A characterization of rank in modules over an integral domain

Let R be an integral domain and let M be a (left, unital) R-module.

  1. Suppose M has rank n, and that S = \{m_i\} is a linearly independent subset of M having maximal cardinality. Let N be the submodule of M generated by S. Prove that N is a free R-module having free rank n. Prove also that M/N is torsion.
  2. Conversely, prove that if there exists a submodule N \subseteq M which is free and has free rank n and such that M/N is torsion, then M has rank n.

  1. Note that N is free since its generating set S is R-linearly independent, and that S is a free generating set. Thus N has free rank n. Now let x+N \in M/N. Note in particular that x and S = \{m_i\} are linearly dependent in M, since S was a linearly independent set of maximal cardinality. Thus there exist s,r_i \in R such that at least one of the s,r_i is nonzero and sx + \sum r_im_i = 0. Suppose s = 0; then in fact \sum r_i m_i = 0, so that r_i = 0 for all i– a contradiction. Thus s \neq 0. Now sx \in N, so that s(x+N) = 0 in M/N. Thus x+N is a torsion element. Since x+N was arbitrary, M/N is torsion.
  2. Let S \subseteq N be a free generating set (of cardinality n). It is certainly the case that S is linearly independent in M, so that the rank of M is at least n. Now let T = \{t_i\}_{i=1}^{n+1} be a set of n+1 elements in M. Since M/N is torsion, for each t_i, there exists a nonzero element r_i \in R such that r_it_i \in N. If we have r_it_i = r_jt_j for some i and j, then T is linearly dependent. Suppose this is not the case, so that \{r_it_i\} \subseteq N is a subset having cardinality n+1. Since N is free of free rank n, there exist nonzero s_i such that \sum s_ir_it_i = 0. In particular, T is linearly dependent in M. Hence the rank of M is at most n. Thus the rank of M is precisely n.
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