LOWER BOUNDS FOR THE NUMBER OF SEMIDUALIZING COMPLEXES
OVER A LOCAL RING
SEAN SATHER-WAGSTAFF∗
Abstract
We investigate the setᑭ(R)of shift-isomorphism classes of semi-dualizingR-complexes, ordered via the reflexivity relation, whereRis a commutative noetherian local ring. Specifically, we study the question of whetherᑭ(R)has cardinality 2nfor somen. We show that, if there is a chain of lengthninᑭ(R)and if the reflexivity ordering onᑭ(R)is transitive, thenᑭ(R)has cardinality at least 2n, and we explicitly describe some of its order-structure. We also show that, given a local ring homomorphismϕ:R→Sof finite flat dimension, ifRandSadmit dualizing complexes and ifϕis not Gorenstein, then the cardinality ofᑭ(S)is at least twice the cardinality ofᑭ(R).
1. Introduction
Throughout this work (R,ᒊ) and (S,ᒋ) are commutative noetherian local rings.
A homologically finite R-complex C issemidualizing if the natural ho- mothety morphism R → RHomR(C, C) is an isomorphism in the derived categoryD(R). (See Section 2 for background material.) Examples of semi- dualizingR-complexes includeRitself and a dualizingR-complex when one exists. The set of shift-isomorphism classes of semidualizingR-complexes is denotedᑭ(R), and the shift-isomorphism class of a semidualizingR-complex Cis denoted [C].
Semidualizing complexes were introduced by Avramov and Foxby [2] and Christensen [4] in part to investigate the homological properties of local ring homomorphisms. Our interest in these complexes comes from their potential as tools for answering the composition question for local ring homomorph- isms of finite G-dimension. Unfortunately, the utility of the semidualizing R-complexes is hampered by the fact that our understanding ofᑭ(R)is very limited. For instance, we do not even know if the setᑭ(R)is finite; see [5] for some recent progress.
∗The author was supported in part by a grant from the NSA.
Received 17 March 2009.
We are interested in the following question, motivated by results from [7], wherein|ᑭ(R)|is the cardinality of the setᑭ(R).
Question1.1. IfRis a local ring, must we have|ᑭ(R)| = 2nfor some n∈N?
Each semidualizingR-complex C gives rise to a notion of reflexivity for homologically finiteR-complexes. For instance, each homologically finiteR- complex of finite projective dimension is C-reflexive. On the other hand, a semidualizingR-complex C is dualizing if and only if every homologically finiteR-complex isC-reflexive. We orderᑭ(R)using reflexivity: write [C] [B] wheneverB isC-reflexive. This relation is reflexive and antisymmetric, but we do not know if it is transitive in general. Achaininᑭ(R)is a sequence [C0] [C1] · · · [Cn], and such a chain has lengthn if [Ci] = [Cj] wheneveri =j.
The main result of this paper, stated next, uses the lengths of chains in ᑭ(R)to provide a lower bound of the form 2non the cardinality ofᑭ(R). It is part of Theorem 3.3 which also contains the analogous result for the set of isomorphism classes of semidualizingR-modules.
Theorem1.2.Assume that the reflexivity ordering onᑭ(R)is transitive. If ᑭ(R)admits a chain of lengthn, then|ᑭ(R)| ≥2n.
An alternate proof of this result is given in Corollary 4.9. One advantage of this second method is that we can describe all the reflexivity relations between these complexes in terms of combinatorial data; see Theorem 4.8.
Using the ideas from Theorem 1.2, we also prove the following comparison result which is a special case of Theorem 3.5.
Theorem1.3.Letϕ:R →Sbe a local ring homomorphism of finite flat dimension. IfR andSadmit dualizing complexes and ifϕ is not Gorenstein, then|ᑭ(S)| ≥2|ᑭ(R)|.
2. Complexes and local ring homomorphisms
This section contains definitions and background material for use in the sequel.
Definition2.1. AnR-complex is a sequence ofR-module homomorph- isms
X= · · ·−→∂n+1X Xn
∂nX
−→Xn−1
∂n−1X
−→ · · ·
such that ∂n−X1∂nX = 0 for each integern. The nth homology module of X is Hn(X):= Ker(∂nX)/Im(∂n+X1). The complex Xishomologically bounded when Hn(X) = 0 for |n| 0. It is homologically finite if the R-module
n∈ZHn(X)is finitely generated. We frequently identifyR-modules withR- complexes concentrated in degree 0.
