Completions of Uncountable Local Rings with Countable Spectra
aa r X i v : . [ m a t h . A C ] M a y COMPLETIONS OF UNCOUNTABLE LOCAL RINGS WITHCOUNTABLE SPECTRA
S. LOEPP AND TERESA YU
Abstract.
We find necessary and sufficient conditions for a complete local (Noetherian)ring to be the completion of an uncountable local (Noetherian) domain with a countablespectrum. Our results suggest that uncountable local domains with countable spectra aremore common than previously thought. We also characterize completions of uncountable ex-cellent local domains with countable spectra assuming the completion contains the rationals,completions of uncountable local unique factorization domains with countable spectra, com-pletions of uncountable noncatenary local domains with countable spectra, and completionsof uncountable noncatenary local unique factorization domains with countable spectra. Introduction
Examples of uncountable Noetherian rings of Krull dimension zero or one that have count-able spectra are plentiful. Surprisingly, however, the existence of uncountable Noetherianrings with countable spectra in higher dimensions was not known until 2016, when Col-bert constructed in [4] an uncountable, n -dimensional Noetherian domain with a countablespectrum for any n ≥
2. Loepp and Michaelsen extended this result in [10] by showingthe existence of an uncountable Noetherian domain with a countable spectrum, but withstronger conditions on the ring: they showed, for all n ≥
0, the existence of an uncountable, n -dimensional, excellent regular local (Noetherian) ring with a countable spectrum.Given that uncountable Noetherian rings with countable spectra in higher dimensionswere not known to exist until very recently, one might expect for such rings to be quite rare.In this paper, we show that these rings are more “common” than perhaps one might haveanticipated, as we find that “most” complete local rings with a countable residue field are,in fact, the completion of an uncountable local domain with a countable spectrum. Moreprecisely, in Section 4, we prove the following theorem, which characterizes exactly when acomplete local ring is the completion of an uncountable domain with a countable spectrum. Theorem 4.7.
Suppose T is a complete local ring with maximal ideal M . • If dim T = 0, then T is the completion of an uncountable local domain with acountable spectrum if and only if T is an uncountable field. • If dim T = 1, then T is the completion of an uncountable local domain with acountable spectrum if and only if(1) no integer of T is a zero divisor, and(2) M / ∈ Ass( T ). • If dim T ≥
2, then T is the completion of an uncountable local domain with acountable spectrum if and only if(1) no integer of T is a zero divisor,(2) M / ∈ Ass( T ), and(3) T /M is countable.
Surprisingly, the necessary and sufficient conditions on a complete local ring of dimensionat least two to be the completion of an uncountable local domain with a countable spectrumare the same conditions necessary and sufficient for it to be the completion of a countablelocal domain [2, Corollary 3.7]. This is more evidence suggesting that uncountable localdomains with countable spectra are more common than expected. In order to prove thatthe conditions of Theorem 4.7 are sufficient, we construct uncountable local domains fromcountable local domains while ensuring that the spectra of these rings are order isomorphicwhen viewed as partially ordered sets.Natural extensions of this result include characterizing completions of uncountable localrings with countable spectra that satisfy certain properties. In this paper, we provide severalextensions of this kind by studying excellent domains, unique factorization domains (UFDs),noncatenary domains, and noncatenary UFDs (Theorems 4.16, 4.24, 4.29, and 4.33). Ingeneral, we are able to construct uncountable local domains with countable spectra andproperty X by beginning with a countable local domain with property X and constructingfrom it an uncountable domain whose spectra is order isomorphic to the countable domain’sspectra. We then show that the uncountable domain also has property X . The countablelocal domain with property X with which we begin comes from previous results in theliterature (for example, [6, Theorem 8] and results from [11]).The outline of this paper is as follows. In Section 2, we provide background. In Section 3,we present a construction of uncountable local domains from countable local domains. Usingthis construction, we then in Section 4 characterize completions of uncountable local domainswith countable spectra, completions of uncountable excellent local domains with countablespectra such that the completion contains the rationals, completions of uncountable localUFDs with countable spectra, completions of uncountable noncatenary local domains withcountable spectra, and completions of uncountable noncatenary local UFDs with countablespectra. 2. Background
All rings in this paper are commutative with identity. We say a ring is quasi-local if ithas exactly one maximal ideal but is not necessarily Noetherian, and we say a ring is local if it has exactly one maximal ideal and is Noetherian. We denote a quasi-local ring R withunique maximal ideal M by ( R, M ), and we denote the completion of a local ring (
R, M )with respect to its maximal ideal by b R .In [8], Lech proves the following result, which characterizes completions of local domains. Theorem 2.1 ([8], Theorem 1) . A complete local ring (
T, M ) is the completion of a localdomain if and only if(1) no integer of T is a zero divisor, and(2) unless equal to (0), M / ∈ Ass( T ).A number of more recent results characterize completions of local domains with certaincharacteristics. Most recently, in [2], the authors characterize completions of domains withcertain cardinalities and domains with spectra that have certain cardinalities. In particular,they prove the following results. Theorem 2.2 ([2], Theorem 2.13) . Let (
T, M ) be a complete local ring such that(1) no integer of T is a zero divisor, and OMPLETIONS OF UNCOUNTABLE LOCAL RINGS WITH COUNTABLE SPECTRA 3 (2) unless equal to (0),
M / ∈ Ass( T ).If T /M is infinite, then T is the completion of a local domain A such that | A | = | T /M | . If T /M is finite, then T is the completion of a countable domain. Theorem 2.3 ([2], Corollary 3.7) . Let (
T, M ) be a complete local ring with dim T ≥ T is the completion of a local domain with a countable spectrum if and only if(1) no integer of T is a zero divisor,(2) M / ∈ Ass( T ), and(3) T /M is countable.Throughout this paper, we show that rings we construct have certain given completions.In order to do this, we make use of the following results, which give sufficient conditions.
Proposition 2.4 ([7], Proposition 1) . If (
R, R ∩ M ) is a quasi-local subring of a completelocal ring ( T, M ), the map R → T /M is onto, and IT ∩ R = IR for every finitely generatedideal I of R , then R is Noetherian and the natural homomorphism b R = T is an isomorphism. Proposition 2.5 ([11], Corollary 2.5) . Suppose (
T, M ) is a complete local ring, and (
R, R ∩ M ) is a local subring of T such that b R = T . Let ( A, A ∩ M ) be a quasi-local subring of T such that R ⊆ A , and such that, for every finitely generated ideal I of A , IT ∩ A = IA .Then A is Noetherian and b A = T .Note that if R is a local ring and b R = T , then T is a faithfully flat extension of R . Itfollows that if I is an ideal of R , then IT ∩ R = I . In addition, T being a faithfully flatextension of R implies that R and T satisfy the going-down theorem, so if P ∈ Spec( T ),then ht( P ∩ R ) ≤ ht( P ). We also have the following relationship between prime ideals of aring and prime ideals of the ring’s completion. Lemma 2.6 ([3], Lemma 2.6) . Let (
T, M ) be the completion of a local ring (
A, A ∩ M ) andlet P be a prime ideal of A . Then for Q ∈ Min(
P T ), we have Q ∩ A = P .We end this section with two results on the cardinalities of local rings and their quotientrings. Proposition 2.7 ([11], Proposition 2.10) . Let (
T, M ) be a local ring. If
T /M is finite, then
T /M is finite. If T /M is infinite, then | T /M | = | T /M | .Let c denote the cardinality of R . Lemma 2.8 ([5], Lemma 2.2) . Let (
T, M ) be a complete local ring with dim T ≥
1. Let P be a nonmaximal prime ideal of T . Then, | T /P | = | T | ≥ c .3. Construction
Given a countable local domain (
S, S ∩ M ) with completion ( T, M ), we show in this sectionhow to construct an uncountable local domain (
B, B ∩ M ) such that S ⊆ B ⊆ T , ideals of B are extended from S , and b B = T . In the following section, we show how this constructioncan be used to prove our main results. In particular, we provide specific countable domainsfrom which to begin the construction, and show that the resulting uncountable local domainsatisfies certain desirable conditions, including having a countable spectrum.Before we begin our construction, we introduce some definitions, inspired by definitionsfrom [10]. S. LOEPP AND TERESA YU
Definition 3.1.
