# Tensor product of vector spaces by the "noncommutative definition"

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.

• There are no bases in general, for modules. Apr 23 '17 at 16:28
• @egreg: Sure, but what I am asking is simpler: Take a $k$-vector spaces, think of $k$ as a general ring. Apply the definition of tensor product of modules over general rings. What do we get? Apr 23 '17 at 16:31
• Vector spaces over fields are, in a natural way, bimodules. With this approach, you get exactly the same thing. Apr 23 '17 at 16:35
• @egreg: Yes, I am aware of that. This allows me to make $V\otimes W$ a $k-k$-bimodule as well. But is it $nm$ dimensional as in the commutative case? I cannot see it from the definition. Apr 23 '17 at 16:37
• @egreg: When you say "with this approach", do you mean that we must take the universal property with respect to balanced maps which respect the bimodule structure of the vector spaces? I think this indeed gives the sake result as in the commutative case, but this seems to be different than applying the definition for general rings. Do we then get two different notions of "tensor product over $k$" which really are non-isomorphic? Apr 23 '17 at 16:40

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.
• I think everything can be clarified for me if I see the definition of tensor product in the context of bi-modules ($R-S$-module $\otimes$ $S-T$-module). What is the exact universal property? Apr 24 '17 at 10:29
• @NathanRikard You take $S$-balanced maps, linear in $R$ and $T$, with codomain an $R$-$T$-bimodule. Apr 24 '17 at 10:53