# How can I prove that $(m_1+m_2)\otimes n=m_1\otimes n+m_2\otimes n$?

Let $M,N$ two $R-$module. Let $R^{M\oplus N}=Span\{e_{m,n}\}_{(m,n)\in M\oplus N}$. We define $K$ su submodule of $R^{M\oplus N}$ generated by the element of the form $$e_{m_1,n}+e_{m_2,n}-e_{m_1+m_2,n}$$ $$e_{m,n_1}+e_{m,n_2}-e_{m,n_1+n_2}$$ $$e_{rm,n}-re_{m,n}$$ $$e_{m,rn}-re_{m,n}$$

and denote $M\otimes N:=R^{M\oplus N}/K$. Let $$\pi: R^{M\oplus N}\longrightarrow M\otimes N$$ defines by $$\pi(e_{m,n}):=m\oplus n.$$

How can I prove that $$(m_1+m_2)\otimes n=m_1\otimes n+m_2\otimes n\ \ ?$$

My idea was since $e_{m_1,n}+e_{m_2,n}-e_{m_1+m_2,n}\in K$, we have that $$0=\pi(e_{m_1,n}+e_{m_2,n}-e_{m_1+m_2,n})=\pi(e_{m_1,n})+\pi(e_{m_2,n})-\pi(e_{m_1+m_2,n}),$$ and thus

\begin{align*} (m_1+m_2)\otimes n&=\pi(e_{m_1+m_2,n})\\ &=\pi(e_{m_1+m_2,n})+0\\ &=\pi(e_{m_1+m_2,n})+\pi(e_{m_1,n})+\pi(e_{m_2,n})-\pi(e_{m_1+m_2,n})\\ &=\pi(e_{m_1,n})+\pi(e_{m_2,n})\\ &=m_1\otimes n+m_2\otimes n. \end{align*}

Does it work ?

Your answer works perfectly. I am a bit more concerned with the notation in your definition part, however. Up to and including $M\otimes N:=R^{M\oplus N}/K$ everything's fine. After that, I think it would be more natural to turn things around from what you have and instead say "Let $\pi: R^{M\oplus N}\to M\otimes N$ be the canonical projection map, and define the notation $m\otimes n := \pi(e_{m, n})$."
Also, you say "since $e_{m,n_1}+e_{m,n_2}-e_{m,n_1+n_2}\in K$", but it's really $e_{m_1, n}+e_{m_2, n}-e_{m_1 + m_2, n}$ you're using. That's probably just a typo, though.
• What is you definition of $m\otimes n$ ? That is a bilinear map ? It's not may definition, and my definition is precisely the one I wrote ! So your answer doesn't answer my question. – user386627 Jan 20 '17 at 17:33