Here is the question I wanna solve:
Suppose $R$ is commutative and $N \cong R^n$ is a free $R-$module of rank $n$ with $R-$module basis $e_1, \dots , e_n.$
$(a)$ 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 $i = 1, \dots , n.$
$(b)$ Show that if $\sum m_i \otimes n_i = 0$ in $M \otimes N$ where the $n_i$ are merely assumed to be $R$-linearly independent then it is not necessarily true that all the $m_i$ are $0.$ [Consider $R = \mathbb Z, n = 1, M = \mathbb Z/ 2\mathbb Z,$ and the element $1 \otimes 2.$]
Here is by proof for part(a):
Let $N \cong R^n $ be a free $R$ module of rank $n$ with $R-$module basis $e_1, \dots , e_n,$ let $R$ be a commutative ring, we want to show that, for every non-zero $R$ module $M,$ $M \otimes N$ can be written uniquely by $\sum m_i \otimes e_i$ where $m_i \in M.$
Now, let $x\in M \otimes N.$ Then, $x=\sum_{j = 1}^p (m^{'}_j \otimes n_j)$ where $n_j= \sum_{i=1}^n r_{ji}e_i$ with $r_{ji} \in R$ (by the definition on pg.$354$ because $N \cong R^n $ is a free $R$ module of rank $n$ with $R-$module basis $e_1, \dots , e_n$). So,
\begin{align*}
x &= \sum_{j = 1}^p (m^{'}_j \otimes (\sum_{i=1}^{n} r_{ji} e_i))\\
&= \sum _{j=1}^p(\sum_{i=1}^{n} r_{ji}m^{'}_j) \otimes e_i \\
&= \sum _{i=1}^n ((\sum_{j=1}^{p} r_{ji}m^{'}_j) \otimes e_i) \\
&= \sum _{i=1}^n m_i \otimes e_i \text{ where $m_i \in M.$ }
\end{align*}
as required. Where the second equality is by the third property of tensor product given in eqn.$10.4$ on pg.$360.$ And the last equality is because $M$ is a nonzero $R-$module.
Note that this form is unique because the $r_{ji}'s$ in the definition of the free module $N$ are unique by the definition of a free module.
Now, assume that $\sum_{i = 1}^{n} m_i \otimes e_i = 0 $ in $M \otimes N$ then because $e_1, \dots , e_n,$ is a basis for $N$ then each of them is nonzero (more explicitly $e_i = (0,0, \dots , 0, 1, 0, \dots, 0)$ where the $1$ appears in the position $i.$) and so for $\sum_{i = 1}^{n} m_i \otimes e_i = 0 $ to be satisfied we must have that $m_i = 0$ for all $i = 1, \dots , n$ as $1 \otimes 0 = 0$ by Example $2$ on pg.$363$ and because tensor product is commutative on commutative rings by proposition $20$ on pg. $374$.
My questions are:
The pages in my proof are those of Dummit and Foote, Abstract Algebra, third edition.
1- Is my proof correct regarding to uniqueness part and the proof of the form or should I construct a map like here Dummit and Foote 10.4.10 like this "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." ?
2- Is my justification for the last part of $(a)$ correct? or What specifically makes $m_i = 0$ for all $i$ in part $(a)$?why when the $n_i's$ are merely assumed to be $R$-linearly independent it is not necessarily true that all the $m_i$ are $0?$ I feel like I am using only the idea of $e_i's$ being basis in my argument.
3- In part$(b),$ what is $N$?
Could anyone help me in answering these questions please?
Best Answer
I don't think your proof is really complete. It is true that it follows from the uniqueness of the expansion, but there are steps in between, such as in your link. In that case the map $b$ is well defined because of the uniqueness of the representation.
The second part of $a$ follows immediately from the first since
$$\sum_{i=1}^n m_i\otimes e_i = 0 = \sum_{i=1}^n 0\otimes e_i$$
and you just showed that such representations are unique, so $m_i=0$ for each $i$.
$$1\otimes 2 = 1 \otimes (2\cdot1)=(2\cdot1)\otimes 1=0\otimes 1=0$$
so we have that $1\otimes 2=0$ but $m=1\neq0$.