Let [tex]R[/tex] be a ring with [tex]1_R[/tex]. If [tex]M[/tex] is an

[math_grl]

Let $$R$$ be a ring with $$1_R$$. If $$M$$ is an R-module that is NOT unitary then for some $$m \in M$$, $$Rm = 0$$.

I'm pretty sure $$Rm = \{ r \cdot m \mid r \in R \}$$. While M being not unitary means that $$1_R \cdot x \neq x$$ for some $$x \in M$$. I'm thinking this problem should be an obvious and direct proof but I can't see it.

 

[adriank]

If m ≠ 1m, then consider m - 1m ≠ 0.

 

[math_grl]

that's embarassing.