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Finite derived quotients of rational complexes vanish

PDF p. 6: Finite-coefficient comparison

Statement

Let $C$ be an object of $D (Z)$ isomorphic to the underlying complex of a complex of $Q$ -vector spaces, and let $n > 0$ be an integer. Then $C \otimes_{Z}^{L} Z / n$ is zero in $D (Z)$ . The derived tensor product is the total tensor product computed with a K-flat replacement; zero in $D (Z)$ means an acyclic complex.

Facts & Assumptions

Given: $C$ and $n$ as stated.

Goal: Show $C \otimes_{Z}^{L} Z / n ≃ 0$ .

[F1]

Multiplication by $n$ is an isomorphism on the underlying complex of a complex of $Q$ -vector spaces. Integer multiplication on rational complexes

[F2]

The derived tensor product with $Z / n$ is represented by the mapping cone of multiplication by $n$ . Its cohomology lies in the natural exact sequence of Derived finite coefficients and the Bockstein sequence.

Proof

1.1F1given

Transporting [F1] across the given isomorphism shows that multiplication by $n$ induces an automorphism of every $H^{i} (C)$ .

2.1F2step 1.1∎

In [F2], both $H^{i} (C) / n H^{i} (C)$ and $H^{i + 1} (C) [n]$ therefore vanish. The exact sequence gives $H^{i} (C \otimes_{Z}^{L} Z / n) = 0$ for every $i$ . Hence the derived quotient is zero.

Depends on

Used by · 1

Rational cofiber implies finite-coefficient equivalence

Dependency tree · 3 levels

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