Maximal Chains of Prime Ideals of Different Lengths in Unique Factorization Domains
aa r X i v : . [ m a t h . A C ] J a n Maximal Chains of Prime Ideals of Different Lengths inUnique Factorization Domains
S. Loepp and Alex SemendingerJanuary 10, 2019
Abstract
We show that, given integers n , n , . . . , n k with 2 < n < n < · · · < n k , thereexists a local (Noetherian) unique factorization domain that has maximal chains ofprime ideals of lengths n , n , . . . , n k which are disjoint except at their minimal andmaximal elements. In addition, we demonstrate that unique factorization domains canhave other unusual prime ideal structures. One important open question in commutative algebra is: given a partially ordered set X , is X isomorphic to the prime spectrum of some Noetherian ring? While this question itselfhas not been answered, progress towards it has been made. For instance, in [6], Hochsteranswered the question of when X is isomorphic to the prime spectrum of a possibly non-Noetherian commutative ring.One approach to the problem is to consider what posets may arise as finite subsets ofSpec R for a Noetherian ring R . It was once thought that these would necessarily possess“nice” properties: most notably, it was at one point thought that all Noetherian rings mustbe catenary. This was disproven in 1956 in [9], when Nagata constructed a noncatenary1oetherian integral domain. In 1979, Heitmann proved a much stronger result: in [4], heshowed that, for every finite partially ordered set X , there exists a Noetherian domain R suchthat X can be embedded into Spec R with an embedding that preserves saturated chains.It is natural to ask if Heitmann’s theorem holds for other classes of rings, such as Noethe-rian unique factorization domains (UFDs). One might expect that the additional structureof a UFD would place restrictions on its prime ideal structure. More precisely, we ask thefollowing open question: Question 1.1.
For every finite partially ordered set X , does there exist a Noetherian UFD A such that X can be embedded into Spec A with an embedding that preserves saturated chains? It has long been thought that this question could be answered in the negative. Forinstance, after 1956, it was conjectured that all Noetherian unique factorization domains(and more generally, all integrally closed domains) must be catenary. This was disprovedby Heitmann in [5], in which he constructed a noncatenary Noetherian UFD. This was theonly example of a noncatenary Noetherian UFD in the literature until 2017, when Avery, et.al. characterized the completions of noncatenary UFDs in [1], which generated a new classof examples. As a corollary of this result ([1], Proposition 4.2), Avery, et. al. showed thatthere is no limit on “how noncatenary” a local UFD can be, in the sense that it can posessmaximal chains of prime ideals with an arbitrarily large difference in lengths. Inasmuchas this result furnishes more examples of noncatenary Noetherian UFDs, it provides mildevidence in favor of an affirmative answer to Question 1.1.In this paper, although we do not provide an answer to Question 1.1, we further developthe techniques from [1] and give explicit examples of noncatenary partially ordered sets whichcan be embedded in a saturated way into the prime spectrum of some Noetherian UFDs. Asa result, we not only greatly expand our collection of examples of noncatenary local UFDs,but we also provide additional evidence in favor of the analogue of Heitmann’s 1979 theorem2olding for UFDs.Our main result (Theorem 3.4) is that, given integers n , n , . . . , n k , with 2 < n < n < · · · < n k , there exists a local (Noetherian) unique factorization domain that has maximalchains of prime ideals of lengths n , n , . . . , n k which are disjoint except at their minimal andmaximal elements. To prove this result, we first start with a complete local ring T that hasmaximal chains of prime ideals of lengths n , n , . . . , n k . We then use the construction in [5]to find a local UFD A whose completion with respect to its maximal ideal is T and suchthat the intersections of the chains of prime ideals of T with A results in maximal chains in A of lengths n , n , . . . , n k . Finally, we show that, after some adjustment of the height oneprime ideals in the chains, the chains are disjoint. Throughout this paper, all rings are assumed to be commutative with unity. Additionally,the symbol ⊂ will indicate strict containment. Definition. If R is a Noetherian ring with exactly one maximal ideal M , then we say ( R, M )is a local ring . If R is not necessarily Noetherian but has one maximal ideal, then we sayit is quasi-local .If A is a local ring, then b A will denote the completion of A with respect to its maximalideal.In order to formally state the motivating question of our research and our results, weneed to develop some terminology regarding partially ordered sets (posets). Definition.
