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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?

Thank you in advance :)

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?

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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.

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  • $\begingroup$ Would you check my edit? $\endgroup$ – Rubertos Mar 8 '15 at 17:02
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    $\begingroup$ Ah, the marvellous I-answer-but-the-question-changes-somehow-could-you-answer-to-the-edit-situation... ;-) $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Mar 8 '15 at 17:06
  • $\begingroup$ @RobertGreen So do you think it's better to post another one for the edit? I was hesitating since the new post would have the same title.. $\endgroup$ – Rubertos Mar 8 '15 at 17:08
  • $\begingroup$ @Rubertos I honestly don't know, but I know that it could be sometimes a little bit annoying to answer clearly to something and to discover that the question changes etc... Anyway, Matt's answer is the only good answer that can be given, I'll only add this : look at Bourbaki, Algèbre, chapitre II, paragraph 3, section 1, and following sections, you will find everything you want to know about your question. $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Mar 8 '15 at 17:11
  • $\begingroup$ @RobertGreen Thank you for the reference of a great book. It's weird that I never tried to look at it even though I'm a fan of Bourbaki.. $\endgroup$ – Rubertos Mar 8 '15 at 17:18

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