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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.

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  • $\begingroup$ Take a module $P$ which has two distinct structures of an $R$-module. ($R=P=\mathbf C$ is an acceptable choice.) Then it seems that a bilinear map for the first structure is only balanced for the second one. $\endgroup$ – ACL Dec 10 '14 at 9:50
  • $\begingroup$ @ACL: yes, two different module structures give a counterexample. The question seems quite easy now:) $\endgroup$ – mathuser77 Dec 10 '14 at 11:25
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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.

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I guess I have a counterexample.

Let $G$ be an abelian group and form the group algebra $kG$ which is commutative. Pick any $kG$-module $M$ and define a structure of module on the dual $M^*$ via $[g.f](m)=f(mg)$. Actually does not matter on which side one multiplies by $g$ since the algebra is commutative. Then the evaluation map $M\times M^* \rightarrow k$ is balanced but not bilinear when $k$ is regarded as the trivial module, $g.1=1$ for any $g\in G$.

I guess it should also be a more elementary counterexample. When teaching tensor products the students might not be acquainted with group algebras.

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  • $\begingroup$ This example seems prety elementary if you take G to be the three element group, in which case kG is isomorphic to three copies of k (at least for char k not 3) $\endgroup$ – Yemon Choi Dec 10 '14 at 9:49
  • $\begingroup$ yes, I guess one can somehow "hide" the group structure in the algebra. $\endgroup$ – mathuser77 Dec 10 '14 at 10:30
  • $\begingroup$ I would prefer to say that you are hiding some kind of group structure in the action of $k^3$ on $k$. $\endgroup$ – Yemon Choi Dec 10 '14 at 12:25
  • $\begingroup$ @ Yemen Choi: exactly! $\endgroup$ – mathuser77 Dec 10 '14 at 13:01

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