Alex Martsinkovsky Functor Categories, Model Theory, and - - PowerPoint PPT Presentation

▶
alex martsinkovsky
SMART_READER_LITE
LIVE PREVIEW

Alex Martsinkovsky Functor Categories, Model Theory, and - - PowerPoint PPT Presentation

Feb 09, 2024 672 likes •1.15k views

D ECONSTRUCTING A USLANDER S F ORMULAS Alex Martsinkovsky Functor Categories, Model Theory, and Constructive Category Theory Prnu July 17, 2019 A LEX M ARTSINKOVSKY D ECONSTRUCTING A USLANDER S F ORMULAS J ULY 17, 2019 1 / 46 P LAN

slide-1
SLIDE 1

DECONSTRUCTING AUSLANDER’S FORMULAS Alex Martsinkovsky

Functor Categories, Model Theory, and Constructive Category Theory Pärnu July 17, 2019

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 1 / 46

slide-2
SLIDE 2

PLAN OF THE TALK

Part 1. Auslander’s formulas Part 2. Zeroth derived functors Part 3. Stabilization of additive functors Part 4. An extension of the Auslander-Reiten formula Part 5. R0F and L0F revisited

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 2 / 46

slide-3
SLIDE 3

AUSLANDER’S FORMULAS

Part 1. Auslander’s formulas

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 3 / 46

slide-4
SLIDE 4

AUSLANDER’S FORMULAS

THE AUSLANDER TRANSPOSE

Let Λ be a ring and A a finitely presented Λ-module. Choose a projective presentation P1 Ý Ñ P0 Ý Ñ A Ý Ñ 0 and dualize it into Λ. The resulting exact sequence 0 Ý Ñ A˚ Ý Ñ P˚

0 Ý

Ñ P˚

1 Ý

Ñ Tr A Ý Ñ 0 defines the Auslander transpose Tr A of A. Dualizing again, we recover A, up to projective equivalence. N.B. The last statement is not true if A is not finitely presented.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 4 / 46

slide-5
SLIDE 5

AUSLANDER’S FORMULAS

THE AUSLANDER-REITEN FORMULA

Λ := artin algebra over a commutative artinian ring k, I := the injective envelope of k viewed as a module over itself, D :“ Homkp , Iq, A1 and B := finitely generated Λ-modules. THEOREM There is a bifunctorial isomorphism D Ext1pA1, Bq » pB, D Tr A1q. REMARK The only needed assumption here is that A1 be finitely presented. If A1 is not finitely presented, the proof breaks down.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 5 / 46

slide-6
SLIDE 6

AUSLANDER’S FORMULAS

THE BIDUALIZATION MAP

Let Λ be an arbitrary ring, A a finitely presented Λ-module, and eA : A Ý Ñ A˚˚ the canonical map. Then there is an exact sequence

Ext1pTr A, Λq A

eA A˚˚

Ext2pTr A, Λq

This is the Λ-component of the exact sequence of functors

Ext1pTr A,

q

A b

ρAb pA˚,

q

Ext2pTr A,

q REMARK This sequence is not exact if A is an infinite projective, even over a field (cardinality!).

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 6 / 46

slide-7
SLIDE 7

AUSLANDER’S FORMULAS

THE COMPANION TRANSFORMATION µpA, q

Assume again that A is finitely presented. Then there is an exact sequence

Tor2pTr A,

q

A˚ b

µpA,

q pA,

q

Tor1pTr A,

q REMARK This sequence is not exact if A is an infinite projective, even over a field (the identity map cannot factor through a finitely generated projective!).

