The Generalized Auslander-Reiten Conjecture and
Derived Equivalences
Kosmas Diveris
Syracuse University
(Joint work with Marju Purin)
AMS meeting at UNL
16 October 2011
K. Diveris (Syracuse Univ.)
16 October 2011
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Introduction
K. Diveris (Syracuse Univ.)
16 October 2011
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Introduction
Conventions: R is a left Noetherian ring and all R-modules are finitely
generated left R-modules. All complexes are given a lower grading:
M = · · · → Mn+1 → Mn → Mn−1 → · · ·
K. Diveris (Syracuse Univ.)
16 October 2011
2/1
Introduction
Conventions: R is a left Noetherian ring and all R-modules are finitely
generated left R-modules. All complexes are given a lower grading:
M = · · · → Mn+1 → Mn → Mn−1 → · · ·
We will work with the bounded derived category, Db (R).
K. Diveris (Syracuse Univ.)
16 October 2011
2/1
Introduction
Conventions: R is a left Noetherian ring and all R-modules are finitely
generated left R-modules. All complexes are given a lower grading:
M = · · · → Mn+1 → Mn → Mn−1 → · · ·
We will work with the bounded derived category, Db (R).
• Obj(Db (R)) = R-complexes M such that H(M) is finitely generated.
K. Diveris (Syracuse Univ.)
16 October 2011
2/1
Introduction
Conventions: R is a left Noetherian ring and all R-modules are finitely
generated left R-modules. All complexes are given a lower grading:
M = · · · → Mn+1 → Mn → Mn−1 → · · ·
We will work with the bounded derived category, Db (R).
• Obj(Db (R)) = R-complexes M such that H(M) is finitely generated.
• Db (R) is a triangulated category with (ΣM)n = Mn−1 .
K. Diveris (Syracuse Univ.)
16 October 2011
2/1
Introduction
Conventions: R is a left Noetherian ring and all R-modules are finitely
generated left R-modules. All complexes are given a lower grading:
M = · · · → Mn+1 → Mn → Mn−1 → · · ·
We will work with the bounded derived category, Db (R).
• Obj(Db (R)) = R-complexes M such that H(M) is finitely generated.
• Db (R) is a triangulated category with (ΣM)n = Mn−1 .
• R-mod ⊂ Db (R) : · · · → 0 → 0 → M → 0 → 0 → · · ·
K. Diveris (Syracuse Univ.)
16 October 2011
2/1
Introduction
Conventions: R is a left Noetherian ring and all R-modules are finitely
generated left R-modules. All complexes are given a lower grading:
M = · · · → Mn+1 → Mn → Mn−1 → · · ·
We will work with the bounded derived category, Db (R).
•
•
•
•
Obj(Db (R)) = R-complexes M such that H(M) is finitely generated.
Db (R) is a triangulated category with (ΣM)n = Mn−1 .
R-mod ⊂ Db (R) : · · · → 0 → 0 → M → 0 → 0 → · · ·
If M, N ∈R-mod, then ExtiR (M, N) = HomDb (R) (M, Σi N).
K. Diveris (Syracuse Univ.)
16 October 2011
2/1
Introduction
Conventions: R is a left Noetherian ring and all R-modules are finitely
generated left R-modules. All complexes are given a lower grading:
M = · · · → Mn+1 → Mn → Mn−1 → · · ·
We will work with the bounded derived category, Db (R).
•
•
•
•
•
Obj(Db (R)) = R-complexes M such that H(M) is finitely generated.
Db (R) is a triangulated category with (ΣM)n = Mn−1 .
R-mod ⊂ Db (R) : · · · → 0 → 0 → M → 0 → 0 → · · ·
If M, N ∈R-mod, then ExtiR (M, N) = HomDb (R) (M, Σi N).
If M and N are quasi-isomorphic complexes, then M ∼
= N in Db (R).
