Let $k$ be a field. Let $V$ and $W$ be vectors spaces over $k$ with bases $v_1,\dotsc,v_m$ and $w_1,\dotsc,w_n$, respectively.
I know the definition of a tensor product over a commutative ring (say, over $k$): It is a $k$-module $V\otimes W$ with a $k$-bilinear map $\alpha:V\times W\rightarrow V\otimes W$ such that every $k$-bilinear map $f:V\times W\rightarrow U$ factors through a unique $k$-linear map $V\otimes W\rightarrow U$.
Now, such $k$-bilinear maps $f$ satisfies: $f(\sum x_i\cdot v_i,\sum y_j\cdot w_j)=\sum_{i,j}x_i\cdot y_i\cdot f(v_i,w_j)$. So, $f$ is determined by its values on pairs $(v_i,w_j)$, and I know how to continue from here to realize $V\otimes W$ as the $nm$-dimensional $k$-vector space with basis $\{v_i\otimes u_j\}_{i,j}$.
Now I am looking at the definition of a tensor product of modules over noncommutative rings, and I want to apply it in the same setting as before to check if I get the same thing (I assume I should get the same thing). I am reading in wikipedia.
So, I'm thinking of $V$ as a right $k$-module and of $W$ as a left $k$-module, and I am considering $k$-balanced maps $f:V\times W\rightarrow A$ (where $A$ is an abelian group).
This time I only get $f(\sum x_i\cdot v_i,\sum y_j\cdot w_j)=\sum_{i,j}f(v_i\cdot x_i,y_i\cdot w_j)$. I can move the $x_i$ and $y_j$ between the left and right components, but I can't get the scalars out of $f$. So, it doesn't seem like $f$ is determined by its values on basis pairs, leaving me perplexed.
Is this tensor product in the noncommutative setting the same module as in the commutative setting? Am I interpreting the definition correctly? Am I looking at the right definition (my goal is to understand whether the definition for noncommutative rings generalizes the definition for commutative rings).
The next thing I want to do is check what happens over division rings, but first I must resolve this.
More generally, if $R$, $S$ and $T$ are rings and we have bimodules ${}_RM_S$ and ${}_SN_T$, then the tensor product $$ M\otimes_S N $$ is in a natural way an $R$-$T$-bimodule ${}_R(M\otimes_S N)_T$.
When $S$ is a commutative ring, a module $M_S$ can be considered as an $S$-$S$-bimodule with $axb=x(ab)=(ba)x$. With this convention, a balanced map is also a bilinear map, if you think to be “corestricted” to its image, which is in a natural way a module over the ring.
So $V\otimes_kV$ is the same thing for both approaches.