# On Presented Dimensions of Modules and Rings

Authors: Dexu Zhou; Zhiwei Gong
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

We define the presented dimensions for modules and rings to measure how far away
a module is from having an infinite finite presentation and develop ways to compute
the projective dimension of a module with a finite presented dimension and the right
global dimension of a ring. We also make a comparison of the right global dimension,
the weak global dimension, and the presented dimension and divide rings into four
classes according to these dimensions.


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## Body

## 1. Introduction

LetR be a ring and n a nonnegative integer. Following [1, 2], a right R-module M is called n-presented in case it has a finite n-presentation, that is, there is an exact sequence of right R-modules(1.1)Fn→Fn-1→⋯→F1→F0→M→0,
where each Fi is a finitely generated free, equivalently projective, right R-module. A module is 0-presented (resp., 1-presented) if and only if it is finitely generated (resp., finitely presented), and each m-presented module is n-presented for m≥n. A ring R is called rightn-coherent in case every n-presented right R-module is (n+1)-presented. It is easy to see that R is right 0-coherent (resp., 1-coherent) if and only if it is right Noetherian (resp., coherent), and every n-coherent ring is m-coherent for m≥n.As in [1, 3], we set λR(M)=sup{n∣M has a finite n-presentation} and note that λR(M)≥n is a way to express how far away a module M is from having an infinite finite presentation. Clearly every finitely generated projective module M has an infinite finite presentation, that is, λR(M)=∞. The lambda dimension of a ring R is the infimum of the set of integers n such that every R-module having a finite n-presentation has an infinite finite presentation. It was studied extensively by Vasconcelos in [3], where it was denoted by λ-dim(R). Note that R is right n-coherent if and only if λ-dim(R)≤n and if and only if every n-presented module has an infinite finite presentation.Ng [4] defined the finitely presented dimension of a module M as f.p.dim(M) = inf{n∣ there exists an exact sequence Pn+1→Pn→⋯→P0→M→0 of R-modules, where each Pi is projective, and Pn+1,Pn are finitely generated}, which measures how far away a module is from being finitely presented. Motivated by this, we define a dimension, called presented dimension, for modules and rings in this paper. It measures how far away a module is from having an infinite finite presentation and how far away a ring is from being Noetherian. In Section 2, we give the definitions and show the properties of presented dimensions. In Section 3, using strongly presented modules, we give the structure of modules with presented dimensions ≤1 and develop ways to compute the projective dimension of a module with a finite presented dimension and the right global dimension of a ring. In Section 4, we define the presented dimension of a ring, make a comparison of the right global dimension, the weak global dimension, and the presented dimension, and divide rings into four classes according to these dimensions. In Section 5, we provide the properties of presented dimensions of modules and rings under an almost excellent extension of rings.Throughout rings are associative with identity, modules are unitary rightR-modules, and homomorphisms are module homomorphisms. The notations pd(M),id(M), and fd(M)denote the projective, injective, flat dimension of M, and rgD(R), wD(R) denote the right global dimension, weak global dimension, respectively. For other definitions and notations in this paper we refer to [5, 6].


