North-HollandMathematicalLibrary Board ofAdvisory Editors:
M. Artin, H. Bass, J . Eells, W. Feit, P. J. Freyd, F. W. Gehring, H. Halberstam, L. V. Hormander, M. Kac, J. H. B. Kemperman, H. A . Lauwerier, W. A. J. Luxemburg, F. P. Peterson, I. M. Singer and A . C . Zaanen
VOLUME 30
NORTH-HOLLAND PUBLISHING COMPANY AMSTERDAM . NEW YORK . OXFORD
RIESZ SPACES I1 A. C. ZAANEN
Leiden University, Leiden, The Netherlands
1983
NORTH-HOLLAND PUBLISHING COMPANY AMSTERDAM . NEW YORK . OXFORD
@
North-Holland Publishing Company, 1983
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the copyright owner.
ISBN: 0 444 86626 4 Published by: NORTH-HOLLAND PUBLISHING COMPANY AMSTERDAM . NEW YORK . OXFORD Sole distributors for the U.S.A. and Canada: ELSEVIER SCIENCE PUBLISHING COMPANY, INC. 52, VANDERBILT AVENUE NEW YORK, N.Y. 10017
PRINTED IN THE NETHERLANDS
PREFACE
This i s Volume I1 of a r a t h e r d e t a i l e d work on Riesz s p a c e s ( v e c t o r l a t t i c e s ) . Volume I , w r i t t e n j o i n t l y w i t h W . A . J .
Luxemburg, appeared i n
1971. The c o n s i d e r a b l e t i m e i n t e r v a l between t h e appearance of t h e s e volumes
i s caused mainly by t h e f a c t t h a t i m p o r t a n t p a r t s of t h e t h e o r y were p u t i n t o a n e l e g a n t and more o r less f i n a l form o n l y v e r y r e c e n t l y . This i s t r u e f o r Chapter 13 on k e r n e l o p e r a t o r s ( i n t e g r a l o p e r a t o r s ) and s t i l l more f o r Chapter 18 on compact o p e r a t o r s . To mention one example, i t w a s unknown n o t s o l o n g ago t h a t , even i n a f a m i l i a r s p a c e l i k e a n L 2 - ~ p a c e , any p o s i t i v e o p e r a t o r m a j o r i z e d by a k e r n e l o p e r a t o r is i t s e l f a k e r n e l o p e r a t o r and any p o s i t i v e o p e r a t o r m a j o r i z e d by a compact o p e r a t o r i s i t s e l f compact. Chapters and s e c t i o n s i n t h e two volumes are numbered through. The p r e s e n t volume b e g i n s , t h e r e f o r e , w i t h Chapter 1 1 , s e c t i o n 76. T h i s does n o t imply t h a t e v e r y t h i n g i n Volume I i s needed f o r Volume 11. For t h e u n d e r s t a n d i n g of Chapters 12-20 i t i s s u f f i c i e n t t o b e f a m i l i a r w i t h Chapters 2-4 and p a r t of Chapter 6 ( F r e u d e n t h a l ' s s p e c t r a l theorem). Much of t h i s can a l s o b e found i n t h e f i r s t p a r t of Chapter I1 o f t h e book by H.H.
Schaefer ( [ I ] ,
1974) a s w e l l a s i n Chapters 1 and 4 o f t h e book by E . de Jonge and A.C.M.
van Rooij ( [ 1 ] , 1 9 7 7 ) .
,
Chapter 1 1 i s a n e x c e p t i o n ; i n t h i s c h a p t e r
w e d i s c u s s e x t e n s i o n s and g e n e r a l i z a t i o n s of what was proved i n Chapter 5 (and s e c t i o n 52 of Chapter 7) about prime i d e a l s i n d i s t r i b u t i v e l a t t i c e s and i n R i e s z s p a c e s . Chapter 12 c o n t a i n s a g e n e r a l i n t r o d u c t i o n t o o r d e r bounded o p e r a t o r s i n R i e s z s p a c e s . I n Chapter 13 t h i s m a t e r i a l i s immediately p u t t o u s e i n an i n v e s t i g a t i o n of k e r n e l o p e r a t o r s i n s p a c e s o f measurable f u n c t i o n s . T h i s c u l m i n a t e s i n A.V.
Buhvalov's n e c e s s a r y and s u f f i c i e n t c o n d i t i o n f o r a n
Operator t o b e a k e r n e l o p e r a t o r . Chapters 14-16, mainly about normed Riesz s p a c e s ( i n p a r t i c u l a r Banach l a t t i c e s ) have t h e i r r o o t s i n a s e r i e s of n o t e s in t h e Proceedings o f t h e N e t h e r l a n d s Academy of S c i e n c e (1963-65;Notes
V
1-13
vi
PREFACE
by W . A . J . Luxemburg and A . C . Zaanen, Notes 14-16 by W . A . J . Chapter 17, based p a r t l y on c l a s s i c a l work of H.F.
Luxemburg). I n
Bohnenblust and S. Kaku-
tan5 around 1940, we d e a l with a b s t r a c t L -spaces, i n p a r t i c u l a r with ALP spaces and AM-spaces. Chapter 18 i s b u i l t around the r e c e n t r e s u l t s of P.G.
Dodds and D.H.
Fremlin on compact operators i n Banach l a t t i c e s . I n the
f i r s t p a r t of Chapter 19 we discuss O r l i c z spaces. In the p r o p e r t i e s of these spaces, g e n e r a l i z a t i o n s of the f a m i l i a r L -spaces, we recognize
P
b e a u t i f u l i l l u s t r a t i o n s of much t h a t was observed about a b s t r a c t normed Riesz spaces i n the e a r l i e r chapters. The second p a r t of Chapter 19 i s devot e d t o band-irreducible
operators ( i . e . ,
o p e r a t o r s t h a t leave no band in-
v a r i a n t except t h e n u l l band and the entkre space). Generalizations Of c l a s s i c a l eigenvalue theorems of 0. Perron, G. Frobenius (matrices with non-negative
e n t r i e s ) and R. Jentzsch ( p o s i t i v e k e r n e l o p e r a t o r s ) a r e de-
rived. These g e n e r a l i z a t i o n s (and even some s t i l l more g e n e r a l r e s u l t s ) a l s o occur i n the abovementioned book by H.H.
Schaefer; our p r e s e n t a t i o n
i s somewhat d i f f e r e n t because we have made an attempt t o keep a l l proofs f r e e of any r e p r e s e n t a t i o n methods. The attempt i s almost s u c c e s s f u l , with one exception i n Theorem 1 3 6 . 8 . A representation-free
(and Zorn-free) proof
of t h i s theorem i s d e s i r e d . F i n a l l y , Chapter 20 on orthomorphisms ( i . e . , o r d e r bounded operators t h a t a r e band preserving) contains r e s u l t s t h a t a r e mostly of very r e c e n t o r i g i n . Some o l d e r r e s u l t s t h a t were f i r s t proved
(1969-71) by m a n s of r e p r e s e n t a t i o n theorems a r e now presented i n a
representation- f r e e manner .
The p r e s e n t a t i o n of the m a t e r i a l i n some of the chapters has been s i g n i f i c a n t l y influenced by the work of s e v e r a l of my s t u d e n t s ( e i t h e r t h e s i s work o r otherwise). This holds f o r A.R. p a r t of Ch. and B.
1 9 ) , W.J.
Claas (Ch. 17), W.K.
Schep (Ch. 13 and the second
Vietsch (Ch. l a ) , C.B.
Huijsmans
de P a g t e r (Ch. 2 0 ) , E. de Jonge ( s e c t i o n s 119 and 133) and J.J. Grob-
l e r ( s e c t i o n 129). Furthermore, many d e t a i l s i n the t e x t come from conv e r s a t i o n s w i t h s t u d e n t s and colleagues. I n p a r t i c u l a r , the f r e q u e n t cont a c t s with W i m Luxemburg over a p e r i o d of many years have been very import a n t , a s e v i d e n t a l s o from the b i b l i o g r a p h i c a l r e f e r e n c e s . Several books dealing with Riesz spaces have been published since 1971. There i s some overlap, b u t not too much, of the p r e s e n t book w i t h the book on Banach l a t t i c e s and p o s i t i v e o p e r a t o r s by H.H. book by E. de Jonge-A.C.M.
Schaefer ( [ 1 ] , 1 9 7 4 ) .
The
van Rooij, very s u i t a b l e a s an i n t r o d u c t i o n t o
t h e theory, is from 1977. In the books by D.H.
Fremlin ( [ 1 ] , 1 9 7 4 )
and
v ii
PREFACE
C.D.
Aliprantis-0.
Burkinshaw ([ 11,1978) t o p o l o g i c a l R i e s z s p a c e s , more
g e n e r a l t h a n n o r m d o n e s , are i n v e s t i g a t e d . The books by H.E. 1974) and by J. Lindenstrauss-L.
T z a f r i r i ([11,1979),
Lacey (c11,
both about classical
Banach s p a c e s , a r e r e l a t e d . The same h o l d s f o r t h e book on k e r n e l o p e r a t o r s by M.A.
K r a s n o s e l s k i i , P.P.
(C11,1976).
Zabreiko, E . I .
P u s t y l n i k and P.E.
F i n a l l y , Yau-Chuen Wong and Kung-Fu Ng (C1],1973)
Sbolevskii discuss
p a r t i a l l y ordered topological vector spaces. Thanks ame due t o Sonja Wassenaar who p r e p a r e d t h e typed m a n u s c r i p t and t o t h e s t a f f a t t h e North-Holland
P u b l i s h i n g Company f o r t h e i r
cooperation. December 1982
A.C.
Zaanen
T ABLE OF CONTENTS
v
PREl' ACE
CHAPTER 11. PRill£ ll>EAL EXTENSlO'
6
16. Prime ideal separ ation 77. The hull-ke-rnel t opol ogy and the d\L&l hull-kernel t.opol ogy
19
78 . The unique priae ideal extension property 79 . Strongly order dense Rieu subs~cea
46
80. Extent iO'Il theorcas
S2
81 • PrUDe ideal extension and t he p rojection property 82. Hor.al and extrenally disconne c ted lattice•
57
32
63
89
CHAPTER 12. ORDER BOUNDED OPEAATORS
95
83, Order bounded opera t or s 84. Order cone i nuoua o pe rators
123
65 . The orde r dual of a Riesz sptlcre
131
86 . Integrals o n ideals o f
me~s ura!ble
f unction•
81. lntearah tmd singular: l i near function•h
88 . The largect. ideal on which every order bo~ded oper4tor ie nrder continuous 89. Annihilator• and weak topologies 90. The carrier of an order bounded linear luncliooal
141 14 6 158
169 176
91 . CO.plex lie•z spaces
187
92. Coaplox order bounded operators
201 209
CHAPTER 1 3. KEMtL OPEAATORS
93. Kernel oper,tors 94. The b4nd ot absol ute kernel ope rators 9S . The b ~nd guncrutcd by the kor nel operators ot f in i t e rank 96 . 8uhva l ov •s
theo r~
91 . Adjoint opera tors
ix
21 2 214
228 234 248
TABLE OF CONTENTS
98. Dunford'• theorc.
257
99.
262
Ctnerali~•d ca~ leaan opera~ors
CIIAP'tl:R 14. NORIUID R IESZ SPACES
279
100. Normed Riesz s p~ces 10 1. Banach lattices
28 1
102. The Banach dual
310
302
CHAPTER 15 . ORDER CONTlNUOUS NORMS
333
10). OrdQr continuous •lornas
335
104.
Meycr-Ni~berg ' s 1~
and disjoint sequence•
353
105. Ando'a thcoreQ and the order ~opology 1 0~. Order continuity and tbe ordeY dual
372 381
CIIAPTEI\ 16. EI
107. Riea& of
L•
in
n
349
and order deniity of
t~
L-
385
n
about Riesz seainores 109. Embedding of L into the order biduol
394
110. Perfect R\eoz &paces
409
1 1 I. Perfect Banach l attices
4 13
11 2. 8taon.cb funct i on spaces
1115
I I 3. Tho Patou property
42 1
I tt.. Ecbedding in the Sanach bid ual; reflexivity
429
115. Adjoint operators
437
116. Di•joint stquences and order concinuout nonDt
44 1
11 7. Copiu
446
108.
~rc
404
455
CIW"l'ER 17. AliSUW:f Lp- SPACES
t18.
Rie•~
apaces with p- additive
456
no~
465
119. Seo.i-H-ap..aces
IZO. J.pretentation by a space of 12:1. Repretentation by a space of
~eaaurabl e (ontin~u•
(unctions functions
471
490 495
CIIAPTER I 8 . COHPACT OPERATORS
12:2. C.ompo.c: t tcu . pr ecompoact set a: and alaao.tt order boundl!d
499
seta 12 3. The bnnd of
AM-co~pact
opera~ors
124. 1'heortmt of Dodd&-i:'rem.l in andl A lipra n tis-Bur k int~ha.w
504 514
'[ABLE 01'
125.
Coc:~pac tness
CONIE~'TS
• AM- compactness and semi -compac tness
xi
524
126. Semi- compac t operators and or de r continuity of the operator
534
norm 127 . St!mi- compa c t operator s and d isj oin t sequences
5 44
128. Semi-compact operators and o r der project ions
550
1Z9. Scl'lli -compact operators and indi ces
556
C1~P ~ER
19 . ORLICl SPACES AND I RREDUCIBLE OPERATORS
571
t30 . Youn.g' s inequality
575
131 . Youog classes
579
132. The
3Ssoeia t~ s p~ ce
of
~n O rl ic~
space
586
133 . The Ban ach d ual of an Odicz space
593
134. The spectrum of an operator in Banach s pace
60 1
13$. The Krein- Rutma n theorem and the spe c tral radius
60$
136. Irreducible operators
613
137 . Spectra l p roper t i es o f
c ~pac t
i rreducib le o perators
13 8. The peripheral spectrum of irreducible pos it i ve op e rato rs CHAl'TER 20. OR·r nOM0Rl'1HSMS AND f - ALCEBRAS
139. &l ementary prope rc iea o f orthomorphi sms
140 . The s pace
Orth(L)
626 63~
645 648
651
141 . E~amp l es of orthomorphisms 142. Pcoperties of f- algobras
663 667
143. The stabi..li zer of a Rics2 spac e
676
1~ 4 .
681
Orthormorphisos in a normed Riesz s pace
145 . Theor ems of Radon- Nikodym type
683
146 . The range of an orthomorphism
692
6 1BLIOGRAPI!Y
70 3
SUBJECT INDEX
717
CHAPTER 1 1
PRIME IDEAL EXTENSION
In the present chapter we consider extensions and generalizations of certain results for prime ideals proved in Volume 1 . Readers interested primarily in the theory of linear mappings between Riesz spaces, in particular when it concerns normed Riesz spaces, may immediately proceed to Chapter 12. (X,e)
Throughout the whole chapter with smallest element
6
,
will denote a distributive lattice X
and L will denote a Riesz space. In section 76
we first introduce the notion of a dual ideal (also called a filter) in (X,O)
. This
x,y E F y E F
implies x
. The
at least of
A
y E F
dual ideal F x
and
y
e ,
X , not containing
is a non-empty subset of
x E F
and such that
is a member of
F
and y
x v y E F
is called prime if
2
x
such that implies
implies that one
. It follows easily that
prime dual ideal if and only if its set theoretic complement w
=
F
is a
X-F
is
a proper prime ideal. The definition in a Riesz space is similar but not quite the same. The non-empty subset F the zeroelement is not in F furthermore u,v E F+ and
ideal F
sup(u,v) E F+
X
=
L-w
implies af E X
is called a dual ideal if
if and only if
w
0
i
If1 E F ;
u 5 v
and
u E F+
is called prime if it follows from
that one at least of
Here we have the theorem that only if
f E F
and
L
implies inf(u,v) E F+ and
. The dual
implies v E F+ u,v E L+
of
u
and v
is a proper prime ideal in L
is in F+
is a prime dual ideal with the extra property that for all real a # 0
. The main
prime ideal separation theorem, stating that if
F is a lower sublattice in exists a prime ideal w
(X,e)
such that
such that
Ic
w
f E X
theorem in section 76 is the I
is a proper ideal and
I rl F
and w r l F
theorem i s due to G . Birkhoff (1933) and M.H.
.
if and
is empty, then there
is empty. This
Stone (1937),
and it was
proved again independently by several other mathematicians (J. Hashimoto, 1952; S . J .
Bernau, 1972). A similar theorem holds in a Riesz space. The
prime ideal extension theorem (cf. Theorem 51.8 for for L ) is an easy corollary.
(X,e)
and Theorem 52.2
CHAPTER I 1
2
In section 7 7 we consider the set A (X,e)
. For any
with
A
,
x E X
let
of all prime dual ideals in
(X:AEA,xEA)
=
. Thesets
, together
,
itself, form a base for the closed sets of a topology in A
the
hull-kernel topology. It will be proved that this is a T -topology such that
A
0
{AIX
and all
are compact. Denoting by
prime ideals, the mapping
X
correspondence between A
and
+
w = X-A
fi
fi
the set of all proper
establishes a one-one
. The hull-kernel topology
R ; the resulting topology in
transferred, therefore, to
in A
can be
is called the
8
dual hull-kernel topology. Some properties of this topology will be discussed. We also introduce the Boolean topology in fi ; this is the weakest topology stronger than the hull-kernel topology and the dual hull{wIx
kernel topology. The collection of all
{wjY , for x,y € X
and all
,
is a subbase for the open sets of the Boolean topology. (X,e)
In section 78 we first prove that every proper prime ideal in
X
has a unique extension to a prime ideal in the Boolean ring generated by (the Boolean ring generated by X
X
is the smallest Boolean ring containing
as a sublattice). The sublattice
(X',8)
(X,8)
of the distributive lattice
is said to have the monotone prime ideal extension property whenever
for every ideal satisfying I'
I' c w '
.
I' in X'
nX
and every proper prime ideal
u)
there exists a prime extension w '
cw
It was proved in Theorem 51.8 that X
extension property with respect to X'
if X
of
in X such that
w
has the unique prime ideal is an ideal in X'
now make this result more precise by showing that X
. We
shall
has the monotone and
unique prime ideal extension property if and only if X
is an ideal in X ' .
A certain special kind of sublattice, called a p-sublattice, has the
monotone prime ideal extension property. By way of example, every subring of a Boolean ring is a p-sublattice. A second example is obtained by considering the lattices X(L) Riesz spaces L L'
.
and
L'
In this situation X(L)
the ideal generated by
and X(L')
L
in L '
. Then L'
ideal P
.
.
L
IL be
Now, let
has the unique prime ideal
u' E L'
,v E
X(L) = X ( I L ) L
.
satisfying
there exists a number o, > 0 and an element u E L+ such that In this case, the unique extension of a given proper prime 2 u
0 < U' 2 v
IL
is a Riesz subspace of
if and only if
Explicitly, this means that for any pair U'
L
is a p-sublattice of X(L')
extension property with respect to
au 2
of all principal ideals in the
respectively, where
.
in L
to a prime ideal in
In the special case that L
IL
is the ideal generated by
is Archimedean and L'
P
in
is the Dedekind
ch. 1 1 1
completion of X(L)
3
PRIME IDEAL EXTENSION
and
L we have
X(L')
L'
=
IL ,
so
X(L')
. The
X(1L)
=
are equal is now exactly the condition introduced by
J.J. Masterson for different purposes (cf. Exercise 3 7 . 1 6 ) . that if
L'
condition that
is the Dedekind completion of
,
L
We shall prove
then the unique prime ideal
extension property is strictly weaker than Dedekind completeness of to the situation that (X,e)
(X,e)
L
(X',8) , we show that
is a sublattice of
has the unique prime ideal extension property with respect to
if and only if the Boolean ring generated by ring generated by
X
(X',f3)
is an ideal in the Boolean
. The results on unique prime
X'
and
. Finally, returning
strictly stronger than the projection property for L
extension are from a
paper by W.A.J. Luxemburg (1973) According to the definition in section 2 1 the ideal A Archimedean Riesz space L
is order dense whenever Add
extended to an arbitrary non-empty subset D dense whenever L
Ddd
=
L
.
of the Archimedean space
{u'Idd n L
. This can be
L
L ; the set D
of
is order
In section 79 we introduce, for a Riesz subspace L'
,
the notion of strong order denseness.
This means, by definition, that for every L'
=
in the
u' > 0
in L'
the intersection
does not consist only of the zeroelement. For an ideal L
this is the same asorderdenseness, but if
L
in
,
is not an ideal in L'
then strong order denseness is a stronger condition than order denseness.
It will be proved that for every
0 < v
5
u'
L
u' > 0 in L'
is strongly order dense in L' there exists an element v
if and only if
in L
such that
holds, and also if and only if for every u ' > 0 in L'
intersection AU, f l L
the
does not consists only of the zeroelement, where u'
denotes the ideal generated by
in L'
. This last result is due to
A. Bigard (1972). In section 80 we assume, for simplicity, that L' = IL (i.e., L Riesz subspace of
L'
such that the ideal generated by
itself). For any ideal I in L in L' L
.
in L'
is a is
is prime in L'
P'
in L
the ideal M'
is strongly order dense in L'
property, then P'
is prime in L'
the particular case that L'
and
L
L'
.A
=
IL
, L'
is
has the projection
for every prime ideal P
is the Dedekind completion of
L
in L
. For
, this result
(due in this case to J.J. Masterson) was already indicated in Exercise 37. 17.
I
in
if and only if for any minimal prime
is a minimal prime ideal in L'
sufficient condition for this is indicated, as follows. If Archimedean, L
L'
I' the ideal generated by
It is not difficult now to prove that for any prime ideal P
the ideal
ideal M
we denote by
L
4
CHAPTER 1 1
Ch. 1 1 1
In section 81 the results in section 80 are rounded off by the L' = IL
following main theorem. If
L
is strongly order dense in L'
L ), then P '
completion of
L
L
if and only if
L'
,
L'
has the projection property and
(all this holds if
is prime in L'
L'
is the Dedekind
for every prime ideal P
in
has the projection property. For the special case that
is the Dedekind completion of
L
, this
result is due to K.K. Kutty
and 3 . Quinn (1972).
In section 82 we return to the lattice
(X,e)
and we first prove
another prime extension theorem. Every proper prime ideal in a sublattice
Y X
of the form Y
.
The lattice
in X
=
(x:x>x ) 0
(X,e)
can be extended uniquely to a prime ideal in
is called normal whenever every proper prime ideal
contains a unique minimal prime ideal. It will be shown that X is d x A y = 0 implies X = {x) v {yjd , and also if
normal if and only if and only if X
holds for any pair of different minimal prime p2 d ideals. Similar conditions (i.e., L = {u} + {vId if u A v = 0 and
L
= p 1 + p2 )
(X,e,e)
= p
1 v
hold for normality of a Riesz space L
with smallest element
8
and largest element
.
The lattice e
is called
extremally disconnected whenever every proper prime ideal in X
is
contained in a unique maximal ideal. We shall prove that a topological space is normal or extremally disconnected (in the usual topological sense)
if and only if the lattice of its closed subsets is normal o r extremally disconnected in the sense as defined here. The lattice
(X,e)
is called completely extremally disconnected
whenever every principal ideal in X for a Riesz space L proper ideal in X
.A
(or
is extremally disconnected. Similarly
necessary and sufficient condition is that any
L ) containing a prime ideal is prime itself.
Another one is that the family of all proper prime ideals containing a given proper prime ideal is linearly ordered by inclusion. Every Riesz space is completely extremally disconnected, but not every lattice (X,e)
is called completely normal whenever every sublattice Y
Y = (x:x?x ) 0
. Finally,
(X,e)
of the form
is normal. A necessary and sufficient condition is that the
family of a l l prime ideals contained in a given proper prime ideal is linearly ordered by inclusion. Another condition is that X = w 1 V w 2 holds for any pair of incomparable proper prime ideals. Complete normality in a Riesz space is defined by these last conditions. Finally, if
(X,e)
i s normal and completely extremally disconnected, then X is completely norma.1. Similarly for a Riesz space. Hence, any normal Riesz space is
Ch. 111
5
PRIME IDEAL EXTENSION
completely normal. It will be shown by examples that all situations considered in the theory can really occur. Some of the results in this section are due to A.A. Monteiro (1954) and some (independently) to W.H. Cornish (1972). It is not without some importance, finally, to observe that the analogy with commutative rings goes still farther than indicated in section 52 (52.6,
52.14 and 52.15) and section 78 (78.16 and 78.17).
I n particular,
regular semi-prime rings are of interest in this respect (the commutative there exists an element r E R
is regular if for every x E R
ring R such that
x
2
x r
=
and R
R has no non-zero
is called semi-prime if
nilpotent elements). The following theorems hold. A. R
is regular if and only if R
ideal I in R
is semi-prime and every principal
is a direct summand (i.e., R = I
B. R is regular if and only if R in R
J
@
for some ideal J ) .
is semi-prime and every prime ideal
is a maximal ideal.
C. If R
is semi-prime and has a unit element, then the following
conditions are equivalent. (i)
Every proper prime ideal in R
is a maximal ideal.
(ii)
Every proper prime ideal in R
is a minimal prime ideal.
(iii) The base sets
{PIr of the hull-kernel topology in the set P
of all proper prime ideals are open and closed. (iv)
The hull-kernel topology in P
(v)
The hull-kernel topology in
(vi)
R
P
is Hausdorff. is a T -topology. 1
is regular.
D. If R
is semi-prime and has a unit element, then the following
conditions are equivalent. (i)
Every proper prime ideal in R
contains a unique minimal prime
ideal. (ii)
Every maximal ideal in R
(iii) R
=
prime ideals. (iv)
M 1 + M2
If r,s E R
(v)
If
and
and M2
rs = 0 , then R
=
of different minimal
+
N(r)
r (i.e., N(r)
=
(x:rx=O) ) .
r,s E R , then N(rs)
=
N(r) + N(s)
is the annihilator of Compare C
contains a unique minimal prime ideal.
for any pair MI
, where
N(r)
.
and D with Theorems 37.6 and 82.5-82.7.
details we refer to C.B. Huiismans ("21, 1974).
N(s)
For proofs and further
[Ch. 11,576
PRIME IDEAL SEPARATION
6
76. Prime i d e a l s e p a r a t i o n A s i n Chapter I ,
element subset
8
. We
of
S
be a d i s t r i b u t i v e l a t t i c e w i t h s m a l l e s t
not containing
X
implies t h a t
x,y E S
X
let
s h a l l i n d i c a t e t h i s by x
d e f i n e d i n s e c t i o n 5 , any
i s c a l l e d a lover s u b l a t t i c e of
8
. According t o t h i s d e f i n i t i o n of X . The lower s u b l a t t i c e S
y E S
A
s e t i s a l s o a lower s u b l a t t i c e
muxima2 whenever
. As
(X,e)
X
if
t h e empty is called
i s n o t p r o p e r l y included i n any o t h e r lower s u b l a t t i c e .
S
D
The non-emuty s u b s e t
of
...
...,
.
x,, x E D that x A A x # 8 n 1 lower s u b l a t t i c e has t h e f i n i t e i n t e r s e c t i o n
property i f i t follows from Obviously, any non-empty
i s s a i d t o have t h e f i n i t e i n t e r s e c t i o n
X
property. There i s another fundamental n o t i o n which we s h a l l d e f i n e now. The subset
F
of
i s c a l l e d a dual i d e a l o r a f i l t e r i f
X
F
s a t i s f i e s the
following c o n d i t i o n s : (i)
F
i s non-empty,
(ii)
8
i s not i n
( i i i ) x,y E F (iv)
x E F
,
F
implies and
x
S
x y
y E F
A
implies
,
.
y E F
I n o t h e r words, a dual i d e a l i s anon-empty property ( i v ) . In algebraic discussions
lower s u b l a t t i c e w i t h t h e e x t r a F
i s usually called a dual ideal,
b u t i n s e t theory one more o f t e n speaks about a f i l t e r . Note t h a t t h e i n t e r s e c t i o n of a family of dual i d e a l s i s e i t h e r empty o r it i s again a dual ideal.
F
The dual i d e a l
i s c a l l e d maximal whenever i t i s n o t p r o p e r l y i s a maximal dual i d e a l i f
included i n any o t h e r d u a l i d e a l . Obviously, F and only i f
F
i s a maximal lower s u b l a t t i c e .
The d u a l i d e a l
i s c a l l e d a prime dual i d e a l i f
F
t h a t one a t l e a s t of
x
and
y
. Proper
F
i s a member of
x v y E F
implies
prime i d e a l s and
prime d u a l i d e a l s a r e r e l a t e d , a s shown by t h e following theorem. THEOREM 76.1.
The proper i d e a l
s e t t h e o r e t i c complement
F = X-I
minima2 prime idea2 i f and only i f
I in
X
i s prime i f and only i f i t s
i s a prime dual i d e a l . Hence, I F = X-I
is a
i s a maximal dual i d e a l (ef.
Theorem 5.4 ( i ) ) . PROOF. We sketch one h a l f of t h e proof. Let w i t h complement
F
. Evidently,
F
I
i s n o t empty and
be a proper prime i d e a l
8
i s not i n
F. I f
Ch. 11.§761
,
x,y E F (because
then
x
that
x
x,y E I
and
x E I (because
would imply
y E I
A
x
7
would imply 5
y
implies
x E I
or
,
y E F
y E I
since
i s an i d e a l ) . This shows a l r e a d y
I
i s a dual i d e a l . Finally, x
F
since
, since
y E F
A
i s prime). Also, x E F
I
would imply
y E I
F
PRIME IDEAL EXTENSION
y E F
V
x v y E I
implies
(because
or
x E F
,
y E F
i s an i d e a l ) . Hence,
I
i s prime. If
g e n e r a t e d by
i E I
f o r some
X
i s an i d e a l i n
I
(I,yo)
and
I
. Note
n o t c o n t a i n i n g t h e element yo
that
c o n s i s t s of a l l
(I,yo)
is x -maximal
I
contained i n
(I,yo)
f o r any
yo
not i n
( i t ) t h a t any x -maximal i d e a l i s prime. 0 Dually, i f F i s a dual i d e a l i n X denote t h e s e t of a l l
(F,yO)
x
satisfying
i s e i t h e r t h e whole s e t
x
, and
so
A
yo
A
i v y
S
s a y now
0
i s maximal
I
xo
,
yo
f o r some
yo = 8
x
is
was proved i n Theorem 5.1
not c o n t a i n i n g f
2
X (if f
xo
the ideal
. To
X
means t h a t
I
. It
I
,
yo
satisfying
can be t h e whole s e t
f o r a given xo n o t i n 0 w i t h r e s p e c t t o t h e p r o p e r t y of not c o n t a i n i n g that
x
f o r some
let
f E F
(F,yO)
. Then
f E F )
or
i t i s a g a i n a d u a l i d e a l which i s t h e n a p p r o p r i a t e l y c a l l e d t h e d u a l i d e a l and
yo
. The d u a l
i d e a l not containing
xo
.
F 'is now c a l l e d x -maximal f o r a 0 n o t i n F whenever xo i s c o n t a i n e d i n (F,yO) f o r any y 0 I n o t h e r words, F i s not p r o p e r l y included i n any o t h e r d u a l
generated by given
xo
F
not i n
F
.
ideal
A e-maximal d u a l i d e a l i s t h e same a s a maximal
dual i d e a l . We g e n e r a l i z e t h e d e f i n i t i o n s of an x -maximal i d e a l and an x -maximal 0 0 I and t h e non-empty lower s u b l a t t i c e F such
dual i d e a l . Given t h e i d e a l that I
I
n
F
i s empty, we s h a l l s a y t h a t
I
i s an F-maximal idea2 whenever
i s n o t p r o p e r l y included i n any o t h e r i d e a l
empty. Note t h a t i f
such t h a t
i s t h e d u a l i d e a l g e n e r a t e d by
F'
F ' = (y:y2x f o r some x E F) ), t h e n i s F'-maximal.
J
I
J
n
F
is
F (i.e.,
I
is t h e F-maximal i f and only i f
I n o t h e r words, we may j u s t a s w e l l assume t h a t
F
is a dual
i d e a l , which makes t h e d e f i n i t i o n more symmetric. This becomes even more e v i d e n t i f we compare t h e d e f i n i t i o n of an F-maximal
i d e a l w i t h t h e now
f o l l o w i n g d e f i n i t i o n of an I-maximal d u a l i d e a l . Given a g a i n t h e i d e a l and t h e dual i d e a l
F
such t h a t
I-moxima2 dual i d e a l whenever ideal
G
such t h a t
I
n
G
F
I
n
F
i s empty, we s h a l l say t h a t F
1 is an
i s not p r o p e r l y i n c l u d e d i n any o t h e r d u a l
i s empty.
PRIME IDEAL SEPARATION
8
Let
THEOREM 76.2 (i)
X
any F-maximal idea2 i n
be a non-empty lower s u b l a t t i c e i n
F
i s prime.
.
(ii) Let 1 be a proper i d e a l i n X i n X is prime. (iii) Let
idea2 and x
.
y E I
A
For
I
, there
.
X
Then
Then any I-maximal dual i d e a l
.If
X
I be a proper i d e a l i n
is not i n F
x
Ch. 11,5761
i s an I-maximal dual
F
y E F
e x i s t s an element
such t h a t
{ e l t h i s i s the weZZ-known f a c t t h a t if F i s a x i s not i n F , there e x i s t s y E F such
=
maximal lower s u b l a t t i c e and that
x
A
y
PROOF.
.
e
=
(i) Let P
xo (cf. Theorem 5.1 (ti)).
, not in
z
P
. The proof
be an F-maximal ideal in X
prime is almost exactly as in the case that F If
P
but such that y
that P
is
consists of only one point
is not prime, there exist elements y
E P
z
A
. The
P
ideal generated by
contains at least one point of the dual ideal F'
.
F
generated by
and
and
,
so
y
.
p, v y E F ' for some p1 E P Similarly, p2 v z E F ' for some p 2 E P Setting p3 = p1 v p 2 , we have p E P with p v y and p3 v z in F' 3 Hence
On the other hand
. Hence
P
P n F'
p3
and
p3 v (yhz)
is in P
(F,y)
at least one point of 2
A
z E I for some
f3
A
y
f 3
and
. Hence, f3
It follows that (iii) If
f3
(yvz)
A
x
F
I ,
so
fq E F in
z
A
On the other hand F
so
P
,
z
. Contradiction, since
is prime.
not in F
I
and
.
f
I
A
.
v
f3
for some
.
,
so
f3
is not prime, y v z E F
f E F
f, E F
= f l A f2
are in F
z
is in I n F
If F
but such that y
A
y E I for some
Setting
. Hence
y
f
v (yhz) is also in
p3
as well as in F'
of all elements majorizing
and
,
be an I-maximal dual ideal in X
there exist elements y
f
are in P
z
A
is empty. It follows that
(ii) Let F set
y
. Similarly,
, we have f3
A
(yvz)
Contradiction, since
. The
contains E F
with
is also in i s empty.
I I7 F
is prime. is not in the I-maximal dual ideal F , the set
contains at least one element of
I , i.e., x
A
.
y E I
f o r Some
(F,X)
y E F
.
