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Reduction of an exact free complex by a nonzerodivisor stays exact above degree one
Statement
Let be a local ring, let be a nonzerodivisor, and let be a finite complex of finite free -modules. If for every , then for every . Equivalently, the reduced complex is exact at all its terms of degree at least two. The degree-one homology can appear from -torsion in the original cokernel.
Facts & Assumptions
Given: The local ring, nonzerodivisor, finite free complex, and exactness in positive degrees.
A short exact sequence of complexes gives a long exact sequence in homology (The long exact sequence in homology).
Proof
Because each is free and is a nonzerodivisor on , multiplication by is injective on every . Thus is a short exact sequence of complexes.
By [F1], for the segment is exact. Both outer groups vanish by hypothesis, so . At the last group is , which need not be -torsion-free; no degree-one conclusion is claimed.
Depends on
Used by
Dependency tree · two levels
7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- The Stacks Project, Algebra, Lemma 10.102.7 (tag 00MZ), reduction of exact free complexes (standard reference, not scraped)