypothesis
(×- k')+k
x+K.
it is a submodule closed
then
exists
we have
the existence
of
In particular,
K
we
is a submodule.
proving
that
the hypothesis
is
Thus
H c
x+K.
Conse-
$
has
the finite
that
[] F.
Ft$ 111.5.10 the finite
LEMMA.
If
V
An open base
for the n e i g h b o r h o o d
[B(vl,v2,...,Vn;0,0,...,0) is linearly Let section closed
then
V*
is linearly
compact
i__~n
~
be a family of closed
subspace
Then each
of
= if( + S ~ } o ~ A , by all
j = 1,2,...,n
I vi E V}
system of
0
of the sets
each of which
consists
is a subspace
of
V*
and thus
topologized.
property.
generated
space,
topology.
PROOF.
V*
is a v e c t o r
V*.
where the
with
Letting each
SC~.
subvarieties
element
Se Let
of
~
of
V*
S : T m, we obtain is a subspace xei C Sei,
x l+Xc~2+...+x
of
f+T V.
Let
i = 1,2,...,m
= x
am
having
is of the form
+x
~l
finite where
interT
= f + T z z = f + S ±. W and
+... +x
~2
the
f+T,
.
~n
is a Hence
be the subspace x~j E S~j, Let
of
V
135 m n g E [ Q (f + S ~ )] N [ O, (f~ + S _~ )]. i=l ~i i j=l j Dj
It follows that
g ~ f
+ S t , whence (~i
~i and similarly
= f gls~j
E S~ ~i
so that gls(~ ' = f~iiS
Consequently
f + ... + x f = x g + ... + x g ~i ~i ~m ~m ~i ~m
(x i + ... + x It follows that the function +...+x
is single valued.
(~i
~j IS~j"
x = x
(x
g-f
)g = (x
f
+ ... + X ~ m ) g = x~if~l + ''" +x~mf~m"
defined on
)f = x
f
W
+...+x
by f
if
x i E S i
Further, we obtain
[Y(X~l+'''+X~m)+6(X~l+'''+X~n)Jf (~{X~l) f~l + ... + (~{X~m )f ~m + (~X~l) f~l + "'" + (6X~n) f~n = 7(xc~ifC~l+...+x = ~[(x
and hence V.
f
Then
l+...+x
is linear.
Let
f
also denote any linear extension of
f 6 V* and for every (~ E A, fls
f E f + S ±. C~
f )+6(x~if~l+...+x~ f~ ) c~m C~m Pn Pn )f] + 6 [ ( x + ...+x )f] ~m ~1 ~n
Consequently
= f 1S so (y f~ f~ (f¢ +S~).
f E
Ct
f- f
f
to all of
~ S ~,(~ that is
c~A
We may now put together some of the results on linearly compact vector spaces as follows. III.5.11 THEOREM. (g,V,T)
The followin$ conditions on a topological vector space
are equivalent.
i)
V
is linearly compact.
ii)
V
is weakly topolo$ized
iii)
V
i~s algebraically
topology, where PROOF.
U
V/K
since
K
mensional,
i) = ii).
Let
g
is also closed.
The completeness
of
isomorphic
K 6 E V
to
U*
with
K
K ~ K, in the quotient is open, and that
But then 5.8 asserts
follows from 5.9.
V/K
that
has finite codimenaion
the finite
4.
be an open base for the neighborhood
For any
is discrete since
so every
and topologically
is some right vector space over
consisting of subspaces. that
and complete.
topology of is linearly
V/K
and thus
system of
0
V/K, we have topologized
must be finite diV
is weakly topologized
136 ii) = iii). completion f
maps
of
V
By 5.2,
(V,T)
onto
U*
iii) = i).
This follows
EXERCISES.
i)
V
continuous ii)
V'
semilinear
Show
topology.
and
immediately
be weakly
that in any vector
topologized of
For information ([i], §i0);
vector
into
V~
V, the finite
V
in
on linearly
£u(V).
We consider
III.6.1 DEFINITION. VV
and Gross
of
yields
that
isomorphism.
spaces.
Show that every
is uniformly
continuous.
topology of
V*
is a weak
linear forms on
V*
V**, where
coinV**
is
of
spaces,
to
A TOPOLOGY
FOR
[5],
£u(V)
is the finite
topology of
VV
relativized
of this topology particularly
£u(V).
£u(V)
spaces
is the finite
at the beginning
£u(V)
consult Dieudonn~
[i] and the Appendix.
For a pair of dual vector
relativized
From the observations finite topology
Now the
and topological
and is dense in
a few simple properties
to the ring structure
topology of
V**
compact vector
see also Fischer
The topology we have in mind here to
V*.
topology.
III.6
relative
V
space
image of
of
from 5.10.
Also show that the space of all continuous
cides with the canonical
U
as in 4.2 which by our hypothesis
the needed algebraic
transformation
endowed with the finite
K~the
for some t-subspace
(f,(U*,~))
and provides
III.5.12 Let
T = Tu(V)
is
of Section
(U,V),
topology
the finite
~u(V)
of
£u(V).
i, we see that a base for the
is given by the sets
B (Xl,X2,.'-,Xn;Yl,Y2,...,Yn) = {a ~ ~ u ( V ) where as in 1.5, the set In particular
the sets
i = 1,2 .....
I x i a = Yi'
~Xl,X2,...,XnJ B(x;y), with
n],
can be taken to be linearly
independent.
x ~ 0, form a subbase and n
B(xl,X2,''',Xn;Yl,''',Y n) An open base for the neighborhood B
= {b E ~u(V) (Xl,X2,...,Xn)(a)
and for
system of
=
NB.x,
i=l ( i'Yi)"
a E £u(V) I xi a
is thus given by the sets
xib , i = 1,2, =
n} ...,
a = O, we have
B(xl,x2 . . . . . Xn)(O ) = (b E gu(V) / x i b = O, i = 1,2 . . . . . n},
,
137 which shows that an open base of the neighborhood system of of annihilators of all finite dimensional subspaces of
V.
0
is given by the set
Another way of ex-
pressing the above neighborhood system can be obtained as follows. 111.6.2 PROPOSITION. is minimal.
Then
PROOF. For any
Let
a ~ £u(V)
and
c =
~ [ik,~k,~k] ~ ~u(V), where k=l
n
B(Vxl'VXk'''''Vhn)(S) = {b E £u(V) I ca = cb}.
First note that
u. ,u. ,...,u i l I i2 n
are linearly independent by 1.2.5.
b E £u(V), we obtain b E B(v~I,VX2 ,..,,vXn)(a ) ~
vxib = vhia
for
i = 1,2,...,n
n k~=lUik~k(Vxk(b
- a),u)
= 0
for
all
u E U
n ~
k~lUik~k(VXk,(V= c*(b-a)*
In particular,
- a)*u)
= 0 ~
for
= 0
c(b-a)
for
= 0 ~
s = 0, we have
~u(V).
O.
We introduce some more notation.
,v
i if
IvXl,v 2 ,..,,vhn } A ring
group of
~
~
~u(V)
It follows
form an open base for the
For
a E ~u(V)
and
c E ~u(V), we let
Now 6.2 yields ,.
k2
(a)
" " 'V~n)
is a basis for
Vc, ~c(0) = ~r(C).
provided with a topology
(~,T).
III.6.3 LEMMA.
For
ca
T
is a top>logical rin~ if the additive
is a topological group and the multiplication is jointly continuous,
be denoted by
PROOF.
cb =
This gives an intrinsic characterization of the topology
tle(a ) = ib ~ ~u(V) I ca = cb}. ~ c (a) = B(v~
u E U
B(v l,Vk2 ,...,vXn)(O) = ~r(C).
that the right annihilators of elements of neighborhood system of
all
Let
d E ~c(a)
We also require that (~u(V),#u(V))
be Hauadorff.
is a topolo$ical ring.
a,b ~ ~u(V), c ~ ~u(V) and
T
to
and consider the basic open set
~c(ab).
e C flca(b), we obtain c(de) = (cd)e = (ca)e = (ca)b = c(ab)
which shows that
de ~ ~c(ab).
Hence
plication is jointly continuous.
~c(a)~ca(b) ~ ~c(ab)
~c(a) -~c(b) ~ ~c(a -b), proving that the additive group of group.
Suppose next that
a # b.
by 1.3.3 implies the existence of ca # cb; if
Then
and the topology
a -b
c E ~u(V)
d ~ ~c(a) n ~c(b), then
~c(a) ~ ~c(b) = ~
so that the multi-
A similar argument shows that ~u(V)
such that
c(a -b) # 0.
ca = cd = cb, a contradiction. ~u(V)
is a topological
is a nonzero element of
is Hausdorff.
~u(V) Hence Thus
which
138 III.6. 4 LEMMA.
In
~u(V),
we have
i)
~Z(~T(V))
= ~ ( T ~)
if
T
is a subspace o f
U,
ii)
~r(~u(S))
= ~s~(V)
if
S
i_.~s~ subspace o f
V.
PROOF. suppose
We will prove i); the proof of ii) is left as an exercise.
that
U
is a t-subspace
sequence of equivalent
(bU* ~ T
~
b E ~u(V),
v
6 V
b*U ~ T
Item i) follows
a~T(V ) = 0 ~
implies
Only the last equivalence and
V*.
from the following
statements
a E ~%(~T(V))
u.i E T
of
We may
ab = 0) ~
needs a proof.
be such that and
ab = 0
b E ~T(V)
Va ~ T m. bU* ~ T
Suppose =
(v~,u i)
b*u i = u i.
for all
V # 0
Hence for any
implies
and let v ~ V
ab = 0.
Let
b = [i,y-i,x]. using
Then
the hypothesis,
we obtain (va,ui) = (va,b*ui) and since
= (v,(ab)*ui)
= (v,O) = 0
u. % T is arbitrary, we have (vast) = 0 for all t E T i Conversely, suppose that Va ~ T ~. Then for any b E £u(V)
Va ~ T ±.
b*U ~ T, we immediately since
= (v,a*b*ui)
b*u E T
and
obtain
Va c T m.
III.6.5 COROLLARY. we have in
for any Thus
For any
u E U, v E V, (v(ab),u)
v(ab) = 0
c E gu(V)
for all
and
v E V
so that for which
= (va,b*u)
= 0
and hence
ab = 0.
[Vl,V2,...,Vn]
a basis of
Vc,
£u(V),
~(vc)~(v) = ~r(~(vc)) = B(v l,v 2,...,vn)(O) = ~c(0) = ~r(C). Note
that in view of 2.7,
a base for the neighborhood
the family of all open subspaces
system of
0; an analogous
situation
of
V
constitutes
occurs
in
~u(V),
viz. III.6.6 COROLLARY. of
~u(V)
constitutes
topology
T
of
U
V.
~T(V)
IT
open subspace
of
UJ
system of
of right i d e a l 0
for the
#u(V).
PROOF.
of
The family
an open base for the neighborhood
Interchanging
the roles of
is open if and only if For each such
from 1.2.3-2.5. the family
~r(C)
S
T = S~
there exists
Hence by 6.5, where
U
c
and
V
is 2.7, we infer that a subspace
for some finite dimensional c ~ ~u(V)
such that
the family in the present
varies over
Vc = S
corollary
subspace
S
as follows
coincides
with
~u(V), which by the above discussion
has
the required properties. III.6.7 EXERCISES. i)
Find necessary
and sufficient
conditions
in order that
~u(V)
be discrete.
139 ii) S
of
For a dual pair
~u(S),
and finite dimensional ~u(V)
subspaees
T
relativized
to
U
and
~T(V)
of
and
respectively?
iii)
Let
bilinear
(U,V)
satisfy all the conditions
form may be degenerate•
the bilinear iv) T~ =
(U,V)
V, what can be said about the topology
Let N
~r(v) V)
What are necessary
form in order that (U,V)
of a dual pair except
~u(V)
be a dual pair.
and sufficient
that the
conditions
on
be Hausdorff. Show that for any subspace
T
of
U, we have
Na .
