Prove $I=\{r\in R\vert rL\subseteq N\}$ is an ideal of $R$.

Question

Let $R$ is commutative ring with identity $1$, let $M$ is module over $R$, and let $N$ and $L$ are submodules of $M$. Prove $$I=\{r\in R\vert rL\subseteq N\}$$ is an ideal of $R$.

To prove this statement, let $a,b\in I$ and $r\in R$. We must show: (1) $a-b\in I$ and (2) $ra\in I$.

This is my proof.

Let $a,b\in I$, we have $aL\subseteq N$ and $bL\subseteq N$. I confused to show $a-b\in I$. Is it right because $N$ is submodule, so $$aL-bL=(a-b)L\subseteq N?$$ So I have $a-b\in I$?

Answer 1

It would make a little more sense to write $(a-b)L\subseteq aL+bL\subseteq N$. (The subtraction sign is not really important here since I guess you are assuming rings with identity anyhow.)

Hopefully you have also included in your proof why $(ra)L\subseteq N$ for any $r\in R$.

Also, you must either specify you're working with commutative rings, or else justify why $(ar)L\subseteq N$ for all $r\in R$.