K. Diveris (Syracuse Univ.)
16 October 2011
2/1
Introduction
Conventions: R is a left Noetherian ring and all R-modules are finitely
generated left R-modules. All complexes are given a lower grading:
M = · · · → Mn+1 → Mn → Mn−1 → · · ·
We will work with the bounded derived category, Db (R).
•
•
•
•
•
Obj(Db (R)) = R-complexes M such that H(M) is finitely generated.
Db (R) is a triangulated category with (ΣM)n = Mn−1 .
R-mod ⊂ Db (R) : · · · → 0 → 0 → M → 0 → 0 → · · ·
If M, N ∈R-mod, then ExtiR (M, N) = HomDb (R) (M, Σi N).
If M and N are quasi-isomorphic complexes, then M ∼
= N in Db (R).
K. Diveris (Syracuse Univ.)
16 October 2011
2/1
Introduction
Conjecture (Auslander-Reiten Conjecture)
If ExtiR (M, M) = 0 = ExtiR (M, R) for all i > 0, then M is projective.
K. Diveris (Syracuse Univ.)
16 October 2011
3/1
Introduction
Conjecture (Auslander-Reiten Conjecture)
If ExtiR (M, M) = 0 = ExtiR (M, R) for all i > 0, then M is projective.
Does not hold for all Noetherian rings, but open for commutative rings
and Artin algebras.
K. Diveris (Syracuse Univ.)
16 October 2011
3/1
Introduction
Conjecture (Auslander-Reiten Conjecture)
If ExtiR (M, M) = 0 = ExtiR (M, R) for all i > 0, then M is projective.
Does not hold for all Noetherian rings, but open for commutative rings
and Artin algebras.
Conjecture (Generalized Auslander-Reiten Conjecture)
If ExtiR (M, M) = 0 = ExtiR (M, R) for all i 0, then pd(M) < ∞.
K. Diveris (Syracuse Univ.)
16 October 2011
3/1
Introduction
Conjecture (Auslander-Reiten Conjecture)
If ExtiR (M, M) = 0 = ExtiR (M, R) for all i > 0, then M is projective.
Does not hold for all Noetherian rings, but open for commutative rings
and Artin algebras.
Conjecture (Generalized Auslander-Reiten Conjecture)
If ExtiR (M, M) = 0 = ExtiR (M, R) for all i 0, then pd(M) < ∞.
If R satisfies the Gen. AR Conj., it satisfies the AR Conj., but they are not
equivalent (at least not for Artin algebras).
K. Diveris (Syracuse Univ.)
16 October 2011
3/1
Introduction
Conjecture (Auslander-Reiten Conjecture)
If ExtiR (M, M) = 0 = ExtiR (M, R) for all i > 0, then M is projective.
Does not hold for all Noetherian rings, but open for commutative rings
and Artin algebras.
Conjecture (Generalized Auslander-Reiten Conjecture)
If ExtiR (M, M) = 0 = ExtiR (M, R) for all i 0, then pd(M) < ∞.
If R satisfies the Gen. AR Conj., it satisfies the AR Conj., but they are not
equivalent (at least not for Artin algebras).
Open for commutative rings.
K. Diveris (Syracuse Univ.)
16 October 2011
3/1
Introduction
Conjecture (Auslander-Reiten Conjecture)
If ExtiR (M, M) = 0 = ExtiR (M, R) for all i > 0, then M is projective.
Does not hold for all Noetherian rings, but open for commutative rings
and Artin algebras.
Conjecture (Generalized Auslander-Reiten Conjecture)
If ExtiR (M, M) = 0 = ExtiR (M, R) for all i 0, then pd(M) < ∞.
If R satisfies the Gen. AR Conj., it satisfies the AR Conj., but they are not
equivalent (at least not for Artin algebras).
Open for commutative rings.
K. Diveris (Syracuse Univ.)
16 October 2011
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Question
K. Diveris (Syracuse Univ.)
16 October 2011
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Question
Motivation: J. Wei has shown that both conjectures are preserved under
a tilting equivalence of Artin algebras.