Suppose (
T, M ) is a complete local ring and (
S, S ∩ M ) is a countable localdomain such that S ⊆ T and b S = T . A ring R is an S -subring of T if S ⊆ R ⊆ T , and, forany r ∈ R ∩ M , there exist c ∈ S ∩ M and a unit d ∈ T such that r = cd . Definition 3.2.
Suppose (
T, M ) is a complete local ring and (
S, S ∩ M ) is a countablelocal domain such that S ⊆ T and b S = T . A CS -subring R of T is a countable quasi-local S -subring of T with maximal ideal R ∩ M .Note that, if ( T, M ) is a complete local ring and (
S, S ∩ M ) is a countable local domainsuch that S ⊆ T and b S = T , then S is itself a CS -subring of T . In addition, the unionof an ascending chain of S -subrings of T is an S -subring of T , the countable union of anascending chain of CS -subrings of T is also a CS -subring of T , and the uncountable unionof an ascending chain of CS -subrings is a quasi-local S -subring of T .The next lemma establishes that S -subrings are in fact domains. Lemma 3.3.
Suppose (
T, M ) is a complete local ring and (
S, S ∩ M ) is a countable localdomain such that S ⊆ T and b S = T . Let R be an S -subring of T . Then, R is an integraldomain. Proof.
To prove this, we show that if P ∈ Ass( T ), then P ∩ R = (0). Suppose that r ∈ P ∩ R ⊆ M ∩ R . Since R is an S -subring, there exist c ∈ S ∩ M and d , a unit in T , such that r = cd . Since d is a unit, we have that c ∈ P , and so c ∈ P ∩ S . However, S is a domainwhose completion is T , so P ∩ S = (0), and therefore we have that c = 0. Thus, r = 0, so P ∩ R = (0), and it follows that R is an integral domain. (cid:3) Given a countable local domain (
S, S ∩ M ) with completion ( T, M ), we construct anascending chain of CS -subrings of T , starting from S , by adjoining elements u ∈ T of aspecific form. Every time we adjoin such a u , we will show that we not only obtain another CS -subring of T , but also that we are indeed adding a new element from T . If we do thisuncountably many times and then take the union of these rings, we obtain an uncountable S -subring of T that we call ( A, A ∩ M ). Finally, we adjoin elements of T to A so that weobtain an S -subring of T that is an uncountable local domain with completion T and suchthat its ideals are extended from S .We begin by describing the elements that we adjoin uncountably many times to the count-able local domain S . Let R be a CS -subring of T , and consider u of the form u = 1 + A z + A z z + · · · + A k z z · · · z k + · · · , where ( A i ) i ∈ Z + ⊆ R ∩ M and ( z i ) i ∈ Z + ⊆ S ∩ M . Since A i , z i ∈ M for all i , we have that the k th term in the series is in M k for k ≥
2, so u is indeed an element of T . In addition, sinceevery term except for 1 is an element of M and 1 / ∈ M , we have that u is a unit in T . For k ∈ Z + , define M k := 1 + A z + · · · + A k − z · · · z k − and K k := A k z · · · z k − + A k +1 z · · · z k − z k +1 + · · · . Note that we can now express u as M k + z k K k for any k ∈ Z + . The next lemma, whichis a slight modification of [10, Lemma 5.1], describes how to adjoin such an element u toa CS -subring R of T so that the resulting ring is a CS -subring of T . We achieve this viaan algorithm by choosing the elements of the sequence ( z i ) i ∈ Z + given a sequence ( A i ) i ∈ Z + satisfying a specific property. OMPLETIONS OF UNCOUNTABLE LOCAL RINGS WITH COUNTABLE SPECTRA 5
Lemma 3.4.
Suppose (
T, M ) is a complete local ring with dim T ≥ S, S ∩ M ) is acountable local domain such that S ⊆ T and b S = T . Let ( R, R ∩ M ) be a CS -subring of T .Then, for any ( A i ) i ∈ Z + ⊆ R ∩ M satisfying the property that whenever i > j , there exists a k such that A i ∈ M k and A j / ∈ M k , there exists a sequence ( z i ) i ∈ Z + ⊆ S ∩ M with z i = 0 forevery i such that, if u ∈ T is of the form u = 1 + A z + A z z + · · · + A k z z · · · z k + · · · , then R [ u ] ( R [ u ] ∩ M ) is a CS -subring of T . Proof.
We first use the sequence ( A i ) to define the sequence ( z i ), so we obtain an element u .We then show that adjoining u to R and localizing results in a CS -subring of T .Let R ′′ := R [ X ] for an indeterminate X . Notice that R ′′ is countable, since R is a CS -subring and is thus countable as well. Use the positive integers to enumerate the nonzeroelements of R ′′ , and consider the i th element in the well-order, denoted G i ( X ). Substitutingany u of the form u = 1 + A z + · · · A z z + · · · + A k z z + · · · z k + · · · into the polynomial G i ( X ) for any i ∈ Z + , we have G i ( u ) = r ℓ,i u ℓ + · · · + r ,i u + r ,i , where r m,i ∈ R . Since we can express u as M k + z k K k for any k ≥
1, we can rewrite G i ( u ) as G i ( u ) = r ℓ,i ( M k + z k K k ) ℓ + · · · + r ,i ( M k + z k K k ) + r ,i . Expanding this polynomial, we obtain G i ( u ) = r ℓ,i ℓ X n =0 (cid:18) ℓn (cid:19) M ℓ − nk z nk K nk + · · · + r ,i X n =0 (cid:18) n (cid:19) M − nk z nk K nk + r ,i ( M k + z k K k ) + r ,i = r ℓ,i M ℓk + ℓ X n =1 M ℓ − nk z nk K nk ! + · · · + r ,i ( M k + z k K k ) + r ,i = G i ( M k ) + r ℓ,i ℓ X n =1 M ℓ − nk z nk K nk + · · · + r ,i z k K k = G i ( M k ) + z k ℓ X m =1 r m,i m X j =1 (cid:18) mj (cid:19) z j − k K jk M m − jk ! . (1)We will define the sequence ( z i ) i ∈ Z + recursively using G j ( M i ) so that z i ∈ S ∩ M for every i ∈ Z + . Since we define each z i to be in S ∩ M , the second of the two terms in (1) is anelement of M . Thus, for any k ≥ G i ( u ) ∈ M if and only if G i ( M k ) ∈ M . In other words,for any k ≥ G i ( u ) is a unit in T if and only if G i ( M k ) is a unit in T .By hypothesis, each A i ∈ R ∩ M , so since each M i is defined to be M i = 1 + A z + · · · + A i − z · · · z i − and each term of this sum is in R , we have that M i ∈ R . Thus, since G j ( X ) ∈ R [ X ], we see that G j ( M i ) ∈ R for all i, j ∈ Z + .Notice that S ∩ M = (0), since b S = T and dim T ≥
