# Definition of tensor product

Here is the standard dedonition of the tensor product of two modules:

Definition for tensor products of two modules:

Let $R$ be a ring and $M$ be a right module and $N$ be a left module.

Let $F(M\times N)$ be the free module on $M\times N$ and $G$ be the submodule of $F(M\times N)$ generated by elements of the form $(m+m',n)-(m,n)-(m',n), (m,n+n')-(m,n)-(m,n'), (mr,n)-(m,rn)$.

Define $M\otimes_R N := F(M\times N)/G$ and call it the tensor product of $M$&$N$.

However, I have seen some notations such as $\otimes_{i=1}^n M_i$ (as in Lang). I'm curious how to expand the above definition of tensor product of two modules to tensor product of multiple modules. How do I construct it?

EDIT:

Lang defines the tensor product of multiple modules as follows:

Let $R$ be a commutative ring and $M_1,...,M_n$ be (left) R-modules. Let $F(\prod_{i=1}^n M_i)$ be the (left) free $\mathbb{Z}$-module on $\prod_{i=1}^n M_i$ and take the quotient by the $R$-submodule generated by following elements:

$(x_1,...,x_i+x_i',...,x_n)-(x_1,...,x_i,...,x_n)-(x_1,...,x_i',...,x_n)$ and $(x_1,...,ax_i,...,x_n)-a(x_1,...,x_i,...,x_n)$.

Then call this quotient module as the tensor prodduct of $M_1,...,M_n$.

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So to make tensor product of multiple modules is it necessary to take $R$ to be commutative? Moreover, I don't understand why these two constructions are equivalent for two modules.

That is, in the above construction $a(n,m)=(n,am)$ need not hold while $a(n,m)=(n,am)$ holds for Lang's construction.

Are these different objects in general? And do we call both of these products as tensor product? Or is there a name to distinguish these two cases?

Or I guess "bimodule" property is a substitue to a hypothesis that $R$ is commutative. Am I right?

This sort of tensor product cannot be extended to more than two modules because in general the tensor product is no longer an $R$-module, only an abelian group. If $N$ is an $(R,R)$-bimodule, then the tensor product is a right $R$-module and we may take the tensor product with a left $R$-module. Again this will only be an abelian group in general. I'm sure you can see how to generalize this to any finite number of factors, or a countably infinite number when the index set is order isomorphic to $\mathbb{N}$ or $\mathbb{Z}$. Without an ordering, though, this kind of tensor product doesn't make sense because of the interplay between the left and right module structures.