Extension closure of relative syzygy modules

Abstract

In this paper we introduce the notion of relative syzygy modules. We then study the extension closure of the category of modules
consisting of relative syzygy modules (resp. relative к-torsionfree modules).

Full text

Vol . 46 N o. 5 SCIENCE IN CHINA (Series A) September 2003
Extension closure of relative syzygy modules
HUANG Zhaoyong (

)
Department of Mathematics, Nanjing University, Nanjing 210093, China (email: huangzy@nju.edu.cn)
Received May 21, 2002; revised October 29, 2002
Abstract In this paper we introduce the notion of relative syzygy modules. We then study the extension
closure of the category of modules consisting of relative syzygy modules (resp. relative k-torsionfree modules).
Keywords: extension closed, ωω-k-syzygy modules, ωω-k-torsionfree modules.
DOI: 10.1360/02ys0169
1 Introduction
Throughout this paper Λis a left noetherian ring and Γis a right noetherian ring, mod Λ
(resp. mod Γop ) is the category of finitely generated left Λ-modules (resp. right Γ-modules). All
modules considered are finitely generated.
Let ΛωΓbe a (Λ,Γ)-bimodule with Λωin mod Λand ωΓin mod Γop.
Definition 1.1. Let A∈mod Λ(resp. mod Γop )andia non-negative integer. We say that
the grade of Awith respect to ω, written gradeωA, is greater than or equal to iif Extj
Λ(A, ω)=0
(resp. Extj
Γ(A, ω)=0)forany0j<i. We say that the strong grade of Awith respect to ω,
written s.gradeωA, is greater than or equal to iif gradeωBifor all submodules Bof A.
Definition 1.2. Let A∈mod Λ(resp. mod Γop )andka positive integer. We call Aa
ω-k-syzygy module if there is an exact sequence 0 →A→X0→X1→ ··· fk−1
−→ Xk−1with all
Xiin addΛω(resp. addωΓ), where addΛω(resp. addωΓ) denotes the full subcategory of mod Λ
(resp. mod Γop ) consisting of all modules isomorphic to the direct summands of finite direct sums
of copies of Λω(resp. ωΓ). We further call Cokerfk−1aω-k-cosyzygy module. We use Ωk
ω(Λ)
(resp. Ωk
ω(Γop)) and Ω−k
ω(Λ)(resp. Ω−k
ω(Γop)) to denote the full subcategory of mod Λ(resp.
mod Γop) consisting of ω-k-syzygy modules and ω-k-cosyzygy modules, respectively.
For any A∈mod Λ, there is an exact sequence P1
f
−→ P0→A→0inmodΛwith P0
and P1projective. Then we have an exact sequence 0 →Aω→Pω
0
fω
−→ Pω
1→X→0, where
()ω=Hom(,ω)andX=Cokerfω.
Definition 1.3[1].Suppose that the natural maps Λ→End(ωΓ)andΓop →End(Λω)are
isomorphisms and Exti
Γ(ω, ω) = 0 for any 1 ik.LetAand Xbe as above. Ais called a
ω-k-torsionfree module if Exti
Γ(X, ω) = 0 for any 1 ik.WeuseTk
ω(Λ)todenotethefull
subcategory of mod Λconsisting of ω-k-torsionfree modules.
Remarks. (1) If Λis a two-sided noetherian ring and ΛωΓ=ΛΛΛ, then the notions in Defini-
tions 1.1—1.3 are just the grade, the strong grade, k-syzygy modules and k-torsionfree modules

612 SCIENCE IN CHINA (Series A) Vol. 46
in the usual sense[2,3] respectively. Particularly, in this case Ω−k
ω(Λ)=modΛfor any k1.
(2) The definition of ω-k-torsionfree modules above is well-defined[1].
(3) Let σA:A→Aωω be the canonical evaluation homomorphism. Ais called a ω-torsionless
module if σAis a monomorphism; and Ais called a ω-reflexive module if σAis an isomorphism.
By Lemma 4 of ref. [1], Ais ω-torsionless (resp. ω-reflexive) if and only if Ais ω-1-torsionfree
(resp. ω-2-torsionfree).
A full subcategory Xof mod Λis said to be extension closed if the middle term Bof any
short exact sequence 0 →A→B→C→0isinXprovided that the end terms Aand Care in
X. For any positive integer k, we showed in ref. [1] that a ω-k-torsionfree module is a ω-k-syzygy
module and so Tk
ω(Λ)⊂Ωk
ω(Λ). In this paper, we mainly discuss the extension closure of Ωk
ω(Λ)
and Tk
ω(Λ). This paper is mainly motivated by the work of Auslander and Reiten[3].
