Explain \(\Ext^1_{\ZZ}\) for abelian groups, where it appears, and compute \[ \Ext^1_{\ZZ}(\ZZ/m\ZZ,\ZZ/n\ZZ) \quad\text{and}\quad \Ext^1_{\ZZ}(\ZZ/m\ZZ,\ZZ). \]
Solution
For abelian groups, \(\Ext^1_{\ZZ}(A,B)\) is the first right derived functor of \(\Hom_{\ZZ}(A,-)\); equivalently, it classifies extensions \[ 0\to B\to E\to A\to0 \] up to the usual equivalence.
Use the free resolution \[ 0\to\ZZ\xrightarrow{\times m}\ZZ\to\ZZ/m\ZZ\to0. \] Applying \(\Hom_{\ZZ}(-,B)\) gives \[ 0\to\Hom(\ZZ/m\ZZ,B)\to B\xrightarrow{\times m}B\to \Ext^1_{\ZZ}(\ZZ/m\ZZ,B)\to0. \] Hence \[ \Ext^1_{\ZZ}(\ZZ/m\ZZ,B)\cong B/mB. \]
Taking \(B=\ZZ/n\ZZ\) gives \[ \Ext^1_{\ZZ}(\ZZ/m\ZZ,\ZZ/n\ZZ) \cong (\ZZ/n\ZZ)/m(\ZZ/n\ZZ) \cong \ZZ/\gcd(m,n)\ZZ. \] Taking \(B=\ZZ\) gives \[ \Ext^1_{\ZZ}(\ZZ/m\ZZ,\ZZ) \cong \ZZ/m\ZZ. \]