I am working through Cohomology of Groups (Kenneth Brown) and I noticed that in the beginning chapters when tensor products came up, he would emphasize the following: the tensor product $N \otimes_R M$ is defined over a ring $R$ whenever $N$ is a right $R$-module and $M$ is a left $R$-module.
In other textbooks I have studied (in both algebra and algebraic topology), both modules are assumed to be left-modules. I am wondering if there is ever a clash in theory with these two different requirements, or if anyone could elaborate on why someone would choose one definition over the other. Thanks.
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$\begingroup$Suppose $M$ and $N$ are both left modules and we define $M\otimes_R N$ as the span of pure tensors $m\otimes n$ satisfying the additivity relations and $rm\otimes n = m\otimes rn$.
Then for $r,s\in R$, you get $$(rs)(m\otimes n) = (rs)m\otimes n = r(sm)\otimes n = sm\otimes rn = m\otimes (sr)n = (sr) (m\otimes n).$$
So you get $(rs)(m\otimes n) = (sr)(m\otimes n)$ and so the relation $(rs-sr)M\otimes_R N = 0$ is forced on you. This is bad, since for example $R\otimes_R R$ will not be isomorphic to $R$ with this definition unless $R$ is commutative.
If you take $M$ to be a right $R$-module, this problem doesn't arise. The point is that when you 'shift' multiplication between $M$ and $N$ in the tensor product, you reverse the order of multiplication, so they both can't be modules on the same side.
Of course none of this matters if $R$ is commutative, since in that case any $R$-module is a module both on the left and on the right. In contexts where $R$ is always commutative, it makes sense to just describe everything in terms of (left) $R$-modules, since the distinction would not serve a purpose.
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