K. Diveris (Syracuse Univ.)
16 October 2011
4/1
Question
Motivation: J. Wei has shown that both conjectures are preserved under
a tilting equivalence of Artin algebras. A tilting equivalence of Artin
algebras is a special case of a derived equivalence.
K. Diveris (Syracuse Univ.)
16 October 2011
4/1
Question
Motivation: J. Wei has shown that both conjectures are preserved under
a tilting equivalence of Artin algebras. A tilting equivalence of Artin
algebras is a special case of a derived equivalence.
Question: Are the conjectures preserved under any derived equivalence of
Noetherian rings?
K. Diveris (Syracuse Univ.)
16 October 2011
4/1
Question
Motivation: J. Wei has shown that both conjectures are preserved under
a tilting equivalence of Artin algebras. A tilting equivalence of Artin
algebras is a special case of a derived equivalence.
Question: Are the conjectures preserved under any derived equivalence of
Noetherian rings?
Goal: To show that the answer is yes for the Generalized AR Conjecture.
This gives 1/2 of Wei’s result as a special case.
K. Diveris (Syracuse Univ.)
16 October 2011
4/1
Question
Motivation: J. Wei has shown that both conjectures are preserved under
a tilting equivalence of Artin algebras. A tilting equivalence of Artin
algebras is a special case of a derived equivalence.
Question: Are the conjectures preserved under any derived equivalence of
Noetherian rings?
Goal: To show that the answer is yes for the Generalized AR Conjecture.
This gives 1/2 of Wei’s result as a special case.
Remark: This has also been shown independently by S. Pan and J. Wei.
K. Diveris (Syracuse Univ.)
16 October 2011
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Preliminaries
K. Diveris (Syracuse Univ.)
16 October 2011
5/1
Preliminaries
We will be considering Db (R) as a triangulated category, so we proceed
with preliminary remarks for any triangulated category T with suspension
Σ.
K. Diveris (Syracuse Univ.)
16 October 2011
5/1
Preliminaries
We will be considering Db (R) as a triangulated category, so we proceed
with preliminary remarks for any triangulated category T with suspension
Σ.
Notation We say M, N ∈ T are eventually orthogonal and write M ⊥ N if
HomT (M, Σi N) = 0 for |i| 0.
K. Diveris (Syracuse Univ.)
16 October 2011
5/1
Preliminaries
We will be considering Db (R) as a triangulated category, so we proceed
with preliminary remarks for any triangulated category T with suspension
Σ.
Notation We say M, N ∈ T are eventually orthogonal and write M ⊥ N if
HomT (M, Σi N) = 0 for |i| 0.
For any class of objects C ⊆ T, we set
⊥C
= {M ∈ T | M ⊥ C for all C ∈ C}
C⊥ = {N ∈ T | C ⊥ N for all C ∈ C}
K. Diveris (Syracuse Univ.)
16 October 2011
5/1
Preliminaries
We will be considering Db (R) as a triangulated category, so we proceed
with preliminary remarks for any triangulated category T with suspension
Σ.
Notation We say M, N ∈ T are eventually orthogonal and write M ⊥ N if
HomT (M, Σi N) = 0 for |i| 0.
For any class of objects C ⊆ T, we set
⊥C
= {M ∈ T | M ⊥ C for all C ∈ C}
C⊥ = {N ∈ T | C ⊥ N for all C ∈ C}
If B, C ⊆ T are any classes of objects, we write B ⊥ C if B ⊥ C for all
B ∈ B and C ∈ C.
K. Diveris (Syracuse Univ.)
16 October 2011
5/1
Preliminaries
We will be considering Db (R) as a triangulated category, so we proceed
with preliminary remarks for any triangulated category T with suspension
Σ.
