A Structural Invariant On Certain Two-Dimensional Noetherian Partially Ordered Sets
aa r X i v : . [ m a t h . A C ] F e b A STRUCTURAL INVARIANT ON CERTAIN TWO-DIMENSIONALNOETHERIAN PARTIALLY ORDERED SETS
CORY H. COLBERT
Abstract.
If ( X, ≤ X ) is a partially ordered set satisfying certain necessary conditions for X tobe order-isomorphic to the spectrum of a Noetherian domain of dimension two, we describe a newposet (Str X, ≤ Str X ) that completely determines X up to isomorphism. The order relation ≤ Str X imposed on Str X is modeled after R. Wiegand’s well-known “P5” condition that can be usedto determine when a given partially ordered set ( U, ≤ U ) of a certain type is order-isomorphic to(Spec Z [ x ] , ⊆ ) . introduction In 1978, R. Wiegand proved in [[3], Theorem 1] that there is only one partially ordered set( X, ≤ X ), up to isomorphism, that satisfies the following five properties:P1 X is countable and has a unique minimal element.P2 X has dimension 2.P3 For each element x of height one there are infinitely many elements y > x. P4 For each pair x, y of distinct elements of height one, there are only finitely many elements z such that z > x and z > y. P5 Given a finite set S of height-one elements and a finite set T of maximal elements, there isa height-one element w such that (a) w < t for each t ∈ T, and (b) if s ∈ S and s < x and w < x then x ∈ T. In particular, if R and S are commutative Noetherian rings, and both (Spec R, ⊆ ) and (Spec S, ⊆ )satisfy Conditions P1, . . . , P5, then (Spec R, ⊆ ) and (Spec S, ⊆ ) are isomorphic as partially orderedsets. Theorem 2 in [4] asserts that if k is a field, and A is a two-dimensional domain that is finitelygenerated as a k -algebra, then (Spec A, ⊆ ) and (Spec Z [ x ] , ⊆ ) are isomorphic if and only if k iscontained in the algebraic closure of a finite field. In particular, (Spec F p [ x, y ] , ⊆ ) and (Spec Z [ x ] , ⊆ )are isomorphic for all prime numbers p, but (Spec Q [ x, y ] , ⊆ ) is not isomorphic to (Spec Z [ x ] , ⊆ )because Q is not contained in the algebraic closure of a finite field.In the following years, work was done to understand what other Noetherian spectra satisfy (anddo not satisfy) the five conditions, and since the spectrum of many commutative Noetherian ringsalready satisfies the first four conditions, the work is typically reduced to showing that a candidatespectrum satisfies Condition P5 in order to conclude that the spectrum is isomorphic, as a partiallyordered set, to (Spec Z [ x ] , ⊆ ) . It was conjectured in [4] that if A is a two-dimensional domain finitelygenerated as a Z -algebra, then (Spec A, ⊆ ) is isomorphic to (Spec Z [ x ] , ⊆ ) . Important advancements
Mathematics Subject Classification.
Primary 13E05. Secondary 06A06.
Key words and phrases.
Noetherian spectrum, partially ordered set, invariant, ascending chain condition, descend-ing chain condition.The author was partially supported by a Lenfest grant from Washington and Lee University. have been made in the direction of that conjecture, including the work of A. Saydam and S. Wiegand,who showed in [[2], Theorem 1.2] that (Spec D [ x, g , . . . , g n ] , ⊆ ) satisfies Condition P5, where D isthe order in an algebraic number field, x is an indeterminate, and g , . . . , g n are nonzero elementsof the quotient field of D [ x ] . While it is known that if two partially ordered sets ( X, ≤ X ) and ( Y, ≤ Y ) satisfy all five conditionsthen they must be isomorphic, little can be said if they do not satisfy P5; they could be isomorphicor they could be very different. For instance, while it is known that (Spec Q [ x, y ] , ⊆ ) does notsatisfy Condition P5, there is no known variation of Condition P5 that would completely classifythe spectrum of all Noetherian rings isomorphic to (Spec Q [ x, y ] , ⊆ ) . This inspired us to attachto (most) partially ordered sets satisfying Conditions P1 through P4 a new partially ordered setStr X whose order relation is inspired by Condition P5. The set Str X largely consists of elementsof the form ( S, T ) , where S is a finite, nonempty subset of the height-one nodes of X, and T isa nonempty subset of the height-two nodes of X. To get a sense of the order relation on Str X, fix a poset ( X, ≤ X ) satisfying Conditions P1 through P4 (for instance, X = Spec k [ x, y ]), take afinite, nonempty, subset S of height-one nodes of X and a finite, nonempty, subset T of height-twonodes of X. If there is a height-one node w ∈ X satisfying Condition P5 with respect to S and T, then we will say ( S, T ) < Str X ( S ∪ { w } , T ) . While the definition of ≤ Str X is more relaxed thanrequiring the existence of a single element w, we show, in our first main result, that Str X has thesame classifying effect that Condition P5 does in the sense that if (Str X, ≤ Str X ) ∼ = (Str Y, ≤ Str Y ) , then ( X, ≤ X ) ∼ = ( Y, ≤ Y ) . In the context of (Spec Q [ x, y ] , ⊆ ) , this means that if R is a commutativeNoetherian domain such that(Str Spec R, ≤ Str Spec R ) ∼ = (Str Spec Q [ x, y ] , ≤ Str Spec Q [ x,y ] )then (Spec R, ⊆ ) ∼ = (Spec Q [ x, y ] , ⊆ ) . More precisely, if ( X, ≤ ) is a poset that satisfies certain necessary conditions in order for there toexist an isomorphism from ( X, ≤ X ) onto (Spec k [ x, y ] , ⊆ ) for some field k, we call ( X, ≤ X ) a J -poset.In Section 6 (Theorem 6.12), we prove: Theorem. If X and Y are J -posets, then X ∼ = Y if and only if Str X ∼ = Str Y. Specifically, if ρ : X → Y is an isomorphism, then the map ϕ : Str X → Str Y given by ϕ ( A, B ) = ( ρ ( A ) , ρ ( B )) for all ( A, B ) ∈ Str X is an isomorphism, and conversely, if ϕ :Str X → Str Y is any isomorphism, then there is an isomorphism ρ : X → Y such that ϕ ( A, B ) = ( ρ ( A ) , ρ ( B ))for all ( A, B ) ∈ Str X. In Sections 2 and 3, we set up basic definitions and notation for the objects of study. In Section 4,we define Str X and we provide some basic examples. In Section 5, we begin a deeper study of Str X, and we show Str X is an invariant (i.e., X ∼ = Y = ⇒ Str X ∼ = Str Y ; see Proposition 5.4). Section6 is entirely devoted to proving the second part of the main theorem, the proof of which is brokeninto two major components: constructing the map ρ : X → Y and showing ρ is an isomorphism (seeTheorem 6.4), and then showing ϕ ( A, B ) = ( ρ ( A ) , ρ ( B )) (Theorem 6.11). Lastly, we study a subset STRUCTURAL INVARIANT ON CERTAIN TWO-DIMENSIONAL NOETHERIAN PARTIALLY ORDERED SETS3
Str F X of Str X, which consists only of elements ( A, B ) ∈ Str X such that both A and B are finite,and we prove, in Theorem 6.15, that if Str F X is endowed with the same order relations as Str X, then every appearance of Str X and Str Y in Theorem 6.12 can be replaced with Str F X and Str F Y, respectively. 2. Preliminaries
In this section, we introduce basic notation and definitions that will be used extensively through-out the paper.Recall that a partially-ordered set ( X, ≤ X ) is a pair of a set X with a binary relation ≤ X thatis reflexive, antisymmetric, and transitive. Specifically, this means that for all x, y, z ∈ X : x ≤ X x (reflexivity); x ≤ X y and y ≤ X x = ⇒ x = y (antisymmetry); and x ≤ X y and y ≤ X z = ⇒ x ≤ X z (transitivity). We will typically refer to a partially ordered set as a “poset.” Notation 2.1.
