Explore BrainMass
Share

Explore BrainMass

    Ring Theory/Largest two-sided ideal

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    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.

    © BrainMass Inc. brainmass.com May 20, 2020, 2:14 pm ad1c9bdddf
    https://brainmass.com/math/ring-theory/ring-theory-largest-two-sided-ideal-81774

    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.

    $2.19

    ADVERTISEMENT