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 7 / 46

slide-8
SLIDE 8

AUSLANDER’S FORMULAS

THE GOAL OF THIS TALK

The goal of today’s talk is to extend the above formulas to arbitrary modules over arbitrary rings.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 8 / 46

slide-9
SLIDE 9

ZEROTH DERIVED FUNCTORS

Part 2. Zeroth derived functors

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 9 / 46

slide-10
SLIDE 10

ZEROTH DERIVED FUNCTORS

PROJECTIVE RESOLUTIONS

Blanket assumption: all functors are from modules to abelian groups F : ModpΛq Ý Ñ Ab and are additive. Given a module M, an exact sequence . . . Ý Ñ P1 Ý Ñ P0 Ý Ñ M Ý Ñ 0 (excluding M itself), where the Pi are projective, is called a projective resolution of M. THEOREM Any two projective resolutions of M are homotopy equivalent.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 10 / 46

slide-11
SLIDE 11

ZEROTH DERIVED FUNCTORS

INJECTIVE RESOLUTIONS

Given a module M, an exact sequence 0 Ý Ñ M Ý Ñ I0 Ý Ñ I1 Ý Ñ . . . (excluding M itself), where the Ii are injective, is called an injective resolution of M. THEOREM Any two injective resolutions of M are homotopy equivalent.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 11 / 46

slide-12
SLIDE 12

ZEROTH DERIVED FUNCTORS

NOMENCLATURE: DERIVED FUNCTORS

Given an additive functor F, apply it to a resolution. Since resolutions are homotopically unique, the homology groups of the resulting complex are unique up to isomorphism. Projective resolutions Injective resolutions Covariant F LiF RiF Contravariant F RiF LiF N.B. For a contravariant F, subscripts and superscripts are flipped.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 12 / 46

slide-13
SLIDE 13

ZEROTH DERIVED FUNCTORS

THE CASE i “ 0: L0F

Zeroth derived functors are of special interest to us: . . .

FpP1q FpP0q

  • L0FpMq
  • D! λF pMq
  • FpMq

PROPOSITION L0F is right-exact (i.e., preserves cokernels). COROLLARY λF : L0F Ý Ñ F is an isomorphism if and only if F is right-exact.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 13 / 46

slide-14
SLIDE 14

ZEROTH DERIVED FUNCTORS

THE CASE i “ 0: R0F

FpMq

D! ρF pMq

  • R0FpMq

FpI0q FpI1q . . .

PROPOSITION R0F is left-exact (i.e., preserves kernels). COROLLARY ρF : F Ý Ñ R0F is an isomorphism if and only if F is left-exact.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 14 / 46

slide-15
SLIDE 15

ZEROTH DERIVED FUNCTORS

THE CASE i “ 0: A SUMMARY

Projective resolutions Injective resolutions Covariant F L0F

λF

Ý Ñ F F

ρF

Ý Ñ R0F Contravariant F F

ρF

Ý Ñ R0F L0F

λF

Ý Ñ F The λF are isomorphisms if and only if F is right-exact; The ρF are isomorphisms if and only if F is left-exact.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 15 / 46

slide-16
SLIDE 16

ZEROTH DERIVED FUNCTORS

QUESTION: WHY IS THE CASE i “ 0 INTERESTING?

Projective resolutions Injective resolutions Covariant F L0F

λF

Ý Ñ F F

ρF

Ý Ñ R0F Contravariant F F

ρF

Ý Ñ R0F L0F

λF

Ý Ñ F Answer: because all arrows are universal: the λF – with respect to natural transformations from right-exact functors to F, the ρF – with respect natural transformations from F to left-exact functors.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 16 / 46

slide-17
SLIDE 17

ZEROTH DERIVED FUNCTORS

THE UNIVERSAL PROPERTY OF λ Rex

@

  • D !
  • L0F

λF F

REMARK This diagram shows that the subcategory of right-exact functors is coreflective in the category of all additive functors and λ is a coreflector, i.e., a counit of the adjunction ι % L0

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 17 / 46

slide-18
SLIDE 18

ZEROTH DERIVED FUNCTORS

THE UNIVERSAL PROPERTY OF ρ F

ρF @

R0F

D !

