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I was reading something about tensor product (balanced product) of modules (over an arbitrary ring $R$), but I cannot realize why we need a left and a right module. Would it be the same with two right (or left) modules? What make this situation so special?

Edit: Another question

Is the tensor product also a module over R or it just a left (right) one? Why?

The definition I've read is the one that uses factor groups of free abelian groups.

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Is your ring commutative or arbitrary? –  Seth Apr 13 at 23:20
    
Arbitrary. Anyway if one can make clear what happen in both case I will appreciate that. –  W4cc0 Apr 13 at 23:21
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If you have a commutative ring then the left/right module stuff doesn't matter because they are equivalent. For a general ring there is a difference between a left and right module. The tensor product is in general an abelian group. If your first module is also an $(S,R)$-bimodule then the tensor product is a left $S$ module. –  Seth Apr 13 at 23:22

1 Answer 1

up vote 2 down vote accepted

The problem is simple: if $A$ and $B$ are both left $R$-modules, we'd want to equate $(ra,b)$ with $(a,rb)$ in the tensor product. But then we'd have $(rsa,b)=(a,rsb)$ and $(rsa,b)=(sa,rb)=(a,srb)$! So this construction forces $R$ to act commutatively on $A\otimes B$.

In general the tensor product of a right and a left $R$-module is merely an abelian group. There's no module structure for essentially the same reason as above: for instance the natural left module structure would come from $B$, but we'd be forced again to say $$(a,rsb)=r(a,sb)=r(as,b)=(as,rb)=(a,srb)$$

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