$𝑅$ -tensor product of modules

Unlike in the special case of the $K$ -tensor product of modules, 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 $𝑀 \otimes 𝑅 𝑁$ is an abelian group such that the $𝑅$ -balanced maps from $𝑀 \times 𝑁$ are in correspondence with the group homomorphisms from $𝑀 \otimes 𝑅 𝑁$ , 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 $𝑀 \otimes 𝑅 𝑁$ together with an $𝑅$ -balanced map $(\otimes) : 𝑀 \times 𝑁 \to 𝑀 \otimes 𝑅 𝑁$ such that any $𝑅$ -balanced map $𝜑 : 𝑀 \times 𝑁 \to 𝐺$ factorizes uniquely through $(\otimes)$ #m/def/module

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such that $―― 𝜑$ is a group homomorphism.

Construction

Let $ℤ (𝑀 \times 𝑁)$ be a free $ℤ$ -module free abelian group on $𝑀 \times 𝑁$ with the natural inclusion function $𝜄 : 𝑀 \times 𝑁 ↪ ℤ (𝑀 \times 𝑁)$ . Let $𝐾$ denote the $ℤ$ -Submodule (subgroup) of $ℤ (𝑀 \times 𝑁)$ 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 \cdot 𝑟, 𝑛) - 𝜄 (𝑚, 𝑟 \cdot 𝑛);

for any $𝑚, 𝑚' \in 𝑀$ , $𝑛, 𝑛' \in 𝑁$ , $𝑟 \in 𝑅$ . We construct the tensor product as the quotient $ℤ$ -module

𝑀 \otimes 𝑅 𝑁 = ℤ (𝑀 \times 𝑁) / 𝐾

with its natural projection $𝜋 : ℤ (𝑀 \times 𝑁) ↠ 𝑀 \otimes 𝑅 𝑁$ , so that the map

(\otimes) = 𝜋 \circ 𝜄 : 𝑀 \times 𝑁 \to 𝑀 \otimes 𝑅 𝑁

Proof of the universal property

By construction $(\otimes)$ is $𝑅$ -balanced. Let $𝜑 : 𝑀 \times 𝑁 \to 𝐺$ 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 $𝐾 \leq k e r ˜ 𝜑$ , 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 $𝑀 \otimes 𝑅 𝑁$ is naturally equipped with the structure of a $(𝑇, 𝑆)$ -bimodule. If $𝑅$ is commutative, then we recover the $K$ -tensor product of modules by considering $𝑅$ -bimodules $𝑀$ and $𝑁$ this way.

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