A totally ordered subset of a partially ordered set X is called a chain . Definition.
Let x < x < · · · < x n be a finite chain of elements in a poset X . We say thatthis chain has length n . 3f the above chain has the property that, for each i ∈ { , . . . , n − } , there are no elements y ∈ X such that x i < y < x i +1 , then we say the chain is saturated .If the above chain is saturated and, additionally, x is a minimal element of X and x n isa maximal element, then we say the chain is maximal . Definition.
Let X be a poset. The height of x ∈ X is the length of the longest saturatedchain of elements x < x < · · · < x and is denoted ht x . The dimension of X is given bydim X := sup { ht x : x ∈ X } .Recall that, when we refer to the height of a prime ideal of a ring R , we mean its heightas a member of the poset (Spec R, ⊆ ). Furthermore, dim R = dim Spec R and is the Krulldimension of the ring R . Definition.
We say a poset X is catenary if, for any x, y ∈ X such that x < y , all saturatedchains of elements between x and y have the same length.This definition is usually applied to rings, where we say a ring R is catenary preciselywhen the poset (Spec R, ⊆ ) is catenary. Definition.
Let X and Y be posets. An injective order-preserving function φ : X → Y willbe called a saturated embedding if φ sends saturated chains in X to saturated chainsin Y . If, additionally, dim( X ) = dim( φ ( X )), we will say φ is a dimension-preservingsaturated embedding. Our motivating question can now be formally stated: for which partially ordered sets X does there exist a local UFD A and a dimension-preserving saturated embedding from X to Spec A ? In the next section, we show that the set of posets for which this can be done islarger than previously known.We now state results which will be used in later proofs. First, we present two very usefulprime avoidance lemmas from [5] and a “cardinality” lemma from [2].4 emma 2.1. ([5], Lemma 2)
Let ( T, M ) be a complete local ring, C a countable set of primeideals of T such that M / ∈ C , and D a countable set of elements of T . If I is an ideal of T which is contained in no single P ∈ C , then I S { ( r + P ) : P ∈ C, r ∈ D } . Lemma 2.2. ([5], Lemma 3)
Let ( T, M ) be a local ring. Let C ⊂ Spec T , let I be an idealof T such that I P for all P ∈ C , and let D ⊂ T . Suppose | C × D | < | T /M | . Then I S { ( r + P ) : P ∈ C, r ∈ D } . Lemma 2.3. ([2], Lemma 2.2)
Let ( T, M ) be a complete local ring of dimension at least one.Let P be a nonmaximal prime ideal of T . Then | T /P | = | T | ≥ | R | . The construction in [5] starts with a subring of a complete local ring T which must satisfyseveral properties. Any ring satisfying these properties is called an N-subring, which we nowdefine. Definition.
Let (
T, M ) be a complete local ring. A quasi-local subring (
R, M ∩ R ) of T isan N-subring if R is a UFD and1. | R | ≤ sup( ℵ , | T /M | ), with equality only when T is countable,2. Q ∩ R = (0) for all Q ∈ Ass T , and3. if t ∈ T is regular and P ∈ Ass(
T /tT ), then ht( P ∩ R ) ≤ Lemma 2.4. ([7], Lemma 11)
Let ( T, M ) be a complete local ring and let R be an N-subringof T . Suppose C ⊂ Spec T satisfies the following conditions:1. M / ∈ C ,2. { P ∈ Spec T : P ∈ Ass
T /rT for some = r ∈ R } ⊂ C , and . Ass T ⊂ C .Let x ∈ T be such that x / ∈ P and x + P is transcendental over R/ ( R ∩ P ) as an element of T /P for every P ∈ C . Then S = R [ x ] ( R [ x ] ∩ M ) is an N-subring of T properly containing R and | S | = sup {ℵ , | R |} . Here and elsewhere, we tacitly make use of the natural injection R/ ( R ∩ P ) → T /P defined by r + ( R ∩ P ) r + P , where R, T , and P are as defined in Lemma 2.4. Under thisinjection, we can think of elements of T /P as being algebraic or transcendental over the ring R/ ( R ∩ P ). Determinations of this sort will prove crucial in Lemma 3.3 and in Theorem 3.4.The construction of specific posets within Spec A starts by carefully choosing chains ofprime ideals of T = b A , which the next lemma allows us to do. These chains, when takentogether and intersected with A , will essentially form the desired posets. Lemma 2.5. ([1], Lemma 2.8)
