Module theory MOC

Tensor product of modules over a noncommutative ring

Unlike in the special case of the tensor product of modules over a commutative ring, the general tensor product of modules may itself lack module structure. Let 𝑅 be a (noncommutative) ring, 𝑀 be a right 𝑅-module and 𝑁 be a left 𝑅-module. The tensor product 𝑀 ⊗𝑅𝑁 is an abelian group such that the 𝑅-balanced maps from 𝑀 ×𝑁 are in correspondence with the group homomorphisms from 𝑀 ⊗𝑅𝑁, as defined by the Universal property.

Universal property

Let 𝑀 be a right 𝑅-module and 𝑁 be a left 𝑅-module. The tensor product is a pair consisting of an abelian group 𝑀 ⊗𝑅𝑁 together with an 𝑅-balanced map ( ⊗) :𝑀 ×𝑁 →𝑀 ⊗𝑅𝑁 such that any 𝑅-balanced map 𝜑 :𝑀 ×𝑁 →𝐺 factorizes uniquely through ( ⊗) module

such that ――𝜑 is a group homomorphism.

Construction

Let ℤ(𝑀×𝑁) be a free ℤ-module free abelian group on 𝑀 ×𝑁 with the natural inclusion function 𝜄 :𝑀 ×𝑁 ↪ℤ(𝑀×𝑁). Let 𝐾 denote the ℤ-Submodule (subgroup) of ℤ(𝑀×𝑁) generated by any elements of the form

𝜄(𝑚,𝑛+𝑛′)−𝜄(𝑚,𝑛)−𝜄(𝑚,𝑛′);𝜄(𝑚+𝑚′,𝑛)−𝜄(𝑚,𝑛)−𝜄(𝑚′,𝑛);𝜄(𝑚⋅𝑟,𝑛)−𝜄(𝑚,𝑟⋅𝑛);

for any 𝑚,𝑚′ ∈𝑀, 𝑛,𝑛′ ∈𝑁, 𝑟 ∈𝑅. We construct the tensor product as the quotient ℤ-module

𝑀⊗𝑅𝑁=ℤ(𝑀×𝑁)/𝐾

with its natural projection 𝜋 :ℤ(𝑀×𝑁) ↠𝑀 ⊗𝑅𝑁, so that the map

(⊗)=𝜋∘𝜄:𝑀×𝑁→𝑀⊗𝑅𝑁

Proof of the universal property

By construction ( ⊗) is 𝑅-balanced. Let 𝜑 :𝑀 ×𝑁 →𝐺 be 𝑅-balanced. By the universal property of the free module we have a unique ℤ-linear map ˜𝜑 such that the following commutes:

and by the 𝑅-balance of 𝜑 it follows 𝐾 ≤ker⁡˜𝜑, so by the universal property of the quotient module ˜𝜑 factors uniquely through 𝜋, yielding the commutative diagram

as required.

Tensor product of bimodules

Note that if 𝑀 is a (𝑇,𝑅)-bimodule and 𝑅 is a (𝑅,𝑆)-bimodule then 𝑀 ⊗𝑅𝑁 is naturally equipped with the structure of a (𝑇,𝑆)-bimodule. If 𝑅 is commutative, then we recover the Tensor product of modules over a commutative ring by considering 𝑅-bimodules 𝑀 and 𝑁 this way.

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