Over an integral domain, rank N + rank M/N = rank M

Let R be an integral domain, let M be a (left, unital) R-module, and let N \subseteq M be a submodule. Prove that \mathsf{rank}(N) + \mathsf{rank}(M/N) = \mathsf{rank}(M).


Suppose S = \{s_i\} \subseteq N is a cardinality-maximal linearly independent set, and choose T = \{t_i\} \subseteq M such that \overline{T} \subseteq M/N is a cardinality-maximal linearly independent set and T and \overline{T} have the same cardinality. Note in particular that S \cap T is empty. We claim that S \cup T is linearly independent in M. To see this, suppose \sum r_i s_i + \sum r_i^\prime t_i = 0. Mod N, we see that \sum r_i^\prime \psi(t_i) = 0, so that r_i^\prime = 0. But then \sum r_is_i = 0, so that r_i = 0. Thus S \cup T is linearly independent in M. In particular, \mathsf{rank}(M) \geq \mathsf{rank}(N) + \mathsf{rank}(M/N).

Now let A \subseteq M be the submodule generated by S \cup T. We claim that M/A is torsion. Suppose to the contrary that M/A is not torsion; that is, that there exists y \in M/A such that if ry \in A, then r = 0. In other words, if ry + \sum r_i s_i + \sum r_i^\prime t_i = 0, then r = 0, and hence r_i = r_i^\prime = 0. Thus S \cup T \cup \{y\} is linearly independent. If y \in N, then we violate the maximality of S, and if y \notin N, then we violate the maximality of T. So M/A must be torsion. By this previous exercise, \mathsf{rank}(M) = \mathsf{rank}(N) + \mathsf{rank}(M/N).

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