Let ( T, M ) be a local ring with M / ∈ Ass T and let P ∈ Min T with dim( T /P ) = n . Then there exists a saturated chain of prime ideals of T , P ⊂ Q ⊂· · · ⊂ Q n − ⊂ M , such that, for each i = 1 , . . . , n − , we have Q i / ∈ Ass T and P is the onlyminimal prime ideal of T contained in Q i . The main result from [1] demonstrates conditions for exactly when a complete local ringis the completion of a noncatenary local UFD. Because all of the posets we will consider inthis paper will be noncatenary, it will be necessary to ensure that the complete local ring T we begin with satisfies these conditions. Theorem 2.6. ( [1], Theorem 3.7)
Let ( T, M ) be a complete local ring. Then T is thecompletion of a noncatenary local UFD if and only if the following conditions hold:1. No integer of T is a zero divisor,2. depth T > , and . There exists P ∈ Min T such that < dim( T /P ) < dim T . We finally present a theorem from algebraic geometry which will be used to find a com-plete local ring with a specific minimal prime ideal structure.
Theorem 2.7. ( [3], Theorem 8.18 (Bertini’s Theorem))
Let X be a nonsingular closedsubvariety of P nK , where K is an algebraically closed field. Then there exists a hyperplane H ⊆ P nK , not containing X , and such that the scheme H ∩ X is regular. If dim X ≥ , then H ∩ X is connected, and therefore irreducible, as well. Furthermore, the set of hyperplaneswith this property forms an open dense subset of the complete linear system | H | , consideredas a projective space. We begin by constructing a complete local ring satisfying the conditions of Theorem 2.6, aswell as other desired conditions. Note that condition 6 of Lemma 3.1 implies that no integerof T is a zero divisor. We then use a modified version of the construction from the proof ofthat theorem to obtain a local UFD with arbitrarily many disjoint chains of different lengths. Lemma 3.1.
Given any k -tuple ( n , n , . . . , n k ) of distinct positive integers all greater than2, there exists a complete local ring T with minimal prime ideals P , . . . , P k such that1. dim( T /P i ) = n i for each i = 1 , . . . , k ,2. T /P i is a regular local ring for each i = 1 , . . . , k ,3. Min T = Ass T = { P , . . . , P k } ,4. If x ∈ T is not a zero divisor and Q ∈ Ass(
T /xT ) (considered as a subset of Spec T ),then ht Q = 1 , . depth T ≥ , and6. T contains Q .Proof. Without loss of generality, suppose n < n < · · · < n k . Let R = C [ x , . . . , x n ], where n = n k . By Bertini’s theorem, there exists an open collection of hypersurfaces in P n suchthat, if H is an element of the open collection, then H is a smooth, irreducible subvariety.To obtain a subvariety of dimension n − j we apply the theorem j + 1 times to obtain smoothirreducible subvarieties H ⊃ ( H ∩ H ) ⊃ · · · ⊃ ( H ∩ · · · ∩ H j ) = H . We then let Q bethe ideal corresponding to H ; that is, Q = I ( H ) ∈ Spec R . By Bertini’s Theorem, H isirreducible, hence Q is a prime ideal. Furthermore, dim R/Q = n − j . Because all of thesevarieties are smooth, for each resultant prime ideal Q , the ring R/Q is regular.We apply this result to obtain prime ideals Q , . . . , Q k such that dim R/Q i = n i and R/Q i is regular for each i . To ensure that they are all incomparable, note that Q i ⊆ Q j if and onlyif V ( Q j ) ⊆ V ( Q i ). Therefore, we need only choose varieties