Show that
(~u(V),$u(V))_
has no proper closed ideals.
For the origin of the results of this section as well as for further discussion, see Jacobson
[5],([6], Chapter
IX, Section 6),([7], Chapter
111.7 The topology in question
is the topology
henceforth will not be mentioned properties or
U
of one-sided
~.
there exists
Let
for some
#u(V)
relativized
to
~u(V)
and
We study here a few topological
The closure of s subset
c E ~u(V).
such that
u.1 6 U, we have
a 6 ~(S).
left ideal of
and
Consequently
(~v)a and thus
~u(V)
The next two propositions
Eyery
a E ~u(S)
b E ~(S)
= Veb ~ vb ~ S.
~u(V).
18).
A
of
~u(V)
• t are due to D1eudonne
[2],
[5]•
III.7.1 PROPOSITION. PROOF.
explicitly.
ideals of
will be denoted by
see also Jacobson
A TOPOLOGY FOR
IV, Section
cb = ca.
Vca ~ S
Hence
III.7.2 PROPOSITION.
~(S)
is closed.
hc(a ) Fi ~ ( S )
Hence
for all
(vh,ui) = ~ ~ 0 = ~vk[i,o-l,~]a
~u(V) Then
Vb ~ S
c E ~u(V).
# ~
and hence
so that Let
~v
Vca ~ V; then
so that
E S
c ~u(S).
For every right ideal
R
o__ff ~ = ~u(V), we have
= ~r,$~%,~(R). PROOF.
First note that
~r,~,$(~T(V))
Let
= ~T ~m(V)
= ~ (V) by 6.4 and 2.2. T independent vectors in V. Then
a 6 ~_(V) and Xl,X2,...,x n be linearly T s*U ~ ~ and letting S = [Xl,X2,...,Xn], for any
u 6 U, we have
Further, we can write
minimal.
a =
~ ' k=l [~k'Vk'kk ]
with
m
( a * u + S ~) A T # @.
It follows
that
m (k~=lUik~k(Vkk,U) +S±) •
By I 2.3,
there exist
t
/
Ul u2''"
.
u
!
N T # ~
' m £ U
such that
(u 6 U). (vAi,u!)j
=
6iJ
for
1 _< i,j _< m
140 since infer
V%l,VX2,...,VXm
are linearly
of
the existence
g k E S±
independent
such
that
by 1.2.5.
Computing
t k = U i k ' Y k + g k E T.
a Uk, we
Letting
m t k = u .Jk "rk
and
b ~ ~ l[Jk,Tk,Xk] " , we obtain
m ~ u. T. (v
b*u =
k=l
]k k
m ~ t (v
,u) = )'k
k= 1 k
,u) E T )~k
and m
Xpb = k~__l(Xp,Ujk)TkV x k = k~= l (Xp ' t k ) v/"k m ~ (x ,u.
=
k=l for
1 < p < n
k ~'k
B(Xl,X2,.-.,x n)
(a) fi ~T(V)
(1)
a E ~T(V).
Conversely, vectors
= Xpa
)~.v
~k
so that
b E and thus
P
in
V
Xkb = xka
let and
for
a E ~T(V).
Let
u ~ U, Xl,X2,...,x n
S = [Xl,X2,...,Xn].
1 ~ k < n
and
There
b*u ~ T.
Let
exists
be linearly
b
t = b'u;
satisfying then
independent
(i),
so
t = a ' u + (b ~ a)*u,
where (Xk,(b - a)*u) = (Xk(b - a),u) = (0,u) = 0. Thus
t C
( a * u + S m) N T
which
proves
that
a*u E ~
and hence
a*U ~ T,
that is
a E g_(V). T III.7.3
COROLLARY.
For any subspace
gT(V)
= ~_(V) T
It is clear that the set is always
a right
generated III.7.4
by
ideal,
= ~ ~ T
of
U, we have
(V) = ~ , ~ , ~ ( a T ( V ) ) .
~r,~(A),
and that
T
where
~r,~(A)
A
is a nonempty
= ~r,~(L),
where
L
subset
of a ring
A.
COROLLARY
c(~.
2.5).
Th____eefollowing
conditions
on a subspace
T
of
are equivalent. i)
T
is closed.
ii)
~T(V)
is closed
iii)
~T(V)
is ~ right
PROOF.
This corollary subspaces
of
corollary
is valid
111.7.5
annihilator.
Exercise.
U
~,
is the left ideal of
implies
onto closed
that the lattice right
ideals
for open subspaces,
COROLLARY.
isomorphism
of
Su(V).
see 7.10,
Every principal
risht
X
in 1.3.10 maps
A slight variation
exercise
ideal of
iv). Su(V)
closed
of this
is closed.
U
141 PROOF.
Exercise.
An important kind of right ideal is provided by the following. III.7.6 DEFINITION.
Let
R
be a right ideal of a ring
is a modular
left identity of
~
relative to
such a case
R
R
if
~.
r - ar E R
An element for all
a E
r ~ ~; in
is a m o d u l a r risht ideal of
The next three results are new. 111.7.7 PROPOSITION. if
T
A right ideal
~T(V)
o_~f ~u(V)
contains an open subspace. PROOF.
r E ~u(V). u = uto.
Necessity. Then
s =
For every
By hypothesis
r - ar E ~T(V)
n
There exist
and
a E ~u(V)
and all
u E
and
(Va) ~
n
Vl,V2,...,v n E V
vk
Let
(k~=l(V,Uik)NkV~k,Ut) = k=l ~ (v,ulk)~k(VXk,Ut) . such that
(v.,u. ) = ~jk J ~k
1.2.3, and hence the last equation implies ~
for some
~ [ik,Nk,~k] , w h e r e n is minimal. k=l v E V, we have (va,u) = 0 and thus
0 = (va,ut)=
Let
is modular if and only
be such that
that
N ( v k , u t) = o.
u = ut~ = ut~(v
for
i ~ j,k < n ---
(Vkk,Ut) = 0
for
by
k = 1,2,...,n.
Then
,ut) n
= [t,~,k]u t - ( ~ [ik,~k,kk])[t,~,~.]u t E T k =i = by hypothesis.
Consequently
Sufficiency.
Since the subspaces
the n e i g h b o r h o o d system of
dimensional subspace Ve = S
implies
S
For any
V. Thus
dim Va < oo
S ~, where
0, the hypothesis
of
(e.g., use 1.2.3). u E T.
(Va) ¢ c_ T, where
implies that
(Ve) ¢ C T, that is and
(Va) m
is o p e n
dim S < ~, form an open base for S mC
There exists an idempotent
v E V, u E U
and thus
T
for some finite
e ~ ~u(V)
(ve,u) = 0
such that
for all
v E V
r C Su(V), we obtain
(ve,(r - er)*u) = (ve,r*u - e*r*u) = (ve,r*u) - (ve,e*r*u) = (ve,r*u) - (ve,r*u) = 0 so that
(r - er)*U c_ T
III.7.8 COROLLARY.
and therefore ~u(V)
r - er E ~T(V).
has an idempotent m o d u l a r
left identity relative
to every m o d u l a r right ideal. PROOF.
This follows from the last part of the proof of 7.7.
111.7.9 T H E O R E M (cf. 2.9).
The followin$ conditions on a subspace
are equivalent. i)
T
is a closed hyperplane.
T
of
U
142 ii)
~T(V)
is a closed maximal risht ideal.
iii)
~T(V)
is a modular maximal risht ideal.
PROOF.
i) ~
i) = iii).
This follows from 7.4 and 1.3.4.
This follows from 7.7 in view of 2.9. n Let e = ~ [ik,~k,X k] E ~u(V) be such that k=l
iii) ~ i). r E ~u(V).
all
ii).
Let
u i E U\T
plane, it follows that
and
r = [j,6,v].
u. = t + u . T J i
for some
Since
t E T
T
and
r-er is
E ~(V)
evidently
T E A.
for
a hyper-
Hence
(r - er)*u = ([j,6,~] - k=l~ [ik,~k(Vxk,Uj)~,v])*u n - ~ u. ?k(V. , u . ) ) ~ ( v ,u) (uj k=l ik ~k J n n = I t - k=l~U.lk~,(vmA k ' t ) - ['~ luU iik ~ )k ( kV X k= for all
t E T, • E A, u E U. n
-ui]T}6(v~'u)
E T
Consequently n
u i Vk(V I ,t) + [ ~ u i Vk(V k ,ui) - ui]T E T k=l k "k k=l k k for all
t E T, T E A.
For
t = 0
and
T ~ 0, we obtain
n
(1)
k~=lUik~k(Vkk,Ui ) - u i E T and for
T = 0, n
k=l
(t E T).
u i Vk(V k ,t) E T k k
By 1.14, we have Hence
(1) yields
T = gZ for some g 6 U* relative n g ( ~ u. ~ , ( v ,u.) - ui) = 0 whence k= I i k K Ak i n gu i =
and b y ( 2 ) ,
(2) to the dual pair
(U,U*).
(3)
~ (gu i )~k(Vk ,ui), k=l k k
we o b t a i n n
(4)
(gu i )~k(Vk ,t) = O. Now s u p p o s e and
(4)
that
k=l
k
(Vxk,U)
= 0
k for
k = 1,2,...,n.
Then
u
gives gu = g(t +uiT ) = gt + (gui)T
= k=if(guik)Vk(Vkk,Ui,) = k=l~(guik)Vk(Vxk,U)
= (gui)T n
= k~=l(gUik)Vk(Vkk,U - t)
= O.
=
which by (3)
t+u.T i
143
It follows
that for
S = [fvxl'fvx2'''"fVkn]
g E by 2.3.
Consequently
T = g~ = v z
so that
S ± ~ gZ
[g] = g± ~ ~ S ~ ~ = S
g E nat V
and hence
, we have
T
and thus
is open since
g = fv
for some
v E V.
v ~ = fvlO, and thus
T
But then
is also closed.
III.7.10 EXERCISES. i)
Show that a maximal
ii) if
V
right ideal of
Prove that every right ideal of
~u(V)
~(V)
is either
closed or dense.
is a right annihilator
if and only
is finite dimensional.
iii)
Show that for any right ideal
R
of
Su(V),
we have
= ~a E ~u(V) I n N r c_ Naj. rER iv) of
For a dual pair U
v)
(U,V),
prove that
the following
conditions
on a subspace
are equivalent. a)
T
is open,
b)
~T(V)
is open.
c)
~T(V)
is the right annihilator
Show that in
~(V),
provided with
of some element
the finite
of
topology,
~u(V). every left ideal is
a left annihilator. vi)
In any ring
~
(not necessarily
(i - s ) ~
~r - ar
~}
Ir E
a E ~. Show that for e = e , we have (i - e)~ = ~r(e). 2 where 6 ~u(V), we have ( i - e)~u(V ) = $ ( l _ e ) , u ( V )
Also show thst
e = e
(i - e ) * U = ~ u - e*u l u E U} relative vii)
to
~T(V)
For any
viii)
Let
a E ~u(V)
The principal
we
let
Mr(S ) T
left identity
=
~(Va)X
for
~u(V)
(v).
be a subspsce
~T(V) = ~r(S)
for some
of
U.
