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For an ideal $I$ of a ring $R$ with identity, let $r(I)=\{r\in R: Ir=0\}$ and $l(I)=\{r\in R: rI=0\}$.

Question: If for any two ideals (two-sided ideal) $I, J$ of $R$, we have $l(I)+l(J)=l(I\cap J)$, does $r(I)+r(J)=r(I\cap J)$?

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well you shd clarify what you mean by the sum of two ideals. if the ring is noncommutative, then the usual definition of sum of two ideals namely all linear combinations of the form $a_1x_1+a_2x_2$ for $x_1\in I$ and $x_2\in J$ does not make any sense ... for all you know the two ideals might be right ideals. the only way to make sense of sum of two ideals would be to just take all the elements of the form $x_1+x_2$ ... suppose you mean it then it is clear that $r(I)+r(J)\subset r(I\cap J)$ .. similarly for the left ideal case. so your question reduces to if $l(I\cap J) \subset l(I) + l(J)$ th – magguu Feb 28 '13 at 18:54
In the noncommutative setting, if $I$ and $J$ are left/right/two-sided ideals, then $I+J = \{x+y \mid x\in I, y\in J\}$ is a left/right/two-sided ideal. It is certainly a subgroup, and $r(x+y)\in I+J$ if $I$ and $J$ are both left ideals, $(x+y)r\in I+J$ if $I$ and $J$ are right ideals. So I'm not sure what you are going on about... – Arturo Magidin Feb 28 '13 at 19:13
A ring $R$ is called a "right Ikeda-Nakayama ring" (or right IN-ring) if for any two right ideals $I$ and $J$, $l(I)+l(J)= = l(I\cap J)$. Having a name to add to the property may be useful (though you seem to be asking a bit less than right IN if you require $I$ and $J$ to be two-sided ideals. – Arturo Magidin Feb 28 '13 at 21:14
Dear Professor Magidin, Yes any IN-ring has this property, but for $R=\begin{pmatrix} F & F \\ 0 & F \\ \end{pmatrix}$, we have $r(I)+r(J)=r(I\cap J)$ for any two ideals $I, J$ and also $l(I)+l(J)=l(I\cap J)$ but $R$ is not an IN-ring. – Ali Mar 1 '13 at 6:39
@Ali Taherifar: I thought it might be strictly weaker; still, you may want to take a look at the literature on IN-rings. – Arturo Magidin Mar 1 '13 at 16:12

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