that are not contained in oneanother, which is possible since the set of candidates is an open dense set, and containmentis a closed condition.Now let I = T ki =1 Q i and let S = b R ∼ = C J x , . . . , x n K , where b R denotes the completion of R with respect to the maximal ideal ( x , . . . , x n ). The desired ring is given by T = S/IS ,where P i = Q i S/IS (we will see shortly that this is in fact a prime ideal of T ). It is clearthat T contains Q .To see that T satisfies the desired properties, observe first that Q ∩ Q ∩ · · · ∩ Q k is aminimal primary decomposition of the ideal I of R . Now, T /P i is a regular local ring as aresult of the fact that the completion of a regular local ring is itself a regular local ring, and \ ( R/Q i ) ∼ = S/Q i S ∼ = S/ISQ i S/IS ∼ = T /P i . P i is a prime ideal (since a regular local ring is a domain) and thatdim( T /P i ) = dim( R/Q i ) = n i .Since Q i is a prime ideal of R for every i = 1 , , . . . k , we have √ I = I and so R/I isreduced. Since R is excellent, R/I is excellent and so its completion,
S/IS , is also reduced.It follows that T = S/IS has no embedded associated prime ideals, and so Min T = Ass T .We now show that Min T = { P , P , . . . , P k } . Suppose P is a prime ideal of S suchthat IS ⊆ P ⊆ Q i S . Then, intersecting with R , we have I ⊆ ( P ∩ R ) ⊆ Q i . Since Q i isminimal over I , P ∩ R = Q i . Now, ( P ∩ R ) S ⊆ P ⊆ Q i S . It follows that Q i S ⊆ P ⊆ Q i S and so P = Q i S . Hence, P i ∈ Min T for all i = 1 , , . . . , k . On the other hand, let P be a minimal prime ideal over IS . Then, intersecting with R , we have I ⊆ P ∩ R . SinceAss( R/IR ) = Min(
R/IR ) = { Q , Q , . . . , Q k } , it must be the case that I ⊆ Q i ⊆ P ∩ R for some i . It follows that IS ⊆ Q i S ⊆ ( P ∩ R ) S ⊆ P and so P = Q i S . Therefore,Min T = { P , P , . . . , P k } .Any regular ring A satisfies Serre’s criterion ( S ) (by [8], Theorem 23.8), which impliesthat if P ∈ Ass(
A/rA ), for any regular r ∈ A , then ht P ≤
1. Suppose x ∈ T is not azerodivisor and let Q ∈ Ass(
T /xT ). Let P i be a minimal prime ideal of T contained in Q .Define T = T /P i and denote the image of Q in T by Q . Since Q ∈ Ass(
T /xT ), by [10],Remark 4.23, we have that Q ∈ Ass(
T /xT ). Therefore ht
Q/P i ≤
1. Since this is true foreach minimal prime ideal P i of T such that P i ⊆ Q , we must therefore have ht Q = 1 asdesired.To see that depth T >
1, we will find a T -regular sequence of length 2. We start byobserving that, if M is the maximal ideal of T then M / ∈ Ass T , and so there exists x ∈ M which is not a zero divisor. If Q ∈ Ass
T /xT , then ht Q = 1. However, by construction,dim T ≥
2, so any height two prime ideal of T will contain a T /xT -regular element, call it y . Then x, y is a regular sequence of length 2 as desired. (cid:4) Lemma 3.2.
Let ( A, M ) be a local UFD with dim A > and let k be a positive integer.Suppose that A contains maximal chains of prime ideals (0) ⊂ P ⊂ P i, ⊂ · · · ⊂ P i,n i − ⊂ M for i = 1 , . . . , k , where P a, = P b, whenever a = b . Then there exist height one prime ideals ˜ P , . . . , ˜ P k ∈ Spec A such that, for i = 1 , . . . , k , ˜ P i ⊂ P i, and ˜ P a = ˜ P b whenever a = b .Proof. By the Prime Avoidance Theorem, there exists z i ∈ P i, such that z i / ∈ P j, for j = i .Let q i be an irreducible factor of z i that is in P i, . Then q i A ⊂ P i, but q i A P b, for b = i .Now set ˜ P i = q i A . Then ˜ P , . . . , ˜ P k ∈ Spec A are height one prime ideals of A and ˜ P a = ˜ P b when a = b . (cid:4) The following technical lemma will prove instrumental in the proof of the main theorem,and, along with Lemma 3.2, is among the primary techniques that allow us to have somecontrol of the prime ideal structure of the resultant UFD.
