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?

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    $\begingroup$ 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. $\endgroup$ – 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 $B$.


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