Let $R$ be a ring (not necessarily commutative), let $M$ be a right $R$-module and let $N$ be a left $R$-module. Then the tensor product $M \otimes_R N$ is an abelian group satisfying the universal property that for every abelian group $Z$ and every bilinear map $f \colon M \times N \to Z$ (that is a map which is additive in both variables and such that $f(mr,n) = f(m,rn)$ ) there exists a unique group homomorphism $f^* \colon M \otimes_R N \to Z$ such that $f^*(m \otimes n) = f(m,n)$.
If $R$ is commutative then an $R$-module structure can be put on $M \otimes_R N$. It is quite clear that this $R$-module satisfies the usual universal property of the tensor product of $R$-modules: that is for any $R$-module $C$ and any bilinear map $f \colon M \times N \to C$ there is a unique $R$-linear map $f^* \colon M \otimes_R N \to C$ such that $f^* (m \otimes n) = f(m,n)$.
My problem is going the other way however. How can I prove the first universal property from the second? The first property is almost a special case of the second, since abelian groups are $\mathbb{Z}$- modules, but I can't work out how to induce the existence of a map from $M \otimes_R N \to Z$ since $M \otimes_R N$ is not necessarily the tensor product over $\mathbb{Z}$ of $M$ and $N$.
Can anyone help me out? Thanks very much.
Best Answer
Let $R$ be a commutative ring. Let $M$ and $N$ be $R$-modules. Let $M\otimes_R N$ be the tensor product in the former sense. Let $M\tilde\otimes_R N$ be the tensor product in the latter sense. We can regard $M\otimes_R N$ as an $R$-module in the obvious way. Then there exists the unique $R$-linear map $\psi\colon M\tilde\otimes_R N \rightarrow M\otimes_R N$ such that $\psi(m\tilde\otimes n) = m\otimes n$.
Let $Z$ be an abelian group. Let $f:M\times N \rightarrow Z$ be a bilinear map such that $f(rm, n) = f(m, rn)$.
There exists the unique group homomorphism $f^* \colon M \otimes_R N \to Z$ such that $f^*(m \otimes n) = f(m,n)$.
Then $f^*\psi\colon M\tilde\otimes_R N \rightarrow Z$ is a group homomorphism such that $f^*\psi(m \tilde\otimes n) = f(m,n)$. Since the set of $m\tilde\otimes n$ generates $M\tilde\otimes_R N$ as a group, The uniquness of $f^*\psi$ as a group homomorphism is clear.