Notation2.2. We work in the derived categoryD(R). References on the subject include [9], [11], [13], [14]; see also [12]. Given twoR-complexesX andY, the derived homomorphism and tensor product complexes are denoted RHomR(X, Y ) and X ⊗LR Y. Isomorphisms in D(R) are identified by the symbol , and isomorphisms up to shift are identified by∼.
Definition2.3. For each integern, thenthBass numberofRisμnR(R)= rankR/ᒊ(ExtnR(R/ᒊ, R)), and theBass seriesofRis the power seriesIRR(t)=
∞
n=0μnR(R)rn.
Letϕ:R →Sbe a local ring homomorphism of finite flat dimension, that is, such thatSadmits a bounded resolution by flatR-modules. TheBass series of ϕ is a formal Laurent series Iϕ(t) with nonnegative integer coefficients such thatISS(t) = Iϕ(t)IRR(t); see [3, (5.1)] for the existence of Iϕ(t). The homomorphismϕisGorensteinatᒋifIϕ(t)=tdfor some integerd.
Example2.4. Letϕ:R →Sbe a local ring homomorphism of finite flat dimension. Whenϕis flat, it is Gorenstein if and only if the closed fibreS/ᒊS is Gorenstein. Also, ifϕis surjective with kernel generated by anR-sequence, then it is Gorenstein.
Semidualizing complexes, defined next, are our main objects of study.
Definition2.5. A homologically finiteR-complexC issemidualizingif the natural homothety morphismχCR:R→RHomR(C, C)is an isomorphism inD(R). AnR-complex Disdualizingif it is semidualizing and has finite injective dimension. Letᑭ(R)denote the set of shift-isomorphism classes of semidualizingR-complexes, and let [C] denote the shift-isomorphism class of a semidualizingR-complexC.
WhenC is a finitely generatedR-module, it is semidualizing if and only if Ext≥R1(C, C)= 0 and the natural homothety mapR →homR(C, C)is an isomorphism. Letᑭ0(R)denote the set of isomorphism classes of semidual- izingR-modules, and let [C] denote the isomorphism class of a semidualizing R-moduleC. The natural identification of anR-module with anR-complex concentrated in degree 0 provides a natural inclusionᑭ0(R)⊆ᑭ(R).
Remark2.6. Letϕ:R →Sbe a local ring homomorphism of finite flat dimension, and fix semidualizingR-complexes B, C. The complexS⊗LRC is semidualizing forSby [4, (5.7)]. The complexS⊗LR Cis dualizing for S if and only ifCis dualizing forRandϕis Gorenstein by [1, (5.1)]. We have S⊗LRB S⊗LRCinD(S)if and only ifB CinD(R)by [6, (1.10)]. Hence,
the functionᑭ(ϕ):ᑭ(R)→ᑭ(S)given by [C]→[S⊗LRC] is well-defined and injective.
The next definition is due to Christensen [4] and Hartshorne [11] and will be used primarily to compare semidualizing complexes.
Definition2.7. LetCbe a semidualizingR-complex. A homologically fi- niteR-complexXisC-reflexivewhen theR-complexRHomR(X, C)is homo- logically finite, and the natural biduality morphism δXC:X → RHomR(RHomR(X, C), C)is an isomorphism inD(R). Define an order on ᑭ(R)by writing [C][B] whenBisC-reflexive. Also, write [C][B] when [C][B] and [C]=[B]. For each [C]∈ᑭ(R)setᑭC(R)= {[B]∈ᑭ(R)| [C][B]}.
Remark2.8. LetA,BandCbe semidualizingR-complexes.
1. IfBisC-reflexive, thenRHomR(B, C)is semidualizing andC-reflexive by [4, (2.12)]. Thus, the map C:ᑭC(R) → ᑭC(R) given by [B] → [RHomR(B, C)] is well-defined. By definition, this map is also an invol- ution (i.e., 2C = idᑭC(R)) and hence it is bijective. From [6, (3.9)] we know that C is reverses the reflexivity ordering: if [A],[B] ∈ ᑭC(R), then [A] [B] if and only if C([B]) C([A]), that is, if and only if [RHomR(B, C)][RHomR(A, C)].