Lemma 3.3.
Let ( T, M ) be a complete local ring as constructed in Lemma 3.1. Furthermore,let Q ∈ Spec T be such that ht Q > and let R be a countable N-subring of T .Then there exists a countable N-subring S of T such that R ⊂ S and S contains agenerating set for Q .Proof. Let Q = ( x , . . . , x n ). We will find elements ˜ x , . . . , ˜ x n ∈ Q such that Q = (˜ x , . . . , ˜ x n )and such that we have a chain of N-subrings R = R ⊂ R ⊂ R ⊂ · · · ⊂ R n = S R i = R i − [˜ x i ] M ∩ R i − [˜ x i ] for each i = 1 , . . . , n .We begin by finding an element ˜ x ∈ Q such that R = R [˜ x ] M ∩ R [˜ x ] is an N-subringand Q = (˜ x , x , . . . , x n ). We can accomplish this through the use of Lemma 2.4, for whichit will suffice to find an element ˜ x ∈ Q such that ˜ x + P is transcendental over R / ( R ∩ P )as an element of T /P for all P ∈ C = { P ∈ Spec T : P ∈ Ass(
T /rT ) , = r ∈ R } ∪ Ass T .First, by Lemma 3.1, M C and Q P for all P ∈ C , so by Lemma 2.1 there exists y ∈ Q such that y / ∈ P for all P ∈ C . We will set ˜ x = x + α y , where α ∈ T remainsto be chosen.Fix some P ∈ C . Note that | R / ( R ∩ P ) | ≤ | R | , and R is countable. As a result,the algebraic closure of R / ( R ∩ P ) in T /P is countable. However, by Lemma 2.3,
T /P is uncountable. Since we have chosen y / ∈ P , every distinct choice of t + P ∈ T /P givesa different x + ty + P . There are thus uncountably many such choices, only countablymany of which make x + ty + P algebraic over R / ( R ∩ P ). In order to find a valueof α such that ˜ x + P is transcendental over R / ( R ∩ P ) for all P ∈ C simultaneously,we make use of another application of Lemma 2.1. For each P ∈ C , let D ( P ) be a setconsisting of one element from each coset of T /P which makes x + ty + P algebraic over R / ( P ∩ R ). Then set D = S P ∈ C D ( P ) . We can now use Lemma 2.1 to find α ∈ M suchthat α / ∈ S { ( r + P ) : P ∈ C , r ∈ D } . This ensures that x + α y + P is transcendentalover R / ( R ∩ P ) for all P ∈ C , which was our goal. We can now set ˜ x = x + α y andlet R = R [˜ x ] M ∩ R [˜ x ] . Note that R is countable.We must also check that Q = (˜ x , x , . . . , x n ). Since y ∈ Q , we can write y = β x + · · · + β n x n for some β i ∈ T . Then we have˜ x = x + α y = (1 + α β ) x + α β x + · · · + α β n x n . α ∈ M , the element 1 + α β is a unit. We can therefore write x = (1 + α β ) − (˜ x − α β x − · · · − α β n x n )and conclude that x ∈ (˜ x , x , . . . , x n ). Therefore we indeed have that Q = (˜ x , x , . . . , x n ).We now outline the process to obtain R . Let C = { P ∈ Spec T : P ∈ Ass(
T /rT ) , = r ∈ R } ∪ Ass T, which we note is a countable set. By Lemma 3.1, M C and Q P for all P ∈ C . Wecan therefore use Lemma 2.1 to find an element y ∈ Q such that y / ∈ P for all P ∈ C .Now fix an element P ∈ C and let D ( P ) be a set containing one element of each coset of T /P that makes x + ty + P algebraic over R / ( P ∩ R ). Setting D = S P ∈ C D ( P ) , wecan find α ∈ M such that ˜ x = x + α y + P is transcendental over R / ( P ∩ R ) for every P ∈ C . We can now let R = R [˜ x ] ( M ∩ R [˜ x ]) . As above, R is a countable N-subring and Q = (˜ x , ˜ x , x , . . . , x n ).We continue with this procedure to find R , . . . , R n , at each step setting C i = { P ∈ Spec T : P ∈ Ass(
T /rT ) , = r ∈ R i − } ∪ Ass T , using Lemma 2.1 to find y i ∈ Q suchthat y i / ∈ P for all P ∈ C i , and setting D i = S P ∈ C i D ( P ) , where D ( P ) is a set containing oneelement from each coset in T /P that makes x i + ty i + P algebraic over R i − / ( P ∩ R i − ).Through another use of Lemma 2.1, we find α i ∈ M such that ˜ x i = x i + α i y i + P istranscendental over R i − / ( P ∩ R i − ) for every P ∈ C i . We then set R i = R i − [˜ x i ] M ∩ R i − [˜ x i ] .We proceed thus for i = 3 , . . . , n to obtain the desired chain of N -subrings R = R ⊂ R ⊂ · · · ⊂ R n = S , completing the proof. (cid:4) We are now in a position to construct the desired UFD. As alluded to before, the proofof Theorem 3.4 is largely based on that of Theorem 2.6 from [1], which in turn is based on12he proof of the main theorem in [5].