Ch. 11,5761
PRIME IDEAL EXTENSION
9
The prime ideal separation theoremwhich follows now is an immediate consequence. The theorem occurs in the famous paper by M.H. Stone ([11,1937, Theorem 6) on distributive lattices. The result stated in the theorem is essentially due to G. Birkhoff ([l1,1933,Theorem 2 1 . 1 ) .
This result has
been rediscovered, in one form or another, by several other mathematicians (for example, J. Hashimoto [11,1952; S . J . Bernau C21,1972). We first recall
I is given, then (by section 5) the prime ideal M
that if the ideal
is said to be I-minimal if any prime ideal M' satisfies M'
=
M
THEOREM 76.3 (Prime ideal separation). I f
and
F
i s a lower sublattice i n
e x i s t s a prime ideal
w
in
X such that
I
(l
F
and
Ic w
also
i s a proper idea2 in X
I
such that
X
I c M' c M
satisfying
.
I
2
i s empty, then there w
n
F
i s empty.
Actually, there e x i s t s an F-maximal prime ideal satisfying these conditions, m d there also e x i s t s an I-minimal prime ideal satisfying these conditions. PROOF. Consider the set of all ideals J
such that
J
3
I and J n F
is empty. It follows from Z o m ' s lemma that this set has a maximal element w
. By
the preceding theorem w
ideal containing
I
.
is prime, i.e., w
Consider the set of all dual ideals G
is an F-maximal prime
such that G
3
F
and
I
n
G
is empty. It follows from Zorn's lemma that this set has a maximal element
A
. By
the preceding theorem X
ideal containing F such that
is prime, i.e., A
. It follows that
w =
X-A
is an I-maximal dual
is an I-minimal prime ideal
n F is empty.
w
Stone's original formulation of the theorem is as follows. Given the proper ideal
I and the dual ideal F
exists a prime ideal w w a I
and
such that
w
such that
I n F is empty, there X = X-w satisfy
and its complement
A 2 F .
The prime ideal extension theorem, already proved in Theorem 5 1 . 8 , may be regarded as an immediate corollary of the prime ideal separation theorem. THEOREM 76.4 (Prime ideal extension). I f
d i s t r i b u t i v e Lattice (X','d) there e x i s t s a prime ideal
w'
and i f w in X'
i s a sub2attice of the i s a proper prime ideal i n X , such that w ' n x = w (X,t?)
.
Ch. 11,5761
PRIME IDEAL SEPARATION
10
F = X-w
PROOF. The s e t generated i n
I
x' 5 p
by
X'
f o r some
i s a lower s u b l a t t i c e . Furthermore, t h e i d e a l
w (i.e.,
t h e s e t of a l l
p E w ) satisfies
x ' E X'
,
I 7 I X = w
satisfying
I ll F
so
i s empty. Hence,
by t h e prime i d e a l s e p a r a t i o n theorem, t h e r e e x i s t s a prime i d e a l such t h a t
X' and
w'
ll (X-w)
w'
Note t h a t
w'
= w'
w ' ll F
and
I
3
i s empty. It follows from
that
ll F = pi
ll X
w'
. This
= w
can be chosen F-maximal o r I-minimal
=J
0
element
of
S
i f desired.
. The
lower s u b l a t t i c e
. As
L
not c o n t a i n i n g t h e zero-
L+
L
i s c a l l e d a lower s u b l a t t i c e i n
inf(u,v) E S
13w
concludes t h e proof.
We b r i e f l y d i s c u s s t h e corresponding n o t i o n s i n a Riesz space d e f i n e d i n s e c t i o n 3 3 , any s u b s e t
in
w'
w'
if
u,v € S
implies
i s maximal i f it i s n o t p r o p e r l y
S
included i n any o t h e r lower s u b l a t t i c e . We a l s o i n t r o d u c e t h e n o t i o n of a
F
dual i d e a l . The s u b s e t
L
of
i s c a l l e d a dual i d e a l o r a f i l t e r i f
F
s a t i s f i e s t h e following c o n d i t i o n s : (i)
F
i s non-empty,
(ii)
0
is not i n
( i i i ) u,v E F+
0 < u
(iv)
S
F
implies
v
and
f E F
and u
Obviously t h e p o s i t i v e p a r t
i n f ( u , v ) E F+
E F+ implies
, v E F+
,
If1 E F
i f and only i f
.
of a d u a l i d e a l i s a non-empty
F+ = F ll L+
lower s u b l a t t i c e w i t h t h e e x t r a p r o p e r t y ( i v ) . The d u a l i d e a l
F
is
maximal whenever it i s n o t p r o p e r l y included i n any o t h e r dual i d e a l . Obviously, F
i s a maximal d u a l i d e a l i f and only i f
lower s u b l a t t i c e . The dual i d e a l follows from is i n
F+
.
u,v E L+
and
i s a maximal
F+
i s c a l l e d a prime dual idea2 i f i t
F
u v v E F+
t h a t one a t l e a s t of
u
and
v
I t can be proved s i m i l a r l y a s i n Theorem 76.1 t h a t t h e complement
X
= L-w
of a proper prime i d e a l
property t h a t if
X
then
F
f E X
implies
i s a prime dual i d e a l w i t h t h e e x t r a
w
af E
i s a prime dual i d e a l such t h a t w = L-X
f E A
Hence, any maximal dual i d e a l
a J: 5
.
Given t h e i d e a l
171 F
implies
. Conversely,
af E
X
for
o
# 0 ,
i s a proper prime i d e a l . It was proved i n Theorem 3 3 . 7 t h a t
i s a maximal d u a l i d e a l i f and only i f
any
a # 0
f o r any r e a l
X
I
F
M = L-F
i s prime and
and t h e non-empty
i s empty, w e say t h a t
I
i s a minimal prime ideaL
f E F
implies
lower s u b l a t t i c e
F
i s a n F-mmimaZ i d e a l whenever
p r o p e r l y included i n any o t h e r i d e a l
J
such t h a t
J
n
F
af E F
for
such t h a t
I
is n o t
i s empty- This
i s t h e same d e f i n i t i o n as i n t h e l a t t i c e case. The proof t h a t any F-maximal
Ch. 11,1761
PRIME IDEAL EXTENSION
11
i d e a l i s prime i s a l s o t h e same a s i n t h e l a t t i c e case. Given t h e i d e a l we say t h a t
F
and t h e dual i d e a l
I
such t h a t
F
i s an I-maximal dual i d e a l whenever
included i n any o t h e r dual i d e a l
such t h a t
G
n F i s empty,
I
i s not properly
F
i s empty. A maximal
I fl G
dual i d e a l i s t h e same a s a 10)-maximal dual i d e a l , where
{O} i s t h e
i d e a l c o n s i s t i n g of t h e zeroelement only. The next theorem, s t a t i n g t h a t dual i d e a l i s prime, i s now analogous to p a r t s ( i i ) and ( i i i )
any I-maximal
of Theorem 76.2. THEOREM 76.5. Let
dual i d e a l and 0
S
v E F
is prime, and
F
. ALSO,
a # 0
i f
u
i s not i n
PROOF. It w i l l be s u f f i c i e n t t o show t h a t the complement
I c P
a prime i d e a l . It i s evident t h a t and
0
5 u S
v
with
it is clear that i f
that
inf(u,w) E I
u+v E P
. By what
such t h a t
v E P u
2
. Now,
u
i s not i n let
u,v
2
,
f
E
F
and
, there
, we
have
w3 E F+
since
are in
I
w,
and
inf(u+v,w ) s inf(u,w ) + inf(v,w3) E 1 3 3 so
u+v
i s not i n
inf(u+v,w ) 3 consequence
F ( s i n c e otherwise
F
i n common). It follows t h a t
P
and
I
u+v E P
i s an i d e a l , and e v i d e n t l y
P
S
. We
and
. Hence,
inf(u,w ) 3
with
F
0
exists
u,v E P
0 with
inf(v,w2)
f E F
P = L-F
i f and only i f
P
. Also,
E P
was j u s t observed, t h e r e e x i s t
inf(u,wl)
w3 = inf(w ,w ) 1 2 B u t then
0
implies
for
, there e x i s t s an element
F
.
I-maximal
af E F
has the property t h a t
F
0
2
inf(u,v) E I
such t h a t
. Then m y
be a proper i d e a l i n L
I
is
,
If1 E P
is Imaximal, such
w E F
show t h a t
w2
in
F+
writing inf(v,w ) 3
in
I.
,
would have t h e element
. As
an immediate
i s then prime.
The prime i d e a l s e p a r a t i o n theorem and t h e prime i d e a l extension theorem follow now, e x a c t l y a s i n t h e l a t t i c e case. The analogy between i d e a l s ( i n p a r t i c u l a r p r i m e i d e a l s ) i n a d i s t r i b u tive lattice
(X,e)
and i n a Riesz space
t h e following remarks. Given a Riesz space
a l l principal ideals i n
L
. Let
4
L
i s made even more evident by
L
, let
X(L)
be t h e mapping from
be t h e s e t of L
onto
X(L)
f E L
assigning to each
. Note
$ ( f ) = Af
satisfying A g
=
the principal ideal Af
that the inverse image $-'(Af) Af
. For
$(inf(u,v))
=
$(sup(u,v))
=
where AU+Av
is the algebraic sum of
It follows easily that $ onto Q(x
g E L
AU
and
A
V
and A
V
=
A
0
(note that AU+Av
is the
) . Evidently, X(L)
is
as smallest element.
= {O}
maps, in a one-one manner, ideals of
L
onto
, maximal ideals onto maximal ideals, x0 -maximal ideals
X(L) 0
f, i.e.,
Au n Av ' $(u+v) = AU + A V '
smallest principal ideal containing AU
ideals of
generated by consists of all
u,v E L+ we have
therefore a distributive lattice with $ ( O )
)-maximal ideals and prime ideals onto prime ideals. Also, rhe
of a lower sublattice F
image $(F) X(L)
Ch. 11,5761
PRIME IDEAL SEPARATION
12
and the inverse image Q-I(G)
is a lower sublattice in L prime dual ideal in X(L)
in L
is a lower sublattice in
of a lower sublattice G
in X(L)
. Finally, note that the inverse image of
is a prime dual ideal in L
a
containing any
non-zero multiple of each of its elements.
The analogy is not complete, however. In theorem 52.4 it was proved that prime ideal extension for Riesz spaces is monotone in the following sense. If
Q,
and
are proper prime ideals in the Riesz subspace K
Q,
satisfying Q, c Q ,
the Riesz space L
extension P, , then Q,
and if Q ,
has a prime extension P2
of
has a given prime ideal
.
satisfying P, c P,
A similar result does not always hold in the lattice case. Indeed, let w1
and
be proper prime ideals in the distributive lattice
w,
such that w I
is properly included in w2
. Assume, furthermore, that
is embedded as a sublattice in the Boolean ring element i3 w;
in
. Then
B , but
w1
and
w ' c w;
1
w2
(X,e)
X
B with the same smallest
can be extended to prime ideals w '
1
and
is impossible because prime ideals in a Boolean
ring are maximal ideals and
so
one maximal ideal cannot be properly included
in another one. That a distributive lattice is embedded in a Boolean ring (even i n a Boolean algebra) is exactly the situation that occurs in the Stone representation theorem, where a given distributive lattive embedded in the Boolean algebra of all subsets of the set i2 prime ideals in X
. For a
(X,e)
of all proper
simple explicit example we refer to Exercise
76.14 at the end of the present section.
is
Ch. 11,5761
13
PRIME IDEAL EXTENSION
EXERCISE 76.6.
The subset
of the commutative ring
S
a rnuZtipZicative s e t if the zeroelement is not in
. Show
xy E S
that if
S
S-maximal ideal in R and
R
is called
x,y E S
and
implies
is a non-empty multiplicative set, then any
is prime. Next, show that if
exists a prime ideal P
in R
I
is an ideal in R
I n S
a non-empty multiplicative set such that
S
S
is empty, there
I c P and P n S
such that
is empty.
This prime ideal separation result is known as Iirull's lemma. EXERCISE 76.7.
The proper ideal
I in the distributive lattice (X,e) (Ia:a€{a)) of ideals
is called completeZy p r i m e whenever for any family it follows from I = n I
that
I
=
I
ideal I is completely prime if and only if EXERCISE 76.8.
Show that an
space as in earlier chapters. The set
. The set
is not in P
is, therefore, the set of all
{PIu
in the Riesz space L {PIu
such that the given element
is assumed to have the (relative)
hull-kernel topology. We recall that, by Corollary 3 5 . 4 , the kernel d Now, let satisfies k({P} ) = (AU) = {uId k({P)U) = n (P:PE{P}u) {PIu
a subset of
.
{PIu
N
w >
o ,
. We have
Po
. Write
w
=
kfl)
=
N n {PIw
EXERCISE 76.9. by
Q,
k(QU)
=
{ u p . Show that
and let {PIv n {PIu
. Since o,
w >
{uId
.
Let
,
{PIw
=
h'
N be
is dense in
be a (relative) neighbor-
i.e., w
{PIv n {PIu is not in
that there exists P E
is non-empty, we have
{uld
.
It follows, on
N not containing w ,
u > 0 be given in the Riesz space L
the set of all u-maximal ideals in L =
.
to show that this neighborhood contains a point of
inf(u,v)
inf(w,u)
so
account of i.e., P E
satisfying k(N) =
Po E {PIu
HINT: Let hood of
.
,
and denote
Show that the kernel
n (Q:Q€Qu) consists of all f E L such that either inf(lf1,u)
(i.e., f E {ujd) Hence, k(QU) Show that
=
Qu
.
I is x-maximal for some x
We use the same notations for prime ideals in a Riesz
proper prime ideals P u € L+
.
for some a E {a)
or
inf(lf1 , u )
is infinitely small with respect to
{ u } ~ holds for all u > 0 if and only if is dense in
{PIu
for every
u > 0
=
u
.
0
L is Archimedean.
if and only if
L
is
Archimedean. HINT: The intersection Q' = Q fl A, principal ideal A
is an ideal i n
A,
of an u-maximal ideal Q that does not contain u
.
and the Q'
is
14
PRIME IDEAL SEPARATION
u-maximal in AU
.
Ch. 11,9761
Indeed, if not, there is a properly larger ideal Q"
in
AU that is u-maximal. Hence, Q" is a prime ideal in A, , and so can be This extended (even uniquely by Theorem 52.2 (ii)) to a prime ideal in L prime ideal is properly larger than Q diction. Hence, Q'
is u-maximal in A, , i.e., Q'
AU
the unique extension, every u-maximal ideal
. Conversely, by
can be extended to a u-maximal ideal Q 0
5
v E Q
consists of the zeroelement and all
.
with respect to u Qu
inf ( I f I , u )
such that
is dense in
{PIv fl {PIu with
{PIu
f E AU
is a maximal ideal in
. By
Q fl AU
=
=
0 or
Q'
in Au
Theorem 52.2 (ii),
,
for all w E A,
Now, observe that
Q'
fl
that are infinitely small with
It follows that k(QU)
u (cf. Theorem 27.5).
f E L
all
.
inf(v,u) E Q'
if and only if
respect to
in L
inf(v,w) E Q'
holds if and only if
i.e., v E Q
. . Contra-
and does not contain u
inf ( 1 f 1 , u )
fl
=
Q consists of
is infinitely small
if and only if every relative open set
inf(v,u) > 0 contains at least one element of
'
QU
i.e., if and only if there i s at least one u-maximal ideal not containing the element
inf(v,u) > 0
follows that Qu k(Qu)
=
{uId
is dense in
for every
EXERCISE 76.10. by
Qu,v
. Note
Let
u > 0
that
{PIu
.
0< u 5 v
.
that are u-maximal as well as k(Q
'
U,V
) = {uId
HINT: We have to show that for any w > 0 not in an ideal Q {vId c {uId
{,Id
.
Hence, since w
is not in
{uId
is not in
,
{vId
.
Since Q
EXERCISE 76.11.
is an ideal, Q
and
w
Q
. But
not in
not containing nor w
.
We generalize the preceding exercise. Let u,v
set of all u-maximal ideals in L
, and
that
then, by the
contains neither u
positive elements in the Archimedean space L
is
u,v
. Note
i.e., inf(w,u)
preceding exercise, there exists a v-maximal ideal Q inf(w,u)
that are
and
{u}~ there exists
that is v-maximal and does not contain u
, it follows that inf(w,u)
. Denote
in the Archimedean space L
. Show that
v-maximal and do not contain u
. It
{uId
is the set of all ideals in L
v-maximal. Equivalently, Q,,,
{PIu
is not in
for every u > 0 if and only if
the set of all ideals in L
dense in
inf(v,u)
denote by
that do not contain v
.
S
UIV
be the
Note that
S may be empty .(for example, if u I v ) . Show that u,v k(SU,v) = {inf(u,v)Id , where it is understood that the kernel of the empty
Ch. 1 1 , 5761
L
set is
15
PRIME IDEAL EXTENSION
. Show that
S
is dense in
u,v
and HINT: If u I v , then S u ,v therefore, that w = inf(u,v) > 0 , so
"'inf(u,v)
*
{PIinf (u,v) 0
<
w
5 u
are empty. Assume,
. Observe now
that
(i.e., the set of all u-maximal ideals not containing v ) is the
S
U,V same as
Qu,w
(i.e., the set of all u-maximal ideals not containing w ) .
EXERCISE 76.12.
Let
u > 0 and v t 0 in the Archimedean space L
,
R the set of all u-maximal ideals Q such that v is u,v contained in any ideal properly larger than Q (it is possible that v is
and denote by
already contained in Q ) . Show that k(R in
. The result that
{PIu
k(Ru,v)
=
U,V
{uld
) = {uId and R is dense u,v is due to A. Bigard and
K. Keimel (Cl1,Lemma 4 ) . R is the set of all u-maximal ideals Q such that u,v is contained in any ideal properly larger than Q , i.e.,
HINT: Note that w
=
sup(u,v)
R u,v
is the set of all ideals that are u-maximal as well as w-maximal. EXERCISE 76.13.
Show that
Let F
inf(lf1:fEF)
HINT: If
L
= 0
.
be a maximal dual ideal in the Riesz space I,
.
is Archimedean, this is trivial since n-lu E F+
for any
,... . In the general case, assume that u > 0 is a . There are two cases: u in F+ or u not in F+ . If u E F+ , then u = inf(v:vEF+) . But this is impossible since 1. < u and 1. is still in F+ . If u is not in F+ , then inf(u,v) = 0 for some v E F+ . But u 5 v since u is a lower bound of F+ . Hence u = inf(u,v) = 0 , contradicting u > 0 . It follows that u = 0 is the only lower bound of F+ in L+ . u
E F+ and n
= 1,2
lower bound of
F+
EXERCISE 76.14. different elements
Let X
be the distributive lattice consisting of five
(f3,x0,x1,x2,e) with
8
as smallest element, e
largest element, and furthermore xo < x 1 and page 16). Show that
w
=
as
xo < x2 (see the diagram,
{ e l is a prime ideal. Show that there are two
0 more proper prime ideals w I
and w 2
such that neither
w 1 = w2 nor Show that by adding yo,yl and y2 as in the diagram, X is 2 c w1 embedded in (obviously) the smallest Boolean algebra B(X) containing X
w
.
Determine the prime ideals in B(X) ideal in B(X)
~
and observe that, although X
each of the prime ideals in B(X)
.
is not an
is a unique extension of
PRIME IDEAL SEPARATION
16
.
one of the prime ideals in X wo
F
EXERCISE 76.15. Let
f E F
extra property that C
Note that
is not included in the extension of
= F+
, but
wo c w 1
the extension of
.
w1
be a dual ideal in the Riesz space implies af E F
. Define
is a cone in L
U {O}
Ch. 11,5761
f
05
g
to mean
that, with respect to this partial ordering, (i)
C
L with the
. Show that that g-f E C . Show
a # 0
for all
is a generating cone
L , (ii) L has the Riesz decomposition property, (iii) L is an anti-lattive (i.e., sup(f,g) exists if and only if f 20 g or f 0 5 g for
0 05 u 05 vl+v2 with
HINT: For (ii), let u = o
and
, v1
0
=
p and u
5
, v2
= 0
or
are all in F+
v +v -U 1 2
=
v1,v2 E C
. The cases
u = v1+v2 are trivial, s o assume that
. Then
).
that
u,vI,v2
inf(v +v -u,iu,ivl,iv2) E F+ I 2
v +v -p , so 1 2 u-p
5
.
(v,-p) + (v -p) 2
By the ordinary Riesz decomposition property in L
there exist elements
q 1,q2 such that
0 Hence, u 1
=
2
q 1 5 v -p
q,+ip
1
and
,05
u 2 = q2+:P
u = u1+u2 with
Since
~ , ~ ~ 2 ” J I -and UI
q2
ip
v2-u2
5
v2-p
5
and
u-p = q1+q2
satisfy uI
5
vI-ip and
fp
5
u 2 5 v2-iP
are all greater than or equal to
4P
,
all
Ch. 11,1761
PRIME IDEAL EXTENSION
these elements are in F+
. Hence
For (iii), bound of
assume that
u1 and
is equal to one of
u1
or
v
05
05
u1
v1 and
05
and
.
u2 0 5 v
0 C
are in the cone
u1,u2
, so ul
u2
0
17
and
u
05
v
.
v2
is an upper
Prove that, unless v
u2 0s u l
u2 (in which case
05
or
u1
05
u2 ),
there is a properly smaller upper bound, for example v-fp for p
=
.
inf(v-ul,v-u2) EXERCISE 76.16.
In Exercise 15.14 we presented an example of an
ordered vector space with the Riesz decomposition property, but not a Riesz space. Here is another example. Let
-
L
be the ordered vector space of all
real differentiable functions on the interval that
a
=
- m
and (or)
b
=
difference if we take the interval decomposition property, but
. It
(x:a<x
is allowed
, and for a and b finite it will make
.
(x:aSxsb)
no
L has the Riesz
Show that
is not a Riesz space. This is Problem 5882
L
in the Am. Math. Monthly (1972), due to E.S. Langford. HINT: Given that u1+u2 ;nvl;2 existence of C.W.. = v 1 1~
w.. (i,j=1,2) 1J
(j=1,2)
j
define w.. 1J
.
13
.
such that
Set u 1 + u 2
w.. (x)
by
=
, we
in L+
= u.
1
(i=1,2)
and
in Exercise 15.14,
if w(x) # 0 and w. .(x)
u. (x)v.(x)w-'(x) 1
C.w J ij
. Exactly as
v1+v2 = w
=
have to prove the
J
1J
=
0
if w(x) = 0 It remains to prove that w is everywhere differentiable. ij Hence, let If w(xo) # 0 , then w ij is differentiable at xo a < x < b and w(xo) = 0 , s o ui(xo) = v.(x ) = 0 and u!(x ) = 0 and
.
0
v!(x ) J o w(xo+h)
=
0 for
# 0
i,j
, we 0 2
wij(x +h) O
0
h as
h i 0
at
x
.
h
J
> 0
o
,
1
0
and assuming that
have
U. (X +h) 1
. Then, for
= 1,2
-
h
-
w.. (x ) 1J
0
-
ui (X0 +h)
=
u. (x +h) v. (xo+h) 1
ui (xO )
h
Similarly for h < 0
0
J
5
hw (xo+h)
+
u!(x ) 1 0
=
. This shows that
w
0
ij
is differentiable
0 '
EXERCISE 76.17.
Now, let L be the ordered vector space of all real
continuously differentiable functions on is still continuous. Show that
L
(a,b)
,
i.e., the first derivative
has the Riesz decomposition property. It
Ch. 11,9761
PRIME IDEAL SEPARATION
18
can be proved that the same holds for the space of twice continuously differentiable functions on
(a,b)
and even for the space of three times
differentiable functions, but not for the space of three times continuously differentiable functions (A. Nagel and W. Rudin [ I ] ,
1974).
HINT: Determine the derivative w! in a point x where w(x) lj by means of the ordinary differentiation formula, and show that if w(xo)
=
0 then w!.(x) 13
EXERCISE 76.18.
tends to zero as x
tends to
xo
# 0
*
We mention another formulation of the Riesz de-
composition property which makes it more evident why the property is also called the Riesz interpolation property (cf. Exercise 15.12). ordered vector space L
Show that the
has the Riesz decomposition property if and only
if L has the property that for any systems fl,...,f and g l ,...,gm n satisfying g. 5 f for i = I , m and j = I , ,n there exists an 1 j element f E L such that g. 5 f 5 fj for all i and j
...
...,
.
0
HINT: By induction it is sufficient to indicate the proof for m
n = 2
=
gi
5
v2
=
. Assume
that L has the decomposition property, and let
.
f for i,j = 1,2 Set u . = f.-g. (i=1,2) , v I = fl-g2 and j 1 1 1 f2-gl Then u1+u2 = vl+v2 , so there exist wij (i,j=l,2) such that
.
W I I + W I 2 = u1
,
w,I
w21 + w22
,
WI2 +
=
u2
(see the diagram). It follows easily that
+
w21
=
v1
,
w22 = v2
f o = g1
+ w I 2 is the desired
element. The proof in the converse direction will also be evident.
Ch. 11,1771
PRIME IDEAL EXTENSION
19
77. The hull-kernel topology and the dual hull-kernel topology
Let R
be the set of all proper prime ideals in either the dis(X,e)
tributive lattice
. For reasons of
or in the Riesz space L
notationalconvenience we shall discuss the lattice case; the discussion for
R will be denoted by
a Riesz space is quite similar. The elements of the elements of w
X by
x,y,...
. We
recall that
not containing x ; union and intersection of and
{w}
XVY the kernel
{w}xhy
of all w
{w}x
and
respectively. For any non-empty subset S
. For any
satisfying w
2
D
subset D
. The
of
,
are
Y of
R
in S , and the kernel of
is the intersection of all w
k(S)
the empty set is X
w
is the set of all
X , the hull h(D)
,x
family of all
E X
is the set
,
is a base
for the hull-kernel topology (briefly hk-topology) in R ; the closure of any given subset in R
S
of
R
. The hull-kernel
is the set h(k(S))
is a T -topology, the base sets
topology
of which are open and compact
0
(cf. Theorem 7.3 for the lattice case and Theorem 36.3 for the case of a Riesz space). Let A
be the set of all prime dual ideals A
a Riesz space L , let A
in X
. In the case of
be the set of all prime dual ideals A
with the extra property that
f E A
implies af E h
in L
for all real
a # 0
.
This extra condition is included here since we desire to have the situation that the set theoretic complement of
A
is a proper prime ideal w
lattice case this holds without any extra condition, by Theorem 76.1. return to the lattice case. For every x E X { h I x = (A:hEA,xEh)
Note immediately that
Furthermore, if and only if
c {A}’
x = y
{A]@
,
. In the We
let
. is empty, and
holds if and only if x
5
. It follows that the mapping
the Boolean algebra of all subsets of
A
y ,
so
x + {XIx
=
from X
into
is a lattice isomorphism.
We define hulls and kernels similarly as for prime ideals. Explicitely, if
S
is a subset of
A
,
the kernel k ( S )
is defined by
HULL-KERNEL TOPOLOGY
20
k(S)
=n (X:XES)
for
non-empty,
S
and t h e k e r n e l of t h e empty s e t i s s u b s e t of
.
X
Fcr any s u b e t
h(k(S))
3
subset
D
and
of
.
X
Theorem 6 . 2 ) t h a t
h(D)
i s d e f i n e d by
of
S
immediately t h a t A
implies
We i n t r o d u c e a topology i n
together with
the hull
X
,
and
k(h(D))
h(k(4)) = 3
D
6 ,
f o r every
I t can be proved now e x a c t l y as f o r prime i d e a l s ( c f . S = h(D)
.
k(h(D)) = D
is, therefore, a
. Note
h(X) = $
f o r every s u b s e t
S
k(S)
.
h(D) = (h:A3D) Hence, h ( $ ) = A
X.Note t h a t
of
D
Ch. 11,1773
A
h(k(S)) = S
and
D = k(S)
implies
by choosing a l l s e t s of t h e form
{XIx
i t s e l f as a subbase f o r t h e c l o s e d s e t s i n t h e topology.
A
A c t u a l l y t h i s i s a b a s e f o r t h e topology, s i n c e any f i n i t e union of s e t s of t h e form the hull
i s a g a i n of t h e same form. For any s u b s e t
{XjX
is either
h(D)
h(D) =
so
n
(if
A
A
s e t i s t h e h u l l of some
o r a n i n t e r s e c t i o n of s e t s D c X
.
THEOREM 77.1. For any s u b s e t For this reason, fhk-topology)
in
A
h(k(S))
1)
, the
c l o s u r e of
is h(k(S)).
S
h(k(S))
. But
h ( k ( S ) ) ; it i s s u f f i c i e n t ,
i s i n c l u d e d i n any c l o s e d s e t
. Then
i . e . , h(D)
A
every closed
.
of t h i s kind i s of t h e form S
of
i s included i n t h e closed set
PROOF. S
3
S
{XIX , s o
t h e topology is called t h e h u l l - k e r n e l topology
t h e r e f o r e , t o prove t h a t h(D)
,
X
i s c l o s e d in t h e j u s t i n t r o d u c e d topology. Conversely, every
h(D)
S,
of
D
i s empty) o r e l s e
,
({hlx:x€D)
closed set i s e i t h e r
Any
D
h(D) = S1
h(D)
f o r some s u b s e t
,
S,
so
3
h(k(S))
.
D
of
S
1
X
I>
S
,
.
so
PRIME I U I ~ A LEXTENSION
Ch. 11,1771
THEOREM 77.2. The hull-kernel topology in
itself and a l l sets
that A
,
X I # X2
PROOF. I f
there e x i s t s a point i n h2
,
contains
X2 XI
b u t not
X2
are compact.
{XIx
. Similarly
XI
b u t not i n
XI
t h e topology i s a T -topology. 0
We prove now t h a t x E X
,
A
,
. If
X2
but not i n
is in
x
i f there i s a point i n
XI
s e t s of
XI
XI
or
but not i n
, i . e . , t h e set t h e o r e t i c complement of f X I x , , so ({X}x)c i s a neighborhood of X2 not
containing
. Hence,
is a T 0-topology such
A
there e x i s t s a point i n
({XIx)‘
t h e open s e t
21
i s compact. The s e t s
A
X2 not contained i n
{XIx
, being
a r e then a l s o compact. Note f i r s t t h a t t h e s e t s
form a base f o r t h e open sets i n t h e hk-topology;
closed sub(iXfx)c
,
f o r t h e compact-
ness proof i t i s s u f f i c i e n t , t h e r e f o r e , t o show t h a t any covering of
by
h
these base s e t s has a f i n i t e subcovering. L e t
be such a covering of Then t h e family any i n t e r s e c t i o n to
{XIz
for
Z
t h a t t h e subset easily that
D
,
A
and assume t h a t no f i n i t e subfamily covers
((XIy:y€D)
ny D
i s non-empty.
{XIyi
= y,
A
of
... A
X
A
.
has t h e f i n i t e i n t e r s e c t i o n property, i . e . ,
, SO
yn
Z
But t h i s i n t e r s e c t i o n i s equal =
y1
A
... A
yn > 0
. This
shows
has t h e f i n i t e i n t e r s e c t i o n property. It follows
i s included i n a t l e a s t one maximal lower s u b l a t t i c e
.
lo ’
Hence, i s then a maximal dual i d e a l , we have D C Xo E A Xo the i n t e r s e c t i o n n ({XIY:y€D) contains ho , s o t h i s i n t e r s e c t i o n i s
and s i n c e
non-empty.
Hence, A Let
This c o n t r a d i c t s
i s compact. Am
be t h e subset of
The r e l a t i v e hk-topology i n
XI,A2
Am
A
c o n s i s t i n g of a l l maximal dual i d e a l s .
i s c a l l e d t h e hk-topology i n
be maximal dual i d e a l s such t h a t
i s properly l a r g e r than
XI
and
X2
,
XI
so
# X2
. Then
X, U h 2
f i n i t e i n t e r s e c t i o n property. Hence, t h e r e e x i s t
Am
. Let
t h e union
X I U X2 does not have t h e
x 1 and
x2
in
X I U X2
22 xI I x2
such that
Let x1 E X I XI
and
{XIx1 X1
Ch. 11,8771
HULL-KERNEL TOPOLOGY x 1 E X, U X,
, we have x I E
XI
.
or x I E X p
since x 1 n x2 = e
cannot be in X I
Then x2
is a lower sublattice. Hence, we have x2 E X2 and {XIx2 are closed sets in the hk-topology of
and
{XIx2
, say.
. Since
. It X
follows that , containing
m X
X i respectively, such that the intersection of {XI and i s empty. Hence, {XIx1 is a closed set containing and not XI
containing X2 containing X I
. It follows that
the intersection of all closed sets
consists only of the point
hk-topology in Am
XI
itself, and so the
is a T -topology. Furthermore, the space Am 1
in its hk-topology; the proof is the same as above (note that Xo last proof is an element of A
is compact in the
). We have, therefore, the following theorem.
m
THEOREM 77.3. The hull-kernel topology i n
Am i s a compact
T -topology. 1
From Theorem 7 6 . 1 it follows immediately that the mapping
JI: X
+ w = X-X
is a one-one correspondence between the elements of A
R
the elements of by
J1
. The subset
onto the subset Rm
$({XIx)
of
mapped onto
{XIx
{oIX U { w )
hk-topology of A with
R
il
,
to
Y
of all maximal dual ideals is mapped
of all minimal prime ideals. The image
is exactly
{ w l X f l {wlY = {wlMY
Am
and
. Hence
{wIX
= {w}xvy
. The mapping
and
JI
{XIx
U {XIy = {XIxvy
{XIx fl {XIy = {XIxAy
is
onto
makes it possible to transplant the
R by defining the sets {wIX , x E X , together
to be a base for the closed sets. The topology generated thus in
is called the dual hull-kernel topology in il (briefly the dhk-topology). THEOREM 7 7 . 4 . The dual hu1l-ksrne1 topology i n
i s a compact , are then To-topoZogy. The s e t s { w l x , x E X , being closed subsets of dhk-compact. Furthermore, the dual hull-kernel topology i n the s e t Rm of il
a l l minimal prime ideaZs i s a compact T1-topoZogy, ueaker then the huZ1 kernel topology i n om
.
PROOF. We have only to show that in Rm
the dhk-topology is weaker
then the hk-topology, i.e., we have to show that every dhk-closed set i s hk-closed. Any dhk-closed set in Rm
is either
am
the sets f p I x
itself (and
Qm
.
is
{ u l x = {oIx fl Rm But are closed (as well as open) in the hk-topology of Rm
indeed hk-closed) or else an intersection of sets
PRIME IDEAL EXTENSION
Ch. 11,5773
(cf. Theorem 7.4(ii)),
23
any intersection of these sets is hk-closed. This
so
is the desired result. In Exercise 77.13 it will be indicated how to prove that the hk-topology and the dhk-topology in
{uIX
if the set of all
,
x E X
For any
the set theoretic complement of {uIx
will be indicated by
are identical if and onIy
sk,
is a Boolean algebra.
