Let G(X) and G(Y) be Gorenstein subcategories induced by an admissible balanced pair (X,Y) in an abelian category A. In this paper, we establish Gorenstein homological dimensions in terms of these two subcategories and investigate the Gorenstein global dimensions of A induced by the balanced pair (X,Y). As a consequence, we give some new characterizations of pure global dimensions and Gorenstein global dimensions of a ring R.
Citation: Haiyu Liu, Rongmin Zhu, Yuxian Geng. Gorenstein global dimensions relative to balanced pairs[J]. Electronic Research Archive, 2020, 28(4): 1563-1571. doi: 10.3934/era.2020082
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Let G(X) and G(Y) be Gorenstein subcategories induced by an admissible balanced pair (X,Y) in an abelian category A. In this paper, we establish Gorenstein homological dimensions in terms of these two subcategories and investigate the Gorenstein global dimensions of A induced by the balanced pair (X,Y). As a consequence, we give some new characterizations of pure global dimensions and Gorenstein global dimensions of a ring R.
Let
Let
Let
Theorem 1.1. (see [4,Theorem 4.1]) Let
(1) For any
(2) For any
(3)
Moreover,
sup{G(P)−dimM∣M∈ModR}=sup{G(I)−codimM∣M∈ModR}. |
In this case, we say that
The main goal of this paper is to generalize Theorem 1.1 to Gorenstein subcategories
Theorem 1.2. Let
(1) For any object
(2) For any object
(3)
Moreover,
sup{G(X)−dimM∣M∈A}=sup{G(Y)−codimM∣M∈A}. |
The common value of the last equality is called the Gorenstein global dimension of the abelian category
The proof of the above results will be carried out in the next section.
Throughout this section, we always assume that
Definition 2.1. (see [3,Definition 1.1]) A pair
(BP0) the subcategory
(BP1) for each object
(BP2) for each object
We say that a contravariantly finite subcategory
Lemma 2.2. If the short exact sequence
(1) If
(2) If
(3) If
(4) If
Proof. We just prove (1) and (2) since (3) and (4) follow by duality.
(1) The first statement follows from [1,Proposition 2.13(1)]. One can prove that
(2) The "only if" part is clear. For the "if" part, since
sup{G(X)−dimM∣M∈A}=sup{G(Y)−codimM∣M∈A}. |
By [3,Propsotion 2.2], all rows and columns are
Recall that the
sup{G(X)−dimM∣M∈A}=sup{G(Y)−codimM∣M∈A}. |
If there is no such an integer, set
Lemma 2.3. The following are true for any object
(1) If
(2) If
Proof. We just prove (1) since (2) follows by duality. If
0→Pn→Pn−1→⋯→P1→P0→A→0 |
with all
0→K−j→Q−j−1→K−j−1→0 |
where
We let
Proposition 2.4. The following are true for any
(1) If
In this case,
(2) If
In this case,
Proof. We just prove (1) since (2) follows by duality.
Proposition 2.5. The following are true for any
(1) If
0→K→G→A→0 and 0→A→L→G′→0 |
such that
(2) If
0→A→G→K→0 and 0→G′→L→A→0 |
such that
Proof. According to Lemma 2.2(1), the results follows by an argument similar to that of Proposition 3.3 in [17].
Corollary 2.6. The following are true for any object
(1) If
(2) If
Proof. We just prove (1) since (2) follows by duality. It is clear
0→A→G→K→0 and 0→G′→L→A→0 |
where all rows and columns are
Proposition 2.7. If
sup{Y−cores.dimP∣P∈X}=sup{X−res.dimI∣I∈Y}⩽n. |
Proof. Suppose
Let
Proof of Theorem 1.2.
Let
0→A→I0→I1→⋯ |
is a
0→Ki→Ii→Ki+1→0,i⩾0. |
Here
0→Ki→Ii→Ki+1→0,i⩾0. |
with
Ωn(Ki)→Ωn(Ii)→Ωn(Ki+1),i⩾0, |
together, there is a
0→Ωn(A)→Ωn(I0)→Ωn(I1)→⋯ |
with
⋯→Ωn(P1)→Ωn(P0)→Ωn(A)→0 |
with
⋯→Ωn(P1)→Ωn(P0)→Ωn(I0)→Ωn(I1)→⋯ |
with each term in
Dually, one can prove
The last equality is immediate from above equivalences and Proposition 2.7.
Let
Corollary 2.8. Let
sup{pidRP∣P∈PP(R)}=sup{ppdRI∣I∈PI(R)}⩽n. |
Let
Corollary 2.9. Let
sup{G(I)−codimP∣P∈G(P)}=sup{G(P)−dimI∣I∈G(I)}⩽n. |
Let
Corollary 2.10. Let
(1) For any finitely generated left
(2) For any finitely generated left
(3)
Moreover,
sup{G(P(F))−dimM∣M∈modΛ}=sup{G(I(F))−codimM∣M∈modΛ}. |
The authors thank the referees for their careful reading and excellent suggestions.
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