In sec. 2 we give some lemmas which will be used later. In sec. 3 we show that
gradeωExti+1
Λ(M,ω)ifor any M∈Ω−(i+1)
ω(Λ)and1ik−1 if and only if Ωi
ω(Λ)=Ti
ω(Λ)
for any 1 ik(Theorem 3.1), which is applied to showing that s.gradeωExti+1
Λ(M,ω)i
for any M∈Ω−i
ω(Λ)and1ikif and only if Ωi
ω(Λ) is extension closed for any 1 ik
(Theorem 3.3). These are generalizations of Proposition 2.26 in ref. [2] and Theorem 1.7 in
ref. [3], respectively. In sec. 4 we deal with the extension closure of ω-torsionless modules and
ω-reflexive modules. If k2, then Ti
ω(Λ) is extension closed for any 1 ikif and only if
gradeωExti
Γ(C, ω)ifor any C∈mod Γop (or Ω−i
ω(Γop)) and 1 ik(Theorem 4.1).
In the following, kis a positive integer, ΛωΓis a faithfully balanced bimodule, that is,
the natural maps Λ→End(ωΓ)andΓop →End(Λω) are isomorphisms, satisfying Exti
Λ(ω, ω)=
0=Ext
i
Γ(ω, ω) for any 1 ik. Under the assumption of ΛωΓbeing faithfully balanced, it is
easy to see that any projective module in mod Λ(resp. mod Γop) and any module in addΛω(resp.
addωΓ)areω-reflexive.
2 Some lemmas
In this section we give some lemmas which will be used later.
Lemma 2.1. Let 0 →Af
−→ Bg
−→ C→0and0→Xα
−→ Ybe two exact sequences in
mod Λ.IfHom
Λ(g, Y ) is an isomorphism, then HomΛ(g, X) is also an isomorphism.
Proof. Since 0 →HomΛ(C, Y )HomΛ(g,Y )
−→ HomΛ(B, Y )HomΛ(f,Y )
−→ HomΛ(A, Y )isexactand
HomΛ(g, Y ) is an isomorphism, HomΛ(f,Y ) is a zero homomorphism.
Consider the following exact commutative diagram:
0HomΛ(C, Y )Hom
Λ(B, Y )Hom
Λ(A, Y )
0Hom
Λ(C, X)Hom
Λ(B, X)Hom
Λ(A, X)
- - -
- - -
000
? ? ?
? ? ?
HomΛ(g, Y )Hom
Λ(f, Y )
HomΛ(g, X)Hom
Λ(f, X)
HomΛ(C, α)Hom
Λ(B, α)Hom
Λ(A, α)

No. 5 EXTENSION CLOSURE OF RELATIVE SYZYGY MODULES 613
Since HomΛ(f, Y ) is a zero homomorphism and HomΛ(A, α) is a monomorphism, HomΛ(f,X)
is also a zero homomorphism. By the exactness of the upper row in the above diagram, HomΛ(g, X)
is an epimorphism and hence an isomorphism.
Lemma 2.2. Let Aand Bbe in mod Λ.IfBis ω-torsionless and Aω=0,thenHom
Λ(A, B)
=0.
Proof. Since Bis ω-torsionless, there is an embedding 0 →B→ωnwith na positive
integer, and an exact sequence 0 →HomΛ(A, B)→[HomΛ(A, ω)]n=(Aω)n. Hence we are done.
Let A∈mod Λ(resp. mod Γop)andP1
f
−→ P0→A→0 an exact sequence in mod Λ(resp.
mod Γop )withP0and P1projective (or P0and P1in addΛω(resp. addωΓ)). Then we get an
exact sequence
0→Aω→Pω
0
fω
−→ Pω
1→X→0
in mod Γop (resp. mod Λ), where X=Cokerfω.
Lemma 2.3 (Lemma 2.1 in ref. [4]). Let Aand Xbe as above. Then we have the
following exact sequences:
0→Ext1
Γ(X, ω)→AσA
−→ Aωω →Ext2
Γ(X, ω)→0,
0→Ext1
Λ(A, ω)→XσX
−→ Xωω →Ext2
Λ(A, ω)→0.
The following Lemmas 2.4 and 2.5 have analogous proofs to Lemmas 2.6 and 2.12 of ref. [15]
respectively.
Lemma 2.4. Let 0 →A→Hf
−→ Bbe an exact sequence in mod Λ(resp. mod Γop)with
Hω-reflexive and Bω-torsionless. Then A∼
=(Cokerfω)ω.
Lemma 2.5. For any A∈mod Λ(resp. mod Γop ), the following statements are equivalent.
(1) Aωis ω-reflexive;
(2) Aωω is ω-reflexive;
(3) (CokerσA)ω=0.
Lemma 2.6. A module M∈Ω2
ω(Λ) if and only if there is a module N∈mod Γop such
that M∼
=Nω.
