Equality of two notions of tensor products over a commutative ring 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.
 A: 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 uniqueness of $f^*\psi$ as a group homomorphism is clear.
A: [This answers a question raised in the comments above]
Let $R$ be a (not necessarily commutative) ring and let $M_R$ and ${}_RN$ be a right and a left $R$-module, respectively. Pick a free presentation $$F_1\to F_0\to M\to 0$$ of $M$ as a right $R$-module, that is, a short exact equence of the form above with $F_0$ andd $F_1$ free right $R$-modules and $R$-linear maps. This amounts to writing $M$ in terms of generators and relations.
Since the tensor product $(\mathord-)\otimes_RN$ is a right exact functor, applying it to the sequence above preserves exactacts, so we get a short exact sequence $$F_1\otimes_RN\to F_0\otimes_RN\to M\otimes_RN\to 0$$ This means, precisely, that $M\otimes_RN$ is the quotient of $F_0\otimes_RN$ by the (image of) $F_1\otimes_RN$.
A: Keeping track of the full categorical characterization of things clarifies this: in either category, there is also $j:M\times N\rightarrow M\otimes_R N$ which is bilinear and $j(rm\times n)=j(m\times rn)$. In the category of $R$-modules ($R$ commutative) there is the further requirement that $M\otimes_R N$ be an $R$-module and that $j(rm\times n)=r\cdot j(m\times n)$. 
In fact, even in the category of abelian groups, there is the obvious canonical $R$-module structure on $M\otimes_R N$, namely, that induced from $r\cdot (m\otimes n)=rm\otimes n=m\otimes rn$. 
Further, given a bilinear map $B:M\times N\rightarrow Z$ with $B(rm\times n)=B(m\times rn)$ in the larger category, inducing unique $\mathbb Z$-linear $\beta:M\otimes_R N\rightarrow Z$, if it should happen that $Z$ is an $R$-module and $B(rm\times n)=r\cdot B(m\times n)$, then provably $\beta$ is $R$-linear: $\beta(rm\otimes n)=B(rm\times n)=r\cdot B(m\times n)=r\cdot \beta(m\otimes n)$.
Edit: in response to the question "how to put $R$-linear structure on $Z$?", I think one does not attempt to do so. Rather, if $Z$ "happens to have" this structure, then the discussion goes through.
A: Let $M\otimes_RN$ be be the tensor product in the first sense (the abelian group construction). As you pointed out, $M\otimes_RN$ has an $R$-module structure, and it is a tensor product in the second sense (the $R$-module construction). Being a universal element, any tensor product in the second sense must then be isomorphic to $M\otimes_RN$, so must have the universal property of the first kind (the abelian group construction).
