Linearity defect, Castelnuovo-Mumford Regularity and Rate...

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Linearity defect, Castelnuovo-MumfordRegularity and Rate of modules

Rasoul Ahangari Maleki

June 18, 2014

Rasoul Ahangari Maleki Linearity defect, Castelnuovo-Mumford Regularity and Rate of modules

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Contents

(1) History(2) What I did.(3) What I am going to do.


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HistoryWhat I did

Contents

(1) History(2) What I did.(3) What I am going to do.


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In this presentation rings are commutative noetherian local (orstandard graded algebras over a field), and modules are finitelygenerated (or graded finitely generated).


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In this presentation rings are commutative noetherian local (orstandard graded algebras over a field), and modules are finitelygenerated (or graded finitely generated).


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Let (R,m, k) be a local ring (or standard graded k -algebra), Man R-module and F→ M be the minimal free resolution of M.

(1) PRM(t) =

∑i dimk TorR

i (M, k)t i

(2) HM(z) =∑

i∈Z dimk (Mi)z i (M is graded)

(3) regR(M) = sup{j − i : and TorRi (M, k)j 6= 0, j ∈ Z, i ∈ N}

(M is graded).


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TorR(k , k) has a graded algebra structure and TorR(M, k) is agraded module over TorR(k , k). So

PRM(t) = HTorR(M,k)(z)


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TorR(k , k) has a graded algebra structure and TorR(M, k) is agraded module over TorR(k , k). So

PRM(t) = HTorR(M,k)(z)


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Rationality of PRM(t)

Definition

PRM(t) is said to be rational if PR

M(t) = f (t)/g(t) ∈ C(t),

A rational expression for PRM(t) has practical applications.

(1) it provides recurrent relation for Betti numbers that can beuseful in constructing a minimal resolution.

(2) it allows for a efficient estimate of asymptotic behavior ofBetti sequences.


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Question(Serre, Kaplasky, Shafarevich,...)Dose PR

k (t) is a rational function?

D. Anick (1982): Gave a negative answer to the question.∃(R,m)s.t m3 = 0, PR

k (t) is irrational.J. Tate (1957): If R is a complete intersection, then PR

k (t) isrational.C. Jacobsson (1985) shows that the rationality of PR

k (t) dosenot imply the rationality of PR

M(t) for all R-module M.


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Question(Serre, Kaplasky, Shafarevich,...)Dose PR

k (t) is a rational function?

D. Anick (1982): Gave a negative answer to the question.∃(R,m)s.t m3 = 0, PR

k (t) is irrational.J. Tate (1957): If R is a complete intersection, then PR

k (t) isrational.C. Jacobsson (1985) shows that the rationality of PR

k (t) dosenot imply the rationality of PR

M(t) for all R-module M.


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However counterexamples do not seem abound.


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However counterexamples do not seem abound.


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Problems

(1) Over which R do all modules have rational poincare series.(2) Do all rational PR

M(t) over R have a common denominator?(3) (1) and (2).

G. Levin (1985) For the problems (1) and (2) it suffices toconsider modules of finite length.


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Problems

(1) Over which R do all modules have rational poincare series.(2) Do all rational PR

M(t) over R have a common denominator?(3) (1) and (2).

G. Levin (1985) For the problems (1) and (2) it suffices toconsider modules of finite length.


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Castelnuovo-Mumford Regularity

By definition regR(M) can be infinite.

QuestionOver which R, regR(M) <∞ for all R-module M.

The following are equivalent:

(Avramov-Eisenbud-Peeva).The following conditions are equivalent:(1) regR(M) is finite for every R-module M;(2) regR(K ) is finite;(3) R is Koszul (regR(K ) = 0 ).


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R is a Koszul algebra⇔ PRk (t) = 1/HR(−t) (so PR

k (t) isrational)


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DefinitionLet F be the minimal free resolution of M. We can construct acomplex of graded modules.

lin(F) : · · · → F gn (−n)→ F g

n−1(−n + 1) :→ · · ·F g0 (0)→ Mg → 0

where Ng = grm(N)

ldR(M) := sup{i ∈ N0 : Hi(lin(F)) 6= 0}(linearity defect).

M is said to be Koszul if ldR(M) = 0. This is equivalent to saythat Mg has a linear resolution.


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(1) ldR(M) ≤ d ⇔ Syzd(M) is Koszul(2) ldR(M) ≤ ∞→ PR

M(t) = q(t)/DR(t)


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Question(1) Over which ring R all modules have finite linearity defect.(2) What is the relation between ldR(M) and regR(M)

DefinitionR is said to be absolutely Koszul if ldR(M) for all R-module M.