Let X be a poset, and let A ⊆ X. (1) If B ⊆ X, we say B ≤ X A if and only if for all b ∈ B, and for all a ∈ A, we have b ≤ X a. If B = { b } is a singleton, and B ≤ X A, we will write b ≤ X A instead of { b } ≤ X A. A similarconvention will apply if A is a singleton. The notations B < X A, A ≤ X B, A < X B aredefined similarly.(2) Define the following quantities: G X ( A ) := { x ∈ X : A ≤ X x } and G ∗ X ( A ) := G X ( A ) \ A. L X ( A ) := { x ∈ X : x ≤ X A } and L ∗ X ( A ) := L X ( A ) \ A. (3) The “minimal upper bound set” of A is defined to bemub X A := min G X A. Definition 2.2.
Let ( X, ≤ X ) , ( Y, ≤ Y ) be posets, and let f : X → Y be a function of sets. We say f : ( X, ≤ X ) → ( Y, ≤ Y ) is a poset map if and only if for all x, x ′ ∈ X we have x ≤ X x ′ = ⇒ f ( x ) ≤ Y f ( x ′ ) . If f : ( X, ≤ X ) → ( Y, ≤ Y ) is a poset map, we say f is an order-embedding if and only if for all x, x ′ ∈ X we have f ( x ) ≤ Y f ( x ′ ) = ⇒ x ≤ X x ′ . Lastly, we say a poset map f is an isomorphism ofposets (or simply isomorphism) if f is a surjective, order-embedding from X onto Y. If there is anisomorphism from ( X, ≤ X ) onto ( Y, ≤ Y ) , we write ( X, ≤ X ) ∼ = ( Y, ≤ Y ) or simply X ∼ = Y if the orderrelations are clear. Remark 2.3. If f is an order-embedding and f ( x ) = f ( x ′ ) , then f ( x ) ≤ Y f ( x ′ ) and f ( x ′ ) ≤ Y f ( x ) , so x ≤ X x ′ and x ′ ≤ X x, thus x = x ′ . In particular, every order-embedding is necessarily injectiveon the level of sets.
Definition 2.4. If X is a poset, we define the dimension of X, denoted dim X, to besup { n ∈ Z : There are x , . . . , x n ∈ X such that x < X x < X . . . < X x n } . If there is no bound on the length of chains of nodes in X, we define dim X = ∞ . If x ∈ X, we definethe height of x in X to be ht X x := dim L X ( x ) . For each integer i ≥ , we define X i ⊂ X to be thenodes of height i in X. CORY H. COLBERT J -Posets Definition 3.1.
Let X be a countable partially ordered set with a single minimal node. We say X is a J - poset if and only if the following conditions are met:J1 dim X = 2 and mub X A is finite for all nonempty A ⊆ X. J2 If x ∈ X , there are infinitely many z ∈ X such that z > X x. J3 If m ∈ X , and F ⊂ X is finite, then there is a finite set K ⊂ X disjoint from F such thatmub X K = { m } . J4 If T ⊂ X is a finite set, then there is t ∈ X such that t < X T. Proposition 3.2.
Let X be a J -poset. If S ⊂ X , T ⊂ X are nonempty, finite sets, then there is t ∈ X \ S such that t < X T. In particular, there are infinitely many w ∈ X such that w < X T. Proof.
We first claim that if F is a nonempty, finite set, then X = ∪ f ∈ F G ∗ X ( f ) . Since X is infinite,there is u ∈ X \ F. By Condition J2, G ∗ X ( u ) is infinite, and by Condition J1, the set ∪ f ∈ F mub X { u, f } ⊂ X is finite, so there is m ∈ G ∗ X ( u ) \ ∪ f ∈ F mub X { u, f } . In particular, m ∈ X , and m / ∈ ∪ f ∈ F G ∗ X ( f ) , because otherwise we would have m ∈ mub X { u, f } for some f ∈ F. Let
S, T be as in the statement of the proposition. By the work in the previous paragraph, thereis v ∈ X \ ∪ s ∈ S G ∗ X ( s ) . Set T ′ := T ∪ { v } . By Condition J4, there is t < X T ′ . So t < X T, and t / ∈ S because t < X v and no element of S is less than v in X. (cid:3) Items J2 and J4 may at first appear rather restrictive, but many Noetherian spectra satisfy bothof those properties. For instance, if k is any field, then the following result of S. McAdam in [1]implies that (Spec k [ x, y ] , ⊆ ) satisfies J4, so is a J -poset: Theorem 3.3. [[1], Corollary 11] Let R be a Noetherian domain, and let x, y be indeterminatesover R. If M , . . . , M m are height 2 primes in R [ x, y ] , then there are infinitely many primes of R contained in M ∩ · · · ∩ M m . Condition J3 is satisfied by many Noetherian rings as well, as the next proposition shows.
Proposition 3.4. If R is a Noetherian ring and P is a prime ideal of R such that ht P ≥ , and F is a finite set of height-one prime ideals of R, then there is a finite set K of prime ideals such that K is disjoint from F, and mub Spec R K = { P } . Proof.
Since (Spec R, ⊆ ) ∼ = (Spec R/ √ , ⊆ ) , we may assume R is reduced. Let M be the set of allfinite sums of height-one prime ideals contained in P but not an element of F. Since ht P ≥ R is Noetherian and reduced, there is a regular element p ∈ P \ ∪ f ∈ F f , and thus a height-one primeideal p with p ∈ p ⊂ P by the principal ideal theorem. Since p / ∈ F, we have M 6 = ∅ . Since R isNoetherian, M has a maximal element J = q + . . . + q j . We claim J = P. If J = P, then there is x ∈ P \ J. If x / ∈ ∪ f ∈ F f , then x is an element of aheight-one prime ideal q contained in P outside F, so q + J ∈ M and therefore J = q + J bymaximality, a contradiction. Therefore, x ∈ ∪ f ∈ F f . Let y ∈ q \ ∪ f ∈ F f . Enumerate the prime idealsin F as f , . . . , f k , f k +1 , . . . , f r so that x ∈ ∩ ki =1 f i and x / ∈ ∪ ri = k +1 f i . For each 1 ≤ i ≤ k, let f i = 1 . STRUCTURAL INVARIANT ON CERTAIN TWO-DIMENSIONAL NOETHERIAN PARTIALLY ORDERED SETS5 If k + 1 ≤ i ≤ r, then choose a regular element f i ∈ f i and not in each f s such that s = i. Then x + f · · · f r y ∈ P \ J, and if x + f · · · f r y ∈ f i for some 1 ≤ i ≤ k, then one of f , . . . , f r , y mustreside in f i , an impossibility. If x + f · · · f r y ∈ f i for k + 1 ≤ i ≤ r, then f i ∈ f i implies x ∈ f i , whichis contrary to how we ordered the prime ideals in F. So x + f · · · f r y is in a height-one prime ideal q ′ not in F, and therefore q ′ + J ∈ M so that q ′ + J = J by maximality, a contradiction. Therefore, J = P, and if K = { q , . . . , q j } , the result is proved. (cid:3) The Structure Poset
Throughout the rest of the paper, X and Y will be J -posets. If Z is any set, we will use P ( Z )to denote the power set of Z. Definition 4.1.
Define Str X ⊂ P ( X ) × P ( X ) to be the set of all pairs ( A, B ) of a set A ⊂ X with a set B ⊂ X satisfying the following properties:(1) A and B are both finite and nonempty, or ( A, B ) = ( x, G ∗ X ( x )) for some x ∈ X , and(2) there is a ∈ A such that a < X B. Definition 4.2.
Let (
A, B ) , ( C, D ) ∈ Str X. We say (
C, D ) dominates (
A, B ) via set W if and onlyif the following conditions are met:E1 A ( C and B ⊇ D. E2 W ⊆ C is nonempty and W < X D. E3 If a ∈ A and a < X m and W < X m, then m ∈ D. If (
A, B ) ∈ Str X, and A = { a } is a singleton, then we will write ( a, B ) instead of ( { a } , B ) . Asimilar convention holds if B = { b } is a singleton. Definition 4.3.
