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 Permanent Link:
 http://ufdc.ufl.edu/AA00022668/00001
Material Information
 Title:
 Generalizations of ideal theory
 Creator:
 Kenoyer, David B., 1953
 Publication Date:
 1982
 Language:
 English
 Physical Description:
 v, 73 leaves : ; 28 cm.
Subjects
 Subjects / Keywords:
 Algebra ( jstor )
Arithmetic ( jstor ) Conceptual lattices ( jstor ) Graduates ( jstor ) Integers ( jstor ) Lattice theory ( jstor ) Mathematical theorems ( jstor ) Modularity ( jstor ) Quotients ( jstor ) Ring theory ( jstor ) Dissertations, Academic  Mathematics  UF Ideals (Algebra) ( lcsh ) Lattice ordered groups ( lcsh ) Lattice ordered rings ( lcsh ) Mathematics thesis Ph. D Rings (Algebra) ( lcsh )
 Genre:
 bibliography ( marcgt )
nonfiction ( marcgt )
Notes
 Thesis:
 Thesis (Ph. D.)University of Florida, 1982.
 Bibliography:
 Bibliography: leaf 72.
 General Note:
 Typescript.
 General Note:
 Vita.
 Statement of Responsibility:
 by David B. Kenoyer.
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 University of Florida
 Holding Location:
 University of Florida
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 Copyright [name of dissertation author]. Permission granted to the University of Florida to digitize, archive and distribute this item for nonprofit research and educational purposes. Any reuse of this item in excess of fair use or other copyright exemptions requires permission of the copyright holder.
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Full Text 
GENERALIZATIONS OF IDEAL THEORY
BY
DAVID B. KENOYER
A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL
OF THE UNIVERSITY OF FLORIDA IN PARTIAL
FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
1982
ACKNOWLEDGEMENTS
The author would like to thank his Supervisory
Committee Chairman Jorge Martinez in particular, and all
of the members of his supervisory committee for their
helpful comments, criticisms, and suggestions. He also
would like to acknowledge the cooperation of his typist,
Sharon Bullivant, who did such quality work in such a short
time.
TABLE OF CONTENTS
Page
ACKNOWLEDGEMENTS ...................................... ii
ABSTRACT .............................................. iv
INTRODUCTION.......................................... 1
CHAPTER
I. PRELIMINARIES .................................. 4
II. COMMUTATIVE MULTIPLICATIVE IDEAL LATTICES ...... 1
III. MULTIPLICATIVE SUBSETS, LATTICES OF EXTENSIONS,
AND LOCALIZATIONS .............................. 22
IV. INVERTIBLE ELEMENTS ............................ 39
V. GENERALIZATIONS OF VALUATION RINGS AND PRUFER
DOMAINS ........................................ 48
VI. GENERALIZATION OF DEDEKIND DOMAINS ............. 60
VII. MODULARITY..................................... 70
BIBLIOGRAPHY ............................................ 72
BIOGRAPHICAL SKETCH ................................... 73
iii
Abstract of Dissertation Presented to the Graduate Council
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy
GENERALIZATIONS OF IDEAL THEORY
By
David B. Kenoyer
August 1982
Chairman: Dr. Jorge Martinez
Major Department: Mathematics
The lattice of ideals of a commutative ring and the
lattice of convex subgroups of a latticeordered group
fit into the more general setting of a milattice, which is
an algebraic lattice with an associative multiplication that
distributes over joins, respects compactness, and satisfies
the ideal multiplication property that ab s a A b. In this
context, the terms primary element, residual quotient,
multiplicative element, cancellative element, and invertible
element are defined, and generalizations of classical
results of commutative ring theory are obtained. A primary
decomposition theorem for modular milattices with ACC and
enough multiplicative elements is given. Multiplicative
subsets, rings of quotients, and extensions and contractions
of ideals are generalized, as are commutative rings with
identity, integral domains, and valuation rings. Several
characterizations of the analogue of PrUfer domains are
given in terms of localizations and arithmetic relation
ships. Dedekind domains are generalized and characterized.
INTRODUCTION
This dissertation is an attempt to place the results
of multiplicative ideal theory in a latticetheoretic
setting which will include .other lattices of subobjects
as well. The two motivating examples we will keep in mind
are the lattice of ideals of a commutative ring and, to a
lesser extent, the lattice of convex subgroups of a
latticeordered group.
We consider an algebraic lattice L endowed with a
multiplication which distributes over all joins, respects
compactness, and satisfies the ideal multiplication property
that ab a A b. A major work in this area is that of
Keimel [5]. He indicates that this approach applies to a
wide class of known examples, including the lattices of
ideals of commutative semigroups, distributive lattices,
and frings. He has developed the theory of minimal primes
to a great extent, and so in Chapter I, after we introduce
the basic concepts, we list his major results, and then turn
to other classical results of commutative ring theory. Our
terminology differs slightly from that of Keimel, and
beginning in Chapter II, we assume that our multiplication
is associative and commutative, as it is in our two models.
Modularity is only assumed for one result, and we do not
know that it is necessary there.
1
2
Chapter II considers primary elements and primary
decompositions, ending with Theorem 2.14, the analogue of
the Primary Decomposition Theorem for commutative Noetherian
rings.
Chapter III is an investigation of multiplicative sets
and our version of rings of quotients. We consider the
"extension" a of an element a in our lattice relative to a
multiplicative subset M of compact elements, and show that,
in passing from the lattice L to the lattice of extensions,
LM, we preserve compactness, primes, primary elements,
radicals, and almost all finite arithmetic operations,
including all operations with compact elements. Theorem
3.22 establishes the fundamental relationship between an
element a and the extensions of a relative to the maximals
exceeding a.
In Chapter IV, we discuss invertibility, including
various arithmetic facts involving invertible elements and
their existence in L. We conclude Chapter IV with a dis
cussion of various interpretations of the concept of an
integral domain. Chapter V looks at results from the theory
of Prefer domains, including the characterizations in terms
of arithmetic relations, which comprise Theorem 5.10. We
also consider our version of a valuation ring, and its
relation to Priifer domains. Chapter VI deals with our
generalization of Dedekind domains, with the main result
being Theorem 6.3. Finally, Chapter VII briefly looks at
the question of modularity, giving a result which leads to
certain conditions that guarantee modularity.
As general references for the multiplicative ideal
theory of commutative rings, we suggest Gilmer [3] or
Hungerford [4]. For the theory of lattices, we refer to
Birkhoff [2], while for the theory of latticeordered
groups, we use Bigard et al. [11, which is written in
French.
Concerning notation, we will use Z to stand for the
set of natural numbers, and R to stand for the set of real
numbers.
CHAPTER I
PRELIMINARIES
Let L be a complete lattice with minimum element 0 and
maximum element 1 ; 0. An element c of L is said to be
n
compact if whenever c < v a. it must be that c < v a. .
iEI k=l 1k
0 is automatically compact, and the join of any finite set
of compact elements is again compact.
A lattice L is algebraic if it is complete and every
element of L is the join of a (possibly infinite) set of
compact elements. In an algebraic lattice, an element c is
compact if and only if whenever c = v a. it must be that
n iEIl
c v a.
k=l k
1.1. Definition: A lattice L is a multiplicative
ideal lattice, or milattice, if L is an algebraic lattice
endowed with an associative binary operation (a,b) ab
that satisfies the following conditions:
(1) ab s a A b for all a,b E L.
(2) a( v bi) = v (ab.) and ( v bi)a = v (bia) for
iEI iEI  i ieI
all a E L and for all families {bili E I} in L.
(3) If x and y are compact in L, then xy is compact
in L.
1.2. Example: Let L be the lattice of ideals of a
ring R. L is a milattice if we take the meet to be inter
section, the join to be the sum, and multiplication to be
the ordinary product of ideals. Here, the compact elements
are the finitely generated ideals.
1.3. Example: A latticeordered group is a group G
endowed with a lattice order which is compatible with the
group operation in the sense that a(b v c) = ab v ac,
(b v c)a = ba v ca, and the dual conditions involving meets
also hold. An ksubgroup of G is a subgroup H which is
also a sublattice; i.e., if a and b are in H, then a v b and
a A b are also in H. An subgroup C of G is convex if
whenever a and b are in C and x E G satisfies a x b,
then x E C. A theorem of Bigard et al. [i], states that
the lattice C(G) of convex Zsubgroups is completely dis
tributive, if we let the join of two convex Zsubgroups be
the convex ksubgroup generated by their union. Thus, if
we let the product and meet coincide, C(G) is a milattice.
Again, the compact elements are the finitely generated ones,
and in this case, these are generated by a single element.
1.4. Definition: A lattice is said to have the
ascending chain condition, or ACC, if every ascending chain
breaks off after a finite number of steps; i.e., if
x 1 x2 : x3 : ..., then there is a positive integer n with
xk = xn for all k n. A result from lattice theory
(Birkhoff [21) is that the following conditions are equiva
lent for an algebraic lattice L:
(a) L has ACC.
(b) Each element of L is compact.
(c) Each nonempty subset of L has a maximal element.
In both Examples 1.2 and 1.3, there is a notion of a
prime element. In Example 1.2, recall that a prime ideal
is a proper ideal P of a ring R satisfying the condition that
if A and B are ideals and AB c P, then A c P or B c P. In
Example 1.3, a prime convex subgroup is a proper convex
isubgroup P such that if A and B are convex ksubgroups of
G and if A n B S P, then A c P or B c P. These two notions
of primes motivate the following definition.
1.5. Definition: If L is a milattice and p E L, then
p is prime if p < 1 and whenever ab p, we must have a p
or b p. Notice that if we want to show that an element is
prime, it is enough to show that if a and b are compact and
ab p, then a 5 p or b p. Conversely, if p is prime,
then certainly the definition of a prime applies when a and
b are compact elements, so that we may assume a and b are
compact in the definition.
An element m of a lattice L with 1 is a maximal if it
is maximal in the set of elements strictly below 1; i.e., if
m a < 1, then m = a. Recall that in a commutative ring
with identity, every proper ideal is contained in a maximal
ideal, and each maximal ideal is prime. In our context, we
have the next three results.
1.6. Proposition: In a lattice L with 1, if 1 is com
pact, then each element a < 1 lies below a maximal of L.
Proof: Let a < 1, and let A = {bfa b < 1}. A ,
since a E A. A is partially ordered as a subset of L. If
C = {cili E I} is a nonempty totally ordered subset of A,
then let c = v c. We have a c, so suppose c 4 A. Then
iEI
n
c = 1, so 1 = v c.. Since 1 is compact, 1 = v c. with
iEI i k=l 'k
ik E I for 1 k s n. But C is totally ordered, so
1 = c0 E C, which is impossible. Thus c E A, so by Zorn's
Lemma, there is an element m which is maximal in A. But if
m : b < 1, then b a, so b e A and b = m. Hence m is a
maximal of L.D
1.7. Example: The converse to Proposition 1.6 is, in
general, false. The following example and ideas are found
in Bigard et al. [1]. Let X be a locally compact space
which is not compact, and let L(X) = {f: X Rjf is locally
constant with compact support}. Then the only prime convex
Zsubgroups of L(X) are those of the form C = {f E GIf(x) =0}
for some x E X. Each of these is maximal, and each convex
Zsubgroup of L(X) is contained in one of these. Also, each
f E L(X) must be in Cx for some x E X, since X is not compact,
and so L(X) cannot be generated by a single element; i.e.,
in C(L(X)), 1 is not compact. A latticeordered group G in
which all primes are maximal is called hyperarchimedean,
and an element a e which generates a latticeordered group
8
G is called a strong order unit of G. Any hyperarchimedean
latticeordered group without a strong order unit will be
a counterexample.
2
1.8. Proposition: In a milattice L, if 1 = 1, then
each maximal is prime. If each a < 1 lies below a maximal,
2
and if each maximal is prime, then 1 = 1.
2
Proof: If 1 = 1, let m be a maximal of L. Since
m < 1, we suppose x and y are in L with xy m, but x m.
Then l(m v y) = (m v x)(m v y) = m2 v xm v my v xy 5 m < 1,
so m v y < 1. But m v y m, so m v y = m and y < m. Thus
m is prime. On the other hand, if each a < 1 lies below a
maximal, and if each maximal is prime, then 12 cannot lie
2
below any maximal, so 1 = l.D
1.9. Definition: The radical of an element a, denoted
rad(a), is defined as rad(a) = A{plp is prime and a 5 p}.
By convention, we set Ap = 1 and v p = 0. A result of
PE petp
Keimel [5] is:
1.10. Lemma: If M is a nonempty set of nonzero compact
elements of a milattice L, and if M is closed under multi
plication, and if a E L with m $ a for each m E M, then
Sa = {blb > a and m $ b for each m E M} must have maximal
elements, and each of these is prime in L.
1.11. Proposition: If L is a milattice and a E L,
then rad(a) = v{xfx is compact and xn < a for some positive
integer n}.
Proof: We show that {xlx is compact and x s rad(a)} =
{xjx is compact and for some positive integer n, xn 5 a}.
First, if x is compact and xn s a, then x S p for each prime
p a, so {x compactlxn S a for some n} S {x compactjx
rad(a)}. But if x is compact and if, for each n E Z ,
X n$ a, then {xnin 11} is a multiplicatively closed set of
nonzero compact elements, and so if Sa = {b E Lib a .and
na
x n b for all n E Z }, then S has a maximal element p which
a
is prime, by Lemma 1.10. Hence p0 a and x t p0, so
{x compactjx s rad(a)} = {x compact for some n > 1, xn s a}.D
Keimel [5] defines a msemiprime element of a milattice
L to be an element x such that for all t E L, t $ x implies
2
t 2 x. He shows that each msemiprime element in a mi
lattice is the meet of the set of prime elements exceeding
it (Theorem A), and so these are the elements that occur as
radicals in milattices. As a corollary, he shows in a
milattice L that the following conditions are equivalent:
(1) Each element of L is the meet of primes, so that
rad(a) = a for all a E L.
(2) Each a E L is semiprime.
2
(3) For each a E L, a = a.
(4) For all a,b E L, ab = a A b and L is distributive.
10
(5) For all a,b E L, ab = a A b and L is completely
distributive.
(6) For all a,b E L, ab = a A b and L is Brouwerian.
In ring theory, these conditions on ideals of a commutative
ring R are equivalent to R being von Neumann regular, by
a result of Kaplansky. Keimel then looks at semiprime mi
lattices, which are milattices in which 0 is semiprime.
He defines the pseudocomplement of an element, and mpre
filters and mfilters, and uses these and the GratzerSchmidt
isomorphisms between elements of L and ideals of the set of
compact elements of L to set up a bijection between maximal
mfilters and minimal primes of a milattice L which is
semiprime and satisfies the property that for all r, s, and
t E L, r A s = r A t = 0 implies (r v s) A t = 0. This gives
him several characterizations of a semiprime pseudocomple
mented lattice. He goes on to look at the set of minimal
primes and develops the hullkernel topology, showing that
it gives a completely regular space, with a basis of closed
andopen sets determined by the compact elements of L. He
finishes with a discussion of zelements, which area gener
alization of zsubgroups of a latticeordered group, and uses
them to describe certain classes of milattices, including
those in which the space of minimal primes is compact. Keimel
actually does all this in a more general setting, frequently
not requiring that the product of compact elements be compact,
or that the product distribute over joins, or that it be
associative.
CHAPTER II
COMMUTATIVE MULTIPLICATIVE IDEAL LATTICES
For the remainder of this dissertation, L will always
represent a milattice with a commutative multiplication.
We will consider the analogues of some classical results of
multiplicative ideal theory.
2.1. Definition: The residual quotient of a by b, de
noted by [a:b], is defined by: [a:b] = v{clbc < a}. Note
that if x is compact and x [a:b], then bx a, so
[a:b] = v{x compactlbx a}.
2.2. Proposition: The following properties of residual
quotients hold in L:
(a) [a:b] z a, [ab:b] a, and b[a:b] < a A b, for all
a,b E L.
(b) If b a, then [a:b] = 1.
(c) [[a:b]:c] = [a:bc] for all a,b,c e L.
(d) [a:b] = [a:a v b] = [a A b:b], for all a,b E L.
(e) [a: v b.i] = A [a:b.], for all a E L and all families
iEI ieI 1
(bi)E L.
(f) [( A ai):b] = A [ai:b], for all b E L and all
iEI iEI
families (a.). a L.
(g) [ab:c] a[b:c], for all a,b,c E L.
(h) [a: A b ] v [a:b.] for all a E L and families
ieI iE I
(bi)i eI c L.
(i) [( v a.):b] v [a.i:b] for all b E L and families
iEI iEI
ie l 1 e L.
(a) el C L.
Proof: (a) ab < a, so a < [a:b]; ab < ab, so
a < [ab:b]; b[a:b] = b(v{clbc < a}) = v{bclbc < a} < a, and
b[a:b] < b.
(b) First bl < b, so if b < a, b1 : a, and [a:b] = 1.
(c) [[a:b]:cl = v{dlcd < [a:b]} = v{dlbcd < a} = [a:bc].
(d) We have (a v b)c < a 4> ac v bc a 4 bc < a <=>
bc a A b.
(e) [a: v b.]( v bi) s a, so for each j E I,
iI ilEI
[a: v b.]b. a, and [a: v b. i < A [a:b.]. But
il J el 1 il
i iEI i iEI
A [a:bi ]) ( v b.) = v (( A [a:b ])bj) < v ([a:bib. ) a,
iEl 1 jEI jEl iel jEl I
and so A [a:b.] = [a: v bi.].
iel iel 1
(f) First, b[( A a) :b3 < A a. : a. for each j E I,
iEI iEl 1 3
so [( A a.):b] < A [a.:b]. But b( A [a.:b]) b[a.:b] < a.
iel =el iel 1 J
for each j E I, so [( A a.):b] = A [a.:b].
iel 1 ie I
(g) This is true since a[b:c] = a(v{dlcd < b}) =
v{adlcd < b} < v{adlcad s ab} [ab:c].
(h) ( A b ) ( v [a:b.]) = v (( A b.)[a:b.])
iel jEI jeI iEI
v (b.[a:b.i1) a.
jEI J J
(i) Since b( v [a.i:bl) = v (b[a.i:b]) v a., we have
iEI iEI iEI
1e el xilET
v [a :b] : [ai:b] < [( v a.):b].D
iel11il 1
EI iEI
2.3. Definition: An element a E L is multiplicative if
a[b:a] = b for each b < a; i.e., if a divides every element
below it. Note that if a is multiplicative, then al = a.
An element a E L is said to be powermultiplicative
n +
if a is multiplicative for each n E Z
Considering our two examples of milattices, if R is a
commutative ring, all principal ideals are multiplicative,
and in fact powermultiplicative. If G is a latticeordered
group, then since multiplication is the meet operation,
every convex ksubgroup is multiplicative, and hence power
multiplicative. Thus, in the two underlying models, each
element of the lattice is the join of a set of compact,
powermultiplicative elements. For this reason, we will
assume this property to obtain one nice result later in
this section, Theorem 2.14, without losing touch with our
motivations.
The first question that arises is: When is the product
of two multiplicative elements again a multiplicative element?
The answer does not seem to be that it is always true, but a
partial answer is offered.
2.4. Lemma: If a and b are multiplicative and if
b = [ab:a], then ab is multiplicative.
Proof: Let c ab. Then c < a, so a[c:a] = c. Thus
a[c:a] ab, so since b = [ab:a], [c:a] < b. Hence
[c:a] = bd, where d = [[c:al:b] = [c:ab], and
ab[c:ab] = abd = a[c:a] = c.0
We now press on toward a primary decomposition result.
Recall that in a commutative ring R, an ideal Q c R is
primary if whenever xy E Q with x and y elements of R, then
x E Q or yn E Q for some positive integer n. Since a compact
ideal is finitely generated, we offer the following.
2.5. Definition: An element q in a milattice is primary
if q < 1 and whenever x and y are compact with xy 5 q, then
either x 5 q or yn s q for some n E Z In the example of
n
a latticeordered group, C= C for each n, so primary is
equivalent to prime.
2.6. Lemma: If q is primary and p = rad(q), then p is
prime.
Proof: Let x and y be compact, with xy 5 p and y 4 p.
Since xy is compact, by our proof of Proposition 1.11 we
have (xy)n <_ q for some n E Z and we assume n is the
smallest positive integer with this property. If n = 1,
15
k +
xy q, so x < q < p or y _< q for some k E Z But y $ p,
k +
so y k q for any k E Z so x < p, and we are done. If
n > 2, then (xy) n = (xy)n xy < q, so again since yk q
for any k E Z (xy) n x < q. But by choice of n,
(xy) nI q, so xm < q for some m E Z and x < rad(q) = p.D
This justifies the following terminology: If q E L is
primary and p is prime with p = rad(q), then q is pprimary,
and p is the prime associated with q.
Note: If q is pprimary, and x and y are compact, with
xy < q, then x < q or y s p; i.e., one of x and y is below q,
or they both lie below p, since L is commutative.
2.7. Lemma: If q is pprimary in L, and a $ p, then
[q:a] = q.
Proof: As always, q < [q:a]. If x is compact with
ax q, then since a $ p, there is y compact with y < a
and y $ p, and so xy < q. Now, y $ p, so x < q, and so
q > v{x compactlax < q} = [q:a].D
2.8. Lemma. If p,q E L satisfy the conditions:
(1) p > q
(2) If x is compact and x < p, then xn < q for some
n E Z+
(3) If x and y are compact and xy 5 q, then x q or
y < p
then p is prime and q is pprimary.
16
Proof: If x and y are compact and xy q, then by (3),
x : q or y 5 p, so by (2), x q or yn q for some n E Z+J,
and q is primary. Next, by (2), p rad(q), by Proposition
1.11. Now, if x is compact and xm q for some m E ,
assume m is the least positive integer with this property.
If m = 1, we are done by (1). If m 2, then xm = xmlx q,
but xm1 is compact and xm1 $ q, so x p by (3). Thus
p = rad(q) and so p is prime.
2.9. Proposition: If p is prime and ql,q2,... ,qn are
n
all pprimary, then A q. is pprimary.
i=l 1
n
Proof: Let q = A q.. First, p > ql, so p a q. If x
i=l
is compact and x p, then for each integer i with 1 < i n,
m.
there is a positive integer m. with x < q., so if
m = max{mil
with xy s q and y $ p. Then xy s qi for each i, and so
x < qi for each i, 1 i s n. Hence x 5 q, and q is
pprimary.0
2.10. Proposition: If q E L is pprimary and a t q,
then [q:a] is pprimary.
Proof: Let x0 be compact, with x0 < [q:a]. Since
a $ q, there is a y0 compact with y0 a and y0 t q. Now,
Y0X0 < a[q:a] q, so since y0 $ q, x0 s p. Thus
[q:a] 5 p. If x is compact and x < p, then xn < q < [q:a]
for some n E Z. If x and y are compact and xy 5 [q:a],
with y $ p, then for each z compact with z a,
(zx)y axy < q, so zx s q. Thus ax : q, and x 5 [q:a].
