ring with a condition

Is there any ring $R$ with essential right ideal $I$ such that
$(I:I)\cap \{ t\in R \mid t(I:I)t \subseteq I \} \neq 0$ and for every non-zero $x,y\in R$, $\{ r\in R \mid xr\in(I:I)\}\cap\{ r\in R\mid xry\notin I \}\neq \emptyset$ ?

where $(I:I)=\{ r\in R\mid rI\subseteq I\}$

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Why do you want to know? and what can you prove so far? mathoverflow.net/howtoask – Yemon Choi Nov 26 '12 at 7:53
The first condition $(I:I) \cap${$t\in R| t(I:I)t\subseteq I$}$\neq \emptyset$ is trivially always satisfied as $0$ belongs to both sets, if you want to have a meaningful condition you should ask $\neq${$0$}. A part this, you should notice, editing your question, that for commutative rings (or more generally when $I$ is two-sided) your second condition cannot be satisfied. This may help... – Simone Virili Nov 26 '12 at 9:51
another observation. How is it possible to satisfy your second condition for all $x,y\in R$? In particular, if $x\in I$, then, as $I$ is a right ideal, $xry$ belongs to $I$ for all $r$ and all $y$. So... for $x\in I$, the set {$r\in R|xry\notin I$} is always empty. For example, take $x=0$ or $y=0$, you will see that your second condition is impossible to satisfy. So the answer is "NO!" there is no such ring:) – Simone Virili Nov 27 '12 at 11:03

As I was remarking in the comments to your question, it is impossible to construct such ring. In fact, as you want that, for all $x,y\in R\setminus\{0\}$, the intersection $\{r\in R:xr\in (I:I)\}\cap\{r\in R:xry\notin I\}\neq \emptyset$, in particular you need that $\{r\in R:xry\notin I\}\neq \emptyset$ for all $x,y\in R\setminus\{0\}$. Notice that, if $x\in I$, then, as $I$ is a right ideal, $xry=x(ry)\in I$ for all $r$ and $y\in I$, so this set is always empty.
Furthermore, if you have $r$ such that $xr\in (I:I)$, then, by definition of $(I:I)$, $xry\in I$ provided $y\in I$. So, if $y\in I$, then $\{r\in R:xr\in (I:I)\}\cap\{r\in R:xry\notin I\}= \emptyset$.
Maybe you have some hope looking in a non-commutative ring and taking $I$ to be a right but not left ideal and imposing your second condition just for $x,y\in R\setminus I$. Furthermore, another necessary condition is that $I\neq (I:I)$. Anyway, this is another question!
Finally, let me also add that the first condition $(I:I)\cap \{t\in R:t(I:I)t\subseteq I\}\neq \{0\}$ is implied by $I\neq\{0\}$ in fact $I$ is always contained in this intersection.
well... hard to believe... even if you changed your question, still your condition cannot be verified when $x\in I$. I strongly suggest you explain us context and motivation! – Simone Virili Nov 27 '12 at 14:38