Show that
a E Su(V).
for this section are Behrens [5],([6], Chapter
([i], Chapter
IX, Section 8),([7], Chapter
II, Section I, Section 3
18). III.8
~,
is a modular
(Ve) ± ~ T.
show that
references
[2], Jacobson
IV, Section
For a set
e
be a dual pair and
if and only if
7), Dieudonn~ and Chapter
and that
if and only if
(U,V)
codim T < ~
E
write
2
for any for
=
with identity),
of functions
ANOTHER TOPOLOGY on a set
X
FOR
£u(V)
into a uniform space
(Y,~),
for every
144
W(~) : [(f,g) E $ x $ Then the family uniformity
I (f(x),g(x)) E ~
[W(~) I ~ E ~}
of uniform
is a base
convergence;
for all
x E X].
for a uniformity
the corresponding
on
topology
$
called the
is the topology of
uniform convergence. We now apply this construction to the set i.
£u(V)
of functions
An open base
family
[T~ I T
2.
v E A
The uniformity
V 3.
for which According
convergence where
there exists
associated with we proceed
system of
U, dim T < ~}
~
on
V
0
in
Tu(V)
and
as follows.
V
is given by the
so that a set
A c V
a finite dimensional
induced by
for
on
Tu(V )
is a ~u(V)-neighborhood
~(A) = {(x,y) Ix - y ~ A].
on
Specifically
is open if
subspace
T
of
U
v + T z c A.
[~(A) I A where
V.
for the neighborhood
is a subspace of
and only if for every such thst
to the uniformity
on
Consequently
there exists
£u(V)
of ~
a neighborhood
to the above construction, induced by
~
has for a base the family 0}
consists A
of
0
of all binary relations such that
the uniformity
has as a base
~
~(A) ~
~.
of uniform
the family
[W(~) I ~ E ~ ,
~ = ~(A),
~('L((A)) = [ ( a , b ) E £u(V))<~,u(V ) /V(a - b ) ~_ A}. 4.
For any neighborhood W(~(A))[a]
A
of
= [b E £u(V)
and thus the topology of uniform consists
of those sets
neighborhood
A
of
C
0
0
with
and
a E ~u(V),
IV(a-b)
convergence
such that
£u(V)
(simply uniform
that for every
W(~(A))[s]
For every finite dimensional
~ A}
on
the property
we have
subspace
topology)
a E C, there exists a
c C.
T
of
U, let
= ~a E ~u(V) I Va ~ T ~} . 111.8.1
PROPOSITION.
The family
an o~en base of the neighborhood PROOF. have
Let
Vs c T ~
T
iT IT
system of
is a subspace o f 0
be a finite dimensional
U, di___~mT < ~}
for the uniform subspace
of
U.
topology For any
on
forms
~u(V).
a E T, we
so that
W(~(TZ))[a]
= [b E £u(V) I V(a-b) ~ r ~} = [b E £u(V)
I -vb E - v a + T ±
for all
v E V}
= [b E ~u(V) I Vb ~ T m} = and thus for every Since
A
T
is open.
Next
let
a E C, there exists is a neighborhood
of
C
be an open set in the uniform
a neighborhood 0, there exists
A
of
0
topology.
such that
a finite dimensional
Then
W(~(A))[a] ~ C. subspace
T
of
145
U
such that
T ~ c A.
Hence
a + T = ~ a + b Ib E ~u(V), Vb 6- T ±} = ic E £u(V) IV(c- a) 6_ T -~} = [c E ~u(V) Iv(a-c) 6_ T ~] = W(~(T±))[a] ~ C as required. III.8.2 COROLLARY.
The family of sets a)
P(ul,u2,...,Un)( where
{Ul,U2,...,Un]
constitutes
ranses
an open base
{b E
=
Ib*u k
~u(V)
over all linearly
for the neighborhood
a*u k
=
for --
independent
system of
k = 1,2, finite
..
.,n}
subsets
a E ~u(V)
of
U,
in the uniform
topology. PROOF.
It suffices
~Ul,U2,...,Un}
to show that
is a basis of
b E a+T
T.
~=~ b = a + c ,
(v(b-a),uk)
= 0
(v,(b*-a*)uk) b*u k
a*u k
vc E T m for all
= 0 for
i__~sminimal, PROOF.
v E V
for all
and
c E ~u(V),
PROPOSITION. we have
where
v E V
k = 1,2,...,n
and
.,n ~
and all
k = 1,2,...,n
b E
(a) P(ul,u2 ,..,,u n)
"
Me(a ) = {b E £u(V) I bc = ac]. n a E £u(V) end c = ~ [ik,~k,X k] E ~u(V), k=l
For
let
where
n
[~c(a) = P(Uil'Ui2'' "''Uin)(a)"
Note that
b E £u(V),
c E £u(V)
and
v E V
k = 1,2,
v
,v ~i
any
for some
. .
a E £u(V)
III.8.3
= p(u I 'u2'''" ,Un)(a),
We have
=
For
a+T
,...,v X ~2
are linearly
independent
by 1.2.5.
For
n
we obtain
b ~ P(Uil,Ui2 ' ...,ui
)(a)
~
b*u k = a~'~uk
for
k = 1,2,...,n
n
n k=l~ ( v , ( b * - a * ) u k ) ~ k V ~ k ~= ~ (v(b - a),Uk)~kVkk k=l ~= v ( b - a)c = 0 bc = ac ~ In the special
case
for all
= 0 = 0
for all for all
hood system of
= 0
b E ~c(a). a = 0, we have
the uniform 0
v ~ V
v E V ¢~ (b -a)c
P(Uil,Ui2,...,Uin)(0 ) = ~c(0) Consequently
v ~ V
consisting
topology
on
= ~(c).
£u(V)
of left annihilators
has an open base of elements
of
for the neighbor~u(V).
Since
the
146 corresponding annihilators immediately
statement
of
Su(V),
is intrinsic
for the finite
topology
it follows
obtain all the statements
sponding ones
nonzero
for the finite
of elements
£u(V)
for the uniform
topology.
and can be defined
for
topology
topologies
on
£u(V)
of the socle.
have been defined
they turned out to be symmetric these remarks,
system of
to the left-right
that the following
we
topology
ring with a
0
It is interesting
in quite different ways,
relative
it is not surprising
duality,
from the corre-
simply as a topology on a primitive
of elements
for right
This also shows that the uniform
socle having an open base for the neighborhood
of left annihilators
is valid
that by the left-right
consisting
that these two but,
duality.
connection
in the end, In view of
between
the two
is valid. III.8.4 PROPOSITION. homeomorphism o f topolosY
£u(V)
with
a - a*
the finite
induced by the V - t o p o l o g y on
PROOF. ~v(U),
The mapping
Further,
~c(a)m
for
topolosy onto
isomorphism
£v(U)
with
and hence
the uniform
U.
We know that the mapping
see 1.1.2.
is a uniform
~:a - a*
a E ~u(V)
is an isomorphism
and
c E SU(V),
of
~u(V)
onto
we have
ca]m
= {h ~ £u(V) feb =
= [ 6* ~ £V (U) I c'b* = c'a*] = ~Lc.(a* ) which proves
that
~
is also a homeomorphism.
Furthermore,
we have
%(mr(C))m = ~(x,y) Ix-y E ~r(C)]m = [(x~,ym) Ix-y ~ ~r(C)] = ~(z~p,y~) I xR0 -yq0 E ~r(C~)} which shows
that
~
is also a uniform
Some of the results Section
18), others
If we consider
COMPLETE
a primitive
ring
~
~u(V) ~ ~ c ~u(V)
on
£u(V)
questions:
(i) What are the completions
the answer
We consider
to the first question,
111.9.1 LEMMA. £(V)
IV,
If
£(V)
i__~sdense i f and only i f
RINGS
with a nonzero
on
and sufficient
U
of
~, and since
V*.
Hence
they are
we may ask the following
of the uniform
first the finite
spaces
conditions
induced by these for completeness
topology on
£u(V)
of
and using
reply to the second.
i__~s$iven S
socle as a ring of linear
for some t-subspace
for some uniformities,
(2) What are necessary spaces?
PRIMITIVE
two topologies
uniform
topologies?
topologies
induce
necessarily
these uniform
([7}, Chapter
are new.
then
the two topologies
isomorphism.
in this section can be found in Jacobson
111.9
transformations,
= ~(~r(C~))
the finite
is n-fold
topology,
transitive
then a subset
for ever7 positive
S
of
integer n.
147 PROOF.
This follows easily from the description
of s base of the finite
topology in Section 6. Note that 9.1 says that the "density" density in the finite topology of III.9.2 PROPOSITION. PROOF. considers of
V*
~(V)
defined in 1.2.9 coincides with the
~(V). i_!s complete
in the finite topolosy.
The reasoning here is quite analogous ~(V)
as a subset of
as a subset of
V V, where
AV, where
A
V
to that in 4.1; indeed, one
has the discrete
has the discrete
topology,
topology.
instead
The details of
the proof are left as an exercise. The next result is due to Dieudonn~ III.9.3 COROLLARY. t-subspace i) ii)
U
of
Let
V*.
(a ~ e*,
ii)
i)
(~(V),~))
(~(U),~)) By 9.1, ~
The mapping
morphism of
~u(V)
be a rin$ such that
is the completion o__~f ~
is the completion o__~f ~ is dense in
~:a ~ a*
in
Hence
~
and
(~,(~(U),~)),
(£(U),~)
in the uniform topology.
Sv(U) ~
~v(U)
~(U)
is complete in the finite so by 9.1, ~
~
is dense
in the uniform topology.
and thus this completion
is unique up to a
([i], p. 197).
We have seen in Sections 6 and 8 that both topologies on ~ In view of this,
and uniform iso-
with the finite topology
~ ~v(U),
must be a completion of
are Hausdorff,
uniform isomorphism by Kelley
in the finite topology.
is an algebraic
The dual of 9.2 shows that
On the other hand, we have
Both
for some
which is complete by 9.2.
topology onto
topology. £(U).
(~(V),~)
(a E £u(V))
with the uniform
by the dual of 8,4.
~u(V) ~ ~ ~ ~u(V)
Then
(identity map,
PROOF.
~
[2].
it makes sense to introduce
the following
as in 9.3 are intrinsic.
two topologies
directly
on an abstract ring. III.9.4 DEFINITION. topology on
~
[~r(C) I c E ~}
Let
~
be a primitive
is the risht socle topology
the topology on
is the left socle topology for In light of the results is a uniform isomorphism, topology,
topology.
~; dually, 0
the family
The
the family
{~(c)
I c~]
~.
in Sections
6 and 8, the canonical
and thus also a homeomorphism
onto a subring of ~
for
system of
0
~.
system of
having as an open base for the neighborhood
well as of
ring with a nonzero socle
having as an open base for the neighborhood
~u(V)
containing
of
isomorphism
~ - £(V)
~, with the right socle
~u(V), with the finite topology,
as
with the left socle topology onto the same subring with the uniform
The above corollary
thus yields
148 III.9.5 COROLLARY. ~:~ ~ £u(V) i)
Let
(~,(~(V),~))
ii)
(~(a ~ a*),(~(U,~)))
III.9.6 COROLLARY.
A primitive
~
PROOF. second)
onto
~
into
a left primitive ~(V)
carries
for any element N
in the left socle topology.
with ~ nonzero socle is complete
follows immediately
V.
~
i__!sisomorphic
Every algebraic
in t_~o
is___~o-
from the first (respectively
statement
can also be proved directly
for we know that in view of the hypothesis, left ideals.
(respectively
~(c))
in the socle of
~
onto the uniform neighborhoods
isomorphism
~
~
right) vector space
ring with minimal
c
ring
Note that the bracketed £v(U),
~r(C)
in the right socle topology,
is uniform and hence a homeomorphism.
The first statement
part of 9.5.
by mapping
~
left) socle topology if and only if
~(V)
and let
Then
i_!s the completion of
for some left (respectivel~
morphism of
in
be a primitive ring with a nonzero socle,
isomorphism.
is the completion of
the right (respectively ~(V)
~
be the canonical
Any isomorphism
onto
~r(C~)
and thus carries in
£(V).
