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Math Help - Proving Rings (sorry for the Tex error)

  1. #1
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    Proving Rings (sorry for the Tex error)

    <br />
Prove that $$\left<\frac{R}{I}\,,+\,,\cdot\right>$$ is a ring.<br />
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by DanielThrice View Post
    <br />
Prove that $$\left<\frac{R}{I}\,,+\,,\cdot\right>$$ is a ring.<br />

    Perhaps you mean to prove that if R is a conmutative ring, I\subset R ideal of R, then R/I is a commutative ring with the standard operations (x+I)+(y+I)=(x+y)+I and (x+I)(y+I)=xy+I . If so, start proving that the sum is well defined that is, if x+I=x'+I and y+I=y'+I then, (x+I)+(y+I)=(x'+I)+(y'+I) .


    Fernando Revilla
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    Sorry, (I was half asleep last night) here is what we already know about the problem:

    Let R = < R, +, *> (* means multiplication) be a ring, and let I < R be an additive subgroup of < R, + >.
    Consider the set of cosets

    R/I = {a + I: a is an element of R}

    equipped with its own operations + and * defined by

    (a + I) + (b + I) = (a + b) + I
    (a + I) (b + I) = ab + I

    Assume that the operations + and are well-defined <---> the
    additive subgroup I satises the following conditions:

    ab is an element of I for all a in R and b in I

    ba is an element of I for all a in R and b in I
    Last edited by DanielThrice; February 17th 2011 at 05:56 AM. Reason: Forgot something
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