Define the following relation ≤ Str X on Str X : Declare ( A, B ) ≤ Str X ( C, D ) if andonly if either A = C and B = D or ( C, D ) dominates (
A, B ) via W for some W ⊆ C. Proposition 4.4. (Str X, ≤ Str X ) is a poset. Proof.
Reflexivity and antisymmetry are clear, so we need only show that if ( A , B ) < Str X ( A , B )and ( A , B ) < Str X ( A , B ) , then ( A , B ) < Str X ( A , B ) . If ( A , B ) dominates ( A , B ) via W, we claim it also dominates ( A , B ) via W. Indeed, A ( A and B ⊇ B , and if a ∈ A and a < X m and W < X m, then a ∈ A and a < X m and W < X m, so m ∈ B . (cid:3) We will say (
C, D ) dominates (
A, B ) via W and ( A, B ) ≤ Str X ( C, D ) via W interchangeably. Example 4.5.
Suppose a, b, c ∈ X , d, e ∈ X and { a, b } < X { d, e } while c < X d and c < X e. If A = { a, b, c } , and D = { d, e } then ( A, D ) ∈ Str X because { a, b } < X D. However, ( c, D ) / ∈ Str X because there is no element in { c } that is less than all of D. Now, ( c, d ) ∈ Str X because c < X d. Note that although ( c, D ) / ∈ Str X, we do have ( { a, c } , D ) ∈ Str X because a < X D. Example 4.6.
Suppose S = { s , s } and T = { t } , where s < X t , but s < X t . Suppose thereis w ∈ X such that mub { w, s } = ∅ and mub { w, s } = { t } . Then the pair
S, T satisfies P5 withrespect to w. Indeed, if m > X s and m > X w, then m ∈ mub X { s , w } because dim X = 2 . So m = t ∈ T. If m > X s and m > X w, then m is vacuously a member of T because there are nosuch elements by assumption. In particular, ( S ∪ { w } , T ) dominates ( S, T ) via W = { w } . Therefore,(
S, T ) < Str X ( S ∪ { w } , T ) . CORY H. COLBERT
Example 4.7.
Suppose (
A, B ) ∈ Str X, and b ∈ B. There is K ⊂ X disjoint from A such thatmub X K = { b } by Condition J3. We claim ( A, B ) < Str X ( A ∪ K, b ) . Indeed,
K < X b becausemub X K = { b } . Also, if a ∈ A and K < X m and a < X m, then m ∈ G ∗ X ( K ) , so { m } ≥ X { b } = min G X ( K ) . Since m ∈ X and dim X = 2 , we have m = b. Since A ( A ∪ K, we have ( A ∪ K, b ) dominates (
A, b )via K. Caution 4.8.
We caution the reader that if (
C, D ) dominates (
A, B ) via W, it need not be thecase that W is disjoint from A. In fact, in many cases, W will be a subset of A. For instance, if A = { a, b, c } and mub X A = { d, e } := D, then ( a, D ) < Str X ( A, D ) because (
A, D ) dominates ( a, D )via A. Likewise, (
A, D ) < Str X ( A ∪ { f } , D ) , where f ∈ X \ A, because ( A ∪ { f } , D ) dominates( A, D ) via A as well. 5. Basic Facts and Definitions Regarding
Str X. Throughout this section, X and Y will always be J -posets. Lemma 5.1.
Let (
A, B ) ∈ Str X. The following items are true.(1) ht
Str X ( a, G ∗ X ( a )) = 0 for all a ∈ X . In particular, if ht(
A, B ) > , then B is finite.(2) If ( C, D ) ∈ Str X, and b ∈ B ∩ D, then there is J ⊂ X such that ( J, b ) ∈ Str X. and( A, B ) < Str X ( J, b ) , and ( C, D ) < Str X ( J, b ) . (3) We have ht Str X ( A, B ) > B = mub K for some K ⊆ A. Proof.
We prove each part individually.i) If (
S, T ) ≤ Str X ( a, G ∗ X ( a )), then S = { a } because S ⊆ { a } and is nonempty. Since a < X T, and T ⊇ G ∗ X ( a ) , we must have T = G ∗ X ( a ) , since the latter set consists of all elements of X thatexceed a. Therefore, ht
Str X ( a, G ∗ X ( a )) = 0 . Consequently, if B is infinite, then by definition of Str X, we have ( A, B ) = ( a, G ∗ X ( a )) for some a ∈ X , so ht Str X ( A, B ) = ht
Str X ( a, G ∗ X ( a )) = 0 , as was justshown.ii) Choose K ⊂ X disjoint from both A and C such that mub X K = { b } . By the argument inExample 4.7, if we set J := K ∪ A ∪ C, then ( J, d ) dominates (
A, B ) via K and ( J, d ) dominates(
C, D ) via K. iii) Suppose first that ht Str X ( A, B ) > . Let ( A , D ) ∈ Str X such that ( A , D ) < Str X ( A, B ) , and let a ∈ A with a < X D. Note a < X B. Let W ⊆ A be such that ( A, B ) extends ( A , D ) via W. If W = { a } , then ( A, B ) extends ( A , D ) via the single element a. Since a ∈ A , Condition E3 wouldimply B = G ∗ X ( a ) , which is infinite. Thus W = { a } . Set K = W ∪ { a } . Then
K < X B. Suppose m > X K. Then m > X W and m > X a so m ∈ B by Condition E3. Therefore, mub X K = B. Conversely, if B = mub X K for some K ⊆ A, then | K | ≥ B = K ⊂ X . Thus, if k ∈ K, then ( A, B ) dominates ( k, B ) via K because { k } 6 = A, and mub X K = B so Condition E3 is satisfied. In particular,( k, B ) < Str X ( A, B )so ht
Str X ( A, B ) > . (cid:3) STRUCTURAL INVARIANT ON CERTAIN TWO-DIMENSIONAL NOETHERIAN PARTIALLY ORDERED SETS7
Definition 5.2. If B ⊂ X is finite and nonempty, define (Str X ) B to be all elements of Str X whose second ordinate is B. That is, (Str X ) B is the set of all elements in Str X of the form( A, B ) . Let (Str X ) B inherit the partial ordering on Str X so that ((Str X ) B , ≤ Str X ) is a subposetof (Str X, ≤ Str X ) . Notation 5.3.
If (
A, B ) ∈ (Str X ) B , then we will write L B ( A, B ) instead of L (Str X ) B ( A, B ) , and if B = { b } is a singleton, we will similarly write L b ( A, b ) instead of L (Str X ) b ( A, b ) . In a similar vein,we will write ht B ( A, B ) instead of ht L (Str X ) B ( A,B ) ( A, B ) , and likewise we will write ht b ( A, b ) insteadof ht L (Str X ) b ( A,b ) ( A, b ) . Finally, if B = { b } is a singleton, we will write (Str X ) b instead of (Str X ) { b } . Proposition 5.4. If ρ : X → Y is an isomorphism of posets, then the map ϕ : Str X → Str Y givenby ϕ ( A, B ) = ( ρ ( A ) , ρ ( B ))for all ( A, B ) ∈ Str X is an isomorphism of posets. Moreover, ϕ ((Str X ) B ) = (Str X ) ρ ( B ) for allfinite, nonempty B ⊂ X . Proof.
Since ρ is an isomorphism, it is also a height-preserving bijection of sets. Moreover, for anysubsets C, C ′ of X, we have ρ ( C ) ≤ Y ρ ( C ′ ) ⇐⇒ C ≤ X C ′ . It is a straightforward exercise tocheck that (
A, B ) ≤ Str X ( A ′ , B ′ ) ⇐⇒ ( ρ ( A ) , ρ ( B )) ≤ Str Y ( ρ ( A ′ ) , ρ ( B ′ )) and that ϕ satisfies theconclusion of the proposition. (cid:3) Example 5.5.
