Let \(R\) be an integral domain.
- Prove that every free \(R\)-module is torsion-free.
- Show that \(\QQ\) is torsion-free but not free as a \(\ZZ\)-module.
Solution
1. Free modules over a domain are torsion-free.
Let \(F\) be free with basis \(\{e_i\}_{i\in I}\), and let \[ 0\ne x=\sum_{i\in I} r_i e_i \] be a finite linear combination. Choose \(j\) with \(r_j\ne0\). If \(0\ne a\in R\) satisfied \(ax=0\), then \[ 0=ax=\sum_i ar_i e_i. \] Linear independence gives \(ar_i=0\) for every \(i\), in particular \(ar_j=0\). Since \(R\) is a domain and both \(a,r_j\) are nonzero, this is impossible. Hence no nonzero element is torsion.
2. \(\QQ\) is torsion-free but not free over \(\ZZ\).
If \(0\ne n\in\ZZ\) and \(q\in\QQ\) satisfy \(nq=0\), then \(q=0\), so \(\QQ\) is torsion-free.
Suppose \(\QQ\) were free over \(\ZZ\) with basis \(\mathcal B\). Choose \(0\ne b\in\mathcal B\). Since \(\QQ\) is divisible, there exists \(x\in\QQ\) with \[ 2x=b. \] Write the finite basis expansion \[ x=\sum_{c\in\mathcal B} n_c c, \qquad n_c\in\ZZ. \] Then \[ b=2x=\sum_c 2n_c c. \] Uniqueness of basis coordinates forces the coefficient of \(b\) to satisfy \[ 1=2n_b, \] impossible in \(\ZZ\). Therefore \(\QQ\) is not a free \(\ZZ\)-module.