1. Let x denote a non-zero element of S ∩ M .Starting with j = 1 and i = 1, we use G j ( M i ) to define z i as follows:(1) If G j ( M i ) is a unit, then let z i = x , and use G j +1 ( M i +1 ) to define z i +1 . S. LOEPP AND TERESA YU (2) If G j ( M i ) = 0, then let z i = x , and use G j ( M i +1 ) to define z i +1 .(3) If G j ( M i ) is nonzero and not a unit, then since G j ( M i ) ∈ R , which is a CS -subring,there exist c ∈ S ∩ M and d a unit in T such that G j ( M i ) = cd . Since G j ( M i ) = 0,we have that c = 0. Let z i = c , and use G j +1 ( M i +1 ) to define z i +1 .Notice that, by this definition, z i = 0 and z i ∈ S ∩ M for all i .Using our given ( A i ) sequence and this definition for the sequence ( z i ), define the element u in the previously specified form, and let R ′ := R [ u ]. We show that R ′ ( R ′ ∩ M ) is a CS -subring of T . Notice that, since R ′ is countable and S ⊆ R ⊆ R ′ , it is enough to show thatif r ∈ R ′ ∩ M , then r is of the form r = cd , where c ∈ S ∩ M and d is a unit in T . If r = 0,then c = 0, d = 1 works.Suppose that r ∈ R ′ ∩ M is a nonzero element of R ′ , so r = G i ( u ) for some i ∈ Z + and r is not a unit. Then G i ( M k ) is not a unit for all k ∈ Z + by the claim above. We show that G i ( u ) = cd for c ∈ S ∩ M and d ∈ T a unit. In order to do so, we first need to show thatevery M i is distinct. Suppose not, and that M j = M i for some i > j . Then, M i − M j = A j z · · · z j + A j +1 z · · · z j +1 + · · · + A i − z · · · z i − = 0 . Since every z i is nonzero and R is a domain, we can cancel z · · · z j , obtaining A j + A j +1 z j +1 + · · · + A i − z j +1 · · · z i − = 0 . By our assumption on the sequence ( A i ), there exists a k such that A j +1 ∈ M k but A j / ∈ M k .Then, subtracting A j from both sides above, we see that A j +1 z j +1 + · · · + A i − z j +1 · · · z i − = − A j / ∈ M k . However, A j +1 ∈ M k , so we have that A ℓ ∈ M k for every ℓ ≥ j + 1; thus, A j +1 z j +1 + · · · A i − z j +1 z j +2 · · · + z i − ∈ M k , which is a contradiction. Thus, all of the M i ’s are distinct.Since 0 = G i ( X ) ∈ R [ X ] and R is an integral domain, it must be that G i ( X ) has at mostdeg( G i ) roots in R . Each M i ∈ R and all of the M i ’s are distinct, so there are infinitelymany distinct M i ’s, not all of which can be one of the finitely many roots of G i ( X ). Thus,there exists some k ∈ Z + such that G i ( M k ) = 0. By case (3) in the algorithm above, wehave that G i ( M k ) = z k d ′ , where d ′ is a unit. Substituting into (1), we have G i ( u ) = z k d ′ + ℓ X m =1 r m,i m X j =1 (cid:18) mj (cid:19) z j − k K jk M m − jk ! . (2)Notice that z k ∈ S ∩ M by definition. In addition, recall that K k is defined to be in M aswell, so since d ′ is a unit in T , the element d ′ + ℓ X m =1 r m,i m X j =1 (cid:18) mj (cid:19) z j − k K jk M m − jk is a unit in T . Thus, r = G i ( u ) can be written as cd , where c ∈ S ∩ M and d ∈ T is a unit.It follows that R ′ localized at R ′ ∩ M is a CS -subring of T . (cid:3) The next lemma shows that there always exists a choice of ( A i ) i ∈ Z + ⊆ R ∩ M such thatthe ring R [ u ] R [ u ] ∩ M from Lemma 3.4 is not equal to R . The statement and proof of thislemma are very similar to those of [10, Lemma 5.2] OMPLETIONS OF UNCOUNTABLE LOCAL RINGS WITH COUNTABLE SPECTRA 7
Lemma 3.5.
Suppose (
T, M ) is a complete local ring with dim T ≥ S, S ∩ M ) is acountable local domain such that S ⊆ T and b S = T . Given a CS -subring ( R, R ∩ M ) of T ,there exists a CS -subring ( R ′ , R ′ ∩ M ) of T where R ( R ′ ⊆ T . Proof.
Since dim T ≥ b S = T , we have that dim S ≥ S ∩ M = (0). Let x ∈ S ∩ M ⊆ R ∩ M with x = 0. Define A i = x q ( i ) , where q : Z + → Z + is a strictlyincreasing function. Notice that ( A i ) i ∈ Z + ⊆ R ∩ M . Suppose that i and j are positiveintegers such that j < i . Then q ( j ) < q ( i ). Since x = 0 and S is a domain, x s = 0 forall positive integers s . Let ℓ be the largest integer such that x q ( j ) ∈ M ℓ , and note that ℓ exists because T ∞ i =1 M i = (0). We then have that x q ( j ) M ℓ +1 , but x q ( j )+1 ∈ M ℓ +1 . As q ( i ) ≥ q ( j ) + 1, we have that x q ( i ) ∈ M ℓ +1 . This shows that our sequence of A i ’s satisfiesthe needed conditions for Lemma 3.4: we have that ( A i ) i ∈ Z + ⊆ R ∩ M , and whenever i > j ,there exists a k ∈ Z + such that A i ∈ M k but A j / ∈ M k . Thus, by Lemma 3.4, there exists asequence ( z i ) i ∈ Z + ⊆ S ∩ M with z i = 0 for every i such that, if u ∈ T is of the form u = 1 + A z + A z z + · · · + A k z z · · · z k + · · · , then R [ u ] R [ u ] ∩ M is a CS -subring of T .We now show that there are uncountably many choices for the element u . First, noticethat, by a diagonal argument, there are uncountably many choices for the function q . Weshow that distinct choices for the function q yield distinct u ’s.Let u = 1 + A z + · · · and u = 1 + B z ′ + · · · , where A i = x q ( i ) and B i = x p ( i ) , with p, q : Z + → Z + both strictly increasing. Suppose that u = u . We show that A i = B i and z i = z ′ i for all i .First, referring back to the algorithm described in the proof of Lemma 3.4, M = 1 forall choices of u . Since z and z ′ are both defined by the algorithm using G ( M ) = G (1),they are the same and both nonzero. Then, equating u and u , we have A + A z + · · · = B + B z ′ + · · · , so then x q (1) + x q (2) z + · · · = x p (1) + x p (2) z ′ + · · · . Without loss of generality, suppose that q (1) ≤ p (1); then, cancelling by x q (1) , we have1 + x q (2) − q (1) z + · · · = x p (1) − q (1) + x p (2) − q (1) z ′ + · · · . Since z i ∈ S ∩ M for all i and 1 / ∈ M , the left-hand side is not in M . Thus, the the right-hand side is not in M either. However, all but the first term are in M , since z ′ i ∈ S ∩ M .Thus, x p (1) − q (1) / ∈ M , implying that p (1) − q (1) = 0, so p (1) = q (1); thus, A = B . Forthe algorithm in the proof of Lemma 3.4 the definitions of z i (resp., z ′ i ) depend only on A j and z j (resp., B j and z ′ j ) for all j < i . Since we have shown that z = z ′ and A = B , wecan show inductively that z i = z ′ i and A i = B i for all i ≥
1. There are uncountably manychoices for the function q , so there are uncountably many choices for u such that R [ u ] R [ u ] ∩ M is a CS -subring of T . Since R is a CS -subring of T , it is countable, and thus there exists a u ∈ T \ R such that R [ u ] R [ u ] ∩ M is a CS -subring of T . Hence, R ′ = R [ u ] R [ u ] ∩ M is the desired CS -subring of T . (cid:3) The next theorem, which is a generalization of [10, Theorem 5.3], guarantees the existenceof an uncountable quasi-local S -subring of T . S. LOEPP AND TERESA YU
Theorem 3.6.
Suppose (
T, M ) is a complete local ring with dim T ≥ S, S ∩ M ) is acountable local domain such that S ⊆ T and b S = T . There exists an uncountable quasi-local S -subring of T , denoted ( A, A ∩ M ). Proof.