By Lemma 2.8, [q:a] is pprimary.D
2.11. Proposition: If {p.illi < n} is a finite set of
primes in L, and if for each integer i between 1 and n, qi
n
is a p.primary, and if a = A q, then each prime p a
1 i=l
n
exceeds some pi (1 : i 5 n), and so rad(a) = A pi.
i=l
n n
Proof: First, II qi A q. 5 p, so for some i with
i=l i=l
1 < i 5 n, qi 5 p, since p is prime. Hence pi = A{primes
exceeding q.} p.D
We will say that an element a has a primary decomposition
n
if a = A q., where each q. is primary. For each i between
i=l
1 and n, let pi = rad(qi). If a has a primary decomposition
n
a = A q., and if for each i and j with 1 i n and
i=l
1 j < n, i j implies pi p., and if for each i with
n
1 i n, q. i A qk' then we call a = A q. a normal
1 ki i=l
primary decomposition of a. A refinement of a primary
n
decomposition a = A q. is a primary decomposition
i=l
n
a = A r., where for each j with 1 5 j s m, there is an
j=l 3
integer ij, 1 < i. n, with r. j q. In light of
3
J J J 1 .
n
Proposition 2.9, if a = A q. is any primary decomposition
i=l 1
of a, we can group together those qi's that have the same
radical, and their meet is again a primary with the same
m
radical, so we obtain a = A r., where if j, j,2' then
j=l 1 2
rad(r.j ) ;rad(r.j2 ), and each r. q. for some i. Now, if
for some J0, 1 < J0 < m, r. 2 Aj r., then we can exclude
30 j ;C
r. from the decomposition, and so we see that every primary
3O
decomposition has a normal refinement.
m n
2.12. Theorem: If a = A q. and a = A r. are two normal
i=l j=l 3
primary decompositions for a E L, where each qi is p.primary
and each r. is tjprimary, then m = n and, after suitable
rearrangement, pi = t. for each i (1 i 5 m).
Proof: If a = 1, then m = n = 0, and we are done. If
a < 1, then from {pl,...,p } u {tl,...,tn}, we choose an
element not strictly below any other. We may assume without
loss of generality that after rearranging terms of the
decompositions, we have chosen p m Suppose p {tl' f t }t .
Then qm $ t. for any 1 j n. By Lemma 2.7, [r. :qm] = r.
n n
for each 1 5 j : n, so [a:qm] = A [r.:q ] = A r. = a.
=j=l j= 3
Also, pm $ pi if 1 < i m1, so qm $ pi for each 1 < i m1,
and Eqi:qm3 = qi for 1 i s ml. But [qm:qm] = 1, so
m mi mi
a = [a:qm] = A [qi:qm] = A qi, and qm A qi, contradicting
i=i i=l i=l
m
the assumption that a = A q. is a normal decomposition (if
i=l
m = 1, then a = [a:q ] = 1, a contradiction). Thus
Pm e {t ,...,t n}, and we may assume that with a suitable
rearrangement, Pm = tn. Let q = qm A r By Proposition
2.9, q is p primary, and [q.:q] = q. if i s m1, and
mi ni
[r.:q] = r. if j n1. Thus [a:q] = A qi = A rj, and both
J J i=l j=i
are normal decompositions. Hence we may continue the process,
and we need only show that m = n. Suppose m < n. After
nm nm
m steps, Pm = tn'...,p = tn l and 1 = ^ r. A t. < 1
tn 1 nrn+i'j=1 j=1 I
since t, prime. This is impossible, and so m = n.0
We now proceed to the primary decomposition. As stated,
we will assume in Theorem 2.14 that each element of L is the
join of a set of compact, powermultiplicative elements, and
we will also assume that L is a modular lattice; i.e., for all
a,b,c E L, if a b, then a A (b v c) = b v (a A c). Again,
the motivating examples satisfy this condition, although we
only use it to show the existence of a decomposition, and
we are not certain that it is needed there.
We say that an element a E L is Aprime (or finitemeet
irreducible, or irreducible) if a = b A c implies that
a = b or a = c.
2.13. Lemma: If L is a milattice with ACC, then each
n
a E L can be written as a = A a., where each a. is Aprime.
i=l 1
Proof: Let S = (a E LI a has no decomposition as
n
a = A a., with each a. an Aprime}. The lemma states that
i=l 1
S = P; we suppose not. By the ACC, there is an element a
which is maximal with respect to S. But then a = b A c,
with a < b and a < c, and so b J S and c J S. Thus
m n
a = ( A b.) A ( A c.), where each b. and each c. is Aprime.
i=l j=l 3 1
This is impossible, and so S = .D
2.14. Theorem: Let L be a modular, commutative mi
lattice in which each element of L is the join of a set of
powermultiplicative elements. If L has ACC, then each
element of L has a normal primary decomposition, which is
unique up to order and the set of radicals of the primaries.
Proof: In light of Lemma 2.13, Theorem 2.12, and the
fact that every primary decomposition has a normal refinement,
we need only show that if a is Aprime, then a is primary.
Suppose a E L is not primary. Then there are compact elements
n +
b and c with bc : a but c $ a and b a for each n E Z.
We may assume that b is powermultiplicative, for b = v bi,
idI
where each bi is compact and powermultiplicative, and since
m n +
b is compact, b = v b Since b $ a for each n E Z
k=l k
there must be one of these b.k 's, which we denote b, such
n +
that b0 $ a for each n E Z Then b0c s bc < a, and we may
replace b with b0. Now, bc a, so [a:b] a v c > a. Also,
[a:bn] [[a:bn]:b] = [a:b n+1], so we have the ascending
chain a < [a:b] [a:b2] < .... This chain must break off
by the ACC; i.e., there is a n E Z+ with [a:b] I = [a:b] I
for each k > n. Now, [a:bn I > a, and also since bn $ a,
n n n
we have a < a v bn. Thus a < [a:b ] A (a v bn). Let
d = [a:bn] A (a v bn). Then d < [a:bn], so dbn < a. Now,
bn is multiplicative, so if r = [(d A bn ):b n], then
n n
b r = d A b. Thus, since d > a, we have
a v b r = a v (d A b ) = d A (a v b ) = d, by modularity.
Hence a bn d = bn a v b2n r, so b2nr a, so
r : [a:b2n] = Ca:bn], so bnr < a, and d = a v bnr = a. Thus
a is not Aprime, and so every Aprime element is primary.0
CHAPTER III
MULTIPLICATIVE SUBSETS, LATTICES OF
EXTENSIONS AND LOCALIZATIONS
We now turn to the idea of localizations, which play an
important role in valuation rings and Priifer domains, as well
as in algebraic geometry. A localization is a special type
of the more general object known as a ring of quotients. The
lattice of extended ideals in a ring of quotients preserves
most of the essential information below a given set of primes
in the lattice of ideals of the original ring, including
prime and primary elements, joins, meets, products, and
compactness. In a commutative ring with identity, each ideal
A is the intersection of the set {AIX E A}, where A. is the
contraction of the extension of A relative to the maximal
ideal MX (A is the indexing set for the set of maximal
ideals), and it is this fact that makes much of the ideal
theory of valuation rings and Priifer domains work.
3.1. Definition: A nonempty subset M of a milattice
L is called a multiplicative subset of L if each m E M is
compact, 0 4 M, and whenever m1 and m2 E M, we must have
m mm2 E M.
One way of constructing a multiplicative subset of a
milattice L is to take {p,X E A}, a nonempty set of
pairwise incomparable primes, and take a subset M0 of
compact elements with the properties:
(1) For each X E A and m E M0, m $ pX.
(2) If q is a prime and, for each X E A, q $ p., then
there is a m E M0 with m 5 q.
n
Set M = { m I n 1 and for each i, 1 i n, mi E M0}.
i=l
M0 S M, so M (. Each element of M is compact, since each
element of M0 is. Since {pi E Al 4, let p be an element.
Since 0 p, no finite products of elements of M0 can be 0,
for no m E M0 satisfies m : p, and p is prime. Thus 0 4 M,
and clearly any finite product of elements of M is again in
M. The fact that every multiplicative subset of L arises in
this fashion is the content of Lemma 1.10. Recall the nota
tion that if a E L and for each m E M, m 4 a, then
Sa = {b E Lia b and for each m E M, m 4 b}. Notice that
if a1 a2, then S a Sa so that each Sa is contained in
j. &a1 a2 a
S0. We let {p,  E A} be the set of elements maximal in S0.
This is a nonempty pairwise incomparable set of primes, and
if b E L such that for each A E A we have b $ p., then there
must be a m E M with m b. We say that {piX E A} is the
set of primes associated with M (or the set of associated
primes of M).
If M is a multiplicative subset of L, and {pIX E A}
is the set of associated primes of M, then let M = {m E Lf
m is compact and for each X e A, m t p,}. M c M and
0 4 Ms, so since Ms is closed under multiplication, it is a
multiplicative subset of L containing M, with the same set
of associated primes. M is called the saturation of M,
S5i"""" '" 
and we say that a multiplicative subset M of L is saturated
if M = M s. Clearly, Ms is saturated, and it contains every
multiplicative subset of L which has {pIX E A} as its set
of associated primes.
In commutative ring theory, if R is a commutative ring
and M c R is a multiplicative set not containing zero, then
when the ring of quotients is formed, all ideals of RM are
extended ideals; i.e., they are all ideals generated by the
subset of RM corresponding to an ideal of R under the
canonical homomorphism D of R into RM. Also, each ideal of
RM meets the image of R in an ideal of that subring, and the
inverse image of these intersections are called contracted
ideals. This is all put forth in Gilmer [3], and it is
shown that all extended ideals are of the form
Be = {$(b)/U(m) I b E B and m E M}
while all contracted ideals are of the form
Bec = {x E Rxm e B for some m E M}.
With this in mind as a motivation, we proceed to our next
definition.
3.2. Definition: Let M be a multiplicative subset of L,
and let {p,[X E Al be its set of associated primes. If
a E L, then set
*
a (M)= v{x compact xm < a for some m E M}.
*
Notice that a (M) > a.
3.3. Lemma: If a E L and x is compact, then
*
x < a (M) xm < a for some m E M.
Proof: Certainly if xm < a for some m E M, and x is
*
compact, then x < a (M). Conversely, suppose x is compact and
*
x < a (M). Then x < v{y compact ym < a for some m E M}, so
n
since x is compact, x v Yk' where for each k < n, Yk is
k=l
n
compact and there is a mk E M with ykmk < a. If m = II m,
k=l
n
then m E M and m < mk for each k < n, so xm < v Ykmk < a.0
k=l
*
3.4. Lemma: a (M) = a (M ).
*
Proof: Since M c Ms, we have a (M) < a (M ). But if x
is compact and x 5 a (Ms), then there is a m E Ms with xm a.
Now, for each X E A, m $ p,, so by the discussion following
Definition 3.1, there must be a m0 E M with m0 < m. Hence
*
xm0 < xm < a, so x < a (M), and a (M) = a (Ms).D
When it is clear from the context just which multipli
cative set we are referring to, we will write a* rather
*
than a (M). Further, we may always assume that we are using
a saturated multiplicative subset. This has no effect upon
our results, but it is convenient, and by Lemma 3.4, it is
justified.
3.5. Definition: If M is a multiplicative subset of L,
we define an equivalence relation on L by:
a ~ b 4> a* = b*.
(If it is necessary to refer to M, write a M b.) Let a
represent the equivalence class of a under ~((a)M, if
necessary). We will say a is the extension of a, and a* is
the contraction of a.
A /
3.6. Proposition: a* = v{blb E a}, a* E a, and
(a*)* = a*.
Proof: Since b < b* for all b e L, we have
a* v{blb e a}, and a* 5 (a*)*. Let x be compact with
x < (a*)*. Then there is an m1 e M with m1x < a*. But m1 is
compact, so m1x is compact, and there must be a m2 e M with
m2mlx 5 a. m1m2 E M, so x < a* and (a*)* = a*. Thus
a* ~ a, so a* e a, and a* = v{blb e a}.D
3.7. Definition: We now want to create a new milattice
with these equivalence classes. To this end, we set
a 5 b 4> a* < b*.
27
A
This is a partial order on {aj a E L}. We need it to be a
lattice order which gives us an algebraic lattice.
3.8. Lemma: Let a,b E L. The following are equivalent:
(1) a < b.
(2) There is an a0 E a with a0 < b.
(3) There is a b0 E b with a < b0.
A% A A%
Proof: (1) => (3): If a < b, then a < a* < b* E b.
(3) => (): Suppose b0 E b, with a < b0. Let x be
compact, x 5 a*. Then there is a m E M with mx < a, so
mx 5 b0, and x < b* = b*, so a* < b*, and a < b.
(1) => (2): If a < b, then a* < b*, so for each x com
pact with x < a*, we choose mx E M with m x < b. Let
a0 = v{m xjxi compact, x < a*}. Since a0 < b, we only need
0 E: a; i.e., a* = a*. First, a* < a* by choice of a0. Let
y be compact, y _< a*. Then there is m E M with
my a0 = v{m xx compact, x < a*}. Now, my is compact, so
n
my < v mkxk, where for each k < n, xk is compact, xk < a*,
k=l
and mk = mk as chosen above. But since xk < a*, mkxk < a*
Xk
with mkxk compact, so we can find pk E M with Pkmkxk < a.
n n n
Let m0= m HI k. We have m0 E M, and m0y < ( H "k) ( v mkX) <
k=l k=l k=l k k
n
v (kmkXk) < a, and so y < a*. Thus a* = a*, so a0 E a.
k=1
A
(2) => (1): Suppose a0 e a with a0 < b. If x is compact
and x < a*, then there is a m E M with mx a0 < b, so x < b*.
A A
Thus a* < b*, and a < b.
3.9. Proposition: If {a.ili E I} is any family in L,
A
let a = v a.. Then a = v a., and so ( v a.)* v (at).
iEl iel 1 iEI iEI
Proof: First, since a a. for each i E I, we have
A A A As
a ai for each i E I. Suppose c 2 a. for each i E I. Then
1^
for each i E I, there is b. E ai with c b.. Thus
v b. c, so if b= v b., b < c, and we need to show that
iel 1 iEI
b = a. If x is compact and x 5 a*, then there is m E M with
mx a. Thus mx s v{y compactly < a. for some i E I}, so
n
since mx is compact, mx < v Yk' Yk compact and for some
k=l
ik E I, Y ai a? For each k : n, choose mk E M with
kk 1k k
n
mikyk < b. Then (mm1m2...m n) E M and mml...m X v Ykmk <
lkn k=l
n
v b. v b., so x b*, and a* b*. A similar argument
k=l k ieI
shows that b* < a*, so a = b < c.D
Thus joins are preserved going from L to {ala E L}, and
{ala E L} is closed under (finite and infinite) joins. The
next result yields the fact that we have constructed a
complete lattice.
3.10. Proposition: If {aili E I} is a family in L, let
A A A
a = A (at). Then a = A a., and a = a*, so {vlv E L} is
iel 1 ieI
complete. Further, if r and s are in L, with t = r A s,
then t = r A s and t* = r* A s*.
29
A A A
Proof: First, a : a* E a. for each i E I, so a a.
1 1 1
for each i E I. Suppose c a. for each i E I. Then c at
11
' A A
for each i E I, so c a. Thus c a, and a = A a.. Now,
iEl 1
if x is compact and x 5 a*, then for some m E M,
mx a= A at, so mx < at for each i E I. But then for
iel 1
each i E I, there is m. E M with m.mx ai, since mx is
compact, and so x at for each i E I since m.m E M. Hence
1 1
A A
x a, so a = a*. Let t = r A s, and let d E r A s. By
what we have just shown, d* = r* A s*. If x is compact and
x r* A s*, then there are elements m and m2 in M with
m1x : r and m2x s. Thus m1m2 E M with mlm2x : r A s = t,
so d* = r* A s* s t*, and since t 5 r* A s*, we have
A A A,
t* = r* A s* = d*, and t = r A s.[
A A^
3.11. Lemma: For each a E L, a is compact in {blb E LI =
there is an element a0 E a with a0 compact in L.
A /
Proof: If a0 is compact in L, then suppose a0 < A c..
By Proposition 3.10, a0 a* < A (c), so since a is
0 0 lEI 1
n
compact, a0 5 A c which we will label c. Then
k=l 'k
A A n ^ A^
a0 < c = A c. ,so a0 is compact in {blb E L}.
0 k=l 'k
A A
Conversely, if a is compact in {bjb E L}, then since L
is algebraic, a = v{xfx is compact and x a}. By Proposition
A ^A n ^
3.9, a = v{xlx is compact and x < a}. Thus a = v Xk, where
k=l
for each k < n, xk is compact and xk a. But then if
n
a0 = v xk, a0 is compact and again by Proposition 3.9,
k=l
^ n ^
a0 = v xk = a.0
k=l
Propositions 3.9 and 3.10 together with Lemma 3.11,
^
imply that {a[a E L} is an algebraic lattice, which we will
denote LM. We tie all of this together in the next result.
3.12. Theorem: If L is a milattice and is a (satu
rated) multiplicative subset of L, then LM = {ala e L} is an
algebraic lattice, with least element 0 and greatest element
A ^
1 = {b E L I there is some m E M with m < b}. Thus M c 1,
and 1 is compact.
Proof: Propositions 3.9 and 3.10 show that LM is a
complete lattice, and Proposition 3.9 and Lemma 3.11 show
that a = v{xlx is compact and x a}, so that LM is algebraic.
A^ A A,
0 < a 5 1 for each a E L, so 0 a < 1. To see that
1 = {b E Lib m for some m E M}, suppose b m for some
m E M. Then for each x compact in L, mx b, so b* = v{x E LI
x is compact} = 1. On the other hand, if b* = 1, then let
m 0 E M. Since m0 is compact and m0 5 1, there is a m1 E M
^
with m0m1 < b, and so we set m = m0m1 E M. Then 1 = {b E LI
b 2 m for some m E M}, and in particular, M E 1. Since
M e ( and each element of M is compact, 1 is compact.D
Now we must introduce a product on L There is a per
AA A
fectly natural way to do this, namely ab = (ab), but we
need to check that this is welldefined.
3.13. Lemma: If a and b are in L, then ab ~ a*b*.
Proof: Clearly (ab)* < (a*b*)*, so we need only show
a*b* (ab)*. But a*b* = v{x1x21x1 and x2 are compact,
xI < a*, x2 s b*} = v{x1x21x1 and x2 are compact, and there
are ml,m2 E M with m1x1 a and m2x2 b} S V{x1x21x1 and x
are compact, and for some m E M, mx1x2 ab} s (ab)*.D
Thus if a = a2 and bI = b2, then albI ~ a*b* = a*b* ~
a2b2, so we define ab = ab.
3.14. Theorem: If L is a milattice and M is a multi
plicative subset of L, then LM is also a milattice, in which
AA A A
1 is compact and 1a = a for each a E L.
Proof: In light of Theorem 3.12, we need only consider
the product on LM. It is both associative and commutative,
since the product on L is. Since ab 5 a A b for each a,b E L,
we have ab S a A b by Proposition 3.10. If a E L and
{bili E I} is a family in L, then let b = v b. andc=ab. Then
iEI
a( v b.) = ab = c= v ab. If x and y are compact in LM,
iEI iEI
A
then there are compact elements x0 E x and y0 E y, so that
A A A A
xy = x0Y0 is compact since x0y0 is compact. Since l*a 5 a,
A< A A
we have la a. But if x is compact and x < a*, then there
2 2
is a m E M with mx a. Now, m E M and m x s ma 5 la, so
a* < (la)*, and 1a = a.D
The last part of Theorem 3.14 is an expected result if
this is to parallel commutative ring theory, for if R is a
commutative ring and if M is a multiplicative subset of R
with 0 4 M, then RM is a commutative ring with identity.
Now, the following result is part of an exercise in Gilmer
[3], and is easily verified.
3.15. Proposition: If S is a commutative ring, then S
has an identity if and only if SA = A for each ideal A of S
and S is finitely generated as a Smodule.
Thus, for our version of a ring with identity, we let
L have the properties that 1 is compact and la = a for all
elements a E L.
Now, we would like this lattice of extensions LM to
reflect all of the information below the set of associated
primes of M, including prime and primary elements, radicals,
and residual quotients by compact elements. First, we
consider the primes.
3.16. Lemma: If {pXIA E A} is the set of primes
A
associated with the multiplicative subset M, then {pXIA E A}
A
is the set of maximals in LM. If X1 d X21 then P, and p2
are incomparable, and for each A E A, pX = p* and pA is prime
in LM.
^. A
Proof: p. < 1 for each A E A, for if not, then for
some X E A, p, E 1, and p X m for some m E M, which is
A
impossible. Also, if a E L and a > p, for some A E A, then
there is an a0 E a with a0 > p., which is maximal in
S0 = {b E LIfor each m E M, m b}. Thus there is a m E M
,% A A
with a m, so since M c 1, a = 1, and PA is maximal in LM.
As 2\ A\ A A\
Since (1) = 1, p. is prime for each X E A. If a 1, then
since M c 1, for each m E M, a t m. Thus a E So, so a p,
for some A E A, and a pX. This says {piX E A} is the
set of maximals of LM. Now, if XAX2 E A with A1 tX 2, then
p and p are incomparable, so let x1 and x2 be compact with
P1 P2
x 1 PXl x2 P2 x 1 p 2 and x2 $ p .l For each m E M,
2?1 2 2 1*
m pA2, so mxI x pA and mx2 $ pA Thus xI 1 pl and
x p* so p* and p* are incomparable, and so are P
2 X2 X1 X2 X1
A A\ A A
p2. If X E A, pX ; 1, so since p* E pX, p* does not exceed
2*
any m E M, and p* : p0 for some X0 E A. Thus pA : p? < p0
Xnd 00 X 0
an0 = since the p.'s are pairwise incomparable.0
and s PA 0
A
3.17. Theorem: The primes of LM are those classes p
where p is prime in L and p pX for some p. in the set
^0 ^0
{pXX E A} of primes associated with M. If p is prime and
p p. for some X E A, then p = p* and there are no other
primes of L in p.
A A A
Proof: Suppose a is prime in L,. Then a < 1, so
a p, for some X E A, and a* < pO = pO If bc < a*,
0 0 X0 0
.A A\ A\ A\ A A A
then bc a, so b a or c : a, and so a* is prime since
b a* or c a*. If q E a with q prime in L, and if x is
compact with x < a*, then for some m e M, mx < q. But
q a* < p 0, so m $ q, and so x < q. Thus q = a*, and a*
0A
is the only prime in a. On the other hand, suppose p is
prime in L, with p < p. for some X E A. If ab = p, then
ab : p*. Suppose b $ p*. Then there is a x compact with
x $ p* but x 5 b. For each y compact with y 5 a, xy is
compact and xy < p*, so there is an element m E M with
mxy p. Now, p py so m $ p, and so xy < p. But x $ p*,
1
A A A\
so y 5 p, and so a p. Thus a p, so p is prime, and by the
first part of the proof, p = p* is the only prime in p.E
The theorem establishes a onetoone correspondence
which preserves order between primes of LM and primes of L
that do not exceed any elements of M. We now want to do the
same for primary elements, but first we consider radicals.
3.18. Proposition: If a E L, and if r = rad(a), then
/% /A
r = rad(a) and r* = rad(a*).
A /%
n+
Proof: First, r = v{xl x is compact andxn 5 a for some E Z +}.
n /n A
If x is compact and x < a, then x is compact and (x) < a.
A A
Conversely, if (x) < a with x compact, there is x0 E x with
x compact in L. Now, n < a*, so sincexn is compact, there is
<0 a.0 T h
a m E M with mx _< a. Thus (mx)0Jn 5 mxn < a, mx0 is compact,
A/ A A A A A% A / j s opc ad%
and mx0 = lx0 = x0 = x, so r = v{xjx is compact and xn < a
+ A A
for n E Z } = v{xlx is compact and (x) < a for some
+ /
n E } = rad(a). If x is compact and x < r*, then there is
m E M with mx < r, so for some n E Z +, (mx)n < a, and
AA y A // /A/A A% /A /A
(mx) < a. But mx = lx = x, so (x)n < a, and xn < a* with
x compact. Thus r* < rad(a*), but if rad(a*) = s, then
/% /% /%, /%
s = rad(a*) = rad a = r, so r* = rad(a*).D
3.19. Theorem: Let M be a multiplicative subset of L
with {pI E A} the set of associated primes of M. Let p be
a prime in L, with p < p. for some X E A. If q is pprimary,
then q is pprimary in LM with q = q* the only primary in q,
/% /%
for p or any other prime. Conversely, if a is pprimary,
/%
then a* is p*primary, and a* is the only primary in a.
