On the linearity defect of the residue field Liana S ega - PowerPoint PPT Presentation

On the linearity defect of the residue field Liana S ¸ega University of Missouri, Kansas City October 16, 2011

Linearity defect: definition ( R, m , k ) = commutative local Noetherian ring; m � = 0 M =finitely generated R -module; R g = ⊕ m i / m i +1 M g = ⊕ m i M/ m i +1 M. and

Linearity defect: definition ( R, m , k ) = commutative local Noetherian ring; m � = 0 M =finitely generated R -module; R g = ⊕ m i / m i +1 M g = ⊕ m i M/ m i +1 M. and Consider a minimal free resolution of M : d n F = · · · → F n +1 − → F n → · · · → F 0 → 0 and the filtration of F given by the subcomplexes: · · · → F n +1 → F n → · · · → F i → m F i − 1 → m 2 F i − 2 · · · → m i F 0 → 0

Linearity defect: definition ( R, m , k ) = commutative local Noetherian ring; m � = 0 M =finitely generated R -module; R g = ⊕ m i / m i +1 M g = ⊕ m i M/ m i +1 M. and Consider a minimal free resolution of M : d n F = · · · → F n +1 − → F n → · · · → F 0 → 0 and the filtration of F given by the subcomplexes: · · · → F n +1 → F n → · · · → F i → m F i − 1 → m 2 F i − 2 · · · → m i F 0 → 0 The associated graded complex is a complex of R g -modules: F g = · · · → F n +1 g ( − n − 1) → F n g ( − n ) → · · · → F 0 g → 0 (Herzog and Iyengar): the linearity defect of M is the number: ld R ( M ) = sup { i ∈ Z | H i ( F g ) � = 0 } .

Connections to regularity ⇒ F g is a minimal free resolution of M g . • ld R ( M ) = 0 ⇐ In this case, reg R g ( M g ) = 0 . We say that M is a Koszul module .

Connections to regularity ⇒ F g is a minimal free resolution of M g . • ld R ( M ) = 0 ⇐ In this case, reg R g ( M g ) = 0 . We say that M is a Koszul module . ⇒ R g is a Koszul algebra. • ld R ( k ) = 0 ⇐ We say that R is a Koszul ring .

Connections to regularity ⇒ F g is a minimal free resolution of M g . • ld R ( M ) = 0 ⇐ In this case, reg R g ( M g ) = 0 . We say that M is a Koszul module . ⇒ R g is a Koszul algebra. • ld R ( k ) = 0 ⇐ We say that R is a Koszul ring . • ld R ( M ) < ∞ iff M has a syzygy which is Koszul.

Interpretation If i > 0 , let µ n i ( M ) denote the natural map Tor R i ( m n +1 , M ) → Tor R i ( m n , M ) induced by the inclusion m n +1 ⊆ m n .

Interpretation If i > 0 , let µ n i ( M ) denote the natural map Tor R i ( m n +1 , M ) → Tor R i ( m n , M ) induced by the inclusion m n +1 ⊆ m n . Theorem. Let i > 0 . Then: H i ( F g ) = 0 ⇐ ⇒ µ n i ( M ) = 0 = µ n i − 1 ( M ) for all n > 0 .

Interpretation If i > 0 , let µ n i ( M ) denote the natural map Tor R i ( m n +1 , M ) → Tor R i ( m n , M ) induced by the inclusion m n +1 ⊆ m n . Theorem. Let i > 0 . Then: H i ( F g ) = 0 ⇐ ⇒ µ n i ( M ) = 0 = µ n i − 1 ( M ) for all n > 0 . ⇒ µ n • ld R ( M ) ≤ d ⇐ i ( M ) = 0 for all i ≥ d and all n > 0 . ⇒ µ n • ld R ( M ) = 0 ⇐ i ( M ) = 0 for all i and n

The Graded case When R is a standard graded k -algebra and M is a graded R -module, one can use the same definitions, with m = R � 1 . Herzog and Iyengar: ld R ( M ) < ∞ = ⇒ reg R ( M ) < ∞ In particular: ld R ( k ) < ∞ = ⇒ reg R ( k ) < ∞ , hence R is a Koszul algebra (Avramov and Peeva) and ld R ( k ) = 0 .

The Graded case When R is a standard graded k -algebra and M is a graded R -module, one can use the same definitions, with m = R � 1 . Herzog and Iyengar: ld R ( M ) < ∞ = ⇒ reg R ( M ) < ∞ In particular: ld R ( k ) < ∞ = ⇒ reg R ( k ) < ∞ , hence R is a Koszul algebra (Avramov and Peeva) and ld R ( k ) = 0 . An analysis of the proof reveals that a weaker hypothesis suffices: µ 1 Proposition. ≫ 0 ( M ) = 0 = ⇒ reg R ( M ) < ∞ . In particular, µ 1 ≫ 0 ( k ) = 0 = ⇒ R is Koszul. i : Tor R i ( m 2 , M ) → Tor R (Recall that µ 1 i ( m , M ) .)

Questions Back to the local case. • If ld R ( k ) < ∞ does it follow that ld R ( k ) = 0 ? (Herzog and Iyengar)

Questions Back to the local case. • If ld R ( k ) < ∞ does it follow that ld R ( k ) = 0 ? (Herzog and Iyengar) ≫ 0 = 0 does it follow that µ n = 0 ? • For any n : If µ n

Questions Back to the local case. • If ld R ( k ) < ∞ does it follow that ld R ( k ) = 0 ? (Herzog and Iyengar) ≫ 0 = 0 does it follow that µ n = 0 ? • For any n : If µ n • If ld R ( M ) < ∞ for every finitely generated R -module ( R is absolutely Koszul ), does it follow that R is Koszul?

