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I am confused of the definition of tensor product of modules over a non commutative ring.

First let $R$ be a commutative ring and let $M$ and $N$ be two $R$ modules. Let $F_{R}(M\times N)$ be the free $R$ module over the set $M\times N$ and let $K$ be the submodule generated by elements of the form $(x,y_1+y_2)-(x,y_1)-(x,y_2)$, $(x_1+x_2,y)-(x_1,y)-(x_2,y)$, $(rx,y)-r(x,y)$, $(x,ry)-r(x,y)$. Then the tensor product is $F_{R}(M\times N)/K$. My doubt is why cant we do the same if R is a non commutative ring? Where is the commutativity of $R$ is used above? Please help me.

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    There's no reason why you can't construct the corresponding tensor product for noncommutative rings (though it makes more sense to do so by the universal property than with an explict construction, and it's not as crucial as it is in the commutative case); it's just that tensor products are usually introduced in introductory commutative algebra classes. – 2017-01-10

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You can do the same thing for a noncommutative ring; it's just not as useful and so is not a standard definition. Notice that these relations imply that $$rs(x\otimes y)=r(sx\otimes y)=sx\otimes ry=s(x\otimes ry)=sr(x\otimes y)$$ for any $r,s\in R$ and any $x\in M$, $y\in N$. So $R$ will act "commutatively" on the tensor product "$M\otimes N$" defined in this way: the action will factor through the quotient $R/[R,R]$ by the commutator ideal. So constructing tensor products in this way loses all information about the noncommutativity of $R$ (and of its action on the modules $M$ and $N$). This is rarely useful when thinking about noncommutative rings.

(Indeed, even if you do want to talk about this construction for a noncommutative ring, you don't need to, since you can define it just using the tensor product of modules over a commutative ring. For the tensor product of $M$ and $N$ defined in this way is naturally isomorphic to the tensor product $M/[R,R]M\otimes_{R/[R,R]} N/[R,R]N$ of $R/[R,R]$-modules, with its natural $R$-module structure.)

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    Thank you so much for a clear reply. – 2017-01-10
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    What do you mean by $rs(x\otimes y)$? It seems to me that if $R$ is noncommutative, then the tensor product of two $R$-modules is not necessarily an $R$-module (unless one has the structure of a $R$-bimodule). Even if $M$ is a right module and $N$ a left module, if you try to force an $R$-action on $M\otimes_R N$ (defined in the usual way by identifying $(mr,n) = (m,rn)$) via the $R$-module structures on $M$ or $N$, one finds for the same reason you wrote that the action will factor through $R/[R,R]$. So, I'm still confused why one wants to only tensor right modules by left modules.... – 2018-11-07
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    ...and correspondingly quotient by the relation $(mr,n) = (m,rn)$... – 2018-11-07
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    @oxeimon: I am following the definition proposed in the question, in which $M$ and $N$ are both left $R$-modules and $M\otimes_R N$ is defined to be a left $R$-module. You are right that for the usual definition of $M\otimes_R N$ when $M$ is a right $R$-module and $N$ is a left $R$-module, the tensor product has no action of $R$. – 2018-11-07
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    @EricWofsey But then I don't understand the reason why we restrict to tensoring right modules by left modules. If we alternatively define the tensor product of two left $R$-modules (for $R$ noncommutative) by quotienting out by $(rm,n) = (m,rn)$, what is lost? (We're not losing any $R$-module structure on the tensor product since we didn't really have it to begin with) – 2018-11-07
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    @oxeimon: If you quotient out by that relation, then you get exactly the same definition as proposed in the question. Indeed, notice that in that case $r(m,n)=(rm,n)=(m,rn)$ will give a well-defined left action of $R$. – 2018-11-07
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    Another way to say it is that you can do the argument I gave in my answer without ever pulling the scalars all the way out: $$rsx\otimes y=sx\otimes ry=x\otimes sry=srx\otimes y$$ – 2018-11-07
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    @EricWofsey Ahh! Perfect. Thanks. – 2018-11-07