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Question(1) Over which ring R all modules have finite linearity defect.(2) What is the relation between ldR(M) and regR(M)

DefinitionR is said to be absolutely Koszul if ldR(M) for all R-module M.


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Paper 1 Aim: Study the regularity and Koszulness of modulesover local rings


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Let (R,m, k) be a local ring which has an element x 6= 0 = x2

and xm = m2. Then R is absolutely Koszul (Avramov- Iyengar-Sega)

1 M is Koszul if xM = 0;2 regR(M) ≤ 1;3 mM is Koszul;4 For every ideal I of R the quotient ring R/I is a Koszul ring.


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Let (R,m, k) be a Cohen-Macaulay local ring of dimensiond > 0 and of minimal multiplicity. Then any R-moduleannihilated by a minimal reduction of m is a Koszul module. Inparticular, R is a Koszul ring.


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Let (R,m, k) be a Cohen-Macaulay local ring of dimensiond > 0 and of minimal multiplicity. Then any R-moduleannihilated by a minimal reduction of m is a Koszul module. Inparticular, R is a Koszul ring.


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Let (R,m) be a local ring and M be an R-module. IfAnn(x)m = m2, for all x ∈ m \m2, then the following hold.(i) M is Koszul if there exists x ∈ m\m2 such that Ann(x)M=0;

in particular, R is a Koszul ring.(ii) If x1, · · · , xs are in m \m2, then (x1, · · · , xs)M is Koszul;(iii) regR(M) ≤ 1;(iv) for every ideal I ⊆ m the quotient ring R/I is Koszul;


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Paper 2 (Joint with Rossi) Aim: study the relation betweenldR(M) and regR(M) and answer to an question asked byHerezog and Iyengar.


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(Rossi,-)Let R be a standard graded algebra and let M be a gradedR-module with D(M) = {i1, · · · , is}. Assume that M is Koszul.Then for each n ≥ 1 we have

TorRn (M, k)j = 0 for j 6= i1 + n, · · · , is + n.

Furthermoreif for some n ≥ 1 and 1 ≤ r ≤ s, we have TorR

n (M, k)ir+n = 0,then TorR

m(M, k)ir+m = 0, for all m ≥ n.


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(Rossi,-)Let R be a standard graded algebra and let M be a gradedR-module with D(M) = {i1, · · · , is}. Assume that M is Koszul.Then for each n ≥ 1 we have

TorRn (M, k)j = 0 for j 6= i1 + n, · · · , is + n.

Furthermoreif for some n ≥ 1 and 1 ≤ r ≤ s, we have TorR

n (M, k)ir+n = 0,then TorR

m(M, k)ir+m = 0, for all m ≥ n.


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(Rossi,-)Let R be a standard graded algebra and let M be a R-module.If ldR(M) = d <∞ then

regR(M) = max{ti(M)− i : 0 ≤ i ≤ d}.

In particular, if M is Koszul then regR(M) = t0(M).


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(Herzog, Iyengar)

Let R be a standard graded k -algebra. If ld(k) <∞ thenld(k) = 0.


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QuestionHerzog, Iyengar (2005) Let (R,m, k) be a a local ring. Ifld(k) <∞ then is ld(k) = 0?.


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positive Answers

(i) (Sega 2013) m3 = 0(ii) R is complete intersection and Rg is CM(iii) (Rossi,- (2013)) R is of homogeneous type.


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DefinitionBackelin Let R be a standard graded k-algebra

Rate(R) := sup{(tRi (K )− 1)/i − 1 : i ≥ 2}

Herzog , Aramova:

rateR(M) := sup{tRi (M)/i : i ≥ 1}.

A comparison with Bakelin’s rate shows thatRate(R) = rateR(m(1)). The rate of any module is finite.


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Backelinfor all d ≥ 1

Rate R(d) ≤ dRate(R)/de

Eisenbud, Reevse and Totaro started their work (initial ideals,veronese subrings,...) from a request by George Kempf for asimpler proof. They showed for d >> 0 R(d) admits aquadratic initial ideals.


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Jahangiri,-for all d ≥ 1

rateR(d)(M(d)) ≤ max{drateR(M)/de, dtR0 (M)/de}

A direct consequence of this is Backelin’s Theorem.If d ≥ rateR(M), then M(d) is a Koszul module over R(d).


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Thanks .


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