/< /%
Proof: Suppose a is pprimary in LM, with p prime in L
A /A
and p < p. for some X E A. Then p = rad(a), and p = p* =
(rad(a))* = rad(a*). If x and y are compact with xy < a*,
t A A an y A A A /A a
then x and y are compact and xy < a, so x < a or (y)n < a
Z+ n
for some n E Z Thus x a* or y < a*, so since a* < p < ,
a* is pprimary. If q E a with q primary, then if x is com
pact and x < q*, there must be m E M with mx < q. But since
m ,mn frecn 7+ n
m E M, mn E M for each n E Z and so m n q. Thus x < q,
so q = q* = a*. Conversely, if q is pprimary in L, with
p < p for some X E A, then q p., so q < p, < 1. Now,
/\ ^ A /\
p = rad(q), so p = rad(q). If x and y are compact with
xy q, then let x0 and y0 be compact in L with x0 e x,
A
Y0 e y. Since x0y0 q*, there is a m E M with mx0y0 q.
Z+ mn mn
Again, for each n E Z mn E M, so mn $ q. Thus x0y0 < q,
n +A A A
so x0 < q or y0 < q for some n E Z and so a = a+0 q or
A A ,
(b) = (b0) n q. Therefore q is pprimary and by the
first part, q* is pprimary and is the only primary in q, so
q = q*.D
Finally, we turn our attention to residual quotients.
3.20. Proposition: Let a,b E L. Then
(1) If d = [a:b], then d [a:bl.
(2) If b is compact and d = [a:b], then d = [a:b].
(3) ([a:b])* < [a*:b*].
(4) ([a*:b*])* = [a*:b*].
(5) If c = [a:b], then c* = [a*:b*].
A A. At AA^ A A A
Proof: (1) Ford=v{xlbx< a} < v{xjbx a} = [a:b].
(2) Let b be compact in L, and let x be compact with
bx A a. There is a x0 E x with x0 compact. Then bx0 is
bx a. There is a Xoe x with x0, compact. Then bx0. is
compact, with bx0 < a*, so for some m E M, mbx0 A a.
A A\ A A A A\ AS
mx,0 = lx = x, so x 5 v{yly is compact and by a a} = d, so
by part (1), d = [a:b].
(3) For b*[a:b]* <(b[a:b])*
(4) As always, [a*:b*] a ([a*:b*])*. Let c = ([a*:b*3)*.
Now c= v{xx compactand xm [a*:b*] for some m e M} =
v{xjx compact and b*xm a a* for some m E M}. If x is compact
and b*xm a a* for some m e M, then for any y compact with
A A\ A A A\ A
y a b*, we have yxm a a*. Thus yxm = yx < a, so yx < a*
so b*x a a*. Therefore, c = v{xlx compact and b*xm s a* for
some m E M} 5 v{xlx compact and b*x 5 a*}, so c = [a*:b*].
(5) Let c = [a:b] = v{ylby o a}. If x is compact with
x < c*, then x < c, so bx < a, and b*x* < (bx)* < a*. Thus
x a x* A [a*:b*], so c* a [a*:b*]. Conversely, if x is com
A\A A\
pact and x < [a*:b*], then b*x < a*, so bx < a. Thus
A. A A\ A
x a [a:b] = c, so x < c* and c* = [a*:b*].D
3.21. Definition: If p is prime in L and M = {mlm is
compact and m I p}, we say that LM is the localization at p,
P
p
and use the notation L = LM .
p
In commutative ring theory, if R is a commutative ring
with 1, then each ideal Ais the intersection of the contractions
of all of its extensions in the localizations at the maximals
exceeding A. This allows one to assume a unique maximal ideal
when trying to establish an arithmetic result, since by
passing to the localization, all finite joins, meets, and
products are preserved. We obtain the same result here.
3.22. Theorem: If L is a milattice with 1 compact and
1la = a for each a E L, then let {p. i E I} be the set of
1
maximals of L. For a E L, let (a)i be the class of a in
L i, and let a. = v{blb E (a).} = a* with respect to Li
Let J = {j E I la p.}. Then a = A a. = A a..
J iEI 1 jEJ 3
Proof: Since a < a. for each i < I, a A a. < A a..
1 iEI i jEJ 3
Let x be compact with x < A a.. For each j E J, x < a., so
jej 3
there is a compact element m. with m. $ p. and m.x a.
Then [a:x] 2> mi, so for each j E J, [a:x] 4 p.. Since 1 is
compact and a < [a:x], [a:x] = 1. Thus x = 1x < a, so
A a. :_ a.0i
jEJ 3
CHAPTER IV
INVERTIBLE ELEMENTS
We now consider the analogue of invertible ideals of
commutative ring theory. Recall that in a commutative ring
R, if an ideal A contains a regular element (an element that
is not a zerodivisor), then A is invertible if and only if
it satisfies two properties:
(1) A[B:A] = B for each ideal B c A.
(2) [AB:A] = B for each ideal B of R.
An ideal satisfying (1) is multiplicative, while an ideal
satisfying (2) is cancellative. We have already mentioned
that every principal ideal, and thus every power of a princi
pal ideal, is multiplicative. If a principal ideal is
generated by a regular element, then the ideal is cancellative
as well, and thus is invertible. For these reasons, when we
consider properties of principal ideals or elements of the
ring, it is often enough to look at multiplicative and invert
ible compact elements. One thing should be mentioned here:
when dealing with the lattice of convex Zsubgroups of a
latticeordered group, if a < 1, then 1a = aa = a since
multiplication is intersection, and so one will never en
counter any elements besides 1 that could be called
cancellative or invertible in the sense of (1) and (2) above.
40
4.1. Definition: If L is a commutative milattice, an
element a E L is cancellative if [ab:a] = b for each b E L;
equivalently, if ab = ac, then b = c, so that multiplication
by a is an injection. An element a E L is invertible if a
is both cancellative and multiplicative (Definition 2.3);
i.e., multiplication by a is an orderpreserving bijection
of L onto [0,a] = {b e LJ0 b a}. We now list some
results concerning the nice properties and behavior of
invertible elements.
4.2. Corollary: If a is invertible and b is multiplica
tive, then ab is multiplicative.
Proof: Since [ab:a] = b, the result follows from
Lemma 2.4.D
n
4.3. Lemma: Let a = I a.. Then a is cancellative =>
i=l
each a. is cancellative.
1
Proof: If a is cancellative, let i0 n. For each
b E B, [ab:a] = b, so b = [ab:a] = [[ab:a. i]: ii a.] i
10 i i0
[(( n a.)[a. ib:a. 1]): Ili a. ] [a. b:a. i] b, so
imi 0 0 0 ii 0 0 0
b = [ai0 b:a. ], and each a.i is cancellative. Conversely,
suppose al,...,an are cancellative. We proceed by induction
on n. If n = 1, a = a1 is cancellative. Suppose that for
each positive integer k : n1, if C,1...,ck are cancellative,
k
then II c. is cancellative. Then for b E L,
j=l ni ni ni ni
[ba:a] = [[((b a.)an) :an ]: II a.] = [(b H a.) : n a.] = b
i=l1 i=li1 i=1 i=li1
by induction, so a is cancellative.0
n
4.4. Proposition: If a = H a., then a is invertible
i=l 1
Each a. is invertible.
1
n
Proof: If a = n a. is invertible, then a is cancella
i=l 1
tive, so by Lemma 4.3, each a. is cancellative. If n = 1,
1
we are done, so let n > 1. Let i0 5 n, and suppose b = ai .
10
Then b H a. : a, and b = [(b H a.): I a.] by Lemma 4.3.
i 1i0 i .i0 i ei0
i^i .O 1 ^O1
Thus a. [b:a. ] = E(a[b:a. ]): I a.] since H a. is can
1 ii 1 1~i 1
1i0 0 0 1
cellative, and so a. i[b:a. i = [(a[[(b IT a.): H a.]:a. ]) :
10 10 0 1 1
Sa.] = [(a[(b n a.) :a]): n a. ] = [(b H a.) : IT a.] =b,
i^i i0 ii0 ii0 i0i0 ii0
so each a. is multiplicative, and thus invertible. Converse
10
ly, if al,...,a are invertible elements of L, then by Lemma
n
4.3, a = R a. is cancellative. We proceed by induction on n.
i=l 1
If n = 1, a = a1 is invertible. Suppose that for each
positive integer k nl, whenever cl,...,ck are invertible
k ni
elements we must have I c. invertible. Then H a. is
j=l 3 i=l 1
invertible and a is multiplicative, so by Corollary 4.2, a
is invertible.0
4.5. Proposition: If any a E L is invertible, then for
each b E L, lb = b, so that 1 is invertible.
Proof: Let a be invertible. Then a is multiplicative,
so a = a[a:a] = a*l. Then bl = [(bla):a] = [ba:al = b.0
4.6. Proposition: If a is invertible, then
a( A bi) = A abi.
iEI idI
Proof: First, A b. = A [b.a:a] = [( A b.a):a] by
iEI iEII iEI
by Proposition 2.2, part (f), so a( A b.) = a[( A bia):a] =
iEI iEI
A (b.a) since A b.a < a.0
iEI i 1I
1e ielI
This result states that if a is invertible in L, then
multiplication by a is a lattice isomorphism of L onto [0,a]
which preserves meets and joins of infinite as well as finite
families.
n
4.7. Proposition: If a is invertible and a = a for
some n 2, then a = 1.
Proof: First, l.a = a = an = (a nl)a, and a is cancel
lative, so anl = 1. Then n 2 implies a 2 an = 1, so
a = l.0
4.8. Lemma: If a is invertible and {a.ii I} is a
family in L with a. < a for each i E I, then
( v a.) :a] = v [a :a].
iEI iEI
Proof: First, a[( v a.):a] = v a. = v (a[a.:a]) =
iEI iEI ieI
a(v [a.:a]). Then [( v a.):a] = v [a. :a] since a is
i EI iEI iEI
cancellative.
4.9. Proposition: Let L contain an invertible element.
Then the following conditions are equivalent:
(1) 1 is compact.
(2) Every invertible element in L is compact.
(3) There exists an invertible element a E L with a
compact.
Proof: (1) => (2): Let b be any invertible element of
L. Since L is algebraic, b = v b., where each b. is compact.
iEI 1
Then 1 = [b:b] = [( v b.):b] = v [bi:b]. Since 1 is compact,
iEI iEI
n
1 = v [b. :b]. Then b = 1b = b( v [b. :b]) =
k=l k k=l k
n n
v (b[b. i:b]) = V b. and so b is compact.
k=l k k=l k
(2) = (3): Obvious since L contains an invertible
element.
(3) => (): Let a be compact and invertible in L, and
1 = v c.. Then a = a*l = v aci, so since a is compact,
iEI iEI
n n
a = v ac. = a v c. Since a is cancellative and a1l = a,
k=l lk k=l k
n
1 = v c. so 1 is compact.0
k=l k
In the theory of commutative rings there are several
conditions that are equivalent to the condition that a ring
R is an integral domain. We now consider some of these in
the context of milattices. From the theory of commutative
rings with identity, the following conditions are equivalent:
(a) 0 (as an ideal) is prime.
(b) Every nonzero ideal exceeds a cancellative ideal.
(c) Every nonzero ideal exceeds an invertible ideal.
(d) Every nonzero ideal is the join of invertible
principal ideals.
In the language of milattices, L is a milattice in which 1
is compact, invertible, and:
(1) 0 is prime.
(2) Every nonzero element of L exceeds a cancellative
element.
(3) Every nonzero element of L exceeds an invertible
element.
(4) Every nonzero element of L is the join of compact
invertible elements.
Of course, the implications (4) => (3) => (2) => () are obvious.
We give examples to show that no other implications hold
in general, even if we assume ACC and modularity.
(1) # (2): L = {0,x,l}, 0 < x < 1, with x = x for
all n E Z and 1a = a for all a E L. Here, 0 is prime,
since if a > 0 and b > 0, then a x and b x,, so
2
ab x = x > 0. However, 0 is never cancellative, and
since xx = xl, x is not cancellative, so x does not exceed
a cancellative element. See Figure 4.11.
(2) # (3): L = {0} u {pnqmlm,n 0} where p = q = 1,
n m n m rqs
1x = x for all x E L, 0 = ^ p = ^ q and p q > pq <=>
n=l m=l
m + n < r + s or m + n = r + s n + s.
Here, l,p, and q are cancellative, and so by Lemma 4.3,
every nonzero element of L is cancellative. However,
2 2 2
p[q:p] = p q, and q[p :q] = pq < p so neither p nor q is
invertible, and so no element below 1 can be invertible, in
light of Proposition 4.4. See Figure 4.12.
(3) # (4): Let 1 = {pk k E Z, k 0} u tr}, with
Pk p* P if k Z Z, and Pk e r for all k E Z, k > 0. Order
by setting Pk pt if k < Z, with r = ^ P Set pk*PZ = Pk+'
k=l
2 2 {k
and pk r = rpk = r = r. Let = {qIk E Z, k 2 0} u {s},
with order given by qk > qt if k < , and s = A qk Set
k=0
2
qk'q = qk+k and q ks = sqk = = s. Then
.4
1i 2 = {(a,b)la E Q1, b E Q2} with (al,b1) < (a2,b2) 2
(b1 < b2) or (b1 = b2 and a1 a2), and (al,b1)(a2,b2) =
(a1a2,blb2). Let L = [(P0,S)),(p0,q0)3 = {(a,b) E Xi Q 221
(a,b) (P0,S)}, with the inherited order and
(a1a2,b1b2) if bI 1 s and b2 s
(al,bl)'(a2,b2) =
(P0,S) if b = s or b = s.
Relabel as follows: (p0,q0) = 1, (pq0) = pk (r,q0) = r,
k kYk
POqk ) qk (k) = p q (pos) = 0, (r,qk) = rq. Then
multiplication is done in the obvious commutativee) way, with
1x = x for all x E L, pkr = r = r2, x0 = 0 for all x E L,
and p = q = 1. L is a milattice with ACC; it is totally
ordered, so it is modular. Now, p is not multiplicative,
since p[q:p] = pq < q, so p is not invertible. But if
p = V a. with each a. E L, then p = a. for some i0 E I, and
iEI 1
so p is not the join of a family of invertible elements.
However, q is invertible, so q is invertible for each k > 1,
and every element except 0 exceeds some power of q. See
Figure 4.13.
Thus, we have many possible choices for a generalization
of the lattice of ideals of an integral domain. We choose
the strongest, since it is used the most.
4.10. Definition: A milattice L is a domain lattice
if 1 is compact and each nonzero element is the join of a
set of invertible elements (which are compact by Proposition
4.9).
0
FIGURE 4.11
co
n
A q=
n=l
1
p
q,
2
p
pq
2
q
3
p
2
P q
pq2
3
q
4
p
3
p q
2 2
P q
3
pq
4
q
5
p
00
n
A p =
n=l
00o
n
r = A p = pr'
n=l q
pq
Pq
2
P q
3
p q
qr
2
q
2
Pq
2 2
P q
3 3
p q
2r
g r4
FIGURE 4.12
0o
n
A qg
n=l
FIGURE 4.13
= 04
x I
CHAPTER V
GENERALIZATIONS OF VALUATION RINGS AND PREFER DOMAINS
In commutative ring theory, a valuation ring D is an
integral domain with identity in which the set of ideals is
totally ordered. This motivates our next definition.
5.1. Definition: A vlattice is a domain lattice L
which is totally ordered. The following theorem lists some
basic properties of valuation rings that carry over to
vlattices.
5.2. Theorem: If L is a vlattice, let a E L with a < 1.
Then
(1) If a e 0 and a is compact, then a is invertible.
(2) Rad(a) is prime in L.
n k k+l
(3) If p0 = A a then p0 is prime. If a =a
n=l
for some k 1, then a = a, and a is prime.
no
(4) If p is prime and p < a, then p < p0 = A a .
n=l
(5) If b E L and a < rad(b), then ak b for some
k E Z+.
Proof: (1): If a t 0 and a is compact, then write
n
a = v a. where each a. is invertible. Then a = v a. since
iEI k=l k
a is compact, so that a = a. since L is totally ordered, and
n0
a is invertible.
(2): Rad(a) = A{plp is prime and a < p}. But
{pjp is prime and a < p} is totally ordered, so by 1.7 of
Keimel [5], rad(a) is prime.
(3): Let x and y be compact elements with x 4 p0 and
+ mn
y J po. Then for some m and n E Z x $ am and y a Thus
am < x and a < y, and since y is compact and nonzero, it
is invertible by (1). Hence amy < xy, and am+n _< amy, so
am+n < xy, and xy $ P0. Therefore, p0 is prime. Now, if
k k+l k
a = a for some k > 1, then p0 = a so that a = p0 and
a is prime.
(4): If p is prime and p < a, then p < a for each
k E Z so p < P0.
(5): If for each k E Z +, b t ak then b < ak for each
+ n k
k E Z so b : p0 = A q and rad(b) < p0 < a. Thus
k=l
a $ rad(b).L
5.3. Definition: If L is a milattice and p is a prime,
then p is a branched prime if there is a pprimary q e p. We
say p is unbranched if p is the only pprimary in L. As for
valuation rings we have:
5.4. Theorem: If L is a vlattice and p is a nonzero
prime of L, then let S = {qXIX E A} be the set of pprimaries.
(1) If q is pprimary and if x is compact with x $ p,
then q = qx. If q is compact, then p is maximal.
(2) S is closed under multiplication, so
{pk k E Z+} < S. If p p2 then S = {pkk E Z+.
(3) If p is branched and q f p with q pprimary, then
CO
A qn = A{qlX E A}.
n=l
(4) p0 = A{qXIX E A} is prime, and there are no primes
a satisfying p0 < a < p.
Proof: (1) q p < x, and since x is nonzero and
compact, x is invertible, so q = ax. q is pprimary and
x p, so a q, and thus a = q, so q = qx. If q is compact,
then q is invertible since q r 0, and if x is compact with
x J p, then xq = lq = q, so x = 1, and p is maximal.
(2): If q, and q2 E S, then rad(qlq2) = p. If x and y
_n +
are compact with xy < qlq2, and if xn $ p for each n E Z,
then q, = qlx by part (1), so xy < qlq2 = xqlq2. Thus qlq2
2
is pprimary. Now, suppose p e p and let q be a pprimary.
2 2
Then p < rad(q), so q exceeds a power of p and hence a
power of p, by Theorem 5.2, part (5). Let n be the least
positive integer for which pn < q. If n = 1, then p = q. If
n1 nI
n 2 2, then since p > q let y be compact with q < y 5 p
q < y, so since y is invertible, q = ay. Since q is pprimary
and q < y, a p, so q = ay py < n <_ q, and q = p .
(3): Clearly A q A q" Now, q < p = rad(q.) for
n=l XEA
n +
each A E A, so for each X E A, q < q. for some n E Z Thus
^ qAn = ^{qjX E A}.
n=l
(4): If p is unbranched, we are done. If p is
CO n
branched, p0 = A q for each pprimary q < p, so by
n=l
Theorem 5.2 part (3), p0 is prime and each prime a < p
satisfies a p0.D
5.5. Theorem: If L is a vlattice and p is a nonzero
prime, then the following are equivalent:
(1) p is branched.
(2) There is an element a t p with rad(a) = p.
(3) There is an invertible element x with rad(x) = p.
(4) p > v{blb is prime and b < p}.
(5) There is a prime r < p with no primes a satisfying
r < a < p.
Proof: (1) => (2): obvious.
(2) => (3): Choose x invertible with a < x p. This
can be done since a < p, each nonzero element of L is a join
of invertibles, and L is totally ordered. Then rad(x) = p.
(3) => (4): If x is invertible with rad(x) = p, then
x > b for each prime b < p, so x 2 v{blb is prime and b < p}.
But x is invertible and 1 is compact, so x is compact, and
n
so if x = v{blb is prime and b < p}, then x = v bk, and
k=l
since L is totally ordered, x = b0, a prime strictly below p,
and rad(x) e p, contradicting our choice of x.
(4) => (5): Let r = v{blb < p and b prime}. If x and y
are compact with xy 5 r, then since xy is compact,
n
xy i v bk, and since L is totally ordered, xy b0, where
k=l
b0 < p and b0 prime. Thus x < b0 or y : b0, so x : r or
y 5 r, and r is prime.
(5) => (1): We proceed by considering the localization
at p. First, if x is invertible in L, we show that x is
invertible in L If b < x, let b0 E b with b0 < x, and set
p 0
d = [b0:x]. Since x is compact, d = [b0:x] by Proposition
3.20, part (2). Thus x [b0:x] = xd = b0, and x is multipli
A A\ As A* A\ A
cative. Now, suppose that xb = xc, with b,c E L Then for
each y compact with y :b, xy xc, so since xy is compact,
there is a mn0 E Mp = {mjm is compact and m $ p} such that
xm0y 5 xc. Thus m0y c since x is cancellative, and so
b 5 c. Similarly, c b, so x is cancellative, and thus
invertible. Now, in L p is the unique maximal, and L is
totally ordered. Also, r < p and r is prime, with no primes
between r and p. Let a be an invertible element in L with
r < a 5 p. Then a is invertible, and r < a 5 p < 1. Since
..A A 2 A A A 2 ^
a is invertible and a 1, (a) a, so r < (a) < p. Clear
2* A5 AS A. AAs A^ 2
ly, rad((a) ) = p. If x and y are compact with xy < (a) ,
then suppose (y) 4 (a) for any n E Z Since L is a
p
vlattice, we must have y p = rad((a)2), by Theorem 5.2,
A A A, A
part (5). But since y is compact and y 0, y is invertible
A A A2 A
by part (1) of Theorem 5.2, so if y = p, then (y) < y, and
00 A n
A (y) P= p0 is prime with P0 < p. Thus, p0 o r, so for some
n=l
(y)< r < (a), contradicting our choice of y.n
n E Z I (y) < r < (a) contradicting our choice of y.
A ^ 2 ^ 22 "
Therefore, y > p, so y = 1, and x (a) Hence (a) is
"2 2*
pprimary in L p, with (a) d p, and by Theorem 3.19, (a )
2 *
is pprimary with (a ) < p.D
5.6. Definition: A vlattice L is said to be a discrete
vlattice if each primary element is a power of its radical.
This definition agrees with the corresponding definition in
commutative ring theory (Gilmer [3]), and as in that case, a
result of Theorem 5.4, part (2), is that a vlattice L is
discrete if and only if no branched prime p is idempotent
2
(that is, p < p for all branched primes).
5.7. Proposition: Let L be a vlattice with unique
maximal p > 0. The following conditions are equivalent:
(1) L has ACC.
(2) Every element a with 0 < a < 1 is a power of p.
(3) L is discrete, and p is the only nonzero prime.
Proof: (1) = (2): Let a E L with 0 < a < 1. Since L
has ACC, p is compact, and by Theorem 5.2, part (1), p is
invertible. Since a p, a = p[a:p]. Let a = [a:p3.
Recursively, let an+l = [a n:p] for each n E Z. We have the
ascending chain
a = a0 < a 1 a2 : a ... .
By the ACC, there is a least integer n such that a = ak for
all integers k n. Thus an = a n+ = a n:p]. If an p
= n lpn
then an = pan, and since a > a > 0, a is compact and
therefore invertible. Thus, p = 1, which is impossible.
Therefore, an $ p, so an = 1, and we have a = pa = p2a2 =
na n
p = p n = p .
(2) => (1): This is clear.
(2) => (3): This is clear.