  • Lex

REMARK This diagram shows that the subcategory of left-exact functors is reflective in the category of all additive functors and ρ is a reflector, i.e., a unit of the adjunction R0 % ι

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 18 / 46

slide-19
SLIDE 19

ZEROTH DERIVED FUNCTORS

AN INTRINSIC CHARACTERIZATION OF λ

LEMMA If F is covariant (resp., contravariant), then λF : L0F Ý Ñ F (resp., λF : L0F Ý Ñ F) evaluates to an isomorphism on projectives (resp., injectives). PROPOSITION If F is covariant (resp., contravariant), then λF : L0F Ý Ñ F (resp., λF : L0F Ý Ñ F) is the unique natural transformation from a right-exact functor to F which evaluates to an isomorphism on projectives (resp., injectives).

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 19 / 46

slide-20
SLIDE 20

ZEROTH DERIVED FUNCTORS

AN INTRINSIC CHARACTERIZATION OF ρ

LEMMA If F is covariant (resp., contravariant), then ρF : F Ý Ñ R0F (resp., ρF : F Ý Ñ R0F) evaluates to an isomorphism on injectives (resp., projectives). PROPOSITION If F is covariant (resp., contravariant), then ρF : F Ý Ñ R0F (resp., ρF : F Ý Ñ R0F) is the unique natural transformation from F to a left-exact functor which evaluates to an isomorphism on injectives (resp., projectives).

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 20 / 46

slide-21
SLIDE 21

ZEROTH DERIVED FUNCTORS

EXAMPLE: L0 OF THE COVARIANT HOM

EXAMPLE F :“ pA, q, where A is finitely generated. In this case, L0F

λ

Ý Ñ F is the canonical transformation A˚ b Ý Ñ pA, q because it is an isomorphism on projectives and A˚ b is right-exact. Hence LipA, q » ToripA˚, q for all i.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 21 / 46

slide-22
SLIDE 22

ZEROTH DERIVED FUNCTORS

EXAMPLE: R0 OF THE TENSOR PRODUCT

EXAMPLE F :“ A b , where A is finitely presented. In this case, F

ρ

Ý Ñ R0F is the canonical transformation A b Ý Ñ pA˚, q because it is an isomorphism on injectives and pA˚, q is left-exact. Hence RipA b q » ExtipA˚, q for all i.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 22 / 46

slide-23
SLIDE 23

ZEROTH DERIVED FUNCTORS

RECOGNIZING L0F

Let F : Mod pΛq Ý Ñ Ab be an additive covariant functor. The natural transformation FpΛq b

τ

F

FpΛq b M

τM

FpMq

x b m ✤

Fprmqpxq

where x P FpΛq, m P M, and rm : Λ Ý Ñ M : λ ÞÑ λm, evaluates to the canonical isomorphism on Λ. Whence

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 23 / 46

slide-24
SLIDE 24

ZEROTH DERIVED FUNCTORS

RECOGNIZING L0F

PROPOSITION If F commutes with coproducts, then τ : FpΛq b Ý Ñ F evaluates to an isomorphism on projectives, and therefore FpΛq b » L0F This explains the example with L0pA, q.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 24 / 46

slide-25
SLIDE 25

ZEROTH DERIVED FUNCTORS

FIRST APPLICATIONS: RECOGNIZING THE TENSOR

PRODUCT

As a consequence, THEOREM (EILENBERG, WATTS) If a covariant functor F commutes with coproducts and is right-exact, then F » FpΛq b .

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 25 / 46

slide-26
SLIDE 26

ZEROTH DERIVED FUNCTORS

RECOGNIZING R0F

Let F : Mod pΛq Ý Ñ Ab be an additive contravariant functor. The natural transformation F

σ p

, FpΛqq FpMq

σM pM, FpΛqq

x ✤

  • ˆ

m ✤

Fprmqpxq

˙ where x P FpMq, m P M, and rm : Λ Ñ M : λ ÞÑ λm, evaluates to the canonical isomorphism on Λ. Whence

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 26 / 46

slide-27
SLIDE 27

ZEROTH DERIVED FUNCTORS

RECOGNIZING R0F

PROPOSITION If a contravariant functor F converts coproducts into products, then σ : F Ý Ñ p , FpΛqq evaluates to an isomorphism on projectives, and therefore p , FpΛqq » R0F

