Let be an integral domain and let be a (left, unital) -module. Recall that the rank of a module is the largest possible cardinality of a linearly independent subset. is the set of all such that for some nonzero .
- Prove that has rank 0.
- Prove that and have the same rank.
Let be a nonempty set. Note that for each , there exists a nonzero element such that . In particular, cannot be linearly independent. Thus has rank 0.
Let be a linearly independent set of maximal order. Note in particular that no element of can be in (each subset of is linearly independent, in particular the singleton subsets are linearly independent.) We claim that the subset is linearly independent. To see this, suppose for some . Then there exists a nonzero element such that in . Since is linearly independent, we have in . Since is an integral domain, and , we have for all . Thus is linearly independent. We claim also that and have the same cardinality. It suffices to show that for all and , . To that end, suppose such that . Since and are linearly independent, . Thus and have the same cardinality. In particular, the rank of is at least as great as the rank of .
Conversely, let and suppose is a linearly independent subset of maximal order. We claim that is linearly independent in . To see this, suppose . Now in . Thus , and so is linearly independent. Certainly the cardinality of is at least as great as the cardinality of . Thus the rank of is at least as great as the rank of .
Hence and have the same rank.