I have read that for a ring $ R $ in general, right $ R $-modules are not the same things as left $ R $-modules.

  • Why do we say that a right $ R $-module is equivalent to a left $ R $-module only when $ R $ is commutative?

  • I feel that the commutativity of $ R $ is strictly an internal property of $ R $, so how can the commutativity of $ R $ affect scalar multiplication on an $ R $-module $ M $, which is an external operation?

  • 1
    $\begingroup$ Perhaps because $R$ and its ideals are all modules over $R$? $\endgroup$ – Prahlad Vaidyanathan Oct 22 '13 at 3:14
  • 1
    $\begingroup$ Prahlad I think, those are the special cases. What if module M over R is not a subset of ring R? $\endgroup$ – Pitt HarmanN - FreshmaN Oct 22 '13 at 3:22


a right R-module is equivalent to a left R-module only when R is commutative

This statement has a pair of serious problems to resolve.

  1. Because "equivalent" is undefined, it's unclear what the statement means. (Discussed briefly below.)

  2. It uses "only when," but that is the wrong logical direction: it should use just when. There are in fact not-commutative rings where left and right modules are pretty much the same, so "only when" is not really appropriate.

I think a better version of the statement is:

A right $R$ module structure can be used as a left $R$ module structure when $R$ is commutative

Modules over commutative rings

Suppose you've defined a right $R$ module for a commutative ring so that $mr$ makes sense for $m\in M$ and $r\in R$. Then naively one might say "oh, well that must be the same thing as $rm$."

Actually, it's not immediately clear that that is a legitimate thing to do, but yes, using commutativity of $R$, you can verify that the new action $rm:=mr$ satisfies all the module axioms. After that is done, we would have used the right $R$ module structure as a left $R$ module structure.

Symmetrically, left module structures can be used as right module structures, and in fact if you switch sides twice, you wind up with the original module you started with. That's why we say when $R$ is commutative, right and left modules are "the same."

Modules over rings in general

It turns out that a left $R$ module structure on an abelian group $M$ amounts to a ring homomorphism of $R\to End(M)$, where $End(M)$ is the set of group endomorphisms of $M$.

On the other hand, a right $R$ module structure on $M$ amounts to a ring homomorphism from $R\to End(M)^{op}$, where $End(M)^{op}$ is the opposite ring of $End(M)$. (Equivalently, you could use a ring homomorphism from $R^{op}\to End(M)$.)

This whole "opposite ring" business is what makes it necessary to keep track of module sides: the homomorphisms of $R$ into $End(R)$ might be completely different from those into $End(R)^{op}$. That's just the way things are.

Finally, if $R\cong R^{op}$, then something nice happens! Since $R$ and $R^{op}$ can't be distinguished, you can in fact use a left $R$ module structure as a right $R$ module structure. Formally, if you had the homomorphism $R\to End(M)$, you could simply compose it with the isomorphism $R^{op}\to R\to End(M)$ obtaining $R^{op}\to End(M)$, a right $R$ module structure.

This is obviously the case for commutative rings, since the identity map is an isomorphism of a commutative ring with its opposite ring. But this applies more generally, since there are not-commutative rings isomorphic with their opposite rings.

Your second question

Your question is based on good observations :) The previous paragraph supplies a partial explanation of why the internal property of commutative rings can influence the two categories of modules. The reason is that since commutativity makes the ring "symmetric," its right modules and left modules are going to look alike.

If you want to deepen your understanding of how the properties of a ring and its two module categories interact, then you will have a good time studying module theory because that is one of module theory's main topics.

  • 2
    $\begingroup$ Thank you so much for this detailed answer. $\endgroup$ – Pitt HarmanN - FreshmaN Oct 25 '13 at 3:42
  • 1
    $\begingroup$ I don't see any proof about your claim "using commutativity of R R , you can verify that the new action rm:=mr r m := m r satisfies all the module axioms" $\endgroup$ – onurcanbektas May 14 '17 at 7:06
  • $\begingroup$ @Leth It is a very straightforward proof. What part did not go well for you? $\endgroup$ – rschwieb May 14 '17 at 11:05
  • 1
    $\begingroup$ I'm still try to digest the fact that we are defining $rm$ as $mr$.I mean when I wrote that comment, which was 5 hours ago, I thought defining such thing causes some problem, but right not I can see there is no problem, but nevertheless, I still don't like it. $\endgroup$ – onurcanbektas May 14 '17 at 12:20
  • 1
    $\begingroup$ @Leth it does cause a problem sometimes if $R$ isn't commutative... namely the requirement that $(rs)m=r(sm)$ $\endgroup$ – rschwieb May 14 '17 at 13:21

