# Left and right R-modules

OK, this question may be induced by my highly commutative mind and the wee midnight hours. Why do we bother with right R-modules vs left R-modules in view of the fact that given, say, a left R-module (M,+,$*$) (with scalar multiplication $*$) it is instantly a right R-module (M,+,$\circ$) with same addition and right multiplication $m\circ r =: r*m$ -- I believe the axioms for a right module check. In other words, in this way we have an isomorphism of the categories of right and left R-modules: RMod $\cong$ ModR ??? Thus, whatever we say about one of these categories (or their objects and morphisms) is instantly true for the other? No? I may be wrong, but do people use a phrase "let us assume that M is an R-R bimodule" or something like that? Could they mean M is a left R-module and a right R-module with some secret scalar multiplication on the right, other than the one I defined above?

Now that @Qiaochu et al found an error in my thinking, pointing out that the categories of right and left R-modules may be quite different, let alone isomorphic, then why is it, that one waives one's hands often saying "This argument works exactly the same way for right R-modules as the one we just exhibited for the left R-modules." ? This in spite of the fact that the two categories may be totally different. Do we have any fair mechanisms enabling us to know when the proofs are left-right independent, or is it that we can say that only after we actually check the proofs one line at a time...?

• You wrote that you believe the axioms for a right module are satisfied, but clearly you have not checked! Do yourself a favor, and try to check, and that way you will see what goes wrong. – Mariano Suárez-Álvarez Feb 28 '15 at 6:37
• @MarianoSuárez-Alvarez I have now written it down and proved my commutative brain wrong: $m\circ(rs)=r*(s*m)$ whereas $(m\circ r)\circ s=s*(r*m)$, which, as Qiaochu Yuan pointed out, works, only for commutative R. Sorry to annoy you for not checking. I checked it in my head, but it obviously was switching r and s automatically for me. – Rado Feb 28 '15 at 16:05

That only works if $R$ is commutative. If $R$ is noncommutative the correct statement is that left $R$-modules are the same as right $R^{op}$-modules, but it's annoying to have to write ops everywhere. Also, yes, even if $R$ is commutative, it's interesting to talk about bimodules.