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Ring Theory/Largest two-sided ideal

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Let I be a right ideal of a ring R and let A = {r in R: (R/I)r = 0}.
Prove that A is the largest two-sided ideal of R contained in I.

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Solution Preview

Proof:
First, I show that A is a subring of R
For any r,s in A, we have (R/I)r=0, (R/I)s=0, then
(R/I)(r+s)=(R/I)r+(R/I)s=0+0=0, this impies r+s is in A
(R/I)rs=((R/I)r)s=0*s=0, this implies that rs is in A
So A is a subring.
Second, I show that A is a two-sided ideal.
For any r in R, s in A, we have (R/I)s=0, ...

Solution Summary

This solution is comprised of a detailed explanation to prove that A is the largest two-sided ideal of R contained in I.

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