Given a noncommutative ring $R$, and two (left) $R$-modules $M$ and $N$, how does one define a left action on the the vector space tensor product $M \otimes N$? Multiplying on just the first factor of the tensor product seems a little unnatrual, but I can't see what else to do.



Left-Module Structure on the Tensor Product ofTwo Left  Modules
Let $R, S$ be two (unital and associative to be safe) algebras over a commutative ring $k$ and let $M, N$ be respectively a left $R$-module and a left $S$-module. Then we can define the tensor product $M \otimes\_k N$ by the usual universal property, and it is naturally a left $R \otimes\_k S$-module by functoriality. If $M, N$ are both left $R$-modules, then $M \otimes\_k N$ is a left $R \otimes\_k R$-module, so defining a left $R$-module structure on the tensor product which is compatible with its $k$-module structure is tantamount to giving a morphism of $k$-algebras


$$\Delta : R \to R \otimes\_k R.$$


Note that any such morphism gives a left $R$-module structure to the tensor product of two left $R$-modules, but it won't behave the way you expect it to with respect to multiple tensor products unless $\Delta$ is coassociative and cocommutative. I will be ignoring this.


Now, the ring $R \otimes\_k R$, by its universal property, naturally admits two morphisms $R \to R \otimes\_k R$, namely the two inclusions $i\_1, i\_2 : R \to R \otimes\_k R$, which correspond to the two $R$-module structures described by Peter Samuelson in the comments. If you don't want to just provide $\Delta$ as extra data, then your question can be reinterpreted as follows: 



> 
> What other natural morphisms $R \to R \otimes\_k R$ are there?
> 
> 
> 


Suppose that $\Delta\_R : R \to R \otimes\_k R$ is a family of morphisms which is natural in $R$; that is, it defines a natural transformation. Applying the forgetful functor to $\text{Set}$, we get a natural transformation from the forgetful functor $U : k\text{-Alg} \to \text{Set}$ to the functor


$$U^2 : k\text{-Alg} \ni R \mapsto U(R \otimes\_k R) \in \text{Set}.$$


The functor $U$ is representable by the $k$-algebra $k[x]$, so by the Yoneda lemma, natural transformations $U \to U^2$ can be naturally identified with elements of


$$U^2(k[x]) \cong U(k[x, y]).$$


More concretely, any polynomial in two variables $f(x, y) = \sum f\_{ij} x^i y^j \in k[x, y]$ over $k$ induces a natural map of *sets*


$$R \ni r \mapsto \sum f\_{ij} r^i \otimes r^j \in R \otimes\_k R$$


and these are the only such natural maps. Now the question reduces to the following:



> 
> Which of the maps $r \mapsto \sum f\_{ij} r^i \otimes r^j$ are always morphisms of $k$-algebras? 
> 
> 
> 


The short answer is that it depends a lot on $k$. For starters, letting $R = k[x], r = x, R \otimes\_k R = k[x, y]$ and requiring compatibility with scalar multiplication gives


$$cr \mapsto c f(x, y) = f(cx, cy) \in k[x, y]$$


for all $c \in k$. Comparing coefficients of $x^i y^j$ gives


$$(c - c^{i+j}) f\_{ij} = 0$$


for all $i, j$ and for all $c$. This suggests the possibility of other natural morphisms $R \to R \otimes\_k R$ in characteristic $p$. In particular, if $k = \mathbb{F}\_p$ then we have $c^{p^r} = c$ for all $r \ge 0$, hence $f$ can have components of degree $p^r$ and some of these give rise to genuine natural morphisms such as


$$R \ni r \mapsto r^p \otimes 1 \in R \otimes\_k R.$$


Assume for simplicity that $k$ is an integral domain of characteristic $0$ (these hypotheses can probably be considerably weakened). Then the above condition implies that $f$ is homogeneous and linear, so our desired morphism can only have the form


$$R \ni r \mapsto f\_{10} r \otimes 1 + f\_{01} 1 \otimes r \in R \otimes\_k R.$$


Compatibility with multiplication now requires (taking $R = k[x, y], R \otimes\_k R = k[x, y, z, w]$)


$$xy \mapsto f\_{10} xy + f\_{01} zw = (f\_{10} x + f\_{01} z)(f\_{10} y + f\_{01} w)$$


and comparing coefficients gives


$$f\_{10}^2 = f\_{10}, f\_{01}^2 = f\_{01}, f\_{10} f\_{01} = 0.$$


Our hypotheses then imply that exactly one of $f\_{10}$ or $f\_{01}$ can be equal to $1$ and the other must be equal to $0$. (If $k$ is not an integral domain then other things can happen; in general a choice of $f\_{01}, f\_{10}$ above is equivalent to a direct product decomposition $k = k\_1 \times k\_2$ with respect to which $f\_{10} = (1, 0), f\_{01} = (0, 1)$.) In other words, when $k$ is an integral domain of characteristic $0$ then the only natural morphisms $R \to R \otimes\_k R$ are


$$R \ni r \mapsto r \otimes 1 \in R \otimes\_k R$$


resp. 


$$R \ni r \mapsto 1 \otimes r \in R \otimes\_k R$$


and the corresponding natural left $R$-module structures on $M \otimes\_k N$ are given by multiplication on the first resp. the second factor. In particular, 



> 
> If $k$ is a field of characteristic $0$, then the only natural morphisms $R \to R \otimes\_k R$ are the two obvious ones.
> 
> 
> 


Group algebras are highly misleading here: any group algebra $k[G]$ (more generally any monoid algebra $k[M]$) is naturally equipped with a comultiplication given by extending


$$\Delta : k[G] \ni g \mapsto g \otimes g \in k[G] \otimes\_k k[G]$$


and $\Delta$ induces the usual tensor product of representations of $G$ over $k$. It is worth mentioning that this map is not defined using any of the structure maps of the group $G$; more generally, for any set $S$ there is a canonical diagonal map $\Delta : S \to S \times S$ given by sending $s$ to $(s, s)$, and this induces a canonical coalgebra structure on the free module $k[S]$ given by extending


$$k[S] \ni s \mapsto s \otimes s \in k[S] \otimes\_k k[S].$$


But a generic $k$-algebra is not a free vector space over any distinguished set. 


