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Let $M$ and $N$ be modules over a ring $R$. Generally, the tensor product $M\otimes N$ is defined to be an abelian group with a balanced map $j:M\times N\to M\otimes N$ such that for any abelian group $G$ with a balanced map $i:M\times N\to G$ there is a unique homomorphism $\phi:M\otimes N\to G$ such that $\phi\circ j=i$.

My question is why not to define the tensor product as following:

Suppose $M$ and $N$ are left $R$-modules, the tensor product $M\otimes N$ is defined to be a left $R$-module with a bilinear map $j:M\times N\to M\otimes N$ such that for any $R$-module $G$ with a bilinear map $i:M\times N\to G$ there is a unique homomorphism $\phi:M\otimes N\to G$ such that $\phi\circ j=i$.

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    $\begingroup$ How do you define a bilinear map when both $M$ and $N$ are left $R$-modules? (Also, in the first definition, please clarify the left/rightness of the modules -- that might explain some things.) $\endgroup$ – darij grinberg May 5 '15 at 15:59
  • $\begingroup$ @darij grinberg: $f: M \times N \to G$ is bilinear, if $f(-,n), f(m,-)$ are $R$-linear for all $m \in M, n \in N$. $\endgroup$ – tj_ May 5 '15 at 23:16
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One reason is this leads to modules that behave poorly with respect to commutators. For example, let $m\in M$, $n\in N$, and suppose $r,s\in R$ don't commute. Then $$rm\otimes sn=rs(m \otimes n)=sr(m \otimes n)$$ So any commutator $rs-sr$ annihilates the whole tensor product.

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  • $\begingroup$ How do you choose $i,\phi$ to obtain the identity $rs(m \otimes n) = sr(m \otimes n)$ ? $\endgroup$ – tj_ May 5 '15 at 23:32
  • $\begingroup$ $\phi(rm,sn)=r\phi(m,sn)=rs\phi(m,n)$. But you could also pull $s$ out first then $r$. This would happen with any bilinear map. $\endgroup$ – Matt Samuel May 6 '15 at 0:55
  • $\begingroup$ Yes, thanks for the clarification. $\endgroup$ – tj_ May 6 '15 at 7:51

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