2. Assume that C is a semidualizing R-module. Using [6, (3.5)] we see that, ifBisC-reflexive, thenBis isomorphic up to shift with a semidualizing R-module, and hence so isRHomR(B, C). In particular, we haveᑭC(R) ⊆ ᑭ0(R).
3. IfDis a dualizingR-complex, then [D][C] by [11, (V.2.1)], i.e., we haveᑭD(R)=ᑭ(R).
4. LetXbe anR-complex such that Hi(X)is finitely generated for eachi and Hi(X) = 0 fori 0. IfC⊗LRC⊗LRX is semidualizing, thenC ∼ R by [7, (3.2)].
5. By [10, (3.3)], given a chain [Cn][Cn−1] · · · [C0] inᑭ(R), one has Cn C0⊗LRRHomR(C0, C1)⊗LR· · · ⊗LRRHomR(Cn−1, Cn).
Remark2.9. Letϕ:R → S be a local ring homomorphism of finite flat dimension. The map ᑭ(ϕ):ᑭ(R) → ᑭ(S) from Remark 2.6 respects the reflexivity orderings perfectly by [6, (4.8)]: if [B],[C] ∈ ᑭ(R), then [C] [B] if and only ifᑭ(ϕ)([C])ᑭ(ϕ)([B])that is, if and only if [S⊗LRC] [S⊗LRB].
The next definition is due to Avramov and Foxby [2] and Christensen [4].
Definition2.10. LetCbe a semidualizingR-complex. A homologically boun-dedR-complexXis in theBass classwith respect toC, denotedBC(R), when theR-complexRHomR(C, X)is homologically bounded and the natural evaluation morphismξXC:C⊗LR RHomR(C, X) → X is an isomorphism in D(R). A homologically boundedR-complexXis in theAuslander classwith respect toC, denotedAC(R), when theR-complexC⊗LRXis homologically bounded and the natural morphism γXC:X → RHomR(C, C ⊗LR X) is an isomorphism inD(R).
Remark 2.11. Let B and C be semidualizingR-complexes. Then B ∈ BC(R)if and only if [B] [C]; see [8, (1.3)]. One hasB ∈ AC(R)if and only ifB⊗LRCis a semidualizingR-complex by [8, (4.8)].
3. Bounding the number of elements inᑭ(R)
This section contains the proofs of Theorems 1.2 and 1.3 from the introduction.
We begin with two lemmas.
Lemma3.1.LetA,BandCbe semidualizingR-complexes such thatBand CareA-reflexive andB isC-reflexive. IfC ∼A, thenRHomR(B, A)is not C-reflexive.
Proof. Remark 2.8.1 implies thatRHomR(B, A)andRHomR(C, A)are semidualizingR-complexes and thatRHomR(C, A)isRHomR(B, A)-reflex- ive. In the next sequence, the first isomorphism is from Remark 2.8.5:
RHomR(B, A)
RHomR(C, A)⊗LRRHomR(RHomR(C, A),RHomR(B, A)) RHomR(C, A)⊗LRRHomR(RHomR(C, A)⊗LRB, A) RHomR(C, A)⊗LRRHomR(B,RHomR(RHomR(C, A), A)) RHomR(C, A)⊗LRRHomR(B, C).
The second and third isomorphisms are Hom-tensor adjointness, and the fourth isomorphism comes from the fact thatCisA-reflexive.
SetX=RHomR(B, C)⊗LRRHomR(RHomR(B, A), C), and suppose that the complex RHomR(B, A) is C-reflexive. Remark 2.8.5 explains the first isomorphism in the next sequence, and the second isomorphism is from the previous display
C RHomR(B, A)⊗LRRHomR(RHomR(B, A), C)
RHomR(C, A)⊗LRRHomR(B, C)⊗LRRHomR(RHomR(B, A), C) RHomR(C, A)⊗LRX.
Similarly, this yields the next sequence A RHomR(C, A)⊗LRC
RHomR(C, A)⊗LRRHomR(C, A)⊗LRX.
It follows from Remark 2.8.4 thatRHomR(C, A)∼Rand hence C RHomR(RHomR(C, A), A)∼RHomR(R, A) A sinceCisA-reflexive. This contradicts the assumptionC ∼A.
