why $RM \neq 0$

Question

Definition: we said that $M$ is simple if $RM \neq 0$ and $M$ has no proper submodule.

i couldnt understand why it must be $RM \neq 0$

could you please explain why ? Thank you for your helping.

Answer 1

This is the definition for "simple module" adopted for rings which do not necessarily have identity and/or modules which are not assumed to be unital. For rings with identity, we usually assume that $1\cdot m=m$ for all $m\in M$, and in that case, the definition can omit the part about $RM\neq \{0\}$.

I'm not sure where you encountered the definition, but I remember I learned it in Jacobson's Structure of rings (page 5). It turns out that you can find an isomorphism $M\cong R/L$ for a maximal left ideal (which is also a modular left ideal) of $R$ iff $RM\neq 0$ and $M$ has no nontrivial submodules.

($\Leftarrow$) If $RM\neq\{0\}$, pick $r, m$ such that $rm\neq 0$. Then the map $r\mapsto rm$ from $R\to M$ is onto $M$, and the first isomorphism theorem retrieves $L$ such that $R/L\cong M$.

($\Rightarrow$) Let $L$ be a maximal left ideal which is also modular. This means that there exists $e\in R$ such that for all $x\in R$, $x\equiv ex$ in $R/L$. So if you take any nonzero $y\in R/L$ and conjure up the $e\in R$ that does this, $y\equiv ey\neq 0$ in $R/L$, proving that $R(R/L)\neq \{0\}$.

Other modules which only have the two trivial submodules, but unfortunately also have $RM=\{0\}$ do not have this nice property, so they are not as important. One example of such a module is the $2\Bbb Z$ module $2\Bbb Z/4\Bbb Z$.