# Tensor product of modules

Like the free group and the free module, the tensor product of two -modules is another construction satisfying an interesting universal property, but with a richer structure than these two. Namely, it is (when possible) an -module, denoted such that every bilinear map from to another -module uniquely factors through

Given a unitary ring a right -module and a left -module we will first construct the tensor products of and as an abelian group. We can then, when possible, give it a module structure.

**Definition**: The tensor product of and over the ring , noted or is the quotient of the free -module (or equivalently the free abelian group) over by the subgroup generated by all elements of the form ()

;

;

for any and Elements of are called tensors and the cosets, denoted , represented by are called simple tensors.

Note the similarity between this definition and that of the free group or free module: we took all couples in and quotiented out by the minimal relations giving us the most general module possessing the properties we wanted, namely

;

;

for all (Note the similarity between these and the definition of a bilinear map.) It is easy to see that every tensor is a finite sum of simple tensors and that is an abelian group (for addition of tensors).

**Definition:** For and two rings, an –bimodule is a module which is a left -module and a right -module.

Suppose is an -bimodule, then can be given a left -module structure by letting

An important remark is that for a commutative ring and a -module letting the left and right actions of on be equal by gives a natural -bimodule structure and thus gives a left -module structure. To simplify matters, I will therefore assume in the rest of this post that is commutative and that has this natural bimodule structure.

Theorem (Universal property):Let be a commutative and unitary ring, and three -modules and a bilinear map. Then there exists a unique -module homomorphism such that ie such that this diagram commutes :

(Where )

**Proof:** By the universal property of free abelian groups (identifying abelian groups with -modules) the map defines a unique -module homomorphism from the free -module on to such that Since is bilinear, sends every element of the form () to so contains the subgroup by which we quotiented to obtain at the beginning of this post. Letting be thus gives us a well-defined group homomorphism such that Moreover, we have

so is also an -module homomorphism. Since generates (every tensor is a finite sum of simple tensors) we can conclude by linearity that is the unique -module homomorphism satisfying these properties.

QED

Note that for every -module homomorphism the map is an -bilinear map. So by the last theorem, there is a bijection between the sets and

**Reference:** D. S. Dummit, R. M. Foote, Abstract Algebra, 3rd ed.

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