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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.

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I'm having trouble seeing what two notions you are thinking of. As you said, a right $R$ module $M$ and a left $R$ module $N$ can be tensored over $R$. If $M$ happens to be an $S-R$ bimodule, then $M\otimes_R N$ has (the obvious) left $S$ module structure. If $N$ is an $R-T$ bimodule, then $M\otimes_R N$ has a right $T$ module structure. If $R$ is commutative then $R=S$ or $R=T$ are possiblities. Are these really different definitions? They all stem from the definition of the group $M\otimes_R N$. –  rschwieb Aug 20 '12 at 13:45
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Dear Paul: I think you're right: the first universal property is a priori stronger than the second. The only way I see to prove that the second implies the first is to check that a solution to the second is indeed a solution to the first. $+1$ –  Pierre-Yves Gaillard Aug 20 '12 at 13:55
    
the methods for constructing the modules explicitly are different in both cases... in the first case you take a free abelian group and quotient out by relations and in the second you take a free R-module and quotient out by relations... maybe the two ideas are not really compatible? –  Paul Slevin Aug 20 '12 at 14:20
    
It seems that it can be done by hand by just checking that the standard free construction of the tensor product of modules satisfies the first universal property, checking generators and well-definedness and so forth. Incidentally it implies that if $F$ is the subgroup of $M \otimes_Z N$ generated by elements of the form $(xr) \otimes_\mathbb Z (y) - x \otimes (ry)$ Then $$\left(M \otimes_\mathbb{Z} N \right) / F \cong M \otimes_R N$$ as abelian groups. –  Paul Slevin Aug 20 '12 at 16:08
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@Paul, that the second construction, using a free presentation, gives the same thing as the 1st one is just a consequence of the fact that the tensor product is right-exact. –  Mariano Suárez-Alvarez Aug 20 '12 at 18:19

4 Answers 4

up vote 4 down vote accepted

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.

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+1 I think your answer is the clearest. –  M Turgeon Aug 20 '12 at 21:49
    
This is exactly what I was looking for. Thank you. –  Paul Slevin Aug 20 '12 at 21:55

[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) $F_1\otimes_RN$.

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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.

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How does one put $R$-linear structure on $Z$? –  Paul Slevin Aug 20 '12 at 19:03

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).

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It is not clear what you are saying. There are two classes of maps being talked about, and each of them has a "universal element" in your terminology. They have no reason to be isomorphic just by virtue of being "universal" (in their own class). –  Uday Reddy Aug 25 '12 at 21:14

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