## 2. Presented Dimensions of Modules

Definition 2.1.
LetM be a right R-module, define the presented dimension of M as follows:
(2.1)FPd(M)=inf{m∣thereexistsaprojectiveresolution⋯→Pm+j→⋯→Pm→Pm-1→⋯→P0→M→0suchthatPm+i,i=0,1,2,…arefinitelygenerated}.
If there is no such resolution, then define FPd(M)=∞.In particular, ifFPd(M)=0, then M has an infinite finite presentation. In this case, we call M a strongly presented module. Consequently, we may regard the presented dimension as a measure of how far away a module is from having infinite finite presentation.Clearly,R is right n-coherent if and only if FPd(M)=0 for each (n-1)-presented right R-module M, if and only if every (n-1)-presented module has infinite finite presentation.Proposition 2.2.
LetM be a right R-module, then FPd(M)≤pd(M)+1.Proof.
Directly by Definition2.1.We remark thatFPd(M) can be much smaller than pd(M). Take R=ℤ4. The ideal 2ℤ4 has projective dimension ∞ while FPd(2ℤ4)=0 for R is Noetherian.Proposition 2.3.
No finitely generated rightR-module has presented dimension 1.Proof.
Suppose thatM is a finitely generated right R-module with FPd(M)=1. There is a projective resolution
(2.2)⋯→P2→P1→P0→d0M→0,
where P1,P2,… are finitely generated; it follows that kerd0 is finitely presented and hence finitely generated. Note that 0→kerd0→P0→M→0 is exact and M is finitely generated, thus P0 is finitely generated, so FPd(M)=0, a contradiction.It is known that every finitely presented flat rightR-module is projective, that is, if f.p.dim(M)=0, then(2.3)fd(M)<f.p.dim(M)+1⇒pd(M)<f.p.dim(M)+1.
For the presented dimensions of modules, we give a general result as follows.Theorem 2.4.
Assume thatFPd(M)=m<∞ and t≥0 is an integer, then
(2.4)fd(M)<FPd(M)+tiffpd(M)<FPd(M)+t.Proof.
Whenm=0, this is trivial. Now suppose that 0<m<∞, then there is a projective resolution
(2.5)⋯→Pm+n→dm+n⋯→Pm→dmPm-1→⋯→P0→d0M→0,
where Pm,…,Pm+n,… are finitely generated.
We only need to prove the necessity. LetKi=kerdi, i=0,1,…,n+m,…, and K-1=M. For each integer t≥0, there is an exact sequence
(2.6)0→Km+t-2→Pm+t-2→⋯→Pm→⋯→P0→M→0,(2.7)0→Km+t-1→Pm+t-1→Km+t-2→0.
Since fd(M)<m+t, we have that Km+t-2 is flat from (2.6). Note that Pm+t is finitely generated and projective, thus Km+t-1=Imdm+t is finitely generated. From (2.7) and [7], it follows that Km+t-2 is projective. Thus pd(M)≤m+t-1 from (2.6), so pd(M)<m+t.In particular, we have the following corollary.Corollary 2.5.
M is a projective right R-module if and only if FPd(M)≤1 and M are flat.Proof.
(⇒). Immediately from Proposition 2.2.
(⇐). If FPd(M)=0, then M is finitely presented, Thus M is projective.
