Over an integral domain, an ideal is free as a module exactly when it is principal.Suggest a slogan
Let \(R\) be an integral domain and \(I \subseteq R\) an ideal. Then \(I\) is a free \(R\)-module if and only if \(I\) is a principal ideal.
Over a ring with zero divisors, a principal ideal need not be free. In \(\ZZ/6\ZZ\), the principal ideal \(\gens{2}\) is not a free \(\ZZ/6\ZZ\)-module, since \(3 \cdot 2 = 0\) with \(3 \neq 0\), whereas a nonzero element of a basis of a free module has zero annihilator.