Note that the hypothesis ᑭC(R) ⊆ ᑭA(R) from the next result is satis- fied when eitherAis dualizing forR or the reflexivity ordering on ᑭ(R)is transitive.
Lemma3.2.LetAandC be semidualizingR-complexes such thatCisA- reflexive andC ∼A. Assume thatᑭC(R)⊆ᑭA(R), e.g., that the reflexivity ordering onᑭ(R)is transitive. The injectionA:ᑭA(R)→ᑭA(R)given by [B] → [RHomR(B, A)] mapsᑭC(R) intoᑭA(R)ᑭC(R). In particular
|ᑭA(R)| ≥2|ᑭC(R)|.
Proof. The first conclusion is a reformulation of Lemma 3.1; see also Remark 2.8.1. For the second conclusion, note that A is injective by Re- mark 2.8.1, so |A(ᑭC(R)) = |ᑭC(R)|. Since A(ᑭC(R)) ⊂ ᑭA(R) ᑭC(R), we conclude that ᑭC(R) and A(ᑭC(R)) are disjoint subsets of ᑭA(R)such that|ᑭC(R)| = |A(ᑭC(R))|. The second conclusion now fol- lows.
The next result contains Theorem 1.2 from the introduction.
Theorem3.3.Assume thatᑭ(R)admits a chain[Cn][Cn−1] · · · [C0] such that ᑭC0(R) ⊆ · · · ⊆ ᑭCn−1(R) ⊆ ᑭCn(R). Then |ᑭ(R)| ≥ 2n. If [Cn]∈ᑭ0(R), then|ᑭ0(R)| ≥2n.
Proof. For the first statement, we show by induction onjthat|ᑭCj(R)| ≥ 2j. For j = 0 this is straightforward. For the inductive step assume that
|ᑭCj(R)| ≥ 2j. Lemma 3.2 implies that|ᑭCj+1(R)| ≥ 2|ᑭCj(R)| ≥ 2j+1as desired.
When [Cn] ∈ ᑭ0(R), we have ᑭCj(R) ⊆ ᑭCn(R) ⊆ ᑭ0(R) for j = 0, . . . , nby Remark 2.8.2. Thus, the second statement is proved like the first statement.
We next provide an explicit description of the 2nsemidualizing complexes that are guaranteed to exist by Theorem 3.3. A second description (in the case C0 R) is given in Theorem 4.8.
Remark3.4. Assume thatᑭ(R)admits a chain [Cn][Cn−1] · · · [C0] such thatᑭC0(R)⊆ · · · ⊆ ᑭCn−1(R)⊆ ᑭCn(R). GivenR-complexesC and B, setC†B =RHomR(C, B).
The next diagram shows the stepsn=0,1,2,3 of the induction argument in the proof of Theorem 3.3, with some of the reflexivity relations indicated with edges:
C0
C0
C0 C0
C0†C1 C0†C1 C0†C1†C2 C0†C1 C0†C1†C2
C0†C2 C0†C2†C3 C0†C2
C0†C1†C2†C3 C0†C1†C3
C0†C3.
More generally, for each sequence of integersi = {i1, . . . , ij}such that j ≥ 0 and 1 ≤ i1 < · · · < ij ≤ n, the R-complexCi = C
†Ci1†Ci2†Ci3...†Cij 0
is semidualizing. (Whenj = 0 we haveCi = C∅ = C0.) The classes [Ci] are parametrized by the allowable sequencesi, of which there are exactly 2n. These are precisely the 2nclasses constructed in the proof of Theorem 3.3.
If theCiare all modules, then we may replaceRHom with Hom to obtain a description of the semidualizing modules that the theorem guarantees to exist.
The next result contains Theorem 1.3 from the introduction.
Theorem3.5.Letϕ:R →Sbe a local ring homomorphism of finite flat dimension. Assume that ᑭ(R) has a unique minimal element[A]. (For in- stance, this holds whenRadmits a dualizing complex.)IfSadmits a dualizing complexDS and ifϕis not Gorenstein atᒋ, then|ᑭ(S)| ≥2|ᑭ(R)|.
Proof. Letᑭ(ϕ):ᑭ(R) →ᑭ(S) be the induced map from Remark 2.6.
