1
$\begingroup$

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 $r(I)+r(J)=r(I\cap J)$?

$\endgroup$
5
  • $\begingroup$ 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 $\endgroup$
    – magguu
    Feb 28, 2013 at 18:54
  • $\begingroup$ 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... $\endgroup$ Feb 28, 2013 at 19:13
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 28, 2013 at 21:14
  • $\begingroup$ 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. $\endgroup$
    – Ali
    Mar 1, 2013 at 6:39
  • $\begingroup$ @Ali Taherifar: I thought it might be strictly weaker; still, you may want to take a look at the literature on IN-rings. $\endgroup$ Mar 1, 2013 at 16:12

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.