Let k be an algebraically closed field, and let x, y be indeterminates over k. ByTheorem 3.3, (Spec k [ x, y ] , ⊆ ) satisfies Condition J4, so (Spec k [ x, y ] , ⊆ ) is a J -poset. Let m and n be two maximal ideal of k [ x, y ] . By the Nullstellensatz, m = h x − a , y − b i and n = h x − a , y − b i , forsome a , a , b , b ∈ k. In particular, there is an automorphism σ : k [ x, y ] → k [ x, y ] carrying m onto n , and this induces an isomorphism σ ∗ : (Spec k [ x, y ] , ⊆ ) → (Spec k [ x, y ] , ⊆ ) such that σ ∗ ( m ) = n . By Proposition 5.4, (Str Spec k [ x, y ]) m ∼ = (Str Spec k [ x, y ]) n . Definition 5.6.
If (
A, B ) ∈ Str X, define ℓ ( A, B ) := |{ a ∈ A : a < X B }| , and η ( A, B ) := | A | − ℓ ( A, B ) . Note that ℓ ( A, B ) ≥ A, B ) ∈ Str X because each pair ( A, B ) is required to have anelement a ∈ A such that a < X B in order to be a member of Str X. Lemma 5.7.
Let (
A, B ) ∈ (Str X ) B such that ht B ( A, B ) > . Then(1) | L B ( A, B ) | = (2 ℓ ( A,B ) − η ( A,B ) , and(2) B = mub A if and only if | L B ( A, B ) | is odd. Proof.
We prove each part individually.i) If ht B ( A, B ) > , then B = mub X K for some K ⊆ A by Lemma 5.1. Thus, if C ⊆ A such thatthere is c ∈ C with c < X B, we have ( C, B ) ≤ Str X ( A, B ) via K. There are a total of 2 | A | subsetsof A, of which 2 η ( A,B ) are subsets D of A such that ( D, B ) / ∈ Str X. This is because the sets D areprecisely those subsets of A such that there is no d ∈ D with d < X B. That is, | L B ( A, B ) | = 2 | A | − η ( A,B ) = 2 ℓ ( A,B )+ η ( A,B ) − η ( A,B ) = (2 ℓ ( A,B ) − η ( A,B ) . CORY H. COLBERT ii) If B = mub X A, then A < X B. So η ( A, B ) = 0 , and therefore | L B ( A, B ) | is odd. Conversely, if | L B ( A, B ) | is odd, then η ( A, B ) = 0 . Since ht B ( A, B ) > , we have B = mub X K for some K ⊆ A. Since
A < X B and B = mub X K, it follows that B = mub X A as well. (cid:3) Definition 5.8.
For each integer r ≥ , let ( I r , ≤ I r ) be the poset with exactly r height-zero nodes u , . . . , u r and exactly one height-one node m exceeding each height-zero node. Example 5.9.
Let a, b ∈ X , and suppose mub X { a, b } = B. Then L B ( { a, b } , B ) is completelydescribed by the following figure: ( a, B ) ( b, B )( { a, b } , B ) { b } { a } Figure 1.
The labels of sets K above the linesmean that the higher node dominates the lowernode via K. The reader will observe that the poset in Figure 5.9 is order-isomorphic to I , and this is nocoincidence. Indeed, if L B ( A, B ) ∼ = I for some ( A, B ) ∈ (Str X ) B , then | L B ( A, B ) | = 3 , which isodd, so B = mub X A by Lemma 5.7. Since | A | ≥ , and ( a, B ) < Str X ( A, B ) via A for all a ∈ A, itfollows that A = { a, b } for some distinct a, b ∈ X . That is, L B ( A, B ) ∼ = I if and only if A = { a, b } for distinct a, b ∈ X and B = mub X A. Example 5.10.
Recall that in any poset we say b covers a if a < b and no element of the poset isboth greater than a and less than b. Let X be a J -poset that satisfies Conditions J4 and P5, suchas (Spec Z [ x ] , ⊆ ) . Let u ∈ X and m ∈ X such that u < X m. Then there exists v ∈ X such that( { u, v } , m ) covers ( u, m ) in Str X, and L m ( { u, v } , m ) ∼ = I . Example 5.11.
This example is intended to demonstrate how Str X detects a failure to enjoyCondition P5. Let k = Q – where Q is the algebraic closure of Q – let x, y be indeterminates over k, let R = k [ x, y ] , and let X = Spec R. Let P = h x − y i R, and let m = h x − a, y − b i for a, b ∈ k bea maximal ideal of R, different from h x, y i , such that P ⊂ m. By [[5], Example 2.1], we know thereis no height-one prime ideal Q such that mub X { P, Q } = { m } . Now, (
P, m ) ∈ Str X. Let µ ( P, m ) be the cardinality of the smallest subposet Λ of (Str X ) m , containing ( P, m ) , of positive dimension and of the form L m ( A, m ) for some (
A, m ) ∈ Str X. If L m ( A, m ) has positive dimension for some A, then | L m ( A, m ) | ≥ µ ( P, m ) ≥ . If µ ( P, m ) = 3 , then Λ ∼ = I , so mub X { P, Q } = { m } for some Q ∈ Spec k [ x, y ] by Example 5.9, acontradiction. Therefore, µ ( P, m ) ≥ . Of course, µ ( P, m ) ≥ x, m ) is any height zero node in(Str X ) m , where x ∈ X , then there will fail to exist y ∈ X such that mub X { x, y } = { m } if andonly if µ ( x, m ) ≥ . STRUCTURAL INVARIANT ON CERTAIN TWO-DIMENSIONAL NOETHERIAN PARTIALLY ORDERED SETS9 Proof of Main Theorem
We now turn our attention to the main result of this paper, which is that Str X completelydetermines X up to isomorphism. In other words, we seek to prove that if ϕ : Str X → Str Y is anisomorphism of posets, where Y is a J -poset, then there is an isomorphism ρ : X → Y such that ϕ ( A, B ) = ( ρ ( A ) , ρ ( B ))for all ( A, B ) ∈ Str X. First, we show that if m ∈ X , then there is a unique n ∈ Y such that ϕ restricts to anisomorphism of posets from (Str X ) m onto (Str Y ) n . This fact will demonstrate that there is abijection of sets ρ : X → Y . Having determined where the “points on the surface” go (i.e., themaximal nodes of X – thinking of X as the spectrum of a Noetherian ring), we then have a generalsense of where the “irreducible curves on the surface” should go (i.e., the height-one nodes of X )since each x ∈ X is determined by G ∗ X ( x ) by Condition J1. In other words, ρ will induce a setbijection ρ : X → Y . Of course, there is no mystery as to where the “generic point” (i.e., theunique minimal node of X ) should go, so the map ρ : X → Y is clear. We then set ρ := ρ ∪ ρ ∪ ρ and show that ρ has the desired properties.The second part of the argument involves a deeper study of not only the structural informationabout X contained within Str X, but how well ϕ is carrying that information over to Str Y. Lemma 6.1.
Let ϕ : Str X → Str Y be an isomorphism of posets. If ϕ ( A , m ) , and ϕ ( A , m ) ∈ (Str Y ) n for some m , m ∈ X , n ∈ Y , and sets A , A ⊂ X , then m = m . In particular, if ϕ ( A, M ) ∈ (Str Y ) n , then M = { m } is a singleton. Proof.
Let ϕ ( A , m ) = ( C , n ) and let ϕ ( A , m ) = ( C , n ) . By Lemma 5.1, there is ( C , n ) ∈ (Str Y ) n such that both ( C , n ) < Str Y ( C , n ) and ( C , n ) < Str Y ( C , n ) . Therefore, both ( A , m ) < Str X ϕ − ( C , n ) := ( A , B ) and ( A , m ) < Str X ( A , B ) . Now, B ⊆ { m } ∩ { m } , and since B is nonempty, we have m = m . Suppose (
A, M ) ∈ (Str X ) such that ϕ ( A, M ) = (
C, n ) . If m, m ′ ∈ M, there are sets K, K ′ ⊂ X such that both ( A, M ) < Str X ( K, m ) and (
A, M ) < Str X ( K ′ , m ′ ) . Therefore,(
C, n ) < Str Y ϕ ( K, m ) := (
J, n )and (
C, n ) < Str Y ϕ ( K ′ , m ′ ) := ( J ′ , n ) . Note that we know the second ordinate in the pair (
J, n )(resp. ( J ′ , n )) is { n } because it must be a nonempty subset of { n } . By the previous paragraph, m = m ′ . (cid:3) Lemma 6.2. If m ∈ X , then the restriction ϕ | (Str X ) m : (Str X ) m → Str Y is an isomorphism from (Str X ) m onto (Str Y ) n for some n ∈ Y . Proof.