In general, a left module over a ring $R$ is equivalent to a right module over the ring $R^{\operatorname{op}}$, whose multiplication is the same as that in $R$, but in the opposite direction (i.e. $a\cdot_{\operatorname{op}} b = b \cdot a$). This is because in a left module, say $M$, we multiply by elements on the left, so, if by $\lambda_a: M\mapsto M$ we denote the function defined by left-multiplication by an element $a \in R$, we have $$[\lambda_a \circ \lambda_b](m) = ab \cdot m = \lambda_{ab}(m).$$ In contrast, in a right module, if we let $\rho_a$ denote right multiplication by $a\in R$, we have $$[\rho_a \circ \rho_b](m) = m \cdot ba = \rho_{ba}(m).$$

Therefore, since composition goes in the opposite direction in a right module as in a left module, by writing everything backwards, we can turn a right $R$-module into a left $R^{\operatorname{op}}$-module. If $R$ is commutative, $R\cong R^{\operatorname{op}}$, so right and left modules are equivalent.

Your second question touches on an interesting idea - far from being isolated, facts intrinsic to a ring $R$ and facts about the category of $R$-modules are in fact very strongly linked, and often in very interesting ways.


The "external multiplication" of scalars from a ring $R$ is modelled (if tacitly) on the internal multiplication of scalars within the ring. I will discuss two angles of evidence for this viewpoint.

Semisimplicity. Consider semisimple $R$-modules $M$. Then $M$ decomposes as a direct product of simple modules as $M\cong M_1\times\cdots\times M_n$. Simple modules are characterized as being quotients of the scalar ring ($N\cong R/I$) up to isomorphism. This tells us very directly that the scalar operation is modeled very closely off of the multiplication operation within $R$. More generally modules can be more complicated than semisimple ones, but "more complicated" is no reason to expect the ring's multiplication operation to stop having control or influence over the module structure.

Action maps. An $R$-module $M$ is essentially encoded as a homomorphism $R\to{\rm End}(M)$. Thus the scalar operation is encoded by a homomorphic image of $R$, which is obviously dependent on the nature of $R$'s internal operations.

In the above I do not mention left versus right; fill in the blanks. That multiplication in $R$ is different between left v. right perspectives means the ideal theory may differ as well.

Categorical thoughts. If $R$ is commutative then all ideals are two-sided automatically, and any left $R$-module can be turned canonically into a right $R$-module (simply define the same action!). From a high-brow viewpoint with the right language we can say there is a natural equivalence between the categories of left $R$-modules and right $R$-modules. However in the noncommutative setting the most we can claim is an equivalence between left $R$-modules and right $R^{\rm op}$-modules, where $R^{\rm op}$ stands for the opposite ring. In general $R\not\cong R^{\rm op}$ (tough examples here).


For a commutative ring $R$, whatever you can say (in module theoretic terms) about the category of left modules over $R$ can be said about the category of right modules, because there's a “canonical” way to turn any diagram of left modules into a diagram of right modules.

In the general case of noncommutative rings this is not true any more: rings can be left noetherian (artinian, coherent, perfect, …) and not right noetherian (artinian, coherent, perfect, …). All these properties can be defined in module theoretic terms via diagrams: a ring is left perfect when every left module has a projective cover, for instance; it is left noetherian when every direct sum of injective left modules is injective.

In case the ring $R$ has an involution, that is, a map $\varphi\colon R\to R$ such that

  • $\varphi^2$ is the identity,
  • $\varphi(r+s)=\varphi(r)+\varphi(s)$,
  • $\varphi(rs)=\varphi(s)\varphi(r)$,

then we are in a situation similar to the commutative case (where we can take $\varphi$ to be the identity): any left module $M$ can be turned into a right module with $xr=\varphi(r)x$ and it's easy to show that left module morphisms become right module morphisms. However, this is the exception rather than the rule for noncommutative rings.

Some module theoretic properties are left-right symmetric, though: semisimplicity and semiperfectness are examples.

The fact that every left module can be seen as a right module over the opposite ring simply tells that general properties of modules hold both for left modules and right modules: for instance, every module has an injective envelope, because a proof given for left modules doesn't depend on special properties of the ring.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.