Here is the question I want to answer:
A module is simple if it is not the zero module and it has no proper nonzero submodule.
$(a)$ Let $M$ be an $R-$module. Show that the following conditions are equivalent.
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$M$ is a simple $R-$module.
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for any nonzero $x,y \in M$ there exists $r \in R$ such that $rx = y.$(i.e. the action of $R$ on $M$ is transitive)
I managed to prove $1 \implies 2.$ For $2 \implies 1,$ here is my trial (after getting some help from kind volunteers here):
Assume that $M$ is an $R-$module and assume that for any nonzero $x,y \in M$ there exists $r \in R$ such that $rx = y$ (i.e. the action of $R$ on $M$ is transitive). We want to show that $M$ is a simple $R-$module i.e. it is non-zero and it has no non-zero proper submodules.
First, $M \neq 0$ because the $R-$action on $M$ is a non-zero action. Is this reasoning correct?
Second, Suppose that $S \subset M$ is a non-zero submodule. We want to show that $S = M.$ If $S \neq M,$ then $\exists m \in M \setminus S$ But then I do not know how to complete this argument, could anyone help me in completing it please?
Best Answer
Hint: to show that $M$ is simple, you can equivalently prove that $M \neq 0$ but $M = Rx$ for all nonzero $x \in M$.
Can you take it from there?