Hence
~
~
of
is also ~
(respectively
onto ~(c~))
the uniform neighborhoods ~
must be a uniform
and thus also a homeomorphism.
We will now derive a topology for a semiprime extension of its nonzero socle.
The procedure
prime rings with a nonzero socle whose as linear transformations
ring which is an essential
is somewhat
similar to that for
topology was derived
on a vector space.
from its representation
We will now go via a complete direct
sum. III.9.7 LEMMA.
Let
[~¢~A
be a family of prime rings with a nonzero socle
each endowed with the right socle topology. the product
topology has the famil X
neighborhood
and
system o f
O, where
PROOF.
Elements of
a E ~.
Then there exist
i = 1,2,...,n
and
~
c~ = 0
P~ = Ib E N I b~ ~ P~}.
~
Then the ring
i~r(C) I c ~ ~}
~ ~ endowed with o~fA ~ as an open base for the
is the socle of
~.
will be written in the form ~l,~2,...,~n ~ A
otherwise.
By definition
forms a subbase for the product
~ =
a = (a).
such that
c i ~ ~
For every open set the family of sets
topology
in
~.
Let
P~ ~
in
c ~
for ~,
I ~E A, P~
let open in ~ ]
With the notation just introduced,
we assert
a+~r(C) For let with
b E a +~r(C).
c~i d ~i = 0
b = a+ d
Then
and thus
(1)
= i=l~s~i +~r(C~i)
b
where
cd = 0.
E a(~i + ~/r(cc~i) •
Hence
Further,
b
= a
+d
a + ~ r ( C ) ~ a i +~r(C~i )
149
and thus also Then
b~i = a ~ i + d ~ i ,
# ~i
~
~ a +~r(C). i=l ~i "
where
[I a +~r(C i=l ~i
Now if exists
i"
c~ ~ ~
P~
Furthermore,
product
(i) shows
is an open set in ~-- P~"
~
Now let socle
~.
~
be a semiprime
By 11.5.9,
and an isomorphism contains
system of
~
Conse-
and there
for which
c
= 0
topology.
forms a base for the
{~r(C) I c ~ ~}
forms an open
O.
~
extension of its nonzero
, a E A, with nonzero socles
into a complete direct sum
N ~
such that
~ ~ . Giving each ~ the right socle topology and c~:A ~ topology, ~ can be given a topology for which ~ is a homeo~.
In particular, ~0
has the relativized
and thus has an open base for the neighborhood annihilators
system of
of the elements of the socle by 9.7.
which shows that the topology is intrinsic, representation
~.
~.
0
Let
~
The topology on
This property
the family
topology
of right
carries over to
i.e., does not depend upon the particular
be a semiprime ~
product
consisting
It is clear that we can proceed directly
III.9.8 DEFINITION. nonzero socle 0
if
the socle of
n ~ the product c~A ~ morphism of ~ onto
system of
~ P~
is open in the product
ring which is an essential
mapping
a
c 6 •
I a 6 ~, c ~ ~}
there exist prime rings
~
d~ = b~ - a~
(i).
~ , then
and thus the family of sets
base for the neighborhood
let
) ~ P~.
a +~r(C)
[a+~r(C)
n ~ i=l ~i +~r(c~i)'
bE
cd = (c d ) = 0.
Hence for
= a +~r(C
that each
the family of sets
topology of
with
which establishes
~ # ~, we obtain by (i), a + ~ r ( C )
let
i = 1,2,...,n;
b = a+d
a + ~ r (c)
such that
Conversely, for
We obtain
) ~ a+~r(C) --
a 6 P~, where
Consequently
~
c~id~i = 0
fOrn i = 1,2,...,n.
quently
if
a+~r(C)
as follows
ring essential
(cf. 9.4).
extension of its
having as an open base for the neighborhood
{~r(C) Ic ~ ~}
is the right socle topology
for
~.
The
left socle topolog~ is defined dually. The following result is new. 111.9.9 THEOREM. of it___ssnonzero socle containing For each containing (a)~
Let ~.
the socle of ~ E A, .let ~U ( V )
= (a ~ ).
Let
~
~ Let
be a semiprime ~
~ ~ , where each
~
be the ~
is an essential
(U ,V ).
~
extension
~ ~A ~ is a prime ring with a nonzero socle.
be an isomorphism __°f ~
for some dual pair T
ring which
be an isomorphism of
onto a subring of
onto a subring Define
socle topology on
~
~J~
__°f £U ( V )
on
~ ~ by 0~A ~ ~, give each £(V )
the
150 finite
topology
(q09,( I~ ~ ( V
and let
),@))
~
be the ~roduct
is the completion o f
topology on
~ ~(V ¢6A
).
Then
(~,~).
~A PROOF. of
9
First note that the existence
by II •2.8 and 1.2.13.
uniform isomorphism the product cussion
~
it.
~p
a uniform
By Kelley
of
q0
is guaranteed
by II.5.9
the right socle topology,
onto a dense subspace
topology makes
following
Giving
of
~(V )
by 9.5.
isomorphism
([I], p. 183),
~
9~
and that
becomes
Further,
a
giving
II
in view of 9.7 and the dis-
is a uniform
isomorphism
of
II
onto a subring of ring of
onto a subII ~ ( V ). Hence q0~ is a uniform isomorphism of ~ ~A ). In view of 9.2, each £(V ) is complete so that by Kelley ([i],
II £(V
~6A p. 194),
II ~ ( V
)
is complete.
It is easy to see that
~
maps
the socle of
~A onto the socle of
~I ~ ( V
).
By 9.1 and
i.2.13, we know that the socle
gu
(v)
of
~A
~ ~
is dense in
exists
~(V
c i E Pi [~ ~U
)• (V
c~. l
a direct that
for
is an open subset of
i = 1,2,...,n.
Since
~(V
), then there
the socle of
II ~ ( V
)
is
c~:A
sum of the socles of all
£(V
c~i = c.i
for
), as it easily i = 1,2,...,n
follows
from II.5.8, we have
c~ = 0
and
otherwise,
is an
of the socle of
II ~ ( V o~A is also an element of
c
that
)
Pi
i
c = (c~), where
element
Hence if
dense in
~I ~(V )
). With the notation of the proof of 9.7, we have n n P.. This shows that the socle of ~I £(V ) is i= 1 i ~A which completes the proof of the theorem•
~A III.9.10 COROLLARY. topology i f a n d only i f
A ring
~
as in 9•9 i_~s complete
~
is isomorphic
all linear transformations
on left vector
in the right socle
to a complete
direct
spaces,
~ ~(V
say
sum of the rings o f ).
Furthermore,
if
a6A both
~
and
~ £(V
)
are provided with
the topologies
as in 9.9,
then every
~A algebraic
isomorphism
of
~
onto
II ~ ( V
)
is a uniform
isomorphism
and hence a
o~A homeormorphism. PROOF.
Exercise.
III.9.11 EXERCISES. i) family
Let
~
be a ring satisfying
{~r(C) Ic E ~}
satisfies
the hypothesis
the requirements
in 9.9.
for the neighborhood
system of
0
there exists
such that
~r(C) ~ ~r(a) ~ ~r(b).
ii)
c E ~
What are necessary
hypothesis topologies?
in 9.9 in order
for a topology on
and sufficient that
~
conditions
be complete
Show directly
that the
on a family of sets to be s base ~, i.e.,
on a ring
that for any
~
a,b E ~,
satisfying
the
both in the left and the right socle
151 iii)
Let
~
be a semiprime ring with a nonzero socle
Show that the family
[~r(F) IF ~ ~ + ~
and
F
is finite}
quirements for a base of the neighborhood system of topology.
Show that
essential extension of
(~,T)
found in Leptin
0.
and let
~, how does
T
~ = ~r,~(~).
satisfies the re-
Let
is a Hausdorff topological ring.
A characterization of
Dieudonn~
~
T
denote this If
~
is also an
compare with the right socle topology of
2?
~I ~(V ) as a semisimple linearly compact ring can be c~A ([2], Part I). For more information on complete rings, see
[2],[3]
and Eckstein
[I].
152
APPENDIX
ON LINEARLY COMPACT PRIMITIVE AND SEMISIMPLE RINGS by Richard Wiegandt 0.
INTRODUCTION
In ring theory the celebrsted Wedderburn-Artin significance.
It states that a ring
~
(in this case semisimple means semiprime)
Structure Theorem is of central
with d.c.c, on right ideals is semisimple if and only if
~
is s finite direct sum
of rings of linear transformations on finite dimensional vector spaces over division rings.
It was Leptin
[2] who succeeded in eliminating both finiteness conditions
from this characterization;
he proved that linearly compact semisimple rings are
just complete direct sums of rings of linear transformations on vector spaces over division rings.
The price he had to pay was topological,
mathematician may object that linear compactness conditions.
Nevertheless,
Leptin's result is in a certain sense the best possible
generalization of the Wedderburn-Artin Structure Theorem. rings started, Zelinsky).
and a topologically minded
is actually a kind of finiteness
The study of topological
in fact, earlier (see the cited papers of Kaplansky, Dieudonn~ and
In particular,
[3] and Zelinsky
Wedderburn-Artin-like
theorems were obtained by Ksplansky
[2], but these theorems are consequences of Leptin's result.
It is surprizing that in the flood of mathematical publications only some dozen papers dealing with linearly compact rings have appeared in the last fifteen years since the publication of Leptin's papers.
What is even more surprizing, Leptin's
already classic results are not available in any textbook, monograph or even in lecture notes.
This strange situation may be due to several reasons, one of them is
definitely the fact that the study of linearly compact rings requires certain familiarity both with algebra and topology. It is the purpose of this appendix to discuss Leptin's results concerning linearly compact semisimple rings and several related topics.
The preceding lecture
notes by Professor M. Petrich, particularly Part III, provide an excellent and natural background for this aim. the preceding lecture notes,
Taking into consideration the material presented in the discussion of linearly compact semisimple rings can
be approached easily via linearly compact primitive rings.
This method differs from
153 that of Leptin, Moreover,
so there
our approach
the use of inverse Of course, explained
Finally,
of the theory
appendix.
I express
my
this
to describe
extensive
knowledge
zero socle were
of linearly
further
gratitude to his
results
proofs.
topological;
compact
however,
rings
and recent
to Professor
lecture
notes
can be
developments,
we
M.
Petrich
for the oppor-
and for his critical
remarks
of p r i m i t i v e in what
the following
and II.l.21.
RINGS
of linearly
rings.
So
follows we
and faithful ~ - m o d u l e s
in II.l.20
i.i LEMMA.
MORE ABOUT PRIMITIVE
the structure
considered,
Irreducible
have
the original
this appendix.
In order
defined
For
sincere
appendix
i.
more
with
and less
at the end of this note.
tunity of a t t a c h i n g concerning
algebraic
inevitable.
fragments
in this short
a few coincidences
to be more
limits was
only
give references
are only
seems
far only
shall not
as well
Concerning
compact
primitive
primitive impose
as primitive
irreducible
rings,
we need
rings w i t h non-
this requirement.
rings
have already
and faithful ~ - m o d u l e s
been we
simple
If
M
is an irreducible
faithful ~ - m o d u l e
and
0 # x E M,
then
x~ = M. PROOF.
or
Since
x~ = M.
x9~ ~ M,
Suppose
~
the i r r e d u c i b i l i t y
= 0
and consider
of
M
implies
that either
x~ = 0
the submodule
<x) = {kx I k = 0,+i,+2 .... }. The
irreducibility
0 = <x>~ = ~ ,
of
M
implies
contradicting
<x~ = M,
the assumption
further that
since
M
x~ = 0, we obtain
is faithful.
Hence
~
= M,
as asserted. 1.2 LEMMA. ~(h,V)
of all
Moreover,
if
linear M
over a suitable PROOF.
