Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
3  
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
add comment

1 Answer

up vote 1 down vote accepted

After answering, the question changed so I adapt my answer to the new question.

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.

share|improve this answer
    
many thanks for your answer. I think there exist a ring with this property. –  mohammad Nov 27 '12 at 14:34
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.