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 27 / 46

slide-28
SLIDE 28

ZEROTH DERIVED FUNCTORS

FIRST APPLICATIONS: RECOGNIZING THE

CONTRAVARIANT Hom

As a consequence, THEOREM (EILENBERG, WATTS) If a contravariant functor F converts coproducts into products and is left-exact, then F » p , FpΛqq.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 28 / 46

slide-29
SLIDE 29

STABILIZATION OF ADDITIVE FUNCTORS

Part 3. Stabilization of additive functors

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 29 / 46

slide-30
SLIDE 30

STABILIZATION OF ADDITIVE FUNCTORS

INJECTIVE STABILIZATION OF AN ADDITIVE FUNCTOR

Let F : Λ-Mod Ñ Ab be an additive covariant functor on left modules. DEFINITION The injective stabilization F of F is defined by the exact sequence 0 Ý Ñ F Ý Ñ F

ρF

Ý Ñ R0F REMARK F is additive as a subfunctor of the additive functor F. REMARK F is the largest subfunctor of F vanishing on injectives.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 30 / 46

slide-31
SLIDE 31

STABILIZATION OF ADDITIVE FUNCTORS

COMPUTING THE INJECTIVE STABILIZATION: THREE

EASY STEPS

Let B be a left Λ-module. To compute FpBq: embed B in an injective: 0 Ñ B

ι

Ñ I, apply F compute Ker Fpιq. Thus, FpBq is defined by the exact sequence 0 Ý Ñ FpBq Ý Ñ FpBq

Fpιq

Ý Ñ FpIq

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 31 / 46

slide-32
SLIDE 32

STABILIZATION OF ADDITIVE FUNCTORS

INJECTIVE STABILIZATION OF THE TENSOR PRODUCT

Change of notation: The injective stabilization of F :“ A b will be denoted by A

ã

b . Thus A

ã

b B “ pA

ã

b qpBq Terminology: A is inert, B is active. EXAMPLE Take Λ :“ Z. Then: Z

ã

b Q{Z “ 0 as Q{Z is injective (or Z is projective); Z

â

b Q{Z “ Q{Z (just tensor 0 Ñ Z Ñ Q with Q{Z). N.B. The harpoon always points to the active variable.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 32 / 46

slide-33
SLIDE 33

STABILIZATION OF ADDITIVE FUNCTORS

INJECTIVE STABILIZATION OF THE TENSOR PRODUCT

EXAMPLE If A is finitely presented, then (A-B, 1969) A

ã

b » Ext1pTrA, q

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 33 / 46

slide-34
SLIDE 34

STABILIZATION OF ADDITIVE FUNCTORS

PROJECTIVE STABILIZATION OF pA, q

Similar definitions apply to the remaining three choices for λ and ρ. EXAMPLE Let A be a left Λ-module. The projective stabilization of pA, q is just pA, q, the Hom modulo projectives. If A is finitely presented, then pA, q » Tor1pTr A, q

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 34 / 46

slide-35
SLIDE 35

STABILIZATION OF ADDITIVE FUNCTORS

PROJECTIVE STABILIZATION OF p , Bq

EXAMPLE Let B be a left Λ-module. The projective stabilization of p , Bq is just p , Bq, the Hom modulo injectives (sic!).

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 35 / 46

slide-36
SLIDE 36

STABILIZATION OF ADDITIVE FUNCTORS

STABILIZATION: SUMMARY

Projective resolutions Injective resolutions Covariant L0F

λF

Ý Ñ F Ý Ñ p

  • i

Ý Ñ F

ρF

Ý Ñ R0F Contravariant i Ý Ñ F

ρF

Ý Ñ R0F L0F

λF

Ý Ñ F Ý Ñ p

  • Projective stabilization = the cokernel of the counit λ.

