## A characterization of rank in modules over an integral domain

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

- Suppose has rank , and that is a linearly independent subset of having maximal cardinality. Let be the submodule of generated by . Prove that is a free -module having free rank . Prove also that is torsion.
- Conversely, prove that if there exists a submodule which is free and has free rank and such that is torsion, then has rank .

- Note that is free since its generating set is -linearly independent, and that is a free generating set. Thus has free rank . Now let . Note in particular that and are linearly dependent in , since was a linearly independent set of maximal cardinality. Thus there exist such that at least one of the is nonzero and . Suppose ; then in fact , so that for all – a contradiction. Thus . Now , so that in . Thus is a torsion element. Since was arbitrary, is torsion.
- Let be a free generating set (of cardinality ). It is certainly the case that is linearly independent in , so that the rank of is at least . Now let be a set of elements in . Since is torsion, for each , there exists a nonzero element such that . If we have for some and , then is linearly dependent. Suppose this is not the case, so that is a subset having cardinality . Since is free of free rank , there exist nonzero such that . In particular, is linearly dependent in . Hence the rank of is at most . Thus the rank of is precisely .

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