Show that \(\operatorname{Ext}(A, \mathbb{Q}) = 0\) for all \(A\). [Consider the homology with \(\mathbb{Q}\) coefficients of a Moore space \(M(A, n)\).]
Solution
1. Divisibility and injectivity of \(\mathbb{Q}\):
1.1. An abelian group \(D\) is divisible if for every \(d \in D\) and non-zero integer \(n \in \mathbb{Z}\), there exists \(x \in D\) such that \(nx = d\).\(\mathbb{Q}\) is divisible because \(n(q/n) = q\) for all \(q \in \mathbb{Q}\) and \(n \in \mathbb{Z} \setminus \{0\}\).
arithmetic of rational numbers.
1.2. By Baer’s Criterion, an abelian group (a \(\mathbb{Z}\)-module) is injective if and only if it is divisible.Thus \(\mathbb{Q}\) is an injective \(\mathbb{Z}\)-module.
Baer’s Criterion for modules over principal ideal domains.
2. Vanishing of \(\operatorname{Ext}(A, \mathbb{Q})\) via free resolutions:
2.1. Let \(A\) be any abelian group, and choose a free presentation (resolution):\[ 0 \longrightarrow F_1 \xrightarrow{\;d\;} F_0 \longrightarrow A \longrightarrow 0, \]where \(F_0, F_1\) are free abelian groups.
every abelian group is a quotient of a free abelian group with free kernel.
2.2. Applying the contravariant functor \(\operatorname{Hom}_\mathbb{Z}(-, \mathbb{Q})\) yields the long exact sequence:\[ 0 \longrightarrow \operatorname{Hom}(A, \mathbb{Q}) \longrightarrow \operatorname{Hom}(F_0, \mathbb{Q}) \xrightarrow{\;d^* \;} \operatorname{Hom}(F_1, \mathbb{Q}) \longrightarrow \operatorname{Ext}^1(A, \mathbb{Q}) \longrightarrow 0. \]
derived functor long exact sequence for \(\operatorname{Ext}\).
2.3. Because \(\mathbb{Q}\) is injective, every homomorphism \(\phi: F_1 \to \mathbb{Q}\) extends along the inclusion \(d: F_1 \hookrightarrow F_0\) to a homomorphism \(\widetilde{\phi}: F_0 \to \mathbb{Q}\) such that \(\widetilde{\phi} \circ d = \phi\).Thus the dual map \(d^*: \operatorname{Hom}(F_0, \mathbb{Q}) \to \operatorname{Hom}(F_1, \mathbb{Q})\) is surjective.
definition of injective module.
2.4. By exactness:\[ \operatorname{Ext}^1(A, \mathbb{Q}) \cong \operatorname{coker}(d^*) = \operatorname{Hom}(F_1, \mathbb{Q}) / \operatorname{Im}(d^*) = 0. \]
surjectivity of \(d^*\).
3. Alternative topological perspective via Moore spaces:
3.1. Let \(X = M(A, n)\) be a Moore space with \(\widetilde{H}_n(X) \cong A\) and \(\widetilde{H}_i(X) = 0\) for all \(i \neq n\).By the Universal Coefficient Theorem for cohomology with \(\mathbb{Q}\) coefficients:\[ 0 \longrightarrow \operatorname{Ext}(H_n(X), \mathbb{Q}) \longrightarrow H^{n+1}(X; \mathbb{Q}) \longrightarrow \operatorname{Hom}(H_{n+1}(X), \mathbb{Q}) \longrightarrow 0. \]
Universal Coefficient Theorem for Cohomology.
3.2. Since \(\mathbb{Q}\) is a field, \(H_{n+1}(X; \mathbb{Q}) \cong H_{n+1}(X) \otimes \mathbb{Q} = 0 \otimes \mathbb{Q} = 0\), so \(H^{n+1}(X; \mathbb{Q}) = 0\).The exact sequence forces \(\operatorname{Ext}(A, \mathbb{Q}) = 0\).
exactness with vanishing middle group.
4. Conclusion:\(\operatorname{Ext}(A, \mathbb{Q}) = 0\) for every abelian group \(A\). Q.E.D.
1 through 3.