# 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.

• Isn't the general case done in chapter 3 of Jacobson's Basic Algebra II? – Ivo Terek Apr 28 at 5:23

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).

• Something is wrong about the sidedness in your first Theorem. You could not $R$-tensor $M$ with $N$ if they are both left $R$-modules. – darij grinberg Apr 28 at 2:25