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I have a dumb question on the definition of $R$-modules. I borrow the definitions from Wikipedia.

A left $R$-module $M$ over the ring $R$ consists of an Abelian group $(M,+)$ and an operation $R\times M \rightarrow M$ such that for all $r,s\in R$ and $x,y\in M$, we have

  1. $r(x+y) = rx + ry$
  2. $(r+s)x = rx+sx$.
  3. $(rs)x = r(sx)$.
  4. $1_{R} x = x$ if $R$ has a multiplicative identity $1_R$.

The article later says, a right $R$-module is defined similarly, and then "If $R$ is commutative, then left $R$-modules are the same as right $R$-modules and are simply called $R$-modules."

I have a slight issue with the third item above. How does the above definitions define $rs$? (As we have only defined multiplication of an element of $R$ by an element of $M$). I am assuming that this multiplication operation is the multiplication operation of the ring $R$ itself, say $\circ$, and actually the third line should perhaps read $(r\circ s)x = r(sx)$ instead. Is this true?

And similarly, for the second item we have the addition of two ring elements. This addition operation is the addition operation of the ring $R$, am I right?

In this context, my biggest problem is with this sentence "If $R$ is commutative, then left $R$-modules are the same as right $R$-modules and are simply called $R$-modules." Since we apparently have two different multiplications (one for the definition of the commutative ring $R$, and another for the definition of the module itself), how can the commutativeness of $R$ imply that the left $R$-module above is also a right $R$-module?

Lord Soth
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    To your questions: yes, yes, and every right R-modules is naturally a left by definiing rx:=xr and vice versa. – Alex Youcis Aug 12 '13 at 20:12
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    Indeed, we are lazy, and $rs$ and $rx$ is not the same product. I would prefer $rs$ and $r.x$, denoting the operation of $R$ on $M$ with a dot. – Dietrich Burde Aug 12 '13 at 20:14
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    You have a good point; the multiplication of a "vector"(element in the module) by ring element can take many different forms; from being (most often) the repeated sum of a vector when R=the integers $\mathbb Z$, to many fancy forms, as in a $V_T K$-module, where we take a fixed element T:$V\rightarrow V$ on the elements of V. – FBD Aug 12 '13 at 20:21

1 Answers1

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If we have a left $R$-module $M$, denoted $r\cdot m$, then we can define a right multiplication by $m\star r:=r\cdot m$. Now we must check that this indeed defines a right $R$-module structure, in particular, it must satisfy:

$$m\star(r\circ s)=(m\star r)\star s$$

If you expand the left hand side using the definition, you'll obtain $(r\circ s)\cdot m$, while on the right hand side, you'll obatin $s\cdot(r\cdot m)=(s\circ r)\cdot m$. For these two to be equal for all $r,s\in R$, we require that the ring is commutative.

Jared
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  • Sorry I was away for a while. Thanks for the answer. Then, I guess, if we have a left $R$-module $M$ over a commutative ring $R$, then the assertion "$M$ is also a right $R$-module and thus an $R$-module" does not make sense unless we define what we mean by right multiplication. And, the standard way to define a right multiplication is what you have written; $m \star r:=r \cdot m$. In that case, since $R$ is commutative, this way of right multiplication defines a right $R$ module, and overall we get an $R$-module. Is this correct? – Lord Soth Aug 12 '13 at 21:09
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    @LordSoth Yes, that is correct. – Alex Provost Aug 12 '13 at 21:53
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    @LordSoth More generally what follows from Jared's answer is that the left modules for $R$ are the same as the right modules for $R^{op}$, the ring defined via $r*s:=s\cdot r$. In the case of commutative rings we obviously have $R^{op}=R$. – Julian Kuelshammer Aug 13 '13 at 06:12