# Tensor product of modules: should it be abelian group or module?

If $$A$$ is a right $$R$$-module and $$B$$ is a left $$R$$-module, then the tensor product $$A\otimes_R B$$ is an object $$X$$ with a map $$\theta:A\times B\rightarrow X$$ such that $$\theta$$ is $$R$$-bilinear and $$\theta(ar,b)=\theta(a,rb)$$ for all $$a\in A, b\in B, r\in R$$; further the pair $$(X,\theta)$$ has following universal property. If $$X'$$ is an object and $$\theta':A\times B\rightarrow X'$$ is $$R$$-bilinear map with $$\theta'(ar,b)=\theta'(a,rb)$$ then there is unique $$g:X\rightarrow X'$$ such that $$\theta\circ g=\theta'$$.

In some books, the tensor product $$A\otimes_R B$$ is defined to be an abelian group $$X$$ with map $$\theta$$ satisfying universal properties as stated above. Whereas in some books, it is defined to be an $$R$$-module $$X$$ with a map $$\theta:A\times B\rightarrow X$$ satisfying above universal properties.

What is appropriate choice for tensor product, as module or just as abelian group?

If we accept it as abelian group, then we can give it a left $$R$$-module structure or right $$R$$-module structure; so it seems to me that it should considered as abelian group only (together with map $$\theta$$).

Whereas, the places, where the tensor product is defined to be a module, they do not mention whether it is left $$R$$-module or right $$R$$-module.

For definition of tensor product, we need to take one left $$R$$-module and one right $$R$$-module; so we can not define tensor product to be universal object within some category of left $$R$$-modules or within category of some right $$R$$-modules.

In above write-up, $$\theta$$ can be said to be balanced map; and tensor product is defined to be an object with a balanced map, satisfying universal property.

But my question is concerned about what the object there should be; whether it should be just abelian group of $$R$$-modules?

• If $R$ is commutative, then $A\otimes_R B$ is an $R$-module. But when $R$ is not commutative you cannot define an $R$-module structure: for example, you need $A$ to be a bi-$R$-module then the left action of $R$ on $A\otimes_R B$ makes sense. – Eclipse Sun Apr 18 at 4:36

For commutative $$R$$, the usual construction gives an object $$A\otimes_R B$$ that's an $$R$$-module with action $$r(a\otimes b) = (ra) \otimes b = a\otimes (rb)$$. In the noncommutative case, $$A\otimes_R B$$ is still an abelian group, but the action above doesn't carry over; about the best you can do is to get a left $$R$$-module structure by setting $$r(a\otimes b) = (ra)\otimes b$$ if $$A$$ is an $$R$$-bimodule, or similarly (but distinctly) for $$B$$.