Largest ideal inside an ideal of a ring. Consider a ring $R$ and $I$ be its left ideal. If $I$ is regular, then $(I:R) = \{r\in R\ |\ rR \subset I\}$ is the largest ideal of $R$ that is contained in $I$.
I have tried to prove this by contradiction, assume $\exists\ J \ne I$ an ideal of $R$ such that $(I:R) \subset J$. Then, $\exists\  x \in J$ such that $x \notin (I:R) \Rightarrow xR \nsubseteq I $. From here I cant proceed, if it were a two-sided ideal I could have arrived at a contradiction from here but ring multiplication happens from left and I have a condition with multiplication from the right. Also, I would like to know how regularity plays a role here, I don't see a direct application. 
Thanks.
 A: I don't think that this is true. Consider $R=M_3(\mathbb R)$ and take $I$ to be the set of all matrices with the third column zero:
$$\begin{pmatrix}*&*&0\\*&*&0\\*&*&0\\\end{pmatrix}.$$
$I$ is a left ideal. Also $I$ is regular as $R$ has unity so you can take $e=1_R$. (Btw, I think that your definition of regular left ideal should be: there is $e\in R$ such that $r-re\in I$ for every $r\in R$.)
Let us calculate $(I:R)$; let $A\in(I:R)$. Since $(I:R)\subseteq I$ we may write:
$$A=\begin{pmatrix}a_1&b_1&0\\a_2&b_2&0\\a_3&b_3&0\\\end{pmatrix}.$$
Consider matrices:$$M_1=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\\\end{pmatrix}\mbox{ and }M_2=\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\\\end{pmatrix}.$$
Since $A\in(I:R)$ we have $AM_1,AM_2\in I$, i.e.:
$$\begin{pmatrix}0&0&a_1\\0&0&a_2\\0&0&a_3\\\end{pmatrix},\begin{pmatrix}0&0&b_1\\0&0&b_2\\0&0&b_3\\\end{pmatrix}\in I,$$
which implies $A=0$. Thus $(I:R)=0$. On the other hand we have a regular left ideal $J$ of all matrices with the second and the third column zero: 
$$\begin{pmatrix}*&0&0\\*&0&0\\*&0&0\\\end{pmatrix},$$
and $(I:R)\subsetneq J\subsetneq I$ holds.