,
form a base for the open sets in the hk-topology of {ulx
,
{wIX = (w:x not in w )
. The sets
i.e., {oIX = (u:x€w)
Q
{wIX , x
,
E X
and the sets
together with the empty set, form a base for the open sets in the 61
dhk-topology of
.
The next theorem is due to J. Hashimoto ([11,1952,Theorem THEOREM 77.5. Let
and
S
be subsets of
T
X
4.2).
such that
i s non-
S
empty (T may be empty), and assume that the f a m i l y of s e t s
(
{w
lX, { w'1
:x€S ,y€T)
has the f i n i t e intersection property. Then the f m i Z y has a non-empty intersection. PROOF. Note first that for T
empty this is Theorem 6.8. In the
general case, we begin by observing that since {w),
n
{ w } , ~ = {w)xIAx2
I property. Let
F
,
the subset S of
X
is non-empty and
S
has the finite intersection
be the dual ideal generated by
all upper bounds of finite infima of elements of ideal generated by
T (where it is understood that
S (i.e., F
consists of
S ), and let
I = {el
if
I be the T is
empty). We show that F n I is empty. This is trivially true if If F n I is non-empty, there exist elements x l , y l ,...,y
m
E T
such that
XI A
and so
... A
Xn S y ,
V
...
V
ym
,
...,x
E S
and
I
=
{el
.
24
Ch. 1 1 , 9 7 7 1
HULL-KERNEL TOPOLOGY
contradicting the finite intersection hypothesis. Hence F n I is empty, so
by the prime ideal separation theorem there exists a prime ideal
such that
Ic
and F n wo
wo
=
0
. But
wO
then
This is the desired result. THEOREM 7 7 . 6 . Let
of
62
and T be subsets of
S
X
. Then any
subset A
of the f o m
is campact in the dhk-topology. For
S
non-empty, the set A is also
compact in the hk-topology. PROOF. We may assume that A open covering of A
with
2
is non-empty. Let
in the dhk-topology. Then A
n
U({W}~:Z€Z)
(n{wlz) =
be an
0 , i.e.,
non-empty. Hence, by the preceding theorem, there is a finite
intersection that is already empty (note that this finite intersection must contain at least one
{ w } ~). It follows that for appropriate '1
we have
=
(An{w},;)
0 ,
so
A c Uy{w)
. This shows that
2.
A
* * *
"n
is
dhk-compact. The proof that A
is hk-compact if
is non-empty is similar.
S
R is dhk-compact. is hk-compact.
COROLLARY 77.7. (i) Every hk-closed subset of R
(ii) Every dhk-closed proper subset of PROOF. (i) Every hk-closed subset of is of the form n ({W}~:~€T) with
R
is either the empty set or it
T non-empty. In both cases the set is
dhk-compact. (ii) Every dhk-closed subset A of the form A
=
n
:xES) with S
({w}
be a proper subset of preceding theorem, A
of
Ci
,
then A
=
62
is either D
non-empty. If A
D
is now hk-compact.
itself or else is also given to
is impossible. Hence, by the
25
PRIME IDEAL EXTENSION
Ch. 11,9771
The next theorem, which is of a topological nature, will be of great importance for the proof of the unique prime ideal extension from a distributive lattice
(X,e)
to the Boolean ring generated by
X
. We
shall
have to deal with a T -space possessing a base of open and closed sets. 0
Note that in this case the space is Hausdorff. Indeed, if x
and
y
are
different points in the space, then one of the points, say x , has a neighborhood V
occurring in the base and not containing y
V
complement of
is then an open neighborhood of
THEOREM 77.8.
Let
X
closed and compact subsets of
B(N)
(ii) B(N)
.
N of
be a Hausdorff space possessing a base
B(N)
open, closed and compact s e t s , and l e t
(i)
y
. The
be a coZlection of open,
such that
X
contains all s e t s of N , i s a BooZean ring with respect t o ordinary s e t t h e o r e t i c
union and intersection. Then the Stone representation space of the space
The mapping from each x E X
X
onto
51
i s homeomorphic t o X
x
.
,
PROOF. Given any xo E X
is a proper prime ideal in
consisting of a12 A E B(N)
B(N)
the set of all A E
.
B(N)
Conversely, given the proper prime ideal w the existence of a point
, i.e.,
.
i s homeomorphic t o X
giving r i s e t o the homeomorphism assigns t o
the prime ideal i n
containing the point
xo
B(N)
of a l l proper prime ideals i n B(N)
52
xo E X
such that
B(N)
not
not containing
in B(N) , we shall prove consists exactly of all
w
. For this purpose, let A = B(N)-Iu be the . Since A is a dual ideal it has the finite intersection property, i.e., fly Ai # 0 for A ,,...,A E h . Given AO,AI,...,A E A , the sets .A n A. (i=l,...,n) are relatively closed n subsets of the compact set A. with (A nA.) # 0 . It follows that 0 1 A E
B(N)
not containing xo
dual ideal corresponding to
n (A0nA:A€A)
w
is non-empty, i.e., S
=
n
(A:AEA)
is non-empty. We show that
S
consists of only one point. If not, there exist points xo,yo E S with
Xo
# yo
and
yo
(and so
. Hence, there exist disjoint neighborhoods respectively such that A 1 and
A 1 and Ag 1' =
are in
B(N)
(A:AEB(N),~~EA)
.
A2
A,
and A2
of
xO
are elements of the base
) . Now, let
N
26
Then A ' B
is non-empty (since A
, A 1 and A2 in
A'
3
B(N)
A
. The dual so
xo E S
= I7
.
A1
implies A E A '
, and finally A A'
is not in
,
E A'
is a dual ideal in implies xo E A
A U B E A'
,
so
the prime ideals A
The inclusion is proper, since A ' that has
but not in A
is in A '
X
or
(A:AEA) ,
element (namely, A 1 ) so
), the empty subset of
is prime,(kince
A E A
A c A'
satisfy
A'
B E A ' ) . Furthermore, since
or
it is evident that
E
. This shows already that
ideal A '
A E A'
1
implies A 1 fl A2 E A '
A'
implies B E A '
xo E B ,
A'
Ch. 11,8771
HULL-KERNEL TOPOLOGY
xo
and
contains an
as one of its points but not y
0 '
. Proper inclusion, however, is impossible,
since in a Boolean ring dual prime ideals are maximal. Hence, S
consists
of one point. It has been proved thus that there is a one-one correspondence between the prime ideals w A:AEB(N),x
{wlA
B(N)
and the points
w =
such that
not in A
is such that x
is not in A
. Hence
(w:AEw)
corresponds to (x:x not in A ) . But then This shows that the base sets to the base sets
x E X
, we have A E w if and only if the point x
Given now A E B(N) corresponding to
in
A(AEB(N))
{wIA
{wIA
corresponds to (x:xEA)
of the hk-topology in B
of the given topology in X
,
=
A
.
correspond
i.e., B
and X
are homeomorphic. We return to the situation that smallest element 8
and
B
X is a distributive lattice with
is the set of all proper prime ideals in X
Besides the hk-topology and the dhk-topology in B
.
we introduce the weakest
topology stronger than the hk-topology and the dhlc-topology. This is called the Boolean topology (briefly 5-topology) in B
. The family
is a subbase for the open sets of the 5-topology. Note that { u } ~= R
, so
4
and
$l are
=
Q,
and
included in the subbase. Note also that the
Ch. 11,5771
PRIME IDEAL EXTENSION
27
sets in the subbase are closed as well as open. The family of open and closed sets
is a base for the open sets in the 8-topology. There is a somewhat smaller
base. Indeed, since
we have
for every x
,
and so
is also a base. All sets in this base are open and closed. We prove that they are compact as well. It will be sufficient to show that every
{w}
is compact. For the proof (due to S . J . Bernau,C21,1972) we shall need Alexander's subbase theorem, stating that if the subset A
XO
of a topological
space has the property that every covering by sets of a subbase (for the open sets) has a finite subcovering, than A
is already compact (for a
proof of Alexander's subbase theorem, cf. for example the book by J.L. Kelley, General Topology, Ch. 5,Theorem 6 ) . In our case, let U B be a covering of
Then
{w),
0
{w}
x8
by sets of the subbase
n (n Bc) is empty. This intersection is of the form n{w}xn{wjY:xEs,yET
for certain subsets S
and
T of X
. Hence, this intersection being
empty, a finite intersection must already be empty by Theorem 77.5, i.e., {w),~
has a finite subcovering. The sets in the base
28
Ch. 11,5771
HULL-KERNEL TOPOLOGY
and, therefore, open, closed and compact. Furthermo
he 6- opology is
To
,
S2
with the 6-topology is now homeomorphic to the Stone representation
and hence is Hausdorff. According to the preceding theorem, the space
space of the Boolean ring of its own open, closed and compact subsets. There is a possibly smaller Boolean ring of open, closed and compact sets
N
containing
. The family of
N is a Boolean
all finite unions of sets of
{wlx
ring, the smallest Boolean ring containing all sets
.
is called the Boolean ring generated by the lattice of all
This family S
{wlx (for an
indication of the proof that S is indeed a Boolean ring, cf. Exercise
77.16). Note that all sets in S are open, closed and compact. Hence, il
s.
with the 6-topology is homeomorphic to the Stone representation space of Once more, we make some topological remarks. The non-empty subset Y
is called irreducibZe if for each pair of open
of the topological space X
, O2
such that 0 n Y and 0 n Y are non-empty it 1 2 follows that 0 f7 0 n Y is non-empty. In other words, Y is irreducible 1 2 if and only if for each pair of closed subsets FI,F2 of X such that subsets O 1
X
of
. Evidently, Y is irreducible if and only if for each finite family 0 1 , ...,0 of open subsets of X such that 0. n Y is non-empty for i = I, ...,n it follows it follows that Y c F 1 or Y c F2
Y c F 1 U F2
(nnI
that
0 . ) fl Y
is non-empty or, equivalently, if and only if for each
1
...,Fn
finite family F 1 ,
follows that Y c F.
of closed subsets of
X
for at least one value of
Each subset of X
such that Y c U p i i
=
1,. ..,n
consisting of one point is irreducible. If X
a Hausdorff space and Y
is an irreducible subset, then Y
it
.
is
consists of
one point. Every (relatively) open subset of an irreduaible set is also irreducible. If space X I and Y
Y 1 = $(Y) in X I
Q-](F2)
Q
is a continuous mapping from X is an irreducible subset of
is irreducible in X 1
such that Y 1 c F 1 U F2
included in Q-I(F,)
or
Q
that Y 1 c F 1 or Y 1 c F2
(F2)
, so
,
into the topological then its image
. For the proof, let FI,F2 be closed sets . Then the inverse images Q -1 (F1) and
are closed in X , and Y c $-'(F1) -1
X
,
since Y
U Q-](F2)
. Hence
Y
is
is irreducible. It follows
Y 1 is irreducible.
Ch. 11,5773
29
PRIME IDEAL EXTENSION
THEOREM 77.9. The folZaJing conditions f o r a subset
of a
Y
topological space are equivalent. (i) Y i s irreducible. (ii) The closure Y of Y
is irreducible. 0 i n X the intersection
(iii) For every open s e t
.
empty or dense i n Y (iv)
Every (reZatively) open subset of
is open and 0 n Y and 1 on^^. (iv) Y
0 0
0
n Y is e i t h e r
i s connected ( i . e . , i f
Y
0
n Y non-empty, there cannot e x i s t open 01,02 such t h a t n Y are non-empty, d i s j o i n t and with union equal t o 2
Every f i n i t e f a m i l y of non-empty (reZativelyl open subsets of
has a non-empty intersection. PROOF. The equivalence of (i) and (ii) follows by observing that for
any open 0 the intersections 0 Il Y and
0 n Y-
are simultaneously empty
or non-empty. Now consider condition (iii). To say that, for 0 open, 0 is dense in Y
meets every other non-empty relatively open subset of Y equivalent to (i).
Evidently, (iv) holds if Y
let (iv) hold and assume that Y 01,02
such that 0 n Y 1
Y
,
i.e., (iii) is
is irreducible. Conversely,
is not irreducible. Then there exist open
and 0 fl Y
1
2
are non-empty and disjoint. This
n Y is not connected, which contradicts (iv).
(0 UO )
would imply that
ll
is non-empty, is the same as saying that 0 n Y
0 f'l Y
if
2
Finally, (i) and (v) are evidently equivalent. The set theoretic union of a chain of irreducible subsets of irreducible. Hence, by Zorn's lemma, every irreducible subset of contained in a maximal irreducible set. Evidently (since Y
-
if and only if Y
X X
is is
is irreducible
is irreducible) every maximal irreducible set is closed.
We now consider irreducible subsets of
Q (with respect to the
hk-topology). THEOREM 77.10.
(i) The non-empty subset
and only i f the kernel (ii) The subset
of
k(Y)
Y
of
there e x i s t s a prime ideal ( i . e . , Y consists of a l l
Q
wo w
3
Y
Y
of
a i s irreducible i f
i s a proper prime ideal i n X
.
i s closed and irreducible if and only i f such t h a t Y i s the closure {w0}- of wo 1. Hence, Y i s a rnmirnal irreducible
wo
30
Ch. 11,1771
HULL-KERNEL TOPOLOGY
subset of
i f and on2y i f
0
Y
PROOF. (i) Let first k(Y) non-empty. Then x
and
it follows that x one wo E Y
, the
y
y
A
f o r some minima2 p r i m e idea2 uo
= {pol-
{wlX n Y
be prime. Let
and
y
A
,
is not in wo
A
y
is
are not i n
k(Y)
is not in k(Y)
Y non-empty intersection with such that x
y
A
since Y
Y
and y
y
{
~
= IwIx 3
be closed and irreducible. Then k(Y)
satisfying w
w
wo
. But
wo =
k(Y)
2
2
. It follows that
Y
Y
,
=
h(k(Y))
since Y
being in the hull of
consists of all w
Conversely, if
Y = {wo}-
and k(Y) = wo
is prime,
3
wo
for some prime ideal so
,
are not in k(Y) so
,
then
n~ {wly~ has~ a w E Y
k(Y)
.
is a prime ideal,
. Hence, all
is irreducible. Let us call this prime ideal wo
E Y satisfy
w
x
and
. Hence, there exists a prime ideal . It follows that x A y is not in
Y
is not in w
(ii) Let Y w
. If
n Y are non-empty, and
{wl
and
is irreducible.
be irreducible. In order to show that k(Y)
prime, it will be sufficient to show that if x {wlX n Y
least
i.e.,
and so this intersection is non-empty. It follows that Y
then x
n Y be Y is prime,
. This implies that, for at
is not in k(Y)
element x
Conversely, let Y
{w}
are not in k(Y) and s o , since k(Y)
.
is closed. Hence, every
, is
k(Y)
, i.e., Y = wo , then
{w,}
Y
a member of
-
.
is closed
Y is irreducible.
Note that since a prime ideal may contain several different minimal prime ideals, it may occur that two different maximal irreducible subsets of have a non-empty intersection. EXERCISE 77.11.
{ w } ~c {wIY
,
In the distributive lattice
i.e., { w } ~c {wIx
EXERCISE 77.12.
Let w 1
distributive lattice
(X,e)
the hk-closure of wI
,
of
w
,
and
(X,e)
if and only if x I y w2
,
.
show that
be proper prime ideals in the
. Show that
w 1 c w2
and also if and only if
if and only if w1
w2
is in
is in the dhk-closure
2 '
EXERCISE 77.13.
Let
am
in the distributive lattice minimal prime ideals
~1
be the collection of all minimal prime ideals (X,e)
. By
{pix
we denote the set of all
that do not contain x
. Show that the collection
Ch. 1 1 , § 7 7 1
31
PRIME IDEAL EXTENSION
{ v } ~ is a Boolean ring if and only if the hk-topology and the dhk-topology coincide on each { u l X Hence, the collection of a l l {u)x of all
.
is a Boolean algebra if and only if the topologies coincide on the whole of
Qm
. This result is due
to T.P. Speed (C11,1969).
HINT: Assume that the sets
{ u l x form a Boolean ring. Let {u}xo 0 of {u} is
fixed. We have to show that every hk-open subset (relatively) dhk-open. Any hk-open { u j Y T with
y,
{pix
the sets
for all
xo
5
with respect to
{plxo
T
xO
0 of this kind is a union of sets
. Note now that the complement of
is of the form
{ul YT { u I z T for some z E X , because
form a Boolean ring. Hence, the complement ' 0
respect to
{ulx
dhk-closed,
so
0
Oc
is the intersection of all
{u)zT
. Every
of {P),
0 with
x
5
xo
are dhk-compact by Theorem 7 7 . 4 , s o all sets are (relatively) compact for the dhk-topology in
The dhk-topology and the hk-topology in
{ulX , all
8 5 x 5 xo
{pix
xO
{ulx
Conversely, let the hk-topology and dhk-topology coincide in 5
{u1
0 is (relatively) dhk-open.
Note that all sets
{ulX , 8
is
T
is dhk-closed. Hence 0 is the intersection of
and a dhk-open set, so
, are
be
{pix
0
0
.
{ulx
0 are identical, so all sets
is arbitrary, it follows that xo are hk-compact, and so the collection of all {pix is a Boolean hk-compact. Since
ring by Corollary 8 . 3 . For the proof that the collection of all Boolean algebra if the topologies coincide on
Cim
, note
that
{pix
Qm
is a
is
compact in the dhk-topology (Theorem 7 7 . 4 ) . Hence, if the topologies coincide, then $2, all
{pix
is compact in the hk-topology, and so the collection of
is then a Boolean algebra (Theorem 8 . 4 ) .
EXERCISE 7 7 . 1 4 . Let X
be an infinite point set in which a topology
is introduced by defining that the open sets are the empty set and all subsets of
X possessing a finite complement. Show that X is an
irreducible Tl-space. EXERCISE 7 7 . 1 5 . Let X element 0
. Show that if
X
be a distributive lattice with smallest is linearly ordered, then Ci
is irreducible
in the hull-kernel topology. The converse does not hold; the lattice in Exercise 7 6 . 1 4 is a counterexample. EXERCISE 7 7 . 1 6 . Let and y
.
N be the family of all sets { w l x n {uly for
x
running through the distributive lattice X with smallest element
32
0
. Show that
s
the family
s
Boolean ring (i.e.,
s
Ch. 11,5781
THE UNIQUE PRIME IDEAL EXTENSION PROPERTY
of all finite unions of sets of
N
is a
is a distributive lattice with respect to set
theoretic union and intersection, the empty set is the smallest element of and the set theoretic difference of two sets in
s
is again in S ) .
HINT: It is easy to see that union and intersection of sets in S again i n
. Note now
S
of M
complement MC N
= {w}
N
{wIw E
fl
that if M = {wIx n {w}' with respect to
then N-M
=
N , and hence N-M E S
of two sets i n Then N-B
=
(N-Ni) E S
then A-B
=
U y (NI-B) E S 3
R
N n MC
.
,
are
then the set theoretic
is =
{wIx U { w ) so if Y' {Nn{wlX) U {Nn{wIy} is a union
Now let N E N
. It follows that
.
E N
if
and
A = Um N!
1
J
B
=
U y Ni E
E S and B E
s.
s ,
78. The unique prime ideal extension property
Let 8
,
(X',e)
and let
denote a distributive lattice X'
with smallest element
be a sublattice. It was proved in Theorem 51.8 that
(X,e)
every proper prime ideal w
in X
in X' (i.e.,
the extension is unique if
X'
. The
w'
fl
X
= w );
can be extended to a prime ideal w ' X
is an ideal in
converse is not true, that is to say, the extension can be unique
without X
being an ideal in X
.
The lattice in Exercise 76.14 is an
example; for this very simple lattice every proper prime ideal can be uniquely extended to a prime ideal in the Boolean ring generated by the lattice. This holds generally. We present the proof. THEOREM 78.1.
Every proper pfime idea2 i n the distributive l a t t i c e
has a unique extension t o a prime idea2 i n the Boozean r i n g
(X,e) generated by
X (i.e.,
B(X)
PROOF. As before, let R
B(X)
is the sma2Zest Boolean ring containing X
).
be the set of all proper prime ideals in X
.
As observed in the preceding section, the family of sets
is a base of open, closed and compact sets for the @-topology in R
. Note
now that, by the Stone representation, X is isomorphic to the family of all {wIx Hence, identifying the point x E X with {wlx for every x E X ,
.
Ch.
11,5781
we can identify each point of the Boolean ring of sets
33
PRIME IDEAL EXTENSION
the Stone representation space of explained there). Theorem 77.8,
B(X)
with a finite union
(cf. Exercise 77.16). It follows by Theorem 77.8 that
(wlx fl{wlY
B(X)
is homeomorphic to
Consider now a prime ideal P w1 E R
there exists a point
in
(in the sense
. According to
B(X)
such that the points of
P
are
-
{wjX n {w}'
exactly all finite unions of sets
that do not contain w 1 Conversely, any collection of all finite unions of this kind is the
.
collection of points of a proper prime ideal in B(X)
,
Given wo E Q
let P
be a prime ideal in B(X)
corresponds to a certain w 1 E Q
Then P
i.e., the points of
P
that extends
wo
are exactly all finite unions of sets
{ w ) ~ll
that do not contain w 1
. The points
with the points of
under the identification, i.e., with the points
wo
.
of
*
in the manner indicated above,
P n X = wo
correspond
({~}~:xEw~) On the other hand, these points are the points
exactly
{wlx f l {wly
of the special form
i.e., all
Idx
and so
- wo
w1
satisfying x E w 1
. This shows that
uniquely determined by
wo
-
. Hence
the prime extension P of
wo
is
It may be useful to note that even though in general the points of cannot be uniquely written as finite unions of sets
B(X)
that is needed in the proof is that every point of
e
IwIx n
10)
.
We return to the sithation that (X',&I)
lattice
.
If X
(X,e)
w'
of
X E
X
w
consists of all x' E X'
I' in X'
such that
w
. Indeed, I' l X c w
I' c
w'
X' A
x E w
, so
x' c w '
the following definition.
in X
satisfying x'
I' n X
C w
all
,
the prime extension
A
x E w
for every
in the following sense. Given
, the extension
w'
satisfies
implies that for x' E I' and x E X
. In view of
,
fl { w}'
is a sublattice of the larger
. The extension procedure is monotone
an ideal
{wIx
can be written as
, prime ideal exension i s unique.
i s an ideal in X'
Explicitly, given the proper prime ideal
X
we have
this monotonicity property we present
34
THE UNIQUE PRIME IDEAL EXTENSION PROPERTY
DEFINITION 78.2. The s u b l a t t i c e
Ch. 11,§781
of the d i s t r i b u t i v e l a t t i c e
(X,e)
(X',8) i s said t o have the monotone prime i d e a l extension property whenever f o r every i d e a l I' i n X ' and every proper prime idea2
I' n X c w
X' satisfying
that
there e x i s t s a prime extension
.
I' c w '
(X,e)
Note that if
has this property with respect to
w1,w2 are proper prime ideals in X exists for every prime extension w;
. As
satisfying w' c w; 1
satisfying w1 c w2 of
of
w'
,
such
(X',8)
and
then there
a prime extension
w1
in
w
w
w'
of
w
2
observed before, prime extension can be unique and
at the same time fail to be monotone (cf. Exercise 7 6 . 1 4 ) . For simplicity, assume now first that the ideal generated in the larger lattice X '
by the sublattice X
situation by saying the X in X '
prime ideal w ' w'
fl
X
=
by
@(w')
. As
X'
covers
intersects X
itself. We shall indicate this a consequence, every proper
now in a proper prime ideal. Indeed,
X would imply that the smallest ideal in X '
included in w ' i.e., X'
is X'
= w' = w'
. But this smallest ideal is
, which contradicts ll X , is therefore
X'
containing X
from R'
into R
, , defined
X' c w'
itself, s o
the hypothesis. The mapping
extension theorem then implies that
is
@
. The prime
ideal
is not only into, but also onto f2
@
Furthermore, we have
so
the mapping
@
is hk-continuous. We prove that if, in addition, X
the monotone prime ideal extension property, then $ THEOREM 78.3. I f
X covers
X'
extension property, then the mapping
and
X
has
is also hk-closed.
has t h e monotone prime idea2
$: w' -+ w '
n
X
from
R'
onto
R is
not only hk-continuous, b u t a l s o hk-closed. PROOF. Let F' c R' @(F')
be hk-closed. We have to show that the image
is also closed. Since F' = h(k(F'))
in R'
, we
have
.
Ch.
PRIME IDEAL EXTENSION
11,§781
Every w ' E F' so
the hull of
has, therefore, the property that $ ( w ' ) X
fl
in R
k(F')
0
h Xflk(F')
On the other hand, if
hull of
X
ll
3
$(F')
w
E R
w'
3
3
X fl k(F')
,
and
satisfies
. satisfies X fl k(F') c w
(i.e., w
in the
k ( F ' ) ) , it follows from the monotone prime ideal extension
property that there exists a prime ideal w ' and
35
k(F')
.
Hence w = $ ( w ' )
and
w'
in X' E F'
,
such that w ' ll X w
so
E $(F')
. This
= w
shows that
0
h Xnk(F') It follows that
$(F')
c $(F')
=
.
h(Xflk(F'))
,
so
$(F')
is hk-closed.
We now combine monotonicity and uniqueness of the extension. THEOREM 78.4. ?'he sub2attice (X',e)
(X,e)
of t h e d i s t r i b u t i v e Lattice
has t h e unique and monotone prime idea2 extension property i f and
o n l y if X
i s an idea2 i n X'
PROOF. If X
.
,
is an ideal in X'
the prime ideal extension is unique
and monotone, as already observed earlier in this section. For the proof of the converse, assume that X
has the unique and monotone prime ideal
extension property with respect to X'
. Then
X
property with respect to the ideal generated by just as well assume that X '
x'
X
in X'
is the ideal generated by
X = X'
), and we have to prove now that
is from R'
certainly has the same
. Hence, we may
X (i.e., X
. The mapping
$: w'
+
w'
covers fl X
onto R (as observed above), and by the uniqueness of the
extension the mapping is now one-one. Hence, since the extension is $
monotone, it follows from the preceding theorem that from R' SO
onto
52
. Given
its image $({u'}~~)
x' E X'
,
the set
{w'}~,
is open and compact in R
u ({wIy:yED)
. By
of the form
{wjX for some x E X
is a homeomorphism i s open and compact,
, and
hence of the form
the compactness this is equal to a finite union, and
. Hence
so
36
THE UNIQUE PRIME IDEAL EXTENSION PROPERTY
for some x E X
.
{w')~,
{w'}~
=
,
Ch. 11,9781
But then, by the unique extension, we have and
x'
so
.
x
=
It follows that X
=
.
X'
For later use it will be practical to have available a certain class of sublattices possessing the monotone prime ideal extension property. The following definition will be useful for this purpose. (X,e) of the d i s t r i b u t i v e l a t t i c e i s called a p-sublattice of X ' whenever f o r a l l x,y E X and y' E X' such t h a t y = x v y' and f o r e v e q proper prime i d e a l w ' i n X ' such t h a t y' E w ' there e x i s t s an element z E w ' n X s a t i s f y i n g DEFINITION 78.5. The s u b l a t t i c e
(x',e)
y = x v z . Note that every ideal X z =
y
A
y'
in
(X',8)
is a p-sublattice, since
satisfies the condition mentioned in the definition. It is also
easy to see that the lattice X
in Exercise 76.14 is not a p-sublattice of
.
the Boolean algebra B(X)
X of t h e d i s t r i b u t i v e l a t t i c e
THEOREM 78.6. Every p-sublattice
(X',e) has the montone prime i d e a l extension property. PROOF. Let
I'
be an ideal in X'
such that
I' ll X c w
of
satisfying
w'
o
generated in X'
by
. We have w'
I'
2
I'
and
w
and
a proper prime ideal in X
w
to show that there exists a prime extension
. To
this end, we consider the ideal J'
. We prove first that
J'
nX
= w
,
and
this will be the greatest part of the proof. By means of Zorn's lemma it follows from
I' Il X c w
I' 3 I' and 1; 0 A
y'
E 1;
property of
is prime. If not, there exist x',y' E X'
but
x'
and
y'
are not in
. Hence
which implies
1;
such that
I' nXc 0
x
A
y E 1; n X c w
against our hypothesis that x
=
X-w
. But
and y
.
. From the maximality
such that x
w
w
such that
it follows that there exist elements z ; , z ; E 1;
1;
in the set theoretic difference F y < y' v z;
in X'
is maximal with respect to the property that
We show that 1; x'
that there exists an ideal 1;
is prime,
are in F
= X-w
x' v z i
S
so
.
x E w
and x,y and
or y E w ,
This shows that
Ch. 11,§781
is prime. Assume now that J' fl X
1;
an element x x
=
=
in J' fl X
E I'
z'
. Since
in the ideal
Iw
y v y' v z'
=
y
Hence, since X
. Hence y' v z' . But
=
y v
E
z
w
.
y E w and x E w
It follows that
. Thus we
is not in w
= w ).
exists a prime ideal w '
X
fl w ' c w
w =
X n
w
. On
w'
.
,
is prime.
1;
zE1;)flXcw
z
so
and
w
E 1;
X
fl
x v y =
. This contradicts the hypothesis that = w .
is now a lower sublattice and F fl J'
is
By the prime ideal separation theorem there such that J ' c w '
in X'
It follows immediately that w ' prime extension of that
and
by =
find at last that J' fl X
The non-empty set F = X-w empty (since J' fl X
Hence
E 1;
z'
is
have
x v y
is a p-sublattice, there exists an element
suchthat x v y = y v z . B u t x
E I' implies
z'
, we
x E J'
generated in X'
for some y E w , so
y
5
. Then there
is not equal to w
but not in w
y' v z' for some y'
for some
37
PRIME IDEAL EXTENSION
3
I'
F
and
n
is empty.
w'
. It remains to prove that w ' . It follows from F fl w ' =
i.e., X fl w ' = w
the other hand, J' c w '
is a
0
implies w = X fl J' c X fl w ' .
This concludes the proof.
In theorem 7 8 . 4 it was proved that the sublattice (X,e)
of
(X',0)
has the unique and monotone prime ideal extension property if and only if
X
is an ideal in X'
. The last theorem shows that every p-sublattice has
automatically the monotone prime ideal extension property. The following statement is, therefore, an imediate consequence. THEOREM 78.7. The p-sublattice (X',0)
(X,0)
of t h e d i s t r i b u t i v e l a t t i c e
has the unique prime i d e a l extension property i f and only i f X
is an i d e a l i n X'
.
We recall (cf. section 76) that the distributive lattice of all principal ideals in the Riesz space L
is denoted by
X(L)
. As
observed
before, ideals, maximal ideals and prime ideals in L
correspond to
ideals, maximal ideals and prime ideals in X(L)
L
subspace of the Riesz space L' ideals in L and not in L' a subset of Of
X(L')
X(L')
,
, the so
generated by the elements of
L
. If X(L)
is a proper Riesz are principal
we cannot say immediately that X(L)
. Nevertheless, we
by identifying X(L)
elements of
shall consider X(L)
is
as a sublattice
with the set of all principal ideals in L'
.
38
Ch. 11,9781
THE UNIQUE PRIME IDEAL EXTENSION PROPERTY
The following theorem showsnow that p-sublattices occur frequently. Note first that if we say that (B,9) is a subring of theBooleanring (B',B), this means that elen?ent 9
(B,B)
is a sublattice of
(B',9)
and such that each initial segment
with the same smallest
(b:bEB,B
of
(B,B)
is a Boolean algebra. THEOREM 78.8.
of a BooZean ring
(B,B)
(i) Every subring
(B',9)
is
a p-subZattice.
is a Riesz subspace of t h e R i e s z space L' , then X(L)
(ii) If L a p-sublattice of
.
X(L')
B will be denoted by
PROOF. (i) The elements of B' by
elements of from 8
to b
... . Now,
a',b',
let b
=
x
...
ac
so
that it follows from b
exists and we have ac
=
b
interval in B
,
in B'
b A ac = (avb') A ac =
then ac E w ' u,v E L+
(ii) Let
where Au,Av u,v
and
B
(sac)v
b' E w '
. Hence, B
u ' E (L')'
and
(b'Aac)
5
sup(v,u')
. Then
a sup(v,u')
, u
= a
and generates v
But then u2
5
=
u2
5
v + (u2-v)
-1
u
b' A ac I b'
=
5
V
v+u'
=
A
2
L'
by
sup(v,u')
2
generate the
a > 0 such that supCv,u')
u2 = sup(u,,v)
and u ,
. Then u?-
. Summarizing, we have
is
is
.
with u -v E L
argument as above shows that X(L)
.
is the principal ideal and
satisfies u 1
. Now, let
sup(AU,A
sup(v,u')
,
b
satisfy
it follows that u
still in L and generates AU still in L
a
for some proper prime ideal
same principal ideal. Hence, there exists a number u
x with
that
is a p-sublattice of B'
respectively. Since sup(Av,Au,)
generated by
a v b'
denote the principal ideals generated i n
and A,'
u'
ll
=
a v ac with
=
The last inequality implies that if w'
the
in this interval, let
xc be the complement (in the sense of the Boolean algebra) of
. Note now
,
arb,
. The
a v b'
is a Boolean algebra; for any
respect to b
is
and
0
5 u
2
-v
is a p-sublattice of
u'
X(L')
,
so
.
the same
Ch. 11,9781
39
PRIME IDEAL EXTENSION
For the cases considered in the last theorem it is now a simple matter to determine under which conditions prime ideal extension is unique. THEOREM 7 8 . 9 .
(B,B) of a Boolean ring
(i) A subring
unique prime i d e a l extension property i f and only i f we r e c a l l t h a t i n t h i s case the unique extension prime i d e a l
L
(ti) Let IL
be a Riesz subspace of t h e Riesz space by
t h e i d e a l generated i n L '
.
L
u ' E L'
and an element
a > 0
I n t h i s case, i f
P
,v E
L
The space
X(L) = X(1L)
i d e a l extension property i f and only i f means t h a t f o r any p a i r e x i s t s a number
B i s an i d e a l i n B' of a given proper
.
i s given by
B
in
w
w'
has the
(B',B)
satisfying
L
u E L+
L
. Denote
. Explicitly,
0 c u' 5 v
such t h a t
i s a proper prime i d e a l in
L'
by
has t h e unique prime
au 5
this
there u' 5 u
, it i s s u f f i c i e n t
.
for
the determination of the unique extension t o i n d i c a t e t h e unique extension P"
of
in
IL
P
.
