On the definition of tensor product for modules of commutative and non-commutative rings. Let $A$ be a ring and $M_A$ and  $_AN$ be two $A$-modules. The tensor product $T = M \otimes_A N \in \mathsf{Ab}$ for $A$ non necessarily commutative is an abelian group, where as the tensor product $T' = M \otimes_A N$ of $M$ and $N$ as $A$-modules over a commutative ring has the structure of an $A$-module. 
When $A$ is commutative, can $T$ be given an $A$-module structure such that $T \simeq T'$ in $A-\mathsf{mod}$? Is this a 'canonical' construction? If so, I would highly appreciate a categorical reformulation of this statement.
 A: There are many ways to see this, but one is the following: in general, if $M$ is an $A$-$B$-bimodule and $N$ is a $B$-$C$-bimodule, then we can define $M\otimes_B N$, and it is an $A$-$C$-bimodule. We get the relations $mb\otimes n = m\otimes bn$, $a\cdot (m\otimes n)=(am)\otimes n$ and $(m\otimes n)\cdot c = m\otimes (nc)$.
Now a right $A$-module is the same thing as a $\mathbb{Z}$-$A$-bimodule, and a left $A$ module is the same thing as a $A$-$\mathbb{Z}$-bimodule, so in the setting of the question, we get $M\otimes_A N$, which is a $\mathbb{Z}$-$\mathbb{Z}$-bimodule, which is the same thing as an abelian group.
But when $A$ is commutative, $M$ and $N$ are both $A$-$A$-bimodules in a canonical way (the left and right actions of $A$ are the same, which makes sense because $A$ is commutative), so $M\otimes_A N$ is also an $A$-$A$-bimodule, and it is easy to check that the left and right action of $A$ coincide, so this is just an $A$-module. (Very quickly, $a(m\otimes n)=am\otimes n = m\otimes an = (m\otimes n)a$).
