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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 ?

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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.

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    $\begingroup$ 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. $\endgroup$ – user386627 Jan 20 '17 at 17:33
  • $\begingroup$ @user386627 Usually, tensor product is defined (up to unique isomorphism) as the universal bilinear map, and then we would use your construction to show that it exists. But sure, I'll write something a bit more substantial up. You would need to prove this property for your construction anyhow. $\endgroup$ – Arthur Jan 20 '17 at 17:35
  • $\begingroup$ In fact, I'm exactely studing the proof of the existence of the tensor product... In the theorem, we indeed make such a construction :) $\endgroup$ – user386627 Jan 20 '17 at 17:36
  • $\begingroup$ @user386627 I think my answer is a bit more related to your question this time around. $\endgroup$ – Arthur Jan 20 '17 at 17:48
  • $\begingroup$ Thank you, I corrected it. $\endgroup$ – user386627 Jan 20 '17 at 18:05

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