Proof. Suppose M∼
=Nωwith N∈mod Γop. Because there is an exact sequence Q1→
Q0→N→0inmodΓop with Q0and Q1projective, we have an exact sequence 0 →Nω→
Qω
0→Qω
1with Qω
0,Qω
1∈addΛωand M(∼
=Nω)∈Ω2
ω(Λ). The converse follows from Lemma 2.4.
Lemma 2.7. The following statements are equivalent.
(1) Aωis ω-reflexive for any A∈mod Λ.
(1)op Bωis ω-reflexive for any B∈mod Γop .
(2) [Ext2
Λ(A, ω)]ω=0foranyA∈mod Λ.
(2)op [Ext2
Γ(B, ω)]ω=0for anyB∈mod Γop .
(3) [Ext2
Λ(A, ω)]ω=0foranyA∈Ω−2
ω(Λ).
(3)op [Ext2
Γ(B, ω)]ω=0for anyB∈Ω−2
ω(Γop).
(4) Every module in Ω2
ω(Λ)isω-reflexive.
(4)op Every module in Ω2
ω(Γop)isω-reflexive.

614 SCIENCE IN CHINA (Series A) Vol. 46
Proof. We will prove (1) ⇔(1)op ⇒(2) ⇒(3) ⇒(1)op and (1)op ⇔(4). Then by
symmetry, we are done.
(1) ⇔(1)op The argument for Lemma 2.13 in ref. [5] remains valid here, so we omit it.
(1)op ⇒(2) By Lemma 2.3, for any A∈mod Λthere is an exact sequence XσX
−→ Xωω →
Ext2
Λ(A, ω)→0withX∈mod Γop,andthen0→[Ext2
Λ(A, ω)]ω→Xωωω σω
X
−→ Xωis exact. By
Proposition 20.14 in ref. [6], σω
XσXω=1
Xω,soσω
Xis a split epimorphism and hence Xωωω ∼
=
Xω[Ext2
Λ(A, ω)]ω.By(1)
op,Xωis ω-reflexive, so we have [Ext2
Λ(A, ω)]ω=0.
(2) ⇒(3) It is trivial.
(3) ⇒(1)op For an y B∈mod Γop, there is an exact sequence P1
f
−→ P0→B→0inmod
Γop with P0and P1projective. Then we have an exact sequence 0 →Bω→Pω
0
fω
−→ Pω
1→X→0
in mod Λwith Pω
0,Pω
1in addΛωand X∈Ω−2
ω(Λ), where X=Cokerfω. By Lemma 2.3, we
have exact sequences BσB
−→ Bωω →Ext2
Γ(X, ω)→0and0→[Ext2
Γ(X, ω)]ω→Bωωω σω
B
−→ Bω.
Similar to the above argument we have Bωωω ∼
=Bω[Ext2
Γ(X, ω)]ω.[Ext
2
Γ(X, ω)]ω= 0 by (3),
so Bωωω ∼
=Bωand hence Bωis ω-reflexive.
(1)op ⇔(4) It follows from Lemma 2.6.
Lemma 2.8 (Lemma 4 in ref. [1]). A module in mod Λis ω-torsionless (resp. ω-reflexive)
if and only if it is ω-1-torsionfree (resp. ω-2-torsionfree).
Lemma 2.9. Let k3. Then a ω-reflexive module Ain mod Λis ω-k-torsionfree if and
only if Exti
Γ(Aω,ω) = 0 for any 1 ik−2.
Proof. Let P1
f
−→ P0→A→0 be a projective resolution of Ain mod Λ.Then
0→Aω→Pω
0
fω
−→ Pω
1→X→0(2.9.1)
is exact in mod Γop with Pω
0and Pω
1in addωΓ,whereX=Cokerfω. By Lemma 2.3, Ais ω-
reflexive if and only if Ext1
Γ(X, ω)=0=Ext
2
Γ(X, ω). On the other hand, from the exactness
of the sequence (2.9.1) we get that Exti−2
Γ(Aω,ω)∼
=Exti
Γ(X, ω) for any 1 ik.Nowour
conclusion follows easily.
3 Extension closure of ΩΩk
ω(Λ)
In this section we discuss the extension closure of Ωk
ω(Λ).
Theorem 3.1. The following statements are equivalent.
(1) gradeωExti+1
Λ(M,ω)ifor any M∈Ω−(i+1)
ω(Λ)and1ik−1.
(2) Ωi
ω(Λ)=Ti
ω(Λ) for any 1 ik.
