Dummit and Foote 10.4.10

It is a problem 10.4.10 of Dummit and Foote.

1. 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.

• Did you address the word "uniquely" in your answer to the first part? – Lord Shark the Unknown Dec 20 '18 at 7:57
• @LordSharktheUnknown Oh! I did not. Does that mean I need to change the way to prove that? – Lev Ban Dec 20 '18 at 7:59
• Ensure that the title contains some actual words, and not only LaTeX -- the search functionality of MSE does not interact well with typeset mathematics, making questions comprising only LaTeX (much) harder to find. Furthermore, the MathJax context menu (which pops up when you right-click on some typeset expression) overrides the browser's link context menu, making e.g. opening the question in a new tab difficult. – user10354138 Dec 20 '18 at 8:04

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.