. The
IL
to
PROOF. (i) (B,€i) theorem,
so
extension
P
i s simply t h e i d e a l generated by
P"
is a p-sublattice of
(B',6)
by the preceding
B has the unique prime extension property if and only if B
is an ideal in B ' (by Theorem 7 8 . 7 ) . (ii) The first part of the proof is similar as in part (i). For the determination of generated by
N
and
P"
A,
first that P"
must at least contain the ideal
. Furthermore, on account of
. Hence, for
is equal to
X(L) = X(1L)
,
the
must correspond with the same prime ideal in 0 5 u" E P"
for some u E P
,
, the so
N
u
principal ideal generated by is majorized by a multiple
. This implies that u is contained in the ideal generated by I L . Hence, P" is the ideal generated by P in I L . N
of u in
, note
in I L
P
prime ideals P x(L) = X(IL)
P"
P
We continue with our investigation of the situation that X ( L) = X ( I L ) . First a simple remark. If A
is a band in the Riesz space L
with
Projection property, then it is well-known (cf. Theorem 2 4 . 5 ) that the component u 1
in the band A
of any
u t 0
is given by
40
Ch. 11,5781
THE UNIQUE PRIME IDEAL EXTENSION PROPERTY
I be an ideal in L
Now, let
0
any
5
v E A
and A
the band generated by
is the supremum of all w E I satisfying 0
is obvious then that the component u l
7
=
of
u
in A
I (hence, 5
w
5
v )
. It
is now also given by
sup(w:w€I,o<w
This will be used in the proof of the following theorem. THEOREM 78.10. Let
that
X(L) = X(IL)
more
L'
this, L
L
be a Riesz subspace of the Riesz space
, as i n p a r t
L . If, i n addition t o is an i d e a l i n L' ( i . e . , L = IL ).
has the p r o j e c t i o n property, then so has i s uniformly complete, then
L
I f , instead of t h i s l a s t condition, I L
the Dedekind completion of PROOF. Assume that in L
.
L
i s uniformly complete, then
I in L'
in L'
A
will be denoted by
will be denoted by
a projection band in L' ; given u 2 0 in L u
. As
in A'
(
(1)
. The band
let u'
inf(z,u)
_<
u
. Hence,
(
u' = sup v:v€A,o
,
is
is a projection band in L
= X(IL)
such that
,
z
a
1.
this is now equivalent to showingtbat u' hypothesis that X(L)
A , say
satisfying
, and
L . If u' = 0 , . On account of the
there exists a number a
aw < u' < w
by Theorem 24.5
is a member of
this is obviously the case. Hence, assume that u' > 0 element w E L+
A'
1.
is likewise a majorant in A
We have to show that A
that
I and the
be the component
occurring on the right has a majorant in the band
Then v = inf(z,u) v =
,
A'
observed above, we have
u' = sup v':v'EI,O5v'
Every v'
IL i s
L' has the projection property. Let A be a band
. The ideal generated by
band generated by of
such
L'
(ii) of t h e preceding theorem. I f f u r t h e r
>
0 and an
. It follows then from formula
(I)
41
PRIME IDEAL EXTENSION
Ch. 11,§781
inf(v,aw):vEA,O5viu aw E A , and hence w E A
so
Then v E A , 0 set
5
v
u
5
. On
(v:vEA,O
. Now,
implies v
let u 1
,
5 u1
the other hand
=
so
I This is the desired result, i.e., L
, and
u1
=
complete by Theorem 4 2 . 6 .
=
In order to show now that
. There exist
=
(
sup v:vEL,Oiv5u'
=
This supremum exists in L = u'
. If not, we
element in L
is
L , consider an such that
u' > 0
in
IL
(namely,
1.
since L
is Dedekind complete. We prove that
0 < u'-w E IL
strictly between 0 and
assume instead that and
L'
,
u' E L
so
u'-w
IL
property, so
. It follows that
IL = L
.
is uniformly complete, but
is uniformly complete. Since IL
is an ideal in
IL
Hence, IL is uniformly complete and has the projection is Dedekind complete. The space L
of the Dedekind complete space exist v
and there would be no
. This is impossible, as shown
has the projection property, IL has the projection property
(by Theorem 2 5 . 2 ) .
and
IL
is now the Dedekind completion of L
is a Riesz subspace
such that for any u ' > 0
w in L satisfying 0
The case that L
IL
u E L+
that for any
Finally, we drop the condition that L
1L
,
is Dedekind
satisfying 0 < v i u '
in L
should have
above. Hence u ' = w
L'
so
u' E L
au ). Now, let
w
w
a > 0 and
. This shows in particular
there exists an element v > 0 v
so
is uniformly complete. Then L L
0 < au 5 u ' 5 u
.
has the projection property.
Assume now that, in addition, L
IL
E A
1
is a member of the set, and
u1
uniformly complete and has the projection property, so element u ' > 0 in
u
is an upper bound of the
is the supremum of the set. This shows that u '
u
, so 0 S
inf(u,w)
u1
< v
L
is Archimedean and
.
< u'
L'
i w
. Hence, by
in IL
there
Theorem 3 2 . 6 ,
is the Dedekind completion of
is of particular interest. In this case the ideal
IL is the space L'
itself. The following theorem is, therefore, an immediate consequence. THEOREM 7 8 . 1 1 . Let
.
L'
be t h e Dedekind compZetion o f t h e Archimedean
Then L has t h e unique prime i d e a l extension property W i t h Riesz space L respect t o L' if and only if X(L) = X(L') , i.e., if and onZy if for any
THE UNIQUE PRIME IDEAL EXTENSION PROPERTY
42
o
i
u ' E L'
that
au
5
there e x i s t s a number u' 5 u
. In
t h i s case
The condition that for any have
au
5
u'
5
u
o
a >
and an element
Ch. 11,1781
L+ such
u E
has a ls o t h e p r o j e c t i o n property.
L
u' > 0
for some u E L+
in the Dedekind completion L'
we
is the condition, due to J.J. Masterson,
that occurs in Exercise 37.16. In this exercise we introduced for each ideal A
in L
the ideal A#
ideal in L'
generated by
is of the form A#
for some A
Exercise 37.16 that every ideal in L' if and only if L
in L'
A
.
In general not every
in L
. It was
is of the form A#
shown now in
for some ideal
A
in L
L
is Dedekind complete (i.e., L = L' ), Masterson's condition is trivially
satisfied. If L
satisfies Masterson's condition. Of course, if
satisfies Masterson's condition, then L has the
projection property, as shown by the last theorem. We show that Masterson's condition is strictly between Dedekind completeness and the projection property.
Let
THEOREM 78.12.
Dedekind compZetion to
L denote an Archimedean R i e s z space w i t h
. Then Masterson's
L'
condition f o r
L' I i s s t r i c t l y between Dedekind completeness of
property for
.
L
PROOF. We first present a space L
L
L
(with respect
and the p r o j e c t i o n
satisfying Masterson's condition
without being Dedekind complete. Let L be the Riesz space of all bounded real functions f on the interval C 0 , l l
such that f has an at most
countable range. The Dedekind completion L' is the space of all bounded
.
real functions on C 0 , l l
This shows already that L
complete. For the proof that L
the bounded function u ' ( x ) t 0 on identically zero and
0
2
is not Dedekind
satisfies Masterson's condition, consider
u'(x) < 1
C0,lI
. We may assume that . For n
holds for all x
=
u'
2,3,
is not
...
,
let
and let Em
=
. ...)
(x:u'(x)=O)
of the sets E
...)
(n=2,3,
(n=2,3,
and
u(x)
=
not identically zero and observing that
Since u'
is not identically zero, one at least
2n-l t (n-l)-'
5
2u
holds
=
n
. . The last inequality follows by for n = 2,3, ... . This shows that
0 for x E Em
u 2 u'
-1
for x E E n Then u E L+ , the function u is
is not empty. Now, let u(x)
L
Ch. 11,0781
43
PRIME IDEAL EXTENSION
satisfies Masterson's condition. Next, we present a space L with the projection property, but not satisfying Masterson's condition. Let L be the Riesz space of all real ...)
sequences f = (f(l),f(2), L'
is the sequence space I 1
u' =
There is no u E L+
with finite range. The Dedekind completion
,
so
,...) E L' .
satisfying au
S
u' 5 u
for some a > 0
, and
L
so
does not satisfy Masterson's condition. On the other hand, as proved in Theorem 25.1,
part (iii) of the proof, L has the projection property.
We shall return to the situation that L
L'
is a Riesz subspace of
in the next section. The present section will be concluded by presenting a necessary and sufficient condition for a sublattice to have the unique prime ideal extension property.
(X,e)
THEOREM 78.13. The s u b l a t t i c e
of the d i s t r i b u t i v e Zattice
(XI,@) has t h e unique prime i d e a l extension property w i t h respect t o i f and onZy i f t h e BooZean ring generated by X'
ring generated by PROOF. Let B' sublattice of
. By
B be the subring of
B'
X'
. Then
generated by
X X
is a
. In
consist of all finite suprema of (relative) differences
. Evidently, B
of elements of X
X i s an i d e a l i n the BooZean
be the Boolean ring generated by
. Let
B'
other words, let B X
.
X'
is exactly the Boolean ring generated by
Theorem 78.1 the lattices X
and
X'
have the unique prime ideal
extension property with respect to B and B' respectively. It follows that X has the unique prime ideal extension property with respect to X' if and only if
B has this property with respect to B'
is an ideal in B'
.
,
i.e., if and only if B
Most of the results in the present section were proved by W.A.J.
Luxem-
burg in the paper (C51,1973). EXERCISE 78.14. In Theorem 78.10 it was proved that if subspace of
L'
such that for every
0 _< u' E L'
L is a Riesz
there exists
E L+ and
44
satisfying au < u' < u , then the projection property for L'
a > 0
implies the projection property for L
. This is no
the hypothesis a little weaker. Show that if 0< v
S
u < w
property but
,
in L
to to
longer true if we make
then it is very well possible that L'
without using that X(L)
P to
extension of
,
X(1L)
=
P in IL . This was proved . It can be proved directly, i.e.,
generated by
that if the extension is unique, then the
IL must be the ideal P"
generated by
Assume, for the proof, that the unique extension P#
. Then there exists an element
larger than P" P"
has the projection
In Theorem 78.9 it was proved that if prime extension
IL is the ideal P"
there exists a prime ideal Q
so
satisfying
is unique, then the prime extension of any prime ideal P
L'
after proving first that X(L) = X(1L)
, and , but
L'
is a Riesz subspace of
L has not.
EXERCISE 78.15. from L
L
there exist v,w E L+
such that for every 0 < u' E L'
P"
Ch. 11,9781
THE UNIQUE PRIME IDEAL EXTENSION PROPERTY
is not in Q
u'
unique prime extension
u' > 0
P
in IL
, but
in P#
Q nL
3
P"
.
3
not in
includes
,
L = P
Q n L satisfies (QnL)#
of
P
is properly
in IL such that Q
. It follows that
(QnL)#
of
P#
so
the
. Note
(QnL)# = Q by the uniqueness, and derive a contradiction. Note also, for the extension of Q n L , that Q n L is indeed a proper prime
now that
ideal in L
.
EXERCISE 78.16. proper subring R
We recall that if
,
then R'
R'
is a commutative ring with a
is said to be integrally dependent upon R
whenever for every x' E R' there exists a polynomial n- 1 p(x) = xn + a x + + . a with coefficients an-l,...,ao in R such n- 1 that p(x') = 0 Such subrings R of R' are in some sense analogous to
...
.
the p-sublattices introduced in Definition 78.5.
Prove the following
statements. (i)
If w '
is a proper prime ideal in R'
proper prime ideal in R (ii) such that
If w
.
is a proper prime ideal in R
I' n R c w
satisfies J' Il R c w
,
.
then the ideal J'
a proper prime ideal in R such that
extension w '
such that
I' c
w'
.
and
then w = w ' I'
generated by
(iii) (Monotone prime ideal extension). If w
,
I'
fl
R
n R is a
an ideal in R'
I' and
I'
is an ideal in
c w
,
then w
w
R'
and
has a prime
Ch. 11,1781
Let the sets R
(iv) R'
45
PRIME IDEAL EXTENSION
and
of all proper prime ideals in R and
0'
respectively be provided with the hull-kernel topology. The mapping fl R
$: w ' + w '
is now from R'
,
onto Cl
and
is hk-continuous and
$
hk-closed. HINT: For (ii) and (iii), let w ' w'
3
I' such that
w'
fl R
w'
fl
R
is included in w = w
.
be an ideal in R'
satisfying
is maximal with respect to the property that
w'
. Then, by
It follows that
Exercise 52.14(i),
, so
J' c w '
w'
.
n Rcw
J'
is prime and
A second proof is obtained by considering the quotient ring
Let $: R'
-f
S
be the canonical homomorphism from R'
is integrally dependent upon $(R)
. Hence, by
in $(R)
and
in S
such that w"
I' and
3
Let R and
that prime extension from R x'E
R'
.
w ' fl R = w
to
fl $(R)
R'
be as in the preceding exercise. Show is unique if and only if for each
there exists a finite set
...,x
{xl,
}
of elements in R
containing x'
contains
and conversely (i.e., the radical of the ideal generated by {xl,
...,xn)
such
{x,,
x'
...,x
-+
w'
fl R
).
, we
is now a homeomorphism. Hence, for x' E R'
By the compactness of
{P')xl
...,xn
there exist xl,
have
such that
U~IP'I , so any P' satisfying x' E P' also satisfies i: {x1,...,x 1 c P , and conversely. n For the converse, assume that Pi # Pi , but $(Pi) = $(Pi) It may
I P ' I ~ ~=
.
be assumed that there exists an element x' E R' {P'IX, but that $(Pi)
such that P i
...,x
=
Uy{P')
but not of
. xi
$(P;)
It follows that
...,xn 1
{xI,
. This contradicts
$(Pi)
is in
in R
Pi is not. By hypothesis there exist x l ,
is a subset of
= $(Pi)
.
1,
equals the
HINT: Assume first that prime extension is unique, so the mapping $: w '
.
R'
that every proper prime ideal in R' radical of the ideal generated by
= $(w)
is a prime ideal in
It is not difficult to prove now that w ' = $-'(w-)
EXERCISE 78.17.
S
is a proper prime ideal
$(w)
N
satisfying w '
S = R'/I'.
. Then
the Cohen-Seidenberg theorem (cf. again Exercise
52.14), there exists a prime ideal w R'
onto S
such
46
Ch. 11,1791
STRONGLY ORDER DENSE RIESZ SUBSPACES
75. Strongly order dense Riesz subspaces According to the definition in section 21 the ideal A space L
is said to be order dense in L
equal to L dense in L
if the band generated by
A
. In an Archimedean space L , therefore, the ideal A is whenever Add = L . This definition can be extended to an of the Archimedean space L
arbitrary non-empty subset D said to be order dense i n
L
if
Ddd
=
L
, i.e.,
D
e ( s o {e}d
= {O}
is
is
. By way of example, if the
Archimedean space L possesses a weak unit subset of
in the Riesz
), then any
L containing e is order dense. F o r an ideal this definition of
order denseness agrees quite well with our intuitive ideas of denseness, but not for an arbitrary subset. We shall introduce, therefore, a different notion, the notion of strong order denseness for a linear subspace of an Archimedean Riesz space. For an ideal strong order denseness will be the same as order denseness, but for a more general linear subspace strong order denseness is a much stronger requirement than only order denseness. The definition follows now. The linear subspace D is called strongly order dense i n
of the Archimedean space L
L if, for every
u >
0 in L
,
the
intersection {u}dd n D does not consist only of the zero element. Here we restrict ourselves to the case that D
is a Riesz subspace of
one of the main results will be that the Riesz subspace D order dense in L
if and only if for every u > 0 in L
element v
such that 0 < v
in D
THEOREM 79.1.
5
u
, then
, and
there exists an
holds.
If the linear subspace
strongly order dense i n L
L
is strongly
D o f the Archimedean space
D i s order dense i n
L
L is
, but the converse
i s not always true (not even i f
D is a Riesz subspace). An ideaZ, however, is strongly order dense i f and only i f the ideal i s order dense. PROOF. Let D be strongly order dense in L {uldd
nD
+ COI
for every
d If Ddd # L would hold, then D # {O} in L
satisfying 0 < u E Dd
{ddd 0 D L.
= {O}
. But
o
< u E L+
, so
so
.
there would exist an element u
then we have
. Contradiction. Hence
,
Ddd = L
{ujdd
,
C
i.e., D
Dd
,
so
is order dense in
Ch. 11,5793
47
PRIME IDEAL EXTENSION
TO show that order denseness (for a Riesz subspace) does not always
imply strong order denseness, let L functions on [0,11 Then D
, and
be the Riesz space of all real
let D be the Riesz subspace of all constants.
is order dense, but not strongly order dense.
Finally, let I be an order dense ideal in L
. Since
given in L
I is an ideal and u E Idd = L
element v E I satisfying 0 < v that 0 < v E {uIdd
n I , so
2 u
, and let u > 0 be , there exists an
(cf. Theorem 19.3(iii)).
{uIdd Il I # { O )
.
It follows
I is
This shows that
strongly order dense. For the main theorem in this section, let L be a Riesz subspace of the Archimedean Riesz space L' (the same notations as in the preceding section). For any subset D of L'
,
respect to
L'
Dd
element f
, we
will be denoted by shall write fdd
the disjoint complement of
.
Sometimes, if D
D with
consists of one
. The following theorem
instead of
is essentially due to A. Bigard (C11,1972,Lemma 2). THEOREM 79.2. Given the Riesz subspace
space
L' (i)
have
L of the Archimedean Riesz
, the following conditions are equivalent. L i s strongly dense i n L' , i . e . , for every
Iu'Idd
n L # {Ol
satisfying
0
v
(iii) Every
5
.
u' > 0
(ii) For every u'
.
u' t 0
in
L'
in
1
(
there e z i s t s an element
B'
of a l l bands
B'
in
L'
-+
v
B = B'
n
.
L
A U , ll L
#
in L
i s a one-one mapping from the s e t
onto the s e t of a l l bands
B in L
.
B'
-+
B = B'
i s a l a t t i c e isomorphism between the Boolean algebras o f a l l bands i n L respectively. The inverse mapping i s given by
the bands bands
B'
and
B' Il L and
C'
are d i s j o i n t compZements i n
C' f l L
we
, where A,
{O)
Furthermore, i f these conditions hold, then the mapping and
L'
,
(iv) For every u ' > 0 i n L' we have denotes the idea2 generated by u ' i n L
The mapping
in
L' s a t i s f i e s
u' = sup v:vEL,O
(v)
u' > 0
B' L'
=
Tidd
n
L'
. Finally,
i f and on2y i f the
are d i s j o i n t complements i n L
.
L
48
*
PROOF. (i)
, we
(ii). Given u' > 0 in L'
by hypothesis. Let 0 < p E {u'Idd
fl
. Then
L
n inf(u',p)
.
inf(w,p)
,
zdd r l L # { O }
, which
that w E zd L
>
0 ,
il
so
L # {O) since
such that
so
let 0 < w E zdd fl L
. Finally, let
Then q > 0 , because otherwise it would follow from
0 5 inf(w,z)
of
{u'Idd
s p does not hold, i.e.,
If follows that =
have
inf(u',p)
L' is Archimedean there exists a natural number n
q
Ch. 11,9793
STRONGLY ORDER DENSE RIESZ SUBSPACES
since q
5
inf(nw,np) = nq = 0
5
contradicts 0 < w E zdd
inf(w,p)
=
inf(w,np)
and w
and
. Furthermore, q
are members of
p
L
is a member
. Note now
that
so
(q-n inf(u',p))+ q
so
-1
v = n q
5
=
0
. In other words, we have
n inf(u',p)
5
nu' ,
satisfies v E L
and
0< v
5
u'
. This shows that
condition
(ii) is satisfied. (ii)
*
element v
(iii). Given u' > 0 in L' in L
,
satisfying 0 < v < u'
there exists by hypothesis an
. Hence
u'
is an upper bound of
the non-empty set
).
Mu, = (v:vEL,O
is not the supremum of MU,
satisfying u; < u'
MU, element v1 E L
, and
so
,
there exists an upper bound
1 Of (by hypothesis again) there exists an
satisfying 0 < v 1 5 u'-u' , so by induction v+nv
for every v E Mu'
u'
. But E Mul
then v+vl E Mu,
holds
for every v E Mul
PRIME IDEAL EXTENSION
Ch. 11,1791
n = l,2,
and
... . It follows that
is impossible since L'
(iii)
0 < nv
* (iv). Given
u' > 0
in L'
(iv)
*
(i).
Note that A
L
for n
#
fl L
{u'ydd
is strongly order dense in L'
{O}
nL #
, i.e.,
u' = sup MU,
follows immediately from (iii) u'
0< v
for every for every
{O}
... . This
= l,Z,
. This shows that
u' t 0
holds for every
c
U'
Hence, since by hypothesis A follows immediately that
, it
such that
.
n L # 101
u'
is Archimedean. Hence, we have
that there exists an element v E L A,,
5
1
49
in L'
.
in L' , it
u' t 0
u' > 0
,
in L'
.
i.e.,
It remains to show that (v) is equivalent to the other conditions. In B ' + B'
one direction this is easy. Assume that (v) is satisfied, so
, the
is aone-onemapping. Given u ' > 0 in L'
{ u ' } ~n~ L # {0}
{u'}dd # {0) , so follows that
L
band
{u'}dd
satisfies
since the mapping is one-one. It
is strongly order dense.
Conversely, assume that
L
nL
is strongly order dense in L'
. We shall
prove not only that (v) holds, but also that all the additional properties hold that are mentioned at the end of the theorem. Note first that for any subset D
of
L possessing a supremum do in L , the element do is
also the supremum of an upper bound
db
in L'
D
of
D
satisfying 0 < v < do-db
in
. Indeed, if otherwise, there would exist L' such that db < do . Take v E L
(such an element v
order dense in L' ) . Then d+v s do sup(d:dcD)
= do
in L
implies
exists since L
sup(d+v:dED)
=
d +v 0
the thus established fact that suprema of subsets of the same in L the set B' bands in L implies Bi
and L'
=
so
B'
+
and
n
L
(B;)d
B'
in L'
is one-one, we prove that
, then
one of these, say g'
E (B;)d
,
so
onto
B; # B; B; h'
, is
simultaneously. A l s o , it follows from h' > 0
is contained in B;
contained in B; I7 L
(if existing) are
which is not in the other one. Then
and the strong order denseness of {h'ldd
L
B' I7 L maps bands in L'
inf(1 f'I ,/g'l) > 0 holds for at least one
contained in B;
. Contradiction. Using
follows easily that for any band
. To show that the mapping I7 L # Bi I7 L . If B; # Bi
contains an element f ' h'
, it
n L is a band in L ,
is strongly
for all d E D , but on the other hand
L
that
n L and
but not in B; n L
.
.
{h'ldd n L # { O } The set d (B;) n L , i.e., the set is This shows that B; n L # B; n L
.
50
Ch. 11,9791
STRONGLY ORDER DENSE RIESZ SUBSPACES
It has to be shown next that the mapping B' +B'n L all bands in L form
B
=
B' fl L
,
i.e., we must prove that every band for some band
B'
. Then
bands in L '
, and
Theorem 19.3(ii)). and
C
B IC
in L
is a subset of
C'
. It follows that C , so
is evidently included in B' fl L
In addition, it follows that complements in L' B'
C'
@
, we
@
C
in L
order dense in L' , i.e., B' if
B'
B
B' fl L and
=
and
C'
, which C'
is order dense in L
,
then u and
B'
. On
the other hand dd holds for B' = B
C'
=
.
Cdd are disjoint
0< v
would be disjoint from u'
5
, and
is impossible. Hence, B'
then v @
C'
would is
are disjoint complements. Conversely,
,
then it is evident that
.
If
u t 0 in L dd = B
is disjoint (in L ' ) from B'
u I (B'eC')
i.e., B
holds in L ' (cf.
fl L = B
L are disjoint bands in L
,
, let
C' = Cdd are
is a subset of
B' n L c B
such that
and
is of the in L
B
B' ll L must be included in
are disjoint complements in L' C = C' fl
is disjoint from B fB C dd and from C ' = C , so
L
fl
some u' > 0 in L'
in L
in L
and
B' I C'
. Hence, B'
B' = Bdd and
. Indeed, if
could take v
be disjoint from B
Bdd
=
implies that
the disjoint complement (in L ) of B
B'
(B'flL) I C , since B'
Hence
B
. Given the band
in L'
C be its disjoint complement in L
is onto the set of
. This implies
C
u
=
0
, and
so
B
@
.
are disjoint complements in L
inf(B',B') = B' f l Bi and sup(B',B') = 1 2 I 1 2 hold in the Boolean algebra structure for bands Bi
C
Finally, we recall that =
B;
((Bi)dfl(B;)d)d
in L' (cf. Theorem 22.7).
The mapping
B'
+
and
B' ll L preserves inter-
sections and disjoint complements, s o the mapping preserves suprema and infima, and hence it also preserves the operation of taking the complement in the sense of the Boolean algebra. Since the mapping is one-one, this shows that the mapping is a lattice isomorphism between the Boolean algebras of all bands in L' COROLLARY 79.3. L'
and
space
L' L"
and
L respectively.
(i) If L i s a strongly order dense Riesz subspace of
i s a strongly order dense Riesz subspace of the Archimedean
, then L i s strongly order dense i n
L"
.
(ii) If L i s a strongly order dense Riesz subspace of the Archimedean space L' and B' i s an ideal i n L' , then B = B' f l L i s strongly order dense i n B' .
PROOF. (i) Given u" > 0 in L" , there exists an element u' in L' u' 5 u" Corresponding to u' , there exists an element u in L such that 0 < u 5 u' It follows that 0 < u 5 u" , and so L is
such that 0
.
.
Ch. 11,9791
51
PRIME IDEAL EXTENSION
.
strongly order dense in L"
(ii) We have to show that for any
n
u' > 0 in B '
.
there exists an
L satisfying 0 < v 5 u' Given u ' > 0 in B ' , there exists an element v in L satisfying 0 < v 5 u ' (since L is in B '
element v
strongly order dense in L' ) . From 0 < v < u ' E B'
,
v E B'
v E B'
so
n
L
=
B
. This is the desired
it follows that
result.
Given the Riesz subspace L of the Riesz space L' , we shal
say that
L has the unique band extension property with respect to L' whenever, for every band
B
B' l l L = B
that
in L
.
, there
exists a unique band
B'
in L'
such
For an Archimedean space this property is equivalent to
strong order denseness of
L
.
in L'
THEOREM 79.4. If L i s a Riesz subspace o f the Archimedean space
then L has the unique band extension property with respect t o L
only i f
i s strongly order dense i n L'
.
PROOF. It is evident from the preceding results that if
L
,
L'
i f and
L'
is strongly
order dense in L' , then L has the unique band extension property. Given the band
B in L , the band B '
=
For the converse, assume that
Bdd is the unique extension. L has the unique band extension property.
I01 in L has the band IO} in L' as its unique extension. Hence, given any band B' # { O } in L' , we must have B ' n L # 101 , since The band
otherwise B'
,
L'
the band
{u'jdd
nL#
would also be an extension of {u'jdd {O}
. This shows
{O}
. Given now u' . Hence
>
0 in
satisfies {u'Idd # { O }
in L'
L
that
is strongly order dense in L'
.
EXERCISE 79.5. Let L be a Riesz subspace of the Archimedean space L' Show that if L
is included in the Dedekind completion L#
is strongly order dense in L'
.
L# = (L')#
L
L'
of
In particular, show that if it is given that L# = ( L ' ) #
is strongly order dense in L'
the Archimedean space L'
, then
are not necessarily equal (and
.
so
L
the Dedekind completions L# L'
(c,)
.
,
.
then
is strongly order dense in (L')#
and
is not necessarily included in L#
HINT: For a counterexample in the converse direction, let L' =
, then
and the Dedekind completions satisfy
In the converse direction, show that if
L
L
=
em
1. and
52
Ch. 1 1 , 5 8 ~ 1
EXTENSION THEOREMS
be the Riesz space of all real functions on
EXERCISE 7 9 . 6 . Let L' the interval [0,1] Show that
for every band is equal to
and let L
be the Riesz subspace of all constants.
.
is not strongly order dense in L'
L
in L' , except when B '
B'
.
{O}
EXERCISE 79.7. Let
In particular, show that
, the
L'
=
L' be the Riesz space of all real Lebesgue
measurable functions on the interval [ O , l l
,
and let L
subspace of all continuous functions. Show that L dense in L'
intersection B' n L
be the Riesz
is not strongly order
.
HINT: Similarly as in Exercise 18.14 one can easily construct a function u'(x)
2
0 in L' , not identically zero and such that any satisfying 0
continuous v(x)
v(x)
5
5
u'(x)
is identically zero.
80. Extension theorems In the present section we shall again assume that
.
space of the Riesz space L' by
L
that
By
L
is a Riesz sub-
IL we denote the ideal generated in L'
. Since all results in this section will be
about the various ways
L can be embedded in IL , there is hardly any l o s s of generality in
assuming immediately that L' = IL denote by
.
I in L , we shall now
For any ideal
I' the ideal generated by
I in L'
. Note
that for
I
=
L
this
notation is consistent. Before stating any specific theorems, we make one general remark. Given the prime ideal Q is a prime ideal in L
in L'
,
it is evident that Q n L
. Conversely, given the prime
ideal P
in L , there
is (in view.of the prime ideal extension theorem) at least one prime ideal
Q
in L'
such that Q n L
prime extension Q
of
P
=
P'
Also, even in the case that
extension of
P
.
THEOREM 80.1.
Let
L'
P
=
IL
equivalent. (i) For every prime ideal (ii)
.
It is not necessary, however, that the
is exactly the ideal P'
L'
.
P in L'.
is prime, P' may not be the only prime
. The following P in
conditions are now
L , the ideal
For every minimal prime ideal M
minimal prime ideal i n
generated by
in
L
P'
i s prime in
, the ideal
M'
is a
L'
.
PRIME IDEAL EXTENSION
Ch. ll,§8Ol
(iii) For every minimal prime i d e a l i n L'
i n L , the i d e a l
M
.
53
I f these conditions are s a t i s f i e d , then t h e mapping
M
one mapping from the s e t of a l l minimal prime i d e a l s i n L of a l l minimal prime i d e a l s i n L' . PROOF. (i)
M' Q
*
(ii). Given the minimal prime ideal M
is prime by hypothesis. For the proof that M' is a prime ideal in L '
included in M' fl L
ideal in L
is now an ideal in L' in L'
M'
including M
=
M , and
including M
, we
Q
must have
3
.
M'
i s a one-
onto t h e s e t
,
in L
Q fl L
the ideal
MI
is a prime
. Hence, since
Q fl L = M
so
and since
M'
is minimal, assume that
. Then
satisfying Q c M'
+
i s prime
M'
Q
is the smallest ideal
It follows that
Q
=
M' , i.e.,
is a minimal prime ideal. (ii)
*
(iii). Evident.
6;;)
(i). Given the prime ideal P
prime ideal in L
such that M
prime, s o on account of
L , let M be a minimal
in
is included in P
P' 3 M '
. By
hypothesis M'
is
P' is prime.
the ideal
Now assume these conditions satisfied. It is evident that the mapping M so
II. = M' f l L , so M I = M2 ), 2 all that remains to be proved is that every minimal prime ideal Q in
-t
M'
is one-one (indeed, M'
1
L ' is of the form M'
=
Mi
implies M'
1
ll
for some minimal prime ideal M
the minimal prime ideal Q minimal prime ideal in L
in L '
. We prove
properly included in Q , then M'
. Hence, let
that Q
=
M'
. Indeed, if
M'
is a were
would be a prime ideal (by condition
(iii)) properly included in the minimal prime ideal Q Hence Q = M'
in L
be given. Evidently, M = Q fl L
.
. This is impossible.
It is a natural question now whether there are some simple conditions for L P'
and
in L'
L'
implying that for every prime ideal P
in L
the ideal
is indeed prime. In the following theorem we present some
conditions of this kind. THEOREM 8 0 . 2 . Besides assuming t h a t
and Zet
L
be strongly order dense i n
p r o j e c t i o n property. Then P'
L ' = IL
, l e t L' be Archimedean
. Furthermore,
let i s prime f o r every prime i d e a l L'
L
have the
P
in
L
.
54
EXTENSION THEOREMS
Ch. Il,§801
PROOF. The proof is analogous to Masterson's proof in Exercise 37.17 for the particular case that L'
,
be a prime ideal in L
and let
is the Dedekind completion of inf(u',v')
. For notational convenience, we
or v' E P '
proved that u' E P'
denote the principal ideals generated by respectively. Note that D I E ,
E
The bands
and
. On account of
L
d D n L cP
we have
or
S
w
E Dd
1
w
. Since n
L
, so
n
L
P
shall D
and
E c Dd , in particular v' E Dd
.
by Theorem
. Assume that
d D
n
L c P holds. We
holds, as follows. Take w E L
has the projection property, we have w = w
and w2 E Ddd
L
in L' by
v'
. Let
to be
Ddd fl L are disjoint complements in
L c P
Ddd
prove that in this case v' E P ' v'
so
and
u'
and Ddd are disjoint complements in L'
Dd
79.2 the intersections Dd n L
L
. It has
in L'
= 0
. It follows then from
satisfying
+ w2
with
v' 5 w = w1+w2 with
v',wl in Dd and w2 in Ddd that v' 5 w 1 (indeed, writing s = (v'- wli+ for brevity, it follows from s 2 w2 that inf(s,w 2 ) = s . and w2 E Ddd , so inf(s,w ) = 0 , i.e., s = 0 so v' S w 1 ) . But s E D 2 d This shows that v' E P ' If Hence v' 5 w E D n L c P c P ' Ddd u'
nL
.
1
.
holds, we prove similarly that u' E P'
c P
.
holds. Thus we have
E P' or v' E P' , which is the desired result. In the next theorem we present some conditions
only be satisfied if
L
Dedekind completion of
L'
and
L
,
coincide. For the case that L'
THEOREM 80.3. Besides assuming that
(ii)
an element
u1,u2 E L+
1
L'
2
such that
u
=
=
IL
(iii) L
=
L'
.
is the
such t h a t
, l e t L' be Archimedean.
u;,u; E (L')+
u1 + u2 and
(Interval density property). I f fo E L
strong that they can
the theorem i s due to J. Quinn.
The follozJing conditions are now equivalent. (i) I f 0 5 u < u'+u' with u E L+ and elements
so
f' 5 f o 5 g'
f' < g'
.
ui 5 u!
i n L'
for
, there e x i s t i = 1,2
.
, there e x i s t s
.
Note t h a t i n (i) we have written u
Ch. 11,9801
* (ii). It is sufficient to prove that if
PROOF. (i)
given in L' there exists an element in L Hence, let
0 < v' < u'
0 < v' < u' 5 uo
0
=
. Then (u -vl) 0
+ v'
< (u -v') + u '
0
by hypothesis there exist u1,u2 E L+
u
5
u -v' 0
u2
so
-
u2 5 u '
v'
u -u 1 = u 2 < u ' ,
is an element in L
(ii) If u'
. Then
and
Archimedean and
u'
It follows that
u' E L
If in L' d'.
=
=
u'
.
, we
and
iu' < u'
u'
is an atom, so
.
u'
for
i
1,2
=
sup(d;,d;)
.
i
= 1,2
sup(d;,d;)
Hence
(iii)
5
by hypothesis
. But
u'
L'
is
is now a numerical multiple of
0 and
=
, we have 0
= u'
5
sup(v;,v;) d! < u '
+ sup(-v',-v') 1
By (ii) there exist elements u1,u2 E L+ 2
. .
z
is not an atom, there exist strictly positive elements v;,v;
satisfying inf(v;,v;)
u'-vi
E L such that f u r < z
z
u1+u2 with
have to show that u ' E L
is an atom, we observe that it follows from
that there exists an element
.
such that
such that uo
IL
=
uo E L
is u'
and
,
lying between v'
(iii). Given 0 < u ' E L'
0 < v' < u '
that lies between v'
. First, choose
in L'
so
1
55
PRIME IDEAL EXTENSION
=
u'
* (i).