Proof. Proceed by induction on k. It is not difficult to verify that a module in mod Λis
ω-torsionless if and only if it is in Ω1
ω(Λ). Then by Lemma 2.8 we have Ω1
ω(Λ)=T1
ω(Λ). On the
other hand, when k= 1 the assumption of (1) is empty. So the case for k= 1 is trivial. The case
for k= 2 follows from Lemma 2.7. Now suppose k3.
(1) ⇒(2) By Theorem 1 in ref. [1], Tk
ω(Λ)⊂Ωk
ω(Λ). So we only need to prove Tk
ω(Λ)⊃Ωk
ω(Λ).
Let L∈Ωk
ω(Λ). Then there is an exact sequence 0 →L→Xk−1
f
−→ Xk−2→ ··· → X0→
M→0inmodΛwith all Xi∈addΛω. Since we have assumed (1) at level k,wealsoknow(1)
at level k−1, so by induction assumption we have Ωi
ω(Λ)=Ti
ω(Λ) for any 1 ik−1. Hence
L∈Tk−1
ω(Λ).

No. 5 EXTENSION CLOSURE OF RELATIVE SYZYGY MODULES 615
Let P1
g
−→ P0→L→0 be an exact sequence in mod Λwith P0and P1projective. Then we
have an exact sequence 0 →Lω→Pω
0
gω
−→ Pω
1→X→0inmodΓop with Pω
0and Pω
1in addωΓ,
where X=Cokergω. We will show that Lis ω-k-torsionfree.
Notice that L∈Tk−1
ω(Λ)andk3, so Lis ω-reflexive and hence it suffices to show that
Exti
Γ(Lω,ω) = 0 for any 1 ik−2 by Lemma 2.9.
Put N=Cokerfω. Then, by Lemma 2.4, L∼
=Nωand Lω∼
=Nωω. We claim that Exti
Γ(N,ω)=
0 for any 1 ik−2. If k=3,thenCokerfis a submodule of X0.ButX0is in addΛω,soX0is ω-
reflexive and Cokerfis ω-torsionless. By Lemma 2.3, Ext1
Γ(N,ω)∼
=KerσCokerf=0. Ifk=4,then
Cokerf∈Ω2
ω(Λ)(=T2
ω(Λ)) and Cokerfis ω-reflexive. Thus Ext1
Γ(N,ω)∼
=KerσCokerf=0and
Ext2
Γ(N,ω)∼
=CokerσCokerf=0andthecasefork= 4 follows. If k5, then Cokerf∈Ωk−2
ω(Λ)
and Cokerf∈T
k−2
ω(Λ). Thus Exti
Γ((Cokerf)ω,ω) = 0 for any 1 ik−4 by Lemma 2.9.
It follows from the exact sequence 0 →(Cokerf)ω→Xω
k−2
fω
−→ Xω
k−1→N→0withXω
k−2
and Xω
k−1projective that Exti
Γ(N,ω) = 0 for any 3 ik−2. So Exti
Γ(N,ω) = 0 for any
1ik−2.
By Lemma 2.3, we have an exact sequence
0→Ext1
Λ(Cokerf,ω)→NσN
−→ Nωω →Ext2
Λ(Cokerf,ω)→0.
Then KerσN∼
=Ext1
Λ(Cokerf,ω)∼
=Extk−1
Λ(M,ω)andCokerσN∼
=Ext2
Λ(Cokerf,ω)∼
=Extk
Λ(M,ω).
So we get the following exact sequences:
0→Extk−1
Λ(M,ω)→Nπ
−→ ImσN→0,(3.1.1)
0→ImσN
μ
−→ Nωω →Extk
Λ(M,ω)→0,(3.1.2)
where σN=μπ.SinceExt
i
Γ(N,ω) = 0 for any 1 ik−2 and gradeωExtk−1
Λ(M,ω)k−2,
from the exact sequence (3.1.1) we have Exti
Γ(ImσN,ω) = 0 for any 1 ik−2. Moreover,
since gradeωExtk
Λ(M,ω)k−1, from the exact sequence (3.1.2) we get that Exti
Γ(Nωω,ω)=0
for any 1 ik−2, which yields Exti
Γ(Lω,ω) = 0 for any 1 ik−2.
(2) ⇒(1) Let M∈Ω−k
ω(Λ). Then there is an exact sequence 0 →L→Xk−1
f
−→ Xk−2→
···→X0→M→0inmodΛwith all Xi∈addΛω.By(2),L∈Tk
ω(Λ). By induction assumption,
gradeωExti+1
Λ(M,ω)ifor any 1 ik−2. So it remains to show that gradeωExtk
Λ(M,ω)
k−1. Put N=Cokerfω. From the proof of (1) ⇒(2), we have the following facts:
(i) there is exact sequences 0 →Extk−1
Λ(M,ω)→Nπ
−→ ImσN→0and0→ImσN
μ
−→
Nωω →Extk
Λ(M,ω)→0, where σN=μπ;
(ii) L∼
=Nω;
(iii) Exti
Γ(N,ω) = 0 for any 1 ik−2;
(iv) Exti
Γ(ImσN,ω) = 0 for any 1 ik−2.
