Hom and Ext, Revisited Hom and Ext, Revisited Justin Lyle Lawrence, KS justin.lyle@ku.edu April 28, 2018 JL Hom and Ext, Revisited
Hom and Ext, Revisited • Joint work with Hailong Dao and Mohammad Eghbali JL Hom and Ext, Revisited
Hom and Ext, Revisited • Throughout, let ( R , m , k ) be a commutative Noetherian local ring and M , N finitely generated R -modules. JL Hom and Ext, Revisited
Hom and Ext, Revisited • Throughout, let ( R , m , k ) be a commutative Noetherian local ring and M , N finitely generated R -modules. • There are several results (and open problems!) in the literature which take the form: if Hom R ( M , N ) has some “nice” properties and Ext n R ( M , N ) = 0 for some values of n , then M or N must be free or close to free. JL Hom and Ext, Revisited
Hom and Ext, Revisited Some Examples 1 If R is Gorenstein, dim R = 1, M is Maximal Cohen-Macaulay (MCM) and End R ( M ) is free, then M is free. ([Vas68]) JL Hom and Ext, Revisited
Hom and Ext, Revisited Some Examples 1 If R is Gorenstein, dim R = 1, M is Maximal Cohen-Macaulay (MCM) and End R ( M ) is free, then M is free. ([Vas68]) 2 Suppose R is Cohen-Macaulay (CM), I is a CM ideal of height 1 such that Hom R ( I , I ) is free and Ext 1 ≤ i ≤ d − 1 ( I , I ) = Ext 1 ≤ i ≤ d ( I , R ) = 0. Then M is free. R R ([GT17]) JL Hom and Ext, Revisited
Hom and Ext, Revisited Some Examples 1 If R is Gorenstein, dim R = 1, M is Maximal Cohen-Macaulay (MCM) and End R ( M ) is free, then M is free. ([Vas68]) 2 Suppose R is Cohen-Macaulay (CM), I is a CM ideal of height 1 such that Hom R ( I , I ) is free and Ext 1 ≤ i ≤ d − 1 ( I , I ) = Ext 1 ≤ i ≤ d ( I , R ) = 0. Then M is free. R R ([GT17]) 3 Suppose R is Cohen-Macaulay and is a complete intersection in codimension 1. Furthermore, assume that Q ⊆ R . If M is an R -module that is locally free in codimension one with constant rank, Ext 1 ≤ i ≤ d ( M , M ) = 0, and R Ext 1 ≤ i ≤ 2 d +1 ( M , R ) = 0, then M is free. [HL04] R JL Hom and Ext, Revisited
Hom and Ext, Revisited 2 and 3 are established cases of a famous open conjecture: JL Hom and Ext, Revisited
Hom and Ext, Revisited 2 and 3 are established cases of a famous open conjecture: Conjecture (Auslander-Reiten) If Ext i > 0 R ( M , R ) = Ext i > 0 R ( M , M ) = 0 then M is free. JL Hom and Ext, Revisited
Hom and Ext, Revisited To extend these results, we consider two main questions: JL Hom and Ext, Revisited
Hom and Ext, Revisited To extend these results, we consider two main questions: Question (1) When does Hom R ( M , N ) have a free summand? JL Hom and Ext, Revisited
Hom and Ext, Revisited To extend these results, we consider two main questions: Question (1) When does Hom R ( M , N ) have a free summand? Question (2) When is Hom R ( M , N ) ∼ = N r ? JL Hom and Ext, Revisited
Hom and Ext, Revisited To extend these results, we consider two main questions: Question (1) When does Hom R ( M , N ) have a free summand? Question (2) When is Hom R ( M , N ) ∼ = N r ? Our main technique in answering these questions is to consider two categories which behave nicely with respect to localizing and cutting down a regular sequence. JL Hom and Ext, Revisited
Hom and Ext, Revisited To extend these results, we consider two main questions: Question (1) When does Hom R ( M , N ) have a free summand? Question (2) When is Hom R ( M , N ) ∼ = N r ? Our main technique in answering these questions is to consider two categories which behave nicely with respect to localizing and cutting down a regular sequence. One nice feature of this approach is that we are often able to remove hypotheses such Cohen-Macaulayness and constant rank. JL Hom and Ext, Revisited
Hom and Ext, Revisited Two Key Categories Henceforth, we set t := depth R . We set Deep( R ) := { X | depth M ≥ t } . Our key categories are JL Hom and Ext, Revisited
Hom and Ext, Revisited Two Key Categories Henceforth, we set t := depth R . We set Deep( R ) := { X | depth M ≥ t } . Our key categories are Ω Deep( R ) := { M | ∃ 0 → M → R n → X → 0 exact with X ∈ Deep( R ) } and JL Hom and Ext, Revisited
Hom and Ext, Revisited Two Key Categories Henceforth, we set t := depth R . We set Deep( R ) := { X | depth M ≥ t } . Our key categories are Ω Deep( R ) := { M | ∃ 0 → M → R n → X → 0 exact with X ∈ Deep( R ) } and DF( R ) := { M | ∃ 0 → R → M n → X → 0 exact with X ∈ Deep( R ) } If M ∈ DF( R ) we say M is deeply faithful . JL Hom and Ext, Revisited
Hom and Ext, Revisited Observations • R is in Ω Deep( R ) ∩ DF( R ). JL Hom and Ext, Revisited