Our assumption onAimplies thatᑭ(R)=ᑭA(R). Remark 2.9 provides the first containment in the next sequence while Remark 2.8.3 explains the last equality
ᑭ(ϕ)(ᑭ(R))=ᑭ(ϕ)(ᑭA(R))⊆ᑭS⊗L
RA(S)⊆ᑭ(S)=ᑭDS(S).
Sinceϕis not Gorenstein, Remark 2.6 impliesDS ∼S⊗LRA. The injectivity ofᑭ(ϕ)explains the first inequality in the next sequence
2|ᑭ(R)| =2|ᑭA(R)| ≤2|ᑭS⊗LRA(S)| ≤ |ᑭDS(S)| = |ᑭ(S)|
while the second inequality is from Lemma 3.2.
4. Structure on the Set of SemidualizingR-complexes
This section is devoted to providing a second description of the 2nsemidualiz- ing complexes that are guaranteed to exist by Theorem 3.3. This description is contained in Theorems 4.7 and 4.8. It says, in particular, that these complexes form a dual version of the “LCM lattice” onnformal letters. Note that the version for semidualizing modules has the same form with derived functors replaced by non-derived functors; hence, we do not state it explicitly. We begin with some notation for use throughout the section and five supporting lemmas.
Assumption4.1. Assume throughout this section thatᑭ(R)admits a chain [Cn][Cn−1] · · · [C0]. Fori= 1, . . . , nsetBi =RHomR(Ci−1, Ci). Set B∅=C0. For each sequencei= {i1, . . . , ij}such that 1≤i1<· · ·< ij ≤n andj ≥1, setBi = Bi1⊗LR· · · ⊗LRBij. (Whenj = 1, we haveBi = B{i1} = Bi1.)
Lemma4.2.Under the hypotheses of Assumption 4.1, if1 ≤ i < j ≤ n, then RHomR(Ci−1, Cj−1)⊗LRBj RHomR(Ci−1, Cj).
Proof. Consider the following sequence of isomorphisms:
RHomR(Bj,RHomR(Ci−1, Cj))
RHomR(RHomR(Cj−1, Cj),RHomR(Ci−1, Cj)) RHomR(RHomR(Cj−1, Cj)⊗LRCi−1, Cj) RHomR(Ci−1,RHomR(RHomR(Cj−1, Cj), Cj)) RHomR(Ci−1, Cj−1).
The first two isomorphisms are from Hom-tensor adjointness and the com- mutativity of tensor product. The third isomorphism is from the condition [Cj] [Cj−1]. Since the R-complexes Bj, RHomR(Ci−1, Cj) and RHomR(Ci−1, Cj−1) are semidualizing, it follows from [8, (1.3)] that [RHomR(Ci−1, Cj)] [Bj] so the first isomorphism in the next sequence is from Remark 2.8.5:
RHomR(Ci−1, Cj) Bj ⊗LRRHomR(Bj,RHomR(Ci−1, Cj)) Bj ⊗LRRHomR(Ci−1, Cj−1)
The second isomorphism is from the first displayed sequence in this proof.
Lemma4.3.Under the hypotheses of Assumption 4.1, if1≤i < i+p≤n, then
B{i,i+1,...,i+p} RHomR(Ci−1, Ci+p).
Proof. We proceed by induction on p. The base case is when p = 1.
Setting j = i +1 in Lemma 4.2 yields the first isomorphism in the next sequence
RHomR(Ci−1, Ci+1) Bi+1⊗LRRHomR(Ci−1, Ci) Bi+1⊗LRBi B{i,i+1}. The remaining isormorphisms are by definition.
For the inductive step, assume thatB{i,i+1,...,i+p−1} RHomR(Ci−1,Ci+p−1). This assumption explains the second isomorphism in the next sequence
B{i,i+1,...,i+p} Bi⊗LR· · · ⊗LRBi+p−1⊗LRBi+p
RHomR(Ci−1, Ci+p−1)⊗LRBi+p
RHomR(Ci−1, Ci+p).
The first and third isomorphisms are by definition, and the fourth isomorphism is from Lemma 4.2.
Lemma 4.4.Assume that ᑭCi(R) ⊆ ᑭCj−1(R). Under the hypotheses of Assumption 4.1, if1≤i < j−1≤n−1, then
B{i,j} RHomR(RHomR(Bi, Cj−1), Cj)).