There exists a well-ordered uncountable set C such that every element of C has onlycountably many predecessors. Let 0 denote the minimal element of C . For every element c ∈ C , we inductively define a CS -subring of T that we denote S c .First, let S := S . Suppose 0 < c ∈ C , and assume that S b has been defined for every b < c so that S b is a CS -subring of T . If c has a predecessor, b ∈ C , then define ( S c , S c ∩ M )to be the CS -subring of T obtained from Lemma 3.5 with R = S b so that S b ( S c ⊆ T . Ifinstead c is a limit ordinal, define S c = S b Suppose ( T, M ) is a complete local ring with dim T ≥ S, S ∩ M ) is acountable local domain such that S ⊆ T and b S = T . There exists an uncountable quasi-local S -subring of T , ( B, B ∩ M ), such that, for every principal ideal I of B , IT ∩ B = IB . Proof. Define B = QF ( A ) ∩ T , where ( A, A ∩ M ) is the uncountable quasi-local S -subringof T obtained from Theorem 3.6, and QF ( A ) is the quotient field of A . Then, A ⊆ B , andsince A is uncountable, so is B . Let r ∈ B ∩ M , so r = ab for a, b ∈ A with b = 0. Weshow that r = cd , where c ∈ S ∩ M and d ∈ T is a unit. Since ab ∈ M , there exists m ∈ M such that ab = m , so a = bm , implying that a ∈ A ∩ M . Recall that A is an S -subring, sothere exist c ∈ S ∩ M and d ∈ T a unit such that a = cd . Thus, ab = cdb . First suppose that b ∈ A \ M ; then b is a unit in T , so r = c ( db − ) is of the desired form, since db − ∈ T is aunit.Now suppose that b ∈ A ∩ M , so b = c ′ d ′ for c ′ ∈ S ∩ M and d ′ ∈ T a unit. Notice that c ′ = 0 since b = 0. Then, ab = cdc ′ d ′ ∈ B ∩ M ⊆ T, so, for some v ∈ T , we have that cdc ′ d ′ = v , implying that c = c ′ d ′ d − v . This means that c ∈ c ′ T ∩ S . Recall that the completion of S is T , and so c ′ T ∩ S = c ′ S ; in particular, c = c ′ s for some s ∈ S . Thus, ab = cdc ′ d ′ = c ′ sdc ′ d ′ = sdd ′ = s ( dd ′− ) . If s ∈ S \ M , then ab / ∈ M , which contradicts the assumption that ab ∈ B ∩ M . It must bethat s ∈ S ∩ M , and so r = ab can be written in the desired form. Thus, we have that B isindeed an S -subring of T .We show that B is quasi-local with maximal ideal B ∩ M by showing that the units in B are exactly the elements that are not in B ∩ M . If x ∈ B is a unit in B , then x is aunit in T , so x / ∈ M and x / ∈ B ∩ M . Now suppose that x ∈ B \ ( B ∩ M ). Then, x = 0 OMPLETIONS OF UNCOUNTABLE LOCAL RINGS WITH COUNTABLE SPECTRA 9 and x / ∈ M , so x must be a unit in T . There therefore exists a y ∈ T with xy = 1. Since y = x ∈ QF ( A ) ∩ T = B , we have that x, y ∈ B and x is a unit in B .Finally, we show that if I is a principal ideal of B then IT ∩ B = IB . Notice that I = (cid:0) ab (cid:1) B for some a, b ∈ A with b nonzero. If a = 0, then, IT = (0) and IB = (0), so IT ∩ B = (0) = IB . Now suppose that a = 0. If c ∈ IT ∩ B , then c = ab t for some t ∈ T .In addition, c = a ′ b ′ for some a ′ , b ′ ∈ A with b ′ = 0, since c ∈ B . Now, t = a ′ bb ′ a ∈ B , so c ∈ ab B = IB . It follows that IT ∩ B ⊆ IB . Since IB ⊆ IT and IB ⊆ B , we have that IB ⊆ IT ∩ B . Thus, IT ∩ B = IB , as desired. (cid:3) We now use the properties of the ring B constructed in Theorem 3.7 to show, in the nextthree results, that B Noetherian and has completion T . These results are generalizations of[10, Lemma 6.1, Theorem 6.2, Theorem 6.3]. Lemma 3.8. Suppose ( T, M ) is a complete local ring with dim T ≥ S, S ∩ M ) isa countable local domain such that S ⊆ T and b S = T . Then there exists an uncountablequasi-local S -subring of T , ( B, B ∩ M ), such that finitely generated ideals of B are extendedfrom S , i.e., for any finitely generated ideal J of B , J = ( p , . . . , p k ) B for p i ∈ S . Proof. Let ( B, B ∩ M ) be the uncountable quasi-local S -subring of T constructed in Theo-rem 3.7, such that IT ∩ B = IB for every principal ideal I of B .If J is a finitely generated ideal of B , then J = ( b , . . . , b k ) B for some b i ∈ B . If J = (0),then J is extended from the zero ideal of S . If J = B then J = 1 B and 1 ∈ S . So assumethat b i are nonzero nonunits for all i = 1 , , . . . , k .Then, we have that b i ∈ B ∩ M , so b i = p i u i , where p i ∈ S ∩ M , and where u i is a unit in T for all i = 1 , . . . , k . We show that each u i is also a unit in B . Notice that b i = p i u i ∈ p i T ∩ B .Since p i B is a principal ideal of B , we have that p i T ∩ B = p i B by Theorem 3.7. Thus, p i u i ∈ p i B , and since p i is not a zero divisor in T , we have that u i ∈ B . Since u i / ∈ M , itmust be that u i is a unit in B . Then, the p i ’s and b i ’s generate the same ideals in B sincethey are associates, so J = ( b , . . . , b k ) B = ( p , . . . , p k ) B . Recall that p i ∈ S for all i , so J isextended from S . (cid:3) Theorem 3.9. Suppose ( T, M ) is a complete local ring with dim T ≥ S, S ∩ M ) isa countable local domain such that S ⊆ T and b S = T . Then there exists an uncountablequasi-local S -subring of T , ( B, B ∩ M ), such that, for every finitely generated ideal I of B , IT ∩ B = IB , and finitely generated ideals of B are extended from S . Proof. Let ( B, B ∩ M ) be the uncountable quasi-local S -subring of T whose existence isguaranteed by Lemma 3.8. We have that every finitely generated ideal of B is extendedfrom S . In addition, we always have that IB ⊆ IT ∩ B , so we now show that IT ∩ B ⊆ IB .Let I be a finitely generated ideal of B , which can be written as I = ( p , . . . , p k ) B forsome p i ∈ S by Lemma 3.8. Consider c ∈ IT ∩ B . We will show that c ∈ IB . First, if I = B , then IT ∩ B = BT ∩ B = B = IB , so now suppose I = B . Then, IT ⊆ M , so c ∈ IT ∩ B ⊆ M ∩ B . Note that c = qu for some q ∈ S ∩ M and unit u ∈ T since B is an S -subring of T . Then, qu = c ∈ qT ∩ B = qB . But this implies that u ∈ B and it is a unitin B . Finally, observe that cu − ∈ ( p , . . . , p k ) T , since c ∈ ( p , . . . , p k ) T ∩ B and u − ∈ T ,and recall that cu − = q ∈ S . Also recall that b S = T , and so for any finitely generated ideal I of S , we have that IT ∩ S = IS . Thus, cu − = q ∈ ( p , . . . , p k ) T ∩ S = ( p , . . . , p k ) S ⊆ ( p , . . . , p k ) B = IB. Since u is a unit in B , we have that c ∈ IB . (cid:3) Theorem 3.10. Suppose ( T, M ) is a complete local ring with dim T ≥ S, S ∩ M ) isa countable local domain such that S ⊆ T and b S = T . Then there exists an uncountablelocal domain ( B, B ∩ M ) such that S ⊆ B ⊆ T , b B = T , and ideals of B are extended from S . Proof. By Theorem 3.9, there exists an uncountable quasi-local S -subring of T , ( B, B ∩ M ),such that IT ∩ B = IB for all finitely generated ideals I of B , and all finitely generatedideals of B are extended from S .By Lemma 3.3, we have that B is a domain because it is an S -subring of T . In addition, S ⊆ B and ( S, S ∩ M ) is a local subring of T such that b S = T . Thus, by Proposition 2.5, B isNoetherian with completion T . Since B is Noetherian, all of its ideals are finitely generated,and it follows that all ideals of B are extended from S . (cid:3) Main Results In this section, we use our results from Section 3 to characterize completions of uncountablelocal domains with countable spectra that satisfy various properties, such as being excellent,a UFD, and noncatenary.In order to show many of our results, we view the spectra of the rings in question as partiallyordered sets (posets) under the relation of set containment. For two posets, ( X, ≤ X ) and( Y, ≤ Y ), an order isomorphism from X to Y is a bijective function f : X → Y such thatfor all x, x ′ ∈ X , we have that x ≤ X x ′ if and only if f ( x ) ≤ Y f ( x ′ ). We now show thatif two local domains, one contained in the other, have the same completion and satisfy thecondition that every ideal of the larger one is extended from the smaller one, then the twolocal domains’ spectra are order isomorphic. Proposition 4.1. Let ( T, M ) be a complete local ring, and suppose that ( S, S ∩ M ) and( B, B ∩ M ) are local domains such that S ⊆ B , b S = b B = T , and every ideal of B isextended from S . Then the mapping ϕ : Spec( B ) → Spec( S ) given by P P ∩ S is anorder isomorphism. Proof. Note that if P is a prime ideal of B , then P ∩ S is a prime ideal of S .We show that ϕ is an order isomorphism by showing that ϕ is surjective, and that if P, Q ∈ Spec( B ), then P ⊆ Q if and only if ϕ ( P ) ⊆ ϕ ( Q ). Notice that this latter conditionalso shows injectivity: if ϕ ( P ) = ϕ ( Q ), then we have that ϕ ( P ) ⊆ ϕ ( Q ) and ϕ ( Q ) ⊆ ϕ ( P ),thus ensuring that P ⊆ Q and Q ⊆ P , i.e., P = Q .To show that ϕ is surjective, suppose Q ∈ Spec( S ), and let P be a prime ideal of T with P ∈ Min( QT ). Then, P ∩ B ∈ Spec( B ). Furthermore, ϕ ( P ∩ B ) = ( P ∩ B ) ∩ S = P ∩ S = Q ,by Lemma 2.6. Thus, ϕ is surjective.Suppose P, Q ∈ Spec( B ) with P ⊆ Q . Then P ∩ S ⊆ Q ∩ S , and so ϕ ( P ) ⊆ ϕ ( Q ). Nowsuppose ϕ ( P ) ⊆ ϕ ( Q ), with P = ( x , . . . , x n ) B for x i ∈ S . Then, for all i = 1 , , . . . n , wehave that x i ∈ P ∩ S = ϕ ( P ) ⊆ ϕ ( Q ) = Q ∩ S . It follows that x i ∈ Q for all i , and we havethat P ⊆ Q . Thus, we have that ϕ is an order isomorphism and that Spec( B ) and Spec( S )are isomorphic as posets. (cid:3) If the spectra of two rings are isomorphic as posets, then, for any prime ideal of height k in one ring, there exists a corresponding prime ideal of height k in the other ring. Similarly,if there is a saturated chain of prime ideals of length n between two prime ideals P and Q OMPLETIONS OF UNCOUNTABLE LOCAL RINGS WITH COUNTABLE SPECTRA 11 of one ring, then there exist corresponding prime ideals P ′ and Q ′ in the other ring, as wellas a saturated chain of prime ideals of length n between them. Remark 4.2. Using the isomorphism from Proposition 4.1, we note that the prime idealsof B and S that correspond to each other are generated by the same elements of S . If P ∈ Spec( B ), then P = ( a , . . . , a n ) B with a i ∈ S for all i . Then, since b B = b S = T andsince S ⊆ B , we have the following equalities: ϕ ( P ) = ( a , . . . , a n ) B ∩ S = (( a , . . . , a n ) T ∩ B ) ∩ S = ( a , . . . , a n ) T ∩ S = ( a , . . . , a n ) S. We now focus on characterizing completions of uncountable local domains with countablespectra. In the dimension zero case, this characterization comes as a direct consequence ofthe fact that the completions of fields are themselves, and the fact that all local domains ofdimension zero have exactly one prime ideal. Proposition 4.3. Suppose ( T, M ) is a complete local ring of dimension zero. Then T is thecompletion of an uncountable local domain with a countable spectrum if and only if T is anuncountable field. Proof. First suppose T is the completion of an uncountable local domain with a countablespectrum. This uncountable local domain with a countable spectrum must be dimensionzero, so it must be a field. However, the completion of a field is itself, so it must be T . Thus, T is uncountable and a field.Now suppose T is an uncountable field. Then, T has one prime ideal and b T = T . Thus, T is the completion of an uncountable local domain with a countable spectrum. (cid:3) We use our construction from Section 3 to tackle the sufficient conditions in the case thatthe dimension of the rings in question are at least one. First, we show that the uncountablelocal domain that we construct in the previous section has a countable spectrum. Proposition 4.4. Let ( T, M ) be a complete local ring, and suppose that ( S, S ∩ M ) and( B, B ∩ M ) are local domains such that S is countable, S ⊆ B , b S = b B = T , and every idealof B is extended from S . Then Spec( B ) is countable. Proof. By Proposition 4.1, we have that Spec( B ) and Spec( S ) are isomorphic as posets.In particular, they have the same cardinalities, so since Spec( S ) is countable, Spec( B ) iscountable as well. (cid:3) The following result from [2] characterizes completions of countable local domains. Theorem 4.5 ([2], Corollary 2.15) . Let ( T, M ) be a complete local ring. Then T is thecompletion of a countable local domain if and only if(1) no integer is a zero divisor of T ,(2) unless equal to (0), M / ∈ Ass( T ), and(3) T /M is countableWe use Theorem 4.5 to identify sufficient conditions for a complete local ring of dimensionat least one to be the completion of an uncountable local domain with a countable spectrum. Proposition 4.6. Suppose ( T, M ) is a complete local ring with dim T ≥ T is a zero divisor, (2) M / ∈ Ass( T ), and(3) T /M is countable.Then T is the completion of an uncountable local domain with a countable spectrum. Proof. If T satisfies conditions (1), (2), and (3), then, by Theorem 4.5, T is the completionof a countable local domain. Let this countable local domain be ( S, S ∩ M ) and applyTheorem 3.10. Then, we have that there exists an uncountable local domain ( B, B ∩ M ) suchthat S ⊆ B ⊆ T , b B = T , and ideals of B are extended from S . Finally, by Proposition 4.4,we have that B has a countable spectrum. (cid:3) We are now able to prove one of our main results. Theorem 4.7. Suppose ( T, M ) is a complete local ring. • If dim T = 0, then T is the completion of an uncountable local domain with acountable spectrum if and only if T is an uncountable field. • If dim T = 1, then T is the completion of an uncountable local domain with acountable spectrum if and only if(1) no integer of T is a zero divisor, and(2) M / ∈ Ass( T ). • If dim T ≥ 2, then T is the completion of an uncountable local domain with acountable spectrum if and only if(1) no integer of T is a zero divisor,(2) M / ∈ Ass( T ), and(3) T /M is countable. Proof. First, if dim T = 0, then the desired statement is given by Proposition 4.3.Suppose that ( T, M ) is a complete local ring with dim T ≥ 1, no integer of T is a zerodivisor, M / ∈ Ass( T ), and T /M is countable. By Proposition 4.6 we have that T is thecompletion of an uncountable local domain with a countable spectrum. Now suppose that( T, M ) is a complete local ring with dim T = 1, no integer of T is a zero divisor, M / ∈ Ass( T ),and T /M is uncountable. By Theorem 2.2, T is the completion of an uncountable localdomain A . Since dim A = 1, A has a countable spectrum.Now suppose that ( T, M ) is a complete local ring with dim T ≥ T is a zero divisor and M / ∈ Ass( T ). If dim T ≥ 2, then it must also bethat T /M is countable by Theorem 2.3. (cid:3) We present a corollary of our result in the case that T itself is a domain. Corollary 4.8. Suppose ( T, M ) is a complete local domain. • If dim T = 0, then T is the completion of an uncountable local domain with acountable spectrum if and only if T is uncountable. • If dim T = 1, then T is the completion of an uncountable local domain with acountable