For any $R$-submodule $N$ of $M$, $N=Ma~\text{ if and only if}~ N=MaR,a\in R.$

Question

Let $R$ be a right principal ideal ring with unity and $M_R$ be the right $R$-module. My question is that, is the following formulation valid? For any $R$-submodule $N$ of $M$, $$N=Ma~\text{ if and only if}~ N=MaR,a\in R.$$ In my view, $N=Ma\subseteq MaR$ in general, since $a\in aR$. Also, $aR\subseteq \{ar:r\in R\}$. In particular, $MaR\subseteq Ma(1_R)=Ma$.

Answer 1

The question really is just simply "If $M$ is a right $R$ module and $a\in R$ and $Ma$ is a right $R$ module, then does $Ma=MaR$?

Obviously yes, in any ring (not just a principal ideal ring.) Just as sets you have $Ma\subseteq MaR$. Since you have assumed $Ma$ is a right $R$ module, then by definition $Ma\supseteq (Ma)R$.

You noted the first containment as I did, but your explanation for the second half isn't convincing. Why would you say $MaR\subseteq Ma(1_R)$? You gave no reason other than to refer to the definition of a principal right ideal, which doesn't help. You need to appeal to the fact that $Ma$ is closed under multiplication by $R$ on the right.