(In the notation from Remark 3.4, this readsB{i,j} Bi†Cj−1†Cj Ci−†Ci1†Cj−1†Cj.) Proof. The proof is similar to that for Lemma 4.2. The main point is the following sequence of isomorphisms wherein the last two isomorphisms are from Hom-tensor adjointness and the commutativity of tensor product:
Bi RHomR(RHomR(Bi, Cj−1), Cj−1)
RHomR(RHomR(Bi, Cj−1),RHomR(Bj, Cj)) RHomR(Bj⊗LRRHomR(Bi, Cj−1), Cj) RHomR(Bj,RHomR(RHomR(Bi, Cj−1), Cj)).
The first isomorphism is from the chain [Cj−1][Ci][RHomR(Ci−1, Ci)]
= [Bi], using the containmentᑭCi(R) ⊆ ᑭCj−1(R); see Remark 2.8.1. The second isomorphism is from the condition [Cj][Cj−1] and the definition of Bj.
Lemma4.5.Assume thatᑭC1(R)⊆ · · · ⊆ᑭCn−1(R)⊆ᑭCn(R). Under the hypotheses of Assumption 4.1, each of theR-complexesBi is semidualizing.
Proof. We proceed by induction on n, the length of the chain inᑭ(R). The base casesn= 0,1 are routine. Indeed, when n= 0, we have only one Bi, namelyB∅ = C0. Whenn=1, we have only twoBi, namelyB{1} = B1 andB∅=C0.
Now, assume thatn≥2 and that the result holds for chains of lengthn−1.
Also, assume without loss of generality thati = ∅. We have two cases.
Case 1: Ifi2=i1+1, then the first isomorphism in the next sequence is by definition, and the second isomorphism is from Lemma 4.3:
Bi Bi1⊗LRBi1+1⊗LRBi3⊗LR· · · ⊗LRBij
RHomR(Ci1−1, Ci1+1)⊗LRBi3⊗LR· · · ⊗LRBij
Applying our induction hypothesis to the chain [Cn] · · · [Ci1+1][Ci1−1]
· · ·[C0], we conclude that the tensor product in the final line of this sequence is semidualizing. HenceBi is semidualizing in this case.
Case 2: Ifi2> i1+1, then the first isomorphism in the next sequence is by definition, and the second isomorphism is from Lemma 4.4:
Bi
Bi1⊗LRBi2⊗LRBi3⊗LR· · · ⊗LRBij
RHomR(RHomR(RHomR(Ci1−1, Ci1), Ci2−1), Ci2)⊗LRBi3⊗LR· · · ⊗LRBij
Applying our induction hypothesis to the chain
[Cn] · · · [Ci2][RHomR(RHomR(Ci1−1, Ci1), Ci2−1)] we conclude again thatBi is semidualizing in this case.
Lemma4.6.Assume thatᑭC1(R) ⊆ · · · ⊆ ᑭCn−1(R) ⊆ ᑭCn(R). Under the hypotheses of Assumption 4.1, if s = {s1, . . . , s1} is a sequence of in- tegers such that1 ≤ s1 < · · · < st ≤ nandi ⊇ s, then[Bi] [Bs] and RHomR(Bs, Bi) Bis.
Proof. By Lemma 4.5, theR-complexesBi,BsandBisare semidualizing.
By definition, we haveBs Bi⊗LRBis, so the result follows from [10, (3.5)].
The following theorem compares the complexes constructed in this section to those from the previous section.
Theorem4.7.Assume thatᑭC1(R)⊆ · · · ⊆ᑭCn−1(R)⊆ᑭCn(R). Under the hypotheses of Assumption 4.1, ifC0 R, then theR-complexesBi are precisely the2ncomplexes constructed in Theorem 3.3.
Proof. We proceed by induction on n, the length of the chain inᑭ(R). The base casen= 0 is straightforward: The onlyBi isB∅ =C0 R, which is the only module constructed in the proof of Theorem 3.3.
For the inductive step, assume that n ≥ 1 and that the result holds for the chain [Cn−1] · · · [C0]. Then theR-complexesBi withij ≤ n−1 are precisely the 2n−1complexes constructed in Theorem 3.3 for this chain.