spectrum. • If dim T ≥ 2, then T is the completion of an uncountable local domain with acountable spectrum if and only if T /M is countable. Proof. Suppose dim T = 0. Since T is a domain, it is a field, and so the result follows fromTheorem 4.7. Now suppose dim T ≥ 1. Since T is a domain, we have that no integer of T isa zero divisor and M / ∈ Ass( T ). If dim T = 1, then T is the completion of an uncountable OMPLETIONS OF UNCOUNTABLE LOCAL RINGS WITH COUNTABLE SPECTRA 13 local domain with a countable spectrum by Theorem 4.7. Suppose that dim T ≥ 2. Then, byTheorem 4.7, T is the completion of an uncountable local domain with a countable spectrumif and only if T /M is countable. (cid:3) Example 4.9. By Theorem 4.7, we have that any complete local ring of dimension atleast 2 that is of the form R [[ x , . . . , x n ]] /I or C [[ x , . . . , x n ]] /I is not the completion of anuncountable local domain with a countable spectrum.Using Corollary 4.8, we see that the complete local ring T = Q [[ x , . . . , x n ]] for any n ≥ T /M = Q is countable. Note that T is one of the “nicest” examples of a complete local ring with acountable residue field, so if any ring were to be the completion of an uncountable localdomain with a countable spectrum, it is not that surprising that T would be one. However,we know by Theorem 4.7 that the complete local ring T ′ = Q [[ x, y, z ]] / ( x ) is also thecompletion of an uncountable local domain with a countable spectrum, even though T ′ isnot even a domain itself.4.1. Uncountable Excellent Domains with Countable Spectra. An important class ofrings that are of particular interest to algebraic geometers and number theorists are excellent rings. For any P ∈ Spec( A ), define k ( P ) := A P /P A P . Definition 4.10 ([13], Definition 1.4) . A local ring A is excellent if(a) for all P ∈ Spec( A ), b A ⊗ A L is regular for every finite field extension L of k ( P ), and(b) A is universally catenary.In [9], Loepp characterizes completions of excellent domains in the characteristic zero case. Theorem 4.11 ([9], Theorem 9) . Let ( T, M ) be a complete local ring containing the in-tegers. Then T is the completion of a local excellent domain if and only if it is reduced,equidimensional, and no integer of T is a zero divisor.We characterize completions of uncountable excellent local domains with countable spec-tra, assuming that the completion contains the rationals. We make use of results from [11]to accomplish this. The following lemma identifies sufficient conditions on a local ring to beexcellent. Lemma 4.12 ([11], Lemma 2.8) . Let ( T, M ) be a complete local ring that is equidimensionaland suppose Q ⊆ T . Given a subring ( A, A ∩ M ) of T with b A = T , A is excellent if, forevery P ∈ Spec( A ) and for every Q ∈ Spec( T ) with Q ∩ A = P , ( T /P T ) Q is a regular localring.The following result characterizes completions of countable excellent local domains, as-suming that the completion contains the rationals. Theorem 4.13 ([11], Theorem 3.10) . Let ( T, M ) be a complete local ring with Q ⊆ T .Then T is the completion of a countable excellent local domain if and only if the followingconditions hold:(1) T is equidimensional,(2) T is reduced, and(3) T /M is countable. Given that complete local rings are also excellent, we have the following corollary of Propo-sition 4.3. In particular, Corollary 4.14 characterizes completions of uncountable excellentlocal domains with countable spectra in the dimension zero case. Corollary 4.14. Suppose ( T, M ) is a complete local ring with dim T = 0. Then T is thecompletion of an uncountable excellent local domain with a countable spectrum if and onlyif T is an uncountable field. Proof. Suppose T is the completion of an uncountable excellent local domain with a countablespectrum. By Proposition 4.3, we have that T is an uncountable field.Suppose T is an uncountable field. Then, T is the completion of an uncountable excellentlocal domain with a countable spectrum, namely itself. (cid:3) We now identify sufficient conditions for a complete local ring of dimension at least oneand containing the rationals to be the completion of an uncountable excellent local domainwith a countable spectrum. The following proposition is a modification of [10, Theorem 6.4] Proposition 4.15. Suppose ( T, M ) is a complete local ring such that dim T ≥ Q ⊆ T ,and suppose T satisfies the following conditions:(1) T is reduced,(2) T is equidimensional, and(3) T /M is countable.Then T is the completion of an uncountable excellent local domain with a countable spec-trum. Proof. Since T satisfies conditions (1), (2), and (3), we have by Theorem 4.13 that T is thecompletion of a countable excellent local domain, ( S, S ∩ M ). By Theorem 3.10, there existsan uncountable local domain ( B, B ∩ M ) such that S ⊆ B ⊆ T , b B = T , and ideals of B are extended from S . Finally, by Proposition 4.4, we have that B has a countable spectrum.Thus, it remains to show that B is excellent.Since b B = T and T is assumed to contain Q and be equidimensional, we can applyLemma 4.12. In particular, it suffices to show that, for every P ∈ Spec( B ) and for every Q ∈ Spec( T ) satisfying Q ∩ B = P , we have that ( T /P T ) Q is a regular local ring. Let P ∈ Spec( B ), and suppose Q ∈ Spec( T ) such that Q ∩ B = P . Then, Q ∩ S = ( Q ∩ B ) ∩ S = P ∩ S. Since ideals of B are extended from S , we can write P as P = ( p , . . . , p k ) B for p i ∈ S .Then, P ∩ S = ( p , . . . , p k ) B ∩ S ⊆ ( p , . . . , p k ) T ∩ S = ( p , . . . , p k ) S. Note that ( p , . . . , p k ) S ⊆ P ∩ S , and so ( p , . . . , p k ) B ∩ S = ( p , . . . , p k ) S . It follows that Q ∩ S = ( p , . . . , p k ) S .Now, ( T / ( p , . . . , p k ) T ) Q is a regular local ring, since S is excellent and has completion T .But notice that ( p , . . . , p k ) T = P T , so we have that ( T / ( p , . . . , p k ) T ) Q = ( T /P T ) Q , and( T /P T ) Q is a regular local ring as well. Thus, by Lemma 4.12, B is excellent. (cid:3) We are now able to prove another main result. Theorem 4.16. Suppose ( T, M ) is a complete local ring with Q ⊆ T . • If dim T = 0, then T is the completion of an uncountable excellent local domain witha countable spectrum if and only if T is an uncountable field. OMPLETIONS OF UNCOUNTABLE LOCAL RINGS WITH COUNTABLE SPECTRA 15 • If dim T = 1, then T is the completion of an uncountable excellent local domain witha countable spectrum if and only if(1) T is reduced, and(2) T is equidimensional. • If dim T ≥ 2, then T is the completion of an uncountable excellent local domain witha countable spectrum if and only if(1) T is reduced,(2) T is equidimensional, and(3) T /M is countable. Proof. If dim T = 0 then the desired statement is given by Corollary 4.14.Let ( T, M ) be a complete local ring with dim T ≥ Q ⊆ T , and suppose that T satisfiesconditions (1), (2), and (3). By Proposition 4.15, we have that T is the completion of anuncountable excellent local domain with a countable spectrum. Now suppose ( T, M ) is acomplete local ring with dim T = 1, such that T /M is uncountable and T satisfies conditions(1) and (2). Since T satisfies these conditions and Q ⊆ T (so every nonzero integer is aunit and not a zero divisor), T is the completion of a local excellent domain ( A, A ∩ M ),by Theorem 4.11. Note that dim A = 1 and so A has a countable