Show that \(\ker A \into A\tensor \QQ\) given by \(a \mapsto a\tensor 1\) is the torsion subgroup of \(A\).
Solution
Goal. For an abelian group \(A\), show the kernel of \(A \to A \otimes \QQ\), \(a \mapsto a \otimes 1\), is the torsion subgroup \(T(A)\).
1. Every torsion element lies in the kernel.
1.1. If \(a \in A\) has \(na = 0\) for some \(n > 0\), then \(a \otimes 1 = 0\) in \(A \otimes \QQ\).
\(a \otimes 1 = a \otimes (n \cdot \frac1n) = (na) \otimes \frac1n = 0 \otimes \frac1n = 0\).
1.2. Hence \(T(A) \subseteq \ker\).
1.1 applies to every torsion element.
2. Every element of the kernel is torsion.
2.1. \(A \otimes \QQ \cong S^{-1}A\) where \(S = \ZZ \sm \theset{0}\).
\(\QQ\) is the localization of \(\ZZ\) at the nonzero integers, and \(A \otimes \QQ \cong A \otimes S^{-1}\ZZ \cong S^{-1}A\).
2.2. The map \(A \to S^{-1}A\), \(a \mapsto a/1\), has kernel exactly \(T(A)\).
\(a/1 = 0\) in \(S^{-1}A\) iff \(sa = 0\) for some \(s \in S\), i.e. iff \(a\) is torsion.
2.3. Hence \(\ker = T(A)\).
combine 2.1 and 2.2.
3. Q.E.D.
1 and 2 give both inclusions.