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Thread: Proving Rings (sorry for the Tex error)

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

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

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


    Fernando Revilla
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  3. #3
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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; Feb 17th 2011 at 04:56 AM. Reason: Forgot something
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