Theorem 3.4.
For integers n , n , . . . , n k with < n < n < · · · < n k , there exists a localUFD A such that A has maximal chains of prime ideals of lengths n , n , . . . , n k which aredisjoint except at their minimal and maximal elements.Remark . As this proof is fairly long, we will preface it with a rough outline. We begin byusing Lemma 3.1 to find a complete local ring (
T, M ) in which we can find chains of primeideals of the desired lengths. Our goal will be to use the construction in [5] to create a UFD A ⊂ T such that b A ∼ = T . We would also like A to contain a generating set of each prime idealin the aforementioned chains in Spec T , as this would guarantee that their intersections with A are all distinct. This can only be done for prime ideals P such that ht P >
1, however.We will deal with the case of the height one prime ideals separately.To find such an A , it suffices to create an N-subring R N which contains a generating setof each prime ideal P in the specified chains with ht P >
1. This is accomplished throughmultiple applications of Lemma 3.3.Finally, the height one prime ideals of the chains in T must be dealt with. They arechosen such that their intersection with R N is (0), which will help us ensure that they willall have the same intersection with A . Then other prime ideals can be found which makethe chains disjoint, using Lemma 3.2. This will complete the proof. Proof.
First, using Lemma 3.1, let (
T, M ) be a complete local ring satisfying the six con-ditions of Lemma 3.1. In particular, T has minimal prime ideals P , . . . , P k such thatdim( T /P i ) = n i for each i . Let R be the prime subring of T localized at its intersection with M (note that R ∼ = Q ) and let C = Ass T = Min T . For each P i ∈ Min T , use Lemma 2.5 toconstruct a saturated chain of prime ideals P i = P i, ⊂ P i, ⊂ P i, ⊂ · · · ⊂ P i,n i − ⊂ M suchthat each ideal in the chain contains exactly one minimal prime ideal and none (except P i, )is an associated prime ideal of T . Note that the indices are chosen such that ht P i,j = j .13e will construct A to be a local UFD such that b A ∼ = T . We also want to ensure that, for j >
1, ( P i,j ∩ A ) T = P i,j . It is enough to have A contain a generating set of each of these ideals P i,j . We accomplish this through successive applications of Lemma 3.3. That is, we startwith the N-subring R and, say, the prime ideal P , . By Lemma 3.3, there exists a countableN-subring R ⊃ R which contains a generating set for P , . Applying this lemma in turnfor each P i,j with j > R ⊂ R ⊂ · · · ⊂ R N , where R N contains a generating set of each of the aforementioned prime ideals.We now focus on the height one prime ideals of our chains. We will find prime ideals˜ P , . . . , ˜ P k ∈ Spec T such that, for each i = 1 , . . . , k , we have the following:1. P i, ⊂ ˜ P i ⊂ P i, ,2. ˜ P i contains exactly one minimal prime ideal of T , and3. ˜ P i ∩ R N = (0).Fix some i ∈ { , . . . , k } . We will show that, in fact, almost all prime ideals which satisfy(1) also satisfy (2) and (3). We begin by showing that there are uncountably many primeideals that satisfy (1).Suppose to the contrary that, for some fixed i = 1 , . . . , k , the set B i = { p ∈ Spec T : P i, ⊂ p ⊂ P i, } is countable. Let ˜ T i = T P i, /P i, T P i, . The set B i is in one-to-one correspondencewith the set B ′ i = { p ∈ Spec ˜ T : (0) ⊂ p ⊂ P i, ˜ T } , and so B ′ i is countable. Hence, | B ′ i | < | ˜ T /P i, ˜ T | , so we can use Lemma 2.2 to conclude that P i, ˜ T S p ∈ B ′ i p . But