spectrum. Since thecompletion of A is T , the map A → T /M is surjective (see, for example, Proposition 2.4 of[11]). By Proposition 2.7, we have that | A | ≥ | T /M | = | T /M | . Thus, since T /M is uncountable, A is uncountable as well.Now suppose that ( T, M ) is a complete local ring with dim T ≥ T is reduced and equidimensional. If dim T ≥ 2, then we must also have that T /M is countable by Theorem 2.3. (cid:3) Example 4.17. If T = Q [[ x, y, z ]] / ( xy ), then, by Theorem 4.16, we have that T is thecompletion of an uncountable excellent local domain with a countable spectrum.We now have the following corollary in the case that T is a UFD. Corollary 4.18. Suppose ( T, M ) is a complete local UFD with Q ⊆ T . • If dim T = 0, then T is the completion of an uncountable excellent local UFD with acountable spectrum if and only if T is uncountable. • If dim T = 1, then T is the completion of an uncountable excellent local UFD with acountable spectrum. • If dim T ≥ 2, then T is the completion of an uncountable excellent local UFD witha countable spectrum if and only if T /M is countable. Proof. First, if dim T = 0, then T is a field since it is a domain. The result follows from thecase dim T = 0 of Theorem 4.16.Next, suppose that ( T, M ) is a complete local UFD with dim T = 1 and Q ⊆ T . Then,since T is a domain, it is reduced and equidimensional. By Theorem 4.16, T is the completionof an uncountable excellent local domain with a countable spectrum, ( A, A ∩ M ). Since T is a UFD, A is as well.Suppose that ( T, M ) is a complete local UFD with dim T ≥ Q ⊆ T , and T /M iscountable. Since T is a domain, it is reduced and equidimensional. By Theorem 4.16, T is the completion of an uncountable excellent local domain with a countable spectrum,( A, A ∩ M ). Since T is a UFD, A is as well. If ( T, M ) is the completion of an uncountableexcellent local UFD with a countable spectrum ( A, A ∩ M ), then T /M is countable byTheorem 4.7. (cid:3) Uncountable UFDs with Countable Spectra. In this subsection, we characterizecompletions of uncountable local UFDs with countable spectra. First, we provide resultsfrom [6], which identify necessary and sufficient conditions for a complete local ring to bethe completion of a local UFD. Theorem 4.19 ([6], Theorem 1) . Let R be an integrally closed local domain. Then nointeger is a zero divisor in b R . Moreover, b R is either a field, a DVR, or a ring with depth atleast two. Theorem 4.20 ([6], Theorem 8) . Let ( T, M ) be a complete local ring such that no integeris a zero divisor in T and depth T ≥ 2. Then there exists a local UFD A such that b A ∼ = T and | A | = sup( ℵ , | T /M | ). If p ∈ M where p is a nonzero prime integer, then pA is a primeideal.Notice that Theorem 4.20 provides sufficient conditions for a complete local ring to be thecompletion of a countable local UFD.We now characterize completions of uncountable local UFDs with countable spectra inthe cases of dimension zero and dimension one. Proposition 4.21. Suppose ( T, M ) is a complete local ring. • If dim T = 0, then T is the completion of an uncountable local UFD with a countablespectrum if and only if T is an uncountable field. • If dim T = 1, then T is the completion of an uncountable local UFD with a countablespectrum if and only if T is a DVR. Proof. First suppose dim T = 0 and T is the completion of an uncountable local UFDwith a countable spectrum. By Theorem 4.7, T is an uncountable field. Now suppose T is an uncountable field. Then, T is an uncountable local UFD with one prime ideal, and b T = T . Therefore, T is indeed the completion of an uncountable local UFD with a countablespectrum.Next, suppose dim T = 1 and T is the completion of an uncountable local UFD with acountable spectrum A . Since A is integrally closed, by Theorem 4.19, it must be that b A = T is a DVR (as it cannot be a field or have depth at least 2). Now suppose that T is a DVR.Then, T itself is uncountable by Lemma 2.8. Since a DVR is also a UFD, T is also a UFD.Finally, since b T = T and T has two prime ideals, we have that T is the completion of anuncountable UFD with a countable spectrum. (cid:3) In order to show that the rings we construct are UFDs, we make use of the followingcharacterization of UFDs for Noetherian domains. Theorem 4.22 ([12], Theorem 20.1) . If A is a Noetherian integral domain, then A is a UFDif and only if every height 1 prime ideal is principal.We now use our construction from Section 3 to identify sufficient conditions in the casethat the dimension of the rings in question are at least two. OMPLETIONS OF UNCOUNTABLE LOCAL RINGS WITH COUNTABLE SPECTRA 17 Proposition 4.23. Suppose ( T, M ) is a complete local ring with dim T ≥ T satisfies the following conditions:(1) no integer of T is a zero divisor,(2) the depth of T is at least 2, and(3) T /M is countable.Then T is the completion of an uncountable local UFD with a countable spectrum. Proof. Since T satisfies conditions (1), (2), and (3), we have by Theorem 4.20 that T isthe completion of a countable local UFD. Let this countable local UFD be ( S, S ∩ M ). ByTheorem 3.10, there exists an uncountable local domain ( B, B ∩ M ) such that S ⊆ B ⊆ T , b B = T , and ideals of B are extended from S . Finally, by Proposition 4.4, we have that B has a countable spectrum. Thus, it remains to show that B is a UFD.Let P ∈ Spec( B ) be a height one prime ideal of B . By Theorem 4.22, it is enough to showthat P is a principal ideal. By Proposition 4.1, P corresponds to a height one prime ideal Q of S . Since S is a Noetherian UFD, we have by Theorem 4.22 that Q is a principal ideal, so Q = xS for some x ∈ S . By Remark 4.2, we have that P is generated by the same elementsthat generate Q , so P = xB . Thus P is principal, and so B is indeed a UFD. (cid:3) We are now ready to characterize completions of uncountable local UFDs with countablespectra. Theorem 4.24. Suppose ( T, M ) is a complete local ring. • If dim T = 0, then T is the completion of an uncountable local UFD with a countablespectrum if and only if T is an uncountable field. • If dim T = 1, then T is the completion of an uncountable local UFD with a countablespectrum if and only if T is a DVR. • If dim T ≥ 2, then T is the completion of an uncountable local UFD with a countablespectrum if and only if(1) no integer of T is a zero divisor,(2) the depth of T is at least 2, and(3) T /M is countable. Proof. If dim T = 0 or dim T = 1, the desired statement follows from Proposition 4.21.If dim T ≥ 2, by Proposition 4.23, conditions (1), (2), and (3) are sufficient for T to bethe completion of an uncountable local UFD with a countable spectrum. Now suppose that T is the completion of such a UFD. By Theorem 4.19, we have that no integer of T is a zerodivisor and the depth of T is at least 2. By Theorem 2.3, we have that T /M is countable. (cid:3) Example 4.25. Surprisingly, the ring T ′ = Q [[ x, y, z ]] / ( x ) from Example 4.9 is not onlythe completion of an uncountable local domain with a countable spectrum, but also thecompletion of an uncountable local UFD with a countable spectrum by Theorem 4.24.4.3. Uncountable Noncatenary Domains and UFDs with Countable Spectra. Inthis subsection, we characterize completions of uncountable noncatenary local domains withcountable spectra, and completions of uncountable noncatenary