(3) => (2): In the proof of Theorem 5.5, when we proved
that condition (5) implies condition (1), we actually showed
that in a vlattice with unique maximal p, any invertible
element with radical equal to p is pprimary. Now, every
invertible element a with 0 < a < 1 has rad(a) = p, since
p is the only nonzero prime. Therefore, all invertible
elements below 1 are pprimary, and since L is discrete, they
are all powers of p. Now, if b is any element p of L with
0 < b < 1, then b is the join of a set of invertible elements,
since L is a vlattice. Thus, b is the join of a set of
powers of p, so if n is the least positive integer that
occurs as an exponent in this set of powers of p, then
b = pn. D
We now turn our attention to the analogue of Priifer
domains. These are integral domains with identity in which
every nonzero finitely generated ideal is invertible, so we
consider milattices with 1 compact in which each nonzero
compact element is invertible. We will proceed toward the
many arithmetic relations that characterize Priifer domains
in commutative ring theory, but first we consider localiza
tions in such a milattice.
5.8. Theorem: Let L be a domain lattice. Let
{Pili E I} be the set of all maximals in L. Then the fol
lowing are equivalent:
(1) Lp is a vlattice for each prime p in L.
(2) Lpi is a vlattice, for each i E I.
(3) Every nonzero compact element is invertible.
(4) If a and b are invertible, then a v b is invertible.
2
Proof: (1) => (2): 1 = 1, so each pi is prime.
(2) = (3): Let x be compact, x > 0. Then if (x)i is
the class of x in the localization at pi, (x)i is compact in
A
Lpi, and by Theorem 5.2, part (1), (x)i is invertible in Lpi.
Now, let y be invertible, with y < x in L. Since x is corn
A A A
pact, if z = [y:x], then (z)i = [(y)i:(x)i], and so
A\ A As
(x) i(z) = (y) i, for each i E I. Thus, if at represents the
i1
maximum element of (a) for each a E L, then y = (xz) for
each i E I. By Theorem 3.22, y = xz, and since y is invert
ible, x is invertible by Proposition 4.4.
(3) => (4): If a and b are invertible, then since 1 is
compact, a and b must be compact by Proposition 4.9. Thus
a v b is compact, and a v b : a e 0, so a v b t 0. Hence
a v b is invertible.
(4) => (1): Let p be prime in L. In Lp, 1 is compact,
and in proving (5) => (1) in Theorem 5.5, we showed that if
L is any milattice in which 1 is compact, then if x is
invertible, we must have x invertible in Lp for any prime p
of L. Thus each nonzero element of Lp is the join of a set
of invertible elements, and so we need only show that the set
of invertible elements of Lp is totally ordered. Let a and
b be invertible in Lp, and assume without loss of generality
that a and b are compact in L. Then a v b is compact in L,
n
so a v b = v Ck, where each ck is invertible in L. An
k=l k
A A
induction argument shows that a v b is invertible, so a v b
A A A A A A A A
is invertible. Thus a = (a v b)(x) and b = (a v b)(y), so
A A A A A A A A A A
a v b = (a v b)(x v y), and x v y = 1. But p is the unique
maximal of Lp, so since x v y > p, then x = 1 or y = 1. Thus
a = a v b or b = a v b, so a > b or b a, and Lp is totally
ordered.D
5.9. Proposition: If L is a milattice in which 1 is
compact and every nonzero compact element is invertible,
then L is distributive.
Proof: Let a,b, and c E L. For each maximal p, Lp is
a vlattice by Theorem 5.8, so Lp is totally ordered, and
thus distributive. If d1 = a A (b v c) and d2 = (a A b) v
(a A c), then by Propositions 3.10 and 3.11, for each maximal
^ A *
p of L, d1 = d2 in Lp. Thus, d1 = d2 relative to each
maximal p of L, and by Theorem 3.22, d1 = d2.D
Remark: In commutative ring theory, this is a character
ization of Priifer domains; i.e., an integral domain with
identity is a Priifer domain if and only if its lattice of
ideals is distributive (Gilmer [3]). In our context, we
must leave it as an open question whether the converse of
Proposition 5.9 is true.
The method of proof used in Proposition 5.9 will be
used in the next theorem as well. That is, we assume we are
looking at the localization at a maximal of L, so that L has
a unique maximal, and show the arithmetic relations under
this condition. Again, because lattices ofextensions preserve
all of the arithmetic relations we consider, as well as
compactness and invertibility, we can then generalize.
5.10. Theorem: Let L be a domain lattice. Then the
following conditions on L are equivalent:
(1) Every nonzero compact element of L is invertible.
(2) For all a,b,c E L, a(b A c) = ab A ac.
(3) For all compact elements a,b,c E L, a(b A c) =
ab A ac.
(4) For all a,b E L, (a v b)(a A b) = ab.
(5) For all compact elements a,b E L, (a v b)(a A b) = ab.
(6) If a,b E L and c is compact in L, then
[ (a v b):c] = [a:c] v [b:c].
(7) If a,b,c are compact in L, then
[(a v b):c] = [a:c] v [b:c].
(8) If a,b are compact in L, [a:b] v [b:a] = 1.
(9) If a,b are invertible in L (and so compact, by
Proposition 4.9), [a:b] v [b:a] = 1.
(10) If c E L and a,b are compact in L, then
[c:(a v b)] = [c:a] v [c:b].
(11) If a,b,c are compact in L, then
[c:(a v b)] = [c:a] v [c:b].
Proof: First, (2) => (3) (4) => (5) (6) (7),
(8) => (9), and (10) = (11) easily. We show (1) => (2),
(1) => (4), (1) = (6), (1) = (8), and (1) = (10) by assuming
L is a vlattice, by Theorem 5.8. Then, assuming that L has
a unique maximal, we show that (3) =>(5) => (1), that
(7) => (9) = (1), and finally that (11) => (9).
(1) => (2): Since L is totally ordered, we assume that
b < c. Then a(b A c) = ab = ab A ac.
(1) => (4): Again, assume a Z b, so that (a v b)(a A b)
ab.
(1) => (6): Assuming a b, [(a v b):c] = [b:c] =
[a:c] v [b:c].
(1) => (8): If a < b, then [b:a] = 1, so [a:b] v [b:a] =1.
(1) => (10): Suppose a < b. Then [c:(a A b)] = [c:a] =
[c:a] v [c:b], since if xb < c, then xa < xb < c, so
Cc:b] 5 [c:a].
Now we may only assume that L has a unique maximal p.
(5) => (): Let a and b be invertible in L. Then a and
b are nonzero compact elements, so a v b is nonzero and
compact. But (a v b)(a A b) = ab, which is invertible, so
by Proposition 4.4, a v b is invertible. By Theorem 5.8,
part (4), every nonzero compact element is invertible.
(3) = (5): Let a and b be compact in L. (a v )(a A b) =
a(a A b) v b(a A b) < ab, and because (3) holds, (a v b)(a A b)=
(a v b)a A (a v b)b > ab, so (5) is true.
(9) => (): Let p be the unique maximal of L. We will
show that L is a vlattice. Since 1 is compact and every
nonzero element is the join of a set of invertibles, we
need only show that the set of invertible elements is totally
ordered. Let a and b be invertible in L. [a:b] v [b:a] = 1
by (a), so either [a:b] t p or [b:a] $ p, where p is the
unique maximal of L. Thus [a:b3 = 1 or [b:a] = 1, so
a < b or b < a.
(7) = (9): If a and b are invertible in L, then
1 = [(a v b):(a v b)] = [a:(a v b)] v [b:(a v b)] =
[a:b ] v [b:a] by (7) and Proposition 2.2, part (d).
(11) => (9): If a and b are invertible in L, then
1 = [(a A b):(a A b)] = [(a A b):a] v [(a A b):b] =
[b:a3 v [a:b] by (11) and Proposition 2.2 part (d).A
CHAPTER VI
GENERALIZATION OF DEDEKIND DOMAINS
We now consider milattices in which every nonzero
element is invertible, and 1 is compact.
First, in light of Proposition 4.9, every element of L
is compact, and so L has ACC. Also, 1a = a for each a E L,
and 0 is prime, and L satisfies all of the conditions of
Theorems 5.6, 5.7 and 5.8.
6.1. Proposition: If every nonzero element of L is
invertible, then every nonzero prime is maximal.
Proof: Suppose 0 < p < q 1. Then q[p:q] = p, so if
p is prime, then because q > p, [p:q] < p < [p:q], and so
p = [p:q]. Thus pq = p, and since p is cancellative q = I.D
6.2. Proposition: In a milattice L, if x is invertible
n
and x = n p., where each p. is prime in L, then this repre
i=l
sentation is unique up to the order of the primes.
n m
Proof: Let x be invertible, with x = pi = qj,
i=l j=l
where each pi and each qj is prime. We show that m = n,
and after rearranging, pi = qi for each i n. First, by
60
Proposition 4.4, each pi and each qj is invertible. We
m
proceed by induction on n. If n = 1, then p = l q., so
j=1 I
after rearranging, q, : p since p is prime, and p = ql.
m
If m 2 2, then 1 = (p:ql) = n q., which is impossible, since
j=2
each q. < 1. Thus m = 1 and p = q = x. Now, suppose that
if 1 k < nl, and if r ,r2,...,rk are all primes in L with
k
H r= yinvertible, then this representation of y is unique
k=l
up to the order of the primes. Then if x is invertible and
n m m
x= ilp. = qj, we have p, x = H q., so after rearrang
= j=l j=1 3
ing if necessary, p, > q, since p, is prime. Then
pI[qI:pl] = ql, since p, is invertible. If q, < pI, then
[ql:Pl 5 ql since q, is prime, so by Proposition 2.2, part
(a), ql = [ql:pl], and plql = ql, so since q, is cancellative,
pl = 1, which is impossible. Thus q, = pl, so
n m
x = (pl) ( II pi) = (pl)( n q.), and since p, is cancellative,
i=2 j=2
n m
nlp. = H q.. By induction, m 1 = n 1, and so m = n,
i=2 j=2 3
and after rearranging, pi = qi for each i 5 n.0
We now give the major result of this chapter, which is
similar to the characterizations of Dedekind domains, with
one slight variation, which we will discuss after the proof.
6.3. Theorem: In a milattice L in which 1 is compact,
the following conditions are equivalent:
(1) If a E L and a > 0, then a is invertible.
2
(2) 1 = 1 and all nonzero primes are invertible.
(3) Every nonzero element exceeds an invertible element,
and if 0 < a < 1, then a can be uniquely written
as a finite product of maximals.
(4) L is a domain, each nonzero element of L lies
below only a finite number of maximals of L, and
for each maximal of L, L is a vlattice with ACC
P
(and so L is discrete).
P
(5) L is a domain with ACC, and for each maximal p of
L, L is a vlattice.
P
Proof: (1) => (2): Obvious.
(2) => (I): Let S = {a E Lia is not invertible}. S
since 0 E S, and S is partially ordered with the order from
L. Let C be a nonempty totally ordered subset of S, with
C = {cili E I}. Let c = v c., and suppose that c 4 S. Then
iEI
c is invertible, and by Proposition 4.9, since 1 is compact,
n
c is compact. Thus c = v c. and since C is totally ordered,
k=l k
c = c0 E C, which is impossible. Hence c E S, so by Zorn's
Lemma, S has a maximal element p. Let a and b be in L with
ab p but a $ p. Then a v p > p, so a v p 4 S, and a v p
is invertible. Thus (a v p)d = p, where d = [p:a v p]. Since
p is not invertible, but a v p is, d must not be invertible,
so d E S. But p 5 d, so since p is maximal in S, p = d. Thus
p(a v p) = p, so (a v p)(b v p) = ((a v p)b) v ((a v p)p) =
ab v bp v p = p since ab p. Again, b v p cannot be
63
invertible since a v p is invertible but p is not, so
b v p E S, and so b v p = p. Thus b p, and p is prime.
But p is not invertible, so p = 0, and every nonzero element
of L is invertible.
(1) => (3): Clearly, every nonzero element exceeds an
2
invertible element, in particular, itself. Since 1 = 1,
maximals are prime, so that in light of Propositions 6.1 and
6.2, we only have to show that each element a with 0 < a < 1
can be written as a finite product of primes. Let S = {a E LI
0 < a < 1 and a cannot be written as a finite product of
primes}. If S p, then since L has ACC by Proposition 4.9,
S must have a maximal element b. Since b E S, b is not
prime, and so b is not maximal in L. Let q E L with b < q < 1.
n
Then q 4 S, so q = l p., where each pi is prime. Also, q e 0,
i=l 1
so q is invertible, and b = q[b:q]. Thus Cb:q] cannot be
written as a finite product of primes, or else b could be.
Then [b:q] E S, and b [b:q], so b = [b:q]. Therefore b = bq,
and since b > 0, b is cancellative, so that q = 1. This
contradicts the choice of q, so S is empty.
(3) => (i): First, by Proposition 4.9, 1 is invertible,
2
and so 1 = 1. Now, let p be a maximal in L. If p = 0, we
are done trivially, so suppose p > 0. Then p is prime, and p
n
exceeds an invertible element a. Since a 0, a = II q., where
i=l 1
n
each q. is a maximal of L. Since IT q. = a p, there must
i=l 1
be an integer i0 < n with qi0 < p. But qi0 is a maximal and
0 0'
p < 1, so q. = p. Thus p is a factor of a, so p is invert
10
ible since a is. Therefore each maximal of L is invertible,
so that each nonzero element of L is a product of invertible
elements, and so is invertible.
(1) => (4): By Theorem 5.8, since each nonzero compact
element of L is invertible, then for each maximal p e L, L
P
is a vlattice. Since 1 is compact in L, each element of L
is compact by Proposition 4.9, so L has ACC, and by Lemma
3.11, for each maximal p E L, L has ACC as well. Let a E L
P
with a > 0. If a = 1, then the set of maximals above a is
empty, and is therefore finite. If a < 1, then by the
n
equivalence of (1) and (3) of this theorem, a = II p., where
i=l 1
each pi is a maximal of L, and this representation as a
product of maximals is unique. But if p is any maximal of L
with a : p, then a = p[a:p] since p is invertible. Thus if
[a:p] = 1, then p = a and p is the only maximal exceeding a,
m
while if [a:p] < 1, then [a:p] a > 0, so [a:pl = n q.,
j=l
m
where each qj is a maximal of L, and a = p( H q.). By the
j=l
uniqueness of the representation, p = pi for some i 5 n, and
so the only maximals above a are those in {pl,...,pn}.
(4) => (5): We only have to show that every element of
L is compact. Let a E L with 0 < a < 1, and let {plP2,...,pn}
be the set of maximals exceeding a. For each positive integer
A
i n and for each b E L, let (b)i be the equivalence class
in Lpi containing b. For each i n, (a)i is compact since
Pi_
A A
L has ACC. If a = v{b IX E A}, then (a). = v{(b).iX E A},
p. x 1 Xi
and for each i : n, there is an integer m. such that
A mi
(a). = v (bX ).. Thus, if B = {bl for some i < n and for
k=l k
some k i mi, b, = bX k}, then B is finite and for each i : n,
A A
^\ ^
(a). = v{(bXlIbXE B}. By Theorem 3.22, a = v{b?b, E B}, and
a is compact.
(5) => (i): By the fact that L has ACC, each nonzero
element of L is compact, and so by Theorem 5.8, each nonzero
element is invertible.0
Recall that in commutative ring theory, an integral
domain with identity is a Dedekind domain if and only if
every proper nonzero ideal is uniquely a product of a finite
number of prime ideals (Hungerford [4]). In the context of
milattices, we must insist that each element a with 0 < a < 1
is uniquely expressible as a product of a finite number of
maximals, as the next two examples illustrate.
6.4. Example: Let P and M be disjoint countably infinite
sets, with P = {pklk is a nonnegative integers, M = {MI. is
a nonnegative integer}, pk 1 pk2 if kI k k2, and m.1 mk2
if 9i 1 dk2* Let L0 = {0} u (P x M), and order L0 by setting
0 < a for each a E L0, and (Pkl, ) mZ (Pk ,"m ) if k2 < k1 or k1=k2
1 1 2 2
and Z2 < ZI. L0 is totally ordered, so it is a lattice. For
each family F = {(pk ,m. )i E I}, v (pk.,m) = (Pk,mz),
i ieIl i
where k = min{k.ili E I} and = min{.ilk. = k}. In particular,
v (pk ,mz) E F, so each element of L0 is compact. Also,
iEI 1 1
0, if {kili E I) is unbounded
(pk+l,m0), if k = max{kili E I} and
A (p ,m ) = {U.ilk. = k} is unbounded
iel i ~i1 1
iEI i i
(pk,m),if k = max{bi.i E I} and
Z = max{P.ik. = k}.
Thus, L0 is an algebraic lattice, with 1 = (p0,m0). Define
a product on L0 by setting 0*a = a.0 = 0 for each a E Lo,
and (p m, m) (Pk ,m2 ) = (Pk+2,m1+ + ). It is easily seen
that this product makes L0 a commutative milattice, with 1
2
compact and l*a = a for each a E L0. In particular, 1= 1,
so that (p0,ml), the unique maximal of L0, is prime. The
element (Pl,m0) is also prime, for if a > (pl,m0) and
b > (plm0), then a = (POm1 ) and b = (p0,m 2), so that
ab = (P0,m1 + 2) > (PI,m0). Now, if q > (pl,m0), then either
q = 1, which is not prime, or q = (p0,m) for some Z > 1, in
which case q = (p0,ml) and q is not prime unless Z = 1.
If 0 < q < (plm0), then q = (pkmZ), with k > 1 and with
k 1 if k = 1. Thus, q = (pl,m0) (Pklm,), and q is not
prime. Therefore the only primes of L0 are 0, (p0,ml), and
(pl,m0) and each element a of L0 with 0 < a < 1 can be
uniquely expressed as a product of a finite number of primes.
Also, (pl,m0) is multiplicative, for if a < (Pli,m0), then either
a = 0 or a = (pkmI) with k 1, so a = 0(p1,m0) or
a = (Pku,m2Z)(Plm0). Also, (plm0) is multiplicative, since
(Pl'm0) (Pklmk1) e (Pl'm0)(Pk2 ,m2) if kI 1 k2 or Z1 t Z2'
Thus, (pl,m0) is invertible, and so (Pk,mO) is invertible
for each integer k 1. Therefore each nonzero element of
L0 exceeds an invertible element. However, (p0,ml) is not
invertible, since there is no a E L0 with a(p0,mI) = (pl,m),
and so (p0,m1) is not multiplicative. Hence, not every
nonzero element of L0 is invertible.
6.5. Example: It is not even enough to assume that in
L, each nonzero element is a join of invertibles. To see
this, let P, M, and L0 be the same as in Example 6.4, and
let X be a countably infinite set with X n P = X n M =
X n L0 = c, and X = {xnl n is a nonnegative integer} with
xn r x n2 if n1 n2. Let Ln = (L0 \{0}) x {x n}, identifying
00
(Pkm) with (Pkm,xO), and set L = {0} u ( u Ln). For a
n=0
partial order, set 0 < a for each a E L, and (Pk ,m1,xn ) _
~1 1!
(Pk 2,m ,xn ) if n2 < n1 and k2 < kl, or if n2 < n1 and
k2 = kI and n2 + X2 < nl + k" This is a lattice order on L
which extends the order on L0, and L is an algebraic lattice
with ACC. For a product, set 0*a = a0 = 0 for each a E L,
and P'km1xn )(Pk2 ,m2,xn = (k+k ,m1 +Z 2,xnl+n2) Then
(k1 m,1 n1 2 2 2 1 2 1 2 1 2
L is a milattice, 1 = (p0,m0,x0), 1 is compact and 1a = a
for all a E L. Also, (P0,m0,x0) is the unique maximal of L, and
since 12=, (0,mx0) is prime. The only primes in L are
since 1 =1, (pQOm^.XO) is prime. The only primes in L are
0, (p01,ml,x0), (pl,m0,x0), and (p0,m0,xl), and so each
element a with 0 < a < 1 can be uniquely expressed as a
product of a finite number of primes. The elements
(plm0,x0) and (p0,m0,xl) are invertible, and (p0,.nlx0) =
(pl,m0,x0) v (p0,m0,xl), so since the product distributes
over joins, each nonzero element is the join of a set of
invertible elements. Once again, however, there is no
a e L with a(p0,mlx0) = (Plm0ox0), so (P0,ml,x0) is not
invertible. In Figure 6.6, we label (Pk,m,xn) = p m x
with the convention p = m = x = 1 to simplify the
notation, and L0 from Example 6.4 is labeled.
k
0 = A p =
k=l
00
n
A Xn
n=l
FIGURE 6.6
/ 2
pmx
 
CHAPTER VII
MODULARITY
In Chapter II, we assumed modularity for the milattice
L in order to obtain the result that Aprime elements are
primary, which gave us the primary decomposition of Theorem
2.14. However, throughout the rest of this paper, modularity
was not assumed, and it is not clear that modularity was
needed in Theorem 2.14. The question that arises is whether
we can impose other restrictions on L to guarantee modularity.
Of course, the assumption that L is a milattice is not
enough, for any algebraic lattice, and hence any finite
lattice including the fiveelement nonmodular lattice, is
a milattice with the trivial multiplication a'b = 0 for all
a,b E L. One connection between modularity and invertible
elements is the following result.
7.1. Lemma: If L is a milattice and if a E L is invert
ible, then L can contain no fiveelement nonmodular sub
lattice in which a is the maximum element.
Proof: Suppose L* is a fiveelement nonmodular sub
lattice of L, with L* = {a,b,c,d,e}, satisfying a = b v d =
c v d, e = b A d = c A d, and b > c. If a is invertible in
L, then a = 1a, b = abl, c = acl, d = ad,, and e = ae,
with 1 = bI v dI = c1 v di, e1 = bI A d1 = c1 A dl, and
bI > c1, since multiplication by a is a lattice isomorphism
of L onto [O,a]. Then bl(C1 v dl) = bl(l) = bl, since L
contains an invertible element, but blc1 c1 and
bldI 5 b1 A d1 = el, so that b1c1 v bldI c1 v el = c1 < br
contradicting distributivity of the product.D
This lemma allows us to say that if every nonzero
element of L is invertible, then L is modular. However, we
cannot weaken the amount of invertibility in L beyond this,
for even if L is a milattice in which each nonzero element
is the join of a set of invertible elements, we have the
example at the end of the last section. Here, m = x v p =
x v pm, and px = x A p = x A pm, with pm < p, so L contains
a fiveelement nonmodular sublattice, and is therefore
not modular.
BIBLIOGRAPHY
[1] Bigard, A., Keimel, K., and Wolfenstein, S., Groupes
et Anneaux R4ticules, SpringerVerlag, Berlin (1977).
[2] Birkhoff, G., Lattice Theory, 3rd Edition, Amer. Math.
Soc. Colloq. Publ., vol. XXV, Providence (1967).
[3] Gilmer, R., Multiplicative Ideal Theory, Marcel Dekker,
Inc., New York (1972).
[4] Hungerford, T. W., Algebra, Holt, Rinehartand Winston,
Inc., New York (1974T).
[5] Keimel, K., "A Unified Theory of Minimal Prime Ideals,"
Acta Math. Acad. Sci. Hung., 23 (1972), pp. 5169.
BIOGRAPHICAL SKETCH
David Bruce Kenoyer was born on August 27, 1953,in
Seattle, Washington, and is the only son and second of
three children of Howard W. and Beatrice Kenoyer. David
has spent most of his life in Michigan, attending East
Grand Rapids High School before receiving his B.S. from
Central Michigan University and his M.S. from Michigan
State University. While at Michigan State, he met Susan
Tiffany, whom he later married. David enjoys most sports,
especially those which he shares with Sue, such as golf,
tennis, bowling, and softball, as well as camping and
music.
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
Joy'ge Mdrtinez, Chairman
Associate Professor of
Mathematics
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
XaeWJ 4^. xA&JA
David A. Drake
Professor of Mathematics
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
James E. Keesling J
ofessor of Mathemati&s
I certify that I ha read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
Neil L. White
Associate Professor of
Mathematics
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
Elroy J. Bo duc, Jr.
