It is a problem 10.4.10 of Dummit and Foote.
- Suppose $R$ is commutative and $N\cong R^n$ is a free $R-$module of rank $n$ with $R-$module basis $e_1,e_2,\dots, e_n$.
(a) For any nonzero $R$-module $M$ show that every element of $M\otimes N$ can be written uniquely in the form $\sum_{i=1}^{n}m_i\otimes e_i$ where $m_i\in M$. Deduce that if $\sum_{i=1}^{n}m_i\otimes e_i=0$ in $M\otimes N$ then $m_i=0$ for $i=1,\dots,n$.
Actually I already proved the first part.
Since any element in $M\otimes N$ is a finite sum of simple tensors, for any $\{m^j\}_{j=1}^{p}\subset M$ and $\{n^j\}_{j=1}^p\subset N$ observe that\begin{align*} \sum_{j=1}^{p}m^j\otimes n^j=\sum_{j=1}^{p} \sum_{i=1}^{n}m^j\otimes r_i^je_i=\sum_{i=1}^{n}\sum_{j=1}^{p}(m^jr^j_i)\otimes e_i=\sum_{i=1}^{n}\left( \sum_{j=1}^{p}m^jr^j_i\right)\otimes e_i \end{align*}
where $$n^j=\sum_{i=1}^{n}r^j_ie_i.$$
In other words, $$m_i=\sum_{j=1}^{p}m^jr^j_i.$$
But I cannot see why $\sum_{i=1}^{n}m_i\otimes e_i=0$ implies $m_i=0$ for all $i$. Since the problem says 'Deduce', I think I need to use the result... But how?
I thank to any help in advance.
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$\begingroup$There is a bilinear map $b:M\times N\to M$ defined by$$b\left(m,\sum_{i=1}^n r_ie_i\right)=r_1m.$$This induces a linear map $\beta:M\otimes N\to M$ with$$\beta\left(m\otimes\sum_{i=1}^n r_ie_i\right)=r_1m.$$Then $\beta(m_1\otimes e_1)=m_1$ and $\beta(m_i\otimes e_i)=0$for $i\ne 1$. Then$$\beta\left(\sum_{i=1}^n m_i\otimes e_i\right)=m_1$$so $\sum_{i=1}^n m_i\otimes e_i$ determines $m_1$ etc.
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