2
=
_i
u'
. Writing
and
u ' - inf(v' v') = u ' 1' 2
such that
d! < u . < u' 1
1
sup(ul,u2) i u ' , but on the other hand
. It follows that
If L
and L '
decomposition property in L
=
sup(ul,u2)
=
u'
, and
. for
sup(ul,u2) 2 so u ' E L
.
coincide, then condition (i) is the Riesz
L'
.
In the preceding sections we have met the notions of unique prime ideal extension and unique band extension. We shall now define unique ideal extension. Given again that that L
ever, for every ideal L'
L
is a Riesz subspace of
L' , we shall say
has the unique idea2 estension property with respect to L'
such that
J n L
=
the ideal generated by
I in
.
when-
L , there is a uniquely determined ideal J
I In this case we must have J = I' , i.e., J is I in L' It is also evident that every ideal J
.
in
56
in
EXTENSION THEOREMS
i s now of t h e form
L'
f o r some
I'
in
I
The following theorem, f o r t h e case t h a t
, namely
L
for
.
I = J Il L
i s t h e Dedekind comple-
L'
L , i s due t o J . J . Masterson ( [ l l , c f . E x e r c i s e 37.16).
t i o n of
THEOREM 80.4.
Let
L'
.
a > 0
PROOF. Assume t h a t t o prove t h a t
Then
has t h e unique i d e a l extension
L
if x(L)
i f and only
there e x i s t s an element
0 < u' E L'
f o r some
.
IL
=
L'
property w i t h respect t o every
Ch. I l , § 8 O l
X(L)
=
u
=
, i.e.,
X(L')
E L+ such t h a t
for
au 5 u' 5 u
has t h e unique i d e a l e x t e n s i o n p r o p e r t y . One way
L
i s by o b s e r v i n g t h a t unique i d e a l e x t e n s i o n
X(L')
i m p l i e s unique prime i d e a l e x t e n s i o n , and so
X(L)
=
by Theorem 78.9.
X(L')
A proof which makes no u s e of Theorem 78.9 can b e given a s follows. Given
,
0 < u' E L ' and l e t
U'
A'
ideal
A,'
let
A = A
Il
b e t h e p r i n c i p a l i d e a l g e n e r a t e d by
. By
L
A
generated by
in
so t h e r e e x i s t s an element u E A
,
L'
u E A
u E A' = AU,
implies that
A,'
t h e unique e x t e n s i o n
A,,
so
,
so u
2
. It
= A'
such t h a t
I
in
follows t h a t
J Il L = I
. Given
h y p o t h e s i s an element
u E L+
f o l l o w s from ctu 5 u ' that
u' 5 u
that
I' c J
u' E I '
L
. Hence
J
=
, there
nL
R' =
u
I'
,
, and
= I
in
w'
w' + w'
L'
L'
=
IL
X(L)
=
is besides
E A',
J
X(L') I'
.
no
i s an i d e a l i n
t h e r e e x i s t s by
au 5 u ' 5 u
f o r some
a > 0
. It
J c I'
, and
. Show t h a t
. It
denote by
is evident
R'
the s e t
and hk-closed.
Bedides
.
nL
i s hk-continuous
w e may i n t r o d u c e i n R' t h e T -topology, h a v i n g a l l sets P as a b a s e f o r t h e open s e t s . E v i d e n t l y t h e T -topology i s ({oi}u:uE~+) P weaker t h a n t h e hk-topology. Show t h a t t h e t o p o l o g i e s a r e i d e n t i c a l i f and t h e hk-topology,
only i f
X(L) = x(L')
.
L'
so i t f o l l o w s now from
.
(~w'lu:u€L+)
Show t h a t t h e mapping
J
have proved t h u s t h a t
EXERCISE 80.5. Assume a g a i n t h a t of a l l proper prime i d e a l s
in
u' 2 0
such t h a t u € J
that
. We
any
u'
.
o t h e r i d e a l e x t e n s i o n . Assume, f o r t h i s purpose, t h a t satisfying
L' ,
u' 5 u . On t h e o t h e r hand -1 a u' f o r some a > 0
For t h e proof i n t h e converse d i r e c t i o n , assume t h a t We have t o show t h a t , f o r any i d e a l
in
u'
i s then equal t o t h e
Ch. 11,1811
PRIME IDEAL EXTENSION
57
81. Prime i d e a l e x t e n s i o n and t h e p r o j e c t i o n p r o p e r t y Throughout t h e p r e s e n t s e c t i o n we assume as our b a s i c h y p o t h e s i s t h a t L' = I L
, L'
L
i s Archimedean and
Theorem 80.2 i t has been proved t h a t i f , i n a d d i t i o n , L p r o p e r t y , then
P'
i s prime f o r every prime i d e a l
. In
L'
i s s t r o n g l y o r d e r dense i n
has t h e p r o j e c t i o n
P
in
. It
L
may be
asked whether t h e r e i s a r e s u l t i n t h e converse d i r e c t i o n . The answer i s a f f i r m a t i v e . It w i l l be one of t h e main theorems i n t h e p r e s e n t s e c t i o n t h a t i f , b e s i d e s t h e b a s i c h y p o t h e s i s , we assume t h a t p r o p e r t y , then only i f
i s prime i n
P'
due t o K.K.
We f i r s t make a simple remark. S i n c e
L'
,
u
E L+ has t h e p r o p e r t y t h a t
element
, this
L
L' i s
main r e s u l t i s
i s Archimedean and
L'
L
B
Bdd
i f and only i f
L ' (where, a s b e f o r e , Bdd
t h e second d i s j o i n t complements of if
L i f and
is
L
i t follows from Theorem 79.2 t h a t t h e bands
a r e d i s j o i n t complements i n
C
a r e d i s j o i n t complements i n
u = v+w
in
Kutty and J . Quinn ([11,1972).
s t r o n g l y o r d e r dense i n and
P
has t h e p r o j e c t i o n p r o p e r t y . For t h e s p e c i a l c a s e t h a t
L
t h e Dedekind completion o f t h e Archimedean space
B
has t h e p r o j e c t i o n
L'
f o r every prime i d e a l
L'
and
and
C r i n t h e space
u = v+w w i t h
i s a l s o t h e (unique) decomposition of dd v E Bdd and an element w E C
.
v E B
u
in
Cdd
denote
L ' ). Hence,
and
L'
and
Cdd
,
w E C
then
as a sum of an
THEOREM 81.1. Assume, besides the basic hypothesis mentioned above,
that
L'
has the projection property, and l e t
B
be a band i n L
. Then
i s a projection band in L if and only i f B ' = Bdd holds. I t follows that L has the projection property i f and o n l y i f B ' = Bdd hoZds for
B
every band
B
in L
.
PROOF. Assume f i r s t t h a t
B'
C
Bdd
satisfies
. Given
u' E B '
Observe now t h a t
i s a p r o j e c t i o n band i n
(since
L'
0 < u'
E Bdd , l e t
z
L ) and
z
L
u' 2 0
. Since Bdd
in
satisfy
E L+
h a s an o r d e r p r o j e c t i o n on t h e band
z
a p r o j e c t i o n band i n
Bdd
B
i s e v i d e n t , we have t o prove only t h a t any
u' < z
B (since
B
. is
h a s a l s o an o r d e r p r o j e c t i o n on t h e band
has t h e p r o j e c t i o n p r o p e r t y ) . According t o t h e remark
h e d i a t e l y preceding t h e p r e s e n t theorem, t h e s e p r o j e c t i o n s a r e t h e same, i.e.,
pBz = PBddZ
. on
u' < PBddz, and s o by an element of
account of
u' 5 PBz B
. It
. Note
u' < z
and
now t h a t
follows t h a t
u'
u ' E gdd
P8z E B
E B'
. This
, so
i t follows t h a t u'
i s majorized
is the desired result.
58
11,9811
PRIME IDEAL EXTENSION AND PROJECTION PROPERTY Conversely, assume t h a t
We have t o show t h a t every
. Given
w IB
and
u = v+w
with
L ' ) . I t f o l l o w s from
0 5 v I z
satisfying and
u
v E Bdd
and
satisfying
w E Bd ( s i n c e
B' = B
u = v+w
is certainly so t h a t
0 5 v E Bdd = B '
. We
L
i s of t h e form
u E L+
, it
u E L+
v E Bdd
i s a band i n
B
with
t h a t t h e r e e x i s t s an element
s h a l l prove now t h a t
n
,
and t h e proof w i l l b e complete. For t h e proof of
t h i s w i l l imply then t h a t t h a t i t f o l l o w s from
v
is in
,w E
z E B
,
L
Bd
z E B
. Since
inf(z,u) = v
,
, note
v E B
i s a p r o j e c t i o n band i n
Bdd
L
L = B
.
can b e w r i t t e n a s
u
are i n
inf(z,u) = v
dd
z
so
that
.
i n f ( z , u ) = inf(z,v+w) = i n f ( z , v ) = v This concludes t h e proof.
THEOREM 81.2. Assume, besides the basic hypothesis, t h a t
i s prime i n L'
projection property. Then P' L
i f and only if
L
P'
f o r every prime i d e a l
L
i s prime f o r every prime i d e a l
P'
t o prove, t h e r e f o r e , t h a t i f prime f o r every prime i d e a l
in
L
in
P
L
in
i s properly l a r g e r than
Bdd Bdd
and l e t
is not i n
such t h a t
u'
v = v;+v;
with
we d e r i v e from
i s not i n
B'
Note t h a t B'
not
n
,
v'
1
B'
is n o t i n
L = B c B ' ) , so v;
v E L+
, but
L
. Then
P = Q
n
B'
, so
have only
P
in
. Now
,
L
is
P'
f a i l s t o have t h e p n o j e c t i o n
B
in
L
take
v E L+
.
such u' > 0
t h e r e e x i s t s an element
L
and
,
v ; E Bdd
,
L (because o t h e r w i s e i s likewise not i n
t h e r e exists a prime i d e a l vi
in
satisfying
u' 5 v,
v ' E Bd Taking components i n Bdd, 2 t h a t u' 5 v ' , s o v i i s n o t i n B' (because u' 1 i s an i d e a l i n L' ). R e c a p i t u l a t i n g , w e have
u' I v and
B'
v i E Bdd
v = v'+v' with 1 2
Bdd
. We
i s prime f o r any prime i d e a l
p r o p e r t y . By t h e p r e c e d i n g theorem t h e r e e x i s t s t h e n a band that
P
has the projection P
h a s t h e p r o j e c t i o n p r o p e r t y . Assume, on t h e c o n t r a r y , t h a t
L
then
has the
has the projection property.
PROOF. I t was proved i n Theorem 80.2 t h a t i f property, then
L'
in
L'
i s prime i n
L
Q
v; E Bd
vi L
, v;
not i n
.
B'
would be i n
. Since
such t h a t
Q
vi
is not i n
contains
€5'
and s o , by h y p o t h e s i s , t h e
but
Ch. I l , § S l l
corresponding ideal P'
is prime in L '
not in P'
v; E P'
. Indeed, if
2
v;
, we
get then that v'
v-v' > v
=
1
2 -
inf(w,V,)
-
. We
shall prove now that v;
is
would hold, there would exist (by the
P' ) an element w E P
definition of v
59
PRIME IDEAL EXTENSION
2
inf(w,v)
satisfying w v; 2
0
2
. Since also
v;
, so
.
On account of vi E Bdd the last inequalities imply that
-
v
inf(w,v) E gdd
, so
We have also w E P c Q v - inf(w,v) E Q v; E Q
in P'
= B c Q
inf(w,v) E Q
,
that v E Q
and so
. , and
vi E Q
hence it follows from (since 0
is a statement contradicting the defini-tionof Q
not in P' P' c Q
nL
, so
. The element since v'1
vi
is likewise not in P'
but neither vi
nor v;
It follows that L
vi
v ) . But
S
v;
Q ), vi
is not
. Indeed, we have
is not in Q (by the definition of
either. Thus we arrive at the situation that
5
. Hence,
inf(vi,v;)
= 0
is
,
. This yields a contradiction.
is a member of P'
has the projection property. For a completely different
proof we refer to Exercise 81.7. We shall now consider the situation that L space and L'
the Dedekind completion of L
is an Archimedean Riesz
. The conditions in the basic
hypothesis are then automatically satisfied (i.e., L' Archimedean and L
=
IL
, L'
is
is strongly order dense in L' ) and, moreover, L'
has
the projection property. Hence, Theorem 81.1 and Theorem 81.2 in the beginning of this section become very simple statements. Before discussing these statements, we prove a simple but remarkable theorem. THEOREM 81.3.
space L and compZetion of
If L'
i s the Dedekind eomp2etion of the Arehimedean
I i s an arbitrary idea2 in
I
.
PROOF. Note first that
L' , s o
I'
L
, then
I'
is the Dedekind
is an ideal in the Dedekind complete space
exist v and
I is a Riesz subspace of 1'. there w in I such that 0 < v s u' s w (cf. Theorem 3 2 . 6 ) .
Hence, let u'
P
I'
is Dedekind complete. Evidently
The proof will be complete if we show that for any u' > 0 in I' 0 in I' be given. By the definition of
I'
, there
exists
60
Ch. 11,§811
PRIME IDEAL EXTENSION AND PROJECTION PROPERTY
an element w E I such that
u' 5 w
. Also,
u'
since we have
E L'
and
L' is the Dedekind completion of L , there exists an element v E L+ that 0 < v 2 u ' It follows that v E I' Il L = I Hence 0 < v 2 u '
.
with
v
in I
and w
.
. This is
such w
2
the desired result.
We state and prove the final theorem in this section. THEOREM 81.4. Let
The band
(i)
B
i . e . , i f and only i f
that band
L' be the Dedekind completion of t h e Archimedean
.
L
Riesz space
in
i s a projection band i f and only i f
L
Bdd i s t h e Dedekind comp2etion of
B
. It
B'
follows
L
has the p r o j e c t i o n property i f and only i f i t i s true f o r every
B
in L
(ii)
that
L
and only i f
B
Bdd i s the Dedekind compZetion of
i s a prime idea2 i n L'
P'
dd B ,
=
.
f o r every prime i d e a l
in L
P
i f
has t h e p r o j e c t i o n property.
(iii) The following conditions are equivaZent. (a) L s a t i s f i e s Masterson's'condition i . e . ,
i n L' that
u > 0
there e x i s t s an element (YU 2
u'
5
.
u
(b) For every prime i d e a l
Q
i n L'
Q fl L
such t h a t
=
(c) Every idea2
J
the Dedekind completion of
J
i n L and a number
P (and a c t u a l l y
n
L
.
u' > 0
a > 0
such
there i s only one prime i d e a l
in L
P
i n L'
f o r every
Q
satisfies
i s then equal t o
, i.e.,
J = (JnL)'
P' ). J
is
PROOF. (i) and (ii) follow from Theorem 81.1 and Theorem 8 1 . 2 respectively. The equivalence of (a) and (b) in (iii) was proved already in For the equivalence of (a) and (c), note that by Theorem 8 0 . 4
Theorem 78.11. L
if
has the unique ideal extension property with respect to L
= (JllL)'
EXERCISE 81.5.
holds for every ideal J L' = IL
Assume that
and
L'
(i)
J = (JnL)'
(ii)
X(L)
(iii) L (iv)
Q
=
X(L')
L
and
.
.
in L' L' have the same principal ideals.
holds for every ideal J
, i.e.,
in L '
is Archimedean. Show that
the following conditions are equivalent.
.
if and only
satisfies Masterson's condition, i.e., (a) is equivalent to the
statement that J
L'
L'
has the unique prime ideal extension property with respect to =
(QflL)'
holds for every prime ideal Q
in L '
.
Ch. 11,5813
61
PRIME IDEAL EXTENSION (ii), (iii) follows from Theorem 78.9
HINT: The equivalence of (i),
and Theorem 80.4. It is evident that (i) implies (iv). To show that (iv) implies (iii), let P be prime in L and let Q be an arbitrary prime extension of
P
determined by
P
. Then .
Q
EXERCISE 81.6. Assume that and
L
= P'
(QflL)'
=
L'
by (iv), so
, L'
IL
=
Q
is uniquely
has the projection property
is strongly order dense in L' (note that these conditions are
satisfied if
L'
L ) . Show that the following
is the Dedekind completion of
conditions are equivalent. (i)
L
has the projection property.
(ii)
For every minimal prime ideal M
(iii) Every minimal prime ideal N (iv)
Every band
in L
E
L'
in
in L'
satisfies E
HINT: Condition (i) is equivalent to P' ideal P
the ideal M'
is prime
is then a minimal prime ideal in L' ).
in L' (and actally M'
satisfies N (EflL)'
=
.
being prime for every prime
and this is equivalent to (ii) by Theorem 80.1. For the
in L
equivalence proof of (ii) and (iii), assume that (ii) holds, and let
. Then
a minimal prime ideal in L' easily seen that N fl L
N fl L
and this will imply by (ii) that M'
, which
included in N
(NflL)'
. By
that M
=
,
is minimal).
it follows from (ii) that
is included in N
.
and
N
is
is prime in
.
N = (NflL)'
be a minimal prime ideal
It follows from (iii) that N = (NflL)'
in L'
, so N
For the proof that (i) implies (iv), let (i) hold and let
. Then
B = E rl L
is a band in L (since L'
is strongly order dense in L' , cf. Theorem 79.2), condition (i). But
Bdd
=
(EflL)dd
=
=
M'
such
. This
. Conversely, assume that
in L
=
Bdd is a band in L'
E
=
E
from (iv) that
(EflL)'
, i.e., Bdd
(iv) holds, and let
such that E fl L = B
= B'
, and
E
be a band
is Archimedean and dd so B' = B by
E (cf. again Theorem 79.2),
E = B' = (EflL)'
. Than
properly
(NflL)'
is minimal, so
,
is prime, so (ii) is satisfied.
shows that M' in L'
be
Since N fl L
Theorem 52.3 there exists a minimal prime ideal N N fl L
N
and it is
is prime in L' with M'
Conversely, assume that (iii) holds, and let M in L
,
properly included in N fl L
in L
is impossible since N
a minimal prime ideal in L
. But
is prime in L
is even a minimal prime ideal in L (since other-
wise there is a minimal prime ideal M
L'
.
(NflL)'
=
L
so
B be a band
. It follows
this is equivalent to (i).
Ch. 11,1811
PRIME IDEAL EXTENSION AND PROJECTION PROPERTY
62
EXERCISE 81.7. In Theorem 81.2 it was assumed that
, L'
IL
=
has
L is strongly order dense in L' , and it was
the projection property and is prime i n
proved that P'
L'
for every prime ideal P
L'
in L
if and
only if L has the projection property. Since it was shown already in Theorem 80.2 that P'
in L
is prime for every prime ideal P
if L
has
the projection property, the proof in Theorem 8 1 . 2 was restricted to showing that L has the projection property if P
in L
Assume
. There is a
P'
mapping
+
is prime for every prime ideal
completely different proof for this last statement.
P
prime for every prime
IT:M
P'
. Note
first that by Theorem 80.1 the
is a one-one mapping of the set M
M'
onto the set M'
prime ideals in L
of all minimal
.
of all minimal prime ideals in L'
Note, furthermore, that in L' every prime ideal contains a unique minimal prime ideal (since L'
M'
. Under
set
has the projection property; cf. Theorem 37.ll),
. Now consider the hull-kernel
the same holds in L
the mapping
{M'IU
so
inverse map
-1
is continuous. Thus,
of the compact space continuous on
the image of any open base set
IT
{MIu
morphic mapping from
-1
{MIu
u E L+
onto
{M'IU
,
therefore,
. Note
follows from the hypothesis that L ' (and hence also in
IT
{MIu
,
so
IT
is also
induces a homeo-
that the compactness of
has the projection property
By Exercise 37.14 the hull-kernel topology in M'
{M'IU ) is extremally disconnected, s o by the homeo-
morphism the same holds in {MIu
is the
is a continuous one-one mapping
IT
onto the Hausdorff space
. For every
(cf. again Theorem 37.11).
of
{MIu
so
and
maps open sets onto open sets, which implies that the
IT
IT
topologies in M
{MIu
,
i.e., the closure of every open subset
is open. Furthermore, {MIu
Combining now the compactness of
is compact by the homeomorphism.
{MIu
for every
u 2 0 with the fact that
every prime ideal in L contains a unique minimal prime ideal, we may conclude already that L has the principal projection property (cf. Theorem 37.II). Once having shown that
L
has the principal projection property,
Theorem 31.3 can be applied, i.e., L will have the projection property if it can be shown that the lattice of all principal bands in L complete. Denote by
BU
is equivalent to
, and
{MIu c {MIv (cf. Theorem 35.5).
BU c Bv in the lattice of all principal bands we have {MIU
u E L+
the band generated by
is open and closed, we have
1. c BU
B UT
is Dedekind
note that
Assume now that
. Then, since
Ch. 11,§821
and the set A
, being
well as closed. Since A =
{MI
B
C
uT
If
is compact, it follows from Theorem 37.2 that
{MIu
,
. Note
that
{MI
uT
c{M)
for all
uo
implies
T
B is an upper bound of the set of all uo uO Bv is another upper bound, then {MI C {MIv for all T , s o
A c {MIv
for all
{MIu , is open as
the closure of an open subset of
for some uo E L+
uO B
63
PRIME IDEAL EXTENSION
,
T
so
.
i.e., {MIU c {MIv 0
supremum of the set of all
UT
,
It follows that BuO c Bv
.
B
UT
so
B uO
BU
T
.
is the
82. Normal and extremally disconnected lattices.
Let
(X,e,e)
be a distributive lattice with the smallest element 0
and largest element e (in the terminology of Chapter 1, e
is therefore
the unit of X ) . In this lattice we can introduce, besides the given ordering, the inverse ordering, also called the duaZ ordering. Then X with respect to the dual ordering a distributive lattice with element and 0 as largest element. Evidently, x v y original ordering are the same as x A y and x v y
e
is
as smallest
and x A y in the with respect to
the dual ordering. It follows immediately that a proper prime ideal for one of these orderings is a prime dual ideal for the other ordering, and conversely. Hence, a maximal ideal for one of the orderings is a maximal dual ideal for the other ordering.
It will be evident that the above duality relation can be used to derive dual results from known results. A s a first example, let us consider Theorem 51.8 (ii), the prime ideal extension theorem from an ideal in to the whole of ideal in
(X,O)
(X,0)
. We
and let w
recall the explicit formulation. Let be a proper prime ideal in I in X
exists a proper prime ideal w ' W'
is uniquely determined by
w
satisfying w '
. Explictly, we have
fl
I
(X,€J)
I be an
. Then there . The ideal
= w
Note that this is equivalent to the statement that every prime dual ideal A in I can uniquely be extended to a prime dual ideal A ' in X
. Explicitly,
Ch. 11,1823
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
64
f o r some aEI) =
= (X:fiX,XA&h
(x:x€X,x>b
f o r some bEh
For t h e l a s t e q u a l i t y , n o t e t h a t i f x 2 b
for
b = x
.
E X
a
A
x
a
A
f o r some
E X
Conversely, i f
x t b E
a E I then
x
, then A
b = b E X.
Now, f o r the p a r t i c u l a r case t h a t the l a t t i c e has a l a r g e s t element I
is a principal ideal
I n s t e a d of
I = Ce,x
I
(x:8<x<xo)
=
, we
,
e
and
l e t us d u a l i z e t h i s r e s u l t .
1 0
= (x:8<x<x )
x
) . It follows t h a t i f (with r e s p e c t t o the o r i g i n a l
get then a subset of t h e form
0 Y = Cx , e l = (x:xo<x<e) Note t h a t with r e s p e c t t o the o r i g i n a l ordering 0 Y i s again a l a t t i c e (but now x i s t h e s m a l l e s t element), and with 0 r e s p e c t t o t h e dual ordering Y i s a p r i n c i p a l i d e a l i n X ( t h e p r i n c i p a l i d e a l generated by
0
i s a prime dual i d e a l i n
ordering) h extension
.
X'
to
X
,
A'
and
Equivalently, every proper prime i d e a l extension
w'
to
X
,
and
= (x:x€X,x
then
A
has a unique prime
in
w
Y
has a unique prime
i s given by
w'
w ' = (x:fix,xvatw
,
Y
i s given by t h e formula
f o r some aEY) =
f o r some bEw
We s h a l l prove t h a t t h e same holds i f
X
does not have a l a r g e s t element.
Of course, the proof i s very s i m i l a r t o t h e proof of Theorem 51.8 ( i i ) i n dualized form, but f o r t h e sake of completeness we include t h e proof. LEMMA 82.1.
sublattice dual i d e a l
Let
1'
in X
and l e t X be a prime dual ideal in the can be extended uniquely t o a pz-ime Explicitly,
xo f (X,6)
Y = (x:xtx ) 0
.
. Then
h
x:x€X,xvaEA f o r a l l aEY PROOF. It is e a s i l y v e r i f i e d t h a t
x:x€X,xva€A f o r a l l aEY
Ch. 11,1821
PRIME IDEAL EXTENSION
is a dual ideal in X
. FOK the proof
(xva) v (yva) E X
for all a E Y
so
X
,
is prime in Y
x v xo E A
X'
that
. But
,
is prime, let x v y E A '
X'
that
in particular f o r
for all a E Y ,
then x v a E X
A'
is prime. It is easy to see that
a = xo
n Y
. This
x E X'
so
= A
X'
so
. Since shows
is an
X . Assume that u ' is another prime extension of X . For the proof that X' is included in p' , take a E
extension of
, so
u' n Y
Y-X
=
X
a is not in u ' ) . Every (since u '
is prime and
the proof that
u'
satisfies x v a E X c u ' ,
x E A'
is not in p ' ) . This shows that
a
is included in
,
1'
let x E !.I'
hold, there would exist an element a E Y
X
. But
y
2
x E u'
and y
Hence x € X' , and s o prime extension of
2
p' c
aE Y
X'
, so
. If
such that y y E u' l Y
. It follows that
u'
=
X
= =
,
, say
or y v xo E X
x v xo E X
it follows that
65
so
. For
X' c u '
x E X' x v a
(so
x E !.I'
would not is not in
. Contradiction.
X' , i.e., the
1. is unique.
In the theory of Riesz spaces we have met several times the conditZon that every proper prime ideal contains a unique minimal prime ideal. Indeed, it was shown in Lemma 37.10 that in a Riesz space the condition is equivalent to the condition that disjointness of of the closures
rw),
and
{w];
{wju
, where
and
{w)v
as usual
implies disjointness denotes the set of
all proper prime ideals not containing the element u E L+ closures are in the hull-kernel topology of the set
,
and where the
of all proper prime
ideals. Also, it follows from Theorem 37.11 that the condition is necessary for the Riesz space to have the projection property. It will be shown in the present section that for lattices the corresponding condition is related to normal topological spaces, and the dual condition is similarly related to extremally disconnected topological spaces. DEFINITION 8 2 . 2 . The distKhutive l a t t i c e
whenever every proper p r i m e ideal in
x
(X,e)
i s called noma2
contains a unique minima2 prime
ideal, i . e . , whenever every pKme dual idea2 i s contained i n a unique
maximal dual ideal. Given the proper prime ideal w
in
(X,8)
, we
shall denote by
the intersection of all prime ideals contained in w , so
k(w)
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
66
is an ideal in X. Note that k(w)
Hence, k(w) 8
.
E k(o)
Note also that k(w)
ideals contained in
Ch. 11,9821
is not empty, since
is the intersection of all minimal prime
w ,
LEMMA 8 2 . 3 . Gioen t h e proper prime i d e a l
(X,6) , we have
in
wo
x:xAy=8 for some yEX-w PROOF. For abbreviation, write I
=
f x:uy=8 \
for some yEX-w
. If
w1 = 8
I is an ideal in X
is prime and
I c k(wo)
shows that xo
A
all
satisfying z
h o = X-wo
.
i.e., y
. Note
that
ideal generated by
ho
.
is not in w '
Conversely, assume now that xo A y
2
is a prime ideal for some y E X-wo, Hence,
. This
xo A y = 8 E w 1 , it follows that xo E w '
for all y E X-wo , i.e., xo not in I
y # 0 z
,
implies that y E X-w'
then w ' c wo since w '
'
and if xo E I , i.e., if xo A y
and observe already that contained in wo
1
0)
for some y
,
is not in F
8
and
xo
so
xo E k(wo) , but
. Let
F be the set of
in the prime dual ideal
F
is a dual ideal, the dual
. We prove now that
I and F
are disjoint.
Indeed, if not, we should have xo A y E I for some y E Xo , i.e., (by the definition of I ) x A y A z = 9 for some y E A. and some z E X o But then w
=
y A z E A.
0
,
so
would imply xo E I
xo A w = 0
contradicts the hypothesis that xo
is not in
I
. Hence, I
. This
and
F
.
are
disjoint. It follows by means of the prime ideal separation theorem that there exists a prime ideal w
containing I such that
disjoint. Then w c X-F c X-A
= wo
wo
. Hence, by
the definition of
implies xo E w (since w
so
w
, we
. On the other hand we have
and F
.
are
is a prime ideal contained in have k(wo) c w
xo E F
,
so
xo
,
xo E k(wo) is not in w so
are disjoint). Contradiction. It follows that xo E k(wo)
implies xo E I , i.e., k(wo) c I that k(wo) = I
,
k(wo)
and F
w
. Hence, finally, we obtain the result
The next lemma is an immediate consequence.
Ch. 11,9821
PRIME IDEAL EXTENSION
LEMMA 8 2 . 4 . The proper prime i d e a l
a unique minimal prime i d e a l i f and only if k(w) is a prime i d e a l in X land i n t h i s case k(w) is the unique minimal prime i d e a l auntained i n w 1. k(w)
PROOF. If
.
minimal prime ideal p
Then
u
=
u' ,
u
so
k(w)
k(w)
contains
is obviously the unique minimal contains a unique
w
by the definition of
k(w)
. On
the
contains at least one minimal prime
u
By the uniqueness of
is included in every
, so
k(w) = ~ - i
that
3
w' c w
.
p' c w' c w
so
(x,6)
. Conversely, assume that
other hand, ever prime ideal
u' ,
w
is prime, then k(w)
prime ideal contained in w
ideal
in
67
w' c w
this implies that
, i.e., u c
k(w)
.
It follows
is prime.
We now state and prove the main theorem about normal lattices. We recall d the disjoint complement of x is denoted by {x} .
that for every x E X
THEOREM 8 2 . 5 . The f o l l m i r g conditions f o r the d i s t r i b u t i v e l a t t i c e
are equiva Zent
(X ,6 )
(i)
.
is normal, i . e . , every proper prime i d e a l i n X contains a
X
unique minimal prime ideaZ.
=
If
y = 6 , then
A
x,y E
x
X =
we have
{XI d
v {y}
{xAyld
=
d
.
w
u1
x
.
d 1x1~ v {y~
For any pair o f d i f f e r e n t minimal prime i d e a l s
(v) X
x
For a l l
(iv)
have
is prime f o r every proper prime i d e a l
k(w)
(ii)
(iii) I f
. pl
and
u 2 we
v 1~. 2 '
contains a largest element, then these conditions are equivalent
t o the following condition. (vi) Every maximal i d e a l i n X contains a unique minimal prime i d e a l . PROOF. (i)(ii)
(ii). This was proved in Lemma 8 2 . 4 above.
(iii). Assume that (ii) holds, but (iii) does not hold. Then
there exist
xo,yo E X
proper ideal in X
. It
proper prime ideal w k(w) =
that xo
.
y = 6 and (x0Id v (yo}d is a Od (xo} v {yo}d is contained in some d On account of {x } c w we may conclude from such that xo
A
follows that
0
(x:xhy=tl
is not in k(w)
)
for some yEX-w
. Similarly, {yo} d c w
implies that yo
is not in
Ch. 11,1821
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
68
. On
k(w)
t h e o t h e r hand we have
a t l e a s t of
xo
and
yo
xo
is in
yo = 8
A
and
k(w)
. Contradiction.
k(w)
i s prime, so one
I t follows t h a t ( i i )
implies ( i i i ) .
*
(iii)
{xId
( i v ) . Since
{yJd c {xAyId
V
x , y E X , we d {xhy} , s o
holds f o r a l l
have only t o prove t h a t t h e i n v e r s e i n c l u s i o n h o l d s . Let z E d d (ZAX) A y = 8 By ( i i i ) t h i s i m p l i e s t h a t X = {ZAX} v {y}
.
t h e form and
y
z = p
q = 8
A
Z
with
z A p
that
z
E
=
with
v q
p E {zAxld
. Hence Z A
=
Z
E { x } ~and
d 1x1 v { y l d
(pVq) = (ZAP)
A
Z
z
and
A
V
(ZAq)
,
so
d q 5 q E {y}
. Hence,
any
z
,
q E {yld
i.e.,
E {yId
z A q
,
so
i s of
z
z A x A p =
. This
e
shows
d {xId v {y}
E {xAyId i s a member of
.
This i s t h e d e s i r e d r e s u l t . (iv)
*
(v). E v i d e n t l y ( i v ) implies ( i i i ) , s o i t i s s u f f i c i e n t t o show
ul
t h a t ( i i i ) implies ( v ) . Hence, assume t h a t ( i i i ) h o l d s , and l e t
x
be d i f f e r e n t minimal prime i d e a l s . There e x i s t s an element
i s not i n
that
x
{xJd
i s contained i n
p2
, p2
maximal lower s u b l a t t i c e y
E X-vl
(i.e.,
x
so (since
. Since
x
is in
1 cannot be adjoined t o
p1
X-ul
(i.e.,
x
x A y
= 3'
so
p1 v p2 =
(v)
*
x
€or an element
without d e s t r o y i n g t h e
.
and
p2
. Hence,
w
and
p1
is
c o n t a i n i n g d i f f e r e n t minimal prime i d e a l s
by ( v ) , we have
F i n a l l y , assume t h a t X
ul
. X = 1-1
impossible. Hence ( i ) h o l d s . of
y = 6
A
( i ) . Assume t h a t (v) h o l d s b u t ( i ) does n o t hold. Then t h e r e
e x i s t s a proper prime i d e a l p1
p2
such
is not i n the
p r o p e r t y t o be a lower s u b l a t t i c e ) . It follows then from x d by ( i i i ) t h a t X = 1x1 v { y I d Also, s i n c e y i s n o t i n d Hence prime, we have {y} c u1
.
and pl
i s prime) t h e d i s j o i n t complement
p2
), we must have
X-p
in
X
1
v p
2
c w
, so w
= X
. This
is
has a l a r g e s t element. E v i d e n t l y , normality
i m p l i e s ( v i ) . Conversely, l e t ( v i ) hold, i . e . ,
every maximal i d e a l
c o n t a i n s a unique minimal prime i d e a l . Now, let
w
prime i d e a l c o n t a i n i n g t h e minimal prime i d e a l s
I J ~
contained i n a t l e a s t one m a x i m a l ideal
are minimal prime i d e a l s contained i n
J
J
,
, SO
be an a r b i t r a r y proper
and
p2
i t follows t h a t
. Since ul
and
w
p2 p l = p 2 , by ( v i ) . Hence, w
is
69
PRIME IDEAL EXTENSION
Ch. 11,1821
contains only one minimal prime ideal, i.e,, (i) holds. The equivalence of normality and the condition that x A y = 9 implies {xId v {yld = X is mentioned already in a paper without detailed proofs by A.A. Monteiro (c11,1954). The theorem as stated above was proved by W.H. Cornish (I 11,1972) In E%ercise 8.9'it was indicated that, under a certain additional hypothesis (incomparable proper prime ideals in ( X , e ) have disjoint neighborhoods in the hull-kernel topology), X is normal if and only if and disjointness of {w}x and {w}y implies disjointness of {w};
.