Professor of Subject Specialization
Teacher Education
This dissertation was submitted to the Graduate Faculty of
the Department of Mathematics in the College of Liberal Arts
and Sciences and to the Graduate Council, and was accepted
as partial fulfillment of the requirements for the degree
of Doctor of Philosophy.
August 1982
Dean, Graduate School

Full Text 
CHAPTER V
GENERALIZATIONS OF VALUATION RINGS AND PRUFER DOMAINS
In commutative ring theory, a valuation ring D is an
integral domain with identity in which the set of ideals is
totally ordered. This motivates our next definition.
5.1. Definition: A vlattice is a domain lattice L
which is totally ordered. The following theorem lists some
basic properties of valuation rings that carry over to
vlattices.
Then
5.2. Theorem: If L is a vlattice, let a e L with a < 1,
(1) If a 0 and a is compact, then a is invertible.
(2) Rad(a) is prime in L.
T") V Tr _l_ "1
(3) If pg = a a then p~ is prime. If a = a
n=l
2
for some k Â£ 1, then a = a, and a is prime.
00 n
(4) If p is prime and p < a, then p < pn = a a .
k
n=l
(5) If b e L and a < rad(b), then a^ < b for some
,+
k e Z
Proof; (1): If a 0 and a is compact, then write
n
a = v a. where each a. is invertible. Then a = v a. since
iel
k=l 1k
a is compact, so that a = a. since L is totally ordered, and
10
a is invertible.
48
62
(1) If a e L and a > 0, then a is invertible.
2
(2) 1 =1 and all nonzero primes are invertible.
(3) Every nonzero element exceeds an invertible element,
and if 0 < a < 1, then a can be uniquely written
as a finite product of maximals.
(4) L is a domain, each nonzero element of L lies
below only a finite number of maximals of L, and
for each maximal of L, L is a vlattice with ACC
P
(and so L is discrete).
(5) L is a domain with ACC, and for each maximal p of
L, L is a vlattice.
P
Proof: (1) => (2) : Obvious.
(2) => (1) : Let S = {a e La is not invertible}. S
since 0 e S, and S is partially ordered with the order from
L. Let C be a nonempty totally ordered subset of S, with
C = {c.i e I}. Let c = vc., and suppose that c S. Then
1 iel 1
c is invertible, and by Proposition 4.9, since 1 is compact,
n
c is compact. Thus c = v c. and since C is totally ordered,
k=l 1k
c = Cq e C, which is impossible. Hence c e S, so by Zorn's
Lemma, S has a maximal element p. Let a and b be in L with
ab < p but a Â£ p. Then avp>p, soavp^S, and a v p
is invertible. Thus (a v p)d = p, where d = [p:a v p]. Since
p is not invertible, but a v p is, d must not be invertible,
so d e S. But p < d, so since p is maximal in S, p = d. Thus
P (a v p) = p, so (a v p) (b v p) = ( (a v p)b) v ( (a v p) p) =
ab v bp v p = p since ab < p. Again, b v p cannot be
41
each positive integer k n1, if are cancellative,
k
then II c. is cancellative. Then for b e L,
i=l 3
3 n1 n1 n1 n1
[ba:a] = [[((b U a.)a ) :a ]: H a. ] = [ (b II a.): II a. ] = b
i=l 1 n n i=l 1 i=l 1 i=l 1
by induction, so a is cancellative.
n
4.4. Proposition; If a = II a. then a is invertible
i=l 1
<*> each a^ is invertible.
n
Proof: If a = II a. is invertible, then a is cancella
i=l 1
tive, so by Lemma 4.3, each a^ is cancellative. If n = 1,
we are done, so let n > 1. Let q < n, and suppose b = a.. .
0
Then b II a^ < a, and b = C (b II a^) : II a^] by Lemma 4.3
0
1*1
0
1*1
0
Thus a. [b:a. ] = [(a[b:a. ]): II a.] since II a. is can
0 lQ lQ 1 i:Â£i0 1
cellative, and so a. [b:a. ] = [(a[[(b H a.): II a.]:a. ]):
lQ lQ i:i0 1 1 lQ
II a.] = [ (a[(b II a.):a]): n a.] = [ (b II a.): II a.]=b,
1;i:Lo 17!:Lo x*1o 1 15ilo 1 l:e;Lo
so each a. is multiplicative, and thus invertible. Converse
ly, if a^,...,an are invertible elements of L, then by Lemma
n
4.3, a = 1 a. is cancellative. We proceed by induction on n.
i=l 1
If n = 1, a = a^ is invertible. Suppose that for each
positive integer k < n1, whenever c^,...,c^ are invertible
k n1
elements we must have H c. invertible. Then II a. is
j=l 3 i=l 1
invertible and an is multiplicative, so by Corollary 4.2, a
is invertible.
69
FIGURE 6.6
CHAPTER VII
MODULARITY
In Chapter II, we assumed modularity for the milattice
L in order to obtain the result that Aprime elements are
primary, which gave us the primary decomposition of Theorem
2.14. However, throughout the rest of this paper, modularity
was not assumed, and it is not clear that modularity was
needed in Theorem 2.14. The question that arises is whether
we can impose other restrictions on L to guarantee modularity.
Of course, the assumption that L is a milattice is not
enough, for any algebraic lattice, and hence any finite
lattice including the fiveelement nonmodular lattice, is
a milattice with the trivial multiplication a*b = 0 for all
a,b e L. One connection between modularity and invertible
elements is the following result.
7.1. Lemma; If L is a milattice and if a e L is invert
ible, then L can contain no fiveelement nonmodular sub
lattice in which a is the maximum element.
Proof: Suppose L* is a fiveelement nonmodular sub
lattice of L, with L* = {a,b,c,d,e}, satisfying a = b v d =
cvd, e=bAd=cAd, and b > c. If a is invertible in
L, then a = l*a, b = ab^, c = ac^, d = ad1, and e = ae^,
70
27
A
This is a partial order on {a a e L}. We need it to be a
lattice order which gives us an algebraic lattice.
3.8. Lemma: Let a,b e L. The following are equivalent:
/\ A
(1) a < b.
A
(2) There is an aQ e a with a^ < b.
A
(3) There is a b^ e b with a < bQ.
Proof: (1) => (3) : If a < b, then a < a* < b* b.
A
(3) => (1) : Suppose bQ e b, with a < bQ. Let x be
compact, x < a*. Then there is a m e M with mx < a, so
/V A
mx < bQ, and x < b^ = b*, so a* < b*, and a < b.
A A
(1) =>(2): If a < b, then a* < b*, so for each x com
pact with x < a*, we choose m e M with m x < b. Let
X X
an = v{mxxjx compact, x < a*}. Since an < b, we only need
aQ e a; i.e., a* = a*,
First, at < a* by choice of an. Let
y be compact, y < a*. Then there is m e M with
my < aQ = v{mxxx compact, x ^ a*}. Now, my is compact, so
n
my < v m^x^., where for each k < n, xk is compact, x^ < a*,
k1
and mv = mv as chosen above. But since xu < a*, mkxk < a
x
with mkxk compact, so we can find yu e M with yvmvxu < a.
n
k k k
n n
Let mQ= m II yk We have mQ e M, and mQy < ( II y, ) ( v m,xk) <
k=l k=l k=l
n
V ^kmkxk^ ~ a' and so Y a** Thus a* = a*, so aQ e a.
L 1
(2) => (1) : Suppose aQ e a with aQ < b. If x is compact
and x < a^, then there is a m e M with mx < aQ < b, so x < b*.
A A
Thus a* < b*, and a < b.D
30
for each k
n
a0 = v x,
U k=l K
~ n
a0 = v xk
k=l K
^ n, is compact and x^ < a. But then if
a^ is compact and again by Proposition 3.9,
A
= a.D
Propositions 3.9 and 3.10 together with Lemma 3.11,
imply that {aa e L} is an algebraic lattice, which we will
denote L^. We tie all of this together in the next result.
3.12. Theorem; If L is a milattice and is a (satu
A
rated) multiplicative subset of L, then = {aa e L} is an
/v
algebraic lattice, with least element 0 and greatest element
A /\
1 = {b e L  there is some m e M with m < b}. Thus M Â£ 1,
A
and 1 is compact.
Proof: Propositions 3.9 and 3.10 show that is a
complete lattice, and Proposition 3.9 and Lemma 3.11 show
A A A A A
that a = v{xx is compact and x < a}, so that L^ is algebraic.
AAA
0 s a < 1 for each a e L, so 0 < a < 1. To see that
/\
1 = (b e Lb > m for some m e M}, suppose b Â£ m for some
m e M. Then for each x compact in L, mx < b, so b* = v{x e l
x is compact} =1. On the other hand, if b* = 1, then let
mg e M. Since mg is compact and nig < 1, there is a m^ e M
A
with nigin^ < b, and so we set m = nigin^ e M. Then 1 = {b e l
A
b > m for some m e M}, and in particular, M Â£ l. Since
M 4> and each element of M is compact, 1 is compact.
33
3.16. Lemma: If {p^A e A} is the set of primes
A
associated with the multiplicative subset M, then {p^  X e A}
A A
is the set of maximals in L^. If A^ X^, then p^
are incomparable, and for each A e A, p^
in lM
p* and Px
is prime
Proof: p^ < 1 for each X e A, for if not, then for
A
some X e A, p^ e 1, and p^ > m for some me M, which is
A /V
impossible. Also, if a e L and a > p^ for some X e A, then
A
there is an a^ e a with ag > p^, which is maximal in
Sg = {b e Lfor each m e M, m ^ b}. Thus there is a m e M
/\ /\
with a ^ m, so since M Â£ 1, a = 1, and p^ is maximal in L^.
A 2 A A A A
Since (1) = 1, p^ is prime for each X e A. If a 1, then
A
since M Â£ 1, for each m e M, a Â£ m. Thus a e SQ, so a < p^
A A A
for some X e A, and a < p^. This says J X e A} is the
set of maximals of L^. Now, if A1,A2 e A with A1 X2, then
p^ and p^ are incomparable, so let x^ and x2 be compact with
X1 x2 ~ X1 ^ Pa anc^ x2 ^ Pa For eah m e M,
12 2 1
m $ p^ so mxj^ $ p^ and mx2 Â£ p^ Thus x.^ 1 p* and
2 2 1 1
2
A
Pa
^ i^* and p* are incomparable, and so are p^
2 12 1
x i p* so p
A2
If A e A, p^ 1, so since p^ e p^, p* does not exceed
any m e M, and p* < p^ for some AQ e A. Thus p^ < pÂ£ < p^
and so p^ = p^ = p* since the p^1s are pairwise incomparable.
63
invertible since a v p is invertible but p is not, so
b v p e S, and so b v p = p. Thus b < p, and p is prime.
But p is not invertible, so p = 0, and every nonzero element
of L is invertible.
(1) => (3): Clearly, every nonzero element exceeds an
. 2
invertible element, in particular, itself. Since 1 =1,
maximals are prime, so that in light of Propositions 6.1 and
6.2, we only have to show that each element a with 0 < a < 1
can be written as a finite product of primes. Let S = {a L
0 < a < 1 and a cannot be written as a finite product of
primes}. If S , then since L has ACC by Proposition 4.9,
S must have a maximal element b. Since be S, b is not
prime, and so b is not maximal in L. Let q e L with b < q < 1.
n
Then q it S, so q = np., where each p. is prime. Also, q 0,
i=l 1 1
so q is invertible, and b = q[b:q]. Thus [b:q] cannot be
written as a finite product of primes, or else b could be.
Then [b:q] e S, and b < [b:q], so b = [b:q]. Therefore b = bq,
and since b > 0, b is cancellative, so that q = 1. This
contradicts the choice of q, so S is empty.
(3) => (1): First, by Proposition 4.9, 1 is invertible,
2
and so 1 =1. Now, let p be a maximal in L. If p = 0, we
are done trivially, so suppose p > 0. Then p is prime, and p
n
exceeds an invertible element a. Since a 0, a = II q. where
i1 1
n
each q. is a maximal of L. Since II q. = a p, there must
i=l 1
be an integer i_ < n with q. < p. But q. is a maximal and
0 0
56
of invertible elements, and so we need only show that the set
A
of invertible elements of Lp is totally ordered. Let a and
A
b be invertible in Lp, and assume without loss of generality
that a and b are compact in L. Then a v b is compact in L,
n
so a v b = v c, where each c, is invertible in L. An
k=l K K
induction argument shows that a v b is invertible, so a v b
/\ A A A A A A A
is invertible. Thus a = (a v b) (x) and b = (a v b) (y) so
A A A /\ A A /\ A A A
a v b = (a v b) (x v y) and x v y = 1. But p is the unique
AAA A /\
maximal of Lp, so since x v y > p, then x = 1 or y = 1. Thus
a=avborb=avb, so a > b or b > a, and Lp is totally
ordered.
5.9. Proposition: If L is a milattice in which 1 is
compact and every nonzero compact element is invertible,
then L is distributive.
Proof: Let a,b, and c e L. For each maximal p, Lp is
a vlattice by Theorem 5.8, so Lp is totally ordered, and
thus distributive. If d^ = a a (b v c) and d2 = (a a fc>) v
(a a c), then by Propositions 3.10 and 3.11, for each maximal
~ *
p of L, d^ = d2 in Lp. Thus, d^ = d2 relative to each
maximal p of L, and by Theorem 3.22, d^ = d2*G
Remark: In commutative ring theory, this is a character
ization of Prfer domains; i.e., an integral domain with
identity is a Prfer domain if and only if its lattice of
21
2
chain a< [a:b] < [a:b ] < ... This chain must break off
by the ACC; i.e., there is a n e I+ with [a:bn] = [a:b^]
for each k > n. Now, [a:bn] > a, and also since bn i a,
we have a < a v bn. Thus a < [a:bn] a (a v bn). Let
d = [a:bn] a (a v bn). Then d < [a:bn], so dbn < a. Now,
bn is multiplicative, so if r = [(d a bn):bn], then
bnr = d a bn. Thus, since d > a, we have
a v bnr = a v (d a bn) = d a (a v bn) = d, by modularity.
Hence a > bnd = bna v b2nr, so b2nr < a, so
r < [a:b2n] = [a;bn], so bnr < a, and d = a v bnr = a. Thus
a is not Aprime, and so every Aprime element is primary.
31
Now we must introduce a product on L There is a per
A A A
fectly natural way to do this, namely ab = (ab), but we
need to check that this is welldefined.
3.13. Lemma; If a and b are in L, then ab ~ a*b*.
Proof; Clearly (ab)* < (a*b*)*, so we need only show
a*b* < (ab)*. But a*b* = v{x^x2x^ and x2 are compact,
x^ ^ a*, x2 < b*} = v{x1x2x^ and x2 are compact, and there
are m^,m2 e M with m^x^ < a and m2x2 < b} < v{x^x2lx^ and x2
are compact, and for some m e M, mx^x2 < ab} < (ab)*.D
Thus if a1 = a2 and b^ = b2, then a^b^ ~ ab* = a:Â¡b* ~
A A A
a2b2, so we define ab = ab.
3.14. Theorem: If L is a milattice and M is a multi
plicative subset of L, then is also a milattice, in which
/V A A /V
1 is compact and l*a = a for each a e L.
Proof: In light of Theorem 3.12, we need only consider
the product on L^. It is both associative and commutative,
since the product on L is. Since ab < a a b for each a,b e L,
/\ A A A
we have ab < a a b by Proposition 3.10. If a e L and
{b.jji e 1} is a family in L, then let b = v b. andc = ab. Then
ie I
A A AA/\ A A A A
a( v b.) = ab = c = v ab. If x and y are compact in L,,,
6I 1 iÂ£l 1 M
5
1.2. Example: Let L be the lattice of ideals of a
ring R. L is a milattice if we take the meet to be inter
section, the join to be the sum, and multiplication to be
the ordinary product of ideals. Here, the compact elements
are the finitely generated ideals.
1.3. Example; A latticeordered group is a group G
endowed with a lattice order which is compatible with the
group operation in the sense that a(b v c) = ab v ac,
(b v c)a = ba v ca, and the dual conditions involving meets
also hold. An subgroup of G is a subgroup H which is
also a sublattice; i.e., if a and b are in H, then a v b and
a a b are also in H. An subgroup C of G is convex if
whenever a and b are in C and x e G satisfies a < x < b,
then x e C. A theorem of Bigard et al. [1], states that
the lattice C(G) of convex subgroups is completely dis
tributive, if we let the join of two convex subgroups be
the convex subgroup generated by their union. Thus, if
we let the product and meet coincide, C(G) is a milattice.
Again, the compact elements are the finitely generated ones,
and in this case, these are generated by a single element.
1.4. Definition: A lattice is said to have the
ascending chain condition, or ACC, if every ascending chain
breaks off after a finite number of steps; i.e., if
x^ < X2 x^ ..., then there is a positive integer n with
x^ = xn for all k ^ n. A result from lattice theory
CHAPTER IV
INVERTIBLE ELEMENTS
We now consider the analogue of invertible ideals of
commutative ring theory. Recall that in a commutative ring
R, if an ideal A contains a regular element (an element that
is not a zerodivisor), then A is invertible if and only if
it satisfies two properties:
(1) A[B:A] = B for each ideal B c A.
(2) [AB:A] = B for each ideal B of R.
An ideal satisfying (1) is multiplicative, while an ideal
satisfying (2) is cancellative. We have already mentioned
that every principal ideal, and thus every power of a princi
pal ideal, is multiplicative. If a principal ideal is
generated by a regular element, then the ideal is cancellative
as well, and thus is invertible. For these reasons, when we
consider properties of principal ideals or elements of the
ring, it is often enough to look at multiplicative and invert
ible compact elements. One thing should be mentioned here:
when dealing with the lattice of convex subgroups of a
latticeordered group, if a < 1, then l*a = a*a = a since
multiplication is intersection, and so one will never en
counter any elements besides 1 that could be called
cancellative or invertible in the sense of (1) and (2) above.
39
32
then there are compact elements xQ e x and yQ e y, so that
A A A A
xy = xQyQ is compact since xQy0 is compact. Since l*a < a,
A A A
we have l*a < a. But if x is compact and x < a*, then there
2 2
is a m e M with mx < a. Now, m e M and m x < ma < l*a, so
/\ A A
a* < (l*a)*, and l*a = a.D
The last part of Theorem 3.14 is an expected result if
this is to parallel commutative ring theory, for if R is a
commutative ring and if M is a multiplicative subset of R
with 04M, then R^ is a commutative ring with identity.
Now, the following result is part of an exercise in Gilmer
[3], and is easily verified.
3.15. Proposition; If S is a commutative ring, then S
has an identity if and only if SA = A for each ideal A of S
and S is finitely generated as a Smodule.
Thus, for our version of a ring with identity, we let
L have the properties that 1 is compact and l*a = a for all
elements a e L.
Now, we would like this lattice of extensions to
reflect all of the information below the set of associated
primes of M, including prime and primary elements, radicals,
and residual quotients by compact elements. First, we
consider the primes.
51
(4) : If p is unbranched, we are done. If p is
00 n
branched, Pq = A q for each pprimary q < p, so by
n=l
Theorem 5.2 part (3), p^ is prime and each prime a < p
satisfies a < pQ.D
5.5. Theorem: If L is a vlattice and p is a nonzero
prime, then the following are equivalent:
(1) p is branched.
(2) There is an element a p with rad(a) = p.
(3) There is an invertible element x with rad(x) = p.
(4) p > v{bb is prime and b < p}.
(5) There is a prime r < p with no primes a satisfying
r < a < p.
Proof: (1) => (2) : obvious.
(2) => (3) : Choose x invertible with a < x < p. This
can be done since a < p, each nonzero element of L is a join
of invertibles, and L is totally ordered. Then rad(x) = p.
(3) => (4): If x is invertible with rad(x) = p, then
x > b for each prime b < p, so x > v{bb is prime and b < p}.
But x is invertible and 1 is compact, so x is compact, and
n
so if x = v{bb is prime and b < p}, then x = v b. and
k=l K
since L is totally ordered, x = bg, a prime strictly below p,
and rad(x) p, contradicting our choice of x.
(4) ^ (5): Let r = v{bb < p and b prime}. If x and y
are compact with xy < r, then since xy is compact,
37
compact, with bxn a*, so for some meM, mbxn < a.
A A A A
itiXq = lx = x, so x < v {y I y is compact and by < a} = d, so
/A A A
by part (1), d = [a:b].
(3) For b*[a:b]* < (b[a:bj)* < a* so ([a:b] ) < [a*; b*] .
(4) As always, [a*:b*] < ([a*:b*])*. Let c = ([a*:b*])*.
Now c= v {x  x compact and xm < [a*:b*] for some me M} =
v{xx compact and b*xm < a* for some meM}. Ifxis compact
and b*xm < a* for some meM, then for any y compact with
AAA A A A
y < b*, we have yxm < a*. Thus yxm = yx < a, so yx a*,
so b*x < a*. Therefore, c = v{xx compact and b*xm < a* for
some m e M} s v{xx compact and b*x < a*}, so c = [a*:b*].
A A A A /A /\ A
(5) Let c = [a:b] = v{yby s a}. If x is compact with
/A /A /A /A /A
x < c*, then x < c, so bx < a, and b*x* < (bx)* < a*. Thus
x < x* < [a*:b*], so c* < [a*:b*]. Conversely, if x is com
/A /A /A
pact and x < [a*:b*], then b*x ^ a*, so bx i a. Thus
/A /A /A /A
x < [a:b] = c, so x < c*, and c* = [a*:b*].D
3.21. Definition: If p is prime in L and M^ = {mm is
compact and m $ p}, we say that L^ is the localization at p,
and use the notation L = L
p M
In commutative ring theory, if R is a commutative ring
with 1, then each ideal A is the intersection of the contractions
of all of its extensions in the localizations at the maximals
exceeding A. This allows one to assume a unique maximal ideal
when trying to establish an arithmetic result, since by
passing to the localization, all finite joins, meets, and
products are preserved. We obtain the same result here.
29
Proof: First, a < a* e a. for each iel, so a < a.
11 i
/\ /\
for each i e I. Suppose c < a^ for each iel. Then c < a*
A A A A
for each iel, so c < a. Thus c < a, and a = a a.. Now,
iel 1
if x is compact and x < a*, then for some m e M,
mx < a = a ai, so mx < ai for each iel. But then for
iel 1
each iel, there is m. e M with m.mx < a., since mx is
i li
compact, and so x < a for each iel since m^m e M. Hence
A /V
x a, so a = a*. Let t = r a s, and let d e r a s. By
what we have just shown, d* = r* a s*. If x is compact and
x < r* a s*, then there are elements and m2 in M with
m^x < r and n^x < s. Thus m^m2 e M with m^m2x < r a s = t,
so d* = r* a s* < t*, and since t r* a s*, we have
AAA
t* = r* a s* = d*, and t = r a s.G
A A
3.11. Lemma: For each a e L, a is compact in {bb e L} <=>
/\
there is an element a^ e a with ap compact in L.
Proof: If aQ
By Proposition 3.10
n
compact, a < a c*
U k=l 1
is compact in L, then suppose a
, aQ < a* < a (ct), so since a
iel 1
, which we will label c. Then
k
0
0
<
A C
iel
i *
is
A
<
A
c
n ~ ^
ac. so a. is compact in {bb e L}.
k=l 1k U
Conversely, if a is compact in {b j b e L}, then since L
is algebraic, a = v{xx is compact and x < a}. By Proposition
^ ^ ^ n ~
3.9, a = v{xx is compact and x < a}. Thus a = v x, where
k=l K
20
Proof: Let S = {a e l a has no decomposition as
n
a = a a., with each a. an Aprime}. The lemma states that
i=l 1 1
S = ; we suppose not. By the ACC, there is an element a
which is maximal with respect to S. But then a = b a c,
with a < b and a < c, and so b H and c S. Thus
m n
a= (Ab.) a (ac.), where each b. and each c. is Aprime.
i=l 1 j=l ^ i D
This is impossible, and so S = <}>.