Notation We say M, N ∈ T are eventually orthogonal and write M ⊥ N if
HomT (M, Σi N) = 0 for |i| 0.
For any class of objects C ⊆ T, we set
⊥C
= {M ∈ T | M ⊥ C for all C ∈ C}
C⊥ = {N ∈ T | C ⊥ N for all C ∈ C}
If B, C ⊆ T are any classes of objects, we write B ⊥ C if B ⊥ C for all
B ∈ B and C ∈ C.
K. Diveris (Syracuse Univ.)
16 October 2011
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Preliminaries Con’t
K. Diveris (Syracuse Univ.)
16 October 2011
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Preliminaries Con’t
Definition A nonempty subcategory C ⊆ T is thick if it is a triangulated
subcategory closed under direct summands.
K. Diveris (Syracuse Univ.)
16 October 2011
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Preliminaries Con’t
Definition A nonempty subcategory C ⊆ T is thick if it is a triangulated
subcategory closed under direct summands. If B ⊂ T is any class of
objects, Thick(B) is smallest thick subcategory of T containing B.
K. Diveris (Syracuse Univ.)
16 October 2011
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Preliminaries Con’t
Definition A nonempty subcategory C ⊆ T is thick if it is a triangulated
subcategory closed under direct summands. If B ⊂ T is any class of
objects, Thick(B) is smallest thick subcategory of T containing B.
Example In Db (R), Thick(R) consists of the perfect complexes, i.e., all
complexes that are quasi-isomorphic to bounded complexes of projective
modules.
K. Diveris (Syracuse Univ.)
16 October 2011
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Preliminaries Con’t
Definition A nonempty subcategory C ⊆ T is thick if it is a triangulated
subcategory closed under direct summands. If B ⊂ T is any class of
objects, Thick(B) is smallest thick subcategory of T containing B.
Example In Db (R), Thick(R) consists of the perfect complexes, i.e., all
complexes that are quasi-isomorphic to bounded complexes of projective
modules.
Lemma (Thick-Perp)
For any classes B, C ⊆ T:
and B⊥ are thick, and
1
⊥B
2
B ⊥ C if and only if Thick(B) ⊥ Thick(C).
K. Diveris (Syracuse Univ.)
16 October 2011
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Gen. AR Conj. for Db (R)
K. Diveris (Syracuse Univ.)
16 October 2011
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Gen. AR Conj. for Db (R)
We now state a version of the conjecture Gen. AR Conj. for the derived
category, Db (R):
K. Diveris (Syracuse Univ.)
16 October 2011
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Gen. AR Conj. for Db (R)
We now state a version of the conjecture Gen. AR Conj. for the derived
category, Db (R):
Conjecture (Db (Gen. AR Conj.))
If M ⊥ M and M ⊥ Thick(R), then M is perfect.
K. Diveris (Syracuse Univ.)
16 October 2011
7/1
Gen. AR Conj. for Db (R)
We now state a version of the conjecture Gen. AR Conj. for the derived
category, Db (R):
Conjecture (Db (Gen. AR Conj.))
If M ⊥ M and M ⊥ Thick(R), then M is perfect.
It is immediate that the original Gen. AR Conj. follows from the derived
version. In fact, they are equivalent.
K. Diveris (Syracuse Univ.)
16 October 2011
7/1
Gen. AR Conj. for Db (R)
We now state a version of the conjecture Gen. AR Conj. for the derived
category, Db (R):
Conjecture (Db (Gen. AR Conj.))
If M ⊥ M and M ⊥ Thick(R), then M is perfect.
It is immediate that the original Gen. AR Conj. follows from the derived
version. In fact, they are equivalent.
To show that the derived version follows from the original version, we need
to take a syzygy of a complex.
K. Diveris (Syracuse Univ.)
16 October 2011
7/1
Gen. AR Conj. for Db (R)
We now state a version of the conjecture Gen. AR Conj. for the derived
category, Db (R):
Conjecture (Db (Gen. AR Conj.))
