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...?

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    $\begingroup$ 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. $\endgroup$ – Mariano Suárez-Álvarez Feb 28 '15 at 6:37
  • $\begingroup$ @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. $\endgroup$ – 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.

  • $\begingroup$ It is enough to have an antiautomorphism, actually. $\endgroup$ – Mariano Suárez-Álvarez Feb 28 '15 at 6:36
  • $\begingroup$ Sure, but that's extra structure rather than an extra property. $\endgroup$ – Qiaochu Yuan Feb 28 '15 at 6:47
  • $\begingroup$ It is the property of being isomorphic to its opposite rinG—which is enough for there to be an isomorphism of categories. $\endgroup$ – Mariano Suárez-Álvarez Feb 28 '15 at 8:38
  • $\begingroup$ Fine, let me say what I meant to say. When I said "that only works" I meant "that particular map you're writing down," not "any map whatsoever." $\endgroup$ – Qiaochu Yuan Feb 28 '15 at 8:43

Given R, the categories of left and of right modules may be very different, so that there is not even an equivalence between them. Some of the possible differences can already be seen in the ring itself. For example, a ring may be left noetherian but not right noetherian. The global dimension on the right may be different from the global dimension on the left, and so on.


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