2.14. Theorem: Let L be a modular, commutative mi
lattice in which each element of L is the join of a set of
powermultiplicative elements. If L has ACC, then each
element of L has a normal primary decomposition, which is
unique up to order and the set of radicals of the primaries.
Proof: In light of Lemma 2.13, Theorem 2.12, and the
fact that every primary decomposition has a normal refinement,
we need only show that if a is Aprime, then a is primary.
Suppose a e L is not primary. Then there are compact elements
b and c with be < a but c i a and bn Â£ a for each n e Z+.
We may assume that b is powermultiplicative, for b = v b.,
iel 1
where each b^ is compact and powermultiplicative, and since
, m n +
b is compact, b = v b. Since b Â£ a for each n e Z ,
k=l xk
there must be one of these b. 's, which we denote bn, such
1k U
that bg Â£ a for each n e Z+. Then b^c < be < a, and we may
replace b with bQ. Now, be < a, so[a:b]>avc>a. Also,
[a:bn] [[a:bn]:b] = [a:bn+^], so we have the ascending
57
ideals is distributive (Gilmer [3]). In our context, we
must leave it as an open question whether the converse of
Proposition 5.9 is true.
The method of proof used in Proposition 5.9 will be
used in the next theorem as well. That is, we assume we are
looking at the localization at a maximal of L, so that L has
a unique maximal, and show the arithmetic relations under
this condition. Again, because lattices of extensions preserve
all of the arithmetic relations we consider, as well as
compactness and invertibility, we can then generalize.
5.10. Theorem: Let L be a domain lattice. Then the
following conditions on L are equivalent:
(1) Every nonzero compact element of L is invertible.
(2) For all a,b,c e L, a(b a c) = ab a ac.
(3) For all compact elements a,b,c e L, a(b a c) =
ab a ac.
(4) For all a,b e L, (a v b)(a a b) = ab.
(5) For all compact elements a,b e L, (a v b)(a a b) = ab.
(6) If a,b e l and c is compact in L, then
[(a v b):c] = [a:c] v [b:c].
(7) If a,b,c are compact in L, then
C(a v b):c] = [a:c] v [b:c].
(8) If a,b are compact in L, [a:b] v [b:a] = 1.
(9) If a,b are invertible in L (and so compact, by
Proposition 4.9), [a:b] v [b:a] = 1.
66
where k = min{k.
v (pk ,mÂ£ } e F
iel i i
i e 1} and Â£ = min{Â£.jjk^ = k}. In particular,
so each element of Lq is compact. Also,
.\ =
lei l l
0, if {k.Ii e 1} is unbounded
i1
(Pi i fin.) if k = maxk. I i e 1} and
^k+l 0 i1
{Â£.k. = k} is unbounded
i1 i
(p^,m^), if k = max(b^i e 1} and
Â£ = max(Â£^k^ = k}.
Thus, Lg is an algebraic lattice, with 1 = (p^m^). Define
a product on Lq by setting 0*a = a*0 = 0 for each a e LQ,
and (pm ,m^)(pk ,m ) = (Pk+2'm ) It is easily seen
that this product makes Lq a commutative milattice, with 1
2
compact and l*a = a for each a e LQ. In particular, 1 = 1,
so that (pQ,m^), the unique maximal of LQ, is prime. The
element (p^'^q) is also prime, for if a > (p^itIq) and
b > (p n^) then a = (pn,m0 ) and b = (pn,ma ), so that
0' Â£
ab = (pn,m
0,iÂ£1+Â£2) > (P^itIq) Now, if q > (p^,mQ) then either
q = 1, which is not prime, or q = (pQ,m^) for some Â£ > 1, in
which case q = (p0,m^) and q is not prime unless Â£ = 1.
If 0 < q < (p^,m0), then q = (pk,m^), with k > 1 and with
Â£ > 1 if k = 1. Thus, q = (p^itIq) (Pk_imÂ£) / and <3 is not
prime. Therefore the only primes of Lq are 0, (pQ,m^), and
(P^itIq) and each element a of LQ with 0 < a < 1 can be
uniquely expressed as a product of a finite number of primes.
Also, (p^,mQ) is multiplicative, for if a < (p^,^)/ then either
= (Pk'mÂ£) with k > 1, so a = OCp^itIq) or
a = 0 or a
INTRODUCTION
This dissertation is an attempt to place the results
of multiplicative ideal theory in a latticetheoretic
setting which will include other lattices of subobjects
as well. The two motivating examples we will keep in mind
are the lattice of ideals of a commutative ring and, to a
lesser extent, the lattice of convex subgroups of a
latticeordered group.
We consider an algebraic lattice L endowed with a
multiplication which distributes over all joins, respects
compactness, and satisfies the ideal multiplication property
that ab < a a b. A major work in this area is that of
Keimel [5]. He indicates that this approach applies to a
wide class of known examples, including the lattices of
ideals of commutative semigroups, distributive lattices,
and frings. He has developed the theory of minimal primes
to a great extent, and so in Chapter I, after we introduce
the basic concepts, we list his major results, and then turn
to other classical results of commutative ring theory. Our
terminology differs slightly from that of Keimel, and
beginning in Chapter II we assume that our multiplication
is associative and commutative, as it is in our two models.
Modularity is only assumed for one result, and we do not
know that it is necessary there.
1
54
then a = pa and since a ^ a > 0, a is compact and
n c n n n c
therefore invertible. Thus, p = 1, which is impossible.
2
Therefore, an ^ p, so an = 1, and we have a = pa^ = p =
n n
. . = p a = p .
(2) => (1) : This is clear.
(2) => (3) : This is clear.
(3) => (2) : In the proof of Theorem 5.5, when we proved
that condition (5) implies condition (1), we actually showed
that in a vlattice with unique maximal p, any invertible
element with radical equal to p is pprimary. Now, every
invertible element a with 0 < a < 1 has rad(a) = p, since
p is the only nonzero prime. Therefore, all invertible
elements below 1 are pprimary, and since L is discrete, they
are all powers of p. Now, if b is any element p of L with
0 < b < 1, then b is the join of a set of invertible elements,
since L is a vlattice. Thus, b is the join of a set of
powers of p, so if n is the least positive integer that
occurs as an exponent in this set of powers of p, then
b = pn.D
We now turn our attention to the analogue of Prfer
domains. These are integral domains with identity in which
every nonzero finitely generated ideal is invertible, so we
consider milattices with 1 compact in which each nonzero
compact element is invertible. We will proceed toward the
many arithmetic relations that characterize Prfer domains
in commutative ring theory, but first we consider localiza
tions in such a milattice.
26
than a (M). Further, we may always assume that we are using
a saturated multiplicative subset. This has no effect upon
our results, but it is convenient, and by Lemma 3.4, it is
justified.
3.5. Definition: If M is a multiplicative subset of L,
we define an equivalence relation on L by:
a ~ b a* = b*.
A
(If it is necessary to refer to M, write a b.) Let a
A
represent the equivalence class of a under ~((a)^, if
A
necessary). We will say a is the extension of a, and a* is
A
the contraction of a.
A A
3.6. Proposition: a* = v{bb e a}, a* e a, and
(a*)* = a*.
Proof: Since b < b* for all be L, we have
a* > v{bb e a}, and a* < (a*)*. Let x be compact with
x < (a*)*. Then there is an m^ e M with rn^x < a*. But m1 is
compact, so m^x is compact, and there must be a e M with
it^m^x < a. m^n^ e M, so x < a* and (a*)* = a*. Thus
A /V
a* ~ a, so a* e a, and a* = v{bb e a}.G
3.7. Definition: We now want to create a new milattice
with these equivalence classes. To this end, we set
/N A
a < b a* < b*.
64
p < 1, so q. = p. Thus p is a factor of a, so p is invert
10
ible since a is. Therefore each maximal of L is invertible,
so that each nonzero element of L is a product of invertible
elements, and so is invertible.
(1) => (4): By Theorem 5.8, since each nonzero compact
element of L is invertible, then for each maximal p e L, L
* p
is a vlattice. Since 1 is compact in L, each element of L
is compact by Proposition 4.9, so L has ACC, and by Lemma
3.11, for each maximal p e L, has ACC as well. Let a e L
with a > 0. If a = 1, then the set of maximals above a is
empty, and is therefore finite. If a < 1, then by the
n
equivalence of (1) and (3) of this theorem, a = II p., where
i=l 1
each p^ is a maximal of L, and this representation as a
product of maximals is unique. But if p is any maximal of L
with a < p, then a = p[a:p] since p is invertible. Thus if
[a:p] = 1, then p = a and p is the only maximal exceeding a,
m
while if [a:p] < 1, then [a:p] ^ a > 0, so [a:p] = II q. ,
:=i 3
m
where each q. is a maximal of L, and a = p ( II q.) By the
5 j=l 3
uniqueness of the representation, p = p^ for some i < n, and
so the only maximals above a are those in ip^, .. ,pn> .
(4) ==> (5) : We only have to show that every element of
L is compact. Let a e L with 0 < a < 1, and let ^P]_fP2 Pn^
be the set of maximals exceeding a. For each positive integer
i < n and for each b e L, let (b)^ be the equivalence class
in L containing b.
For each i < n, (a)^ is compact since
13
(h) ( a b.)( v [a:b.]) = v (( a b.)[a:b.]) <
iel jel 3 j el iel 1 3
v (b.[a:b.]) < a.
jel 3 3
(i) Since b( v [a.:b]) = v (b[a.:bl) < v a., we have
iel 1 iel 1 iel
v [a.:b] < [a.:b] < [( v a.):b].D
iel 1 1 iel 1
2.3. Definition: An element a e L is multiplicative if
a[b:a] = b for each b < a; i.e., if a divides every element
below it. Note that if a is multiplicative, then a*l = a.
An element a e L is said to be powermultiplicative
if a11 is multiplicative for each n e Z+.
Considering our two examples of milattices, if R is a
commutative ring, all principal ideals are multiplicative,
and in fact powermultiplicative. If G is a latticeordered
group, then since multiplication is the meet operation,
every convex subgroup is multiplicative, and hence power
multiplicative. Thus, in the two underlying models, each
element of the lattice is the join of a set of compact,
powermultiplicative elements. For this reason, we will
assume this property to obtain one nice result later in
this section, Theorem 2.14, without losing touch with our
motivations.
The first question that arises is: When is the product
of two multiplicative elements again a multiplicative element?
The answer does not seem to be that it is always true, but a
partial answer is offered.
PAGE 1
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PAGE 80
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34
3.17, Theorem: The primes of are those classes p
where p is prime in L and p < p, for some p. in the set
A0 A0
{p^ X e A} of primes associated with M. If p is prime and
p < p^ for some X e A, then p = p* and there are no other
A
primes of L in p.
Proof; Suppose a is prime in L^. Then a < 1, so
A /\
a p, for some Xn e A, and a* < p? = p, If be < a*,
A0 0 A0 A0
A A A A A /\ /\
then be < a, so b < a or c < a, and so a* is prime since
A
b < a* or c < a*. If q e a with q prime in L, and if x is
compact with x < a*, then for some m e M, mx < q. But
q < a* < p. so m $ q, and so x < q. Thus q = a*, and a*
A0
A
is the only prime in a. On the other hand, suppose p is
A A A
prime in L, with p < p^ for some XeA. If ab = p, then
ab < p*. Suppose b 1 p*. Then there is a x compact with
x Â£ p* but x < b. For each y compact with y < a, xy is
compact and xy < p*, so there is an element m e M with
mxy < p. Now, p < p^ so m 1 p, and so xy < p. But x i p*,
so y < p^ and so a < p. Thus a < p, so p is prime, and by the
A
first part of the proof, p = p* is the only prime in p.D
The theorem establishes a onetoone correspondence
which preserves order between primes of and primes of L
that do not exceed any elements of M. We now want to do the
same for primary elements, but first we consider radicals.
53
A A A /\ A A 2 A 2
Therefore, y > p, so y = 1, and x < (a) Hence (a; is
/v /v 2 ^ 2 *
pprimary in L with (a; p, and by Theorem 3.19, (a;
P
2 *
is pprimary with (a ) < p.D
5.6. Definition: A vlattice L is said to be a discrete
vlattice if each primary element is a power of its radical.
This definition agrees with the corresponding definition in
commutative ring theory (Gilmer [3]), and as in that case, a
result of Theorem 5.4, part (2), is that a vlattice L is
discrete if and only if no branched prime p is idempotent
2
(that is, p < p for all branched primes).
5.7. Proposition; Let L be a vlattice with unique
maximal p > 0. The following conditions are equivalent:
(1) L has ACC.
(2) Every element a with 0 < a < 1 is a power of p.
(3) L is discrete, and p is the only nonzero prime.
Proof: (1) => (2) : Let a e L with 0 < a < 1. Since L
has ACC, p is compact, and by Theorem 5.2, part (1), p is
invertible. Since a < p, a = p[a:p]. Let a^ = [a:p].
Recursively, let an+^ = [an:p] for each n e Z + We have the
ascending chain
a = a,
al *
a2 a3 "
By the ACC, there is a least integer n such that a = a, for
n K
all integers k > n. Thus an = an+1 = [an:p]. If an < p,
6
(Birkhoff [2]) is that the following conditions are equiva
lent for an algebraic lattice L:
(a) L has ACC.
(b) Each element of L is compact.
(c) Each nonempty subset of L has a maximal element.
In both Examples 1.2 and 1.3, there is a notion of a
prime element. In Example 1.2, recall that a prime ideal
is a proper ideal P of a ring R satisfying the condition that
if A and B are ideals and AB Â£ P, then A Â£ P or B Â£ P. in
Example 1.3, a prime convex subgroup is a proper convex
subgroup P such that if A and B are convex subgroups of
G and if A n B c p, then A Â£ P or B Â£ P. These two notions
of primes motivate the following definition.
1.5. Definition; If L is a milattice and p e L, then
p is prime if p < 1 and whenever ab < p, we must have a < p
or b < p. Notice that if we want to show that an element is
prime, it is enough to show that if a and b are compact and
ab < p, then a < p or b < p. Conversely, if p is prime,
then certainly the definition of a prime applies when a and
b are compact elements, so that we may assume a and b are
compact in the definition.
An element m of a lattice L with 1 is a maximal if it
is maximal in the set of elements strictly below 1; i.e., if
m < a < 1, then m = a. Recall that in a commutative ring
with identity, every proper ideal is contained in a maximal
ideal, and each maximal ideal is prime. In our context, we
have the next three results.
67
a = (p^^m^) (paying) Also, (p^/irig) is multiplicative, since
(p^ntg) (pk^,mÂ£ ) (plfm0) (pk ,mÂ£ ) if kL k2 or 1 2
Thus, (p^,irig) is invertible, and so (pk,mQ) is invertible
for each integer k > 1. Therefore each nonzero element of
Lg exceeds an invertible element. However, (p^m^) is not
invertible, since there is no a e Lg with a(pg,m^) = (p^,nig),
and so (pg,m^) is not multiplicative. Hence, not every
nonzero element of LQ is invertible.
6.5. Example: It is not even enough to assume that in
L, each nonzero element is a join of invertibles. To see
this, let P, M, and Lg be the same as in Example 6.4, and
let X be a countably infinite set with XnP=XnM=
X n Lg = , and X = (xnn is a nonnegative integer} with
x x if n. n0. Let L = (Ln\{0}) {x }, identifying
1I2 l z no n
00
(pk,m) with (pk,m^,Xg), and set L = {0} u ( u L ). For a
n=0
partial order, set 0 < a for each a e L, and (p, ,m0 ,x ) <
K1 1 nl
(pk ,m^ ,xr ) if n^ < n^ and k2 < k^, or if n2 < n^ and
k2 = k^ and n2 + &2 < n^ + This is a lattice order on L
which extends the order on Lg, and L is an algebraic lattice
with ACC. For a product, set 0*a = a*0 = 0 for each a e L,
and (P^/m^x^) (p^m^x^) = ^kj+k^^+A^nj+n^ Then
L is a milattice, 1 = (pg,mg,Xg), 1 is compact and l*a = a
for all a e L. Also, (pg,m0,Xg) is the unique maximal of L, and
2
since 1 =1, (pg,m^,Xg) is prime. The only primes in L are
CHAPTER III
MULTIPLICATIVE SUBSETS, LATTICES OF
EXTENSIONS AND LOCALIZATIONS
We now turn to the idea of localizations, which play an
important role in valuation rings and Priifer domains, as well
as in algebraic geometry. A localization is a special type
of the more general object known as a ring of quotients. The
lattice of extended ideals in a ring of quotients preserves
most of the essential information below a given set of primes
in the lattice of ideals of the original ring, including
prime and primary elements, joins, meets, products, and
compactness. In a commutative ring with identity, each ideal
A is the intersection of the set {A,J X e A}, where A^ is the
contraction of the extension of A relative to the maximal
ideal (A is the indexing set for the set of maximal
ideals), and it is this fact that makes much of the ideal
theory of valuation rings and Priifer domains work.
3.1. Definition: A nonempty subset M of a milattice
L is called a multiplicative subset of L if each m e M is
compact, 0  M, and whenever m^ and m2 e M, we must have
m^n^ ^ M.
One way of constructing a multiplicative subset of a
milattice L is to take {p^X e A}, a nonempty set of
pairwise incomparable primes, and take a subset MQ of
22
35
3.13. Proposition; If a e L, and if r = rad(a), then
A A
r = rad(a) and r* = rad(a*).
Proof: First, r = v{x x is compact and xn < a for some n e Z + } .
If x is compact and xn < a, then x is compact and (x)n < a.
^ n A
Conversely, if (x) < a with x compact, there is Xg e x with
xQ compact in L. Now, Xg < a*, so sincex^ is compact, there is
a m Â£ M with mx < a. Thus (mXg)n < mxn < a, mxQ is compact,
and mXg = lxQ = Xg = x, so r = v{xjx is compact and xn < a
for ne Z } = v{xx is compact and (x) < a for some
n e Z } = rad(a). If x is compact and x < r*, then there is
m e M with mx < r, so for some n e Z+, (mx)n < a, and
(mx) < a. But mx = lx = x, so (x)n < a, and xn < a*, with
x compact. Thus r* < rad(a*), but if rad(a*) = s, then
A A A A
s = rad(a*) = rad a = r, so r* = rad(a*).D
3.19. Theorem; Let M be a multiplicative subset of L
with {pjJ A e A} the set of associated primes of M. Let p be
a prime in L, with p < p^ for some A e A. If q is pprimary,
A A /V
then q is pprimary in L^ with q = q* the only primary in q,
for p or any other prime. Conversely, if a is pprimary,
v
then a* is p*primary, and a* is the only primary in a.
Proof: Suppose a is pprimary in L^, with p prime in L
A A
and p < p^ for some A e A. Then p = rad (a) and p = p* =
(rad(a))* = rad(a*). If x and y are compact with xy < a*,
/s /V /s /S /s /S A n A
then x and y are compact and xy a, so x a or (y) < a
24
and we say that a multiplicative subset M of L is saturated
if M = M Clearly, M is saturated, and it contains every
s s
multiplicative subset of L which has {p^  A e A} as its set
of associated primes.
In commutative ring theory, if R is a commutative ring
and M c r is a multiplicative set not containing zero, then
when the ring of quotients is formed, all ideals of are
extended ideals; i.e., they are all ideals generated by the
subset of R^ corresponding to an ideal of R under the
canonical homomorphism $ of Rinto R^. Also, each ideal of
R^, meets the image of R in an ideal of that subring, and the
inverse image of these intersections are called contracted
ideals. This is all put forth in Gilmer [3], and it is
shown that all extended ideals are of the form
Be = {$(b)/$(m) b e B and m e M}
while all contracted ideals are of the form
pp .
B ={xeRxmeB for some me M}.
With this in mind as a motivation, we proceed to our next
definition.
3.2. Definition; Let M be a multiplicative subset of L,
and let  A e A} be its set of associated primes. If
a e L, then set
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
This dissertation was submitted to the Graduate Faculty of
the Department of Mathematics in the College of Liberal Arts
and Sciences and to the Graduate Council, and was accepted
as partial fulfillment of the requirements for the degree
of Doctor of Philosophy.
August 1982
Dean, Graduate School
18
integer i., 1 < i. < n, with r. < q. In light of
3 3 3 1 j
n
Proposition 2.9, if a = a q. is any primary decomposition
i=l 1
of a, we can group together those q^'s that have the same
radical, and their meet is again a primary with the same
m
radical, so we obtain a = a r., where if j, j, then
j=l 3 1
rad(r. ) *rad(r. ), and each r. < q. for some i. Now, if
3l 3 2 ^ 4i
for some jA, l a r., then we can exclude
0 J 0 .. v
0 J
r. from the decomposition, and so we see that every primary
II0
decomposition has a normal refinement.
m n
2.12. Theorem: If a = a q. and a = a r. are two normal
i=l 1 jl ^
primary decompositions for a e L, where each q^ is p^primary
and each r^ is t^primary, then m = n and, after suitable
rearrangement, p^ = t^ for each i (1 < i < m).
Proof: If a = 1, then m = n = 0, and we are done. If
a < 1, then from {p^,...,pm} u {t^,...,tn}, we choose an
element not strictly below any other. We may assume without
loss of generality that after rearranging terms of the
decompositions, we have chosen pm Suppose pm 4 {t^,...,tn).
Then q^ Â£ t. for any 1 < j < n. By Lemma 2.7, [r.:q ] = r.
ml 1 3 3
n n
for each l
m jl 3 m jl 3
Also, p^ i p. if 1 < i < m1, so q^ Â£ p^ for each 1 < i < m1,
and [q:q] = q. for 1 < i < m1. But [q :q ] = 1, so
i m ^m ^m
47
x
O
FIGURE
.11
1.
P
q
2
P
pq
2
q
3
P
p2q
pq2
3
q
4
p
p~3q
2 2
p q
3
pq
4
q
p
1
p
2
P
3
P
r = a p
n=l
n
= pr
q
pq
2
p q
3
p q
qr
2
q
2
pq
2 2
p q
3 3
p q
a qn = a = O
n=l
n=l
FIGURE 4.12
2
q r
I
a qn = O <
n=l
FIGURE 4.13
ACKNOWLEDGEMENTS
The author would like to thank his Supervisory
Committee Chairman Jorge Martinez in particular, and all
of the members of his supervisory committee for their
helpful comments, criticisms, and suggestions. He also
would like to acknowledge the cooperation of his typist,
Sharon Bullivant, who did such quality work in such a short
time.
11
28
3.
let a =
9.
v
iel
Proposition;
a.. Then a =
i
If (a^i e 1} is any family in L,
A
v a., and so ( v a.)* > v (a*),
iel 1 iel 1 iel 1
Proof: First, since a s for each i e I, we have
A A A A
a s for each iel. Suppose c Â£ a^ for each iel. Then
A
for each iel, there is b. e a. with c s b. Thus
' li i
A A
v b. c, so if b = vb., b < c, and we need to show that
iel x iel 1
A A
b = a. If x is compact and x < a*, then there is m e M with
mx < a. Thus mx < v{y compactly a^ for some i e I}, so
n
since mx is compact, mx < v yw yv compact and for some
k=l K K
i^ e I, y^ < a^ < a* For each k < n, choose m, e M with
xk 1k
n
m^y^. < b^ Then (mm^n^.. .mn) M and mm^...mnx < v y^m^ <
k k=1
n
vb. s vb., so x < b*, and a* < b*. A similar argument
k=l 1k iel 1
AAA
shows that b* < a*, so a = b c.D
Thus joins are preserved going from L to {aa e L}, and
A
{aa e L} is closed under (finite and infinite) joins. The
next result yields the fact that we have constructed a
complete lattice.
3.10. Proposition: If (a^Ji e 1} is a family in L, let
A A
a = a (a*). Then a = a a., and a = a*, so (vv e L} is
iel iel 1
complete. Further, if r and s are in L, with t = r a s,
AAA
then t = r a s and t* =
r* a s*.
