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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 1 / 1 Introduction K.

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SLIDE 1

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 1 / 1

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SLIDE 2

Introduction

  • K. Diveris (Syracuse Univ.)

16 October 2011 2 / 1

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SLIDE 3

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

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SLIDE 4

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

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SLIDE 5

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

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SLIDE 6

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

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SLIDE 7

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

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SLIDE 8

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 Exti

R(M, N) = HomDb(R)(M, ΣiN).

  • K. Diveris (Syracuse Univ.)

16 October 2011 2 / 1

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SLIDE 9

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 Exti

R(M, N) = HomDb(R)(M, ΣiN).

  • If M and N are quasi-isomorphic complexes, then M ∼

= N in Db(R).

  • K. Diveris (Syracuse Univ.)

16 October 2011 2 / 1

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SLIDE 10

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 Exti

R(M, N) = HomDb(R)(M, ΣiN).

  • If M and N are quasi-isomorphic complexes, then M ∼

= N in Db(R).

  • K. Diveris (Syracuse Univ.)

16 October 2011 2 / 1

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SLIDE 11

Introduction

Conjecture (Auslander-Reiten Conjecture)

If Exti

R(M, M) = 0 = Exti R(M, R) for all i > 0, then M is projective.

  • K. Diveris (Syracuse Univ.)

16 October 2011 3 / 1

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SLIDE 12

Introduction

Conjecture (Auslander-Reiten Conjecture)

If Exti

R(M, M) = 0 = Exti R(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

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SLIDE 13

Introduction

Conjecture (Auslander-Reiten Conjecture)

If Exti

R(M, M) = 0 = Exti R(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 Exti

R(M, M) = 0 = Exti R(M, R) for all i ≫ 0, then pd(M) < ∞.

  • K. Diveris (Syracuse Univ.)

16 October 2011 3 / 1

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SLIDE 14

Introduction

Conjecture (Auslander-Reiten Conjecture)

If Exti

R(M, M) = 0 = Exti R(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 Exti

R(M, M) = 0 = Exti R(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

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SLIDE 15

Introduction

Conjecture (Auslander-Reiten Conjecture)

If Exti

R(M, M) = 0 = Exti R(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 Exti

R(M, M) = 0 = Exti R(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

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SLIDE 16

Introduction

Conjecture (Auslander-Reiten Conjecture)

If Exti

R(M, M) = 0 = Exti R(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 Exti

R(M, M) = 0 = Exti R(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

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Question

  • K. Diveris (Syracuse Univ.)

16 October 2011 4 / 1

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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

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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

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SLIDE 20

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

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SLIDE 21

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

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SLIDE 22

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.)

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SLIDE 23

Preliminaries

  • K. Diveris (Syracuse Univ.)

16 October 2011 5 / 1

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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

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SLIDE 25

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, ΣiN) = 0 for |i| ≫ 0.

  • K. Diveris (Syracuse Univ.)

16 October 2011 5 / 1

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SLIDE 26

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, ΣiN) = 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

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SLIDE 27

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, ΣiN) = 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

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SLIDE 28

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, ΣiN) = 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

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SLIDE 29

Preliminaries Con’t

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SLIDE 30

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.)

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SLIDE 31

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

  • bjects, Thick(B) is smallest thick subcategory of T containing B.
  • K. Diveris (Syracuse Univ.)

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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

  • bjects, 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.)

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SLIDE 33

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

  • bjects, 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:

1

⊥ B and B⊥ are thick, and

2 B ⊥ C if and only if Thick(B) ⊥ Thick(C).

  • K. Diveris (Syracuse Univ.)

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SLIDE 34
  • Gen. AR Conj. for Db(R)
  • K. Diveris (Syracuse Univ.)

16 October 2011 7 / 1

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SLIDE 35
  • 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 7 / 1

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SLIDE 36
  • 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

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SLIDE 37
  • 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

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SLIDE 38
  • 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

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SLIDE 39
  • 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

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SLIDE 40

Syzygy of a complex

  • K. Diveris (Syracuse Univ.)

16 October 2011 8 / 1

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SLIDE 41

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 8 / 1

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SLIDE 42

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 8 / 1

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SLIDE 43

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

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SLIDE 44

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

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SLIDE 45

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

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SLIDE 46

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

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SLIDE 47

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 9 / 1

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SLIDE 48

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

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SLIDE 49

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 → ΣnΩM → ΣQ

  • K. Diveris (Syracuse Univ.)

16 October 2011 9 / 1

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SLIDE 50

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 → ΣnΩM → ΣQ

  • K. Diveris (Syracuse Univ.)

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SLIDE 51

Equivalence of Gen. AR Conj.’s

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SLIDE 52

Equivalence of Gen. AR Conj.’s

Using the triangle Q → M → ΣnΩM → ΣQ and the ‘Thick-Perp’ Lemma one can show

  • K. Diveris (Syracuse Univ.)

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SLIDE 53

Equivalence of Gen. AR Conj.’s

Using the triangle Q → M → ΣnΩM → ΣQ and the ‘Thick-Perp’ Lemma one can show

Proposition

M ⊥ M and M ⊥ Thick(R) iff. Exti

R(ΩM, ΩM) = 0 = Exti R(ΩM, R) for

all i ≫ 0.

  • K. Diveris (Syracuse Univ.)

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SLIDE 54

Equivalence of Gen. AR Conj.’s

Using the triangle Q → M → ΣnΩM → ΣQ and the ‘Thick-Perp’ Lemma one can show

Proposition

M ⊥ M and M ⊥ Thick(R) iff. Exti

R(ΩM, ΩM) = 0 = Exti R(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.)

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SLIDE 55

Roadmap

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SLIDE 56

Roadmap

Goal: To prove

Theorem (Main Theorem)

The Generalized AR Conjecture is stable under derived equivalences between Noetherian rings.

  • K. Diveris (Syracuse Univ.)

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SLIDE 57

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.)

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SLIDE 58

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

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SLIDE 59

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.)

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SLIDE 60

Derived equivalences and perfect complexes

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SLIDE 61

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.

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SLIDE 62

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).

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SLIDE 63

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.)

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SLIDE 64

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.)

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SLIDE 65

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.)

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SLIDE 66

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 13 / 1

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SLIDE 67

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 13 / 1

slide-68
SLIDE 68

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 13 / 1

slide-69
SLIDE 69

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 13 / 1

slide-70
SLIDE 70

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

slide-71
SLIDE 71

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

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SLIDE 72

Gorenstein rings and stable derived categories

The stable derived category of R is: Db

st(R) =

Db(R) Thick(R)

  • K. Diveris (Syracuse Univ.)

16 October 2011 14 / 1

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SLIDE 73

Gorenstein rings and stable derived categories

The stable derived category of R is: Db

st(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 14 / 1

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SLIDE 74

Closing remarks

  • K. Diveris (Syracuse Univ.)

16 October 2011 15 / 1

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SLIDE 75

Closing remarks

Question: Is the AR Conjecture stable under any derived equivalence of Noetherian rings?

  • K. Diveris (Syracuse Univ.)

16 October 2011 15 / 1

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SLIDE 76

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

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SLIDE 77

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

slide-78
SLIDE 78

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

slide-79
SLIDE 79

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

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SLIDE 80
  • K. Diveris (Syracuse Univ.)

16 October 2011 16 / 1