Proving that transitive $R-$action implies $M$ is a simple module.

abstract-algebramodulesproof-explanationsolution-verification

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.

  1. $M$ is a simple $R-$module.

  2. 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 $x \in M$.

Indeed if $M$ is simple and $x$ nonzero, then $(x)$ is a nonzero submodule, and it must coincide with $M$. Conversely, let's show that $M$ is simple. If $S$ is a nonzero submodule, it contains some nonzero $x$. But then $M = (x) \subset S \subset M$, hence $M = S$.

Can you take it from there?

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