Since ϕ : Str X → Str Y is an isomorphism of posets, so is ϕ − : Str Y → Str X. In particular,Lemma 6.1 applied to ϕ − : Str Y → Str X shows that if ϕ ( A, m ) = (
C, N ) for some C ⊂ Y and N ⊂ Y , then N = { n } is a singleton and thus ϕ ((Str X ) m ) ⊆ (Str Y ) n for some n ∈ Y . To see that ϕ ((Str X ) m ) = (Str Y ) n , let ( D, n ) ∈ (Str Y ) n and choose any ( L, n ) ∈ (Str Y ) n thatis in the image of ϕ | (Str X ) m . So (
L, n ) = ϕ ( A , m ) for some ( A , m ) ∈ (Str X ) m . Now, (
D, n ) = ϕ ( A , B ) because ϕ is onto, and by Lemma 6.1, B = { b } for some b ∈ X . By Lemma 5.1, there is (
J, n ) in (Str Y ) n exceeding both ( L, n ) and (
D, n ) . In particular, we have( A , m ) < Str X ϕ − ( J, n ) := ( A , m ) and ( A , b ) < Str X ( A , m )for some ( A , m ) ∈ (Str X ) m . Therefore, ( A , b ) < Str X ( A , m ) so b = m. (cid:3) Note that (Str Y ) n ∩ (Str Y ) n ′ is nonempty if and only if n = n ′ , so the set map ρ : X → Y that sends m → n as in Lemma 6.2 is well-defined. Lemma 6.1 shows that ρ is injective. If n ∈ Y , then Lemma 6.2 applied to ϕ − | (Str Y ) n : (Str Y ) n → Str X shows that ρ is surjective. We nowturn our attention to constructing ρ : X → Y . Lemma 6.3. If x ∈ X , then ϕ ( x, G ∗ X ( x )) = ( s, G ∗ Y ( s )) for some s ∈ Y . Proof.
For each m ∈ G ∗ X ( x ) , let K m ⊂ X be a finite set such that x ∈ K m and mub X K m = { m } . Then ( x, G ∗ X ( x )) < Str X ( K m , m ) for all m ∈ G ∗ X ( x ) . By Lemma 6.1, each ϕ ( K m , m ) = ( J, ρ ( m ))for some finite J ⊂ Y . Since ( x, G ∗ X ( x )) < Str X ( K m , m )for all m ∈ G ∗ X ( x ) , and ϕ is an isomorphism, the node ϕ ( x, G ∗ X ( x )) ∈ Str Y satisfies ϕ ( x, G ∗ X ( x )) < Str Y ( J, ρ ( m ))for all such ( J, ρ ( m )) . Let (
S, T ) = ϕ ( x, G ∗ X ( x )) . By the above paragraph, ρ ( m ) ∈ T for all m ∈ G ∗ X ( x ) . Since ρ : X → Y is a bijection of sets, the set ρ ( G ∗ X ( x )) is infinite and thus T is infinite, so ( S, T ) = ( s, G ∗ Y ( s )) forsome s ∈ Y . (cid:3) Define ρ : X → Y via x → s as in Lemma 6.3. ρ is well-defined, and if ρ ( x ) = ρ ( x ) , then ϕ ( x , G ∗ X ( x )) = ϕ ( x , G ∗ X ( x )) , so x = x because ϕ is injective (see Remark 2.3). That ρ issurjective follows after applying Lemma 6.3 to ϕ − : Str Y → Str X. Let X = { x } and Y = { y } . Define ρ : X → Y by ρ ( x ) = y . Set ρ := ρ ∪ ρ ∪ ρ . Theorem 6.4. If ϕ : Str X → Str Y is an isomorphism, then the map ρ : X → Y as constructedabove is an isomorphism of posets. Proof.
By definition, ρ is height-preserving: ht Y ρ ( x ) = ht X x for all x ∈ X. Additionally, ρ i : X i → Y i is a bijection of sets for i = 0 , , . In particular, each ρ i is a surjective and one-to-one map ofsets from X i onto Y i for i = 0 , , . Suppose a, b ∈ X and a ≤ X b. If a = x , the minimal node of X, then ρ ( a ) = ρ ( x ) = y ≤ Y, STRUCTURAL INVARIANT ON CERTAIN TWO-DIMENSIONAL NOETHERIAN PARTIALLY ORDERED SETS11 so ρ ( a ) ≤ Y ρ ( b ) in that case. If a ∈ X , then a = b and we have ρ ( a ) = ρ ( b ) in that case. Wemay thus assume a ∈ X and b ∈ X . Since a < X b, we have b ∈ G ∗ X ( a ) . Moreover, there is a finite K ⊂ X such that a ∈ K and mub X K = { b } . Therefore,( a, G ∗ X ( a )) < Str X ( K, b ) , and so ( ρ ( a ) , G ∗ Y ( ρ ( a ))) < Str Y ( J ρ ( b ) , ρ ( b )) . Therefore, ρ ( b ) ∈ G ∗ Y ( ρ ( a )) , or equivalently, ρ ( a ) < Y ρ ( b ) . Conversely, assume that ρ ( a ) ≤ Y ρ ( b ) . Similar to above, we may assume that ρ ( a ) ∈ Y and ρ ( b ) ∈ Y and ρ ( a ) < Y ρ ( b ) . Then there is finite J ⊂ Y such that ρ ( a ) ∈ J and( ρ ( a ) , G ∗ Y ( ρ ( a ))) < Str Y ( J, ρ ( b )) . Since ϕ − ( ρ ( a ) , G ∗ Y ( ρ ( a ))) = ( a, G ∗ X ( a )) and ϕ − ( J, ρ ( b )) = ( K ′ , b ) for some K ′ ⊂ X , we have a < X b. (cid:3) We now begin the demonstration that ϕ ( A, B ) = ( ρ ( A ) , ρ ( B )) for all ( A, B ) ∈ Str X. Definition 6.5.
Let (
A, B ) ∈ Str X, and let ϕ, ρ be as in Theorem 6.4. Write ϕ ( A, B ) = (
S, T ) . Theisomorphism ρ restricts to a bijection of sets from X i onto Y i for each i = 0 , , . In particular, since S ⊂ Y , it follows that S = ρ ( A ∗ ) for some unique A ∗ ⊂ X . Likewise, T = ρ ( B ∗ ) for some unique B ∗ ⊂ X . Therefore, given A ⊂ X , B ⊂ X such that ( A, B ) ∈ Str X, we define A ∗ ⊂ X , B ∗ ⊂ X to be those unique subsets of X such that ϕ ( A, B ) = ( ρ ( A ∗ ) , ρ ( B ∗ )) . It is now our goal to show that A ∗ = A and B ∗ = B for all ( A, B ) ∈ Str X. Lemma 6.6.
Let (
A, B ) ∈ Str X, and ϕ, ρ as in Theorem 6.4. Write ϕ ( A, B ) = ( ρ ( A ∗ ) , ρ ( B ∗ )) for A ∗ ⊂ X and B ∗ ⊂ X as in Definition 6.5. Then the following items hold:(1) B ∗ = B, and if B is finite, then the restriction ϕ | (Str X ) B is a poset isomorphism from(Str X ) B onto (Str Y ) ρ ( B ) . (2) We have ℓ ( A ∗ , B ) = ℓ ( ρ ( A ∗ ) , ρ ( B )) (1)and η ( A ∗ , B ) = η ( ρ ( A ∗ ) , ρ ( B )) (2)(3) If ht B ( A, B ) > , then ℓ ( ρ ( A ∗ ) , ρ ( B )) = ℓ ( A, B ) (3)and η ( ρ ( A ∗ ) , ρ ( B )) = η ( A, B ) . (4)(4) If B = mub A, then A ∗ = A. Proof.