If M ⊥ M and M ⊥ Thick(R), then M is perfect.
It is immediate that the original Gen. AR Conj. follows from the derived
version. In fact, they are equivalent.
To show that the derived version follows from the original version, we need
to take a syzygy of a complex.
K. Diveris (Syracuse Univ.)
16 October 2011
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Syzygy of a complex
K. Diveris (Syracuse Univ.)
16 October 2011
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Syzygy of a complex
Let M ∈ Db (R) and P be a projective resolution of M. Set n = sup M and
consider the following truncations of P:
K. Diveris (Syracuse Univ.)
16 October 2011
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Syzygy of a complex
Let M ∈ Db (R) and P be a projective resolution of M. Set n = sup M and
consider the following truncations of P:
P<n = 0 → Pn → · · · → Pinf M → 0
K. Diveris (Syracuse Univ.)
16 October 2011
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Syzygy of a complex
Let M ∈ Db (R) and P be a projective resolution of M. Set n = sup M and
consider the following truncations of P:
P<n = 0 → Pn → · · · → Pinf M → 0
P≥n = · · · → Pi → · · · → Pn+2 → Pn+1 → 0
K. Diveris (Syracuse Univ.)
16 October 2011
8/1
Syzygy of a complex
Let M ∈ Db (R) and P be a projective resolution of M. Set n = sup M and
consider the following truncations of P:
P<n = 0 → Pn → · · · → Pinf M → 0
P≥n = · · · → Pi → · · · → Pn+2 → Pn+1 → 0
which fit into a short exact sequence of complexes
0 → P<n → P → P≥n → 0
K. Diveris (Syracuse Univ.)
16 October 2011
8/1
Syzygy of a complex
Let M ∈ Db (R) and P be a projective resolution of M. Set n = sup M and
consider the following truncations of P:
P<n = 0 → Pn → · · · → Pinf M → 0
P≥n = · · · → Pi → · · · → Pn+2 → Pn+1 → 0
which fit into a short exact sequence of complexes
0 → P<n → P → P≥n → 0
with P<n ∈ Thick(R) and P≥n isomorphic to a stalk complex in Db (R).
K. Diveris (Syracuse Univ.)
16 October 2011
8/1
Syzygy of a complex
Let M ∈ Db (R) and P be a projective resolution of M. Set n = sup M and
consider the following truncations of P:
P<n = 0 → Pn → · · · → Pinf M → 0
P≥n = · · · → Pi → · · · → Pn+2 → Pn+1 → 0
which fit into a short exact sequence of complexes
0 → P<n → P → P≥n → 0
with P<n ∈ Thick(R) and P≥n isomorphic to a stalk complex in Db (R).
K. Diveris (Syracuse Univ.)
16 October 2011
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Syzygy of a complex con’t
We have a short exact sequence of complexes
0 → P<n → P → P≥n → 0
with P<n ∈ Thick(R) and P≥n isomorphic to a stalk complex in Db (R).
K. Diveris (Syracuse Univ.)
16 October 2011
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Syzygy of a complex con’t
We have a short exact sequence of complexes
0 → P<n → P → P≥n → 0
with P<n ∈ Thick(R) and P≥n isomorphic to a stalk complex in Db (R).
We define ΩM to be the module isomorphic to P≥n in Db (R).
K. Diveris (Syracuse Univ.)
16 October 2011
9/1
Syzygy of a complex con’t
We have a short exact sequence of complexes
0 → P<n → P → P≥n → 0
with P<n ∈ Thick(R) and P≥n isomorphic to a stalk complex in Db (R).
We define ΩM to be the module isomorphic to P≥n in Db (R).
The short exact sequence above gives rise to a triangle in Db (R):
Q → M → Σn ΩM → ΣQ
K. Diveris (Syracuse Univ.)
16 October 2011
9/1
Syzygy of a complex con’t
We have a short exact sequence of complexes
0 → P<n → P → P≥n → 0
with P<n ∈ Thick(R) and P≥n isomorphic to a stalk complex in Db (R).
