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Understanding How We Get Koszul Complexes From Regular Sequences

Local-global properties (localization): free, projective, injective, flat, torsion-free, etc?Homology of Chain Complexes from Free ResolutionNeed Counterexample to show Koszul complex is not minimal free resolution?Show that a sequence is a free resolutionMinimal free resolution of ideal generated by three homogeneous polynomialsRegular element of a Noetherian ringCharacterization of sequences which are regular on some moduleTensor product of minimal resolutions is a minimal resolutionregular sequences: proving the geometric interpretationAny module over a regular local ring has finite free resolution

2

$begingroup$

Let $ R $ be a ring, and $ M $ be an $ R $-module. We say that a sequence $ x_1,dots,x_r $ of elements of $ R $ is regular if:

1) $ (f_1,dots,f_r)M neq M, $ and

2) $ f_i $ is a non-zero divisor on $ M/(f_1,dots, f_i-1)M $ for each $ i = 1,dots,r. $

Apparently, it is the case that whenever we have a regular sequence $ f_1,dots,f_r subset R, $ there is an induced Koszul complex which is a free resolution of $ R/(f_1,dots,f_r). $

I'm not quite sure what the form of this complex is, or indeed how to construct it.

abstract-algebra commutative-algebra homological-algebra

share|cite|improve this question

edited Apr 1 at 19:55

user26857

39.5k124284

asked Apr 1 at 15:39

Addled StudentAddled Student

749

$endgroup$

add a comment | 

2

$begingroup$

Let $ R $ be a ring, and $ M $ be an $ R $-module. We say that a sequence $ x_1,dots,x_r $ of elements of $ R $ is regular if:

1) $ (f_1,dots,f_r)M neq M, $ and

2) $ f_i $ is a non-zero divisor on $ M/(f_1,dots, f_i-1)M $ for each $ i = 1,dots,r. $

Apparently, it is the case that whenever we have a regular sequence $ f_1,dots,f_r subset R, $ there is an induced Koszul complex which is a free resolution of $ R/(f_1,dots,f_r). $

I'm not quite sure what the form of this complex is, or indeed how to construct it.

abstract-algebra commutative-algebra homological-algebra

share|cite|improve this question

edited Apr 1 at 19:55

user26857

39.5k124284

asked Apr 1 at 15:39

Addled StudentAddled Student

749

$endgroup$

add a comment | 

$begingroup$

Let $ R $ be a ring, and $ M $ be an $ R $-module. We say that a sequence $ x_1,dots,x_r $ of elements of $ R $ is regular if:

1) $ (f_1,dots,f_r)M neq M, $ and

2) $ f_i $ is a non-zero divisor on $ M/(f_1,dots, f_i-1)M $ for each $ i = 1,dots,r. $

Apparently, it is the case that whenever we have a regular sequence $ f_1,dots,f_r subset R, $ there is an induced Koszul complex which is a free resolution of $ R/(f_1,dots,f_r). $

I'm not quite sure what the form of this complex is, or indeed how to construct it.

abstract-algebra commutative-algebra homological-algebra

share|cite|improve this question

edited Apr 1 at 19:55

user26857

39.5k124284

asked Apr 1 at 15:39

Addled StudentAddled Student

749

$endgroup$

share|cite|improve this question

edited Apr 1 at 19:55

user26857

39.5k124284

asked Apr 1 at 15:39

Addled StudentAddled Student

749

share|cite|improve this question

edited Apr 1 at 19:55

user26857

39.5k124284

asked Apr 1 at 15:39

Addled StudentAddled Student

749

edited Apr 1 at 19:55

user26857

39.5k124284

edited Apr 1 at 19:55

user26857

39.5k124284

asked Apr 1 at 15:39

Addled StudentAddled Student

749

asked Apr 1 at 15:39

Addled StudentAddled Student

749

1 Answer
1

active

oldest

votes

2

$begingroup$

If $x_1,dots,x_r$ is a sequence of elements of $R$, define for each $1le i le r$ the Koszul complex $mathcal K_bullet^(i)(x_i; R)$ by
$$0 to R xrightarrowcdot x_i R to 0,$$
with the right hand copy of $R$ in degree zero, i.e., the right hand copy of $R$ is $mathcal K_0^(i)(x_i; R)$, and the left hand copy in degree 1. Let $d^(i)$ be the differential map on this sequence.

