A fact about the rank of a module over an integral domain

Let R be an integral domain with quotient field F and let M be any (left, unital) R-module. Prove that the rank of M equals the dimension of F \otimes_R M over F.

Recall that the rank of a module over a domain is the maximal number of R-linearly independent elements.

Suppose B \subseteq M is an R-linearly independent set, and consider B^\prime \subseteq F \otimes_R M consisting of the tensors 1 \otimes b for each b \in B. Suppose \sum \alpha_i (1 \otimes b_i) = 0. For some \alpha_i \in F. Clearing denominators, we have \sum r_i ( 1 \otimes b_i) = 0 for some r_i \in R. Now \sum 1 \otimes r_ib_i = 0, and we have 1 \otimes \sum r_ib_i = 0. By this previous exercise, there exists a nonzero r \in R such that \sum rr_ib_i = 0. Since B is linearly independent, we have rr_i = 0 for each i, and since r \neq 0 and R is a domain, r_i = 0 for each i. Thus \alpha_i = 0 (since the denominators of each \alpha_i are nonzero). So B^\prime is F-linearly independent in F \otimes_R M. In particular, we have \mathsf{rank}_R(M) \leq \mathsf{dim}_F(F \otimes_R M).

Now note that if b_i^\prime = \sum \alpha_{i,j} \otimes m_{i,j} \in F \otimes_R M, is a linearly independent set, with \alpha_{i,j} = r_{i,j}/s_{i,j}, then ‘clearing denominators’ by multiplying b_i^\prime by \prod_j s_{i,j}, we have a second linearly independent set with the same cardinality whose elements are of the form 1 \otimes m_i for some m \in M. Suppose there exist r_i \in R such that \sum r_im_i = 0; then \sum r_i(1 \otimes b_i) = 0, and (since the 1 \otimes b_i form a basis) we have r_i = 0. So the m_i are R-linearly independent in M. Thus \mathsf{rank}_R(M) \geq \mathsf{dim}_F(R \otimes_R M).

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