We prove each item individually.
Item 1. If b ∈ B, choose K ⊂ X such that ( A, B ) < Str X ( K, b ) . Upon applying ϕ, we have( ρ ( A ∗ ) , ρ ( B ∗ )) < Str Y ( ρ ( K ∗ ) , ρ ( b )) , by Lemma 6.2 applied to ( K ∗ , b ) ∈ (Str X ) b . In particular, ρ ( b ) ∈ ρ ( B ∗ ) , so b ∈ B ∗ . Conversely, if b ∈ B ∗ , then there is F ∗ ⊂ X such that( ρ ( A ∗ ) , ρ ( B ∗ )) < Str Y ( ρ ( F ∗ ) , ρ ( b )) . Since ϕ is an isomorphism, we must have( A, B ) < Str X ( F, b )for some F ⊂ X . In particular, b ∈ B. So B = B ∗ . To see the second part, we need only show that the restriction ϕ | (Str X ) B is surjective, since theabove work shows that ϕ | (Str X ) B : (Str X ) B → (Str Y ) ρ ( B ) is well-defined, and the map is an order-embedding because ϕ is. Indeed, if ( ρ ( C ∗ ) , ρ ( B )) ∈ (Str Y ) ρ ( B ) , and ϕ ( C, D ) = ( ρ ( C ∗ ) , ρ ( B )) , then, by the above paragraph, ρ ( B ) = ρ ( D ) , so B = D since ρ is a set bijection. Item 2.
Equations 1 and 2 follow from the definitions of the functions ℓ, η : Str X → Z ≥ andTheorem 6.4 since U ≤ X V ⇐⇒ ρ ( U ) ≤ Y ρ ( V )for all subsets U, V of X. Item 3.
By Item 1, the restriction ϕ | (Str X ) B : (Str X ) B → (Str Y ) ρ ( B ) is an isomorphism of posets. Let σ := ϕ | (Str X ) B . Then, | L B ( A, B ) | = | σ ( L B ( A, B )) | = | L ρ ( B ) ( σ ( A, B )) | = | L ρ ( B ) ( ρ ( A ∗ ) , ρ ( B )) | . (5)Since ht B ( A, B ) > , we have | L B ( A, B ) | = (2 ℓ ( A,B ) − η ( A,B ) by Lemma 5.7. Since ϕ is anisomorphism, ht ρ ( B ) ( ρ ( A ∗ ) , ρ ( B ∗ )) = ht ρ ( B ) ϕ ( A, B ) = ht B ( A, B ) > . Therefore, by Lemma 5.7, we have | L ρ ( B ) ( ρ ( A ∗ ) , ρ ( B ∗ )) | = (2 ℓ ( ρ ( A ∗ ) ,ρ ( B ∗ )) − η ( ρ ( A ∗ ) ,ρ ( B ∗ )) . By Equation 5 and the Fundamental Theorem of Arithmetic, we have η ( ρ ( A ∗ ) , ρ ( B ∗ )) = η ( A, B )and ℓ ( ρ ( A ∗ ) , ρ ( B ∗ )) = ℓ ( A, B )as desired.
Item 4.
Suppose B = mub A. As in Item 3, let σ be ϕ | (Str X ) B . By Lemma 5.7 and the fact that σ is an isomorphism from (Str X ) B onto (Str Y ) ρ ( B ) , we haveht B ( A, B ) = ht ρ ( B ) ( ρ ( A ∗ ) , ρ ( B )) > . By Items 2 and 3 and the fact that
A < X B, we have η ( ρ ( A ∗ ) , ρ ( B )) = η ( A ∗ , B ) = η ( A, B ) = 0 (6)and ℓ ( ρ ( A ∗ ) , ρ ( B )) = ℓ ( A ∗ , B ) = ℓ ( A, B ) . (7)By Equations 6 and 7, we have | A ∗ | = | A | . If a ∈ A, then ( a, G ∗ X ( a )) < Str X ( A, B ) via A. By Lemma 6.3,( ρ ( a ) , G ∗ Y ( ρ ( a ))) < Str Y ( ρ ( A ∗ ) , ρ ( B )) . STRUCTURAL INVARIANT ON CERTAIN TWO-DIMENSIONAL NOETHERIAN PARTIALLY ORDERED SETS13 So ρ ( a ) ∈ ρ ( A ∗ ) , and thus a ∈ A ∗ . Since a was arbitrary, we have A ⊆ A ∗ . Finally, since both A and A ∗ are finite sets with | A | = | A ∗ | , we have A = A ∗ as desired. (cid:3) Remark 6.7.
A careful inspection of the proof of Item 1 shows that we only use the fact that ρ : X → Y is a set bijection of sets. Definition 6.8.
If (
C, D ) ∈ Str X and d ∈ D, let N ( C, d ) := { ( A, d ) ∈ Str X : ( C, d ) < Str X ( A, d ) and η ( C, d ) = η ( A, d ) } . Remark 6.9.
If (
C, D ) ∈ Str X and d ∈ D, then ( C, d ) < Str X ( A, d ) if and only if (
C, D ) < Str X ( A, d ) . Corollary 6.10.
Suppose (
C, D ) ∈ Str
X, d ∈ D , and let N ( C, D ) be as in Definition 6.8. If ϕ ( A, d ) = ( ρ ( A ) , ρ ( d )) for all ( A, d ) ∈ N ( C, d ) , then ϕ ( C, D ) = ( ρ ( C ) , ρ ( d )) . In particular, if
C < X d, then ϕ ( C, D ) = ( ρ ( C ) , ρ ( D )) . Proof.
By Lemma 6.6 and Definition 6.5, we may write ϕ ( C, D ) = ( ρ ( C ∗ ) , ρ ( D )) . We claim C = C ∗ . By Condition J3, there exists a finite set F ⊂ X such that F is disjoint from C ∪ C ∗ and mub X F = { d } . Therefore, (
C, D ) < Str X ( C ∪ F, d ) (8)because ( C ∪ F, d ) dominates (
C, D ) via F. Now, η ( C ∪ F, d ) = η ( C, d )because
F < X d. Therefore, ( C ∪ F, d ) ∈ N ( C, d ) , (see Remark 6.9) and so ϕ ( C ∪ F, d ) = ( ρ ( C ∪ F ) , ρ ( d )) . (9)Applying ϕ to (8) and using (9), we see that( ρ ( C ∗ ) , ρ ( D )) < Str Y ( ρ ( C ∪ F ) , ρ ( d )) . Therefore, C ∗ ⊆ C ∪ F, and since F is disjoint from C ∗ , we have C ∗ ⊆ C. Now ( ρ ( C ∗ ) ∪ ρ ( F ) , ρ ( d )) dominates ( ρ ( C ∗ ) , ρ ( D )) via ρ ( F ) because ρ (mub X F ) = mub X ρ ( F ) = { ρ ( d ) } , which follows from the fact that ρ is an isomorphism. Now, there is J ⊂ X such that ϕ ( J, d ) = ( ρ ( C ∗ ∪ F ) , ρ ( d )) . We claim (
J, d ) ∈ N ( C, d ) . Since( ρ ( C ∗ ) , ρ ( D )) < Str Y ( ρ ( C ∗ ) ∪ ρ ( F ) , ρ ( d )) , we have ( C, D ) < Str X ( J, d ) . We need only show that η ( J, d ) = η ( C, d ) in order to conclude that (
J, d ) ∈ N ( C, d ) . Since C ⊆ J, we have η ( C, d ) ≤ η ( J, d ) . By parts 2 and 3 of Lemma 6.6 (note that J ∗ = C ∗ ∪ F ), η ( J, d ) = η ( J ∗ , d ) = η ( C ∗ ∪ F, d ) = η ( C ∗ , d ) ≤ η ( C, d ) . (10) where the last inequality in (10) follows because C ∗ ⊆ C by the above work. Therefore, η ( C, d ) = η ( J, d ) , so ( J, d ) ∈ N ( C, d ) , and thus ϕ ( J, d ) = ( ρ ( J ) , ρ ( d )) = ( ρ ( J ∗ ) , ρ ( d )) = ( ρ ( C ∗ ∪ F ) , ρ ( d )) . Therefore, J = J ∗ = C ∗ ∪ F. Since C ⊆ J, and C is disjoint from F, we have C ⊆ C ∗ so C = C ∗ . To see the final part of this corollary, suppose
C < X d, and suppose ( A, d ) ∈ N ( C, d ) . Then η ( A, d ) = η ( C, d ) = 0 , and ht d ( A, d ) > , so mub X A = { d } by Lemma 5.7. By part 4 of Lemma 6.6, ϕ ( A, d ) = ( ρ ( A ) , ρ ( d )) . By the above work, ϕ ( C, D ) = ( ρ ( C ) , ρ ( D )) . (cid:3) Theorem 6.11.