The Koszul complex $mathcal K_bullet(x_1,dots,x_r; R)$ is defined in degree $n$ by
$$bigoplus_i_1+dots+i_r = n mathcal K_i_1^(1)(x_1; R)otimes_R dots otimes_R mathcal K_i_r^(r)(x_r; R)$$
with differential in degree-$n$ given by the direct sum of the differentials $d_i_1,dots,i_r:mathcal K_i_1^(1)(x_1; R)otimes_R dots otimes_R mathcal K_i_r^(r)(x_r; R) to mathcal K_n-1(x_1,dots,x_r; R)$, with $i_1+ dots + i_r = n$, defined on simple tensors by
$$d_i_1,dots,i_r(u_i_1^(1)otimes dots otimes u_i_r^(r)) = sum_j=1^r (-1)^i_1+dots+i_j-1; u_i_1^(1) otimes dots otimes u_i_j-1^(j-1)otimes d^(j)left(u_i_j^(j)right) otimes u_i_j+1^(j+1) otimesdots otimes u_i_r^(r).$$

It is a theorem that whenever $x_1, dots, x_r$ satisfy

• 
  $x_i$ is a nonzerodivisor on $R/(x_1,dots,x_i-1)$ for $1le i le r$,

then $mathcal K_bullet(x_1,dots,x_r; R)$ is a free resolution of $R/(x_1,dots,x_r)$.

share|cite|improve this answer

edited Apr 1 at 19:56

user26857

39.5k124284

answered Apr 1 at 17:11

cspruncsprun

2,805211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you give me a reference to this theorem?
  $endgroup$
  – Addled Student
  yesterday
  

  
  
  
  

add a comment | 

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2

$begingroup$

If $x_1,dots,x_r$ is a sequence of elements of $R$, define for each $1le i le r$ the Koszul complex $mathcal K_bullet^(i)(x_i; R)$ by
$$0 to R xrightarrowcdot x_i R to 0,$$
with the right hand copy of $R$ in degree zero, i.e., the right hand copy of $R$ is $mathcal K_0^(i)(x_i; R)$, and the left hand copy in degree 1. Let $d^(i)$ be the differential map on this sequence.

The Koszul complex $mathcal K_bullet(x_1,dots,x_r; R)$ is defined in degree $n$ by
$$bigoplus_i_1+dots+i_r = n mathcal K_i_1^(1)(x_1; R)otimes_R dots otimes_R mathcal K_i_r^(r)(x_r; R)$$
with differential in degree-$n$ given by the direct sum of the differentials $d_i_1,dots,i_r:mathcal K_i_1^(1)(x_1; R)otimes_R dots otimes_R mathcal K_i_r^(r)(x_r; R) to mathcal K_n-1(x_1,dots,x_r; R)$, with $i_1+ dots + i_r = n$, defined on simple tensors by
$$d_i_1,dots,i_r(u_i_1^(1)otimes dots otimes u_i_r^(r)) = sum_j=1^r (-1)^i_1+dots+i_j-1; u_i_1^(1) otimes dots otimes u_i_j-1^(j-1)otimes d^(j)left(u_i_j^(j)right) otimes u_i_j+1^(j+1) otimesdots otimes u_i_r^(r).$$

It is a theorem that whenever $x_1, dots, x_r$ satisfy

• 
  $x_i$ is a nonzerodivisor on $R/(x_1,dots,x_i-1)$ for $1le i le r$,

then $mathcal K_bullet(x_1,dots,x_r; R)$ is a free resolution of $R/(x_1,dots,x_r)$.

share|cite|improve this answer

edited Apr 1 at 19:56

user26857

39.5k124284

answered Apr 1 at 17:11

cspruncsprun

2,805211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you give me a reference to this theorem?
  $endgroup$
  – Addled Student
  yesterday
  

  
  
  
  

add a comment | 

2

$begingroup$

If $x_1,dots,x_r$ is a sequence of elements of $R$, define for each $1le i le r$ the Koszul complex $mathcal K_bullet^(i)(x_i; R)$ by
$$0 to R xrightarrowcdot x_i R to 0,$$
with the right hand copy of $R$ in degree zero, i.e., the right hand copy of $R$ is $mathcal K_0^(i)(x_i; R)$, and the left hand copy in degree 1. Let $d^(i)$ be the differential map on this sequence.