IfFPd(M)=1, then fd(M)<FPd(M). From Theorem 2.4, we have pd(M)<FPd(M). Thus pd(M)=0, so M is projective.We recall the mapping cone construction. Suppose thatF:C′→C is a morphism of complexes. Then MC(F) is a complex with MC(F)n=Cn⊕Cn-1′, and the sequence of complexes 0→C→MC(F)→C′(-1)→0 is exact (see [8]).Assume that(2.8)is a commutative diagram in which the vertical maps are projective resolutions. Iff is a monomorphism, MC(F) is a projective resolution of cokerf. If f is an epimorphism, then (2.9)⋯→MC(F)n→MC(F)n-1→⋯→MC(F)2→Z1(MC(F))→kerf→0
is a projective resolution of kerf.Assume thatFPd(M)=m, there is an exact sequence (2.10)⋯→Pm+n→⋯→Pm→Pm-1→⋯→P0→M→0,
where Pi(i=0,…,m+n,…) are projective, and Pm,…,Pm+n,… are finitely generated; we call such an infinite exact sequence a representing sequence of M.Theorem 2.6.
Assume that0→A′→A→A′′→0 is an exact sequence of right R-modules, FPd(A′)=d′,FPd(A)=d,FPd(A′′)=d′′. If two of these are finite, then so is the third. Furthermore,
(2.11)d≤max{d′,d′′},d′′≤max{d,d′+1},d′≤max{d,d′′-1}.Proof.
Suppose thatd′,d′′ are finite. Let P′,P′′ represent sequences of A′,A′′, respectively. There exists a projective resolution P of A such that 0→P′→P→P′′→0 is an exact sequence of complexes. Thus Pm is finitely generated when m≥max{d′,d′′}. So FPd(A)≤max{d′,d′′}.
Suppose thatd′,d are finite. Let P′,P represent sequences of A′,A, respectively, and let F:P′→P cover f:A′→A, thus MC(F) is a projective resolution of A′′. By the definition of MC(F), we have that MC(F)m is finitely generated for each m≥d and m≥d′+1. So FPd(A′′)≤max{d,d′+1}.
Suppose thatd,d′′ are finite. Let P,P′′ represent sequences of A,A′′, respectively, and let G:P→P′′ cover g:A→A′′. Then P′ is a projective resolution of A′, where Pm′=MC(G)m+1(m≥1), P0′=Z1(MC(G)). Thus MC(G)m are finitely generated, whenever m≥d′′ and m≥d+1. So Pm′ is finitely generated for m≥d′′-1 and m≥d. Note that P0′ is finitely generated if d′′≤1 and d=0 by the split exact sequence
(2.12)0→Z1(MC(F))→MC(F)1→MC(F)0→0.
So FPd(A′)≤max{d,d′′-1}.Corollary 2.7.
IfFPd(A1),…,FPd(Am) are finite, then
(2.13)FPd(A1⊕⋯⊕Am)=max{FPd(Ai)∣i=1,…,m}.Proof.
Clearly it suffices to consider the casem=2. Then there exist exact sequences
(2.14)0→A1→A1⊕A2→A2→0,0→A2→A1⊕A2→A1→0.
By Theorem 2.6, we have
(2.15)FPd(A1⊕A2)≤max{FPd(A1),FPd(A2)},FPd(A1)≤max{FPd(A1⊕A2),FPd(A2)-1},FPd(A2)≤max{FPd(A1⊕A2),FPd(A1)-1}.
Suppose that FPd(A1⊕A2)<FPd(A1). Then FPd(A2)≤FPd(A1)-1, thus
(2.16)FPd(A1)≤max{FPd(A1⊕A2),FPd(A2)-1}≤max{FPd(A1⊕A2),FPd(A1)-2}=FPd(A1⊕A2),
which contradicts the hypothesis. So FPd(A1⊕A2)≥FPd(A1). Similarly, FPd(A1⊕A2)≥FPd(A2). Therefore FPd(A1⊕A2)=max{FPd(A1),FPd(A2)}.