The maps µ 1 and the Yoneda algebra Think of µ 1 i as Ext i +1 R ( k, k ) → Ext i +1 R ( R/ m 2 , k ) Set E = Ext R ( k, k ) , with Yoneda product. Set R ! =the subalgebra of E generated by its elements of degree 1 .

The maps µ 1 and the Yoneda algebra Think of µ 1 i as Ext i +1 R ( k, k ) → Ext i +1 R ( R/ m 2 , k ) Set E = Ext R ( k, k ) , with Yoneda product. Set R ! =the subalgebra of E generated by its elements of degree 1 . [J. E. Ross] The following statements are equivalent: • The Yoneda multiplication map E i ⊗ E 1 → E i +1 is surjective. • µ 1 i = 0 .

The maps µ 1 and the Yoneda algebra Think of µ 1 i as Ext i +1 R ( k, k ) → Ext i +1 R ( R/ m 2 , k ) Set E = Ext R ( k, k ) , with Yoneda product. Set R ! =the subalgebra of E generated by its elements of degree 1 . [J. E. Ross] The following statements are equivalent: • The Yoneda multiplication map E i ⊗ E 1 → E i +1 is surjective. • µ 1 i = 0 . We have thus: • µ 1 ⇒ E = R ! > 0 = 0 ⇐ ⇒ E is generated/ R ! by its elements of degree s . • µ 1 � s = 0 ⇐ In particular: If R is a standard graded algebra and E is finitely generated over R ! , then E = R ! and R is Koszul.

Set s ( R ) = inf { i ≥ 1 | a ∩ n i +2 ⊆ na } where � R = Q/ a is a minimal regular presentation of R with ( Q, n ) regular local and a ⊆ n 2 .

Set s ( R ) = inf { i ≥ 1 | a ∩ n i +2 ⊆ na } where � R = Q/ a is a minimal regular presentation of R with ( Q, n ) regular local and a ⊆ n 2 . The following hold: Proposition. (a) If µ 1 4 n − 1 = 0 for some positive integer n , then µ 1 3 = µ 1 1 = 0 (b) µ 1 1 = 0 ⇐ ⇒ s ( R ) = 1

Set s ( R ) = inf { i ≥ 1 | a ∩ n i +2 ⊆ na } where � R = Q/ a is a minimal regular presentation of R with ( Q, n ) regular local and a ⊆ n 2 . The following hold: Proposition. (a) If µ 1 4 n − 1 = 0 for some positive integer n , then µ 1 3 = µ 1 1 = 0 (b) µ 1 1 = 0 ⇐ ⇒ s ( R ) = 1 For the proof: Use the fact that k has a minimal free resolution F with DG Γ algebra structure, obtained by adjoining variables. Then think of µ n i as H i +1 ( F/ m 2 F ) → H i +1 ( F/ m F ) . Thus µ n i = 0 means: If dx ∈ m 2 F i , then x ∈ m F i +1 .

� � � � � � � We have thus: � E = R ! µ 1 > 0 = 0 E is generated over R ! µ 1 � s = 0 by elements of degree s � s ( R ) = 1 µ 1 1 = 0 �

Complete intersection rings Assume R is a complete intersection: � R = Q/ (regular sequence), ⇒ E = R ! . with Q regular local. For these rings: s ( R ) = 1 ⇐

Complete intersection rings Assume R is a complete intersection: � R = Q/ (regular sequence), ⇒ E = R ! . with Q regular local. For these rings: s ( R ) = 1 ⇐ If ld R ( k ) < ∞ , then E = R ! . Proposition.

Complete intersection rings Assume R is a complete intersection: � R = Q/ (regular sequence), ⇒ E = R ! . with Q regular local. For these rings: s ( R ) = 1 ⇐ If ld R ( k ) < ∞ , then E = R ! . Proposition. Under a stronger hypothesis, we obtain a stronger conclusion: Theorem The following statements are equivalent: (a) ld R k = 0 ( R is Koszul) (b) R has minimal multiplicity. Furthermore, if R g is Cohen-Macaulay, then they are also equivalent to (c) ld R k < ∞

Complete intersection rings Assume R is a complete intersection: � R = Q/ (regular sequence), ⇒ E = R ! . with Q regular local. For these rings: s ( R ) = 1 ⇐ If ld R ( k ) < ∞ , then E = R ! . Proposition. Under a stronger hypothesis, we obtain a stronger conclusion: Theorem The following statements are equivalent: (a) ld R k = 0 ( R is Koszul) (b) R has minimal multiplicity. Furthermore, if R g is Cohen-Macaulay, then they are also equivalent to (c) ld R k < ∞ Proof of (c) = ⇒ (b): Reduce first to the Artinian case. Then, a length count.

Artinian rings Theorem Assume R is Artinian with m n +1 = 0 . If µ n − 1 ≫ 0 = 0 , then µ n − 1 > 0 = 0 .

Artinian rings Theorem Assume R is Artinian with m n +1 = 0 . If µ n − 1 ≫ 0 = 0 , then µ n − 1 > 0 = 0 . Corollary If R is Golod and R g is Cohen-Macaulay, then the following statements are equivalent: (a) ld R k = 0 (b) ld R k < ∞ (c) R has minimal multiplicity ( codim R = e ( R ) − 1 )

Artinian rings Theorem Assume R is Artinian with m n +1 = 0 . If µ n − 1 ≫ 0 = 0 , then µ n − 1 > 0 = 0 . Corollary If R is Golod and R g is Cohen-Macaulay, then the following statements are equivalent: (a) ld R k = 0 (b) ld R k < ∞ (c) R has minimal multiplicity ( codim R = e ( R ) − 1 ) The corollary follows from the Theorem, using the fact that an Artinian Golod ring does not have any non-zero small ideals.