Hom and Ext, Revisited Observations • R is in Ω Deep( R ) ∩ DF( R ). • Any t -syzygy module is in Deep( R ) and so any ( t + 1)-syzygy is in Ω Deep( R ). JL Hom and Ext, Revisited
Hom and Ext, Revisited Observations • R is in Ω Deep( R ) ∩ DF( R ). • Any t -syzygy module is in Deep( R ) and so any ( t + 1)-syzygy is in Ω Deep( R ). • It is clear that for any X ∈ DF( R ), X is faithful, and when depth( R ) = 0 deeply faithful and faithful modules coincide. JL Hom and Ext, Revisited
Hom and Ext, Revisited Observations • R is in Ω Deep( R ) ∩ DF( R ). • Any t -syzygy module is in Deep( R ) and so any ( t + 1)-syzygy is in Ω Deep( R ). • It is clear that for any X ∈ DF( R ), X is faithful, and when depth( R ) = 0 deeply faithful and faithful modules coincide. • If M is a semi-dualizing module (that is, if the homothety map R → Hom R ( M , M ) is an isomorphism and Ext i > 0 R ( M , M ) = 0), then M ∈ DF( R ). JL Hom and Ext, Revisited
Hom and Ext, Revisited Observations • R is in Ω Deep( R ) ∩ DF( R ). • Any t -syzygy module is in Deep( R ) and so any ( t + 1)-syzygy is in Ω Deep( R ). • It is clear that for any X ∈ DF( R ), X is faithful, and when depth( R ) = 0 deeply faithful and faithful modules coincide. • If M is a semi-dualizing module (that is, if the homothety map R → Hom R ( M , M ) is an isomorphism and Ext i > 0 R ( M , M ) = 0), then M ∈ DF( R ). • Ω Deep( R ) and DF( R ) behave well with respect to localization and with cutting down a (general) regular sequence. JL Hom and Ext, Revisited
Hom and Ext, Revisited Some Lemmas In particular, JL Hom and Ext, Revisited
Hom and Ext, Revisited Some Lemmas In particular, Lemma (1) Suppose t > 0 . If M ∈ Ω Deep( R ) (resp. M ∈ DF( R ) ) then, for a general R-regular sequence x, we have M ∈ Ω Deep( R ) (resp. M ∈ DF( R ) ). JL Hom and Ext, Revisited
Hom and Ext, Revisited Some Lemmas Lemma (2) i 1 p 1 Let 0 → X − → Y − → Z → 0 be an exact sequence. JL Hom and Ext, Revisited
Hom and Ext, Revisited Some Lemmas Lemma (2) i 1 p 1 Let 0 → X − → Y − → Z → 0 be an exact sequence. 1 If Y ∈ Ω Deep( R ) and Z ∈ Deep( R ) , then X ∈ Ω Deep( R ) . 2 If X ∈ DF( R ) and Z ∈ Deep( R ) , then Y ∈ DF( R ) . JL Hom and Ext, Revisited
Hom and Ext, Revisited Some Lemmas Lemma (2) i 1 p 1 Let 0 → X − → Y − → Z → 0 be an exact sequence. 1 If Y ∈ Ω Deep( R ) and Z ∈ Deep( R ) , then X ∈ Ω Deep( R ) . 2 If X ∈ DF( R ) and Z ∈ Deep( R ) , then Y ∈ DF( R ) . Proof. Examine some pushout diagrams. JL Hom and Ext, Revisited
Hom and Ext, Revisited Some Lemmas Lemma (3) Assume that t = 0 , 0 � = M ∈ Ω Deep( R ) = Ω mod( R ) . The following are equivalent. 1 R | M. 2 M is faithful. 3 Soc( R ) � Ann( M ) . 4 M is not a minimal syzygy on R. JL Hom and Ext, Revisited
Hom and Ext, Revisited Some Lemmas Lemma (4) ( M , R ) = 0 . Then M ∗ free implies M is free. Suppose Ext 1 ≤ i ≤ t R JL Hom and Ext, Revisited
Hom and Ext, Revisited Theorem (A) Let M ∈ Deep( R ) and N ∈ Ω Deep( R ) . Assume that Hom R ( M , N ) ∈ DF( R ) and Ext 1 � i � t − 1 ( M , N ) = 0 . Then R | N. R JL Hom and Ext, Revisited
Hom and Ext, Revisited Theorem (A) Let M ∈ Deep( R ) and N ∈ Ω Deep( R ) . Assume that Hom R ( M , N ) ∈ DF( R ) and Ext 1 � i � t − 1 ( M , N ) = 0 . Then R | N. R This has a nice corollary, which directly extends (1). Corollary Let M ∈ Deep( R ) and N ∈ Ω Deep( R ) . Furthermore, assume that Hom R ( M , N ) is free and Ext 1 � i � t − 1 ( M , N ) = 0 . Then N is free. R JL Hom and Ext, Revisited
Hom and Ext, Revisited Theorem (A) Let M ∈ Deep( R ) and N ∈ Ω Deep( R ) . Assume that Hom R ( M , N ) ∈ DF( R ) and Ext 1 � i � t − 1 ( M , N ) = 0 . Then R | N. R This has a nice corollary, which directly extends (1). Corollary Let M ∈ Deep( R ) and N ∈ Ω Deep( R ) . Furthermore, assume that Hom R ( M , N ) is free and Ext 1 � i � t − 1 ( M , N ) = 0 . Then N is free. R Proof. Induct on µ ( N ), applying the Theorem repeatedly. JL Hom and Ext, Revisited
Hom and Ext, Revisited On the other side of the coin the Theorem gives the following: JL Hom and Ext, Revisited
Hom and Ext, Revisited On the other side of the coin the Theorem gives the following: Corollary Suppose M ∈ Deep( R ) , N ∈ Ω Deep( R ) , Hom( M , N ) is free, and Ext 1 ≤ i ≤ t ( M , N ) = 0 . Then M is free. Proof. = ( M ∗ ) n is The previous corollary implies N is free, so Hom( M , N ) ∼ free. Thus M is free by Lemma 4. JL Hom and Ext, Revisited
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