Since L∈Tk
ω(Λ)andL∼
=Nω,Nωis ω-reflexive and Exti
Γ(Nωω,ω)∼
=Exti
Γ(Lω,ω)=0for
any 1 ik−2 by Lemma 2.9. Since Exti
Γ(ImσN,ω) = 0 for any 1 ik−2andwe
have the exact sequence 0 →ImσN
μ
−→ Nωω →Extk
Λ(M,ω)→0, Exti
Γ(Extk
Λ(M,ω),ω)=0for
any 2 ik−2. On the other hand, Nωis ω-reflexive, so πωμω=σω
Nis an isomorphism by
Proposition 20.14 in ref. [6], and it follows easily that πωand μωare isomorphisms. Moreover,

616 SCIENCE IN CHINA (Series A) Vol. 46
we have a long exact sequence:
0→[Extk
Λ(M,ω)]ω→Nωωω μω
−→ (ImσN)ω→Ext1
Γ(Extk
Λ(M,ω),ω)→Ext1
Γ(Nωω,ω)=0.
So [Extk
Λ(M,ω)]ω∼
=Kerμω=0andExt
1
Γ(Extk
Λ(M,ω),ω)∼
=Cokerμω= 0. Therefore we conclude
that gradeωExtk
Λ(M,ω)k−1.
If Λωis a generator in mod Λ(for example, when Λω=ΛΛ), then Ω−k
ω(Λ)=modΛfor any
k1 and we have the following:
Corollary 3.1. If Λωis a generator in mod Λ, then the following statements are equivalent.
(1) gradeωExti+1
Λ(M,ω)ifor any M∈mod Λand 1 ik−1.
(2) Ωi
ω(Λ)=Ti
ω(Λ) for any 1 ik.
The following theorem is analogous to the result of Theorem 1.1 in ref. [3]. Since the proof
here is similar to that given in ref. [3], we omit it.
Theorem 3.2. Let N∈Tk
ω(Λ). The following statements are equivalent.
(1) s.gradeωExt1
Λ(N,ω)k.
(2) If 0 →L→M→N→0isexactinmodΛwith Lin Tk
ω(Λ), then M∈Tk
ω(Λ).
(3) If 0 →ωn→E→N→0isexactinmodΛwith na positive integer, then E∈Tk
ω(Λ).
The following corollary is an immediate consequence of Theorem 3.2.
Corollary 3.2. The following statements are equivalent.
(1) Ti
ω(Λ) is extension closed for any 1 ik.
(2) s.gradeωExt1
Λ(N,ω)ifor any N∈Ti
ω(Λ)and1ik.
Proposition 3.1. If Ωi
ω(Λ) is extension closed for any 1 ik,thenΩi
ω(Λ)=Ti
ω(Λ)for
any 1 ik.
Proof. Proceed by induction on k. There is nothing to do for the case k=1.
Now suppose k2. Then, by induction assumption, Ωi
ω(Λ)=Ti
ω(Λ), which is extension
closed for any 1 ik−1. For any M∈Ω−(i+1)
ω(Λ)(1ik−1), there is an exact sequence
Xi
fi
−→ ··· →X0→M→0 with all Xjin addΛω.ThenImfi∈Ωi
ω(Λ)=Ti
ω(Λ)(1ik−1) and
Exti+1
Λ(M,ω)∼
=Ext1
Λ(Imfi,ω). So we have s.gradeωExti+1
Λ(M,ω) = s.gradeωExt1
Λ(Imfi,ω)ifor
any 1 ik−1 by Corollary 3.2. Then by Theorem 3.1 we have that Ωk
ω(Λ)=Tk
ω(Λ), which
finishes the proof.
The main result in this section is the following, which is a generalization of Theorem 1.7 in
ref. [3].
Theorem 3.3. The following statements are equivalent.
(1) s.gradeωExti+1
Λ(M,ω)ifor any M∈Ω−i
ω(Λ)and1ik.
(2) Ωi
ω(Λ) is extension closed for any 1 ik.
(3) Ωi
ω(Λ) is extension closed and Ωi
ω(Λ)=Ti
ω(Λ) for any 1 ik.