local UFDs with countablespectra.First, we provide results from [1] and [11], which identify necessary and sufficient con-ditions for a complete local ring to be the completion of a noncatenary local domain andnecessary and sufficient conditions for a complete local ring to be the completion of a count-able noncatenary local domain. Theorem 4.26 ([1], Theorem 2.10) . Let ( T, M ) be a complete local ring. Then T is thecompletion of a noncatenary local domain if and only if the following conditions hold:(1) no integer of T is a zero divisor,(2) M / ∈ Ass( T ), and(3) there exists P ∈ Min( T ) such that 1 < dim( T /P ) < dim T . Theorem 4.27 ([11], Theorem 4.5) . Let ( T, M ) be a complete local ring. Then T is thecompletion of a countable noncatenary local domain if and only if the following conditionshold:(1) no integer of T is a zero divisor,(2) M / ∈ Ass( T ),(3) there exists P ∈ Min( T ) such that 1 < dim( T /P ) < dim T , and(4) T /M is countable.We use the latter result to identify sufficient conditions for a complete local ring to be thecompletion of an uncountable noncatenary local domain with a countable spectrum. Proposition 4.28. Suppose ( T, M ) is a complete local ring that satisfies the followingconditions:(1) no integer of T is a zero divisor,(2) M / ∈ Ass( T ),(3) there exists P ∈ Min( T ) such that 1 < dim( T /P ) < dim T , and(4) T /M is countable.Then T is the completion of an uncountable noncatenary local domain with a countablespectrum. Proof. Since T satisfies conditions (1), (2), (3), and (4), we have by Theorem 4.27 that T isthe completion of a countable noncatenary local domain, ( S, S ∩ M ). Notice that condition(3) guarantees that dim T ≥ 1. Thus, by Theorem 3.10, there exists an uncountable localdomain ( B, B ∩ M ) such that S ⊆ B ⊆ T , b B = T , and ideals of B are extended from S .Finally, by Proposition 4.4, we have that B has a countable spectrum. Thus, it remains toshow that B is noncatenary.Since S is noncatenary, there exists a pair of prime ideals P ( Q of S such that twosaturated chains of prime ideals between P and Q have different lengths. By Proposition 4.1,Spec( B ) and Spec( S ) are order isomorphic, and so there also exist two prime ideals P ′ ( Q ′ of B such that two saturated chains of prime ideals between P ′ and Q ′ have different lengths.Thus, B is noncatenary as well. (cid:3) We now characterize completions of uncountable noncatenary local domains with count-able spectra. Theorem 4.29. Suppose ( T, M ) is a complete local ring. Then T is the completion of anuncountable noncatenary local domain with a countable spectrum if and only if the followingconditions are satisfied:(1) no integer of T is a zero divisor,(2) M / ∈ Ass( T ),(3) there exists P ∈ Min( T ) such that 1 < dim( T /P ) < dim T , and(4) T /M is countable. OMPLETIONS OF UNCOUNTABLE LOCAL RINGS WITH COUNTABLE SPECTRA 19 Proof. If T satisfies conditions (1), (2), (3), and (4), then, by Proposition 4.28, T is thecompletion of an uncountable noncatenary local domain with a countable spectrum. If T isthe completion of such a ring, then, by Theorem 4.26, T must satisfy conditions (1), (2), and(3). Since condition (3) is satisfied, we have that dim T > 2, so, by Theorem 2.3, it must bethat T /M is countable and condition (4) is satisfied. (cid:3) We now use similar arguments as those in Propositions 4.23 and 4.28 to identify sufficientconditions for a complete local ring to be the completion of an uncountable noncatenarylocal UFD with a countable spectrum. The following results from [1] and [11] characterizecompletions of noncatenary local UFDs and countable noncatenary local UFDs. Theorem 4.30 ([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 ) > 1, and(3) there exists P ∈ Min( T ) such that 2 < dim( T /P ) < dim T . Theorem 4.31 ([11], Theorem 4.10) . Let ( T, M ) be a complete local ring. Then T is thecompletion of a countable noncatenary local UFD if and only if the following conditions hold:(1) no integer of T is a zero divisor,(2) depth T > P ∈ Min( T ) such that 2 < dim( T /P ) < dim T , and(4) T /M is countable.We use Theorem 4.31 to identify sufficient conditions on a complete local ring to be thecompletion of an uncountable noncatenary local UFD with a countable spectrum. Proposition 4.32. Suppose ( T, M ) is a complete local ring that satisfies the followingconditions:(1) no integer of T is a zero divisor,(2) depth T > P ∈ Min( T ) such that 2 < dim( T /P ) < dim T , and(4) T /M is countable.Then T is the completion of an uncountable noncatenary local UFD with a countable spec-trum. Proof. Since T satisfies conditions (1), (2), (3), and (4), we have by Theorem 4.31 that T is the completion of a countable noncatenary local UFD, ( S, S ∩ M ). Notice that theconditions of Theorem 3.10 are met, since dim T ≥ B, B ∩ M ) such that S ⊆ B ⊆ T , b B = T , and ideals of B are extended from S . Finally, by Proposition 4.4, we have that B has a countable spectrum.Thus, it remains to show that B is a noncatenary UFD.By Proposition 4.1, Spec( B ) and Spec( S ) are order isomorphic. If P ∈ Spec( B ) is a heightone prime ideal, then it corresponds to a height one prime ideal of S , which is principal since S is a Noetherian UFD. By Remark 4.2, this principal ideal of S and P are generatedby the same elements of S , so P is also principal. Thus, B is a UFD. Similarly, since S is noncatenary, there exist prime ideals P ( Q of S such that two saturated chains ofprime ideals between P and Q have different lengths. Since Spec( B ) and Spec( S ) are order isomorphic, there exist corresponding saturated chains of different lengths between two primeideals of B . Thus, B is noncatenary. (cid:3) We now characterize completions of uncountable noncatenary local UFDs with countablespectra. Theorem 4.33. Suppose ( T, M ) is a complete local ring. Then T is the completion of anuncountable noncatenary local UFD with a countable spectrum if and only if the followingconditions are satisfied:(1) no integer of T is a zero divisor,(2) depth T > P ∈ Min( T ) such that 2 < dim( T /P ) < dim T , and(4) T /M is countable. Proof. If T satisfies conditions (1), (2), (3), and (4), then, by Proposition 4.32, T is thecompletion of an uncountable noncatenary local UFD with a countable spectrum. If T isthe completion of such a ring, then, by Theorem 4.30, T must satisfy conditions (1), (2),and (3). Since condition (3) is satisfied, dim T > 3, so, by Theorem 2.3, we have that T /M is countable and condition (4) must be satisfied as well. (cid:3) Example 4.34. By Theorem 4.29, the ring Q [[ x, y, z, w ]] / ( x ) ∩ ( y, z ) is the completion ofan uncountable noncatenary local domain with a countable spectrum. Similarly, by The-orem 4.33, the ring Q [[ x, y , y , z , z ]] / ( x ) ∩ ( y , y ) is the completion of an uncountablenoncatenary local UFD with a countable spectrum. Acknowledgments We thank the Clare Boothe Luce Scholarship Program for supporting the research of thesecond author. References [1] Chloe I. Avery, Caitlyn Booms, Timothy M. Kostolansky, S. Loepp, and Alex Semendinger, Charac-terization of completions of noncatenary local domains and noncatenary local UFDs , J. Algebra (2019), 1–18, DOI 10.1016/j.jalgebra.2018.12.016. MR3902351 ↑ 17, 18, 19[2] Erica Barrett, Emil Graf, S. 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