We define ΩM to be the module isomorphic to P≥n in Db (R).
The short exact sequence above gives rise to a triangle in Db (R):
Q → M → Σn ΩM → ΣQ
K. Diveris (Syracuse Univ.)
16 October 2011
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Equivalence of Gen. AR Conj.’s
K. Diveris (Syracuse Univ.)
16 October 2011
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Equivalence of Gen. AR Conj.’s
Using the triangle
Q → M → Σn ΩM → ΣQ
and the ‘Thick-Perp’ Lemma one can show
K. Diveris (Syracuse Univ.)
16 October 2011
10 / 1
Equivalence of Gen. AR Conj.’s
Using the triangle
Q → M → Σn ΩM → ΣQ
and the ‘Thick-Perp’ Lemma one can show
Proposition
M ⊥ M and M ⊥ Thick(R) iff. ExtiR (ΩM, ΩM) = 0 = ExtiR (ΩM, R) for
all i 0.
K. Diveris (Syracuse Univ.)
16 October 2011
10 / 1
Equivalence of Gen. AR Conj.’s
Using the triangle
Q → M → Σn ΩM → ΣQ
and the ‘Thick-Perp’ Lemma one can show
Proposition
M ⊥ M and M ⊥ Thick(R) iff. ExtiR (ΩM, ΩM) = 0 = ExtiR (ΩM, R) for
all i 0.
This gives the following:
Theorem
The Gen. AR Conj. holds for R if and only if Db (Gen. AR Conj.) holds for
R.
K. Diveris (Syracuse Univ.)
16 October 2011
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Roadmap
K. Diveris (Syracuse Univ.)
16 October 2011
11 / 1
Roadmap
Goal: To prove
Theorem (Main Theorem)
The Generalized AR Conjecture is stable under derived equivalences
between Noetherian rings.
K. Diveris (Syracuse Univ.)
16 October 2011
11 / 1
Roadmap
Goal: To prove
Theorem (Main Theorem)
The Generalized AR Conjecture is stable under derived equivalences
between Noetherian rings.
So far we’ve shown that Gen. AR Conj. is equivalent to
Conjecture (Db (Gen. AR Conj.))
If M ⊥ M and M ⊥ Thick(R), then M is perfect.
K. Diveris (Syracuse Univ.)
16 October 2011
11 / 1
Roadmap
Goal: To prove
Theorem (Main Theorem)
The Generalized AR Conjecture is stable under derived equivalences
between Noetherian rings.
So far we’ve shown that Gen. AR Conj. is equivalent to
Conjecture (Db (Gen. AR Conj.))
If M ⊥ M and M ⊥ Thick(R), then M is perfect.
It remains to show that this conjecture is stable under derived
equivalences.
K. Diveris (Syracuse Univ.)
16 October 2011
11 / 1
Roadmap
Goal: To prove
Theorem (Main Theorem)
The Generalized AR Conjecture is stable under derived equivalences
between Noetherian rings.
So far we’ve shown that Gen. AR Conj. is equivalent to
Conjecture (Db (Gen. AR Conj.))
If M ⊥ M and M ⊥ Thick(R), then M is perfect.
It remains to show that this conjecture is stable under derived
equivalences.
K. Diveris (Syracuse Univ.)
16 October 2011
11 / 1
Derived equivalences and perfect complexes
K. Diveris (Syracuse Univ.)
16 October 2011
12 / 1
Derived equivalences and perfect complexes
Suppose that R and S are left Noetherian rings and F : Db (R) → Db (S) is
an equivalence of triangulated categories.
K. Diveris (Syracuse Univ.)
16 October 2011
12 / 1
Derived equivalences and perfect complexes
Suppose that R and S are left Noetherian rings and F : Db (R) → Db (S) is
an equivalence of triangulated categories.
