Find a free resolution of \(\QQ\) as a \(\ZZ\)-module.
Solution
1. Let\[ F_0=\bigoplus_{n\ge1}\mathbb Z e_n, \qquad F_1=\bigoplus_{n\ge1}\mathbb Z f_n. \]Define\[ \varepsilon:F_0\to\mathbb Q, \qquad \varepsilon(e_n)=\frac1{n!}, \]and\[ d:F_1\to F_0, \qquad d(f_n)=e_n-(n+1)e_{n+1}. \]
Both \(F_0\) and \(F_1\) are free abelian groups. The formula for \(d\) is chosen from the relation \[ \frac1{n!}=(n+1)\frac1{(n+1)!}. \]
2. The map \(\varepsilon\) is surjective and \(\varepsilon d=0\).
Every rational number \(a/b\) is an integer multiple of \(1/n!\) once \(n\) is large enough that \(b\mid n!\), so \(\varepsilon\) is onto. Also \[ \varepsilon(d(f_n))=\frac1{n!}-(n+1)\frac1{(n+1)!}=0. \]
3. The map \(d\) is injective.
If a finite sum \(\sum_{n=1}^N c_n f_n\) maps to zero, inspect the coefficient of \(e_1\) to obtain \(c_1=0\). Then the coefficient of \(e_2\) gives \(c_2=0\), and inductively every \(c_n=0\).
4. We have \(\ker\varepsilon=\operatorname{im}d\).
Take a finite sum \(x=\sum_{n=1}^N a_n e_n\in\ker\varepsilon\). Modulo \(\operatorname{im}d\), the relations \(e_n=(n+1)e_{n+1}\) imply \[ x\equiv\left(\sum_{n=1}^N a_n\frac{N!}{n!}\right)e_N. \] The condition \(\varepsilon(x)=0\) says \[ 0=N!\varepsilon(x)=\sum_{n=1}^N a_n\frac{N!}{n!}, \] so \(x\equiv0\) modulo \(\operatorname{im}d\). Thus \(x\in\operatorname{im}d\).
5. Therefore\[ \boxed{0\longrightarrow F_1\xrightarrow{d}F_0\xrightarrow{\varepsilon}\mathbb Q\longrightarrow0} \]is a free resolution of \(\mathbb Q\) as a \(\mathbb Z\)-module.
Exactness follows from <1>2–<1>4, and both \(F_0,F_1\) are free by construction.