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    Tor vanishes for torsionfree modules

    E-HAT-3.A-5

    Area
    Topology
    Topics
    Cohomology
    Status
    draft

    From the fact that \(\operatorname{Tor}(A, B) = 0\) if \(A\) is free, deduce that \(\operatorname{Tor}(A, B) = 0\) if \(A\) is torsionfree by applying the previous problem to the directed system of finitely generated subgroups \(A_\alpha\) of \(A\).

    Solution

    Write a torsionfree abelian group \(A\) as the directed union of its finitely generated subgroups: \[ A=\varinjlim_\alpha A_\alpha. \] Each \(A_\alpha\) is finitely generated and torsionfree, hence is a free abelian group. Therefore \[ \operatorname{Tor}(A_\alpha,B)=0 \] for every \(\alpha\).

    By Exercise 4, \(\operatorname{Tor}\) commutes with direct limits in the first variable, so \[ \operatorname{Tor}(A,B) \cong \varinjlim_\alpha\operatorname{Tor}(A_\alpha,B) =0. \]