Every
primitive
transformations
is a faithful division
We have
tins
ring
~
i__~s isomorphic
on a v e c t o r
to a subrin$ o f
space
irreducible ~ - m o d u l e ,
V
then
over M+
th___eetins
a division
is a v e c t o ~
ring
h.
to construct
the d i v i s i o n
ring
h
as well
as the vector
space
V. Since ~ ( M t)
~
is primitive,
of all e n d o m o r p h i s m s
each element
r E ~,
it has a faithful of the additive
irreducible
group
M+
the m a p p i n g r :x - xr
(x E M) +
is clearly
an e n d o m o r p h i s m
of
h.
space
M
.
Consider
~ * = Jr* 6 ~ ( M +) I r ~ ~}
the subset
~-module
is o b v i o u s l y
M.
The set
a ring.
For
154
of
£(M+).
The m a p p i n g
~ : r ~ r*
x(r*+s*)
is a h o m o m o r p h i s m
= xr*+xs*
= xr + x s
from
~
onto
~*,
for
= x(r+s)
and x(r*s*)
hold
for every
every
x E M,
x E M+ i.e.
=
(xr*)s*
and
=
(xr*)s
r,s E ~.
x(r - s) = 0.
If
Since
=
xrs
r* = s*, M
then
xr* = xs*
is faithful,
we have
is valid
r = s.
for
Thus
is an isomorphism. Next,
consider
the set
& = [a E ~ ( M +) I ar* = r*a the so-called
centralizer
of
~
in
for every
~ ( M +)
(g
r* 6 ~},
is,
in fact,
the c e n t r a l i z e r
of
~*
I
in
£(Mt)).
identity. quently
We Let
each
claim
that
&
0 ~ d E g. y E M
has
is a division
Then
there
ring.
is an
&
x E M
is o b v i o u s l y such
that
a ring w i t h
xd ~ 0
and conse-
the form
y = xdr* = xr*d =
(xr)d +
for some some
r E ~
z E M.
which
shows
For every
that
element
d
maps
d
maps
isomorphism must be,
of
~
onto
0
M+
onto
M+
of course,
in
onto
itself.
Suppose
zd = 0
for
r* ~ ~, we have
0 = zdr* = zr*d = Hence
M
(zr)d.
and by I.i it follows and c o n s e q u e n t l y
that
d
z = 0.
Thus
has an inverse
d
in
is an
~ ( M +)
which
g. +
These and by
considerations
the i s o m o r p h i s m
1.3 LEMMA. ~(L,V)
such
mensional x~(W)
that
for every
(0) = ~
The proof
irreducible n >
1
that
M
~ : ~ ~ N*, N ~
b e~
V
primitive
and
element
~(W)
is by induction
x~(W)
and assume
an a r b i t r a r y
xa # 0
which
x # 0
that
N
is a subrin$ o f
If
W
is a finite
of
on the d i m e n s i o n
we have
the v a l i d i t y
x~ = V,
n
W
i__nn ~,
of
W.
&
di-
then
If
for o t h e r w i s e
a E ~(W)
of the statement
where
x E V\W.
is evidently
is a n o n z e r o
such an element
ring
V
n = 0,
then
would
not be
Then
V
~-module.
element
that
ring and suppose
is the a n n i h i l a t o r
space.
that
the d i v i s i o n
~(5,M+).
x E V\W.
W = U+[Xn}
such
space over in
irreducible ~ - m o d u l e .
splits into a direct sum Take
is a v e c t o r
can be embedded
is a faithful
and so for every
a faithful Let
V
subspace o f
= V PROOF.
~
Let
show
submodule does not
U
We shall
equivalent of
V
exist,
to
exhibit ~(W)
and so it must i.e.
for
n - i.
is an (n-l)-dimensional
~(W)
an element
# ~(V). equal
= u% ~(V).
vector a E ~(W)
This will V.
Assume
imply
that
By hypothesis
155 Xn~(U)
= V, so for every element
Xna v = v. thus
If
bv
v E V, there is an
is another element with
av - bv E ~ ( U )
N ~ ( x n) = ~ ( W ) .
av E ~ ( U )
Xnbv = v, then
For any
such that
Xn(a v - b v ) = 0
and
x E V\W, define a mapping
~x:V ~ V
by
V,x The element
xa
xa v
(v E
V).
is not n e c e s s a r i l y unique,
v Indeed,
valued. hence
a
=
= xb
v v show next that ~x
nevertheless
the m a p p i n g
~x
is single-
a - b E ~ ( W ) = ~(V) implies xa - Xbv = x(a v - b y ) = 0, and v v v holds for every x E V. C o n s e q u e n t l y ~x is single-valued. We
is
a homomorphism of
the ~-module
V
onto itself. Since
av+ w
is an element w i t h the property x n av+w = v + w
= Xna v + x n a w = Xn(av + aw)'
we have (v+w)~
Since
x =
Xav+ w
=
xa v + x a
w
= V~x+W%x.
Xnavr = vr = Xnavr , we obtain (vr)~x = Xavr = Xavr = (V~x)r'
Hence
~x
is indeed a module h o m o m o r p h i s m for each
= (vr)~ x
shows that
there is a
v E V xa
w h i c h implies y~(U)
= 0
~x
such that =
Xav
x E U+[Xn]
= V~x
=
(Xna)~x
~Xn}
for every
a t ~A(U).
(V~x)r a E ~(U),
Consequently
By the induction hypothesis,
x - Xn~ x E U ~ W.
generated by
Xn, ~x E &
E W, c o n t r a d i c t i n g the hypothesis
Applying
Moreover,
Since for every
(Xn~n)a
=
y E U, and we obtain
1-dimensional subspace
h.
a = a . v
x a = v, we can write n
(x- Xn~x)a = 0
implies
x E V\W.
belongs to the centralizer
that
Since
implies
x n&
is the
Xn% x £ [Xn] .
Hence
x £ V\W.
1.2 and 1.3 we can easily derive Jacobson's Density Theorem for
primitive rings. 1.4 THEOREM. ring ri9~
£(&,V)
Every primitive ring
i__ssisomorphic to a dense subrin$ of the V
over a division
g. PROOF.
By 1.2 the ring
N
can be embedded into a ring
is an irreducible faithful ~ - m o d u l e Take
~
of all linear transformations on a vector space
n
linearly independent vectors
y l , . . . , y n E V. xlr = Yl
If
n = i, then by
and hence
there is an element differ from
Yn"
~
such that
Xn~ = V.
Xl,...,x n ~ V
Assume
~
xir = Yi
such that in
V
£(V)°
r E ~
such that
is (n-l)-fold transitive. for
i = l,...,n-l, but
~ = ~(Xl,...,Xn).
r + s E £(V)
~
and arbitrary vectors
Hence there exists an element
The linear t r a n s f o r m a t i o n
£(g,V)
is the centralizer of
Xl~ = V, there exists an
is transitive.
r E ~
~
C o n s i d e r the a n n i h i l a t o r
cable and we obtain XnS = Yn"
and
Now, s E ~
Xnr
Then may
1.3 is applisuch that
is now the desired one, namely
156
xi(r + s )
= Yi
holds
ring of
&(V)
for every
for each
i = l,...,n.
n >
i
which means
Thus
~
is an n-fold
transitive
that
~
is dense in
~(V)
It is also true that every dense subring of not use this statement. the usual one theorem,
(cf. Divinsky
consult
Jacobson 2.
[i] and Kert~sz
([7], Chapter
cerning
and semisimple
inverse
complete direct is linearly
rings.
of the density
for the study of linearly compact
We shall prove some elementary
sum of linearly
compact modules,
compact modules
statements
con-
further we shall prove that the
endowed with
the product
topology,
compact.
A
topolo$ies
be a directed
2.1 DEFINITION. Mc~
For a discussion
but we shall
a version of
II).
the preparations
limits of linearly
Only Hausdorff Let
[i]).
is primitive,
1.3 is actually
INVERSE LIMITS AND LINEARLY COMPACT MODULES
In this section we continue primitive
£(~,V)
The proof of 1.4 along with
sub-
and continuous
will be considered.
set of indices.
A system
~ = iMc~,~ ~ I c~> ~ E A}
homomorphisms
~ :c~M c ~ ~ M~
of topological lE-modules
is said to be an inverse
system if
m~ = c ~ holds for every ~ > ~ > ~? and TT~ is the identity mapping on M P? every c~ C A. The inverse limit M = li~m [~ of the inverse system ~ is the topological
submodule
(...,x~, . . . , x , ...) E topology of
M
M
~l M c~A (2
such that
is the product
2.2 PROPOSITION. IIM . c~A ~ PROOF.
of the complete
sum
[~ M consisting all vectors ~ A (2 for every ~ > ~ E A and the
x~c~~ = x
topology
The inverse
direct
limit
We shall show that the set
relativized
to
M = li~m f]
[ M \M
for
M.
is a closed submodule
is open.
of
Take an arbitrary
~A element
(...,x
ponents
K
II M \M. There exist indices cY> ~ such that xc~ ~ # x~. c~A c~ Let K = (...,K~,...,Kc~,...) be an open subset of ~ M such that K~ and K eg:A :~ c~ are open sets with x~ E K~, Xc~ E Kc~ and K ~c~ ~ K~ = ~ and all the other comequal
are Hausdorff clear that Let
,...) E
K M
.
Observe
spaces
does not contain elements be a linearly
M
that the topological
and hence also regular
for the neighborhood ~-modules
M
= M/K
so that such
of
M.
topologized ~ - m o d u l e
system of
0
consisting
form an inverse
system
modules
Hence and
K~ M
does exist.
It is
is closed.
K = [K~o~A
of submodules. ~
under consideration
together with
be an open base
Obviously
the
the homomorphisms
157
M/K ~:M/K
--
if/ K~/K
2 . 3 PROPOSITION ( g e l i n s k y M
i_~s c o m p l e t e , PROOF.
then
M
Taking
~ ~ A, i.e.
x = 0
and
the mapping M
~,
M
forms
~p = ( . . . , x
~ ~ A.
into
li~ ~.
Moreover,
if
t__oo l i ~ . ,...)
with
Further, mp = 0
x E
M
x E M, x
implies
x
=
6 M , 0
for
is Hausdorff, we must have
is complete and consider an arbitrary element
x
6 M
a Cauchy net
For this element
onto
can be embedded
li~ ~.
y = ( .... y , . . . ) 6 Choose elements
~>
an__~d h o m e o m o r p h i c
where
into
for all
Q K . Since the topology of a£A ~ is an isomorphism.
~
Suppose
M
i__ss i s o m o r p h i c
we have a h o m o m o r p h i s m from each
[1]).
~ M/K~
such that
which x
converges
we have
o
~¢-..lim ~.
xff+Kff = yff
to
xo + K
~
for each
some e l e m e n t = y
~ 6 A,
x° 6 M
for each
~ E A.
since
Then
Ix
M
complete.
Hence
is ~
I ff E A]
maps
M
lim ~. Clearly
~
the
topology
is
in
l
Thus A
is
continuous.
discrete ~
and
is
topological ~ - m o d u l e s
if
~ > ~,
the
K
image
E K. Kff~
Since of
in
K~
each is
M
open
a homeomorphism.
be a set and
d i r e c t e d by inclusion.
Take an open submodule K~ ~_ Kff
F
be the set of all finite subsets of
C o n s i d e r the finite direct sums endowed w i t h the product topology.
A.
Then
F
is
n M (f E F) of ~6f ~ If g c f E F, then f
M can be mapped n a t u r a l l y onto ~ M by a p r o j e c t i o n ~ . Now o~f ~ c~g ~ g = [ ~ M ^ , ~ f I g c f E F] forms an inverse system of finite direct sums. ~f ~ ~ -2.4 PROPOSITION. li_m ~ i_!s isomorphic a n d homeomorphic to the complete direct sum
n M ~A ~ PROOF.
endowed with the product topology. Define a m a p p i n g
( .... x .... where
x f = x 1 + . . . +Xffn
for
~: ~ M ~ lim ~ c~A ~ -
by
)~ = (...,xf,...) the
indices
ffl,..,,ffn
E f.
