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Given a ring $R$ and an ideal(two-sided) $I\subset R$, we find an ideal $$[R:I]=\{x\in R| xR\subset I \}$$

It is easy to see that this ideal coincides with the original ideal $I$ if $I$ is a prime ideal. As I can see such ideals give an extension of the original ideal and satisfy an order-reversing property, i.e. $$J\subset I \Rightarrow [R:I]\subset [R:J]$$.

I would like to know more about this ring. I checked the list of ideals on Wikipedia, but it was not helpful.

Secondly, is there a name also for the following ideal? The best I can think of is "annihilator of $I$ in $R$". $$r(I)=\{r\in R| rI=\{0\}\}$$


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Your $r(I)$ is simply $\operatorname{Ann}(I)$, the annihilator of $I$ when $I$ is viewed as an $R$-module. – Brian M. Scott Mar 10 '12 at 17:19
In addition to "ideal quotient", I've heard this called a "colon ideal". – Hurkyl Mar 10 '12 at 17:34
BrianM.Scott and @Hurkyl, thank you. I have also found colon ideals on Wikipedia. – Herband Mar 10 '12 at 17:41
up vote 1 down vote accepted

What you denoted by $[R : I]$ is called an ideal quotient. Three points here: first, the ordering you used is nonstandard, one usually writes $[I : R]$ for that ideal. Second, $[I : R] = I$ always (when $R$ is unital). Third, to make this interesting take another ideal $J \subset R$ and consider $[I : J] = \{ x \in R \ | \ xJ \subset I \}$. This is a general ideal quotient.

Yes, $r(I)$ is called the annihilator of $I$.

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Thank you. I will have to prove the second remark. – Herband Mar 10 '12 at 17:49
I think I have a counterexample $[4\mathbb{Z}:2\mathbb{Z}] = 2\mathbb{Z}$. Your second comment works when $R$ is a ring with unity. Is that right? – Herband Mar 10 '12 at 17:57
If $x \in [I : R]$, then $x = x \cdot 1 \in xR \subset I$. – Justin Campbell Mar 10 '12 at 18:18
please edit your answer accordingly and let me accept the answer. Thanks. – Herband Mar 10 '12 at 18:27
Oh, right. I usually assume people mean ring with unity unless they say otherwise. – Justin Campbell Mar 10 '12 at 18:31

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