When defining tensor products $M\otimes\_R N$ over a commutative ring $R$ one can use a universal property with respect to bilinear maps $M\times N\rightarrow P$, for any $R$-module $P$.


On the other hand, in the general case, for noncommutative rings one has to use balanced maps $M \times N \rightarrow Z$ instead of bilinear. Of course, in the second case $Z$ is just an abelian group.


I recall that $f$ is bilinear if $f(mr, n)=rf(m, n)=f(m, rn)$ while $f$ is balanced if and only if $f(mr,n)=f(m, rn)$.


In the case of a commutative ring, since any bilinear map is also balanced it is clear that the "tensor product with bilinear maps" also satisfies the universal property for balanced maps. Therefore by the unicity of the solution of the universal problem implies that they coincide as abelian groups.


I have the following question:
 Is there an example of a balanced map $M\times N \rightarrow P$ which is not bilinear? I cannot construct one by myself. Here I assume R commutative so I can speak about bilinear maps.



Tensor product-definition-balanced versus bilinear maps
You seem to be asking whether any **group** homomorphism $M\otimes\_RN\to P$ is $R$-linear. There is no reason why this should be true. As a huge overkill, consider $\Bbb R\otimes\_{\Bbb R}\Bbb R=\Bbb R\to\Bbb R$: there are uncountably many group homomorphisms (even $\Bbb Q$-linear maps) that are not $\Bbb R$-linear.


