I'm reading Paolo Aluffi's Algebra: Chapter 0. On tensor product of free modules over commutative rings, he wrote
Note that if $e_1, \dots, e_m$ generate $M$ and $f_1, \dots, f_n$ generate $N$, then the pure tensors $e_i \otimes f_j$ must generate $M \otimes_R N$. In the free case, if the $e_i$'s and $f_j$'s form bases of $R^{\oplus m}$, $R^{\oplus n}$, resp., then the $mn$ elements $e_i \otimes f_j$ must be a basis for $R^{\oplus m} \otimes R^{\oplus n}$. Indeed they generate it; hence they must be linearly independent since this module is free of rank $mn$. In particular, this is all that can happen if $R$ is a field $k$ and the modules are, therefore, just $k$-vector spaces (Proposition VI.1.7). Tensor products are more interesting over more general rings.
It seems to me that he is saying that if the generating $(e_i \otimes f_j)$ is linearly dependent, then we can have a basis of $<mn$ elements, contradicting IBN of commutative rings. I was thinking about taking the maximal linearly independent subset contained in $\{e_i \otimes f_j\}$ and then showing it generates the free module, but this does not seem to hold in general.
By using universal properties to obtain an $R$-linear map from $M\otimes N$ to $R^{mn}$, his assertion can indeed be proved. However I cannot understand his reasoning here. Am I missing something obvious? Thanks in advance.