## 3. Strongly Presented Modules

Theorem 3.1.
FPd(M)≤1 if and only if there are a projective module P0, a free module F0, and a strongly presented module M0 such that M⊕P0=M0⊕F0.Proof.
(⇒). Suppose that FPd(M)≤1. There is an exact sequence 0→K→iP→πM→0, where P is projective and K is strongly presented. Choose a projective module P0 such that P⊕P0 is free, and let F=P⊕P0. Thus we have an exact sequence
(3.1)0→K→iF→σ=π⊕IP0M⊕P0→0.
Suppose that K is generated by the set {g1,g2,…,gn}. Choose a basis {f1,f2,…,fj,…} of F such that i(g1),i(g2),…,i(gn) can be generated by f1,f2,…,fm. Let F1 be generated by f1,f2,…,fm, and F2 generated by fm+1,fm+2,…,fj,…. Then F=F1⊕F2, and i(K)⊆F1. Let M0=σ(F1), F0=σ(F2). Then M0 is strongly presented, F0≅F2, and M⊕P0=M0⊕F0.
(⇐). Suppose that M⊕P0=M0⊕F0, where P0 is a projective module, F0 is a free module, and M0 is a strongly presented module. There is a finitely generated free module F such that the following sequence:
(3.2)0→K→iF0⊕F→fF0⊕M0=M⊕P0→0
is exact and K is strongly presented. Let π:M⊕P0→P0 be the canonical projection. Then we have an exact sequence 0→K′→F0⊕F→πfP0→0, where K′=kerπf. It is clear that i(K)⊆K′. Thus 0→K→iK′→f|k′M→0 is exact, hence
(3.3)kerf|K′=kerf∩K′=i(K)∩K′=i(K),
and f|K′ is epimorphic. Note that K is strongly presented and K′ is projective, thus FPd(M)≤1.Corollary 3.2.
Assume thatR is a ring such that every projective module is free (e.g., R is local). Then FPd(M)≤1 if and only if there are a strongly presented module M0 and a free module F such that M=M0⊕F.Next, we aim to obtain a test for projectivity of modules with finite presented dimensions. In [1, Theorem 1.7], it was proved that pd(M)≤d for every n-presented module M if and only if ExtRd+1(M,N)=0 for every n-presented module M and (n-(d+1))-presented module N. We generalize it as follows.Proposition 3.3.
Assume thatM is a strongly presented module and n≥0 is an integer. Then pd(M)≤n if and only if ExtRn+1(M,N)=0 for every strongly presented module N.Proof.
The necessity is clear. Conversely, we proceed by induction onn. If n=0 and ExtR1(M,N)=0 for every strongly presented module N, there is, an exact sequence
(3.4)0→M1→F→M→0,
where F is finitely generated and free and M1 is strongly presented. Thus ExtR1(M,M1)=0, whence Hom(F,M1)→Hom(M1,M1)→0; is exact, so 0→M1→F→M→0 is split, and M is a direct summand of F, hence projective, that is, pd(M)≤0.
Now suppose thatn≥1 and ExtRn+1(M,N)=0 for every strongly presented module N. Since
(3.5)0=ExtRn+1(M,N)=ExtRn(M1,N),
and M1 is strongly presented, by hypothesis pd(M1)≤n-1, so pd(M)≤n.Corollary 3.4.
Assume thatM is strongly presented and n≥0 is an integer. If pd(M)=n, then ExtRn(M,R)≠0.Proof.
Since pd(M)=n, by Proposition 3.3 there is a strongly presented module N such that ExtRn(M,N)≠0; thus there is an exact sequence 0→N1→P→N→0, where P is finitely generated and projective, and N1 is finitely generated. So we have an exact sequence ExtRn(M,P)→ExtRn(M,N)→0 for pd(M)=n.
Suppose thatExtRn(M,R)=0. Then ExtRn(M,P)=0 for each finitely generated projective module P, so ExtRn(M,N)=0, a contradiction. Therefore ExtRn(M,R)≠0.Lemma 3.5.
Assume thatFPd(M)≤1. Then pd(M)≤n if and only if ExtRn+1(M,N)=0 for every strongly presented module N.Proof.
By Theorem3.1, M⊕P0=M0⊕F, where P0 is projective, F is free, and M0 is strongly presented. Thus pd(M)≤n if and only if ExtRn+1(M,B)=0 for every module B, if and only if ExtRn+1(M0,B)=0 for every module B, if and only if pd(M0)≤n, if and only if ExtRn+1(M0,N)=0 for every strongly presented module N by Proposition 3.3, if and only if ExtRn+1(M,N)=0 for every strongly presented module N.Theorem 3.6.
Assume thatFPd(M)<∞ and n is an integer. Then pd(M)≤n if and only if ExtRn+1(M,N)=0 for every strongly presented module N.Proof.
Suppose thatFPd(M)=m. Then m≤pd(M)+1 by Proposition 2.2, and there is a projective resolution of M(3.6)⋯→Pm+l→dm+l⋯→Pm→dmPm-1→⋯→P0→d0M→0,
where Pm,…,Pm+l,… are finitely generated. Thus kerdm-1 is strongly presented.
Suppose thatn=m-1. Then FPd(kerdm-2)≤1, hence pd(kerdm-2)≤n-m+1 if and only if ExtRn-m+2(kerdm-2,N)=0 for every strongly presented module N by Lemma 3.5. Suppose that n≥m; by Proposition 3.3 pd(kerdm-1)≤n-m if and only if ExtRn-m+1(kerdm-1,N)=0 for every strongly presented module N.
Therefore pd(M)≤n if and only if ExtRn+1(M,N)=0 for every strongly presented module N.Now we obtain a way to compute the right global dimension of a ring.Corollary 3.7.
Assume thatrgD(R)<∞. Then
(3.7)rgD(R)=sup{id(N)∣Nisstronglypresented}.Proof.
By Proposition2.2.  FPd(R/I)<∞ for each right ideal I of R, thus pd(R/I)≤n for each right ideal I of R if and only if ExtRn+1(R/I,N)=0 for each strongly presented module N and each right ideal I of R by Theorem 3.6, if and only if id(N)≤n for each strongly presented module N by the Baer Criterion for injectivity. Therefore the result holds.


