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Page i Acknowledgement Page ii Table of Contents Page iii Abstract Page iv Page v Chapter 1. Preliminaries Page 1 Page 2 Page 3 Page 4 Page 5 Page 6 Page 7 Page 8 Page 9 Page 10 Chapter 2. Commutative multiplicative ideal lattices Page 11 Page 12 Page 13 Page 14 Page 15 Page 16 Page 17 Page 18 Page 19 Page 20 Page 21 Chapter 3. Multiplicative subsets, lattices of extensions, and localizations Page 22 Page 23 Page 24 Page 25 Page 26 Page 27 Page 28 Page 29 Page 30 Page 31 Page 32 Page 33 Page 34 Page 35 Page 36 Page 37 Page 38 Chapter 4. Invertible elements Page 39 Page 40 Page 41 Page 42 Page 43 Page 44 Page 45 Page 46 Page 47 Chapter 5. Generalizations of valuation rings and prufer domains Page 48 Page 49 Page 50 Page 51 Page 52 Page 53 Page 54 Page 55 Page 56 Page 57 Page 58 Page 59 Chapter 6. Generalization of dedekind domains Page 60 Page 61 Page 62 Page 63 Page 64 Page 65 Page 66 Page 67 Page 68 Page 69 Chapter 7. Modularity Page 70 Page 71 Bibliography Page 72 Biographical sketch Page 73 Page 74 Page 75 
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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 