this is impossible,as every element in P i, ˜ T is contained in some height one prime ideal of ˜ T , which implies thatin fact P i, ˜ T ⊂ S p ∈ B ′ i p . As a result of this contradiction, we conclude that B i is uncountable,as desired.We now show that relatively few prime ideals satisfy (1) but not (2) and (3). Note that,given two distinct minimal prime ideals of a Noetherian ring, only finitely many height one14rime ideals can contain them both. As a result, all but finitely many prime ideals satisfyingcondition (1) also satisfy condition (2). As for condition (3), note that, because R N is acountable subring of T , at most countably many height one prime ideals of T will havenonzero intersection with R N . To see this observe that, if we have P ∈ Spec T with ht P = 1and 0 = r ∈ P ∩ R N , then P ∈ Ass
T /rT . As the set Ass
T /rT is finite for each r , the set { P ∈ Spec T : ht P = 1 , P ∩ R N = (0) } is countable. There are hence uncountably manyprime ideals ˜ P i that satisfy conditions (1), (2), and (3).Let ˜ P , . . . , ˜ P k be fixed prime ideals which satisfy conditions (1), (2), and (3) above. Wenow apply Lemma 2.1 with I = T ki =1 ˜ P i and C N +1 = { P ∈ Spec T : P ∈ Ass
T /rT, r ∈ R N } ∪ Ass T to conclude that there exists a nonzerodivisor x ∈ I such that x / ∈ S P ∈ C N +1 P .We now use a similar method to that in the proof of Lemma 3.3 to find an element ˜ x ∈ T such that R N +1 = R N [˜ x ] ( M ∩ R N [˜ x ]) is a countable N-subring. Fix P ∈ C N +1 and note thatdifferent choices of t + P ∈ T /P give different values of x (1 + t ) + P , since x / ∈ P . Since thealgebraic closure of R N / ( P ∩ R N ) in T /P is countable, for all but countably many choices of t ∈ T , the image of x (1 + t ) in T /P will be transcendental over R N / ( P ∩ R N ). Now let D ( P ) be a set containing one element of each coset t + P such that x (1 + t ) + P is algebraic over R N / ( P ∩ R N ) and let D = S P ∈ C N +1 D ( P ) . Using Lemma 2.1, we can find an element α ∈ T such that ˜ x + P = x (1 + α ) + P is transcendental over R N / ( P ∩ R N ) for every P ∈ C N +1 .Finally, by Lemma 2.4, R N +1 = R N [˜ x ] ( M ∩ R [˜ x ]) is a countable N-subring. Note that, by thetranscendental property guaranteed above, ˜ xR N +1 is a prime ideal of R N +1 .At this point, we can use the construction in [5], replacing R in Theorem 8 of [5] with R N +1 from above, to obtain a UFD A ⊂ T such that b A ∼ = T and R N +1 ⊂ A . Therefore, A contains a generating set for each P i,j with j > P i,j ∩ A ) T = P i,j . Therefore, letting p i,j = P i,j ∩ A , for each i = 1 , . . . , k we have a chainof prime ideals p i, ⊂ p i, ⊂ · · · ⊂ p i,n i − ⊂ M ∩ A in A . Furthermore, the constructionguarantees that prime elements in R N +1 remain prime in A . Therefore, ˜ xA ∈ Spec A and15 xA = ˜ P i ∩ A for each i .We claim that A in fact contains disjoint saturated chains of prime ideals as desired.To show this, we need only find appropriate height one elements ˜ p i such that, the chains(0) ⊂ ˜ p i ⊂ p i, ⊂ · · · ⊂ p i,n i − ⊂ M ∩ A are disjoint (except of course at (0) and M ∩ A ).Now, using Lemma 3.2, we find height one prime ideals ˜ p , . . . , ˜ p k ∈ Spec A such that˜ p i ⊂ p i, for each i = 1 , . . . , k , and ˜ p i = ˜ p j whenever i = j . The chains (0) ⊂ ˜ p i ⊂ p i, ⊂· · · ⊂ p i,n i − ⊂ M ∩ A are therefore the desired maximal chains of prime ideals. (cid:4) The main difference between the proof of Theorem 3.4 and that of Theorem 2.6 from [1]is that, in Theorem 2.6, it is enough to construct an N-subring containing a generating setof just one carefully-chosen prime ideal Q , which has the property that dim( T /Q ) = 1 andht
