Tensor products of modules over non-commutative rings I've been learning about tensor products over modules, but where the ring acting on the module is commutative.
When $R$ is non-commutative, we consider a right $R$-module $M$ and a left $R$-module $N$, and look at middle-linear maps instead of bilinear maps. 
I can't find many good resources on the tensor product when $R$ is non-commutative. Does anyone know of any good resources from which I can get a better understanding of it? Thank you.
 A: I think this source looks good.
As far as tensor products over noncommutative rings are concerned, it is crucial to know this short list of results:

Theorem: Let $R$ and $S$ be rings. Let $M$ a right $R$-module. Let $N$ be a left $R$-module and a right $S$-module. Let $L$ be a right $S$-module. Then there is an isomorphism of abelian groups
  $$ \text{Hom}_S ( M \otimes_R N, L) \cong \text{Hom}_R ( M, \text{Hom}_S (N, L))$$
  Which, fixing $N$, is natural in $M$ and $L$.

In, fact, there are two hom tensor adjunctions. One has $\text{Hom}_S (-, -)$ in the adjunction above as maps of left $S$-modules, and another has $\text{Hom}_S (-, -)$ as maps of right $S$-modules.

Corollary: Let $M$ Be an $R$-$S$-bimodule. The functor $\text{Hom}_R (M, -) : R \text{-mod} \rightarrow S \text{-mod}$ commutes with limits (e.g. products, kernels). So does $\text{Hom}_R (M, -) : \text{mod-} R \rightarrow \text{mod-} S$. Similarly, with tensor and colimits (e.g. direct sums, cokernels).

Another corollary of this is the instance when $S$ is an $R$-algebra. Then $\text{Hom}_S (S, -)$ naturally isomorphic to the forgetful functor and $S \otimes_S -$ is naturally isomorphic to the forgetful functor. Then we get an adjunction involving the forgetful functor.
Also it is good to know these properties of tensor:

Theorem: Let $R$ and $S$ be rings. Let $N$ be a right $R$-module, let $M$ be a left $R$-module and a right $S$-module. Let $L$ be a left $S$-module. Then $(N \otimes_R M) \otimes_S L \cong N \otimes_R ( M \otimes_S L)$. 

and

Theorem: Let $M$ be a left $R$-module. Then $R \otimes_R M \cong M$. (similarly for right $R$-modules).

