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My professor wrote: "If $R$ is any ring, $A$ is any abelian group and we define $ra = 0$ $\forall$ r $\in$ $R$ $\forall$ a $\in$ $A$, then $A$ will be a left $R$-module " but this statement is not clear for me, will this be the trivial module {0}, is my understanding right?

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    $\begingroup$ Your professor's statement is not true with the usual definition of module, which required that the multiplicative unit of $R$ act as the identity. $\endgroup$ – Qiaochu Yuan Sep 27 '17 at 8:23
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    $\begingroup$ @QiaochuYuan Your statement assumes that $R$ has a unit. That's not necessarily true (at least I think it's not). $\endgroup$ – Arthur Sep 27 '17 at 8:44
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    $\begingroup$ @QiaochuYuan I think my ring is not with unit or it can not be with unit because of the given condition. $\endgroup$ – Intuition Sep 27 '17 at 9:57
  • $\begingroup$ It's a bad convention to have "ring" mean "not-necessarily-unital ring" without any explanation. Most mathematicians don't use this convention so it's just confusing. $\endgroup$ – Qiaochu Yuan Sep 27 '17 at 17:57
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No. The trivial module has a single element (usually denoted by $0$). Besides that, I don't understand what's not clear about the statement. All module axioms are satisfied, right?!

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  • $\begingroup$ yes all axioms are satisfied....but the given condition leads to that the ring action on the elements of $A$ always leads to 0....right? what is the importance of this type of modules? $\endgroup$ – Intuition Sep 27 '17 at 9:54
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    $\begingroup$ @Intuition You are right. Now, about the importance… That's like asking what's the importance of the null matrix. It is the most basic exemple of module, and it can be defined on any abelian group. $\endgroup$ – José Carlos Santos Sep 27 '17 at 9:59

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