Q < dim T −
1. This ensures that the resulting UFD A will be noncatenary, but it givesus no further information about the structure of Spec A . That we can apply this techniquemultiple times and obtain more information about Spec A is the primary insight of thisresult.By virtue of Lemma 3.2, we have considerable control over the height one elements ofthe chains: for instance, we can leave them “collapsed,” as they are before the application ofthe lemma. Alternatively, we can choose to apply the lemma for only some chains, leavinga few chains “collapsed” at height one and the rest disjoint (see Figure 3 for an example). Corollary 3.6.
Given distinct integers n , n , . . . , n k such that n i > for each i = 1 , . . . , k ,there exists a local UFD A with maximal chains of prime ideals of lengths n , . . . , n k whichshare a height one element but all of whose height two elements and above are disjoint.Furthermore, for any ℓ < k , there are maximal chains of prime ideals of lengths n , . . . , n k such that the first ℓ share a single height one element and the remaining k − ℓ are disjoint.Proof. Construct A as in the proof of Theorem 3.4. Then, the chains (0) ⊂ ˜ xA ⊂ p i, ⊂ . . . p i,n i − ⊂ M ∩ A are disjoint except at their maximal, minimal and height one elements.16inally, we note that by applying Lemma 3.2 to only k − ℓ chains, we can ensure that thefirst ℓ share a height one element and the rest are disjoint. (cid:4) Figure 1: A poset X with maximal chains of lengths 4, 5, 6, and 7, and the correspondingprime ideals in the rings T and A from Lemma 3.1 and Theorem 3.4, respectively. • •••••••••••••••••• • X P , P , P , P , P , P , P , P , P , P , P , P , P , P , P , P , P , P , P , P , P , P , M Spec T (0) ˜ p p , p , p , p , p , ˜ p p , p , p , p , ˜ p p , p , p , ˜ p p , p , M ∩ A Spec A Example 4.1.
Consider the poset X as shown in Figure 1, which contains four maximalchains of lengths 4, 5, 6, and 7 which are disjoint except at their minimal and maximalelements. As a result of Theorem 3.4, there exists a local UFD A such that there is adimension-preserving saturated embedding from X into Spec A . To find this ring A , we firstapply Lemma 3.1 to find a complete local ring T with chains of prime ideals as illustrated.17e then apply Theorem 3.4 to find a local UFD A which contains the illustrated primeideals.Figure 2: A partially ordered set which admits a saturated embedding, but no dimension-preserving saturated embedding, into the prime spectrum of a Noetherian UFD ( X ). ••• •• X ••• ••• X Example 4.2.
Let X , X be as in Figure 2. There is no dimension-preserving saturatedembedding from X to a local UFD, as every dimension-3 local UFD is catenary. However,as a result of Corollary 3.6 with n = 3 and n = 4, there is a dimension-preserving saturatedembedding from X to some local UFD. Therefore, for any local UFD A , every saturatedembedding from X into Spec A fails to preserve dimension. Example 4.3.
Let Y , Y , and Y be the posets illustrated in Figure 3. As a result ofCorollary 3.6, there exist local UFDs A , A , A such that for i = 1 , ,
3, there is a dimension-preserving saturated embedding from Y i to Spec A i .18igure 3: Partially ordered sets with “disjoint” ( Y ), “partially collapsed” ( Y ), and “com-pletely collapsed” ( Y ) height one elements. ••• •••• •••• Y •• •••• •••• Y •• •••• ••• Y Acknowledgements
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