Let ϕ, ρ be as in Theorem 6.4. If (
A, B ) ∈ Str X, then ϕ ( A, B ) = ( ρ ( A ) , ρ ( B )) . Proof.
We claim that it suffices to prove the theorem for all nodes in Str X of the form ( A, b ) whereht b ( A, b ) > . Given that assumption, if (
C, B ) ∈ Str
X, b ∈ B, and ( A, b ) ∈ N ( C, b ) from Definition6.8, then ht b ( A, b ) > A, b ) > Str X ( C, b ) . Thus ϕ ( A, b ) = ( ρ ( A ) , ρ ( b )) , and so by Corollary6.10, we have ϕ ( C, B ) = ( ρ ( C ) , ρ ( B )) . Therefore, let (
A, b ) ∈ Str X such that ht b ( A, b ) > . If mub X A = { b } , then the result holds bypart 4 of Lemma 6.6. Since ht b ( A, b ) > , we have { b } = mub X A if and only if η ( A, b ) = 0 . Thus,assume η ( A, b ) = 0 , and enumerate the elements q ∈ A such that q < b as q , . . . , q η ( A,b ) . For each i = 1 , . . . , η ( A, b ) , let b i ∈ X be a node of X such that q i < X b i , and set B i := { b, b i } . Choose,by Proposition 3.2, x ∈ X such that x < X ∪ η ( A,b ) i =1 B i and x / ∈ A. Define Q i := { x, q i } for each i = 1 , . . . , η ( A, b ) . Recall that for a pair (
A, B ) , we define A L = { a ∈ A : a < X B } . Write A = A L ∪ { q , . . . , q η ( A,b ) } . Note that { b } = mub X A L , and thus ( A, b ) < Str X ( A ∪ { x } , b ) (11)and ( Q i , B i ) < Str X ( A ∪ { x } , b ) (12)for all 1 ≤ i ≤ η ( A, b ) . Let ϕ ( A, b ) = ( ρ ( A ∗ ) , ρ ( b ))and ϕ ( A ∪ { x } , b ) = ( ρ (( A ∪ { x } ) ∗ ) , ρ ( b )) . We claim A ∗ = A. In order to prove this, we first show ( A ∪ { x } ) ∗ = A ∪ { x } and conclude that A = A ∗ . By Lemma 6.6, we have η ( A, b ) = η ( A ∗ , b ) and ℓ ( A, b ) = ℓ ( A ∗ , b ) , STRUCTURAL INVARIANT ON CERTAIN TWO-DIMENSIONAL NOETHERIAN PARTIALLY ORDERED SETS15 so | A | = | A ∗ | . By, our choice of x ∈ X \ A, we have η ( A, b ) = η (( A ∪ { x } ) ∗ , b ) and ℓ (( A ∪ { x } ) ∗ , b ) = 1 + ℓ ( A, b ) , so | ( A ∪ { x } ) ∗ | = 1 + | A | . Moreover, by (11), ρ ( A ∗ ) ⊂ ρ (( A ∪ { x } ) ∗ ) so A ∗ ⊂ ( A ∪ { x } ) ∗ . Since A L < X b and ( A L , b ) < Str X ( A ∪ { x } , b ) via A L , Corollary 6.10 implies ϕ ( A L , b ) =( ρ ( A L ) , ρ ( b )) , so ( ρ ( A L ) , ρ ( b )) < Str Y ( ρ (( A ∪ { x } ) ∗ ) , ρ ( b )) , and thus A L ⊂ ( A ∪ { x } ) ∗ . Fix 1 ≤ i ≤ η ( A, b ) . By the choice of B i and Corollary 6.10, we have ϕ ( Q i , B i ) = ( ρ ( Q i ) , ρ ( B i )) . After applying ϕ to (12) and using the fact that ρ is an isomorphism, we see Q i ⊂ ( A ∪ { x } ) ∗ . Since i was arbitrary, we have ∪ η ( A,b ) i =1 Q i ⊂ ( A ∪ { x } ) ∗ , and by choice of each Q i , it follows that | ∪ η ( A,b ) i =1 Q i | = η ( A, b ) + 1 . Putting all of this together, we see that we have accounted for 1 + | A | elements in ( A ∪ { x } ) ∗ : ℓ ( A, b ) elements from A L and η ( A, b ) + 1 elements from ∪ η ( A,b ) i =1 Q i . Therefore,( A ∪ { x } ) ∗ = A L ∪ { q , . . . , q η ( A,b ) } ∪ { x } = A ∪ { x } . Since A ∗ ⊂ A ∪ { x } , it suffices to show x / ∈ A ∗ . If x ∈ A ∗ , then ρ ( x ) ∈ ρ ( A ∗ ) , so( ρ ( x ) , G ∗ X ( ρ ( x ))) = ϕ ( x, G ∗ X ( x )) < Str Y ϕ ( A, b ) = ( ρ ( A ∗ ) , ρ ( b )) . Since ϕ is an isomorphism, we have ( x, G ∗ X ( x )) < Str X ( A, b ) , so x ∈ A, a contradiction. Therefore, A ∗ ⊂ A, and since both A, A ∗ are finite sets with | A | = | A ∗ | , we have A = A ∗ . (cid:3) The previous theorem, combined with Proposition 5.4, immediately gives us:
Theorem 6.12. If X and Y are J -posets, then X ∼ = Y if and only if Str X ∼ = Str Y. Specifically, if ρ : X → Y is an isomorphism, then the map ϕ : Str X → Str Y given by ϕ ( A, B ) = ( ρ ( A ) , ρ ( B )) for all ( A, B ) ∈ Str X is an isomorphism, and conversely, if ϕ :Str X → Str Y is any isomorphism, then there is an isomorphism ρ : X → Y such that ϕ ( A, B ) = ( ρ ( A ) , ρ ( B ))for all ( A, B ) ∈ Str X. Definition 6.13.
Define the following subposet of Str X :Str F X := { ( A, B ) ∈ Str X : A, B are finite } . If B ⊂ X is finite, define (Str F X ) B := (Str X ) B . As a corollary to Theorem 6.11, we show that an isomorphism from ψ : Str F X → Str F Y inducesan isomorphism ρ : X → Y such that ψ ( A, B ) = ( ρ ( A ) , ρ ( B )) for all ( A, B ) ∈ Str F X. For the restof the paper, let ψ : Str F X → Str F Y be an isomorphism of posets. By Theorem 6.4 and 6.11, itsuffices to find an isomorphism ϕ : Str X → Str Y such that ϕ restricts to ψ on Str F X. Note thatthe proofs of Lemmas 5.7, 6.1, 6.2 apply to ψ and Str F X. Also, by Remark 6.7, the proof of Item 1of Lemma 6.6 applies to ψ and Str F X as well. Lemma 6.14.
