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$R$ is a commutative ring and $M$ is a left $R$ module.

Define

$\cdot : M*R \to M$ as $(x,a) \to x\cdot a$ by,

$$x\cdot a=ax$$

To prove : $M$ is a right $R$ module.

Proof:

Let $x,y \in M$ and $a,b\in R$,

Then the first two properties i.e.

$(x+y)\cdot a = x\cdot a+ y\cdot a$ and $x\cdot (a+b) = x\cdot a+x\cdot b$ are very easy to verify.

I am a little confused about the third one i.e.

$$x\cdot (ab) = (ab)x= a(bx)= a(xb)= (ax)b= (x\cdot a)b$$

Is it correct to use the commutativity of $R$ to switch $bx$ to $xb$, because $x\in M$ but $M$ is a module and hence a vector space over a ring and hence can i say that elements of $M$ are also in $R$ ?

Michael Hardy
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johny
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  • No. $bx = x \cdot b.$ That's how it's define. There is no meaning of $xb,$ unless you write it as a short form of $x \cdot b$. Use the commutativity of $R: (ab)x=(ba)x.$ – Krish Feb 22 '15 at 16:44
  • @Krish I tried using commutativity in this form $(ab)x=(ba)x=b(ax)=b(x\cdot a)$ but even with this how am i going to get $(x\cdot a)b$ ? – johny Feb 22 '15 at 16:49
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    $x\cdot (ab) = (ab)x=(ba)x= b(ax)= b(x\cdot a)=(x \cdot a)\cdot b$. (In the last equality we are using the fact that $x \cdot a \in M$) – Krish Feb 22 '15 at 17:02
  • @Krish Right,my bad. Thanks ! – johny Feb 22 '15 at 17:03

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