Show that \(\operatorname{Ext}(A, \mathbb{Z})\) is isomorphic to the cokernel of \(\operatorname{Hom}(A, \mathbb{Q}) \to \operatorname{Hom}(A, \mathbb{Q}/\mathbb{Z})\), the map induced by the quotient map \(\mathbb{Q} \to \mathbb{Q}/\mathbb{Z}\). Use this to get another proof that \(\operatorname{Ext}(\mathbb{Z}_{p^\infty}, \mathbb{Z}) \approx \widehat{\mathbb{Z}}_p\) for \(p\) prime.
Solution
Apply \(\operatorname{Hom}(A,-)\) and its derived functor \(\operatorname{Ext}(A,-)\) to \[ 0\longrightarrow\mathbb Z\longrightarrow\mathbb Q \longrightarrow\mathbb Q/\mathbb Z\longrightarrow0. \] Since \(\mathbb Q\) is divisible, Exercise 4 gives \[ \operatorname{Ext}(A,\mathbb Q)=0. \] The resulting exact sequence therefore contains \[ \operatorname{Hom}(A,\mathbb Q) \longrightarrow \operatorname{Hom}(A,\mathbb Q/\mathbb Z) \longrightarrow \operatorname{Ext}(A,\mathbb Z) \longrightarrow0. \] Hence \[ \boxed{\operatorname{Ext}(A,\mathbb Z) \cong\operatorname{coker}\bigl(\operatorname{Hom}(A,\mathbb Q)\to \operatorname{Hom}(A,\mathbb Q/\mathbb Z)\bigr).} \]
Now take \(A=\mathbb Z_{p^\infty}\). Since \(A\) is torsion and \(\mathbb Q\) is torsionfree, \[ \operatorname{Hom}(A,\mathbb Q)=0. \] Thus \[ \operatorname{Ext}(\mathbb Z_{p^\infty},\mathbb Z) \cong\operatorname{Hom}(\mathbb Z_{p^\infty},\mathbb Q/\mathbb Z). \] The \(p\)-primary subgroup of \(\mathbb Q/\mathbb Z\) is itself isomorphic to \(\mathbb Z_{p^\infty}\), so every homomorphism lands there and \[ \operatorname{Hom}(\mathbb Z_{p^\infty},\mathbb Q/\mathbb Z) \cong\operatorname{End}(\mathbb Z_{p^\infty}). \] An endomorphism is determined by compatible multipliers modulo \(p^n\) on the unique cyclic subgroup of order \(p^n\). Hence \[ \operatorname{End}(\mathbb Z_{p^\infty}) \cong\varprojlim_n\mathbb Z/p^n\mathbb Z =\widehat{\mathbb Z}_p. \] Therefore \[ \boxed{\operatorname{Ext}(\mathbb Z_{p^\infty},\mathbb Z) \cong\widehat{\mathbb Z}_p.} \]