For an abelian group \(A\) let \(p: A \to A\) be multiplication by \(p\), and let \({}_pA = \ker p\), \(pA = \operatorname{im} p\), and \(A_p = \operatorname{coker} p\) as in the proof of Proposition 3F.12. Show that the six-term exact sequences involving \(\operatorname{Hom}(-, \mathbb{Z})\) and \(\operatorname{Ext}(-, \mathbb{Z})\) associated to the short exact sequences \(0 \to {}_pA \to A \to pA \to 0\) and \(0 \to pA \to A \to A_p \to 0\) can be spliced together to yield an exact sequence
\[\operatorname{Hom}(pA, \mathbb{Z}) \to \operatorname{Ext}(A_p, \mathbb{Z}) \to \operatorname{Ext}(A, \mathbb{Z}) \xrightarrow{p} \operatorname{Ext}(A, \mathbb{Z}) \to \operatorname{Ext}({}_pA, \mathbb{Z}) \to 0\]
where the map labeled “\(p\)” is multiplication by \(p\). Use this to show:
\(\operatorname{Ext}(A, \mathbb{Z})\) is divisible if \(A\) is torsionfree.
\(\operatorname{Ext}(A, \mathbb{Z})\) is torsionfree if \(A\) is divisible, and the converse holds if \(\operatorname{Hom}(A, \mathbb{Z}) = 0\).
Solution
Consider the two short exact sequences \[ 0\to{}_pA\to A\xrightarrow{p}pA\to0, \qquad 0\to pA\to A\to A_p\to0. \] Since \({}_pA\) and \(A_p\) are killed by \(p\), every homomorphism from either group to the torsionfree group \(\mathbb Z\) is zero.
Applying \(\operatorname{Hom}(-,\mathbb Z)\) and \(\operatorname{Ext}(-,\mathbb Z)\) to the first sequence gives the tail \[ 0\to\operatorname{Ext}(pA,\mathbb Z) \to\operatorname{Ext}(A,\mathbb Z) \to\operatorname{Ext}({}_pA,\mathbb Z)\to0. \] Applying them to the second gives \[ 0\to\operatorname{Hom}(A,\mathbb Z)\to\operatorname{Hom}(pA,\mathbb Z) \to\operatorname{Ext}(A_p,\mathbb Z) \to\operatorname{Ext}(A,\mathbb Z) \to\operatorname{Ext}(pA,\mathbb Z)\to0. \] Splicing at \(\operatorname{Ext}(pA,\mathbb Z)\) yields \[ \operatorname{Hom}(pA,\mathbb Z)\to\operatorname{Ext}(A_p,\mathbb Z)\to \operatorname{Ext}(A,\mathbb Z)\to\operatorname{Ext}(A,\mathbb Z)\to \operatorname{Ext}({}_pA,\mathbb Z)\to0. \] The composite of the two middle maps is induced contravariantly by \[ A\xrightarrow{p}pA\hookrightarrow A, \] which is multiplication by \(p\) on \(A\). By functoriality of Ext, the composite is therefore multiplication by \(p\) on \(\operatorname{Ext}(A,\mathbb Z)\). Hence the spliced sequence is precisely \[ \boxed{ \operatorname{Hom}(pA,\mathbb Z)\to\operatorname{Ext}(A_p,\mathbb Z)\to \operatorname{Ext}(A,\mathbb Z)\xrightarrow{p} \operatorname{Ext}(A,\mathbb Z)\to \operatorname{Ext}({}_pA,\mathbb Z)\to0.} \]
1. If \(A\) is torsionfree, then \(\operatorname{Ext}(A,\mathbb Z)\) is divisible.
Torsionfreeness gives \({}_pA=0\). Exactness at the second copy of \(\operatorname{Ext}(A,\mathbb Z)\) says multiplication by \(p\) is surjective for every prime \(p\). Therefore multiplication by every positive integer is surjective, so \(\operatorname{Ext}(A,\mathbb Z)\) is divisible.
2. If \(A\) is divisible, then \(\operatorname{Ext}(A,\mathbb Z)\) is torsionfree.
Divisibility gives \(A_p=0\) for every prime \(p\). Exactness at the first copy of \(\operatorname{Ext}(A,\mathbb Z)\) then says multiplication by \(p\) has zero kernel. Thus the Ext group has no \(p\)-torsion for any prime \(p\), hence is torsionfree.
3. Conversely, if \(\operatorname{Hom}(A,\mathbb Z)=0\) and \(\operatorname{Ext}(A,\mathbb Z)\) is torsionfree, then \(A\) is divisible.
Precomposition with the surjection \(A\xrightarrow p pA\) injects \[ \operatorname{Hom}(pA,\mathbb Z)\hookrightarrow\operatorname{Hom}(A,\mathbb Z)=0, \] so \(\operatorname{Hom}(pA,\mathbb Z)=0\). The spliced exact sequence therefore identifies \[ \operatorname{Ext}(A_p,\mathbb Z) \] with the kernel of multiplication by \(p\) on \(\operatorname{Ext}(A,\mathbb Z)\). This kernel is zero by torsionfreeness.
But \(A_p\) is a vector space over \(\mathbb Z_p\), and if it were nonzero it would admit a nonzero homomorphism to the subgroup \(\mathbb Z_p\subset\mathbb Q/\mathbb Z\). Using \[ \operatorname{Ext}(A_p,\mathbb Z) \cong\operatorname{Hom}(A_p,\mathbb Q/\mathbb Z) \] from Exercise 5 (since \(\operatorname{Hom}(A_p,\mathbb Q)=0\)), this would contradict \(\operatorname{Ext}(A_p,\mathbb Z)=0\). Hence \(A_p=0\), i.e. \(pA=A\), for every prime \(p\). Therefore \(A\) is divisible.