An abelian group \(G\) is defined to be divisible if the map \(G \xrightarrow{n} G\), \(g \mapsto ng\), is surjective for all \(n > 1\). Show that a group is divisible if and only if it is a quotient of a direct sum of \(\mathbb{Q}\)’s. Deduce from the previous problem that if \(G\) is divisible then \(\operatorname{Ext}(A, G) = 0\) for all \(A\).
Solution
First suppose \[ G\cong\left(\bigoplus_{i\in I}\mathbb Q\right)/K. \] A direct sum of copies of \(\mathbb Q\) is divisible coordinatewise. If \(\bar g\) is the image of \(g\) in the quotient and \(n>0\), choose \(h\) upstairs with \(nh=g\); then \(n\bar h=\bar g\). Hence every quotient of a direct sum of \(\mathbb Q\)’s is divisible.
Conversely, let \(G\) be divisible. Choose a generating set \((g_i)_{i\in I}\) for \(G\) as an abelian group. For each \(i\), divisibility lets us choose coherently elements \[ g_{i,1}=g_i, \qquad n\,g_{i,n!}=g_{i,(n-1)!} \] for \(n\ge2\); inductively this defines a homomorphism \[ \phi_i:\mathbb Q\to G \] with \(\phi_i(1)=g_i\) (equivalently, extend the map \(\mathbb Z\to G\), \(1\mapsto g_i\), across the inclusions \(\mathbb Z\subset \frac1{2!}\mathbb Z\subset\frac1{3!}\mathbb Z\subset\cdots\) using divisibility). Taking the direct sum gives \[ \Phi:\bigoplus_{i\in I}\mathbb Q\to G. \] Its image contains every generator \(g_i\), so \(\Phi\) is surjective. Thus \(G\) is a quotient of a direct sum of copies of \(\mathbb Q\).
By Exercise 3, \(\operatorname{Ext}(A,\mathbb Q)=0\) for every \(A\). Ext takes direct sums in the second variable here to the corresponding direct sums for free resolutions of \(A\), so \[ \operatorname{Ext}\!\left(A,\bigoplus_I\mathbb Q\right)=0. \] If \[ 0\to K\to\bigoplus_I\mathbb Q\to G\to0 \] is the quotient sequence, the six-term exact sequence for \(\operatorname{Hom}(A,-)\) and \(\operatorname{Ext}(A,-)\) ends with \[ \operatorname{Ext}\!\left(A,\bigoplus_I\mathbb Q\right) \longrightarrow \operatorname{Ext}(A,G)\longrightarrow0. \] The left group is zero, hence \[ \boxed{\operatorname{Ext}(A,G)=0} \] for every divisible \(G\).