Mixing
Koszul Complex of Powers of Elements
Clash Royale CLAN TAG#URR8PPP
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For a sequence of elements $underlinea= a_1, ..., a_r in R$, let $K^bullet(underlinea;R)$ be the Koszul complex generated by $underlinea$. Let $underlinea^v = a_1^v_1, ..., a_r^v_r in R$.
My question: if $K^bullet(underlinea;R)$ is acylic is $K^bullet(underlinea^v;R)$ acylic?
My thoughts: if $underlinea$ is regular and R is nice enough this is true. When R is nice, regular sequences always give acyclic Koszul complexes and powers of regular sequences are regular. So this means and meaningful counterexamples would have to avoid the case of $R$ Noetherian, local and $R$ Noetherian,graded with $underlinea$ homogeneous of positive degree.
commutative-algebra
asked Jul 20 at 12:51
do_math
365
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up vote
1
down vote
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For a sequence of elements $underlinea= a_1, ..., a_r in R$, let $K^bullet(underlinea;R)$ be the Koszul complex generated by $underlinea$. Let $underlinea^v = a_1^v_1, ..., a_r^v_r in R$.
My question: if $K^bullet(underlinea;R)$ is acylic is $K^bullet(underlinea^v;R)$ acylic?
My thoughts: if $underlinea$ is regular and R is nice enough this is true. When R is nice, regular sequences always give acyclic Koszul complexes and powers of regular sequences are regular. So this means and meaningful counterexamples would have to avoid the case of $R$ Noetherian, local and $R$ Noetherian,graded with $underlinea$ homogeneous of positive degree.
commutative-algebra
asked Jul 20 at 12:51
do_math
365
1
If $R$ is noetherian, then the answer is positive via the Buchsbaum-Eisenbud acyclicity criterion, and if I'm not mistaken this holds for the non-noetherian case, too, due to Northcott's generalization of the above mentioned criterion.
– user26857
Jul 20 at 18:13
This gives me hope but I don't quite see the argument. For the Buchsbaum-Eisenbud acyclicity criterion: if $M_i$ is the matrix of $K(underlinea;R)$ corresponding to the $i$th differential and $N_i$ the matrix of $K(underlinea^v;R)$ corresponding to the $i$th differential, then $N_i$ should be obtained from $M_i$ by taking all elements and raising it to the appropriate power. How should I relate the fitting ideal $F_j(N_i)$ to the fitting ideal $F_j(M_i)$? Am I missing some way to manipulate determinants? (Thanks for your response by the way.)
– do_math
Jul 20 at 18:37
Were you thinking that after accounting for the rank condition the appropriate fitting ideals would be "monomial" ideals? ("Monomial" in the sense that they are generated by products of the $a_i$'s.) Maybe I should test and see if that happens for $r = 3, 4$.
– do_math
Jul 20 at 20:52
I think 1.6.30 from Bruns and Herzog does the job.
– user26857
Jul 21 at 10:06
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For a sequence of elements $underlinea= a_1, ..., a_r in R$, let $K^bullet(underlinea;R)$ be the Koszul complex generated by $underlinea$. Let $underlinea^v = a_1^v_1, ..., a_r^v_r in R$.
My question: if $K^bullet(underlinea;R)$ is acylic is $K^bullet(underlinea^v;R)$ acylic?
My thoughts: if $underlinea$ is regular and R is nice enough this is true. When R is nice, regular sequences always give acyclic Koszul complexes and powers of regular sequences are regular. So this means and meaningful counterexamples would have to avoid the case of $R$ Noetherian, local and $R$ Noetherian,graded with $underlinea$ homogeneous of positive degree.
commutative-algebra
asked Jul 20 at 12:51
do_math
365
For a sequence of elements $underlinea= a_1, ..., a_r in R$, let $K^bullet(underlinea;R)$ be the Koszul complex generated by $underlinea$. Let $underlinea^v = a_1^v_1, ..., a_r^v_r in R$.
My question: if $K^bullet(underlinea;R)$ is acylic is $K^bullet(underlinea^v;R)$ acylic?
My thoughts: if $underlinea$ is regular and R is nice enough this is true. When R is nice, regular sequences always give acyclic Koszul complexes and powers of regular sequences are regular. So this means and meaningful counterexamples would have to avoid the case of $R$ Noetherian, local and $R$ Noetherian,graded with $underlinea$ homogeneous of positive degree.
commutative-algebra
asked Jul 20 at 12:51
do_math
365
edited Jul 20 at 12:59
asked Jul 20 at 12:51
do_math
365
asked Jul 20 at 12:51
do_math
365
asked Jul 20 at 12:51
do_math
365
365
1
If $R$ is noetherian, then the answer is positive via the Buchsbaum-Eisenbud acyclicity criterion, and if I'm not mistaken this holds for the non-noetherian case, too, due to Northcott's generalization of the above mentioned criterion.
– user26857
Jul 20 at 18:13
This gives me hope but I don't quite see the argument. For the Buchsbaum-Eisenbud acyclicity criterion: if $M_i$ is the matrix of $K(underlinea;R)$ corresponding to the $i$th differential and $N_i$ the matrix of $K(underlinea^v;R)$ corresponding to the $i$th differential, then $N_i$ should be obtained from $M_i$ by taking all elements and raising it to the appropriate power. How should I relate the fitting ideal $F_j(N_i)$ to the fitting ideal $F_j(M_i)$? Am I missing some way to manipulate determinants? (Thanks for your response by the way.)