Lemma (Rickard)
The restriction of F gives an equivalence between Thick(R) and Thick(S).
K. Diveris (Syracuse Univ.)
16 October 2011
12 / 1
Derived equivalences and perfect complexes
Suppose that R and S are left Noetherian rings and F : Db (R) → Db (S) is
an equivalence of triangulated categories.
Lemma (Rickard)
The restriction of F gives an equivalence between Thick(R) and Thick(S).
This gives that M ⊥ Thick(R) ⇐⇒ F (M) ⊥ Thick(S).
K. Diveris (Syracuse Univ.)
16 October 2011
12 / 1
Derived equivalences and perfect complexes
Suppose that R and S are left Noetherian rings and F : Db (R) → Db (S) is
an equivalence of triangulated categories.
Lemma (Rickard)
The restriction of F gives an equivalence between Thick(R) and Thick(S).
This gives that M ⊥ Thick(R) ⇐⇒ F (M) ⊥ Thick(S).
K. Diveris (Syracuse Univ.)
16 October 2011
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Proof of Main Theorem (last step)
From the previous slide: If F : Db (R) → Db (S) is an equivalence.
Lemma (Rickard)
The restriction of F gives an equivalence between Thick(R) and Thick(S).
K. Diveris (Syracuse Univ.)
16 October 2011
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Proof of Main Theorem (last step)
From the previous slide: If F : Db (R) → Db (S) is an equivalence.
Lemma (Rickard)
The restriction of F gives an equivalence between Thick(R) and Thick(S).
Suppose that S satisfies the conjecture below, we will show that R does.
Conjecture (Db (Gen. AR Conj.))
If M ⊥ M and M ⊥ Thick(R), then M is perfect.
K. Diveris (Syracuse Univ.)
16 October 2011
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Proof of Main Theorem (last step)
From the previous slide: If F : Db (R) → Db (S) is an equivalence.
Lemma (Rickard)
The restriction of F gives an equivalence between Thick(R) and Thick(S).
Suppose that S satisfies the conjecture below, we will show that R does.
Conjecture (Db (Gen. AR Conj.))
If M ⊥ M and M ⊥ Thick(R), then M is perfect.
K. Diveris (Syracuse Univ.)
16 October 2011
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Proof of Main Theorem (last step)
From the previous slide: If F : Db (R) → Db (S) is an equivalence.
Lemma (Rickard)
The restriction of F gives an equivalence between Thick(R) and Thick(S).
Suppose that S satisfies the conjecture below, we will show that R does.
Conjecture (Db (Gen. AR Conj.))
If M ⊥ M and M ⊥ Thick(R), then M is perfect.
If M ⊥ M and M ⊥ Thick(R) then F (M) ⊥ F (M) and F (M) ⊥ Thick(S).
K. Diveris (Syracuse Univ.)
16 October 2011
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Proof of Main Theorem (last step)
From the previous slide: If F : Db (R) → Db (S) is an equivalence.
Lemma (Rickard)
The restriction of F gives an equivalence between Thick(R) and Thick(S).
Suppose that S satisfies the conjecture below, we will show that R does.
Conjecture (Db (Gen. AR Conj.))
If M ⊥ M and M ⊥ Thick(R), then M is perfect.
If M ⊥ M and M ⊥ Thick(R) then F (M) ⊥ F (M) and F (M) ⊥ Thick(S).
Since the conjecture holds for S we have F (M) ∈ Thick(S).
K. Diveris (Syracuse Univ.)
16 October 2011
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Proof of Main Theorem (last step)
From the previous slide: If F : Db (R) → Db (S) is an equivalence.
Lemma (Rickard)
The restriction of F gives an equivalence between Thick(R) and Thick(S).
Suppose that S satisfies the conjecture below, we will show that R does.
Conjecture (Db (Gen. AR Conj.))