Obviously
~
maps
isomorphically onto li~ ~ and is continuous. Taking into account that the M c~ oEA product topology is the w e a k e s t one such that p r o j e c t i n g onto the components we get back the topology of the components,
it is obvious
that
~
is a homeomorphism.
We continue this section with some elementary properties of linearly compact modules.
The following assertion, w h i c h will be useful later, provides
tunity for another d e f i n i t i o n of linearly compact modules.
the oppor-
158
2.6 PROPOSITION if it is linearly
(*)
~
property,
(Leptin
topologized
PROOF.
Conversely, A
+ K k.
M
_G = {a
and
N
2.7
intersection
N G =
linearly
M.)
O (a
If
N
compact
Let
K = {Ks}c~ A
such that each
s
the intersection
K
a
E (a+Ks)
Q N.
in
(a + K Clearly
a E N
s + (K
PROOF.
= {a S + K S } o ~ A
Ovbiously
G
it
topolo$ized then
N
is
the neighborhood
Take an element
~-module closed
system
a E N.
M
in of
M. 0
in
Now for each
Take also an element
~ N) c ( a + K )
Q N.
and so
For any element
ba E a + (K S N N).
Hence also
Let
~
A} = i ( a + K s )
of
N
0 N Is E A}
with the finite intersection NF
property.
has an element in
O ((a+Ks) NN) c_ q ( a + K ~ ) c~A c~A
N.
Hence
= a
N = N. be a continuous
homomorphism N.
Then
of a linearly compact Mqo
i__sslinearly
compact
topology. topology of
M~
is clearly linear.
is any family of closed subvarieties then
is linearly compact, a nonempty
M.
for
the intersection
~ N)) =
The relativized
section property,
M.
(*), we have
a linearly
into a linearly topolo$ized ~-module
in the relativized
is an open base
of
Moreover,
and thus
2.8 PROPOSITION. M
property.
is always open whenever
topology,
is nonempty.
b s - a~ E Ka N N
is linearly compact, ~aN (as+ ( K
of
relativized
be a n o p e n b a s e
*s a family of open subvarieties
~-module
family
+ K x ) = Q ( a a + Z o) = N F
f = { a s + ( K S N N) I ~
that
a submodule
a submodule
the
) ~ N
) Q N ~ a + (K Q N). -- s s
implies
linearly
and it consists of open subvsrieties
is a submodule of
s
bs E (a +Ke) N N, we have
N
~}~CA
Hence by condition
+Z
is
linearly
M
Since
Take an arbitrary
having the finite intersection
consisting of submodules
property
is closed,
(*).
compact.
PROPOSITION.
PROOF.
(a + K
0
(As it is easy to verify,
is
in any linear topology
+ Z ~ + K k Is E A, % ~ Al, where
system of
an open submodule of
is
which has the finite intersection
suppose that (*) is fulfilled.
0 #
is linearly compact if and only
the following condition:
the validity of condition
for the neighborhood has the finite
M
intersection.
of closed subvarieties
Form the family
and
and satisfies
Since every open submodule implies
f = {a +Z~
a +Z
An ~-module
family of open subvarieties
ha___~snonempty
compactness
contains
[i]).
K -i = ~ (a S
K -i
intersection.
+ K)~- i}c~A
of
M~
If
with the finite inter-
is such a family in
has a nonempty intersection.
Consequently
M. also
Since K
M has
159
2.9 PROPOSITION.
If
topologized ~ - m o d u l e PROOF. homomorphism
M/A
M,
By 2.7, A
A
and
then and
~:M ~ M/A
M
is closed in
is closed.
Hence
differ from k:a~+K~
it is sufficient
varieties of Let
has the form
M~.
Project
~ a ~+K
a
M
).
. c~
E [~F k-~
~
=
ak+K k
~ K a~A c&
Bq0~ I = A + B
Since
M
c~
Let
F_ = [ a % + K k } x E A
M.
-~F = ~a ~ + K
is linearly compact, N F k-c~
and take the element
a
= (...,a o
ak+ ~AK
K %
for each ~EA
of open sub-
is nonempty for each
,...) ~
II M . ¢~A ~
~
Then
a
~.
is o
since only a finite number of K k 's differ
k
and F has the finite intersection property. ~y ~I M is linearly compact. c~A ~ 3.
be such a
where only a finite number of
M
that
is closed in
to consider only families of open
onto the ~-th component
Thus we have a family
obviously contained in each from
is linearly compact in
~ M of linearly compact ~A ~ in the product topology.
_is _ _linearly _ compact
K%
and so the natural
The complete direct sum
In view of 2.6,
Each
~
In view of 2.8, Bq~
subvarieties with finite i n t e r s e c t i o n property. family.
M.
are closed submodules of
is continuous.
2.10 PROPOSITION.
PROOF.
are linearly compact submodules of a linearly
A + B B
and thus again by 2.7, Bop
S-modules
B
Hence
a
~ ~F -
o
proving
LINEARLY C O M P A C T PRIMITIVE RINGS
We now turn to structure theorems for linearly compact primitive rings. 3.1 PROPOSITION. topolosY
continuous image o f ideal of
~
is a primitive rin$ endowed with a linearl~ compact
~.
Moreover
M ~ ~ ~/K, where
K
M~
which is a
is an open maximal right
2.
PROOF.
Since
For any element e E ~
If
T, then there exists an irreducible faithful S - m o d u l e
~
is primitive,
0 ~ m ~ M, we have
such that
me = m.
there is an irreducible faithful ~ - m o d u l e mO~ = M
Consider
M.
and hence there exists an element
the set
J = [y = x - ex Ix ~ ~ }. Then J
J
is obviously a right ideal of
h o m o m o r p h i c a l l y and continuously,
mapping.
As it is easy to verify,
relativized topology.
Hence
Take an open right ideal
maximal excluding
e.
of
~
is Hausdorff,
of
~,
for if
that
K~
J
~.
and
The m a p p i n g ~
x~ = x - ex
relativized to
the topology induced by
J ~
maps
~
is the identical on
J
is just the
is linearly compact and, by 2.7, closed in K
of
~
such that
e ~ K
By Zorn's lemma such right ideal even regular[).
This right ideal
is a right ideal properly containing
and K
K
onto
J c K
does exist
and
~. K
is
(the topology
is s maximal right ideal K, then
e E K~
implying
160
r = (r- er) + e r for every
r ~ ~
and
Consider
~/K
image of
is clearly
topology
system of
~(M)
0
of
K
m E M = ~/K.
We can take
isopen~ s o i s J(m).
a continuous
compact
PROOF.
If
~.
Applying topology ~(V).
2.7, ~
in
is closed
REMARK. weakest
one
of
~(V),
is a linearly of
~
£(V).
and
Let
direct
~(V)
so
x E J(m)
~
onto
compact
means
£(V)
primitive
rx E K.
that
V ~ ~/K
ring,
for some vector
the finite
Since
then there is
space
V
Consequently
V
over a
topology. ~
is a dense subring of
is a vector
£(V)
space over the centralizer topology,
compact
primitive
ring
~ : ~ ~ ~(V)
is also a homeomorphism.
be a vector
space over a division
~
ring
is a
&.
sum
Then
~I J of minimal idempotent £ ( V ) - i s o m o r p h i c aEA ~ topology 7 of I] J induces the finite topology
is a weakest
Take a basis
{x
linearly
I ~ ~ A}
compact
of
V
(Hausdorff)
and consider
topology of
right ~(V) ~(V).
the right annihilators
~J((~l..... (Yn) = {a ~ ~(V) I x(~ a = 0, i = i ..... n} i and
J(~l ..... ~n ) = {a C ~(V) I x~a = O, ~ E A\{~ I ..... ~n ]} for every
finite subset
J(~l,...,~n) J~i
=
J(~i).
By 11.1.5,
{~l,...,~n}
is an ~ - m o d u l e Further,
each
of
A.
and a direct
sum
J~i
It is easy to verify J(~l,...,~n)
is an idempotent
each
J~i
is generated by an idempotent
If the image of
x i
under
e i E ~(V)
by
~ = ~ = £(V).
If the topology of the linearly
The product
PROOF.
the open base
is linearly compact also in this weaker
then the mapping
J .
the topology with
~.
~(V), where
Since
in
is a complete
ideals
m = r+K,
is endowed with
3.4 PROPOSITION. ~(V)
i.e.
of the right ideals
0]
1.4 and 1.2, we obtain
in the finite ~
~i ~
~(V)
A
of
topology
isomorphism
ring
~/K
and
N J(m) = 0 since M is faithful. We have proved mEM The finite topology ~(M) o f ~ is Hsusdorff and weaker
3.3 PROPOSITION.
division
irreducible ~ - m o d u l e
Moreover,
3.2 PROPOSITION. than the linearly
~,
consisting
J(m) = [x~ ~ I m x = for every
faithful
a
~.
the finite
for the neighborhood
~ = K~
K ~ = ~.
The factor module is s continuous
~ L+K
minimal
is given by
e~i,
that
=
~ J of ~-modules i=l ~i right ideal of £(V).
i.e.
d~i = e~i~(V).
161
e xai ~i then
e
2
= e
ai
~
j=l
pjx
implies ai
2 x ie i
=
0 # Xaieai ~j = a i
Hence
i = l,...,n. Now
Consider
a linear = x
x~ke~kae~i q~:e
r ~ e
ak
ae
ak
~ ,
.
J-J-l~jx{jeai
r
xai e ~i = o.x i ~i
transformation
# 0
which
a 6 ~(V)
shows
that
e
~i
is an ~(V)-homomorphism
between
right
~(V)e
ideals
for
xaka
i =
'
X
PkPi a i The mapping
# O. ai
J
and
J
ak Since
(Pi @ 0)
such that
~k
=i
the zero mapping. minimal
=
for one of the ~j 's involving
have
we
(9j ~ &),
~j
, and
~
is not
ai
J
and j are irreducible ~(V)-modules (as they are ak ai ~6(V)), ~ is an isomorphism. Hence all the J 's are iso-
of
morphic £(V)-modules. We prove next
that
£(V)
is a direct
sum of
J(a I .... ,an) n ~(al, ...,an) = O.
Clearly a E ~(V)
a linear
and choose
x~b = We now have
transformation X a. I
if
0
otherwise
b E J(al,...,~ n)
and
J(al,...,an)
Take an arbitrary b E ~(V)
and
linear
~(~l,...,an). transformation
such that
~ ~ [~I ..... ~n }
s - b E ~I(~l,...,a n)
implying
a = b + (a- b) E J(~l .... '~n ) +~(al'''''an)" Thus
~(V)
is the required
direct
J(~l,...,~n) Consider
the inverse ~ = [
~
the direct
J
, ~f I f,g
are finite
~ aiEf
are endowed with
J
topology
~(V)
for neighborhood of
compact
the product
topology
(which
is of
£(V).
finite
system of
By III.9.2,
~(V)
0
subset in
[a I ..... a n} ~ A}
£(V),
is complete
we obtain in
~(V) o
the finite Hence
2.3 and
and we obtain ~(V)
linearly
A},
Choosing
2.4 are applicable
Now 3.2 implies
of
~i
= [~(a I .... ,an) I for every for an open base
subsets
g
sums
course discrete).
~ £(V)/~(a I ..... ~n ).
system
ai~f ai where
sum, and in addition
~
that
~(V)
in
~(V).
l~im Q ~
~ J . ~A ~
is s weakest
topology
of
£(V)
and by 2.10, ~(V)
is
.