Proof. (1) ⇒(2) By (1) and Theorem 3.1 we have Ωi
ω(Λ)=Ti
ω(Λ) for any 1 ik.Let
N∈Ti
ω(Λ)(1ik), then N∈Ωi
ω(Λ) and there is an exact sequence 0 →N→Xi−1→···→
X0→M→0 with all Xjin addΛω.ThenExt
1
Λ(N,ω)∼
=Exti+1
Λ(M,ω)andM∈Ω−i
ω(Λ). So
s.gradeωExt1
Λ(N,ω) =s.gradeωExti+1
Λ(M,ω)i(1 ik) by (1) and hence Ti
ω(Λ)isextension
closed for any 1 ikby Corollary 3.2. Therefore we conclude that Ωi
ω(Λ)isalsoextension
closed for any 1 ik.

No. 5 EXTENSION CLOSURE OF RELATIVE SYZYGY MODULES 617
(2) ⇒(3) By Proposition 3.1.
(3) ⇒(1) By (3) and Corollary 3.2, s.gradeωExt1
Λ(N,ω)ifor any N∈Ti
ω(Λ)=Ωi
ω(Λ)and
1ik.Sos.grade
ωExti+1
Λ(M,ω)ifor any M∈Ω−i
ω(Λ)and1ik.
Corollary 3.3. If Λωis a generator in mod Λ, then the following statements are equivalent.
(1) s.gradeωExti+1
Λ(M,ω)ifor any M∈mod Λand 1 ik.
(2) Ωi
ω(Λ) is extension closed for any 1 ik.
(3) Ωi
ω(Λ) is extension closed and Ωi
ω(Λ)=Ti
ω(Λ) for any 1 ik.
4 Extension closure of Tk
ω(Λ)
In this section we deal with the extension closure of Tk
ω(Λ), especially, of T1
ω(Λ)andT2
ω(Λ).
We use l.idΛ(ω) to denote the left injective dimension of ωas a left Λ-module.
Proposition 4.1. If l.idΛ(ω)k,thenTk
ω(Λ) is extension closed.
Proof. Let 0 →Af
−→ B→C→0 be an exact sequence in mod Λwith Aand C
ω-k-torsionfree. Consider the following exact commutative diagram with last two rows splitting:
000
↑↑↑
0→Af
−→ B→C−→ 0
↑↑↑
0→F0→F0G0−→ G0−→ 0
↑↑↑
0→F1→F1G1−→ G1−→ 0
where all Fiand Giare projective. Then we get the following exact commutative diagram:
00 0
↓↓ ↓
0→Cω→Bωfω
−→ Aω
↓↓ ↓
0→Gω
0→Gω
0Fω
0−→ Fω
0−→ 0
↓↓ ↓
0→Gω
1→Gω
1Fω
1−→ Fω
1−→ 0
↓↓ ↓
ZY X
↓↓ ↓
00 0
It follows from the snake lemma that there is an exact sequence 0 →Cω→Bωfω
−→ Aω→Z→
Y→X→0. Because Cis ω-k-torsionfree, C∈Ωk
ω(Λ) by Theorem 1 in ref. [1]. On the other
hand, l.idΛ(ω)k,soExt
1
Λ(C, ω)∼
=Extk+1
Λ(Ω−k
ω(C),ω) = 0 and hence fωis epic, which induces
an exact sequence 0 →Z→Y→X→0. Since Aand Care ω-k-torsionfree, Exti
Γ(X, ω)=0=
Exti
Γ(Z, ω) for any 1 ik.SoExt
i
Γ(Y,ω) = 0 for any 1 ikand hence Bis ω-k-torsionfree.
ΛωΓis called a cotilting bimodule if l.idΛ(ω)<∞and r.idΓ(ω)<∞[7] .

618 SCIENCE IN CHINA (Series A) Vol. 46
Corollary 4.1. If ΛωΓis a cotilting bimodule with l.idΛ(ω)k,thenTk
ω(Λ)isextension
closed.
Proposition 4.2. The following statements are equivalent.
(1) T1
ω(Λ) is extension closed.
(2) gradeωExt1
Γ(C, ω)1 for any C∈mod Γop .
(3) gradeωExt1
Γ(C, ω)1 for any C∈Ω−1
ω(Γop).
Proof. (1) ⇒(2) Let C∈mod Γop and P1
f
−→ P0→C→0 a projective resolution of C
in mod Γop. By Lemma 2.3, we have an exact sequence:
0→Ext1
Γ(C, ω)→XσX
−→ Xωω →Ext2
Γ(C, ω)→0,
where X=Cokerfω.
Put Y=ImσXand assume that σX=μπ,whereπ:X→Yis an epimorphism and μ:Y→
Xωω is a monomorphism. Since πωμω=σω
Xis an epimorphism by Proposition 20.14 in ref. [6],
πωis also an epimorphism and hence an isomorphism. So, by applying ()ωto the exact sequence
0→Ext1
Γ(C, ω)→Xπ
−→ Y→0, we have KerExt1
Λ(π, ω)∼
=[Ext1
Γ(C, ω)]ω.