## 4. Presented Dimensions of Rings

Definition 4.1.
Define the presented dimension ofR as follows:
(4.1)FPD(R)=sup{FPd(M)∣MisafinitelygeneratedrightR-module}.It is easy to see thatFPD(R)=0 if and only if every finitely generated module has an infinite finite presentation, if and only if every finitely generated module is finitely presented, if and only if R is right Noetherian. Thus we may regard the presented dimension of a ring as a measure of how far it is from being right Noetherian.Proposition 4.2.
FPD(R)≤rgD(R)+1.Proof.
By Proposition2.2, FPD(M)≤pd(M)+1, thus the result follows immediately.Note thatFPD(R) can be much smaller than rgD(R). Take R=ℤ4. Then rgD(R)=∞ while FPD(R)=0 for R is Noetherian.Following Proposition2.3, we have the following corollary.Corollary 4.3.
No ring can have presented dimension 1.In the following, we investigate the relations of the right global, weak global, and presented dimensions of rings.Theorem 4.4.
LetR be a ring.  (1)
IfFPD(R)≤wD(R), then rgD(R)=wD(R).(2)
IfFPD(R)>wD(R), then rgD(R)=FPD(R) or FPD(R)-1.(3)
IfrgD(R)>wD(R), then FPD(R)=rgD(R)+1.Proof.
(1) It suffices to prove that wD(R)≥rgD(R) and suppose that wD(R)=s<∞. Let M be finitely generated. Since FPD(R)≤wD(R)=s, we have FPd(M)=m≤s, thus there is a projective resolution
(4.2)⋯→Pm+n→dm+n⋯→Pm→dm⋯→P0→M→0,
where Pm,…,Pm+n,… are finitely generated. Since wD(R)=s, it follows that kerds-1 is flat. Note that s≥m, hence kerds-1 is finitely presented, whence projective, that is, pd(M)≤s. So wD(R)≥rgD(R).
(2) If FPD(R)=∞, the result follows immediately from Proposition 4.2. Now suppose that FPD(R)=m<∞. Since FPD(R)>wD(R)≥0, by Corollary 4.3 we have m≥2. Let M be finitely generated and fd(M)=k, thus kerdt is finitely presented for each t≥m-1.
Iffd(M)≤FPd(M), then kerdm-1 is flat, hence projective, so pd(M)≤FPd(M).
Iffd(M)>FPd(M), then kerdk-1 is flat, hence projective, so
(4.3)pd(M)≤fd(M)≤wD(R).
ThereforergD(R)≤FPD(R).
On the other hand, by Proposition4.2, FPD(R)≤rgD(R)+1. So rgD(R)=FPD(R) or FPD(R)-1.
(3) From (1) and (2), we have FPD(R)=rgD(R)+1 or FPD(R)=rgD(R). Thus we need only to consider rgD(R)=m<∞ and prove FPD(R)≠rgD(R). Suppose that FPD(R)=rgD(R). Let M be a finitely generated right R-module with FPd(M)=m, then there is an exact sequence
(4.4)0→Pm→Pm-1→dm-1Pm-2→dm-2⋯→P0→M→0,
where Pi is projective and Pm is strongly presented. Let Km-2=kerdm-2. Note that m≠1 and m>wD(R)≥0. We consider the exact sequence
(4.5)0→Pm→Pm-1→Km-2→0.
Since wD(R)<rgD(R)=m, Km-2 is flat. Suppose that Q such that Pm-1⊕Q=F is free. Then
(4.6)0→Pm→F→Km-2⊕Q→0
is exact, and Km-2⊕Q is flat. Let p1,…,pm generate Pm. Using the flatness of Km-2⊕Q, there exists a homomorphism F→Pm such that pi↦pi(i=1,2,…,t). Thus the above short sequence splits, and so F≅Pm⊕Km-2⊕Q. Thus Km-2 is projective, therefore pd(M)≤m-1, and so rgD(R)≤m-1, a contradiction. Hence FPD(R)≠rgD(R), so FPD(R)=rgD(R)+1.Corollary 4.5.
rgD(R)=max{wD(R),FPD(R)-1}.From the foregoing discussion, we can classify rings by the right global dimensions, weak global dimensions, and presented dimensions of rings.In the diagrams, represents two consecutive numbers while   means that the numbers may not be consecutive.


