Show that \(\operatorname{Tor}(A, \mathbb{Q}/\mathbb{Z})\) is isomorphic to the torsion subgroup of \(A\). Deduce that \(A\) is torsionfree if \(\operatorname{Tor}(A, B) = 0\) for all \(B\).
Solution
Use the short exact sequence \[ 0\longrightarrow\mathbb Z\longrightarrow\mathbb Q \longrightarrow\mathbb Q/\mathbb Z\longrightarrow0. \] Since \(\mathbb Q\) is flat over \(\mathbb Z\), tensoring with an abelian group \(A\) yields the exact segment \[ 0\longrightarrow \operatorname{Tor}(A,\mathbb Q/\mathbb Z) \longrightarrow A\otimes\mathbb Z \longrightarrow A\otimes\mathbb Q. \] Identifying \(A\otimes\mathbb Z\) with \(A\), the last map is the canonical localization map \[ A\longrightarrow A\otimes\mathbb Q. \] Its kernel consists exactly of the torsion elements of \(A\): if \(na=0\) then \(a\otimes1=(na)\otimes(1/n)=0\), while if \(a\otimes1=0\) in the localization then some nonzero integer kills \(a\). Therefore \[ \boxed{\operatorname{Tor}(A,\mathbb Q/\mathbb Z)\cong T(A)}. \]
Consequently, if \(\operatorname{Tor}(A,B)=0\) for every abelian group \(B\), then in particular \[ \operatorname{Tor}(A,\mathbb Q/\mathbb Z)=0, \] so \(T(A)=0\) and \(A\) is torsionfree.