9
1.11. Proposition: If L is a milattice and a e L,
then rad(a) = v{xx is compact and xn < a for some positive
integer n}.
Proof; We show that {xx is compact and x < rad(a)} =
{xx is compact and for some positive integer n, xn < a}.
First, if x is compact and xn < a, then x < p for each prime
p > a, so {x compactxn < a for some n} c {x compactx <
rad(a)}. But if x is compact and if, for each n e Z+,
xn i a, then {xnn si} is a multiplicatively closed set of
nonzero compact elements, and so if S = (b e Lb > a .and
xn $ b for all n e Z+}, then S has a maximal element p which
is prime, by Lemma 1.10. Hence Pg > a and x i pg, so
{x compactx < rad(a)} = (x compact for some n s 1, xn < a} .
Keimel [5] defines a msemiprime element of a milattice
L to be an element x such that for all t e L, t x implies
2
t Â£ x. He shows that each msemiprime element in a mi
lattice is the meet of the set of prime elements exceeding
it (Theorem A), and so these are the elements that occur as
radicals in milattices. As a corollary, he shows in a
milattice L that the following conditions are equivalent:
(1) Each element of L is the meet of primes, so that
rad(a) = a for all a e L.
(2) Each a e L is semiprime.
2
(3) For each a e L, a = a.
(4) For all a,b e L, ab = a a b and L is distributive.
CHAPTER VI
GENERALIZATION OF DEDEKIND DOMAINS
We now consider milattices in which every nonzero
element is invertible, and 1 is compact.
First, in light of Proposition 4.9, every element of L
is compact, and so L has ACC. Also, l*a = a for each a e L,
and 0 is prime, and L satisfies all of the conditions of
Theorems 5.6, 5.7 and 5.8.
6.1. Proposition; If every nonzero element of L is
invertible, then every nonzero prime is maximal.
Proof; Suppose 0 < p < q < 1. Then q[p:q] = p, so if
p is prime, then because q > p, Cp:q] ^ p ^ [p:q], and so
P = [p:q]. Thus pq = p, and since p is cancellative q = l.D
6.2. Proposition: In a milattice L, if x is invertible
n
and x = II p. where each p. is prime in L, then this repre
i=l 1 1
sentation is unique up to the order of the primes.
n m
Proof: Let x be invertible, with x = IT p. = H q. ,
i=l 1 j=l 3
where each p^ and each q^ is prime. We show that m = n,
and after rearranging, p^ = q^ for each i
60
BIBLIOGRAPHY
[1] Bigard, A., Keimel, K. and Wolfenstein, S., Groupes
et Anneaux Reticules, SpringerVerlag, Berlin (1977).
[2] Birkhoff, G., Lattice Theory, 3rd Edition, Amer. Math.
Soc. Colloq. Publ., vol. XXV, Providence (1967).
[3] Gilmer, R., Multiplicative Ideal Theory, Marcel Dekker,
Inc., New York (1972) .
[4] Hungerford, T. W., Algebra, Holt, Rinehartand Winston,
Inc., New York (1974) .
[5] Keimel, K., "A Unified Theory of Minimal Prime Ideals,"
Acta Math. Acad. Sci. Hung., 23 (1972), pp. 5169.
72
55
5.8. Theorem: Let L be a domain lattice. Let
{p^li e 1} be the set of all maximals in L. Then the fol
lowing are equivalent:
(1) Lp is a vlattice for each prime p in L.
(2) Lp^ is a vlattice, for each i e I.
(3) Every nonzero compact element is invertible.
(4) If a and b are invertible, then a v b is invertible.
2
Proof: (1) => (2) : 1 = 1, so each p^ is prime.
(2) (3): Let x be compact, x > 0. Then if (x)^ is
A
the class of x in the localization at p^, (x)^ is compact in
 /s
Lp^, and by Theorem 5.2, part (1), (x)^ is invertible in Lp^.
Now, let y be invertible, with y < x in L. Since x is com
A A A
pact, if z = [y:x], then (z)i = C(y)i:(x)i], and so
A A A
(x)^(z)^ = (y)^, for each i e I. Thus, if a* represents the
A
maximum element of (a)i for each a e L, then yi = (xz)i for
each i e I. By Theorem 3.22, y = xz, and since y is invert
ible, x is invertible by Proposition 4.4.
(3) => (4) : If a and b are invertible, then since 1 is
compact, a and b must be compact by Proposition 4.9. Thus
a v b is compact, and a v b > a 0, so a v b 0. Hence
a v b is invertible.
(4) => (1) : Let p be prime in L. In Lp, 1 is compact,
and in proving (5) => (1) in Theorem 5.5, we showed that if
L is any milattice in which 1 is compact, then if x is
A
invertible, we must have x invertible in Lp for any prime p
of L. Thus each nonzero element of Lp is the join of a set
38
3.22. Theorem: If L is a milattice with 1 compact and
l*a = a for each a e L, then let {p^ i e 1} be the set of
A
maximals of L. For a e L, let (a)^ be the class of a in
A
L and let a. = v{bb e (a).} = a* with respect to L
p i x p.
Let J = {j e la < p.}. Then a = a a. = a a..
3 iel 1 jeJ 3
Proof: Since a < a. for each i < I, a < a a. < Aa..
i T i T i
le I jeJ J
Let x be compact with x < a a.. For each j e J, x a., so
jeJ 3 3
there is a compact element m. with m. i p. and m.x < a.
3 3 3 3
Then [a:x] > m^, so for each j e J, [a:x] i p^. Since 1 is
compact and a < [a:x], [a:x] = 1. Thus x = l*x < a, so
a a. < a.D
jeJ 3
17
[q:a] < p. If x is compact and x < p, then xn < q < [q:a]
for some n Â£ Z+. If x and y are compact and xy < [q:a],
with y Â£ p, then for each z compact with z < a,
(zx)y < axy < q, so zx < q. Thus ax < q, and x < [q:a].
By Lemma 2.8, [q:a] is pprimary.D
2.11. Proposition: If {p.l
primes in L, and if for each integer i between 1 and n, q
n
is a p.primary, and if a = a q., then each prime p > a
1 i=l 1
n
exceeds some p. (1 < i < n), and so rad(a) = a p..
1 i=l 1
n n
Proof: First, II q. < a q. < p, so for some i with
i=l 1 i=l 1
l
exceeding q^} < p.D
We will say that an element a has a primary decomposition
n
if a = a q., where each q. is primary. For each i between
i=l 1 1
1 and n, let p^ = rad(q^). If a has a primary decomposition
n
a = a q., and if for each i and j with 1 < i < n and
i=l 1
1 < j < n, i j implies p^ p^, and if for each i with
n
1 < i < n, q. i a q, then we call a = a q. a normal
1 k*i i=l 1
primary decomposition of a. A refinement of a primary
n
decomposition a = a q. is a primary decomposition
i=l X
n
a = a r., where for each j with 1 < j < m, there is an
j=l 3
3
As general references for the multiplicative ideal
theory of commutative rings, we suggest Gilmer [3] or
Hungerford [4]. For the theory of lattices, we refer to
Birkhoff [2], while for the theory of latticeordered
groups, we use Bigard et al. [1], which is written in
French.
Concerning notation, we will use Z+ to stand for the
set of natural numbers, and R to stand for the set of real
numbers.
10
(5) For all a,b e L, ab = a a b and L is completely
distributive.
(6) For all a,b e L, ab=aAb and L is Brouwerian.
In ring theory, these conditions on ideals of a commutative
ring R are equivalent to R being von Neumann regular, by
a result of Kaplansky. Keimel then looks at semiprime mi
lattices, which are milattices in which 0 is semiprime.
He defines the pseudocomplement of an element, and mpre
filters and mfilters, and uses these and the GratzerSchmidt
isomorphisms between elements of L and ideals of the set of
compact elements of L to set up a bijection between maximal
mfilters and minimal primes of a milattice L which is
semiprime and satisfies the property that for all r, s, and
t Â£ L, rAs=rAt=0 implies (r v s) a t = 0. This gives
him several characterizations of a semiprime pseudocomple
mented lattice. He goes on to look at the set of minimal
primes and develops the hullkernel topology, showing that
it gives a completely regular space, with a basis of closed
andopen sets determined by the compact elements of L. He
finishes with a discusssion of zelements, which area gener
alization of zsubgroups of a latticeordered group, and uses
them to describe certain classes of milattices, including
those in which the space of minimal primes is compact. Keimel
actually does all this in a more general setting, frequently
not requiring that the product of compact elements be compact,
or that the product distribute over joins, or that it be
associative.
42
4.5. Proposition: If any a e L is invertible, then for
each b e L, l*b = b, so that 1 is invertible.
Proof; Let a be invertible. Then a is multiplicative,
so a = a[a:a] = a*l. Then b*l = [(b*l*a):a] = [b*a:a] = b.D
4.6. Proposition: If a is invertible, then
a( a b.) = a ab..
iel iel
Proof: First, a b. = a [b.a:a] = [( a b.a):a] by
iel 1 iel 1 iel 1
by Proposition 2.2, part (f), so a( a b.) = a[( a b.a):a] =
iel 1 iel 1
a (b.a) since a b.a < a.D
iel 1 iel 1
This result states that if a is invertible in L, then
multiplication by a is a lattice isomorphism of L onto [0,a]
which preserves meets and joins of infinite as well as finite
families.
4.7. Proposition: If a is invertible and a11 = a for
some n > 2, then a = 1.
Proof: First, l*a = a = a = (an ;a, and a is cancel
n_ ^ n
lative, so a =1. Then n > 2 implies a > a11 = 1, so
a = l.D
CHAPTER I
PRELIMINARIES
Let L be a complete lattice with minimum element 0 and
maximum element 1*0. An element c of L is said to be
n
compact if whenever c < v a. it must be that c < v a. .
iel 1 k=l 1k
0 is automatically compact, and the join of any finite set
of compact elements is again compact.
A lattice L is algebraic if it is complete and every
element of L is the join of a (possibly infinite) set of
compact elements. In an algebraic lattice, an element c is
compact if and only if whenever c = v a. it must be that
n 1
c = v a. .
k=l xk
1.1. Definition: A lattice L is a multiplicative
ideal lattice, or milattice, if L is an algebraic lattice
endowed with an associative binary operation (a,b) * ab
that satisfies the following conditions:
(1) ab < a a b for all a,b e L.
(2) a( v b.) = v (ab.) and ( v b.)a = v (b.a) for
iel iel 1 iel iel
all a e L and for all families {b^i e 1} in L.
(3) If x and y are compact in L, then xy is compact
in L.
4
45
We give examples to show that no other implications hold
in general, even if we assume ACC and modularity.
(1) (2) : L = {0,x,l}, 0 < x < 1, with xn = x for
all n e Z+ and l*a = a for all a e L. Here, 0 is prime,
since if a > 0 and b > 0, then a s x and b > x,, so
2
ab Â£ x = x > 0. However, 0 is never cancellative, and
since x*x = x*l, x is not cancellative, so x does not exceed
a cancellative element. See Figure 4.11.
(2) ;<6>(3): L= {0} u {pnqmm,n s 0} where p^ = = 1,
l*x = x for all xeL, 0= Apn= a qm, and pnqm ^ prqS ^
n=l m=l
m + n
Here, l,p, and q are cancellative, and so by Lemma 4.3,
every nonzero element of L is cancellative. However,
2 2 2
p[q:p] = p q, and q[p :q] = pq < p so neither p nor q is
invertible, and so no element below 1 can be invertible, in
light of Proposition 4.4. See Figure 4.12.
(3) p (4) : Let = {p^k e Z, k s 0} u {r}, with
Pk pÂ£ if k Â£, and p^ r for all k e Z, k Â£ 0. Order
oo
by setting p, > pÂ£ if k < Â£, with r = A p.. Set Pk*PÂ£ = Pk+Â£/
k=l
2
and Pk*r = rp^ = r = r. Let Q2 = {qkk e Z, k Â£ 0} u {s},
OO
with order given by q, > q, if k < Â£, and s = a q Set
K 36 k=0 K
qk*qÂ£ = qk+Â£ and qks = s*qk = s2 = s. Then
^1^ fi2 = i(a'b)la e b e } with (a1,b1) < (a2,b2) >
(bi < b2) or (bj^ = b2 and aj^ < a2) and (a^b^ (a2,b2) =
(aia2,bib2^* Let L = ^(p0,s)'(Po'qo)^ = i(a/b) e ^ n2
(a,b) > (Pg,s)}, with the inherited order and
49
(2): Rad(a) = A{pp is prime and a < p}. But
{pp is prime and a < p} is totally ordered, so by 1.7 of
Keimel [5], rad(a) is prime.
(3): Let x and y be compact elements with x  p^ and
y $ Pq. Then for some m and n e Z + x i am and y $ a11. Thus
am < x and a11 < y, and since y is compact and nonzero, it
is invertible by (1). Hence amy < xy, and am+n < amy, so
am+n < Xy, and xy $ Pq. Therefore, pQ is prime. Now, if
k k+1 k
a = a for some k > 1, then Pq = a so that a = pQ and
a is prime.
(4): If p is prime and p < a, then p < a for each
1 T t
k e Z so p < pQ.
(5): If for each k e Z+, b Â£ a^, then b < a^ for each
4" ^ k
k e Z so b < pn = a q f and rad(b) < p < a. Thus
U k=l u
a i rad(b).
5.3. Definition: If L is a milattice and p is a prime,
then p is a branched prime if there is a pprimary q p. We
say p is unbranched if p is the only pprimary in L. As for
valuation rings we have:
5.4. Theorem: If L is a vlattice and p is a nonzero
prime of L, then let S = {q^X e A} be the set of pprimaries.
(1) If q is pprimary and if x is compact with x $ p,
then q = q*x. If q is compact, then p is maximal.
CHAPTER II
COMMUTATIVE MULTIPLICATIVE IDEAL LATTICES
For the remainder of this dissertation, L will always
represent a milattice with a commutative multiplication.
We will consider the analogues of some classical results of
multiplicative ideal theory.
2.1. Definition: The residual quotient of a by b, de
noted by [a:b], is defined by: [a:b] = v{cbc < a}. Note
that if x is compact and x < [a:b], then bx < a, so
[a:b] = v{x compactbx < a}.
2.2. Proposition: The following properties of residual
quotients hold in L:
(a) [a:b] > a, [ab:b] s a, and b[a:b] < a a b, for all
a,b e L.
(b) If b < a, then [a:b] = 1.
(c) [[a:b]:c] = [a:bc] for all a,b,c e L.
(d) [a:b] = [a:a v b] = [a a b:b], for all a,b e L.
(e) [a: v b.] = a [a:b.], for all a e L and all families
iel 1 iel 1
iÂ£i s L
(f) [( a a.):b] = a [a.:b], for all b e L and all
iel iel
families (a.) T c l.
riel ~
11
7
1.6. Proposition; In a lattice L with 1, if 1 is com
pact, then each element a < 1 lies below a maximal of L.
Proof: Let a < 1, and let A={ba,
since a e A. A is partially ordered as a subset of L. If
C = (c^i e 1} is a nonempty totally ordered subset of A,
then let c = v c. We have a < c, so suppose c A. Then
ie I
n
c = 1, so 1 = vc.. Since 1 is compact, 1 = v c. with
iel 1 k=l xk
ik e I for 1 < k < n. But C is totally ordered, so
1 = Cq e C, which is impossible. Thus c e A, so by Zorn's
Lemma, there is an element m which is maximal in A. But if
m < b < 1, then b > a, so b e A and b = m. Hence m is a
maximal of L.D
1.7. Example; The converse to Proposition 1.6 is, in
general, false. The following example and ideas are found
in Bigard et al. [1]. Let X be a locally compact space
which is not compact, and let L(X) = {f: X * Rf is locally
constant with compact support}. Then the only prime convex
subgroups of L(X) are those of the form Cx = {f e Gf(x) =0}
for some x e X. Each of these is maximal, and each convex
subgroup of L(X) is contained in one of these. Also, each
f e L(X) must be in C for some x e X, since X is not compact,
and so L(X) cannot be generated by a single element; i.e.,
in C(L(X)), 1 is not compact. A latticeordered group G in
which all primes are maximal is called hyperarchimedean,
and an element a s e which generates a latticeordered group
50
(2) S is closed under multiplication, so
{p^k e Z+} < S. If p p^, then S = (p^k e Z+}.
(3) If p is branched and q p with q pprimary, then
oo
a q11 = Aiq^  A e A}.
n=l
(4) pQ = A{qAX e A} is prime, and there are no primes
a satisfying p^ < a < p.
Proof: (1) q < p < x, and since x is nonzero and
compact, x is invertible, so q = ax. q is pprimary and
x ^ p, so a < q, and thus a = q, so q = qx. If q is compact,
then q is invertible since q 0, and if x is compact with
x p, then xq=l*q=q, so x = 1, and p is maximal.
(2) : If q^ and q^ e S, then rad(q^q2) = p. If x and y
are compact with xy < q^q2, and if xn  p for each n e Z + ,
then q^ = q^x by part (1) so xy i q^q2 = xqlq2
Thus
qlq2
is pprimary. Now, suppose p p and let q be a pprimary.
2 2
Then p < rad(q), so q exceeds a power of p and hence a
power of p, by Theorem 5.2, part (5). Let n be the least
positive integer for which pn < q. If n = 1, then p = q. If
n s 2, then since pn ^ > q let y be compact with q < y < pn
q < y, so since y is invertible, q = ay. Since q is pprimary
and q
OO ^ CO
(3): Clearly a q > a q.. Now, q < p = rad(q,) for
n=l Xe A A
each X e A, so for each X e A, q11 < q^ for some n e Z + .
a q11 = a{q  X e A}.
n=l
Thus
25
a (M) = v{x compact xm < a for some me M}.
k
Notice that a (M) Â£ a.
3.3. Lemma; If a e L and x is compact, then
xxm
Proof: Certainly if xm < a for some m e M, and x is
*
compact, then x < a (M). Conversely, suppose x is compact and
*
x < a (M). Then x < v{y compact! ym ^ a for some me M}, so
n
since x is compact, x ^ v y^, where for each k < n, y^ is
k=l
n
compact and there is a m^ e M with y^m^. < a. If m = II m^,
k=l
n
then m e M and m < m, for each k < n, so xm < v y m, < a.D
K k=l K
*
3.4. Lemma: a (M) = a (M ).
s
*
Proof: Since M Â£ Ms, we have a (M) a (Mg). 3ut if x
k
is compact and x < a (M ), then there is a m e M with xm < a.
s s
Now, for each 1 e A, m  p^, so by the discussion following
Definition 3.1, there must be a nig e M with mg < m. Hence
* k k
xmQ < xm < a, so x < a (M), and a (M) = a (Mg).
When it is clear from the context just which multipli
cative set we are referring to, we will write a* rather
12
(g)
(h)
(i)
[ab:c] ^ a[b:c], for all a,b,c e L.
[a: a b.] ^ v [a:b.] for all a e L and families
il 1 iel 1
iÂ£l Â£ L
[( v a.):b] > v [a.:b] for all b e L and families
iel 1 iel 1
(ai>iel Â£ L
Proof: (a) ab < a, so a < [a:b]; ab < ab, so
a < [ab:b]; b[a:b] = b(v{cbc < a}) = v{bcbc < a} < a, and
b[a:b] < b.
(b) First b*l < b, so if b < a, b1 < a, and [a:b] = 1.
(c) [[a:b]:c] = v{dcd [a:b]} = v{dbcd < a} = [a:bc]
(d) We have (a v b) c < a ac v be ^ a <=> be ^ a <=>
be < a a b.
(e) [a: v b.]( v b.) < a, so for each j e I,
iel 1 iel 1
[a: v b.]b. < a, and [a: v b.] < a [a:b.]. But
iel 1 3 iel 1 iel 1
( a [a :b. ]) ( v b.) = v (( a [a:b.])b.) < v ([a:b.]b.) < a,
iel jel J jel iel J jel J 3
and so a [a:b.] = [a: v b.].
iel 1 iel 1
(f) First, b[( a a.):b] < a a. < a. for each jel,
iel 1 iel 1 3
so [( a a.):b] < a [a.:b]. But b( a [a.:b]) < b[a.:b] ^ a.
iel 1 iel 1 iel 1 33
for each j e I, so [( a a.):b] = a [a.:b].
iel 1 iel 1
(g) This is true since a[b:c] = a(v{dcd < b}) =
v{adcd < b} < v{adcad < ab} ^ [ab:c].
59
(3) => (5) : Let a and b be compact in L. (a v ) (a a b) =
a(a a b) v b(a a b) < ab, and because (3) holds, (a v b)(a a b)
(a v b)a a (a v b)b > ab, so (5) is true.
(9) => (1) : Let p be the unique maximal of L. We will
show that L is a vlattice. Since 1 is compact and every
nonzero element is the join of a set of invertibles, we
need only show that the set of invertible elements is totally
ordered. Let a and b be invertible in L. [a:b] v [b:a] = 1
by (a), so either [a:b] i p or [b:a] i p, where p is the
unique maximal of L. Thus [a:b] = 1 or [b:a] = 1, so
a b or b < a.
(7) => (9) : If a and b are invertible in L, then
1 = [(a v b):(a v b)] = [a:(a v b)] v [b:(a v b)] =
[a:b ] v [b:a] by (7) and Proposition 2.2, part (d).
(11) => (9) : If a and b are invertible in L, then
1 = [ (a a b) : (a a b) ] = [ (a a b) :a] v [ (a a b) :b] =
[b:a] v [a:b] by (11) and Proposition 2.2 part (d).
16
Proof; If x and y are compact and xy < q, then by (3),
x < q or y < p, so by (2) x < q or y11 < q for some n e Z + ,
and q is primary. Next, by (2), p < rad(q), by Proposition
1.11. Now, if x is compact and xm < q for some m e Z+,
assume m is the least positive integer with this property.
If m = 1, we are done by (1). If m > 2, then xm = xm ^x < q,
m_i. in.1
but xm is compact and xra i q, so x < p by (3). Thus
p = rad(q) and so p is prime.
2.9. Proposition: If p is prime and q^,q2,...,qR are
n
all pprimary, then a q. is pprimary.
i=l 1
n
Proof: Let q = a q.. First, p ^ q., so p s q. If x
i=l 1
is compact and x < p, then for each integer i with 1 < i < n,
m.
there is a positive integer itu with x < q^, so if
m = maxing 1
with xy < q and yip. Then xy < q^ for each i, and so
x < q^ for each i, 1 < i < n. Hence x < q, and q is
pprimary.
2.10. Proposition: If q e L is pprimary and a i q,
then [q:a] is pprimary.
Proof: Let xQ be compact, with xQ < [q:a]. Since
a i q, there is a y^ compact with y^ < a and yQ i q. Now,
y0xQ < a[q:a] < q, so since yQ i q, xQ < p. Thus
68
/ (Pq/^/Xq), (p1,m0,x0), and (p^m^x^, and so each
element a with 0 < a < 1 can be uniquely expressed as a
product of a finite number of primes. The elements
(pi'mo,xo) and (P0'IVX1) are invertible' and ^Po,ml,X0^
(p1,m0,x0) v (pQ,mQ/x^)/ so since the product distributes
over joins, each nonzero element is the join of a set of
invertible elements. Once again, however, there is no
a e L with a(p0,mlfx0) = (p1,m0,x()), so (p^m^Xg) is not
}c o n
invertible. In Figure 6.6, we label (p^,m^,xn) = p m x
with the convention p^ = m^ = x^ = 1 to simplify the
notation, and Lq from Example 6.4 is labeled.