The Koszul complex $mathcal K_bullet(x_1,dots,x_r; R)$ is defined in degree $n$ by
$$bigoplus_i_1+dots+i_r = n mathcal K_i_1^(1)(x_1; R)otimes_R dots otimes_R mathcal K_i_r^(r)(x_r; R)$$
with differential in degree-$n$ given by the direct sum of the differentials $d_i_1,dots,i_r:mathcal K_i_1^(1)(x_1; R)otimes_R dots otimes_R mathcal K_i_r^(r)(x_r; R) to mathcal K_n-1(x_1,dots,x_r; R)$, with $i_1+ dots + i_r = n$, defined on simple tensors by
$$d_i_1,dots,i_r(u_i_1^(1)otimes dots otimes u_i_r^(r)) = sum_j=1^r (-1)^i_1+dots+i_j-1; u_i_1^(1) otimes dots otimes u_i_j-1^(j-1)otimes d^(j)left(u_i_j^(j)right) otimes u_i_j+1^(j+1) otimesdots otimes u_i_r^(r).$$

It is a theorem that whenever $x_1, dots, x_r$ satisfy

• 
  $x_i$ is a nonzerodivisor on $R/(x_1,dots,x_i-1)$ for $1le i le r$,

then $mathcal K_bullet(x_1,dots,x_r; R)$ is a free resolution of $R/(x_1,dots,x_r)$.

share|cite|improve this answer

edited Apr 1 at 19:56

user26857

39.5k124284

answered Apr 1 at 17:11

cspruncsprun

2,805211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you give me a reference to this theorem?
  $endgroup$
  – Addled Student
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

If $x_1,dots,x_r$ is a sequence of elements of $R$, define for each $1le i le r$ the Koszul complex $mathcal K_bullet^(i)(x_i; R)$ by
$$0 to R xrightarrowcdot x_i R to 0,$$
with the right hand copy of $R$ in degree zero, i.e., the right hand copy of $R$ is $mathcal K_0^(i)(x_i; R)$, and the left hand copy in degree 1. Let $d^(i)$ be the differential map on this sequence.

The Koszul complex $mathcal K_bullet(x_1,dots,x_r; R)$ is defined in degree $n$ by
$$bigoplus_i_1+dots+i_r = n mathcal K_i_1^(1)(x_1; R)otimes_R dots otimes_R mathcal K_i_r^(r)(x_r; R)$$
with differential in degree-$n$ given by the direct sum of the differentials $d_i_1,dots,i_r:mathcal K_i_1^(1)(x_1; R)otimes_R dots otimes_R mathcal K_i_r^(r)(x_r; R) to mathcal K_n-1(x_1,dots,x_r; R)$, with $i_1+ dots + i_r = n$, defined on simple tensors by
$$d_i_1,dots,i_r(u_i_1^(1)otimes dots otimes u_i_r^(r)) = sum_j=1^r (-1)^i_1+dots+i_j-1; u_i_1^(1) otimes dots otimes u_i_j-1^(j-1)otimes d^(j)left(u_i_j^(j)right) otimes u_i_j+1^(j+1) otimesdots otimes u_i_r^(r).$$

It is a theorem that whenever $x_1, dots, x_r$ satisfy

• 
  $x_i$ is a nonzerodivisor on $R/(x_1,dots,x_i-1)$ for $1le i le r$,

then $mathcal K_bullet(x_1,dots,x_r; R)$ is a free resolution of $R/(x_1,dots,x_r)$.

share|cite|improve this answer

edited Apr 1 at 19:56

user26857

39.5k124284

answered Apr 1 at 17:11

cspruncsprun

2,805211

$endgroup$

share|cite|improve this answer

edited Apr 1 at 19:56

user26857

39.5k124284

answered Apr 1 at 17:11

cspruncsprun

2,805211

edited Apr 1 at 19:56

user26857

39.5k124284

edited Apr 1 at 19:56

user26857

39.5k124284

answered Apr 1 at 17:11

cspruncsprun

2,805211

answered Apr 1 at 17:11

cspruncsprun

2,805211