Injective stabilization = the kernel of the unit ρ.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 36 / 46

slide-37
SLIDE 37

AN EXTENSION OF THE AR FORMULA

Part 4. An extension of the AR formula

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 37 / 46

slide-38
SLIDE 38

AN EXTENSION OF THE AR FORMULA

THE ORIGINAL AR FORMULA

REMARK Let Λ be an algebra over a commutative ring R (can be Z). Choose an injective R-module J and let DJ :“ HomRp , Jq. Let A1 be a finitely presented Λ-module. Then there is an isomorphism DJ Ext1pA1, Bq » pB, DJ Tr A1q, functorial in A1 and B.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 38 / 46

slide-39
SLIDE 39

AN EXTENSION OF THE AR FORMULA

AN AR FORMULA FOR ARBITRARY MODULES

PROPOSITION The tensor – covariant Hom adjunction induces an isomorphism DJpA

ã

b Bq » pB, DJpAqq, functorial in A and B.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 39 / 46

slide-40
SLIDE 40

AN EXTENSION OF THE AR FORMULA

PROOF

PROOF. Apply A b the injective envelope 0 Ñ B Ñ I: 0 Ý Ñ A

ã

b B Ý Ñ A b B Ý Ñ A b I Apply the exact functor DJ: pA b I, Jq Ý Ñ pA b B, Jq Ý Ñ pA

ã

b B, Jq Ý Ñ 0 Use the tensor-Hom adjunction: pI, DJpAqq Ý Ñ pB, DJpAqq Ý Ñ HompB, DJpAqq Ý Ñ 0

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 40 / 46

slide-41
SLIDE 41

R0F AND L0F REVISITED

Part 5. R0F and L0F revisited

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 41 / 46

slide-42
SLIDE 42

R0F AND L0F REVISITED

EXTENDING THE DEFINING SEQUENCE

Recall the defining sequence 0 Ý Ñ F Ý Ñ F

ρF

Ý Ñ R0F We want to construct a natural transformation βF : R0F Ý Ñ F ˝ Σ

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 42 / 46

slide-43
SLIDE 43

R0F AND L0F REVISITED

THE CONSTRUCTION

  • FpBq
  • FpBq
  • FpBq

ρF pBq

  • FpBq

Fpιq

  • R0FpBq

j

  • βF pBq
  • FpI0q

Fpd0q Fpeq

  • FpI1q

FpΣBq

κF pΣBq FpΣBq Fpmq FpI1q

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 43 / 46

slide-44
SLIDE 44

R0F AND L0F REVISITED

A FOUR-TERM SEQUENCE FOR R0

The leftmost column gives a sequence

F F

ρF R0F βF F ˝ Σ

0,

PROPOSITION

1

The above sequence is a complex, exact at F and F.

2

If F is half-exact, then the sequence is exact at R0F.

3

If F is epi-preserving, then the sequence is exact at F ˝ Σ, i.e., βF is epic.

4

If F is right-exact, then the sequence is exact.

5

If F is mono-preserving, then the end terms of the sequence vanish and ρF is monic.

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 44 / 46

slide-45
SLIDE 45

R0F AND L0F REVISITED

APPLICATIION

If F “ A b , then F is right-exact and the above sequence is exact. If, furthermore, A is finitely presented, then R0F » pA˚, q, F » Ext1pTr A, q, F ˝ Σ » Ext2pTr A, q, and we recover the formula

Ext1pTr A,

q

A b

ρAb pA˚,

q

Ext2pTr A,

q

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 45 / 46

slide-46
SLIDE 46

R0F AND L0F REVISITED

A FOUR-TERM SEQUENCE FOR L0

By reversing all arrows we get a similar complex 0 Ý Ñ F ˝ Ω Ý Ñ L0F Ý Ñ F Ý Ñ F Ý Ñ 0. If F is left-exact, then this complex is exact. If F :“ pA, q, and A is finitely presented, we recover the exact sequence

Tor2pTr A,

q

A˚ b

µpA,

q pA,

q

Tor1pTr A,

q

ALEX MARTSINKOVSKY DECONSTRUCTING AUSLANDER’S FORMULAS JULY 17, 2019 46 / 46