162
C o m b i n i n g 3.3 and 3.4, we arrive at 3.5 THEOREM. i)
For a rin$
~,
the followin$ statements are equivalent.
~
is a primitive ring endowed with s weakest
~
i__ssisomorphic a n d h o m e o m o r p h i c
linearly compact
(Hsusdorff)
topology. ii)
formations on a vector space with the finite topology iii)
~
~-isomorphic
V
to the rin$
£(•,V)
over a d i v i s i o n rin$
of all linear trans-
A, where
~(&,V)
is endowed
~(V).
i_~s isomorphic and homeomorphic
to a complete direct sum
idempotent minimal risht ideals
product topology and each
J
J , where
~ J of ~6A ~ - is endowed with the
n J
is discrete.
It is now clear that a linearly compact primitive ring always has a nonzero socle, A topological ring
~
will be called topolo$icall~ simple,
its only closed ideals and 3.6 PROPOSITION. PROOF.
1.4.3.
~(V),
to show that any ring
is topologically simple.
Ideals of
It follows from 1.2.10 that every nonzero ideal of
transitive for
0
and
~
are
A linearly compact primitive rin$ i__~st o p o l o g i c a l l y simple.
In view of 3.5, it suffices
the finite topology
if
~ 2 # 0, cf. 1.3.5.
n = 1,2,...,
and so by III.9.1,
~(V),
endowed with
~(V)
were found in
~(V)
is dense in
£(V)
is n-fold in the finite
topology. 4.
L I N E A R L Y COMPACT S E M I S I M P L E RINGS
Before dealing with linearly compact semisimple rings, we will prove some preliminary propositions. if
~/P
primitive ideal of
p
P
of a ring
Hence
~
If
P
is a primitive
is a maximal closed ideal of
PROOF.
~
We will prove that
such that
p
of 3.1, one csn show that
P
~.
M
Taking into account that m E M
0 ~ mU.
w h i c h shows that
implies that
P
= ~
M
~/P M
with
N,
in
is primitive,
can be considered as ~.
As in the proof in the discrete
holds with a suitable open m a x i m ~ l right ideal
that there is an element UnP
Since Then
is a linearly topologized ~ - m o d u l e
3.
Hence
M.
is just the annihilator of
M ~ ~/K
a E ~\P.
ideal, is a
ideal of a linearly compact rin$
is closed in
topology and that Let
0
~.
there exists sn irreducible faithful ~ / P - m o d u l e an ~ - m o d u l e
is called a primitive
is primitive if and only if
N.
4.1 PROPOSITION. then
An ideal
is a primitive ring.
M
ms # 0 P
is a maximal closed ideal of
is a faithful ~ / P - m o d u l e , and a n e i g h b o r h o o d is closed in ~.
~.
U
of
Finally,
K
of
we deduce a 3.6
with
163
A set
~
consisting of primitive ideals
P , ~ E A, of a linearly compact ring
is said to be an idependent system if for every finite subset {P I,...,P } n n n of ~, we have ~/ n P -~ N ~/P . Using Zorn's lemma, it follows that every i=l ~i i=l ~i linearly compact ring ~ has a maximal independent system e 0 of primitive ideals. 4.2 PROPOSITION.
Let
e I = {PI~}c~A
and
@2 = ~P2~}~EB
be two maximal in-
dependent systems of primitive ideals of a lines!l y compaqt rin $ weakest linearly compact topology. PROOF.
N
c~API~
=
N P
~
Suppose that the statement is not ture.
we may assume that
~API~N ~-- ~B~ P2~"
such thatn o~API~n ~_ D 2. D1 =
Then
N P ~ D2° i=l i~ i --
D2
endowed with a
Without loss of generality,
There is a maximal closed ideal
Hence for any finite subset
Since
2~
~
.
i~l,. .°,~n}
of
D 2 = P2~ o E @ 2 A, we hsve
is a maximal closed ideal, by 2.9 we have
D 2 # D I + D 2 = D I + D 2 = 2. Thus
~/D 1 N
D2
splits in a direct sum of
Since the topology of
~
DI/DI[~ D 2 ~ ~/D 2
and
D2/D 1 N D 2 ~ 2/D I.
is a weakest one and the isomorphism
~:~/D I N D 2 ~ ~/D I ~ ~/D 2 is continuous, ~
is a homeomorphism.
system of primitive ideals of
containing
lel,D2}
is an independent
~, contradicting the msximality of
Since for any primitive ideal @
Consequently
p
of
~
@i"
there is a maximal independent system
P, 4.2 immediately implies
4.3 PROPOSITION.
Let
e = ~P
I ~ E A}
primitive ideals of a linearly compact ring
be a maximal independent system of 2.
Then
N p
e~gals the intersection
o6A ~ of all Primitive ideals of 4.4 DEFINITION. primitive ideals of
~.
A ring ~
equals
~
is semisimple if the intersection
>(~)
of all
0.
In a terminology consistent with 11.5.4, a semisimple ring should be called semiprimitive. only if
~
In view of II.5.4, it is clear that a ring is a subdirect sum of primitive rings.
primitive ideals of
~
2
is semisimple if and
The intersection
~(~)
is referred as to the Jacobson radical of the ring
of all ~.
For
more information on this subject, consult Jacobson ([7], Chapter I). Linearly compact semisimple rings have several interesting characterizations. Some of them are presented in the following result.
Characterizations
are due to Leptin [2], whereas iv) can be found in Wiegandt 4.5 THEOREM.
Th___eefollowin$ conditions on a rin$
~
ii) and iii)
[i].
are equivalent.
164 i)
~
compact
is a linearly compact semisimple ring endowed with a weakest
(Hausdorff)
ii)
~ ~
~ £(f~c,V ), algebraically and topologically,
the right hand side is the product ,~.(V ) , (y
C~E
iii)
where the topology of
topology of the finite topologies
~ J , where each
J
v~c ~
~
is an idempotent minimal
right ideal, J
is s linearly compact r e g u l a r ring endowed with _a weakest
(Hausdorff)
PROOF.
topology
i) = ii).
primitive ideals of [ a I .... ,C~n}
of
the
~.
is discrete. linearly
(regularity is m e a n t in the algebraic sense of 1.3.5).
Choose a maximal
independent
By 4.3, we have
A, we have n
system
~ P = 0 a6A ~
~ = iP(~ I ~ 6 A}
of
and for any finite subset
n
~/ ~ p
~
i=l a'i
~ '~/p
i=l
~i
These finite direct sums obviously form an inverse system 111.5.9,
of
v
topology of the right hand side is the product topology and each
compact
~(V )
A.
~ ~
iv)
linearly
topology.
[~.
Taking into account
and 2.3 and 2.4 of this appendix, we deduce
In view of 3.5, ~ / p is weakest,
~--~ lira [~ ~
N/P tEA
~ £(&
holds for each
,V )
the isomorphism
~
11 ~ / P
c~ E A•
Since the topology of
is also a homeomorphism.
a6A ii) = iii).
This follows directly from 3.5.
iii) = iv). that every
J
= e ~
By 11.1.14, and
&
a E J , there is an
(evre?)-i = eyse? a(e
Thus for each
s)a
in =
~ E C, J
~ E C, there is an idempotent
is a division ring with
r E ~
A v. (e
for any
= e ~e
such that
Consequently,
r)(e
s)(e
r)
=
(e
is a regular ring•
~
is also,
Note that the discrete
E J e .
such Hence for
and
e re VV
a E JV
and
e?s E Jr' we have
)(e
se
)r
Thus
has an inverse
= e r = a.
~ J
~c hence
e
a = e r V
for re
the identity
are regular rings and v
topology in each
is a w e a k e s t one,
J V
so the product topology yields again a weakest one. iv) = i).
Consider an arbitrary element
existence of an element
r E ~
such that
z E ~(~).
z = r - zr•
We shall establish
the
To this end, consider
the
right ideal J = j r - zr I r E ~ } . According
to II.7.6, J
is a m o d u l a r right ideal of
lemma insures the existence of a m o d u l a r right ideal
~. M
Suppose
z @ J.
Then Zorn's
which is maximal with
165 respect to the exclusion of ideal
N
z.
properly containing
In fact, M M
r = (r - zr) + z r ~ M + N holds for every
r E ~.
is a maximal right ideal,
contains
Obviously
M
z
for any right
and thus
~ N
is modular.
Consider next the two-sided
ideal P = [y E ~ l~y ~ M}. Since
~
is regular,
it follows that
~/P-module.
there is an
y = yxy E M
Consequently
x E ~
implying
p
with p c M.
yxy = y.
Hence for every
Moreover, ~ / M
is a primitive ideal of
~.
y E P,
is a faithful
By hypothesis, we have
z 6 ~(~) c P c M c o n t r a d i c t i n g the assumption element
r 6 ~
such that
z ~ M.
Consequently
z E J
Again take an arbitrary element
a E ~(~).
x E ~
such that
axa = a.
element
r ~ ~
such that
xa = r - xar, and we obtain
a = axa = ar - a x a r ~(~) = 0, i.e. 4.6 COROLLARY.
~
Then
Since
element
Thus
4.7 C O R O L L A R Y
is semisimple.
A semisimple ring
~ N
endowed with a weakest linearly compact i__~stopologically simple.
(Leptin
[2]).
~ linearly compact ring
of
A ring
is finite for all
~.
~
is topologically simple
Trivial by 4.5 and 4.6. ~
is linearly compact,
a__n_nopen base for the n e i g h b o r h o o d system of
PROOF.
there is an
there is an
N ~ £(~,V).
4.8 COROLLARY.
dim V
By the above,
S t r a i g h t f o r w a r d by 3.6 and 4.5.
if and only i f PROOF.
is regular,
xa E ~(N).
ar - a r = O.
=
topology i~s primitive if and only if PROOF.
and thus there exists an
z = r - zr.
Necessity.
Without
0
semisimple and its ideals f o r m
if and only if
~ ~
n £(&
,V )
and
~ E A. By 4.5, we have
R £ ( V ). ~A loss of generality, we may assume that ~/U ~
~ ~
C o n s i d e r an open ideal
U
~ £ ( V ). finite
Since
H £ ( V ) is discrete, each component ~ ( V ) is discrete in the finite finite ~ n topology. H e n c e 3.5 implies that each ~ ( V ) splits into a finite product ~ J. i= 1 1
of minimal idempotent ~ - i s o m o r p h i c
right ideals
J..
In this case the dimension of
i
each
V
must be finite,
for
dim V
= n .
S u f f i c i e n c y is obvious.
166 4.9 COROLLARY.
A rin$
topology if and only if PROOF. £(V )
~
~ ~
For necessity,
must be finite since
4.10 COROLLARY.
PROOF.
Trivial
Finally,
~
the terms '~primitive"
(cf. Leptin compact.
system
dim V
is finite.
imply
0
~
i__sssimple if and only if
rings are always prime and, by II.i.25, is valid,
and "semisimple" respectively.
in
~, then
Further
can no longer be replaced by the terms For let
~
denote
is considered p
for rings with
for linearly compact rings in 3.5 and 4.5
the ring of all p-adic
as an open base for a
induces a linearly compact
~i = p~' as s closed submodule of
topology on
~, is linearly
(Prove that a closed submodule of a linearly compact S-module
linearly compact in the relativized exactly the ~l-SUbmodules,
~i
to see, ~i
has no primitive
two nonzero
ideals
and thus
where
~ = {pi~ i i = 1,2,...j
[i]).
is again trivial.
is finite.
"prime" and "semiprime",
neighborhood
JK # 0
linearly compact,
~.
and the number of the components Sufficiency
A linearly compact tins
dim V
Even though primitive
If
is finite for all
i__ssprimitive and linearly compact in the discrete
~ ~ ~(~,V)
nonzero socle also the converse
integers.
dim V
is discrete.