65
L has ACC. If a = v{b, X e A}, then (a). = v{(b.).X e A},
P A 1 A 1
*1
and for each i n, there is an integer m. such that
/v mi
(a) = v (b, ) .. Thus, if B = (b. for some i < n and for
1 k=l Ak 1 A
some k < m. b. = b. }, then B is finite and for each i < n,
1 A A,
k
/v A
(a) ^ = v{(b^)^j Â£ B}. By Theorem 3.22, a = v{b^b^ e B}, and
a is compact.
(5) => (1) : By the fact that L has ACC, each nonzero
element of L is compact, and so by Theorem 5.8, each nonzero
element is invertible.
Recall that in commutative ring theory, an integral
domain with identity is a Dedekind domain if and only if
every proper nonzero ideal is uniquely a product of a finite
number of prime ideals (Hungerford [4]). In the context of
milattices, we must insist that each element a with 0 < a < 1
is uniquely expressible as a product of a finite number of
maximals, as the next two examples illustrate.
6.4. Example: Let P and M be disjoint countably infinite
sets, with P = fpkk is a nonnegative integer}, M = l is
a nonnegative integer), p^ p^ if k^ k^, and m^ m^
if 2* Let Lq = {0} u (P x M), and order Lq by setting
0 < a for each a e LQ, and (pk ,mÂ£ ) < (p^ ,m0 ) if k2 < k^ or k^= k2
and 2 Lq is totally ordered, so it is a lattice. For
each family F = {(pk_,m.)i e i), v (p ,mÂ£ ) = (p ,m),
i i iel i i
43
4.8. Lemma; If a is invertible and (a^i e 1} is a
family in L with a^ < a for each iel, then
[( v a.):a] = v [a.:a].
iel iel
Proof: First, a[( v a.):a] = v a. = v (a[a.:a]) =
iel 1 iel 1 iel 1
a( v [a. :a]). Then [( v a.):a] = v [a.:a] since a is
iel iel 1 iel 1
cancellative.Q
4.9. Proposition: Let L contain an invertible element.
Then the following conditions are equivalent:
(1) 1 is compact.
(2) Every invertible element in L is compact.
(3) There exists an invertible element a e L with a
compact.
Proof: (1) => (2) : Let b be any invertible element of
L. Since L is algebraic, b = v b., where each b. is compact
iel 1 1
Then 1 = [b:b] = [( v b.):b] = v [b.:b]. Since 1 is compact
iel 1 iel
n
1 = v [b. :b]. Then b = l*b = b( v [b. ;b]) =
k=l xk k=l xk
n n
v (b[b. :b]) = v b. and so b is compact.
k=l xk k=l 1k
(2) (3) : Obvious since L contains an invertible
element.
Abstract of Dissertation Presented to the Graduate Council
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy
GENERALIZATIONS OF IDEAL THEORY
By
David B. Kenoyer
August 1982
Chairman: Dr. Jorge Martinez
Major Department: Mathematics
The lattice of ideals of a commutative ring and the
lattice of convex subgroups of a latticeordered group
fit into the more general setting of a milattice, which is
an algebraic lattice with an associative multiplication that
distributes over joins, respects compactness, and satisfies
the ideal multiplication property that ab a a b. In this
context, the terms primary element, residual quotient,
multiplicative element, cancellative element, and invertible
element are defined, and generalizations of classical
results of commutative ring theory are obtained. A primary
decomposition theorem for modular milattices with ACC and
enough multiplicative elements is given. Multiplicative
subsets, rings of quotients, and extensions and contractions
of ideals are generalized, as are commutative rings with
identity, integral domains, and valuation rings. Several
IV
61
Proposition 4.4, each and each q. is invertible.
We
m
proceed by induction on n. If n = 1, then p = II q., so
j=l 3
after rearranging, q^ < p since p is prime, and p = q^.
m
If m > 2, then 1 = (pjq.) = II q., which is impossible, since
j=2 3
each qj <1. Thus m = 1 and p = q = x. Now, suppose that
if 1 s k < n1, and if r^,r2,...,r^ are all primes in L with
k
II r = y invertible, then this representation of y is unique
1=1 *
up to the order of the primes. Then if x is invertible and
n m m
x = II p. = II q., we have p, Â£ x = II q. so after rearrang
ed 1 j=l ] 1 j=l 3
ing if necessary, p^ > q^ since p^ is prime. Then
p^Cq^rp^] = q^, since p^ is invertible. If q^ < p^, then
Cqi:Pl] ^1 s:*nce is prime, so by Proposition 2.2, part
(a), q^ = [q^rp^l, and p^q^ = q^ > so since q1 is cancellative,
p^ = 1, which is impossible. Thus q^ = p^, so
n m
x = (p,) ( II p.) = (p.) ( II q.), and since p. is cancellative,
x i=2 1 j=2 1 x
n m
n p. = II q. By induction, m 1 = n 1, and so m = n,
i=2 1 j=2 3
and after rearranging, p^ = q^ for each i < n.D
We now give the major result of this chapter, which is
similar to the characterizations of Dedekind domains, with
one slight variation, which we will discuss after the proof.
6.3. Theorem: In a milattice L in which 1 is compact,
the following conditions are equivalent:
14
2.4. Lemma: If a and b are multiplicative and if
b = [ab:a], then ab is multiplicative.
Proof: Let c < ab. Then c < a, so a[c:a] = c. Thus
a[c:a] < ab, so since b = [ab:a], [c:a] < b. Hence
[c:a] = bd, where d = [[c:a]:b] = [c:ab], and
ab[c:ab] = abd = a[c:a] = c.D
We now press on toward a primary decomposition result.
Recall that in a commutative ring R, an ideal Q c r is
primary if whenever xy e Q with x and y elements of R, then
x e Q or y11 e Q for some positive integer n. Since a compact
ideal is finitely generated, we offer the following.
2.5. Definition; An element q in a milattice is primary
if q < 1 and whenever x and y are compact with xy < q, then
either x < q or y11 < q for some n e Z + In the example of
a latticeordered group, Cn = C for each n, so primary is
equivalent to prime.
2.6. Lemma: If q is primary and p = rad(q), then p is
prime.
Proof: Let x and y be compact, with xy < p and y $ p.
Since xy is compact, by our proof of Proposition 1.11 we
have (xy)n < q for some n e Z+, and we assume n is the
smallest positive integer with this property. If n = 1,
TABLE OF CONTENTS
Page
ACKNOWLEDGEMENTS ii
ABSTRACT iv
INTRODUCTION 1
CHAPTER
I.PRELIMINARIES 4
II.COMMUTATIVE MULTIPLICATIVE IDEAL LATTICES 11
III.MULTIPLICATIVE SUBSETS, LATTICES OF EXTENSIONS,
AND LOCALIZATIONS 22
IV.INVERTIBLE ELEMENTS 39
V.GENERALIZATIONS OF VALUATION RINGS AND PRUFER
DOMAINS 48
VI.GENERALIZATION OF DEDEKIND DOMAINS 60
VII.MODULARITY 7 0
BIBLIOGRAPHY 7 2
BIOGRAPHICAL SKETCH 73
iii
52
n
xy < v b, and since L is totally ordered, xy < bn, where
k=l K 0
bg < p and bg prime. Thus x < bg or y < bg, so x < r or
y < r, and r is prime.
(5) => (1) : We proceed by considering the localization
at p. First, if x is invertible in L, we show that x is
invertible in L If b Â£ x, let b e b with b< x, and set
p 0 0
A A
d = [bg:x]. Since x is compact, d = [bn:x] by Proposition
AAA
3.20, part (2). Thus x [bn:x] = xd = bn, and x is multipli
Q
A A A A
cative. Now, suppose that xb = xc, with b,c e L^. Then for
A A A A
each y compact with y
there is a mg e = {mm is compact and m Â£ p} such that
xirigy < xc. Thus nigy < c since x is cancellative, and so
A A A A A
b < c. Similarly, c b, so x is cancellative, and thus
A
invertible. Now, in L^, p is the unique maximal, and L is
A A /\
totally ordered. Also, r < p and r is prime, with no primes
A /\
between r and p. Let a be an invertible element in L with
A A A A A
r < a < p. Then a is invertible, and r < a < p < 1. Since
A A A A O A A A O A
a is invertible and a 1, (a; a, so r < (a) < p. Clear
A 2 A A A A A A 2
ly, rad((aj ) = p. If x and y are compact with xy < (a) ,
then suppose (y)n i (a)z for any n e Z Since Lp is a
A A A 2
vlattice, we must have y ^ p = rad((a; ), by Theorem 5.2,
/N A
part (5). But since y is compact and y 0, y is invertible
by part (1) of Theorem 5.2, so if y = p, then (y)z < y, and
oo ^ ~ /v/v
A (y) = Pg is prime with pg < p. Thus, pQ < r, so for some
j A A A 2 A
n e Z (y)n < r < (a)z, contradicting our choice of y.
2
Chapter II considers primary elements and primary
decompositions, ending with Theorem 2.14, the analogue of
the Primary Decomposition Theorem for commutative Noetherian
rings.
Chapter III is an investigation of multiplicative sets
and our version of rings of quotients. We consider the
A
"extension" a of an element a in our lattice relative to a
multiplicative subset M of compact elements, and show that,
in passing from the lattice L to the lattice of extensions,
L^j, we preserve compactness, primes, primary elements,
radicals, and almost all finite arithmetic operations,
including all operations with compact elements. Theorem
3.22 establishes the fundamental relationship between an
element a and the extensions of a relative to the maximals
exceeding a.
In Chapter IV, we discuss invertibility, including
various arithmetic facts involving invertible elements and
their existence in L. We conclude Chapter IV with a dis
cussion of various interpretations of the concept of an
integral domain. Chapter V looks at results from the theory
of Prfer domains, including the characterizations in terms
of arithmetic relations, which comprise Theorem 5.10. We
also consider our version of a valuation ring, and its
relation to Prfer domains. Chapter VI deals with our
generalization of Dedekind domains, with the main result
being Theorem 6.3. Finally, Chapter VII briefly looks at
the question of modularity, giving a result which leads to
certain conditions that guarantee modularity.
19
m m1 m1
a = [a:qm] = A = A q f and > a q^, contradicting
i=l i=l 1 i=l
m
the assumption that a = a q. is a normal decomposition (if
i=l 1
m = 1, then a = [aiq^] = 1, a contradiction) Thus
p e {t,,...,t }, and we may assume that with a suitable
rearrangement, p^ = t^. Let q = q^ a r^. By Proposition
2.9, q is p primary, and [q.:q] = q. if i < m1, and
1 m1 n1
[r.:q] = r. if j < n1. Thus [a:q] = a q. = a r., and both
3 3 i=l 1 j=l 3
are normal decompositions. Hence we may continue the process,
and we need only show that m = n. Suppose m < n. After
nm nm
m steps, p = t ,...,p. = t and 1= a r. < At.
m n ^1 nm+1' 3 1 j
since t^ prime. This is impossible, and so m = n.D
We now proceed to the primary decomposition. As stated,
we will assume in Theorem 2.14 that each element of L is the
join of a set of compact, powermultiplicative elements, and
we will also assume that L is a modular lattice; i.e., for all
a,b,c e L, if a > b, then a a (b v c) = b v (a a c). Again,
the motivating examples satisfy this condition, although we
only use it to show the existence of a decomposition, and
we are not certain that it is needed there.
We say that an element a e L is aprime (or finitemeet
irreducible, or irreducible) if a = b a c implies that
a = b or a = c.
2.13. Lemma: If L is a milattice with ACC, then each
n
a e L can be written as a = a a., where each a. is Aprime.
i=l 1 1
58
(10) If c e L and a,b are compact in L, then
[c:(a v b)] = [c:a] v [c:b].
(11) If a,b,c are compact in L, then
[c:(a v b)] = [c:a] v [c:b].
Proof: First, (2) => (3), (4) => (5), (6) => (7),
(8) => (9) and (10) => (11) easily. We show (1) => (2) ,
(1) => (4) (1) (6) (1) => (8) and (1) = (10) by assuming
L is a vlattice, by Theorem 5.8. Then, assuming that L has
a unique maximal, we show that (3) =>(5) =>(1), that
(7) => (9) => (1) and finally that (11) => (9) .
(1) => (2): Since L is totally ordered, we assume that
b < c. Then a(b a c) = ab = ab a ac.
(1) => (4) : Again, assume a s b, so that (a v b) (a a b)
ab.
(1) => (6) : Assuming a < b, [ (a v b) :c] = [b:c] =
[a:c] v [b:c].
(1) (8) : If a < b, then [b:a] = 1, so [a:b] v [b:a] =
(1) => (10): Suppose a < b. Then [c:(a a b) ] = [c:a] =
[c:a] v [c:b], since if xb < c, then xa < xb < c, so
[c:b] < [c:a].
Now we may only assume that L has a unique maximal p.
(5) => (1) : Let a and b be invertible in L. Then a and
b are nonzero compact elements, so a v b is nonzero and
compact. But (a v b)(a a b) = ab, which is invertible, so
by Proposition 4.4, a v b is invertible. By Theorem 5.8,
part (4), every nonzero compact element is invertible.
8
G is called a strong order unit of G. Any hyperarchimedean
latticeordered group without a strong order unit will be
a counterexample.
2
1.8. Proposition: In a milattice L, if 1 = 1, then
each maximal is prime. If each a < 1 lies below a maximal,
. 2
and if each maximal is prime, then 1 =1.
2
Proof; If 1 =1, let m be a maximal of L. Since
m < 1, we suppose x and y are in L with xy < m, but x Â£ m.
2
Then l(m v y) = (m v x)(m v y) = m v xm v my v xy < m < 1,
so m v y < 1. But mvy>m, somvy=m and y < m. Thus
m is prime. On the other hand, if each a < 1 lies below a
2
maximal, and if each maximal is prime, then 1 cannot lie
2
below any maximal, so 1 = l.D
1.9. Definition: The radical of an element a, denoted
rad(a), is defined as rad(a) = A{pp is prime and a < p}.
By convention, we set a p = 1 and v p = 0. A result of
J pe
Keimel [5] is:
1.10. Lemma: If M is a nonempty set of nonzero compact
elements of a milattice L, and if M is closed under multi
plication, and if a e L with m i a for each m e M, then
S = {bib i a and m i b for each m e M} must have maximal
cl
elements, and each of these is prime in L.
GENERALIZATIONS OF IDEAL THEORY
BY
DAVID B. KENOYER
A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL
OF THE UNIVERSITY OF FLORIDA IN PARTIAL
FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
1982
71
with 1 = v = ci v ei = &i a = cj^ a d^, and
> c^, since multiplication by a is a lattice isomorphism
of L onto [ 0,a]. Then h>1 ( i v d1) = b.^1) = b^, since L
contains an invertible element, but b^c^ < c^ and
^1^1 b^ a d^ = e^, so that b^c^ v b^d^ ci v = c^ < b^,
contradicting distributivity of the product.
This lemma allows us to say that if every nonzero
element of L is invertible, then L is modular. However, we
cannot weaken the amount of invertibility in L beyond this,
for even if L is a milattice in which each nonzero element
is the join of a set of invertible elements, we have the
example at the end of the last section. Here, m = x v p =
x v pm, and px=xAp=xA pm, with pm < p, so L contains
a fiveelement nonmodular sublattice, and is therefore
v
not modular.
36
for some n e Z + Thus x < a* or y11 < a*, so since a* < p^ < 1,
a* is pprimary. If q e a with q primary, then if x is com
pact and x < q*, there must be m 6 II with mx < q. But since
me M, mn e M for each n e Z + and so m11 Â£ q. Thus x < q,
so q = q* = a*. Conversely, if q is pprimary in L, with
A A A
p < p^ for some X e A, then q < p^, so q < p^ < 1. Now,
A /S A A
p = rad(q), so p = rad(q). If x and y are compact with
A A A A
xy < q, then let Xq and y^ be compact in L with Xq e x,
A
yQ e y. Since xQy0 < q*, there isa m Â£ M with ihXqYq 5 q.
Again, for each n e Z + m11 e M, so m11 i q. Thus Xgyg < q,
n j a a a
so Xq < q or yg < q for some n e Z and so a = a^ < q or
(b)n = (tig) < q. Therefore q is pprimary and by the
A
first part, q* is pprimary and is the only primary in q, so
q = q*.
Finally, we turn our attention to residual quotients.
3.20. Proposition: Let a,b e L. Then
A A A
(1) If d = [a:b], then d ^ [a:b],
A A A
(2) If b is compact and d = [a:b], then d = [a:b].
(3) ([a:b])* < [a*:b*].
(4) ([a*:b*])* = [a*:b*].
A A A
(5) If c = [a:b], then c* = [a*:b*].
Proof: (1) Ford = v{xbx a} v{xbx < a} = [a:b].
(2) Let b be compact in L, and let x be compact with
A A A /y
bx < a. There is a xQ e x with Xq compact. Then bxQ is
40
4.1. Definition; If L is a commutative milattice, an
element a e L is cancellative if[ab:a] = b for each b e L;
equivalently, if ab = ac, then b = c, so that multiplication
by a is an injection. An element a e L is invertible if a
is both cancellative and multiplicative (Definition 2.3);
i.e., multiplication by a is an orderpreserving bijection
of L onto [0,a] = {b e L0 < b < a}. We now list some
results concerning the nice properties and behavior of
invertible elements.
4.2. Corollary; If a is invertible and b is multiplica
tive, then ab is multiplicative.
Proof: Since [ab:a] = b, the result follows from
Lemma 2.4.
n
4.3. Lemma: Let a = IT a. Then a is cancellative <=>
i=l 1
each a^ is cancellative.
Proof:
If
a is cancellative,
let
^o
^ n.
For each
b f
B, [ab:
a] =
b, so b
= [ab:a] =
[[ab
:a. ] :
n a. ] >
10
[((
n a.)Ca.
V 1
b:a.
]) : n
a.] > [a.
b: a.
]
s b,
so
i*iQ 0 x0 i*iQ x 0 x0
b = [a. b:a. ], and each a. is cancellative. Conversely,
x0 10 10
suppose a^,...,a are cancellative. We proceed by induction
on n. If n = 1, a = a^ is cancellative. Suppose that for
46
(a^a2,b^b2) if s and b2 s
(a^b^ (a2,b2) =
(PQ/s) if b1 = s or b2 = s.
3S follows: (pQfcjg) 1, (Pj,/<3g) =P i = r,
k k S k
p0,qk) = q (pk,qil) = p q / (P0/S) = 0, (r,qk) = rq Then
multiplication is done in the obvious (commutative) way, with
k 2
l*x = x for all xeL, pr=r=r, x*0 = 0 for all x e L,
and p^ = q^ = 1. L is a milattice with ACC; it is totally
ordered, so it is modular. Now, p is not multiplicative,
since p[q:p] = pq < q, so p is not invertible. But if
p = v a. with each a. e L, then p = a. for some in e I, and
iel 10 U
so p is not the join of a family of invertible elements.
However, q is invertible, so q is invertible for each k > 1,
and every element except 0 exceeds some power of q. See
Figure 4.13.
Thus, we have many possible choices for a generalization
of the lattice of ideals of an integral domain. We choose
the strongest, since it is used the most.
4.10. Definition; A milattice L is a domain lattice
if 1 is compact and each nonzero element is the join of a
set of invertible elements (which are compact by Proposition
4.9) .
characterizations of the analogue of Priifer domains are
given in terms of localizations and arithmetic relation
ships. Dedekind domains are generalized and characterized.
v
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
je Ma'rtinez, Chairman
Associate Professor of
Mathematics
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
David A. Drake
Professor of Mathematics
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
James E. Kee
Professor of Mathemati
I certify that I ha^p/read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
Neil L. White
Associate Professor of
Mathematics
BIOGRAPHICAL SKETCH
David Bruce Kenoyer was born on August 27, 1953,in
Seattle, Washington, and is the only son and second of
three children of Howard W. and Beatrice Kenoyer. David
has spent most of his life in Michigan, attending East
Grand Rapids High School before receiving his B.S. from
Central Michigan University and his M.S. from Michigan
State University. While at Michigan State, he met Susan
Tiffany, whom he later married. David enjoys most sports,
especially those which he shares with Sue, such as golf,
tennis, bowling, and softball, as well as camping and
music.
73
15
xy
< q, so x < q <
P
k
or y < q
for some k e
Z+. But
y
i P,
so
y q for any
k
Z SO X
< p, and we
are done.
If
n
s 2, then (xy)n
=
/ \ttl
(xy) xy <
q, so again
since y
i
q
i XI 1
for any k e Z (xy; x < q. But by choice of nf
(xy)n i q, so xm < q for some m e Z + and x < rad(q) = p.D
This justifies the following terminology: If q e L is
primary and p is prime with p = rad(q), then q is pprimary,
and p is the prime associated with q.
Note: If q is pprimary, and x and y are compact, with
xy < q, then x < q or y < p; i.e., one of x and y is below q,
or they both lie below p, since L is commutative.
2.7. Lemma: If q is pprimary in L, and a Â£ p, then
Cq:a] = q.
Proof: As always, q < [q:a]. If x is compact with
ax < q, then since a i p, there is y compact with y < a
and y i p, and so xy < q. Now, yip, so x < q, and so
q v{x compact  ax < q} = [q:a].D
2.8.
(1)
(2)
(3)
Lemma. If p,q e L satisfy the conditions:
p > q
If x is compact and x < p, then xn < q for some
n e Z +
If x and y are compact and xy < q, then x < q or
y < p
then p is prime and q is pprimary.
44
(3) => (1) : Let a be compact and invertible in L, and
1 = vc.. Then a = a*l = v ac., so since a is compact,
iel iel 1
n n
a = vac. = a v c. Since a is cancellative and a*l = a,
k=l 1k k=l 1k
1 = v c. so 1 is compact.
k=l 1k
In the theory of commutative rings there are several
conditions that are equivalent to the condition that a ring
R is an integral domain. We now consider some of these in
the context of milattices. From the theory of commutative
rings with identity, the following conditions are equivalent:
(a) 0 (as an ideal) is prime.
(b) Every nonzero ideal exceeds a cancellative ideal.
(c) Every nonzero ideal exceeds an invertible ideal.
(d) Every nonzero ideal is the join of invertible
principal ideals.
In the language of milattices, L is a milattice in which 1
is compact, invertible, and:
(1) 0 is prime.
(2) Every nonzero element of L exceeds a cancellative
element.
(3) Every nonzero element of L exceeds an invertible
element.
(4) Every nonzero element of L is the join of compact
invertible elements.
Of course, the implications (4) => (3) => (2) => (1) are obvious.
23
compact elements with the properties:
(1) For each X e A and m e Mq, m Â£ p^.
(2) If q is a prime and, for each X e A, q i p^, then
there is a m e Mg with m < q.
n
Set M = { II m. I n > 1 and for each i, 1 < i < n, m. e Mn}.
i=l 1 1 u
Mq Â£ M^ so M (f). Each element of M is compact, since each
element of Mq is. Since {p^X e A} , let p be an element.
Since 0 < p, no finite products of elements of Mq can be 0,
for no m e Mq satisfies m < p, and p is prime. Thus 0 i M,
and clearly any finite product of elements of M is again in
M. The fact that every multiplicative subset of L arises in
this fashion is the content of Lemma 1.10. Recall the nota
tion that if a e L and for each m e M, m a, then
Sa = {b e La < b and for each me M, m $ b). Notice that
if a, < a0, then S s S so that each S is contained in
L ^ q 3.2 3
Sq. We let {p^  X e A} be the set of elements maximal in Sq.
This is a nonempty pairwise incomparable set of primes, and
if b e L such that for each X e A we have b i p^, then there
must be a m e M with m < b. We say that  A e A} is the
set of primes associated with M (or the set of associated
primes of M).
If M is a multiplicative subset of L, and {p^X e A}
is the set of associated primes of M, then let M = {m e l
s
m is compact and for each X e A, m ^ p^}. M Â£ Mg, and
0 i M so since M is closed under multiplication, it is a
multiplicative subset of L containing M, with the same set
of associated primes. Mg is called the saturation of M,