## 5. On Ring Extensions

In this section, assume thatS≥R is a unitary ring extension. We aim to investigate properties of the presented dimensions of modules and rings. We first recall some concepts.(1) The ring S is called rightR-projective [9] in case, for any right S-module MS with an S-module NS, NR∣MR implies NS∣MS, where N∣M means that N is a direct summand of M. For example, every n×n matrix ring Rn is right R-projective [9].(2) The ring extension S≥R is called a finite normalizing extension [10] in case there is a finite subset {s1,…,sn}⊆S such that S=∑i=1nsiR and siR=Rsi for i=1,…,n.(3) A finite normalizing extension S≥R is called an almost excellent extension [11] in case RS is flat, SR is projective, and the ring S is right R-projective.(4) An almost excellent extension S≥R is an excellent extension [9] in case both RS and SR are free modules with a common basis {s1,…,sn}.Excellent extensions were introduced by Passman [9] and named by Bonami [12]. Examples include the n×n matrix rings and the crossed products R*G where G is a finite group with |G|-1∈R. Almost excellent extensions were introduced and studied by Xue [11] as a nontrivial generalization of excellent extensions and recently studied in [2, 13–15].Proposition 5.1.
Assume thatS≥R is a finite normalizing extension and RS is flat. Then for each right R-module MR, we have
(5.1)FPd(M⊗RS)S≤FPd(MR).Proof.
IfFPd(MR)=∞, it is clear. Suppose FPd(MR)=m<∞. There is a projective resolution of M(5.2)⋯→Pm+n→⋯→Pm→Pm-1→⋯→P0→M→0,
where Pm,…,Pm+n are finitely generated. Since RS is flat, there is an exact sequence of right S-modules
(5.3)⋯→Pm+n⊗RS→⋯→Pm⊗RS→⋯→P0⊗RS→M⊗RS→0,
where Pi⊗RS is a projective right S-module, and Pm⊗RS,…,Pm+n⊗RS,… are finitely generated. So FPd(M⊗RS)≤m, therefore FPd(M⊗RS)S≤FPd(MR).Proposition 5.2.
Assume thatS≥R is a finite normalizing extension, RS is flat, and S is right R-projective. Then for each right S-module MS, one has
(5.4)FPd(MS)≤FPd(M⊗RS).Proof.
By [11, Lemma 1.1], MS is isomorphic to a direct summand of M⊗RS. By Corollary 2.7, FPd(M⊗RS)≥FPd(MS).Proposition 5.3.
Assume thatS≥R is an almost excellent extension. Then for each right S-module MS, one has FPd(MR)≤FPd(MS).Proof.
IfFPd(MS)=∞, then it clear. Suppose that FPd(MS)=m<∞. Then there is a projective resolution
(5.5)⋯→Pm+n→⋯→Pm⋯→P0→MS→0,
where Pi(i=0,1,…) are right S-modules and Pm,…,Pm+1,… are finitely generated. Since S≥R is an almost excellent extension, it follows that Pi(i=0,1,…) are projective right R-modules, and Pm,Pm+1,… are finitely generated right R-modules. Thus
(5.6)⋯→Pm+n→⋯→Pm→Pm-1→⋯→P0→M→0,
is a projective resolution of MR. So FPd(MS)≥FPd(MR).Corollary 5.4.
Assume thatS≥R is an almost excellent extension. Then for each right S-module MS, one has FPd(MR)=FPd(MS)=FPd(M⊗RS).Theorem 5.5.