{wIThe additional hypothesis can be dropped, but the proof needs some Y modification. We show how this can be done.
THEOREM 82.6. The d i s t r i b u t i v e l a t t i c e
disjointness of
and
{w},
Cw3-
(X,e) i s normal i f and only i f
and {GI implies disjointness of the closures Y the closures taken i n the hulZ-kernel topology of the s e t
iwIx
Y' fi of a l l proper prime ideals.
PROOF. We first recall Theorem 7.6(ii), stating that for any x # 9 we if and only if there exists a minimal prime ideal u o E {wlx have wo E {w}; Assume now that X is normal and such that uo is contained in wo x,y E x have the property that {wIx and {wIy are disjoint, but {wIx n {w}By Theorem 7.6(ii), as stated and {LO}- are not. Let wo E {w}: Y Y above, there exist minimal prime ideals u l and uz , contained in wo , such that u1 E { w } ~ and p2 E {wIy , so ul # u2 since {wlx and are disjoint. On the other hand we have ul = uz by the normality of X
.
.
.
Contradiction. Hence, if X
is normal, then disjointness of
{wJx
and
implies disjointness of their closures. Conversely, assume the
cwIy
disjointness condition satisfied, but X not normal. Hence, there exists a containing different minimal prime ideals y I and Then w 0 3 y2 E {wIx , be an element in p1 but not in p2 E {wJx , once again by Theorem 7.6(ii). On account of x E 1.1, there
proper prime ideal wo
u2 SO
. Let wo
x
.
.
Hence { w I X exists an element y E X-p1 satisfying x A y = 9 = {wIXAy = {w}~ is empty, which implies by hypothesis that {a};
tl
{w), {w};
empty. But w0 3 ul E {wIY implies wo E { w I i , s o wo E Cwl; n {wly Contradiction. Hence, the disjointness condition implies normality.
=
.
is
70
WORMAL AND EXTREMALLY DISCONNECTED LATTICES
ch- 1
' §a2] 9
I n Lemma 3 7 . 1 0 ( i i ) t h e corresponding theorem i s proved f o r Riesz spaces. The proof i s n o t a p p l i c a b l e t o t h e l a t t i c e c a s e , s i n c e i t i s used t h a t i n a Riesz space we can s u b t r a c t . On t h e o t h e r hand, t h e p r e s e n t proof f o r t h e l a t t i c e c a s e i s a p p l i c a b l e i n t h e Riesz space s i t u a t i o n a s w e l l . A s may b e i n f e r r e d from t h e l a s t remark, analogous r e s u l t s a s f o r l a t t i c e s hold f o r Riesz s p a c e s . The Riesz space e v e r every proper prime i d e a l i n For any proper prime i d e a l
k(w)
w
L
i s c a l l e d normal when-
L
c o n t a i n s a unique minimal prime i d e a l .
i n t h e a r b i t r a r y Riesz s p a c e
L
the ideal
i s d e f i n e d t o b e t h e i n t e r s e c t i o n of all prime i d e a l s contained
i n w ; t h e proof t h a t
k(w)
(f :fEL,inf ( I f I ,g)=O f o r some gEL+-o)
=
i s e x a c t l y a s f o r t h e l a t t i c e case. Also, w i d e a l i f and o n l y i f
k(w)
c o n t a i n s a unique minimal prime
i s prime. The main theorem on normal Riesz s p a c e s
i s t h e n a s follows.
The f o l l d n q c o d i f i o n s for the Riesz space L are
THEOREM 82.7.
equivalent. (i) L i s normal, i.e., every proper prime idea2 i n L contains a unique minima2 prime ideat. ( i i ) k(w) i s prime for every proper prime idea2 w (iii)
= 0
If i n f ( u , v )
a2gebraic swn of the bands (iv)
, then
d L = {u} + {vId
d {v} ) .
{ u } ~and
.
( i . e . , L i s the
~f u , v E L+ , then { i n f ( u , v ) T d = { u l d + {v}d For any p a i r of d i f f e r e n t minima2 prime ideats
.
(v) u1 arid p 2 ue haue L = p 1 + p2 If L contains a strong u n i t , then these conditions are equiuaZent t o the foZ2owing condition.
.
( v i ) Every mo2ima2 idea2 i n
contains a unique minima2 prime i d e a t .
L
PROOF. We r e s t r i c t o u r s e l v e s t o t h e proof t h a t ( i i i ) i m p l i e s ( i v ) . d S i n c e [ u I d + {v} c { i n f ( u , v ) l d h o l d s f o r a l l u , v E L+ , w e have only t o prove t h a t t h e i n v e r s e i n c l u s i o n h o l d s . Let v = 0
(whu)
A
form
w = p+q
w h u h p = O
. By
with and
( i i i ) t h i s implies t h a t 0
5
pE
[WAU)~
and
vhq=O.Hence
0
5
L =
w
E
{WAU)
d
inf(u,v)
+ {vjd,
.
, so so w i s of t h e
0 s q E { v } ~ It f o l l o w s t h a t
w with w that w
=
w
A
p E {uId
A
71
PRIME IDEAL EXTENSION
Ch. 11,9821
w = inf(w,p+q) s inf(w,p) + inf(w,q) and w
A
q
q
S
E {vld ,
so
w
[ u } ~+ [vld
{uId + [vId
. Hence, any
w
q E {vId
A
is majorized by an element in [uld + {vld
element of
=
A
p + w
A
q
. This shows
, i.e., w itself is an
w E {inf(u,v)}d
is a member of
. This is the desired result.
We return to the lattice situation, and before introducing now the dual property of normality we make one more remark about the particular case that
(X,e)
normality that x
has alargestelement e A
y
= 8
implies X =
equivalently by saying that x p E {xld and some q E [yId
A
.A
y =
8
. In this case the condition for
{XI d v
[yId may be expressed
implies e
=
p v q
for some
similar remark holds for the case of a
Riesz space with a strong unit. DEFINITION 82.8. The distributive l a t t i c e
(X,B,e) i s called
extremally disconnected whenever every proper prime ideal i n X i s contained i n a unique maximal ideal, i . e . , whenever every prime dual ideal i n X contains a unique minimal prime dual ideal. It is obvious from the definition that
(X,e,e) is extremally
disconnected if and only if X is normal with respect to the dual ordering. The following theorem follows, therefore, immediately from the main theorem for normal lattices. THEOREM 82.9. The folZ&ing are equivalent.
conditions f o r the distributive l a t t i c e
(X,e,e)
X i s extremally disconnected, i . e . , (i) X i s contained i n a unique maximal ideal. (ii) For m y prime dual ideal
prime dual ideals contained i n
A
, the
intersection
k(A)
of a l l
1 i s prime.
(iii) If x v y = e , there e x i s t elements and x v x = y l v y = e . 1
every proper prime ideal i n
(iv) Every minimal prime ideal i n X maximal ideal.
xl,y,
such that
x1 " y 1 = e
i s contained i n a unique
PROOF. Note that (i), (ii), (iii) correspond with (i), (ii), (iii) the main theorem for normal lattices, and (iv) correspond with (vi).
in
72
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
Ch. 11,§821
C e r t a i n p r o p e r t i e s of a t o p o l o g i c a l space a r e r e f l e c t e d i n t h e p r o p e r t i e s of t h e l a t t i c e of i t s c l o s e d s u b s e t s . We r e c a l l t h a t t h e topolog i c a l space
i s c a l l e d normal whenever f o r every p a i r
E
c l o s e d s u b s e t s t h e r e e x i s t d i s j o i n t open s e t s and
of d i s j o i n t
F1 = O1 t h e s p a c e i s a l s o r e q u i r e d t o b e Hausdorff, b u t we
. Sometimes
F2 c O2
Fl,F2
01,02 such t h a t
s h a l l n o t r e q u i r e t h i s h e r e . A l s o , t h e space
is c a l l e d e x t r e m a l l y d i s -
E
connected whenever t h e c l o s u r e of every open s u b s e t i s open. It i s e a s i l y seen t h a t t h i s i s e q u i v a l e n t t o t h e c o n d i t i o n t h a t f o r every p a i r of d i s j o i n t open s u b s e t s t h e r e e x i s t d i s j o i n t c l o s e d s e t s that
O 1 c F1
and
and t h e open s e t
-
closure
so
of
O1
-
. Indeed,
O2 c F2
i s given, t a k e
O1
O I 'O2 such
i f the l a s t condition is s a t i s f i e d t o be t h e complement of t h e
O2
O1 ; i t f o l l o w s t h a t
i s c l o s e d a s w e l l a s open, and
O2
i s open a s w e l l a s c l o s e d .
O1
( i ) The topological space
THEOREM 82.10.
the l a t t i c e
i f the l a t t i c e
E
O1 fl 0 =
2
b e normal, i . e . ,
E
,
F1 fl F2 = @
0
i s extremally disconnected i f and only
o f i t s closed subsets i s extremally disconnected.
X
PROOF. ( i ) Let satisfying
and
F1 c O1
(i.e.,
F1 ). S i m i l a r l y
F;
F1 VF2 01,02 such t h a t
as well as
. Now,
F2 c O2 and
O1
taking f o r
,
F; fl F2 = @
i.e.,
F; U F' = E
2
. Also,
F; E
. Hence,
Fi
and
r e s p e c t i v e l y , we have
O2
i s a member of t h e d i s j o i n t complement
a r e d i s j o i n t , w e have t h e space
given t h e c l o s e d s e t s
t h e r e e x i s t open s e t s
t h e s e t t h e o r e t i c complements of
F; fl F1 = @
i s normal i f and only i f
E
of i t s closed subsets i s normal.
X
( i i ) The topological space
F;
Fl,F2
since
{ F I I d of O1
and
O2
t h e normality condition f o r
can be expressed e q u i v a l e n t l y by s a y i n g t h a t f o r any p a i r
E
FI,F2
i n the l a t t i c e
Fi ,Fh
in
X
X such t h a t
satisfying
F
1
fl F
, F;E
Fi E [F,}d
X = [F,}
equivalent t o the condition t h a t
=
0
t h e r e e x i s t elements F; U F; = E
and
d
v {F21d
, which
. This
is
i s exsctly
c o n d i t i o n ( i i i ) i n t h e main theorem f o r n o r m a l i t y of a l a t t i c e . Hence, t h e topological space
i s normal i f and only i f t h e l a t t i c e
E
( i i ) The l a t t i c e
X
t h e preceding theorem i f and only i f f o r any p a i r F
1
U F
2
= E
t h e r e e x i s t elemehts
= F; U F2 = E
and
complements of
F,
F; f l F; =
and
i s normal.
X
i s e x t r e m a l l y d i s c o n n e c t e d by c o n d i t i o n ( i i i ) i n
FP
0
F;,F;
. It
in
X
FI,F2 such t h a t
in
X
satisfying
F; U F1 =
i s e a s i l y s e e n by t a k i n g s e t t h e o r e t i c
t h a t t h i s is e q u i v a l e n t t o t h e c o n d i t i o n t h a t
f o r any p a i r of d i s j o i n t open sets
01,02
in
E
there exist disjoint
PRIME IDEAL EXTENSION
Ch. 11,5821
closed sets F;,F;
such that
O1 c F ;
and
73
O2 c F ;
. By
one of the
remarks above, the last condition is equivalent to the condition that E extremally disconnected. Hence, the topological space E connected if and only if the lattice X
is
is extremally dis-
is extremally disconnected.
Before introducing more general concepts, we present the following simple theorem. THEOREM 82.11. (i) The d i s t r i b u t i v e l a t t i c e
only i f every i d e a l in X
i s normal i f and
(X,e)
i s normal, and a l s o i f and only i f every
i s normal.
principal i d e a l i n X
(ii) The d i s t r i b u t i v e l a t t i c e
(X,e,e)
and only i f every s u b l a t t i c e of the form
i s extrernally disconnected i f is eztremally d i s -
(x:xo<x<e)
connected. PROOF. (i) Let x,y E I with x
I
A
y
I be an ideal in the normal lattice (X,O) , and let d 0 Then X = {XI V {yId since X is normal, so
.
=
.
v
=
This shows that
I is normal. It is evident that if every ideal is normal,
then every principal ideal is normal. Now assume that every principal ideal is normal. We show that
X
We have to prove that every xl
A
xo
=
y1
A
yo
principal ideal (ii) (X,e,e)
is normal. Let xo,yo E X
with
E X is of the form
x
z
. This is seen to hold by . 0 0
= €I
z =
x
A
y
0 0 v y 1 with
= 8
.
considering the normal
(x:@<x<x vy vz)
is extremally disconnected if and only if
X
is normal
with respect to the dual ordering, i.e., if and only if every sublattice (x:xolx<e)
i s a normal principal ideal with respect to the dual ordering.
This is equivalent to the condition that every
(x:x <x<e) 0-
-
is extremally
disconnected with respect to the original ordering. In Chapter 5 (cf. Theorem 33.3(ii)) L
it was proved that in a Riesz space
every ideal including a given prime ideal is prime, and the family of all
(Prime) ideals including a given prime ideal is linearly ordered by inclusion. These properties are immediately transferred to the distributive lattice X ( L )
of all principal ideals in L
. It is not
every distributive lattice X with smallest element 8
true, however, that has these properties.
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
74
({e}
is a prime
is not prime; furthermore, {€I}
is contained
The simple lattice in Exercise 7 6 . 1 4 is a counter example ideal, but the ideal
such that neither w1 = w2 nor w2 = 1 It is of interest, therefore, to investigate what distributive
in prime ideals w 1 holds).
and
lattices share with (X,0)
if w
{e,xo}
Ch. 11,5821
in X
w2
X(L)
the above mentioned properties. Note already thaf
has these properties, then it follows that for each prime ideal and each x
satisfying w 1
2
.
w
not in w
there exists a unique x-maximal ideal w 1 (X,e) has a largest element e ,
In particular, if
then each proper prime ideal is contained in a unique maximal ideal, i.e., X
is extremally disconnected. The converse is not true; an extremally
disconnected lattice
(X,e,e)
does not necessarily have the abovementioned
properties, as can be shown by a simple example (cf. Exercise 82.20).
We
first prove a simple lemma. LEMMA 82.12. For any
i d e a l generated by of
(X,e)
A
maximal i d e a l in
A
.
PROOF. (i) Let J
in
w
(since x
is not in J
includes w n Ax
=
n Ax
, and
J
uniqueness of the extension from Ax maximal). It follows that Ax c w ' w'
is x-maximal. Hence, w (ii) Let w
in Ax w ll Ax
,
,
), s o if
so
is not in the is not x-maximal,
properly including w
. But
x E w'
,
n Ax
(since J is x which is impossible because w'
=
A
is x-maximal.
be x-maximal in X
. If the ideal
there exists a maximal ideal J
. By
n Ax is a
J is prime and
x
w
w
this inclusion is proper (by the
to X ), s o so
. Then
. Evidently
to X = w
there exists an x-maximal (prime) ideal w ' then w ' fl Ax
holds now.
, the intersection
X
be a maximal ideal in Ax
has a unique prime extension w
ideal w
denote t h e principal
i n Ax , t h e (unique) prime extension
J
(ti) For any x-maximal ideal
J
Ax
. The following
i s x-maximal.
to X
J
, let
= (y:e
For any maximal i d e a l
(i) w
in
x
, i.e.,
x
in Ax
part (i) the prime extension w '
of
w
n Ax is not maximal
properly including
J
is x-maximal, and
w'
properly includes w (this follows from the explicit formula for the extension in Theorem 51.8(ii);
the formula is cited at the very beginning of
the present section). This is impossible, since w w fl
Ax
is maximal in
Ax
-
is x-maximal. Hence,
Ch. 11,5821
75
PRIME IDEAL EXTENSION
COROLLARY 82.13.
The following uniqueness conditions are equivalent.
(i). Given any prime ideal
i n X and any element
w
there e x i s t s a unique x-maximal prime ideal (ii) For any
x E X and for any proper prime ideal
not i n
x
such that
w1
w1
P
3
w
w
.
,
i n the
principal ideal Ax there e x i s t s a unique mmimal ideal J i n Ax such that J 3 P I n other words, every principal ideal A, i s extremally
.
disconnected.
Ax
.
PROOF. (i)
Given x E X
(ii).
=s
In the lattice A
(with 8'
, let P
be a proper prime ideal in
as smallest and x
as largest element)
there exists at least one x-maximal ideal (i.e., maximal ideal) containing
P
. Let
J1
J 2 be ideals of this kind and let
and
. By
unique prime extensions to X
and w2
w1
the lemma above w 1
and
w2
be their are
x-maximal, so (since they both contain the unique prime extension of it follows from (i) that w1 (ii)
, and hence J 1
= w2
* (i). Similarly.
= J2
.
P )
We now generalize the notion of an extremally disconnected lattice. The condition that the lattice must have a largest element can be dropped. DEFINITION 82.14. The d i s t r i b u t i v e l a t t i c e
(X,e)
i s said t o be
completely extremally disconnected whenever every p2.incipal ideal i n X i s extremally disconnected. In view of Theorem 82.9 the lattice
(X,e)
x,y E X there exist elements v y = x v y I = x v y (note that,
disconnected if and only if for every pair X,,yI E X
such that x
A
and
yI = 8
is completely extremally
x
1 I Compared to the formulas in Theorem 82.9, we have interchanged x1 and y 1 1.
It follows that x 1 < x XI = XI A
=
Similarly y 1
(XVY,)
5
Y
A
y
5
= (X
(x1 Ax) v 0 = x 1
x
AX)
1
. Indeed, x 1 5 x I v V
(X
, and
so
y = x v y 1 implies
Ay ) =
1
1
XI
< x .
.
The folZoLling conditions f o r the d i s t r i b u t i v e l a t t i c e are equivalent.
THEOREM 82.15. (X,e)
and y I
NORMAL AND EXTRFNALLY DISCONNECTED LATTICES
76
Ch. 11,5823
i s completely extremally disconnected.
(i)
X
(ii)
Any proper ideal i n X
containing a prime ideal i s prime i t s e l f .
(iii) The f a m i l y of a l l proper prime ideals containing a given proper
prime ideal closure
i s linearly ordered by inclusion. I n other words, the
w
o f the s e t
iw)-
i n fi i s a linearly ordered s e t .
{w}
(iv) The f a m i l y of a l l prime dual ideals contained i n a given prime dual ideal i s linearly ordered by inclusion. If
(v)
and
w1
hull-kernel topology of
fi
.
and
w,
have d i s j o i n t neighborhoods i n the
w2
For any proper prime ideal
(vi)
and f o r any
w
e x i s t s a unique x-maximal ideal containing PROOF. (i) prime ideal w
Hence, let x A y E I =
x v y1 = x v y
. Since
X
so
x E I
(ii)
such that x 1 A y 1 = 0
x ) and x 1 v y
S
*
(iii). Let w 1
and
=
x v y
, we
.
have
be proper prime ideals containing the
w2
i s included in the other.
is included in the ideal w 1 n w 2 , s o
hypothesis (ii).
w1,w2 such that x 1 is in w l
. Then
but not in w l
is in w2
prime) we have
x1 E w 1 fl w2
definition of x I and containing w
w1
Also, since neither of
there exist x 1 , x 2 E X
x2
x
1
A x2
w2
is prime by
is included in the other,
E wI n
or x2 E w 1 f l w2
n
but not in w2 w2
,
so
and x2
(since w 1
n
. This contradicts the
19
is
. Hence the family of all proper prime ideals
(iii)
*
(iv).
These are equivalent by duality (cf. Theorem 76.1).
(v). Let w 1
and
w2
be proper prime ideals, neither of them
included in the other. We have to prove the existence of elements x in w ) and
w2
is linearly ordered by inclusion.
(iii)
1
=
. This is the desired result.
prime ideal w , and assume that neither of w Then w
and x 1 v y
x 1 A y 1 = 0 it follows that x 1 E w or y 1 E w , x 1 E w c I and x A y E I , so x 1 V (xhy) E I Hence,
x1 v x = x (since x 1
observing that
containing the
is completely extremally disconnected by
. From
. Then
w
there
w
to show that x A y E I implies x E I or y E I.
hypothesis, there exist xI,yI E X say x 1
not i n
x
.
w
I be a proper ideai in X
(ii). Let
. We have
, neither of them
are proper prime ideals i n X
w2
included i n the other, then
y (not in w ) such that 2
{ w ) ~
and
(not
are disjoint. Assume
Ch. 11,5821
PRIME IDEAL EXTENSION
that such x not in w I
and
y
do not exist,
and all y not in w2
{wIx n { w }
non-empty for all x Y Then, by Theorem 77.5, the family of
so
.
77
{wIX n {wIy has a non-empty intersection (note that in the
all these
present application of Theorem 77.5 the family T is the union of all x not in w l the intersection of all these
n ( ~ w ~ ~ not : x in
w
and all y not in w2 ). Let
{wIX ll { w }
1
Y
. Now, since
and
0
be in
wo
)
is the set of all prime ideals contained in w 1 it follows that w c w 1
is empty and the set S
,
wo c w2
so
, and and
w1
similarly for w2
of the family of all proper prime ideals containing wo
w2
’
are both members
. By hypothesis
(iii)
this family is linearly ordered by inclusion. This contradicts the assumption that neither of w2
ideal w
,
w
and w2
w1
are proper prime ideals containing the prime
then either w 1 c w2
imply that w1 2
and
(vi). We first go back from (v) to (iii), i.e., asauming (v) we
prove that if
and
w2
w1
and
w4
(vi)
*
Indeed, if not, (v) would
w
w
(cf. Theorem 7.6 (i)),
. Similarly, every neighborhood of
contains w
or
be a prime ideal, let x be not in w
w cw3
4
,
so
implies (iii). We prove now
be x-maximal ideals containing w
w3 c w4
property that w3 = w4
.
is in the closure of
that (iii) implies (vi). Let we have either
w2 c w 1
. Contradiction. Hence, (v)
contains w
and let w3
or
have disjoint neighborhoods. On the other hand
implies that w1
every neighborhood of w2
.
have disjoint neighborhoods in the hull-kernel topology of (v)
w1
is included in the other. Hence, w 1
w1,w2
,
. By hypothesis (iii)
and so it follows from the x-maximal
. Hence, the x-maximal ideal containing
w
is unique.
(i). This is exactly the statement that part (i) implies part (ij
in Corollary 82.13. In Exercise 8.9 the above theorem was partly proved. In fact, it was Proved that (v) implies (iii) and (vi). We mention some examples. Any Boolean ring X disconnected. Indeed, every proper prime ideal in X
is completely extremally is maximal, and so
condition (ii) is trivally satisfied (i.e., any proper ideal containing a Prime ideal is prime itself). Furthermore, as already noted earlier, the lattice X(L)
of all principal ideals in a Riesz space L
is completely
extremally disconnected (any proper ideal containing a prime ideal is prime
78
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
itself by Theorem 33.3(ii)). interval [O,l]
Ch. 11,1821
The lattice of all closed subsets of the unit
(with its ordinary topology) is not extremally dis-
connected, simply because
CO,l]
is not an extremally disconnected topo-
logical space. For an example of an extremally disconnected lattice that is not completely extremally disconnected we refer to Exercise 82.20. The dual notion of a completely extremally disconnected lattice is a completely normal lattice. We present the definition. (X,e)
DEFINITION 82.16. The d i s t r i b u t i v e l a t t i c e
i s said t o be
completely normal whenever every sublattice of the form
(x:xo<x) i s
norma I . By Theorem 82.11 the sublattice (x:x <x) 0-
every principal ideal in sublattice of the form saying that
(X,e)
is normal if and only if
(x:x <x) is normal, i.e., if and only if every 0(x:xo<x<x ) is normal. This is equivalent to 1
is completely normal if and only if every sublattice
(x:x <x<x ) is extremally disconnected with respect to the dual ordering. 0- - 1 This observation is important for the proof of the following main theorem about completely normal lattices. THEOREM 82.17. The following conditions f o r the d i s t r i b u t i v e l a t t i c e
(X,6> are equivalent. (i)
X
i s completely normal. containing a prime dual ideal i s prime
(ii) Any dual ideal i n X
itself. (iii) The f a m i l y of a l l prime dual i d e a l s containing a given prime
dual ideal i s linearly ordered by inclusion. (iv) The f a m i l y of a l l prime ideals contained i n a given proper prime ideal i s linearly ordered by inclusion. (v)
For any prime dual ideal
A
and f o r any
x
not i n X
there
e x i s t s a unique prime dual ideal A 1 such t h a t X I 3 A and X I i s a maximal prime dual ideal with respect t o the property of not containing (vi)
For any proper prime ideal
w
and f o r any
a unique prime ideal w1 such t h a t w 1 c w and ideal wiht respect t o the property of containing
w1
x
.
x E w
x
there e x i s t s
i s a minimal pr*e
.
PRIME IDEAL EXTENSION
Ch. 11,§821
PROOF. (i)
=s
dual ideal in X xo v yo E F arbitrary
Let X be completely normal, and let F
(ii).
E X
yo E F
or
and set wo
=
(x:€I<x<w )
=
be a
to show that
. For this purpose, choose an .
xo v yo v zo
normal, any sublattice of the form sublattice Y
. We have
containing the prime dual ideal A
implies xo E F zo
79
Since X
is completely
(x:w,<x~wo) is normal, i.e., the
has the property that, with respect to the dual
0
ordering, every principal ideal in the sublattice is extremally disconnected. In other words, Y
(x:€I<xrw )
=
is completely extremally dis-
0
connected with respect to the dual ordering. Note now that the intersections
Y fl F and Y that Y
F
Il
fl
X
are not empty (since
zo
and Y n A
is an ideal in Y
E Y n
A c
Y n F ); note also
is a prime ideal in Y with
respect to the dual ordering. Hence, with respect to the dual ordering, Y is completely extremally disconnected and Y n F
is an ideal containing
. It follows from Theorem 82.15 that Y f l F . Note now that xo and yo are members of Y
the prime ideal Y n A a prime ideal in Y
with respect to the dual ordering, it is given that X Hence xo E Y n F (ii)
*
or yo E Y
(iii).
XI
Let
prime dual ideal
A
the other. Then X
, and
,
y
o
and, ~
~
and
A2
be prime dual
assume that neither of
deals containing the
A ,A2
x1 n
is prime by hypothesis (ii). Also, since neither of x1,x2 E X
is contained in i2 , SO
X2 is contained
AI,A2
such that x 1 is in X I
but not
.
in X 2 and x2 is in A 2 but not in A 1 Then x 1 v x2 is in x 1 n so (since X I n X is prime) we have x I E X I n A 2 or x2 E X I n A 2
-
2
This contradicts the defintion of prime dual ideals containing (iii) (iii)
*
A
. Hence, the family of
x 1 and x2
A
be a prime dual ideal, and let x
respect to the property of not containing x
. Let
kind. By hypothesis (iii) we have either A
cX
follows from the x-maximality property that
A
x
-
x2,
all
is linearly ordered by inclusion.
There exists at least one prime dual ideal containing A
dual ideal A ,
.
(iv). These are equivalent by duality :(cf. Theorem 76.1).
(v). Let
containing A
1
XI
or = A2
(vi)
(vi).
.
and maximal with and
A
be not in A
2
A2
cXI
be of this
,
so it
. Hence, the prime
and maximal with respect to not containing
is unique. (v)
n
as desired.
is contained in the dual ideal
in the other, there exist
~
n F , since Y n F is a prime ideal. It
or yo E F
xo E F
follows that
O
is then
These are equivalent by duality.
* (i). It is sufficient to show that, for an arbitrarily chosen
~
.
80
Ch. 1 1 , § 8 2 1
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
xo E X
,
the sublattice Y
the proper prime ideal
w
(x:x <x)
=
0-
is normal. In other words, given
in Y , we have to show that
unique minimal prime ideal in Y
w
contains a
. Hence, let the proper prime ideal
w
in
Y be given, and let p 1 and p 2 be minimal prime ideals in Y , both of these contained in w By Lemma 8 2 . 1 (and by the remarks preceding the
.
lemma) the prime ideals w , u l u',pi
and
pi
to X
.
and
p2
in Y
have unique prime extensions
Explicitly,
x:xEX,xcb for some bEw and similarly for p i
and
pi
. Evidently, xo E
p i t w1
. We
show that
is a prime ideal, minimal with respect to the property of containing
pi
xo
.
Indeed, if
included in p i included in p 1
11;
,
is a prime ideal in X
then p; fl Y
,
containing xo
and properly
is a non-empty prime ideal in Y
properly
(that the inclusion is proper follows from the explicit
extension formula). This is impossible since
p1
is a minimal prime ideal
. Hence, p i is minimal with respect to containing xo . The same holds for p i . By hypothesis (vi) the prime ideal contained in w ' and minimal with respect to containing xo is unique, s o p ; = p i . Hence in Y
p l = p;
ny
= p1
2
n
i.e., the given prime ideal w
Y
= p
2 '
in Y
contains a unique minimal prime
ideal. We observed already that several of the conditions for normality in Theorem 8 2 . 5 are mentioned by A.A. Montecro ( C 1 1 , 1 9 5 4 ) in a paper without detailed proofs. In this paper are also mentioned conditions for completely normal lattices (Theorem 8 2 . 1 7 ) and for completely extremally disconnected lattices (Theorem 8 2 . 1 5 ) as well as the theorem that a topological space is normal or extremally disconnected if and only if the lattice of its closed subsets is so (Theorem 8 2 . 1 0 ) . There is another condition for complete normality, due to W.H. Cornish ([I 1 , 1 9 7 2 ) . The condition extends the normality condition saying that
X
=
FI 1
V p2
holds for any pair of different minimal prime ideals.
PRIME IDEAL EXTENSION
Gh. 11,9821
THEOREM 82.18. The d i s t r i b u t i v e l a t t i c e i f and only i f
ideals
X
v w2
= w1
(X,O) i s completely normai!
hoZds f o r any pair of incomparable prime
( i . e . , prime ideals
w1,w2
81
w1,w2 such that neither o f them i s
included i n the other). PROOF. Let X be completely normal and let u1,w2 be incomparable prime ideals in X w1 v w2
X
.
. Assume now
Then
w 1 v w2
that the ideal
w 1 v w2
satisfies
is included in a proper prime ideal
w
0 '
By condition (iv) for complete normality in the preceding theorem the prime ideals contained in wo and
w2
are linearly ordered by inclusion, s o (since w
are included in wo ) we have
diction. Hence w, v w 2
X
=
.
w1 c
w2
or
. Contra-
w2 c w 1
The converse is also easy. It is given now that w 1 v w2 incomparable wl,w2
,
and we have to show that X
i.e., we have to show that if
w1
and
in the proper prime ideal wo , then
w2
w
0
=
X
. Contradiction.
for
are prime ideals contained
w 1 c w2
. But
X
is completely normal,
or
w2 c w 1
then X
. Assume this
and incomparable
to be false. Then there exists a proper prime ideal wo prime ideals w1,w2 contained in wo
=
1
= w
= wo '
1
so
There is no connection between normality and the property to be extremally disconnected (cf. Exercise 82.20).
However, if
as well as completely extremally disconnected, then X normal. Similarly, if normal, then X
(X,O,e)
(X,e)
is normal
is completely
is extremally disconnected and completely
is completely extremally disconnected.
THEOREM 82.19.
(i) If the d i s t r i b u t i v e l a t t i c e
(X,e)
i s normal as
well as completely extremally disconnected, then X i s completely normal. (ii) S i m i Z a r l y , i f
(X,e)
has a largest element
is is completely
e and X
extremally disconnected as well as completely normal, then X extrema l l y disconnected. PROOF. (if Let
(X,O)
be normal (i.e., every proper prime ideal
contains a unique minimal prime ideal) and completely extremally disconnected (i.e., the proper prime ideals containing a fixed proper prime ideal are linearly ordered by inclusion). We shall prove that X
is
completely normal (i.e., the prime ideals contained in a fixed proper prime
82
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
Ch. 11,1821
ideal are linearly ordered by inclusion). For this purpose, let w ,w 1
let u0
prime ideals contained in the proper prime ideal wo , and
.
unique minimal prime ideal contained in wo
be
2
be the
It follows immediately that
is also the unique minimal prime ideal contained in w 1 , and similarly
p0
. Hence, w 1 and w2 are proper prime ideals containing the prime . By hypothesis, the prime ideals containing uo are linearly ordered by inclusion, s o or w2 c w 1 . This is the desired result. w1 c w2 for w2
ideal uo
(ii) The proof is similar. A s observed earlier, any Riesz space
connected (equivalently, the lattice X(L)
L
is completely extremally dis-
of all principal ideals in L
is completely extremally disconnected). Hence, if
L
is normal (i.e., if
any proper prime ideal contains a unique minimal prime ideal), then L
is
completely normal. In this case, therefore, we have the situation that if the proper prime ideal ideals containing in wo
, is
wo
wo
is given, then the family of all proper prime
, as well as the family of all prime ideals contained
linearly ordered by inclusion. In other words, the union of
these two families is a chain of proper prime ideals such that member of the chain. Given the proper prime ideals wo corresponding chains are identical if chains are disjoint if set Q
wo
and
wh
wo
and
wh
and
are incomparable. It follows that the
principal projection property (i.e., for any u E L+ {uId
0 '
of all proper prime ideals is the disjoint union of all such chains
minimal prime ideal in the chain. We recall that if =
the
are comparable and the
Q ) of the
Each chain is the closure (in the hull-kernel topology of
L
is a
wo
0'
0
{uIdd ), then L
L
has the quasi
we have
is normal (by Lemma 3 7 . 1 0 ) , and s o
L
is
completely normal in this case. Not every Riesz space is normal; it was indicated in Exercise 3 7 . 1 3 that every maximal ideal in the space C([O,ll) of all real continuous functions on C 0 , l l
contains infinitely many
different minimal prime ideals. EXERCISE 82.20. Show that the distributive lattice
(X,e)
in Exercise
7 6 . 1 4 is completely normal but not extremally disconnected. Hence, with
respect to the dual ordering, X
is completely extremally disconnected but
not normal. Show that by taking the smallest element of X obtains a lattice X'
away, one
that is completely normal as well as extremally
disconnected. Actually, there are only two proper prime ideals in X'
, and
both of these are minimal as well as maximal. Show also that by appropriately
Ch. 11,8821
83
PRIME IDEAL EXTENSION
adding one extra element to X
, one
can obtain a lattice X"
that is nor-
mal as well as extremally disconnected, but neither completely normal nor completely extremallydisconnected.Finally, show thatby adding to X' its "reflection" with respect to its largest element, one obtains a lattice X"' that is neither normal nor extremally disconnected. Note that all possible combinations occur. EXERCISE 82.21. Show that every subset of the topological space E
is
normal (with respect to the relative topology in the subset) if and only if the lattice X
of closed subsets of
E
is completely normal. Formulate
and prove the dual result. HINT: We indicate how to prove that complete normality of that the relative topology in the arbitrary subset E" Let FY = F1 fl E" =
of
E
X
implies
is normal.
