Let \(A\) be a \(\mathbb{Z}\)-module, and suppose \(m \in \mathbb{Z}^+\). Prove that \(A/mA \cong A \otimes_{\mathbb{Z}} (\mathbb{Z}/m\mathbb{Z})\).
Solution
Write \(T=A\otimes_{\mathbb Z}(\mathbb Z/m\mathbb Z)\) and \(\bar r=r+m\mathbb Z\). The inverse isomorphisms are \[ \begin{aligned} \Phi:A/mA&\longrightarrow T, &a+mA&\longmapsto a\otimes\bar1,\\ \Psi:T&\longrightarrow A/mA, &a\otimes\bar r&\longmapsto ra+mA. \end{aligned} \]
1. The displayed formula defines a \(\mathbb Z\)-linear map \(\Psi\).
The map \(\beta:A\times(\mathbb Z/m\mathbb Z)\to A/mA\) given by \(\beta(a,\bar r)=ra+mA\) is independent of the representative \(r\): replacing \(r\) by \(r+mk\) changes \(ra\) by \(m(ka)\in mA\). The map is additive in both variables. For \(s\in\mathbb Z\) it satisfies \[ \beta(sa,\bar r)=r(sa)+mA=(rs)a+mA =\beta(a,s\bar r). \] Thus it is balanced. The defining universal property of the tensor product gives an additive map \(\Psi\) with the asserted formula. Every additive homomorphism of abelian groups is \(\mathbb Z\)-linear, so this is a module map.
2. The formula for \(\Phi\) descends to \(A/mA\), andthe maps are inverse.
The map \(\eta:A\to T\), \(a\mapsto a\otimes\bar1\), is additive. If \(a=mb\), balancing gives \(\eta(a)=(mb)\otimes\bar1=b\otimes\bar m=0\). Thus \(mA\) lies in its kernel, and \(\eta\) descends to the \(\mathbb Z\)-linear map \(\Phi\) on \(A/mA\).
For every coset, one has \(\Psi\Phi(a+mA)=a+mA\). For every elementary tensor, balancing gives \[ \Phi\Psi(a\otimes\bar r)=(ra)\otimes\bar1 =a\otimes\bar r. \] Elementary tensors generate \(T\) additively, so the latter identity holds on all of \(T\). Both composites are the identity, proving the isomorphism. When \(m=1\), the same formulas apply and both modules are zero.