# Prove that $[R:I] =\{r \in R\mid xr \in I\text{ for every }x \in R\}$ is an ideal of $R$ that contains $I$

If $I$ is an ideal in a ring $R$ let $[R:I] =\{r \in R\mid xr \in I\text{ for every }x \in R\}$. How can I show that $[R:I]$ is an ideal of $R$ which contains $I$.

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Welcome to math.SE: since you are a new user, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. – Zev Chonoles Apr 27 '12 at 0:19
Hint: First show that I is contained in here (use the fact that I is an ideal). Then the ideal part should follow immediately, since for $r\in [R : I]$ and any $x\in R$ we have $xr \in I \subset [R:I]$ – Deven Ware Apr 27 '12 at 0:39
You have swapped the "numerator" and the "denominator" in your definition of what is sometimes called a colon ideal. The correct notation is $$(I:R) =\{r \in R\mid xr \in I\text{ for every }x \in R\}$$ – Georges Elencwajg Apr 27 '12 at 15:02

Containment: You need to show that elements of $I$ have the necessary property, ie. for all $x\in I$, $xr\in I$ for each $x\in R$.
Both follow directly from the definitions: once you grok the definition of Ideal and $[R:I]$ it should be pretty quick.
Bonus question: Show that $I=[R:I]$ if $R$ contains $1$.