Assume thatS≥R is a finite normalizing extension and RS is flat.  (1)
IfS is right R-projective and FPD(S)<∞, then FPD(S)≤FPD(R);(2)
IfFPD(R)<∞, then(5.7)FPD(R)≤FPD(S)+max{l,s},
where l=pd(SR) and s=sup{FPd(AR)∣A∈Mod-SandFPd(AS)=0}.Proof.
(1) Suppose that FPD(S)=m, M is a finitely generated right S-module, and FPd(MS)=m. Since S is right R-projective, there is an exact sequence of S-modules
(5.8)0→MS→M⊗RS→C→0,
where FPd(C)≤m for FPD(S)=m. By Theorem 2.6 we have
(5.9)FPd(M⊗RS)S≤max{FPd(MS),FPd(CS)}=m,m=FPd(MS)≤max{FPd(M⊗RS)S,FPd(CS)-1}≤FPd(S)=m,
thus FPd(M⊗RSS)=m. It follows from Proposition 5.1 that FPd(M⊗RSS)≤FPd(MR). So FPD(S)≤FPD(R).
(2) Suppose that FPD(R)=m, M is a finitely generated right R-module, and FPd(MR)=m<∞. Since RS is flat, by [16, Lemma 2.3], there is an exact sequence of R-modules(5.10)0→M→M⊗RS→C→0.
Note that RS is finitely generated, which implies thats M⊗RS and C are finitely generated, thus FPd(C)≤m for FPD(R)=m. By Theorem 2.6, we have
(5.11)m=FPd(MR)≤max{FPd(M⊗RSR),FPd(CR)-1}≤FPD(R)=m,
hence FPd(M⊗RSR)=m. Let FPd(M⊗RS)S=t≤FPD(S). Then there is a projective resolution of the right S-module M⊗RS(5.12)⋯→Qt+1→Qt→Qt-1→⋯→Q0→M⊗RS→0,
where Qt,Qt+1,… are finitely generated. Thus we have the following exact sequences:
(5.13)0→Kt-1→Qt-1→Kt-2→0,0→Kt-2→Qt-2→Kt-3→0,⋯0→K0→Q0→M⊗RS→0,
where Ki=Im(Qi+1→Qi) and FPd(Kt-1)=0. By Proposition 2.2,
(5.14)FPd(Qi)R≤pd(Qi)R+1≤pd(SR)+1=l+1,
and FPd(Kt-1)R≤s. Following Theorem 2.6 and Proposition 2.2, we have
(5.15)FPdR(Kt-2)≤max{FPdR(Qt-1),FPdR(Kt-1)+1}≤max{l+1,s+1}=1+max{l,s}.
Again by Theorem 2.6, we have
(5.16)FPdR(Kt-3)≤2+max{l,s},⋯FPdR(K0)≤t-1+max{l,s},m=FPd(S⊗RM)R≤t+max{l,s}≤FPD(S)+max{l,s}.
Therefore FPD(R)≤FPD(S)+max{l,s}.Note that ifS≥R is an almost excellent extension, then pd(SR)=0, and (5.17)s=sup{FPd(AR)∣A∈Mod-SandFPd(AS)=0}=0.thusCorollary 5.6.
Assume thatS≥R is an almost excellent extension. Then FPD(R)=FPD(S).Proof.
Suppose thatS≥R is an almost excellent extension. Then SR is a finitely generated projective R-module and S is right R-projective. By Theorem 5.5 and Proposition 5.2, we have FPD(R)=FPD(S).To close this section, we give an example of an excellent extensionS≥R, which is provided by Xue in [16]. Let R be a ring graded by a finite group G. The smash product R#G is a free right and left R-module with a basis {pg∣g∈G} and the multiplication determined by (rpg)(r′ph)=rrgh-1′ph, where g,h∈G,r,r′∈R, and rgh-1′ is the gh-1-component of r′.Example 5.7.
LetR be a ring graded by a finite group G with |G|-1∈R. Then
(5.18)FPD(R)=FPD(R#G).Proof.
By [17, Theorem 1.3], we know that G acts as automorphisms on R#G and the skew group ring (R#G)*G≅Rn where n=|G|. Since skew group rings and finite matrix rings are excellent extensions, the result follows.



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*Source: 256267-2010-06-08.xml*