Answer to Question #149663 in Abstract Algebra for Sourav Mondal

Question

Let I be an ideal of a ring R. Define
[R:l]={x belongs to R | rx belongs to I for all r belongs to R}
Prove that
(i) [R: I] is an ideal of R.
(ii) I subset of [R : I]

Answer 1

Let $I$ be an ideal of a ring $R$ and $[R:I]=\{x\in R\ |\ rx \in I\text{ for all }r\in R\}$

(i) Let $a\in[R: I]$ and $b\in[R: I]$. Then $ra \in I$ and $rb \in I$ for all $r\in R$, and therefore, $r(a-b)=ra-rb \in I$ for all $r\in R$. We conclude that $a-b\in[R: I]$, and thus $[R: I]$ is a subgroup of the additive group of a ring $R$.

Let $a\in[R: I]$ and $s\in R$ be arbitrary. It follows that $ra \in I\text{ for all }r\in R$. Since $I$ is an ideal, $s(ra)\in I$ for all $s\in R$ and $r(as)=(ra)s\in I$ for all $r\in R$. Consequently, $ra\in [R: I]$ and $as\in[R: I]$ for all $s,r\in R$, and $[R: I]$ is an ideal of a ring $R$.

(ii) If $a\in I$, then $ra \in I$ for all $r\in R$. It follows that $a\in[R: I]$, and thus $I$ is a subset of $[R : I]$.