Suppose
η:0→ω→Kγ
−→ Y→0
is an element in KerExt1
Λ(π, ω), that is, Ext1
Λ(π, ω)(η) = 0. Then we have the following pull-back
diagram with the first row splitting:
0→ω→Nu
−→ X−→ 0
↓
v↓π
η:0→ω→Kγ
−→ Y−→ 0
So there is a homomorphism u:X→Nsuch that uu=1
Xand hence π=γ(vu). Notice that
Yis ω-torsionless since Yis a submodule of a ω-torsionless module Xωω.Sinceωis ω-torsionless,
Kis also ω-torsionless by (1). So we have an embedding 0 →K→ωnwith na positive integer.
Since πωis an isomorphism, HomΛ(π,ω n) is also an isomorphism. It follows from Lemma 2.1 that
HomΛ(π, K) is an isomorphism. Then there is a homomorphism h:Y→Ksuch that vu=hπ
and so π=γ(vu)=γhπ.Butπis an epimorphism which implies 1Y=γh. So we conclude that
the exact sequence ηsplits, which implies that KerExt1
Λ(π, ω)=0and[Ext
1
Γ(C, ω)]ω=0.
(2) ⇒(3) It is trivial.
(3) ⇒(1) Let 0 →Kβ
−→ Lα
−→ M→0 be an exact sequence in mod Λwith Kand M
ω-torsionless.
Suppose P1
f
−→ P0→L→0 is a pro jective resolution of Lin mod Λ.PutN=Cokerfω.
By Lemma 2.3, KerσL∼
=Ext1
Γ(N,ω). Since N∈Ω−1
ω(Γop), [Ext1
Γ(N,ω)]ω= 0 by (3) and thus
(KerσL)ω=0. NoticethatKis ω-torsionless, so HomΛ(KerσL,K) = 0 by Lemma 2.2. Moreover,
σMis a monomorphism and αωωσL=σMα,soKerσL⊂Kerα∼
=Kand hence KerσL=0,which
implies that Lis ω-torsionless.
Corollary 4.2. If T1
ω(Λ) is extension closed, then Mωis ω-reflexive for any M∈mod Λ.
Proof. Let M∈mod Λand P1
f
−→ P0→M→0 a projective resolution of Min mod Λ.
Put N=Cokerfωand L=Imfω. By Lemma 2.3, CokerσM∼
=Ext2
Γ(N,ω). Since 0 →L→Pω
1→

No. 5 EXTENSION CLOSURE OF RELATIVE SYZYGY MODULES 619
N→0isexactwithPω
1∈addωΓ,[Ext
2
Γ(N,ω)]ω∼
=[Ext1
Γ(L, ω)]ω= 0 by Proposition 4.2. Thus
(CokerσM)ω= 0 and therefore Mωis ω-reflexive by Lemma 2.5.
Proposition 4.3. The following statements are equivalent.
(1) Ti
ω(Λ) is extension closed for 1 i2.
(2) gradeωExti
Γ(C, ω)ifor any C∈mod Γop and 1 i2.
(3) gradeωExti
Γ(C, ω)ifor any C∈Ω−i
ω(Γop)and1i2.
(4) Exti−1
Λ(Exti
Γ(C, ω),ω) = 0 for any C∈mod Γop and 1 i2.
(5) Exti−1
Λ(Exti
Γ(C, ω),ω) = 0 for any C∈Ω−i
ω(Γop)and1i2.
Proof. (1) ⇒(2) Let C∈mod Γop and P1
f
−→ P0→C→0 a projective resolution of C
in mod Γop.PutB=Cokerfω. By Lemma 2.3, KerσB∼
=Ext1
Γ(C, ω)andCokerσB∼
=Ext2
Γ(C, ω).
Put L=ImσBand let σB=μπ,whereπ:B→Lis an epimorphism and μ:L→Bωω is a
monomorphism. By Proposition 4.2, gradeωExt1
Γ(C, ω)1and[Ext
1
Γ(C, ω)]ω= 0. By applying
()ωto the exact sequence 0 →Ext1
Γ(C, ω)→Bπ
−→ L→0, we know that Ext1
Λ(π, ω)isa
monomorphism.
Since σω
B=πωμωand πωis an isomorphism (see the proof of (1) ⇒(2) in Proposition 4.2),
and since σω
Bis an epimorphism by Proposition 20.14 in ref. [6], μωis also an epimorphism.
On the other hand, Ext1
Λ(σB,ω)=Ext
1
Λ(μπ, ω)=Ext
1
Λ(π, ω)Ext1
Λ(μ, ω). By applying ()ωto the
exact sequence 0 →Lμ
−→ Bωω ν
−→ Ext2
Γ(C, ω)→0, KerExt1
Λ(σB,ω)∼
=KerExt1
Λ(μ, ω)∼
=Ext1
Λ
(Ext2
Γ(C, ω),ω).
