Suppose $N \cong R^n $ then for every non-zero $R$ module $M$ show that $M \otimes N$ can be written uniquely by $\sum m_i \otimes e_i $ Suppose $N \cong R^n $ be free $R$ module of rank $n$ with basis $\{e_1,...,e_n\}$ and $R$ is commutative then for every non-zero $R$ module $M$ show that $M \otimes N$ can be written uniquely by $\sum m_i \otimes e_i $
Now for $x\in M \otimes N$ $x=\sum m_i \otimes n_i$ where $n_i= \sum r_{ij}e_j$ then $x=\sum _{j=1}^n(\sum_{i=1}^{k_i} r_{ij}m_i) \otimes e_j=\sum _{j=1}^nm'_j \otimes e_j $ but why this is unique?
 A: We have $R^n \cong R \oplus \ldots \oplus R$, and $\otimes$ distributes over $\oplus$ so in fact $M \otimes R^n \cong (M \otimes R) \oplus \ldots \oplus (M \otimes R)$. Each $M \otimes R$ is just an $M$, and clearly in $M \oplus \ldots \oplus M$ we have each element having a unique expression as claimed.
We can also see this more directly: the construction of the tensor product is as a quotient of the free R-module on generators $M \times R$: we quotient by


*

*$(x+y) \otimes n = x \otimes n + y \otimes n$

*$x \otimes (m + n) = x \otimes m + x \otimes n$

*$rx \otimes n = x \otimes rn$


Since $N$ is free, these allow us to view $M \otimes N$ as a quotient of the free $R$-module on generators $M \times \{e_1, \ldots, e_n\}$. (Using $R$-linearity on the right-hand side.) Having done this, there are no relations we could apply to have $m_1 \otimes e_1$ lying in the span of a sum $\sum_{i > 1} m_i \otimes e_i$, as there are no relations between $m \otimes e_i$ and $m \otimes e_j$ for $i \neq j$.
