Over an integral domain, the modules R and R/Tor R have the same rank

Let R be an integral domain and let M be a (left, unital) R-module. Recall that the rank of a module is the largest possible cardinality of a linearly independent subset. \mathsf{Tor}\ M is the set of all m \in M such that rm = 0 for some nonzero r \in R.

  1. Prove that \mathsf{Tor}\ M has rank 0.
  2. Prove that M and M/\mathsf{Tor}\ M have the same rank.

Let S \subseteq \mathsf{Tor}\ M be a nonempty set. Note that for each m \in S, there exists a nonzero element r \in R such that rs = 0. In particular, S cannot be linearly independent. Thus \mathsf{Tor}\ M has rank 0.

Let S = \{m_i\}_I \subseteq M be a linearly independent set of maximal order. Note in particular that no element of S can be in \mathsf{Tor}\ M (each subset of S is linearly independent, in particular the singleton subsets are linearly independent.) We claim that the subset S + \mathsf{Tor}\ M \subseteq M/\mathsf{Tor}\ M is linearly independent. To see this, suppose \sum r_i m_i \in \mathsf{Tor}\ M for some r_i \in R. Then there exists a nonzero element a \in R such that \sum ar_im_i = 0 in M. Since S is linearly independent, we have ar_i = 0 in R. Since R is an integral domain, and a \neq 0, we have r_i = 0 for all i. Thus S + \mathsf{Tor}\ M is linearly independent. We claim also that S and S + \mathsf{Tor}\ M have the same cardinality. It suffices to show that for all i and j, m_i - m_j \notin \mathsf{Tor}\ M. To that end, suppose a \in R such that am_i - am_j = 0. Since m_i and m_i are linearly independent, a = 0. Thus S and S + \mathsf{Tor}\ M have the same cardinality. In particular, the rank of M/\mathsf{Tor}\ M is at least as great as the rank of M.

Conversely, let S = \{m_i\} and suppose S + \mathsf{Tor}\ M is a linearly independent subset of maximal order. We claim that S is linearly independent in M. To see this, suppose \sum r_i m_i = 0. Now \sum r_i \overline{m_i} = 0 in M/\mathsf{Tor}\ M. Thus r_i = 0, and so S \subseteq M is linearly independent. Certainly the cardinality of S is at least as great as the cardinality of S + \mathsf{Tor}\ M. Thus the rank of M is at least as great as the rank of M/\mathsf{Tor}\ M.

Hence M and M/\mathsf{Tor}\ M have the same rank.

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