The remaining 2n−1 semidualizing R-complexes constructed in Theo- rem 3.3 (according to the proof) for the chain [Cn][Cn−1] · · · [C0] are then of the formRHomR(Bi, Cn). Thus, it remains to show that each of these complexes is of the formBs. This is shown in the following sequence wherein we use the notation [n]= {1,2, . . . , n}:
RHomR(Bi, Cn) RHomR(Bi, B1⊗LR· · · ⊗LRBn) RHomR(Bi, B[n]) B[n]i
The first isomorphism follows from Remark 2.8.5 using the assumptionC0
R. The second isomorphism is by definition. The third isomorphism is from Lemma 4.6.
The next result provides a purely combinatorial description of the reflexivity ordering on the set of [Bi].
Theorem4.8.Assume thatᑭC1(R)⊆ · · · ⊆ᑭCn−1(R)⊆ᑭCn(R). Under the hypotheses of Assumption 4.1, the following conditions are equivalent:
(i) One hasBi Bs;
(ii) One hasBs ∈BBi(R); and (iii) One hasi⊇s.
Proof. The equivalence (i)⇐⇒(ii) is from Remark 2.11; and the implic- ation (iii)⇒(i) is contained in Lemma 4.6.
(i)⇒(iii): Assume that [Bi] [Bs]. Lemma 4.6 implies that [Bs∪i] [Bs] and [Bi] [Bis], and furthermore that RHomR(Bs, Bs∪i) Bis. Since the reflexivity ordering onᑭ(R)is transitive, we have
[Bs∪i][Bs][RHomR(Bs, Bs∪i)].
From [7, (3.3.a)] we conclude that [Bs∪i]=[Bs], and hence the first equality in the next sequence:
[R]=[RHomR(Bs, Bs∪i)]=[Bis].
It is straightforward to show that this implies thatis= ∅, and thusi⊆s.
Corollary 4.9. Assume that ᑭC1(R) ⊆ · · · ⊆ ᑭCn−1(R) ⊆ ᑭCn(R). Under the hypotheses of Assumption 4.1, one has[Bi] = [Bs]if and only if i=s. In particular, there are exactly2nclasses of the form[Bi].
Proof. One implication of the biconditional statement is straightforward.
For the converse, assume that [Bi] = [Bs]. Then [Bi] [Bs] [Bi], so Theorem 4.8 implies thati ⊆s⊆i, that isi=sas desired.
It follows that there are exactly 2nclasses of the form [Bi] because this is the number of possiblei⊆[n].
Remark4.10. Assume thatn=3 and thatᑭC1(R)⊆ᑭC2(R)⊆ᑭC3(R). The next diagram shows the case of the ordered set ofR-complexes of the formBi, with all the reflexivity relations indicated with edges:
B∅
B{1} B{2} B{3}
B{1,2} B{1,3} B{2,3}
B{1,2,3}.
Compare this to the diagram in Remark 3.4.
We conclude with a version of Theorem 4.8 for Auslander classes.
Theorem4.11.Assume thatᑭC1(R)⊆ · · · ⊆ᑭCn−1(R)⊆ᑭCn(R). Under the hypotheses of Assumption 4.1, the following conditions are equivalent:
(i) One hasBi ∈ABs(R); (ii) One hasBs ∈ABi(R);
(iii) TheR-complexBi⊗LRBsis semidualizing; and (iv) One hasi∩s= ∅.
Proof. The equivalences (i)⇐⇒(ii)⇐⇒(iii) are from Remark 2.11.
(iv)⇒(iii): Ifi∩s = ∅, thenBi ⊗LRBs Bi∪s, which is semidualizing by Lemma 4.5.
(iii)⇒(iv): Assume that Bi ⊗LR Bs is a semidualizing R-complex, and suppose thata∈i∩s. It follows that we haveBi ⊗LRBs Ba⊗LRBa⊗LRX
whereX=Bi{a}⊗LRBs{a}. Remark 2.8.4 implies thatBa ∼R, and hence Ca−1 RHomR(RHomR(Ca−1, Ca), Ca)
RHomR(Ba, Ca)
∼RHomR(R, Ca) Ca.
This contradicts the condition [Ca]=[Ca−1] which is part of the assumption [Ca][Ca−1]. Thus, we must havei∩s= ∅.
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