Suppose
ζ:0→ω→Mα
−→ Bωω →0
is an element in KerExt1
Λ(σB,ω), that is, Ext1
Λ(σB,ω)(ζ) = 0. Then we have the following pull-
back diagram with the first row splitting:
0→ω→Nβ
−→ B−→ 0
↓
γ↓σB
ζ:0→ω→Mα
−→ Bωω −→ 0
So there is a homomorphism β:B→Nsuch that ββ=1
Band hence σB=α(γβ). By Corollary
4.2, Bωis ω-reflexive. It follows from Lemma 2.5 that Bωω is ω-reflexive. Since ωis ω-reflexive,
Mis also ω-reflexive by (1). Since σM(γβ)=(γβ)ωωσB,σB=α(γβ)=ασ−1
M(γβ)ωωσB.
So (1Bωω −ασ−1
M(γβ)ωω)σB= 0 and hence Kerν=ImσB⊂Ker(1Bωω −ασ−1
M(γβ)ωω). Then
by Theorem 3.6 in ref. [6] there is a homomorphism δ:Ext
2
Γ(C, ω)→Bωω such that 1Bωω −
ασ−1
M(γβ)ωω =δν. In addition, [Ext2
Γ(C, ω)]ω∼
=(CokerσB)ω= 0 by Lemma 2.5. Then by
Lemma 2.2 HomΛ(Ext2
Γ(C, ω),B
ωω)=0sinceBωω is ω-reflexive. So δ= 0 and hence 1Bωω =
ασ−1
M(γβ)ωω, which implies that the exact sequence ζsplits. Thus KerExt1
Λ(σB,ω)=0and
Ext1
Λ(Ext2
Γ(C, ω),ω) = 0. So we conclude that gradeωExt2
Γ(C, ω)2.
(2) ⇒(3) ⇒(5) and (2) ⇒(4) ⇒(5) are trivial.
(5) ⇒(1) By Proposition 4.2, T1
ω(Λ) is extension closed. Let 0 →K→Lα
−→ M→0bean
exact sequence in mod Λwith Kand Mω-reflexive. Then Lis ω-torsionless by Proposition 4.2.

620 SCIENCE IN CHINA (Series A) Vol. 46
Let P1
f
−→ P0→L→0 be a projective resolution of Lin mod Λ. By Lemma 2.3, CokerσL∼
=
Ext2
Γ(N,ω), where N=Cokerfω(∈Ω−2
ω(Λ)). Consider the following exact commutative digram:
0→K→Lα
−→ M−→ 0
↓β↓σL↓σM
0→Kerαωω →Lωω αωω
−→ Mωω −→ 0
where σMis an isomorphism, σLis a monomorphism and βis an induced homomorphism. By the
snake lemma, we get an exact sequence 0 →Kβ
−→ Kerαωω →Ext2
Γ(N,ω)→0. By Corollary
4.2, Lωis ω-reflexive. It follows from Lemma 2.5 that [Ext2
Γ(N,ω)]ω∼
=(CokerσL)ω=0. By(5),
Ext1
Λ(Ext2
Γ(N,ω),ω) = 0. Thus, by applying ()ωto the last exact sequence, we know that βωis an
isomorphism and then βωω is also an isomorphism. On the other hand, βωωσK=σKerαωω βand σK
is an isomorphism, so σKerαωωβis an isomorphism which implies that σKerαωω is an epimorphism.
Then σKerαωω is an isomorphism since Kerαωω is clearly ω-torsionless. So we conclude that βis
also an isomorphism, which implies that Ext2
Γ(N,ω)=0andCokerσL=0. ThusLis ω-reflexive.
We are now in a position to state the main result in this section.
Theorem 4.1. Let k2. The following statements are equivalent.
(1) Ti
ω(Λ) is extension closed for 1 ik.
(2) gradeωExti
Γ(C, ω)ifor any C∈mod Γop and 1 ik.
(3) gradeωExti
Γ(C, ω)ifor any C∈Ω−i
ω(Γop)and1ik.
(4) Exti−1
Λ(Exti
Γ(C, ω),ω) = 0 for any C∈mod Γop and 1 ik.
(5) Exti−1
Λ(Exti
Γ(C, ω),ω) = 0 for any C∈Ω−i
ω(Γop)and1ik.
Proof. Use Propositions 4.2 and 4.3.
Acknowledgements The author was partially supported by the National Natural Science Foundation
of China (Grant No. 10001017), Scientific Research Foundation for Returned Overseas Chinese Scholars (State
Education Ministry) and Nanjing University Talent Development Foundation.
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