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For a sequence of elements $\underline{a}= a_{1}, ..., a_{r} \in R$, let $K^{\bullet}(\underline{a};R)$ be the Koszul complex generated by $\underline{a}$. Let $\underline{a}^{v} = a_{1}^{v_{1}}, ..., a_{r}^{v_{r}} \in R$.

My question: if $K^{\bullet}(\underline{a};R)$ is acylic is $K^{\bullet}(\underline{a}^{v};R)$ acylic?

My thoughts: if $\underline{a}$ 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 $\underline{a}$ homogeneous of positive degree.

do_math
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    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 '18 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 $K(\underline{a};R)$ corresponding to the $i$th differential and $N_{i}$ the matrix of $K(\underline{a}^{v};R)$ 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 '18 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 '18 at 20:52
  • I think 1.6.30 from Bruns and Herzog does the job. – user26857 Jul 21 '18 at 10:06

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