Show that every free module over a unital integral domain is torsion-free. Show that the converse is false.
Solution
1. Every free module over a unital integral domain is torsion-free.
Let \(R\) be a unital integral domain and let \(F\) be a free \(R\)-module with basis \(B\). Suppose \(0\ne r\in R\) and \(x\in F\) satisfy \(rx=0\). Write the unique finite basis expansion \[ x=\sum_{i=1}^n a_i b_i, \] where the \(b_i\in B\) are distinct. Then \[ 0=rx=\sum_{i=1}^n (ra_i)b_i. \] Linear independence of \(B\) gives \(ra_i=0\) for every \(i\). Since \(R\) is an integral domain and \(r\ne0\), each \(a_i=0\). Thus \(x=0\). Hence \(F\) has no nonzero torsion element.
2. The converse is false.
Take \(R=\ZZ\) and the \(\ZZ\)-module \(\QQ\). It is torsion-free: if \(0\ne n\in\ZZ\) and \(q\in\QQ\) satisfy \(nq=0\), then \(q=0\).
We show that \(\QQ\) is not free as a \(\ZZ\)-module. Suppose it had a basis \(B\), and choose \(b\in B\). Since \(\QQ\) is divisible, there exists \(c\in\QQ\) with \(2c=b\). Express \[ c=\sum_{i=1}^m n_i b_i \] in the basis, where \(n_i\in\ZZ\). Multiplying by \(2\) gives \[ b=\sum_{i=1}^m 2n_i b_i. \] By uniqueness of basis coordinates, the coefficient of \(b\) on the right must be \(1\). But every coefficient there is even, a contradiction. Thus \(\QQ\) is torsion-free but not free.