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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

Ideal: just check directly using the definition of ideal.

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 $1$.

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Suppose that x and y belong to [R:I]. Then for every r in R, rx and ry belong to R, so r(x-y) belong to R, and hence x-y belongs to [R:I]. This shows that [R:I] is a group. Further, if x belongs to [R:I] and r, r' belong to r, then r'(rx) = (r'r)x belongs to [R:I]. This shows that [R:I] is an ideal. This is a quick idea think I'm on the right path? – user30072 Apr 27 '12 at 19:33

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