From Wikipedia:

"If $R$ is commutative, then left $R$-modules are the same as right $R$-modules and are simply called $R$-modules."

The definition of left $R$-module: $M$ is a left $R$-module if $M$ is an abelian group and $R$ a ring acting on $M$ such that

(i) $r(m_1 + m_2) = rm_1 + rm_2$

(ii) $(r_1 + r_2 ) m = r_1 m + r_2 m$

(iii) $1m = m$

(iv) $r_1 (r_2m) = (r_1 r_2) m$

I don't understand what commutativity of $R$ has to do with the module being left and right. If $R$ is commutative it means that $r_1 r_2 = r_2 r_1$. Now how does it follow from that that $rm = mr$? $M$ is not a subset of $R$, it could be anything so how does commutativity of $R$ make elements of $M$ and $R$ commute, too?

  • 1
    $\begingroup$ If you assume a left R module is also a right R module then you will need commutativity of R in order for the last property to hold $\endgroup$
    – Belgi
    May 13 '12 at 9:13
  • $\begingroup$ I believe that any left $R$-module over a commutative ring is isomorphic to a right $R$-module, and vise-versa. $\endgroup$ May 13 '12 at 9:13
  • 1
    $\begingroup$ @AlexBecker: That's somewhat non-trivial to make precise, as the two objects live in different categories. It would be easier to say that the two categories are isomorphic. $\endgroup$
    – Zhen Lin
    May 13 '12 at 9:44

The definition of right $R$ module is similar except that elements of $R$ are written on the right, and the axioms become

(i) $(m_1 + m_2)r = m_1r + m_2r$

(ii) $m(r_1 + r_2 ) = mr_1 + mr_2 $

(iii) $m1 = m$

(iv) $(mr_1) r_2 = m(r_1 r_2)$

If $R$ is a commutative ring and $M$ is a left $R$ module, then we can define a right $R$ module $M'$ that is the same abelian group as $M$ and with essentially the same action, but with the elements of $R$ acting on the right: namely, if $rm$ is the result of $r\in R$ acting on $m\in M$, then we define the action of $R$ on $M'$ by $mr = rm$. (That is, it is not a question of whether it is "true" that $mr=rm$; rather, we define it to be so.) It can then be verified that $M'$ satisfies the axioms, and there is nothing lost going from $M$ to $M'$ or vice versa. Commutativity is needed for axiom (iv). That is what is meant on Wikipedia.

However, one should be careful. If $M$ is a left $R$ module, then $M$, as an abelian group, might be given a right $R$ module action that is not obtained from the left action in this way, even if $R$ is commutative. Therefore in some contexts, where more than one action is being considered, it can be useful to designate left and right modules even of commutative rings.


Given a ring $R$, the opposite $R^{op}$ is the ring with the same underlying group, but with multiplication defined by $(a,b)\mapsto ba$.

Now a right $R$-module can be defined as a left $R^{op}$-module.

Clearly $R$ is commutative if and only if $R=R^{op}$; in this case left and right $R$-modules are 'the same'.

[Edit: my answer adds nothing new to Jonas' answer, but I think the term opposite ring deserves to be mentioned in this thread.]


What Wikipedia wants to say:

If $R$ is a commutative ring and $M$ a right $R$-module, the law $r\cdot m:=mr$ defines a left $R$-module structure on $M$.

So all assertions about right $R$-modules can be turned very easily into assertions about left $R$-modules by writing multiplication on the left. Hence we lose nothing if we study only left $R$-modules.

Make sure you understand that this is generally not the case if $R$ is not commutative.


For a right module, the multiplication is usually written with the scalar on the right, and axiom (iv) becomes $(mr_1)r_2=m(r_1r_2)$. If you were to write the product with the scalar on the left, axiom (iv) would become quite inconvenient: $r_1(r_2m)=(r_2r_1)m$. But of course, if $R$ is commutative, this doesn't matter.


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