(with F,,F2 closed) be relatively
E-such that ; ' F
and F" are disjoint, i.e., 2 F1 n F2 has no points in common with E" By the complete normality
closed subsets of Fo
F ; = F2 fl E"
and
of
X
.
there exist closed sets F;,F;
.
such that F; fl F 1
F;
=
Il
F2
= Fo
Fi U F' = E The set theoretic complements 01,02 of F;,F; are 2 open and disjoint, and the intersections with the complement : F of FO satisfy Fi fl FE c 0; fl : F for i = 1,2 Hence, the intersections with and
.
certainly satisfy Fi fl E" c 0; n E"
the smaller set E"
In other words, the sets 0; = 0. fl E" (i=1,2) 1
the relative topology of E"
1
, and
=
1,2
.
are disjoint and open in
we have FY c 0; 1
for i
for i
1
= 1,2
. This
shows that the topology in E" is normal. EXERCISE 82.22. (i) Let the distributive lattice
{pix = The lattice
(p:p~n~,xnot in p
).
is called dLsjunctive if x # y
implies {plx #
(X,9) is disjunctive if and only if x # y
Show that {XId
(X,8)
be the set of all minimal prime ideals in Om (X,9) and, exactly a s several times before, let
# {y}d
.
I v } ~.
implies
(ii) Let there be given a base for the closed sets in the topological space E
. Show that
E
is Hausdorff if and only if, for any different
points x1,x2 in E , there exist sets FI,F2 in the base such that xi is not in F. for i
=
1,2 and such that F 1 U F2
=
E
.
NORMAL AND EXTREMALLY DISCONNECTED LATTICES
84
( i i i ) Show t h a t t h e d i s j u n c t i v e l a t t i c e
{pix
x E X, t h e s e t
i f , f o r each
Ch. 11,5823
i s normal i f and only
(X,9)
i s Hausdorff i n i t s d u a l h u l l - k e r n e l
topology. d i f and only i f {xId = {y} Y For ( i i i ) , assume f i r s t t h a t a l l { u I X a r e Hausdorff i n t h e
HINT: For ( i ) , n o t e t h a t by Theorem 6.5. dual hull-kernel X
{vIX
= {p}
topology. We have t o prove t h a t any p r o p e r prime i d e a l i n
c o n t a i n s a unique minimal prime i d e a l . Hence, l e t
prime i d e a l i n now t h a t
ul
u2
for
and l e t
x E X b e such t h a t
w
u2
and
p1
. Assume . Then
. The
{pix
sets
form a b a s e f o r t h e c l o s e d s e t s i n t h e dual h u l l - k e r n e l
{uIx . Hence, {uIz , u2
i s not i n
I n o t h e r words, we have
there exist s e t s
{u}
is not i n
and
E
z1
pl
{t~}
, z2 E
. On
z2
p2
d i s j u n c t i v e p r o p e r t y ) . I t f o l l o w s from
zi E
E w
way
i s chosen. C o n t r a d i c t i o n . Hence, w
z1
and
ui
{ u I Z 2 such t h a t
and Iulz
U IuIZ2 =
1
i = 1,2
for
c w
{vIx
.
(by t h e
z 1 v z2 = x
i s not i n
z 1 v z2 = x
t h e o t h e r hand
z1 v z2
x
be a given proper
w
i s not i n
x
a r e d i f f e r e n t p o i n t s i n t h e Hausdorff space z S x
topology of
u1
,
c o n t a i n s d i f f e r e n t minimal prime i d e a l s
w
and
{uIZ
X
that
i n view of t h e
w
c o n t a i n s only one minimal
ideal. Conversely, assume t h a t
{uIX i s
n o t Hausdorff i n i t s d u a l h u l l - k e r n e l
different have
Cul
i s normal and f o r some
X
z1
ul,u2 U {u}
i n {uIx 22 p
such t h a t f o r a l l
. In
# [ulx
the s e t
x E X
topology. Then t h e r e e x i s t and
z1 E pl
o t h e r words, f o r a l l
E u 2 we
z2
z 1 v z2
.
i n the ideal
I n view of t h e and u 2 we have { u L l V z 2 # {uIX d i s j u n c t i v e p r o p e r t y t h i s shows t h a t x i s n o t i n I , so t h e r e e x i s t s a I
g e n e r a t e d by
prime i d e a l
such t h a t
w
and
I c w
x
is not i n
w
. Hence,w
proper prime i d e a l c o n t a i n i n g d i f f e r e n t minimal prime i d e a l s
ul
is a and
u2
.
Contradiction. EXERCISE 82.23.
Let
t h e f a m i l y of a l l s e t s
(X,9)
{xIdd
o r d e r e d by i n c l u s i o n . Show t h a t element
feldd
=
{el
. Show
be a d i s t r i b u t i v e l a t t i c e , and l e t for A
x E X
,
A
be
these s e t s being p a r t i a l l y
i s a d i s t r i b u t i v e l a t t i c e with smallest
t h a t t h e l a t t i c e A i s d i s j u n c t i v e . Formulate
and prove t h e corresponding r e s u l t f o r a Riesz space. HINT: Note t h a t
t h e set of a l l
{uIx
A
, partially
(for
orderdd by i n c l u s i o n , i s isomorphic t o
x E X ) , p a r t i a l l y o r d e r e d by i n c l u s i o n . This
f o l l o w s from Theorem 6.5 ( i i ) , according t o which {xIdd c {yIdd
a r e e q u i v a l e n t (and hence
IwIx
=
{ulx
c
I U ) ~and
{u)
and = {yIdd
Ch. 11,8821
85
PRIME IDEAL EXTENSION
{uIx
are equivalent). The set of all smallest element
{pie
is a distributive lattive with
equal to the empty set (cf. section 6 ) . Hence A
{eldd
is likewise a distributive lattice with Note that the supremum and infimum of
=
{e}
as smallest element.
{xvyIdd and dd { x ~ y } respectively. ~ ~ The disjoint complement of the given element {XI
in A all
is the set of all {y}dd {yIdd
such that
{xlId # {x21d
, and
satisfying y
y E {xld
. Hence, if
{xlIdd and
so
IxIdd and
complements. It follws that A
by
. By
y
i.e., the set of
,
then
{xIdd c I
{yIdd
=
{x}dd
. Show that
(X,e)
I is a
implies y E I
. We
is a d-ideal (cf. Theorem 5.4
be the same lattice as in the preceding exercise. Prime
ideals in X will be denoted by in A
,
in the distributive lattice
recall that any minimal prime ideal in X Let A
= 0
{x,fdd # {x2}dd
{x2Idd have different disjoint
x E I implies
d-ideal if and only if x E I and (iv)).
x
are
is disjunctive.
EXERCISE 82.24. (i) The ideal I is called a d-ideal if
A
{yldd
w
(exactly as before), prime ideals
a d-prime ideal is meant a prime ideal which is also
a d-ideal. Show that the mapping
is a one-one mapping of the set of A l l prime ideals in A all d-prime ideals in X
with minimal prime ideals in X
, and
(ii) Minimal prime ideals in X
,
the set
{uIx
correspond
conversely. Formulate and prove
the corresponding result for Riesz spaces. each x € X
onto the set of
such that minimal prime ideals in A
are denoted by
u
. Show that, for
is Hausdorff i n its dual hull-kernel topology
if and only if every d-prime ideal in X
contains a unique minimal prime
ideal. Formulate and prove the corresponding result for Riesz spaces. HINT: For (ii), note that every d-prime ideal in X
contains a unique
minimal prime ideal if and only if the disjunctive lattice A
is normal.
Apply Exercise 82.22. EXERCISE 82.25. Show that the following conditions for the Riesz space
L
are equivalent.
(i) topology*
For every u E L+
the set
Cu}u
is compact in the hull-kernel
86
Ch. 11,9821
NORMAL AND EXTREMALLY DISCONNECTED LATTICES (ii)
The lattice of all
lattice of all
is a Boolean ring. Equivalently, the
{uldd is a Boolean ring.
(iii) L has theM-complementation property, i.e., for all u,v E L+ satisfying 0 2 v dd dd Iv+v,l = {u}
2
.
u
there exists an,ielement v2 I v
such that
L(iv) For every u E L+ the hull-kernel topology and the dual hullkernel topology coincide on {p}u (v) Every proper d-prime ideal in L is a minimal prime ideal.
.
HINT: The conditions (i), (ii), (iii) are equivalent by Corollary 37.3; the conditions (ii) and (iv) are equivalent by Exercise 77.13. Note that (v) is equivalent to the condition that every proper prime ideal in A
is a
minimal prime ideal. It follows from Theorem 8.5 and Theorem 8.6 that this is equivalent to (ii). EXERCISE 82.26. element of x
if
i.e., 2
.
8 A
Let
Given x E X
be a distributive lattice with smallest
(X,8)
, the
is the largest element disjoint to x ,
x ~ x * = eand x* y = 8
The lattice
implies y
(X,e)
x*
2
. Evidently
(i) 5
e = 8*
Show that
x . (ii)
Show that
complements, then p
(xVy)* = x* A
(iii) Show that and only if x** (iv)
y
x
x
(X,8)
be pseudocomplemented.
y*
A
x**
5
and
. Show that
x* = x***
. Hence, if
p
and
q
for every are pseudo
is a pseudo complement. A
y
A
=
.
y
(xAy)** = x**
if and only if
8
5 z* A
x**
if and only if x**
y**
y = 8
A
y**
A
.
, and 2 z*
ahso if
. Use this
Let x E X be given. Show that the smallest pseudo complement
such that z* t x
z*
and
q
y** = 0
Show that
to prove that (v)
A
is uniquely determined.
is the largest element of X
implies x* 2 y* ; furthermore, x
y
x*
is called pseudocompZemented if every x E X has a
pseudo complement. In the following, let x
is called pseudo complement
element x* E X
satisfies z*
=
. Use this
x**
are pseudo complements, the supremum of x
pseudo complements i s
(xvy)**
(vi) Let D(X) = (x:X*=e)
. Furthermore implies y E D(X) .
x
A
y E D(X)
y
t
x
= =
(x*AY*)*
.
(x:x**=e)
x v x* E D(X)
(vii) Given the proper prime ideal w
to prove that if x
and y
. Show that
for every in X
,
x
in the set of
x,y E D(X)
implies
and x E D(X)
the following
,
PRIME IDEAL EXTENSION
Ch. 11,§821
conditions are equivalent: (a) w
,
is not in w
x E w
(c)
Note that if X
87
is minimal, (b) x E w
,
implies x** E w
implies that
w f l D(X)
(d)
x*
is empty.
is the lattice of all ideals in a distributive lattice
with smallest element or in a Riesz space, then Id is the pseudo
I for any I E X
complement of
HINT: For (ii), write x* Similarly x* so
I
(xvy)*
x**
A
y = 8
first that x** x** w
x**
A
y**
I z*
A
y**
I
w
is minimal
generated by
. Hence
y = t3
z A
y*
A
. For (iii),
. For y
A
=r
y
5 z*
A
y
,
A
x
A
y
A
.
= t3
implies y I x* x
so
z
(xVy)* , i.e.,
z I
= t3
and then that x**
, so
x*
y**
z = 8
A
z = d'
A
.
=
,
x***
It follows
. Hence
=)
X-w
and
,
x (note that
(d)
8
is not in F ; otherwise 0
and hence q I x* E w
proper since x E w-w, (c)
A
z I
, so
z A (xvy) = t3
note that x
= t3
A z
. Then
= z
. Now write ( ~ y ) l * = w , s o x A y w . Then = (ay)** . For (vii) we first prove (a) (b). Assume that and x E w as well as x* E w . Let F be the dual ideal
for some q E X-w
=r
y*
(iv), let x
exists a prime ideal w 1 (b)
A
.
disjoint to F
. This
, which , so w1
q
A
x
is impossible). There
. The
c w
is impossible since w
is easy. For (d)
=
inclusion is
is minimal.
* (a), assume that (d) holds and
w
is not
minimal. Then there exists a prime ideal w 1 c w with an element x E w-w 1 Then x* E w 1 c w , s o x v x* E w But then x v x* E D(X) n w , which contradicts
.
(a).
EXERCISE 82.27.
The pseudocomplemented lattice
lattice if x* v x** = e lattice X*
V
(X,e)
y* = e
for every x E X
is a Stone lattice if and only if x
, i.e.,
if and only if X
(X,e)
. Show that A
y
= t3
is normal. Hence, the pseudocom-
(X,8)
prime ideal in X
contains a unique minimal prime ideal.
x
(X,e)
be a pseudocomplemented lattice. Show that
is a Stone lattice if and only if
...,xn XI A ... A Xl,
Let
(xIA..Ax ) * = xr v n
...
V
x* n
. Equivalently, X is a Stone lattice if and only 8 implies x; v ... v x* = e . n
in X x = n
implies
is a Stone lattice if and only if every proper
plemented lattice
EXERCISE 82.28.
is called a Stone
the pseudocomplemented
for all if
HINT: It is sufficient to show that in a Stone lattice we have (XAy)* = x* v y* Theorem 82.5.
. This can be
proved similarly as (iii)
* (iv) in
CHAPTER 12
ORDER BOUNDED OPERATORS
L
In section 83 we assume that
and M
are Riesz spaces, with M
complete. The linear operator T from L Tf 2 0 holds i n if
M
into M
for all f t 0 in L
and
Dedekind
is called positive if
T i s called order bounded
T maps order bounded sets (in L ) onto order bounded sets (in M ) .
We recall that a subset D exist
f,g E L
such that
of f
5
L d
5
is said to be order bounded if there
g
for all d E D
. Any
positive
T is order hounded if and only if T = T 1-T 2 TI and T2 positive (Jordan decomposition). The vector space of all
operator is order bounded and with
order bounded operators (from L
Lb
. For
TI and
T2 in Lb
positive. This makes
Lb
into M ) is denoted by
we write T
1
5
Lb(L,M)
, briefly
T2 whenever T2-TI is
an ordered vector space with the set of all
positive operators as positive cone. We shall prove that
Lb is a Dedekind
complete Riesz space with respect to this ordering (L.V. Kantorovitch,l936; is the space of real
H. Freudenthal,l936). In the special case that M numbers,
is 'exactly the space L"
Lb
of all order bounded linear
functionals on L
.
theorem that L"
is a Dedekind complete Riesz space goes back to F. Riesz
The space L"
is called the order dual of
(1928) already. Denoting the null operator by T+ = sup(T,8)
of any
+
T u for all u E L+ There exist dua Tf+
=
T E
Lb
. The
the positive part
satisfies
0
sup Tv:o
. Similar formulas hold formulas; for T t 8 =
,
8
L
f E L
0
sup Sf:B<S
/TI = T++T-
.
arbitrary, we have
,
and similar formulas hold for Tfin the last formula.
for T- and for and
and
T(lf1)
. The supremum is attained
90
Ch. 121
CHAPTER 12
defined on a Riesz
Under certain conditions positive operators T subspace K
of
can be extended to positive operators defined on
L
K
for example if the ideal generated by
is majorized by a sublinear operator p (p(f)=p(lfl))
as well as monotone
is the whole space L such that p
(O
M ) . As an application we prove that if Dedekind complete space M , then M
L
in L
,
L
o r if
T
is absolute
implies p(f)
5
p(g)
in
is a Riesz subspace of the
contains the Dedekind completion L-
L (A.S. Veksler, 1969).
of
In many cases (for example, if subset of
L
L
is Archimedean) any order closed
is relatively uniformly closed (ru-closed).
This holds in
particular for ideals. The converse is not true in general; it is rather special if a Riesz space L has the property that every ru-closed ideal in L
is order closed. The space L
Riesz homomorphism from L
has this property if and only if every
into an arbitrary Archimedean Riesz space is a
a-homomorphism (D.H. Fremlin,1977). The relation with norm closed ideals in normed Riesz spaces will be discussed in section 105.
In section 8 4 we introduce order continuous and a-order continuous operators as subsets of continuous if u
J.
Lb
. The operator
T E
Lb is called order
inf(TuTl = 0 holds for any downwards directed system
0 in L and T
is called o-order continuous (sequentially order
continuous) if this property holds for sequences (i.e., u infITu I = 0 ). It will be proved that T only if order convergence of Tfn
to
Tf
in M
fn
to
f
J.
0 implies
is a-order continuous if and
in L
implies order convergence of
. This explains the name. The set of all order continuous
operators will be denoted by
L
(abbreviation for
Ln(L,M)
) and the set
Lc . Evidently Ln is a subset of Lc and Ln are bands in . The disjoint complement of Lc with respect to Lb is denoted by Ls ; the elements of Ls are called singular operators. The disjoint complement of Ln with respect to Lb is denoted by Lsn ; the elements of Lsn are called normal of all 0-order continuous operators by
Lc
. It will be proved
that
singular operators. We have
L If L
b
= L e l c s
= L
n
.
is Archimedean and if order convergence and relatively uniform con-
vergence for sequences in L
-.
for example, if and
e L s n
0 < p <
are equivalent, then
Lc
=
Lb
. This holds,
L ( X , A , u ) , where ( X , A , p ) is o-finite measure space P It does n o t hold for p = m (unless in trivial exceptional
L
=
Ch. 121
ORDER BOUNDED OPERATORS
cases). If L
is order separable, then
L
91
Ln , i.e., every o-order
=
continuous operator is order continuous. In particular, order separability of
L implies that every o-order continuous linear functional on L
order continuous. This holds, for example, if L M(r)(X,A,p)
is
is any ideal in the space
of all real u-measurable functions on X , where again
(X,A,p) is a o-finite measure space. In section 85 we investigate the order dual L" happen that L
is of infinite dimension, but L"
functional. If
(X,A,p)
more closely. It can
contains only the null
L
is a o-finite measure space without atoms and
or L (X,A,p) for 0 < p < 1 , then L" consists P of the null functional only. It can also happen that every linear is either M(r)(X,A,p) functional on
L
L"
is a member of
. This occurs if and only if
Archimedean and every principal ideal in L
(equivalently, by Theorem 61.4, if and only if and every quotient space of
L
L
is
is of finite dimension
L is uniformly complete
is Archimedean). It was proved some years
ago that another equivalent condition is that every ideal in L
is a
projection band (C.B. Huijsmans,1976). In the general case, when it is usually not known whether one of these extreme cases occurs, there is a condition to see if a linear functional on L belongs to L" on L
condition says that the linear functional Q
and only if Q(f ) tends to zero for any sequence n that converges relatively uniformly to zero. The existence.of non zero elements in L-
(fn:n=1,2,
. It is evident that if
the null functional, then p(f)
lQl(lfl)
if
in L
@
E L"
is not
is a non zero Riesz seminorm
. Conversely, if there exists a non zero Riesz seminorm in
L , then contains a non zero element. The proof is based on the Hahn-Banach
in L L"
...)
is related to the existence
of non zero Riesz seminorms in L =
or not. The
satisfies Q E L"
extension theorem. A s a particular case of a general extension theorem for operators in section 8 3 , we get the result that any positive linear functional on a Riesz subspace of L seminorm in L
which is majorized by a Riesz
can be extended to a positive linear functional on
L
. As
an application it will be shown that there exist non zero singular linear
.
functionals on the sequence space Lm If L is a normed Riesz space (i.e., L norm), then the Banach dual L*
of
L
is equipped with a Riesz
is an ideal in the order dual L"
h y normed Riesz space which is norm complete is called a Banach lattice. In a Banach lattice L we have L* = L"
.
.
Ch. 1 2 1
CHAPTER 12
92
Any a-order continuous (order continuous) linear functional on
. The
be called an integral (normal integral) on L
set L:
. In section 86 we
L
will
of all
integrals is a band in L ; the same is true for the set :L integrals on
L
of all normal
investigate integrals on L
, where L
is an ideal in the space M(I) (X,A,p) such that, once more, (X,A,p) is a a-finite measure space. Note that L is order separable, so a-order continuity is the same as order continuity. In particular, every integral on
L
is a normal integral. The subset E
of
X
is called an L-zero
f E L vanishes p-almost everywhere on
set if every
largest L-zero set Em
E
. There exists a
(uniquely modulo sets of measure zero); the set
theoretic difference X-E, is called the carrier of the ideal L integral $
on L
the word. Precisely, for any integral $ real function t
for all
f E L
on
X
on
L
there exists a u-measurable
such that
. The function
-
is unique on the carrier of
t
L (X,A,p) for some p P well-known classical result.
L
case that
. Any
can be written as an integral in the usual sense of
I
satisfying
=
p <
5
L
, we
. In
the
get a
Section 87 is devoted mainly to examples. We first introduce some N
further terminology. The disjoint complements of respect to
Li
L" ) are denoted by
Lc
L i (with
and
and Lin ; the members of
L"
and Lin
are called singular linear functionals and normal singular linear functionals respectively. The point set X
is said to have a measurable cardinal if
there exists a a-additive (non-negative) measure X all subsets of
X
0 < X(X)
such that
or countable subsets E
of
X
.
< =
and
on the collection of
A(E) = 0 for all finite
In all other cases
X
is said to have
a non-measurable cardinal. Not much is known about the existence of sets with a measurable cardinal. There is a conjecture that every set has a nonmeasurable cardinal. It has been proved that if the continuum hypothesis holds, then the set of all real numbers has a non-measurable cardinal. In the first example of section 87 we let L be the space of all real functions on a non-empty point set X
. Then
L" = L z
.
is always true;
has a non-measurable cardinal, then L" = L" = L" In the second c n example we specialize X to be the set of all real numbers and we prove if X
that now
L" = :L
= Lz
, without
again a non-empty point set and
using the continuum hypothesis. If X L
is the space of all bounded real
is
Ch. 121
93
ORDER BOUNDED OPERATORS
functions on X , then L" = L" holds if and only if X has a nonc n measurable cardinal. Furthermore, if L is the space of all real continuous functions on the closed interval If
L
(x:OSx
is super Dedekind complete, then L
.
L-
.
is order separable, and
L" = L" If L is Dedekind o-complete, then L" = L" does not c n c n necessarily hold. It is not known if Dedekind completeness of L implies so
.
If it is true, however, that every set has a non-measurable L" = L" c n cardinal, then L" = L i holds for every Dedekind complete space L (W.A.J. Luxemburg,l967). In section 88 we assume L,M,Lb and its subspaces to be the same as before (in particular, M is still assumed to be Dedekind complete). For any subset A
of
L
the (L,M)-annihilator
A '
is defined by
TELb:Tf=O for all fEA and for any subset
B
of
Lb
the inverse (L,M)-annihilator
'B
is
defined by fEL:Tf=O for all TEB is real number space, then
If M
Lb
=
L"
,
and
so
for A c L
and
B c L"
we get the familiar definitions for annihilator and inverse annihilator. is an ideal in L , then A'
If A
is a band in
Lb
. If
B
is an ideal
Lb , then OB is an ideal in L . We also introduce the ideal La , consisting of all f E L such that If/ 2 un J 0 implies Tun .I. for every positive T in Lb . Obviously, La is the largest ideal in L in
in L
0
on which every order bounded operator is o-order continuous. Similarly, we define Lan as the largest ideal in L
on which every order bounded
and operator is order continuous. It will be proved that La = '(Ls) a 0 = O(Lsn) In the converse direction, (L ) = Ls holds if La is an 0 super order dense in L and (L ) = Lsn holds if Lan is order dense in L
.
Lan
.
In section 89 it is assumed that W and of
W' W
is a real or complex vector space
a linear subspace of the algebraic dual of W the annihilator A'
of
A
is defined by
. For any subset
A
CHAPTER 12
94
Ao
Ch. 1 2 1
($EW':+(f)=O f o r a l l fEA)
=
and f o r any s u b s e t
of
B
W'
the inverse annihilator
O B
of
is
B
d e f i n e d by 'B
(fEW:$(f)=O
=
It i s obvious t h a t O(W')
= {O}
,
.
= I01 ; we assume t h a t
W'
i.e.,
for a l l ~ E B )
s e p a r a t e s t h e p o i n t s of
W'
i s so large that
W'
. This
W
implies t h a t
may b e regarded a s a l i n e a r subspace of t h e a l g e b r a i c d u a l of and
W
a r e i n d u a l i t y , a s i t i s sometimes c a l l e d ) . The weak topology
W'
a(W,W')
is by d e f i n i t i o n t h e weakest topology i n
a r e continuous f u n c t i o n s on
W
. It
such t h a t a l l
W
i s proved t h a t
W
c l o s u r e of any l i n e a r subspace a(W',W) topology i n
Banach d u a l topology
W*
, we
a(w*,tl)
W'
is
A
. If
'(Ao)
in
w*
. Similar
a(W,W')
r e s u l t s hold f o r
i s normed and we choose f o r
W
g e t t h e weak topology
.
We r e t u r n t o t h e s i t u a t i o n t h a t
@ E W'
w i t h t h e topology
i s a Hausdorff t o p o l o g i c a l v e c t o r space and t h a t t h e
a(W,W') the
W
(so
W'
a(W,W*) i n
W
4 E L
i s a Riesz space. For any
L
the
W'
and t h e weak s t a r
the ideal
i s c a l l e d t h e n u l l i d e a l ( o r a b s o l u t e k e r n e l ) of complement
4
and t h e d i s j o i n t
i s c a l l e d t h e c a r r i e r of
C($) =
4 ( n o t t o be confused
w i t h t h e c a r r i e r of an i d e a l of measurable f u n c t i o n s ) . I n s e c t i o n 90 it
i s proved t h a t f o r p o s i t i v e
6
and
The formula
C($A@) = C ( 4 ) flC($)
a t l e a s t of
4
and
(I
J,
in
L" we have
does n o t h o l d g e n e r a l l y . It h o l d s i f one
i s a member of
p r o p e r t y . As observed a l r e a d y , N($)
Lcf
or also i f
i s an i d e a l . I f
is a band; t h e converse does n o t always hold. I f or
L
h a s t h e p r o j e c t i o n p r o p e r t y and i f
if and o n l y i f
N($)
i s a band i n
L
. In
4 E Lz
L
,
t h i s case
L
h a s t h e Egoroff
E L i
, then
N(+)
i s Dedekind a-complete t h e n we have
$
L = N($)
C($)
@
E L :
.
Ch.
12,9831
ORDER BOUNDED OPERATORS
I n s e c t i o n 91 it is proved f i r s t t h a t i f
95
i s Archimedean and
L
uniformly complete, t h e n i t i s p o s s i b l e to d e f i n e i n t h e c o m p l e x i f i c a t i o n
L+iL of
an a b s o l u t e v a l u e
L
I f+igl Equivalently, i f v a l u e of
h
=
If+igi
by
).
sup[f
cose+g sine:O<e<2a
h = f+ig
,
1h.l = supe Re(he-ie)
then
c o i n c i d e s w i t h t h e e x i s t i n g one i f / h l + h 2 / = Ihl I + I h 2 /
t r i a n g l e i n e q u a l i t y and
. The
absolute
i s real, s a t i s f i e s the
h if
Ihl
I I Ih21. Freuden-
t h a l ' s s p e c t r a l theorem can b e extended t o complex Riesz spaces. I f a s u b s e t of
L+iL
Dd =
, the
(hEL+iL: l h l l l f /
i s a band ( i . e . ,
f o r a l l fED
t h e r e a l p a r t i s a band i n
p r o p e r t i e s a s i n t h e r e a l case. F i n a l l y , i f ideal i n
L
, the
h
quotient
h € L + iL
and f o r every
is
D
d i s j o i n t complement
(L+iL)/(A + i A )
r
r
L )
, and
has s i m i l a r
Dd
i s a uniformly c l o s e d Ar admits an a b s o l u t e v a l u e ,
t h e a b s o l u t e v a l u e of t h e equivalence c l a s s of
Ihl
is e x a c t l y t h e e q u i v a l e n c e c l a s s of
.
I n s e c t i o n 9 2 , f i n a l l y , we i n v e s t i g a t e t h e s i t u a t i o n t h a t t h e r e i s M
, which
L+iL
into
a n o t h e r Riesz space operator and
T2
T
from
i s of t h e form
r e a l o r d e r bounded o p e r a t o r s
M ) . This i m p l i e s t h a t
Lb(L+iL,M+iM)
We prove t h a t f o r any complex
IT1
i s Dedekind complete, Every o r d e r bounded M+iM
(i.e.,
TI
and
T = Tl+iT2 T2
map
with
TI
into
L
is t h e c o m p l e x i f i c a t i o n of
Lb(L,M)
.
i n t h e space t h e a b s o l u t e v a l u e
T
satisfies IT/ (u) = sup(lTh1 :h€L+iL, IhlSu)
,
f o r every
u E L+ ( W . J .
Dedekind a-complete,
de Schipper; H.H.
S c h a e f e r ; 1973-1974).
If
L
is
t h e proof can be s i m p l i f i e d (Exercise 92.7).
83. Order bounded o p e r a t o r s We r e c a l l t h a t t h e mapping r e a l v e c t o r space if
W
T
of t h e r e a l v e c t o r space
V
into the
is c a l l e d a linear operator ( o r zinear transformation)
96
Ch. 12,1831
ORDER BOUNDED OPERATORS
0
T af+bg = a Tf + b Tg blds for all f
and
g
in V
and all real numbers
a,b
. We
shall
briefly say operator instead of linear operator. Defining addition of operators (from V
into W ) and multiplication of an operator (from V
into W ) by a real constant by (TI+T2)f = Tlf+T2f ; (aT)f fE V
for all L(V,W)
and all real numbers
of all operators from
V
a Tf
=
,
a
it is evident that the set
into W
is a raal vector space. In the
case that W
is the space of real numbers, we get in this manner the
vector space
v#
of all real linear functionals on
called the (real) algebraic dual of V Let L
and PI
Definition 1 1 . 1 ) ; usual, by
L+
and M+ f
5 g
v
. The space
v#
is
be real ordered vector spaces (as defined in
the positive cones of L
and M will be denoted, as
respectively. The operator T for all u E L+
called positive if Tu E M" and only if
.
in L
implies Tf
5
Tg
from L
. Evidently, T
in M
into M
is
is positive if
. Note that any finite
linear combination with non-negative coefficients of positive operators is a positive operator. In the following we assume about L (i) L
generating (i.e-, every with
and M
that
is an ordered vector space such that the positive cone L+ f E L has at least one decomposition f
=
is
u-v
1,
u,v E L+
(ii) L has the dominated decomposition property (i.e., if u,v,w E L+ satisfy 0 < u u, s v
, u2
(iii) M
< w
5
v+w
, then
there exist u 1 and
u in L+ 2
such that
and u,+u2 = u ) ,
is a Dedekind complete Riesz space.
Observe that by Corollary 15.6 every Riesz space satisfies the conditions (i), (ii). For the equivalence of the dominated decomposition property and the Riesz interpolation property we refer to Exercise 15.12 and Exercise 7 6 . 1 8 . Some authors call any ordered vector space with a generating cone and possessing the dominated decomposition property already a Riesz space (cf. for example L. Fuchs,Cll),
but we shall adhere to the conventions established in the earlier chapters that a Riesz space is the
12,5831
Ch.
ORDER BOUNDED OPERATORS
97
t h e same a s a v e c t o r l a t t i c e . For examples of spaces s a t i s f y i n g ( i ) and ( i i ) w i t h o u t b e i n g a Riesz space we r e f e r t o E x e r c i s e 15.14 (where t h e s p a c e of a l l with
p
and
f ( x ) = p(x)/q(x)
r e a l polynomials and
q
t o E x e r c i s e 76.16 f u n c t i o n s on (Ti;)
on t h e closed r e a l i n t e r v a l
L
(where
q(x) > 0
for a l l
L
is
Ca,bl
x E Ca,bl ) and
i s t h e space of a l l r e a l d i f f e r e n t i a b l e
Ca,bl ) . Returning t o t h e c a s e t h a t c o n d i t i o n s (i),
(ii),
h o l d , t h e r e a d e r i s advised t o keep i n mind t h e important s p e c i a l
case t h a t
i s a Rhesz space and
L
M
i s t h e space of a l l r e a l numbers.
F i r s t of a l l we s h a l l prove a u s e f u l e x t e n s i o n lemma. LEMMA 83.1. Let T(U) t 0
be an additive mapping from
T
u E L+
for a22
T(U+V)= T ( U ) +
and
that
T
i s an extension o f
T
,i.e.,
PROOF. We prove f i r s t t h a t a t 0
and every r e a l a d d i t i v i t y of and
r'
T
. If
such t h a t
r u 5 au 5 r ' u
0
. For a
5
i n t o M+
a
, i.e.,
i n t o M such
.
u E L+
holds f o r every
u E L+
t h i s follows e a s i l y from t h e
i s p o s i t i v e and i r r a t i o n a l , choose r a t i o n a l
r < a < r'
.
u,v E L+
Tu = T ( U ) holds for a22
-r(au) = a-r(u)
rational
from L
T
Then there e x i s t s a unique p o s i t i v e operator
L+
f o r a22
T(V)
. Given
u E L+
,
r
i t follows from
that
so
r f a
Letting
r'
and
+
n o t e t h a t we u s e h e r e t h a t
a ,we have
M
rT(u)
+
aT(u)
and
i s Archimedean ( s i n c e
M
r ' T ( u ) J. a-r(u) ;
i s Dedekind
complete by h y p o t h e s i s ) . It follows then t h a t
SO
-r(au) = aT(u)
.
Now observe t h a t i f
.
u y v , u I , v I E L+
sacisfy
T(U)-T(V) = T ( u ~ ) - T ( v ~ ) For t h e p r o o f , n o t e t h a t T ( U ) + T ( V) = 1
T(U+V
1
) = T ( u ~ + v )=
T(U
1
)+T(v)
, from
u-v = ul-vl
,
u+vl = u I + v
then
, so
which t h e d e s i r e d
98
Ch. 12,8831
ORDER BOUNDED OPERATORS
c o n c l u s i o n f o l l o w s , Hence, Since every f = u-v
position
with
, then
Tf = T(u)-T(v)
decomposition of Tu =
. Note
f
f,g E L
f
i s d e f i n e d now on
f = u-v
and
g = u -v
I
T(au) = aT(u)
that
T ( a f ) = aTf
there exist
, for
T
, i.e.,
L
u,v,uI,vI
T(f+g) = Tf+Tg
L+
in
and
L
such t h a t
I Y S 0
all
=
.