If M ⊥ M and M ⊥ Thick(R), then M is perfect.
If M ⊥ M and M ⊥ Thick(R) then F (M) ⊥ F (M) and F (M) ⊥ Thick(S).
Since the conjecture holds for S we have F (M) ∈ Thick(S).
Another application of Rickard’s Lemma gives M ∈ Thick(R), as desired.
K. Diveris (Syracuse Univ.)
16 October 2011
13 / 1
Proof of Main Theorem (last step)
From the previous slide: If F : Db (R) → Db (S) is an equivalence.
Lemma (Rickard)
The restriction of F gives an equivalence between Thick(R) and Thick(S).
Suppose that S satisfies the conjecture below, we will show that R does.
Conjecture (Db (Gen. AR Conj.))
If M ⊥ M and M ⊥ Thick(R), then M is perfect.
If M ⊥ M and M ⊥ Thick(R) then F (M) ⊥ F (M) and F (M) ⊥ Thick(S).
Since the conjecture holds for S we have F (M) ∈ Thick(S).
Another application of Rickard’s Lemma gives M ∈ Thick(R), as desired.
K. Diveris (Syracuse Univ.)
16 October 2011
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Gorenstein rings and stable derived categories
The stable derived category of R is:
Dbst (R) =
K. Diveris (Syracuse Univ.)
Db (R)
Thick(R)
16 October 2011
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Gorenstein rings and stable derived categories
The stable derived category of R is:
Dbst (R) =
Db (R)
Thick(R)
Corollary
If R and S are stably derived equivalent Gorenstein rings, then the Gen.
AR Conj. holds for R is and only if it holds for S.
K. Diveris (Syracuse Univ.)
16 October 2011
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Closing remarks
K. Diveris (Syracuse Univ.)
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Closing remarks
Question: Is the AR Conjecture stable under any derived equivalence of
Noetherian rings?
K. Diveris (Syracuse Univ.)
16 October 2011
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Closing remarks
Question: Is the AR Conjecture stable under any derived equivalence of
Noetherian rings?
Question: Are the AR Conjecture and the Generalized AR Conjecture
equivalent for commutative rings?
K. Diveris (Syracuse Univ.)
16 October 2011
15 / 1
Closing remarks
Question: Is the AR Conjecture stable under any derived equivalence of
Noetherian rings?
Question: Are the AR Conjecture and the Generalized AR Conjecture
equivalent for commutative rings?
Note that Thick(R) = ⊥ Db (R). Thus, the following statement specializes
to Db (Gen. AR Conj.)) when T = Db (R):
If M ⊥ M and M ⊥ ⊥ T then M ∈ ⊥ T
K. Diveris (Syracuse Univ.)
16 October 2011
15 / 1
Closing remarks
Question: Is the AR Conjecture stable under any derived equivalence of
Noetherian rings?
Question: Are the AR Conjecture and the Generalized AR Conjecture
equivalent for commutative rings?
Note that Thick(R) = ⊥ Db (R). Thus, the following statement specializes
to Db (Gen. AR Conj.)) when T = Db (R):
If M ⊥ M and M ⊥ ⊥ T then M ∈ ⊥ T
Question: Is this statement interesting in any other triangulated
categories?
K. Diveris (Syracuse Univ.)
16 October 2011
15 / 1
Closing remarks
Question: Is the AR Conjecture stable under any derived equivalence of
Noetherian rings?
Question: Are the AR Conjecture and the Generalized AR Conjecture
equivalent for commutative rings?
Note that Thick(R) = ⊥ Db (R). Thus, the following statement specializes
to Db (Gen. AR Conj.)) when T = Db (R):
If M ⊥ M and M ⊥ ⊥ T then M ∈ ⊥ T
Question: Is this statement interesting in any other triangulated
categories?
K. Diveris (Syracuse Univ.)
16 October 2011
15 / 1
K. Diveris (Syracuse Univ.)
16 October 2011
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