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.


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.

  • 2
    $\begingroup$ Thanks, this makes sense. In the book we are considering the ring $\mathbb{Z} G$ where G is any group. In general it will not be commutative, so I can see why Brown uses this definition. $\endgroup$
    – user194928
    Jun 12 '15 at 3:42
  • $\begingroup$ Perhaps this is implicit but as a remark observe that $M\otimes_R N$ is not supposed to be module for an arbitrary pair of right-left $R$-modules $(M,N)$. Indeed, in Brown's book $M\otimes_R N$ is defined as the "balanced quotient" of $M\otimes_\mathbb{Z} N$ (more precisely $U(M)\otimes_\mathbb{Z} U(N)$, where $U$ is that functor which forgets $R$-actions, but never mind forgetting), which is an abelian group . $\endgroup$
    – Alp Uzman
    Jul 3 '18 at 2:59

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