Tf + Tg
and a l l
u E L+
holds f o r a l l
bene proved t h a t
and n o t on t h e p a r t i c u l a r
i s a d d i t i v e on
T
T(u)+T(u,)-T(v)-T(v,) From
T
.
u E L+
. Indeed,
it f o l l o w s t h a t i f we d e f i n e
already that
W e s h a l l prove now t h a t for a l l
h a s a t l e a s t one decom-
depends only on
holds f o r every
T(U)
,
u,v E Lc Tf
f E L
f E L
, it
a 2 0
and a l l r e a l
i s a p o s i t i v e o p e r a t o r from
L
follows e a s i l y
. It
a into
M
has thus which extends
T .
F i n a l l y , f o r t h e proof t h a t o p e r a t o r which extends such t h a t
f = u-v
. Given
T
, and
i s unique, assume t h a t
T
f E L
there exist
i s another
TI
and
u
v
in
L+
hence
T f = T u-T v = T ( u ) - T ( v ) = Tu-Tv = Tf 1 I 1 This shows t h a t
TI = T
, so
We a g a i n assume t h a t p o s i t i v e cone and such t h a t Given of
L
fo
and
go
in
L
i s unique.
T
L
i s an ordered v e c t o r space w i t h a g e n e r a t i n g
L
h a s t h e dominated decomposition p r o p e r t y .
satisfying
o r d e r i n t e r v a l of t h e form
C-u,ul
f o r some
symmetric. I n t h e p a r t i c u l a r c a s e t h a t f
f o 5 go
, the
subset
i s c a l l e d an order i n t e r n a l , and i t i s denoted by
E C-U,~]
i f and only i f
The s u b s e t
D
of
L
If1
<
u
.
L
u E L+
(f:fobf5go)
Cfo,gol
is s a i d t o be
i s a Riesz space we have
i s s a i d t o be bounded (order bounded)
i n c l u d e d i n some o r d e r I n t e r v a l . I n t h i s c a s e o r d e r i n t e r v a l . For t h e p r o o f , assume t h a t
. Any
D
if
D
is
i s included i n a s y m e t r i c
D c Cf , f ' l 0 0
. Since
L+
is
Ch. 12,5831
ORDER BOUNDED OPERATORS
ut-vt for appropriate u1,v' E L+ , so f;, u'+vl . Similarly, -f u-v with u,v E L+ , -f u+v ,
generating, we have f' < 0 -
99
and hence
=
f > -u-v 0 -
[-u-v,u'+vl]
. But
0
then D
< 0 -
so
=
. It follows already that
D
is included in [-w,wl
is included in for w
From these remarks it is evident that the algebraic bounded subsets D1 and D2 DEFINITION 8 3 . 2 . Let
=
.
u+v+u'+v'
sum
of
D,+D2
is again bounded.
L be as before and l e t
be a (not necessari-
M
l y Dedekind complete) Riesz space. The operator T from L i n t o M i s called order bounded i f the image under T of every bounded subset of L i s a bounded subset o f
. In
other words, T i s order bounded whenever the image of every s y m e t r i c order interval i n L i s a bounded subset of
M
. Equivalently,
M
T i s order bounded whenever, f o r every
u
E L+ , the s e t
(ITf 1 :-ur-fr-u)
i s a bounded subset of
M+
. Evidently,
bounded operators from L i n t o M of all operators from
L(L,M)
L
the s e t
. The space
#
L
operator T from L i n t o M and T2 p o s i t i v e .
. I n particular,
into M
a l l order bounded linear functionals on
algebraic duaL
of a l l order
Lb(L,M)
i s a linear subspace of the space
L"
the s e t
of
L"
L i s a linear subspace of the
i s called the order duaL o f
i s called regular i f
L
. The
T = T -T with 1 2
TI
It is obvious that every regular operator is order bounded. In the decomposition theorem which follows now it is shown that for M same. From here on we assume again that M
Dedekind
into M ) are the
complete regular and order bounded operators (from L
is Dedekind complete.
THEOREM 8 3 . 3 . (i) Any positive operator from
L
into M
i s order
bounded. (ii) (Jordan decomposition). The operator
T from
L into M
order bounded i f and o n l y i f there e x i s t positive operators (from L i n t o M
such t h a t
)
PROOF. (i) Let more, let u € L+
T = T -T 1
2
T2
'
T be a positive operator from L
and -u 2 f s u
TI and
is
. Then
-Tu
5
Tf
2
into M
Tu, so
. Further.
iTf/ < Tu
100
Ch. 12,§831
ORDER BOUNDED OPERATORS
This shows that
Tu
is an upper bound of the set
.
(ITf I :-uSf5u)
Hence, T is order bounded. are order bounded, T
TI and T2 positive, then T I and T2 T is order bounded. For the converse, assume that
T = TI-T2 with
(ii) If
so
, and
is a bounded subset of M+ bounded subset of M T(U)
exists in M
, which
,
the set
so
(Tv:OSvSu)
is certainly a
implies that
0 is Dedekind complete. The element TO = 0 is one
since M
t 0
T(U)
the set
= sup Tv:O
of the elements in the set so
u E L+
is order bounded. For any
i.e.,
Given u 1 and u2
T(U)
(Tv:OSv
, we
in L+
T(U1)+~(u2)
=
. We
E M+
of which
prove that
is the supremum,
T(U)
T
is additive on
):o~vl
sup
In the converse direction, if v E L+ v
.
have
satisfies 0
S
S T(U
in L+ such that 0 < v 1 S u 1 , 0 5 v 2 (by thedominated decomposition property), and s o
By taking the supremum on the left for all v
+u ) 1 2
v 5 u1+u2
exist v 1 and v Vl+V2 =
L+
satisfying 0
,
. there
5
u2
and
S
v
u +u 1 2 ’
S
it follows that
It has been shown thus that
T
is an additive mapping from L+
into M+
so by the extension lemma (Lema
83.1) there exists a positive operator
from L
T
into M
which extends
. Write
T2
=
T1-T
. By
definition of
, T
Ch. 12,5831
ORDER BOUNDED OPERATORS
101
, i.e.,
T u = T ( U ) 2 Tu f o r every u E L+ , s o T u t 0 1 2 p o s i t i v e . Hence T = T1-T2 w i t h T I and T2 p o s i t i v e .
we have
is
T2
A s mentioned b e f o r e , we d e n o t e t h e v e c t o r space of a l l o p e r a t o r s from into
L
by
M
o p e r a t o r s by T
T2
2
1
L(L,M)
whenever
and t h e l i n e a r subspace of a l l o r d e r bounded
. For
Lb(L,M)
TI
and
in
T2
Lb(L,M)
we s h a l l w r i t e
i s p o s i t i v e . E v i d e n t l y , t h i s makes
T -T 2 1
ordered v e c t o r space such t h a t t h e p o s i t i v e cone a l l p o s i t i v e o p e r a t o r s from
L
into
. We
M
Lb(L,M)
an
i s t h e s e t of
Li(L,M)
s h a l l prove now t h a t
Lb(L,M)
i s a c t u a l l y a Dedekind complete Riesz space. THEOF34 83.4. The ordered vector space
i s a Dedekind com-
Lb(L,M)
plete R i e s z space. PROOF. The n u l l o p e r a t o r from be an a r b i t r a r y member of
T
into
L
L b (L,M)
w i l l be denoted by
M
. Let
8
TI
and l e t t h e p o s i t i v e o p e r a t o r
be
d e f i n e d a s i n p a r t ( i i ) of t h e preceding theoaem, s o T u = sup/ Tv:O
f o r every
. Then
u E L+
T u t 0 and T u 2 Tu f o r every u E L+ , s o 1 1 and 8 Let T ' b e a n o t h e r upper bound of T
i s an upper bound of
TI
and
8'
. Then
.
T
T'u t T'v t Tv
(
T'u t sup Tv:O
T,
It f o l l o w s t h a t
T1 = sup(T,B) = T+ i s t h e supremum of
0 5 v
for
1
6 u
,
.
= T u 1
i s t h e l e a s t upper bound of
. Hence, T
for
and
T
and
8
,
i.e.,
E L b ( ~ , M ) , t h e mapping
T,S
in t h e
S
so
space
T+(S-T)+
.
L b ( ~ . ~ i) s a
L b ( ~ , ~ )Thus
Riesz space. For t h e proof t h a t
8
2
T
Lb(L,i) element u2
in
4 5 To
. To
in
.
t h i s end, l e t
~ ( u ) exists i n L+
L b ( ~ , ~ i)s Dedekind complete, assume t h a t
L b ( ~ , ~ )We have t o prove t h a t
t h e systems
M+
T(u)
=
sup,, T,,u
since
{T,,ul)
and
M
f o r every
i s Dedekind {Tou2}
sup T
exists i n
uaE L+
. The
complete. F o r
a r e e q u i d i r e c t e d , so
u1
and
ORDER BOUNDED OPERATORS
102
Ch. 12,5831
.c(uI+u2) = sup( Tu ul +T0 u 2
sup T u +sup T u 0 1
by Theorem 15.8(iii).
= T(u~)+T(u~)
This shows that
T
is an additive mapping from L+
into M+ , and so (by the extension lemma) there exists a positive operator T
from L
u E L+
,
of all Tu
.
into M which extends
the operator T
.
T Since Tu = sup T u for every u u is obviously the desired supremum of the system
COROLLARY 83.5. If L i s an ordered vector space w i t h a generating
p o s i t i v e cone and w i t h the dominated decomposition property, then the order dual
L"
L
of
i s a Dedekind complete Riesz space w i t h the s e t of a l l L as p o s i t i v e cone. I n p a r t i c u l a r , the
p o s i t i v e Zinear f u n c t i o n a l s on order dual
L"
L
o f any Riesz space
i s a Dedekind complete Riesz space.
L (L,M) is a Dedekind complete Riesz space is due to b L.V. Kantorovitch ([1],1936) and to H. Freudenthal (C17,1936,section 13). The corollary about L" is due to F. Riesz (Cll,1928;C21,1937). Together The theorem that
with Freudenthal's spectral theorem this was one of the starting points of Riesz space theory. THEOREM 83.6. (i) Let
any (1)
T E Lb(L,M)
and any
Lb(L,M)
be t h e same as before. For
we have
0 0
T+u = sup Tv:OSvSu
-
and
L,M u E L+
(21
-T u = inf Tv:OSv
(3)
lTul 5 ITI(u)
, ,
.
(ii) In p a r t i c u l a r , l e t and any f E L we have
L
be a Riesz space. For any
T E Lb(L,M)
Ch. 12,5831
103
ORDER BOUNDED OPERATORS
and f o r any
we have
u E L+
(7)
in the proof of TI According to that proof, T 1 is an extension of the mapping
PROOF. (i) T+ Theorem 83.3. T
from Lf
for u E L+
is the operator which was called
into M"
, and
T
is defined by
. Hence, the formula
(1)
follows. For the proof of ( Z ) ,
note
that in view of (1) we have
For ( 3 ) , note that Tu
S T+u 5
ITl(u>
and
For the proof of ( 4 ) , write T3 = sup(T 1 ,T2) hence
,
, SO
5
T-u
S
ITI(U)
so
T3
=
(T1-T2)++T2 , and
-Tu
which is the desired result. The proof of (5) is similar. (ii) L and
If1
is now a Riesz space, so for any
f E L
the elements f+,f-
are well defined, and we have
which proves (6). Finally, for the proof of ( 3 ) , note first that
Ch. 12,9831
ORDER BOUNDED OPERATORS
104
Observe now that
0
i
. Hence
(v-wl 5 u
so
v i u
,0
i
w
2
u implies v-w
i
u and w-v
S
u ,
/TI(u) = sup(T(v-w):O
: 1 f 1 iu)
.
Thus formula (7) follows. We present another application of the extension lemma. THEOREM 83.7.
(i) Let
complete, and l e t
positive operator from
Then
T
L
that
8
I
'T . It
T to
TIA (i.e.,
Of
TA
< T
dA f E A
T(A of
M
into M
be Riesz spaces with L
. Furthermore,
. For every
u E L+
let
Dedekind
M
T be a
we s e t
, so there e x i s t s a L i n t o M which extends T . Note and TAf = Tf for a l l f E A , whereas T f = 0 for A f o l Z m s that TA i s an extension of the r e s t r i c t i o n The eztension TA i s the minimal positive extension T I i s another positive extension of T1A , then
i s an additive mapping from L+ i n t o M+
unique positive operator all
L and
A be an ideal i n
A i f
TA from
.
TI
In particular, i f i s a p o s i t i v e Zinear functionaZ on L , then there e x i s t s a positive Zinear functionaZ $ on L such that $ A ( f ) = $(f) for aZZ
f E A and
$,(f)
= 0
f o r aZZ
f /Ad
. The functionaZ
minimaz p o s i t i v e Linear extension of the r e s t r i c t i o n
.
$IA (ii) With the same notations as above, i f the r e s t r i c t i o n
$A i s the
TIA i s
Ch. 12,§831
105
ORDER BOUNDED OPERATORS
sequentiaZZy order continuous on A (i.e., vn E A f o r n = 1 , 2 , ... and v C 0 implies Tv C 0 i n M 1, then the minimal extension TA i s n n sequentially order continuous on L Similarly, i f T I A i s order continuous on A ( i . e . , Tv C 0 holds i n M for any dotlnwards directed system vT C 0 i n A 1, then TA i s order continuous on L
.
.
~ ( u ), a s defined above, e x i s t s f o r every
PROOF. ( i ) Note f i r s t t h a t u E L+ For
by t h e Dedekind completeness of
uI
and
u
in
2
L+
M
. Hence
maps
T
i n t o 'M
L+
.
we have
T(U ) + T(U2) = I
{
sup T(V 1+V 2) :V 1 'v2 EA,0
v E A
satisfies
2 T ( U 1+U
0
4
0
such t h a t
5
v1 5 u1
Tv = Tv +Tv2
1
,0
S
2
v 2 5 u2
, and
L+
T(U
into
opera t o r TIA
any
.
t . By
. Let v E A
TA
L
into
M
prove t h a t
0
T
v
satisfying
u
5
v 2 u1+u2
T
0 2 v S u
n sup TAun e x i s t s i n
, we
have
, i.e.,
,
which extends
. For
TIA
T u 2 T v = Tv
1
1
any
u E L+
, so
= TAu
i s t h e minimal p o s i t i v e l i n e a r e x t e n s i o n of
TA
TIA S
0
i s an a d d i t i v e mapping from
which extends
TJu 2 sup(Tv:v€A,OSv
( i i ) Let
in
v ] +v 2 = v , s o
be another p o s i t i v e e x t e n s i o n of
satisfying
This shows t h a t
v2
t h e e x t e n s i o n lemma (Leuma 83.1) t h e r e e x i s t s a p o s i t i v e
from
TI
there
.
+u2) 5 T ( u ~ + ) T ( u ~ ) Hence
M
v,
T ( U ~+) T ( u ~ )
By t a k i n g t h e supremum on t h e l e f t f o r a l l we g e t
,
and
v 4 uI+u2
e x i s t (by t h e dominated decomposition p r o p e r t y ) elements A
2) '
be s e q u e n t i a l l y o r d e r continuous on f u
sup TAun
M
5
in
L
implies
and s a t i s f i e s
T u 4 TAu An
. It
A
T/A.
. We have
to
i s evident t h a t
T u A
For t h e proof of t h e converse i n e q u a l i t y , l e t
v E A
satisfy
0 < v
5 u
and
106
ORDER BOUNDED OPERATORS
and l e t
Ch. 12,9831
... . Then
v = inf(v,un) f o r n = l , Z , n Tv 4 Tv Since u 2 v implies T u n n n An
.
sup T u An
2
sup Tv
n
= Tv
2
vn 4 v
TAvn = Tvn
,
in
A
,
so
i t follows t h a t
,
and s o
T u = sup TAun , i . e . , T u 4 TAu A An implies o r d e r c o n t i n u i t y of TA
Hence of
. The
TIA
proof t h a t o r d e r c o n t i n u i t y
is similar.
We s p e c i a l i z e t h e l a s t theorem t o t h e c a s e t h a t
A
is a principal
ideal. THEOREM
83.8.
( i ) Let
.
L
M
and
be Riesz spaces with M Dedekind
complete, and l e t v E L+ Furthermore, l e t For every u E L+ we s e t from L i n t o M
.
inf(u,nv)
Then
T~
1
:n=1,2
T
be a positive operator
1.
,...
i s an additive mapping from L+ i n t o M+ , so there e x i s t s a
.
unique positive operator Tv from L i n t o M which extends T~ Note T and Tvf = Tf f o r aZZ f i n the principal ideal A, that 0 S Tv
E (AJd . The operator Tv i s the rmkirnaZ positive linear extension of the r e s t r i c t i o n TIA, . In particular, i f + i s a positive linear functionaZ on L , then there e x i s t s a positive Zinear functional $, on L such t h a t +,(f) = $ ( f ) f o r aZZ f E A, and +,(f) = 0 f o r aZZ f E (Av)d . The functional bV i s the minimal positive Zinear extension of the r e s t r i c t i o n of t o A, ( i i ) If TIAV i s order continuous (or sequentiazzy order continuous) on A, , then Tv i s order continuous (or sequentialZy order continuous) on L . generated by
+
v
, whereas
Tvf = 0
.
PROOF. Given
u E L+
, we
set
for a l l
f
ORDER BOUNDED OPERATORS
Ch. 12,1831
107
and
.
sup W' = T ~ ( U ) It i s e v i d e n t t h a t
Note t h a t
For t h e converse i n e q u a l i t y , n o t e t h a t i f w
that
and s o
I
u
w
I
and
w
nv
I
. It
inf(u,nv)
SUP
T (U) V
W'
Tw E W
f o r some n a t u r a l number follows t h a t
w'
= SUP
w
sup W
5
C
W
sup W'
,so
5
sup W
i s given, we may assume n
(depending on
. Hence
sup W'
w ),
.
A l l d e s i r e d c o n c l u s i o n s f o l l o w t h e r e f o r e from t h e p r e c e d i n g theorem. Once more, l e t and l e t
T E &(L,M)
h o l d s f o r every
L
and
. Then
M
b e Riesz spaces w i t h
M
Dedekind complete
according t o t h e formulas ( 1 ) and ( 7 ) i n Theorem
u E L+
8 3 . 6 . We s h a l l prove now a p a i r of d u a l formulas. For t h e f i r s t of t h e s e formulas and f o r some of t h e f a c t s about t h e o p e r a t o r tional
T
V
(and t h e func-
Qv ) i n t h e l a s t theorem, c f . I. Namioka (C11,Lema 7.7 and
Theorem 7.8). THEOREM 83.9. Let
complete and Zet f E L
T
L
and
M
be Riesz spaces with
be a positive operator from
M
Dedekind
L into M
These suprema are attained, i.e., these suprema are maxima. I n p a r t i c u b , i f 4 i s a positive linear functional on f E L
. For any
wehave
, then
L
and if
.
108
Ch. 12,§831
ORDER BOUNDED OPERATORS
sup(Sf:e<S
Tf+
5
For the converse inequality, let the operator Tf+ be defined as in the preceding theorem i.e.,
1
. Then
for u E L+
8 5
Tf+
5
T
,so
Tf+ is an admissable S
. The operator
yields the desired maximum on account of
Tf+
T f = Tf+fc
f+
-
Tf+f-
=
Tf+ - 0 = Tf+
.
This concludes the proof of the first formula. For
we have
IS1 5 T
lSfl
5
ISl(lfl)
5
T(lfI)
For the converse inequality, consider the operator S
SO
S
is admissable. The thus defined S
, so
=
T +-Tff
. Then
yields the desired maximum on
account of Sf = T f f+
EXAMPLE 83.10. case that L and m
-
T -f f
=
Tf+ + Tf- = T(lf1)
.
As a very simple but illuminating example we consider the and M
are Archimedean Riesz spaces of finite dimension n
respectively (with n
assume then that
L
2 1
and m
2 1
). By Theorem 2 6 . 1 1 we may
is n-dimensional real number space Xin
usual coordinatewise ordering, and similarly M = Xim
with the
. The vectors having
Ch. 12,5831
ORDER BOUNDED OPERATORS
109
one coordinate equal t o one and a l l o t h e r coordinates zero a r e b a s i s v e c t o r s in
Rn
R m ) . Any o p e r a t o r
(and s i m i l a r l y i n
from
T
En
lRm i s
into
represented (with r e s p e c t t o these bases) by a matrix
...,m
( t . . ) : i = 1, 1J where each
....n
; j = I,
,
i s a r e a l number. It i s evident t h a t a l l numbers
tij
+
tij =
[
maxi, i j '
0)
and
tij
+
-
=
max(-tij,O)
.
t. = t -t f o r a l l i and j Hence, t h e ij ij ij o p e r a t o r s T+ a n d T- , having the m a t r i c e s ( t z j ) and (tT.1 r e s p e c t i 1J vely, a r e p o s i t i v e operators (from Rn i n t o Rm') such t h a t T = T+-Ta r e non-negative,
and
.
?Rn
This shows t h a t every o p e r a t o r from
into
i s a l s o evidenet t h a t i n the Riesz space
L (Rn,IRm) b T
i s t h e Jordan decomposition of
T = T+-T-
IT1 = T++T-
negative p a r t s . Hence
i s o r d e r bounded. It
lRm
t h e decomposition
i n t o i t s p o s i t i v e and
(1
has t h e matrix
t.. I )
.
1J This matrix r e s u l t w i l l be considerably generalized i n Chapter 13 t o
, where
L
and
M
a r e Riesz spaces of
measurable f u n c t i o n s on p o i n t s e t s
Y
and
X
r e s p e c t i v e l y and where t h e
o p e r a t o r s from operator
for a l l
T
f
L
into
i s of t h e form
E L
. The
r o l e of t h e matrix the operator IT(x,y)
M
function (tij)
T(x,y)
. One
/TI = sup(T,-T)
, the
k e r n e l of
T
, takes
over t h e
of t h e r e s u l t s which we s h a l l prove i s t h a t i n t h e space
Lb(L,M)
has t h e k e r n e l
I (Theorem 9 4 . 3 ) .
We b r i e f l y consider the case t h a t
L
and
M
a r e normed Riesz spaces.
A more d e t a i l e d d i s c u s s i o n of normed Riesz spaces w i l l follows i n Chapter
14. We r e c a l l t h a t a normed Riesz space i s a Riesz space provided with a Riesz norm. The norm i n t h e Riesz space in
L
implies
llfll
_<
L
i s a Riesz norm i f L
L
into
Ig1
i s a Banach l a t t i c e and
i s a Dedekind complete normed Riesz space, then any o r d e r bounded
o p e r a t o r from
2
IlgIl. Any norm complete normed Riesz space i s c a l l e d
a Bmzach l a t t i c e . It w i l l be proved now t h a t i f
M
1f 1
M
i s norm bounded. F i r s t a simple lemma.
ORDER BOUNDED OPERATORS
110
LEMMA 83.11.
space
L
fn converges i n norm t o
If
fn z g f o r a l l
and i f
n
, then
Ch. 12,§831 f i n the normed Riesz
f t g
.
- -
.
PROOF. For the proof it may be assumed that g = 0 Since If -f I < If-fn 1 , the sequence (fn:n=1,2, ) converges in norm to f But f- = 0 for all n , s o f- = 0 In other words, f 2 0 n
.
...
.
.
THEOREM 83.12. Any order bounded operator
S
T from the Banach l a t t i c e
L
i n t o the Dedekind complete nomned Riesz space M i s norm bounded. In particular, any order bounded linear functional on the Banach l a t t i c e L i s norm bounded, i . e . ,
. Note
the or&r dual
L"
'
i s contained i n the Banach dual
i s not Dedekind complete, then any p o s i t i v e operator from the Banach l a t t i c e L i n t o M i s norm bounded. L*
that i f the nomed Riesz space
M
PROOF. We may assume that T is positive. If T
is not norm bounded,
(un:n=1,2, ...) in L such that Ilunl( = n-2 k for k = 1 , 2 ,... , and /ITunll t n for all n Writing sk = C u 1 n the sequence ( s :k=l,Z, ) is a Cauchy sequence in L , s o s k k 'Onverges in norm to an element s Since sk 2 un for k t n , it follows from the there exist positive elements
... .
last lemma that IITslI t IITunll bounded.
t
.
.
u for all n , so Ts E Tu But then n n n for all n , which is impossible. Hence, T
s
2
is norm
It will be proved later (cf. Exercise 83.29 and Theorem 85.6) any normed Riesz space L the Banach dual L* dual L"
. Hence
L* = L"
if L
that for
is contained in the order
is a Banach lattice.
We present a generalization of the familiar Hahn-Banach extension theorem. THEOREM 83.13.
vector space
V
Let
p be a sublinear mapping from the arbitrary real i n t o the Dedekind complete Riesz space M This means
.
for a12 f,g E v and p(af) = ap(f) real nwRbers a 2 0 . Furthemore, l e t t be an operator from subspace VI of V i n t o M such that t(f> s p(f) for a l l Then there e x i s t s an operator T from V i n t o M such t h a t for a l l f E YJ and Tf 2 p(f) f o r a l l f E V . that
p(f+g)
s p(f)+p(g)
for a l l
the linear f E Vl Tf = t(f)
.
Ch.
12,1831
ORDER BOUNDED OPERATORS
111
PROOF. Almost t h e same a s f o r t h e c l a s s i c a l Hahn-Banach theorem.
COROLLARY 83.14. Let
be an element i n t h e p o s i t i v e cone of the L and l e t vo be a p o s i t i v e element i n the
uo
Dedekind complete Riesz space pyincipal ideal T
operator
= 0
and Tf
in
A(u ) 0
L
for all f
. The
erists a positive Tug = vo
i n t o i t s e Z f ) such that
.
A(u )
complement of
0
0 i vo 5 aOuO f o r an a p p r o p r i a t e number
p ( f ) = a f + for 0 a l l f E L , i s s u b l i n e a r . On t h e l i n e a r subspace of a l l m u l t i p l e s of uQ we d e f i n e tfXu ) = X v Then t(huo) i 0 = p(Xuo) f o r X i 0 and 0 0 ' t(Xu ) = Avo 5 Xaouo = p(AuO) f o r X > 0 , s o t can be extended t o an 0
operator For
T
u E L+
mapping
L
i n the d i s j o i n t
PROOF. Note f i r s t t h a t
a. > 0
. Then there
uo
generated by
( i . e . , T maps
in
L
p from L
such t h a t
we have
T(-u)
i s p o s i t i v e . Replacing restriction of
A(uo)
.
TIA(uO)
Tug = vo
and
0
, so
p(-u)
S
=
Tf 5 p ( f )
Tu
for a l l
. This
0
5
f E L
shows t h a t
.
T
by t h e minimal p o s i t i v e extension of t h e
T
, we
i n t o i t s e l f , defined by
can make
T
vanish on t h e d i s j o i n t complement
i s a l i n e a r subspace Vl L (but not n e c e s s a r i l y a Riesz subspace of L ) such
I n t h e next theorem i t w i l l be assumed t h a t of t h e Riesz space t h a t f o r any
f E L
t h e r e e x i s t s an element
g E V,
satisfying
Note t h a t t h i s i s equivalent t o assuming t h a t f o r any gl
and
g2
in
Vl
satisfying
t h a t t h e i d e a l generated by
g,
5
f 5 g2
, and
t h a t t h e converse i m p l i c a t i o n i s f a l s e ; t a k e f o r continuous f u n c t i o n s on
f i g
L
. Note,
incidentally,
t h e space of a l l r e a l
L
vanishing a t t h e o r i g i n and f o r
[-1,11
.
there e x i s t
t h e condition implies
i s t h e whole space
Vl
f E L
the
V,
space of a l l odd f u n c t i o n s . THEOREM 83.15
(L.V.
Kartorovitch C21,1937/. Let
subspace of the Riesz space element
g E V1
L
such t h a t for any
satisfying f i g
, and l e t
t
from V I i n t o the Dedekind complete Riesz space p o s i t i v e means, therefore, that there e x i s t s a p o s i t i v e i.e.,
Tf = t ( f )
forcall
operator f E VI
V1
f E L
be a Zinear there e x i s t s an
be a positive operator M ( t o say that
t
is
f o r a l l f E L+ n v1 ). Then from L i n t o M uhich estends t
t(f) 2 0
.
T
,
112
ORDER BOUNDED OPERATORS
Ch. 12,9831
f E L+ be given. The set of all g E V 1 satisfying g 2 f
PROOF. Let
t(g) t 0 holds for each
is non-empty by hypothesis and
It follows then from the Dedekind completeness of M
g
in this set.
that
P(f
.
exists in M+ 0
f
5
Note already that p(f)
fg in L
I all of
5
p(f+g)
=
L
implies 0
by defining p(f)
pI(f+g)+}
< -
p(f ++g+ )
f,g E L
. Furthermore it
p(af)
ap(f)
M
=
. We 5
p(f)
for a 2 0
. Hence, p
t(f)
=
p(f)
f+
+
p(g+) (af')
,
so
.
t(f)
on V 1
. Then
f E L
p(f)
+
af+
f E L+
Il
p
V1
a t 0 that
on V1
i.e.,
it is useful to observe
, but
5 f+ 5
by taking the infimum for all
.
into
this is evident, since
f E V1
It follows then from f
to
for all
p(g)
for real
is a sublinear mapping from L
by
p(f+)
5
=
f+ has then a
g E V,
that
g E V 1 with
g t f+
, we
p(f)
theorem there exists therefore an operator T
is an extension of
T
is positive. For
Tf t 0
and
. This concludes the proof that t is dominated the sublinear mapping p . In view of the generalized Hahn-Banach
obtain
so
= =
is dominated by
. For
f E L+ n V 1
for
in M.. We extend p
is not necessarily a member of V 1
in V 1
g
t(g)
5
t(f)
p(f2)
in this case. For an arbitrary
first that majorant
t
= 5
for arbitrary
p(f+)
f E V1
for all
p(f+)
follows from
shall prove now that
t(f)
t(f)
=
p(f,)
2
t
and
f E L+
T
T
from L
is still dominated by
into M p
such that
. We show that
we have
. This concludes the proof.
COROLLARY 83.16. If K
i s a Riesz subspace of
L
such t h a t the i d e a l
i s the whole space L , then any p o s i t i v e operator from K i n t o the Dedekind complete Riesz space M has a p o s i t i v e l i n e a r extension
generated by to
L
K
. I n particular,
l i n e a r extension t o Riesz space and
L
L
any p o s i t i v e linear functionaZ on K has a p o s i t i v e
. This holds
i n particular i f
i s the Dedekind completion of
K
K
.
i s an Archimedean
We have derived the Kantorovitch extension theorem from the generalized Hahn-Banach theorem. It is possible to give
an independent proof of the
Ch. 12,5831
113
ORDER BOUNDED OPRATORS
Kantorovitch theorem which is adapted to the special situation and which of course makes essential use of Zorn's lemma (cf. Theorem X. 3 . 1 in the book by B.Z. Vulikh c 1 1 ) . We continue with still another extension theorem, based again on the Hahn-Banach theorem. We shall deal again with the extension of a positive operator
t
subspace K
from a Riesz
to the whole space L
present situation the ideal generated by
It is assumed now, however, that
K
, but
is not assumed to be
in the
L itself.
is dominated in a spezial manner by a
t
sublinear mapping. THEOREM 83.17.
Let
be a sublinear mapping from the Riesz space
p
L
i n t o the Dedekind complete space
M such t h a t p i s absolute andmonotone ( i . e . , p(f) = p(lf1) f o r a l l f E L and 0 2 f 5 g i n L implies p(f) S p(g) i n M 1. Furthermore, l e t t be a p o s i t i v e operator from the Riesz subspace K of L i n t o M such t h a t lt(f)l 5 p(f) holds f o r all f E K . Then there e x i s t s a p o s i t i v e operator T f r a L i n t o M such t h a t T extends t and 1Tfl S T(lf1) S p(f) holds f o r a l l f E L . PROOF. The mapping for all
f E L
,
pl from L
into M
is sublinear just as well as
, defined by p,(f) p , and we have
= p(ff)
for all f E K . Hence, by the Hahn-Banach theorem, there exists an operator T from L into M which extends t and such that Tf S p 1 (f) holds for all f E L . It follows that for f E L+ we have T(-f) so
Tf 2 0
S
, which
pl(-f)
=
p (-f)
shows that
T
+I
=
p(0)
= 0
,
is positive. Finally, using now that T is
positive, we get ITfl 5 T(lf1)
5
pl((fl)
=
p(lf/) = p(f)
for every
f E L
As an application of Corollary 83.16 we shall prove now that if a Riesz subspace of the Dedekind complete space M , then M
. L
is
contains the L , but not in a unique manner in general. This result is due to A.I. Veksler ([11,1969). Precisely stated, we have the
Dedekind completion L-
of
1 I4
ORDER BOUNDED OPEPATORS
Ch. 12,1831
following theorem. THEOREM 83.18. If L
space
i s a Riesz subspace o f the Dedekind compZete L1 of M such t h a t L 1 i s
, then there e x i s t s a Riesz subspace
M
Riesz isomorphic t o the Dedekind compZetion
LA of
t be the identity mapping on L
PROOF. Let operator T
,
into M
operator from L
from L-
into M
u- > 0
Indeed, for any satisfying 0 < u
which extends
0
observe first that T in L+
.
are of course incomparable. Now choose u
< u 5 u-
.
implies Tu- > 0 in M
,
are strictly positive disjoint elements in L^
and v-
and v-
such that
. We
i s a one-one mapping, we make the following
Before proving now that T remark. If u1
t
Tu- 2 Tu = u > 0 in M
so
i s a positive
t
there exists an element u
in L-
,
u-
2
. Then
by Corollary 83.16 there exists a positive
so
is strictly positive, i.e., u- > 0 in LA
then u-
under an isomorphism
L
L invariant.
t h a t leaves
and v
. Then, in general, u
v-
Z
be disjoint, but it is possible to choose u
and v
and v
and v
in L
will not
in such a manner that
these elements are incomparable. Indeed, if the choice is made completely
0< u < v
arbitrarily, it may happen that Archimedean, the inequality nu
v
2
Take the largest natural number n
,
satisfies v 1 2 v-
= inf(vl+nu,v-)
Hence, the elements u but
v1 2 u
Tu-
2
Tv-
and
, we
Tu- and
choose u
are incomparable. But
u
Tv-
and
Tu
5
We prove now that T
=
vI in L
does not hold, so
immediately that
does not hold for all n u
I v1
inf!v,\v-)
v
5
v-
2
Tu^-Tv^ # 0
Tu-
.
For the proof that
T
5
u-
and
v
Z
1
,
v-
are incomparable. It follows
as indicated, Tv-
5 Tv,
,
so
Tu
=
u
and
Tv
I
=
v
1
so we get a contradiction.
and
Tv-
.
f- < 0 implies
f- has positive and negative parts
respectively that are strictly positive. Then u=
.
= v-
is one-one, that is to say, Tf- # 0 for f^ # 0
. Hence, assume now that
positive and disjoint, so Tf-
... .
= 1,2,
we have
In the beginning it was shown already that f- > 0 or Tf- # 0
is
holds. Then v 1 = v-nu
satisfy 0 < u
and v,
L
are incomparable, because if for example
v,
T