– do_math
Jul 20 at 18:37
Were you thinking that after accounting for the rank condition the appropriate fitting ideals would be "monomial" ideals? ("Monomial" in the sense that they are generated by products of the $a_i$'s.) Maybe I should test and see if that happens for $r = 3, 4$.
– do_math
Jul 20 at 20:52
I think 1.6.30 from Bruns and Herzog does the job.
– user26857
Jul 21 at 10:06
add a comment |Â
1
If $R$ is noetherian, then the answer is positive via the Buchsbaum-Eisenbud acyclicity criterion, and if I'm not mistaken this holds for the non-noetherian case, too, due to Northcott's generalization of the above mentioned criterion.
– user26857
Jul 20 at 18:13
This gives me hope but I don't quite see the argument. For the Buchsbaum-Eisenbud acyclicity criterion: if $M_i$ is the matrix of $K(underlinea;R)$ corresponding to the $i$th differential and $N_i$ the matrix of $K(underlinea^v;R)$ corresponding to the $i$th differential, then $N_i$ should be obtained from $M_i$ by taking all elements and raising it to the appropriate power. How should I relate the fitting ideal $F_j(N_i)$ to the fitting ideal $F_j(M_i)$? Am I missing some way to manipulate determinants? (Thanks for your response by the way.)
– do_math
Jul 20 at 18:37
Were you thinking that after accounting for the rank condition the appropriate fitting ideals would be "monomial" ideals? ("Monomial" in the sense that they are generated by products of the $a_i$'s.) Maybe I should test and see if that happens for $r = 3, 4$.
– do_math
Jul 20 at 20:52
I think 1.6.30 from Bruns and Herzog does the job.
– user26857
Jul 21 at 10:06
1
1
If $R$ is noetherian, then the answer is positive via the Buchsbaum-Eisenbud acyclicity criterion, and if I'm not mistaken this holds for the non-noetherian case, too, due to Northcott's generalization of the above mentioned criterion.
– user26857
Jul 20 at 18:13
If $R$ is noetherian, then the answer is positive via the Buchsbaum-Eisenbud acyclicity criterion, and if I'm not mistaken this holds for the non-noetherian case, too, due to Northcott's generalization of the above mentioned criterion.
– user26857
Jul 20 at 18:13
This gives me hope but I don't quite see the argument. For the Buchsbaum-Eisenbud acyclicity criterion: if $M_i$ is the matrix of $K(underlinea;R)$ corresponding to the $i$th differential and $N_i$ the matrix of $K(underlinea^v;R)$ corresponding to the $i$th differential, then $N_i$ should be obtained from $M_i$ by taking all elements and raising it to the appropriate power. How should I relate the fitting ideal $F_j(N_i)$ to the fitting ideal $F_j(M_i)$? Am I missing some way to manipulate determinants? (Thanks for your response by the way.)
– do_math
Jul 20 at 18:37
This gives me hope but I don't quite see the argument. For the Buchsbaum-Eisenbud acyclicity criterion: if $M_i$ is the matrix of $K(underlinea;R)$ corresponding to the $i$th differential and $N_i$ the matrix of $K(underlinea^v;R)$ corresponding to the $i$th differential, then $N_i$ should be obtained from $M_i$ by taking all elements and raising it to the appropriate power. How should I relate the fitting ideal $F_j(N_i)$ to the fitting ideal $F_j(M_i)$? Am I missing some way to manipulate determinants? (Thanks for your response by the way.)
– do_math
Jul 20 at 18:37
Were you thinking that after accounting for the rank condition the appropriate fitting ideals would be "monomial" ideals? ("Monomial" in the sense that they are generated by products of the $a_i$'s.) Maybe I should test and see if that happens for $r = 3, 4$.
– do_math
Jul 20 at 20:52
Were you thinking that after accounting for the rank condition the appropriate fitting ideals would be "monomial" ideals? ("Monomial" in the sense that they are generated by products of the $a_i$'s.) Maybe I should test and see if that happens for $r = 3, 4$.
– do_math
Jul 20 at 20:52
I think 1.6.30 from Bruns and Herzog does the job.
– user26857
Jul 21 at 10:06
I think 1.6.30 from Bruns and Herzog does the job.
– user26857
Jul 21 at 10:06
add a comment |Â
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1
If $R$ is noetherian, then the answer is positive via the Buchsbaum-Eisenbud acyclicity criterion, and if I'm not mistaken this holds for the non-noetherian case, too, due to Northcott's generalization of the above mentioned criterion.
– user26857
Jul 20 at 18:13
This gives me hope but I don't quite see the argument. For the Buchsbaum-Eisenbud acyclicity criterion: if $M_i$ is the matrix of $K(underlinea;R)$ corresponding to the $i$th differential and $N_i$ the matrix of $K(underlinea^v;R)$ corresponding to the $i$th differential, then $N_i$ should be obtained from $M_i$ by taking all elements and raising it to the appropriate power. How should I relate the fitting ideal $F_j(N_i)$ to the fitting ideal $F_j(M_i)$? Am I missing some way to manipulate determinants? (Thanks for your response by the way.)
– do_math
Jul 20 at 18:37
Were you thinking that after accounting for the rank condition the appropriate fitting ideals would be "monomial" ideals? ("Monomial" in the sense that they are generated by products of the $a_i$'s.) Maybe I should test and see if that happens for $r = 3, 4$.
– do_math
Jul 20 at 20:52
I think 1.6.30 from Bruns and Herzog does the job.
– user26857
Jul 21 at 10:06