Question #149663
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]
1
Expert's answer
2020-12-10T14:35:50-0500

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


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

Let a[R:I]a\in[R: I] and sRs\in R be arbitrary. It follows that raI for all rRra \in I\text{ for all }r\in R. Since II is an ideal, s(ra)Is(ra)\in I for all sRs\in R and r(as)=(ra)sIr(as)=(ra)s\in I for all rRr\in R. Consequently, ra[R:I]ra\in [R: I] and as[R:I]as\in[R: I] for all s,rRs,r\in R, and [R:I][R: I] is an ideal of a ring RR.


(ii) If aIa\in I, then raIra \in I for all rRr\in R. It follows that a[R:I]a\in[R: I], and thus II is a subset of [R:I][R : I].



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