4.6 and 3.5 immediately
where
and l inearl~ compact in the discrete
in view of 3.5.
4.11 COROLLARY. ~ £(&,V)
~
A tins
topolosy if and only if
is semisimple
~ ~(V ) and finite ~ 4.8 is applicable,
J, K
topology.)
Since the ~-submodules
is linearly compact as a ~l-mOdule. ideals
of
(cf. Wiegsndt
~i' we have
which shows that
0
[3]).
J = p~,
of
~i
for some ~i"
In addition,
are
As it is easy
On the other hand,
K = pm~
is a prime ideal of
prime, but is not semisimple.
is always
for any
n,m > i,
Hence
~i
the socle of
is ~
equals
0. For further discussion of semisimple [2], Wiegandt
[2] and Eckstein
An obvious generalization compactness.
of linear compactness
Since every semisimple
socle (see Wiegandt
linearly compact rings, we refer to Leptin
[i]. is the notion of local linear
locally linearly compact ring has a nonzero
[4]), the study of their structure
is not the case in the theory of locally compact rings Linearly
compact nonsemisimple
[3] and Arnautov = ~(~)).
[i], and particular
[i], Fuchs
to Andrunakievi@,
Arnautov
(cf. Skornjakov
rings were investigated
by Leptin
[1],[2]). [2], Wiegandt
attention was paid to radical rings
Further results on linear compactness
Fischer and Gross
can be easily handled, which
[i], and Warner
were obtained by Wolfson
[i].
and Ursu [i], Warner
(where
For recent developments
[2], widiger
[i], we refer
[i] and Wiegandt
[5].
167
CURRENT ACTIVITY From the material
of these Lectures,
we may easily extract the following
categories: i.
Objects:
pairs of dual vector spaces,
Morphisms: 2.
semilinear
Objects: weakly Morphisms:
3.
Objects:
complete
lattices
(U-closed
Objects:
rings with a nonzero socle,
ring isomorphisms;
Morphisms: 5.
adjoint;
topologized vector spaces,
maximal primitive
Objects:
with a surjective
iseomorphisms;
Morphisms: 4.
isomorphisms
satisfying
subspaces of
the double covering condition
V),
lattice isomorphisms;
multiplicative
semigroups
of objects
in 3. above
(characterized
abstractly), Morphisms:
semigroup
isomorphisms.
Further categories may be obtained by taking certain subobjects Under some mild restrictions categories
are either equivalent
relationship
or isomorphic.
semigroups,
by the author entitled rings and lattices".
toward a better understanding
of homomorphisms
transformations, pierce
[i].
see Fajans
Multiplicative
[4], Satyanarayana peinado Eckstein
[I]. [2].
widiger
These Lectures
thus represent a step
among these different branches
of linear transformations
and antiisomorphisms [1],[2],
includes mainly in-
of various
semigroups
Jodeit and Lam [i], Mihalev
semigroups
Ill, a comprehensive
of linear
and Satalova
[i],
of certain rings have been studied by Petrich survey of this subject can be found in
Semigroup methods have been successfully
used in ring theory by
Further classes of rings with unique addition have been recently
found by Martindsle investigated
of vector and projective
geometry.
The recent work on semigroups vestigations
study of the
forms the subject of an un-
"Categories
of the relationship
of modern algebra an~ projective
in 3., 4., and 5.
most of the resulting
A comprehensive
among these and many other categories
published manuscript spaces,
on objects of these categories,
[i] and Stephenson
by Andrunaklevlc,
[i], Wiegandt
[5].
Arnautov
]i].
Linear compactness
and Ursu [i], Arnautov
for rings has been [i], Warner
[1],[2],
168
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175
LIST OF SYMBOLS
(&,v)
1
v, (V,A), &+, V * = (V*,A)
2
g(v) = g(A,V), ~(V) = £(A,V), g(V), ~'(V),
dimV,
A•B,
rank a, N a
3
(U,V) = (U,a,V) = (U,/X,V;~), fu' nat U
4
F(~,v), £u(V), Fu(b,V), gU(~,v), ~n,U(b,V), a*
6
A\B, bX' DO{, S-, ~BA-, [i,'f,X] C(A),
7 ii
iB(A )
16
~2, ~ ( N ) ,
~r (g)
19
~Rn
22
Es , } ( U , V )
23
Ge S / I , J(a),
25
I ( a ) , ¢+, g (A,V)
27
(v) Q ~°(I,G,M;P),
28 Ln(A),
Qn,u(A, V)
29
Xv,(V), (®,a), ( a , v ) ~ (Z~',v')
34
(b ,w)
35
(u,b,v) ~ ( u ' , ~ ' , v ' ) , g~,u(b, v)
~(w,a)
38 43
~g, m v
44
C(S) ,£+
45 46
~, A, M, I
47
g, G(S), ¢9(S),
48
HK
60 + V
49 54 k
~(S), %a' IDa' rra' [[(S), T = T(K:S), ~k , p 2
k
55
S
56
~(s)
6O 62
AB, aB, Ab A n , G£,~(B), AB, A n, aB Ai iE I
65 ~r,~(B)
66 67 68
176 70
AB
74 75
k op, £, ~,
76
~(I,A,A;P) g
80 81 91
+G A i iE I
92
~es ~A
¢~
(a)
95
~ ,~ c~AC~ c~
96 97
rl s c~ a6A
102 (a) r x , X A, (X,~) c~A ~
112 •
,
S(~;y), B((~l,...,~n,X[ . . . , X n ) ¢~AIIX , B(vl, . .,Vn;~l, . . .
' Tu(V)
= ~U(A,V)
113
.,~n ), C(vi,.. .,v n;6l,...,~n )'
SJ-, T z, [A], [Wl,W2,..-,Wn],
v't, uZ
114 115
(b,v ,'0
119
codim S
127
a*
X[A], X[x], (xj~) (S,~), [S n In e D}, (f,(X*,%~*))
128 129 130 136
~U (V)' B(xl,...,Xn;Yl,...,yn)' B(xl,...,Xn)(a)
'O.c(a), 7
137
(~, "r)
W(~), "~(A), ~ ,
139 'r
144 145
p(ul, . . .,Un) (a), ~c(a)
<~), ~(M+), M+, r*, ~*
153
N(w)
154
¢y ~ ' ,e--lira
156
~J(Ul.... ~n )
J(~l ..... ~n )' J
160
~(v)
161
~(~)
163
177
INDEX
adjoint of a linear transformation, adjoint of a semilinear affine transformation, algebra,
35
54
46
annihilator,
60
atomic ring,
69
basis,
5
transformation,
3
bilinear
form, 4
biorthogonal
sets, 81
bitranslation,
55
Boolean ring, 46 Cauchy net, 129 center, 45 centralizer, centralizer
48 of a primitive
characteristic,
circle composition, codimension, compatible
ring,
154
16 62
119 (topology-with
complement,
the group structure),
3
complete direct sum, 96 complete
lattice,
125
complete uniform space, completely
O-simple
129
semigroup,
completion of a uniform space, congruence
71 129
(induced by), 54
conjugate of a linear transformation, conjugate of a semilinear conjugate contains
space,
2
small sets,
converges
(a net
dense extension,
133 ), 129
54
dense set of linear transformations, dense uniform subspace, diagonal,
128
dimension,
3
6
transformation,
129
12
127
178
direct product of rings, 96 direct product of semigroups,
97
direct sum, 91 directed
set, 129
directs
(a binary relation ) ,
discrete
division ring, doubly
129
direct sum, 96 i
transitive,
12
dual pair, 4 dual space,
2
endomorphism, essential
3
extension,
60
eventually
in ( n e t ) ,
extension,
54
129
external direct sum, 96 faithful module,
71
finite intersection
property,
finite topology of
X A, 113
finite topology of
~u(V),
finite topology of
V*,
general
linear group,
generalized generates
homothetic hyperplane,
29 62
), 81
set, 3
Green's relations, homogeneous
136
I13
inner automorphism,
(a set
generating
132
75
component,
94
transformation,
54
115
ideal, 6 idealizer,
ii
idempotent
ideal,
identity
66
relation,
in (net
a set),
128 129
independent
family of subrings,
91
independent
system of primitive
ideals,
induced by a semilinear induced congruence,
54
inner automorphism,
48
inner bitranslation,
55
isomorphism,
41
163
179 inner left (right) translation, 55 inverse limit, 156 inverse system of topological modules,
156
invertible matrix, 53 irreducible module, 71 Isomorphic dual pairs, 38 isomorphic vector spaces, 34 Jacobson radical, 163 large ideal, 60 left annihilator,
19,66,91
left ideal, 27 left ~-module,
73
left row-independent matrix, 76 left socle, 69 left socle Lopology, 147,149 left Lranslation, 55 left vector space, I linear form, 2 linear transformation, linear variety,
2
132
linearly compact module,
132
linearly independent set, 3 linearly topologized module,
131
locally matrix ring, 25 A X I-matrix, 76 maximsl dense extension, 54 maximal subgroup, 25 minimal left (right, two-sided) ideal, 67 modular left identity, 141 modular right ideal, 141 multiplication induced by a scalar, 44 n-fold transitive,
12
natural image, 4 natural isomorphism of
V
into
net, 129 nilpotent element, 74 nilpotent ideal, 66 nondegenerate bilinear form, 4 null space, 3
V**, 26
180
open function, 126 order isomorphism,
23
orthogonal set of idempotents,
15
orthogonal sum, 92 pair of dual vector spaces, 4 permutable translations,
61
prime ideal, 71 prime ring, 71 primitive ideal, 162 primitive idempotent, 81 primitive regular semigroup, 97 primitive ring, 71 principal factor, 27 product of relations, 75 product of topological spaces, 112 product topology, 113 projection homomorphism, 96 proper ideal, 32 quasi-inner sutomorphism, 62 range, 3 rank, 3 Rees matrix ring, 76 Rees matrix semigroup, 29 Rees quotient, 27 regular element (semigroup, ring), 19 relative uniformity,
129
relativization of a uniformity, ~-homomorphism,
129
75
rlght annihilator, 19,66,91 right column-independent matrix, 76 rlght ideal 27 rlght R-module,
70
right socle, 69 rlght socle topology, 147,149 right translation, 55 rlght vector space, 2 rlng of endomorphisms,
3
rl-semigroup, 85 ~-isomorpbic modules, ~-isomorpbism,
75
75
181
r-maximal ring, 86 r-maximal semigroup, 86 ~-module,
70
row finite matrix, 52 ~-suhmodule,
70
sandwich matrix, 29 scalsr multiplication,
2
scalars, 2 semidirect product, 50 semlgroup, 3 aemlgroup of endomorphisms,
3
semlgroup socle, 73 semilinear automorphism, 34 semilinear isomorphism of dual pairs, 38 semilinear isomorphism of vector spaces, 34 semilinear transformation,
34
semiprime ring, 66 semiprime ideal, 71 semiprime semigroup, 67 semisimple ring, 163 simple ring, 19 socle, 68 split extension, 51 subdirect sum (product), 96 subspace, 3 subvariety, 132 sum of subrings~ 68 topological group, 114 topological module, 131 topological ring, 137 topological vector space, 115 topologically simple ring, 162 topology of s uniformity,
128
topology of uniform convergence, 144 total subspace, 4 transitive,
12
translation, 54 translational hull, 55 t-subspace, 4
182
uniform isomorphism,
129
uniform semigroup (ring), 63 uniform space~ 128 uniform subspace, 129 uniform topology, 144 uniformity,
128
uniformly continuous function, 129 uniformly equivslent uniform spsces, 129 uniformity of uniform convergence,
144
unique addition, 45 U-topology,
113
vectors, 2 V-topology,
113
weskly reductive semigroup, 54 weskly topologized vector spsce, 131 zero ring, 19