Let
X, Y be J -posets, let ψ : Str F X → Str F Y be an isomorphism of posets, andlet x ∈ X . Let b , b , b , . . . be an enumeration of G ∗ X ( x ) , and, for each i ≥ , let K i ( x ) be all( K, b i ) ∈ Str F X such that x ∈ K and mub X K = { b i } . Let K Str F X ( x ) = ∪ i =1 K i ( x ) . Then ψ ( K Str F X ( x )) = K Str F Y ( y )for some unique y ∈ Y . Proof. If Z is any J -poset, and z, w ∈ Z are distinct elements such that K Str F Z ( z ) ⊆ K Str Z F ( w ) , then z and w are both less than every element t ∈ Z such that z < Z t. Since z, w are distinct, wehave G ∗ Z ( z ) ⊂ mub Z { z, w } , a contradiction because mub Z { z, w } is finite. The uniqueness part ofthe lemma follows, and it suffices to show that ψ ( K Str F X ( x )) ⊆ K Str F Y ( y ) for some y because wemay apply the same argument to ψ − : Str F Y → Str F X to show that K Str F X ( x ) ⊆ ψ − ( K Str F Y ( y )) ⊆ K Str F X ( x ′ )which implies x = x ′ so ψ ( K Str F X ( x )) = K Str F Y ( y ) . With the enumeration of G ∗ X ( x ) as b , b , . . . , let B i := { b , . . . , b i } for each integer i ≥ . ByItem 1 of Lemma 6.6, we may write, for each integer j ≥ , ψ ( x, B j ) = ( F j , ρ ( B j )) for some finite F j ⊂ Y (see Remark 6.7). Now, ( F j , ρ ( B j )) < Str F Y ( J, ρ ( b i ))for all i ≤ j such that ( J, ρ ( b i )) ∈ ψ ( K i ( x )) . Now, F j ⊆ j \ i =1 \ ( J,ρ ( b i )) ∈ ψ ( K i ( x )) J := W j , so each W j is nonempty. Moreover, W j ⊇ W r for all r ≥ j. Since each W j is finite, there exists j such that W j = W r for all r ≥ j . In particular, \ ( J,ρ ( b i )) ∈∪ ∞ i =1 ψ ( K i ( x )) J = ∞ \ j =1 W j is nonempty. Let y ∈ ∩ ∞ j =1 W j . If (
K, b i ) ∈ K i ( x ) , then ht X ( K, b i ) > η ( K, b i ) = 0 , soht ψ ( K, b i ) > η ( ψ ( K, b i )) = 0 , and therefore mub Y J = { ρ ( b i ) } where ( J, ρ ( b i )) = ψ ( K, b i ) . Since y ∈ ∩ ∞ j =1 W j , we have y ∈ J, so ψ ( K, b i ) ∈ K Str F Y ( y ) . Thus, ψ ( K Str F X ( x )) ⊆ K Str F Y ( y ) . (cid:3) Theorem 6.15. If X, Y are J -posets, then Str F X ∼ = Str F Y if and only if X ∼ = Y. Additionally, if ψ : Str F X → Str F Y is an isomorphism, then there is an isomorphism ρ : X → Y such that ψ ( A, B ) = ( ρ ( A ) , ρ ( B ))for all ( A, B ) ∈ Str X. STRUCTURAL INVARIANT ON CERTAIN TWO-DIMENSIONAL NOETHERIAN PARTIALLY ORDERED SETS17
Proof. If ρ : X → Y is an isomorphism, then ψ : Str F X → Str F Y given by ψ ( A, B ) = ( ρ ( A ) , ρ ( B ))is an isomorphism. To get the other direction, it suffices to define a poset isomorphism ϕ : Str X → Str Y that extends ψ. If (
A, B ) ∈ Str F X, define ϕ ( A, B ) = ψ ( A, B ) . Otherwise, (
A, B ) = ( a, G ∗ X ( a )) for some a ∈ X . Then ψ ( K Str F X ( a )) = K Str F Y ( c ) for some unique c ∈ Y by the previous lemma. Define ϕ ( a, G ∗ X ( a )) = ( c, G ∗ X ( c )) . If ( A , B ) ≤ Str X ( A , B ) , and B is finite, then so is B , so ϕ ( A , B ) ≤ Str X ϕ ( A , B ) because ϕ extends ψ. If B is infinite and B is finite, then write ϕ ( A , B ) = ϕ ( a, G ∗ X ( a )) = ( c, G ∗ X ( c )) , and write ψ ( A , B ) = ( C, D ) . Since D = mub Y J for some J ⊆ C, we have it suffices to show D ⊂ G ∗ Y ( c ) and c ∈ C to conclude ( c, G ∗ X ( c )) < Str Y ( C, D ) . Having shown this, applying similarreasoning to ψ − shows that ϕ − : Str Y → Str X exists and is a poset map. Therefore, ϕ is anisomorphism, and by Theorem 5.1, we get the desired result.If b ∈ B , and ( K, b ) ∈ K Str F X ( a ) , then ϕ ( K, b ) = ( J ′ , ρ ( b )) ∈ K Str F Y ( c ) . So D = ρ ( B ) ⊂ G ∗ Y ( c ) . Let K ⊆ A be a maximal subset of A such that mub X K = B . Note that a ∈ K. Fix b ∈ B , and let K and K be finite subsets of X such that K , K , and K are pairwise disjoint andmub X K ∪ K = mub X K ∪ K = { b } . Now, (
K, B ) < Str F X ( K i ∪ K, b ) for i = 1 , , and if ( K ′ , B ) < Str F X ( K i ∪ K, b ) for i = 1 , K ′ ⊂ X , then K ′ ⊆ ( K ∪ K ) ∩ ( K ∪ K ) , so ( K ′ , B ) ≤ Str F X ( K, B ) because of how we chose K , K and K. In other words, (
K, B ) is thegreatest element of (Str F X ) B that is less than both ( K ∪ K , b ) and ( K ∪ K , b ) . Let ( J i , ρ ( b )) = ϕ ( K ∪ K i , b ) , and let ( S, ρ ( B )) = ϕ ( K, ρ ( B )) . Since ( K ∪ K i , b ) ∈ K Str F X ( a ) , it follows that c ∈ J ∩ J by the previous lemma. Moreover, since ( K, B ) ≤ Str F X ( A, B ) , we have ( S, B ) ≤ Str F Y ( C, ρ ( B )) . We claim c ∈ S. Both ( J , ρ ( b )) and ( J , ρ ( b )) both exceed ( S, ρ ( B )) in Str F Y. Moreover, since ϕ restricts to anisomorphism from the subposet (Str F X ) B ∪ (Str F X ) b onto (Str F Y ) ρ ( B ) ∪ (Str F Y ) ρ ( b ) , the node( S, ρ ( B )) is the greatest element of (Str F Y ) ρ ( B ) that is less than both ( J , ρ ( b )) and ( J , ρ ( b )) . Since c ∈ J ∩ J , and c < Y ρ ( B ) by the previous paragraph, we have ( c, ρ ( B )) < Str F Y ( J i , ρ ( b )) . So ( c, ρ ( B )) < Str F Y ( S, ρ ( B )) , and therefore c ∈ S ⊆ C. (cid:3) Acknowledgements
The author would like to thank his mentor, collaborator, and “big sister” S. Loepp for manyinsightful comments and for greatly helping to improve the quality of the paper. The author alsowishes to thank Washington & Lee University for their support via the Lenfest grant.
References [1] S. McAdam. Intersections of height 2 primes.
J. Algebra , 40:315–321, 1977.[2] A. Saydam and S. Wiegand. Prime ideals in birational extensions of two-dimensional domains over orders.
J. PureAppl. Algebra , 201:142–153, 2005.[3] R. Wiegand. Homeomorphisms of affine surfaces over a finite field.
Journal of the London Mathematical Society ,s2-18(1):28–32, 1978.[4] R. Wiegand. The prime spectrum of a two-dimensional affine domain.
J. Pure Appl. Algebra , 40:209–214, 1986. [5] R. Wiegand and S. Wiegand.
Prime Ideals in Noetherian Rings: A Survey , pages 175–193. Springer, 02 2011.
Email address ::