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First of all, excuse me if the question seems stupid. But I want to understand what is written in this screenshot. The book name is (Extensions of Rings and Modules by Birkenmeier, Park, and Rizvi). The authors say that $Q$ is acting on $E$ from the right. The only possible action of $Q$ on $E$ from the right is given by $e.q=q(e)$ for every $e\in E,q\in Q$. But, unfortunately, this action is not associative (i.e., $e(q_1q_2) \neq (eq_1)q_2$). Am I mistaken?!.

enter image description here

Hussein Eid
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1 Answers1

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The only possible action of $Q$ on $E$ from the right is given by $e.q=q(e)$ for every $e\in E,q\in Q$. But, unfortunately, this action is not associative (i.e., $e(q_1q_2) \neq (eq_1)q_2$). Am I mistaken?!.

That's the right action. Perhaps because you are using the left-hand notation for functions it looks wrong to you.

Writing them on the right, instead:

$e\cdot (q_1q_2):=(e)(q_1q_2)=((e)q_1)q_2=(e\cdot q_1)\cdot q_2$

It is just the mirror image of the more familiar left-hand notation.

By applying the action on the right, the intention is to use the right hand notation and composition for functions to apply them in the correct order to the input.

rschwieb
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  • Ok. But I wonder how is the multiplication operation on $Q$ is defined if we want to use the left-hand notation for functions. It looks like that if $q_1,q_2\in Q$ then $q_1.q_2$ is defined to be $q_2 \circ q_1$. This is the only possible way to define the multiplication operation on $Q$ in order for $E$ to be a right $Q$-module if one wants to use the left-hand notation not the right-hand notation for functions. Is this true?! – Hussein Eid Jun 07 '23 at 15:46
  • Yes, you will have to transfer the right hand $Q$ operation to a left-hand $Q^{op}$ operation. – rschwieb Jun 07 '23 at 18:11