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Math Help - Well-defined, surjective ring homomorphism.

  1. #1
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    Well-defined, surjective ring homomorphism.

    Hey guys.

    I have a commutative ring R, with two ideals, I and J, in R, where I \subseteq J.
    Now I have to show that  \varphi : R / I \rightarrow R / J given by  \varphi ( x + I ) = ( x + J ) is a well defined, surjective ring homomorphism.

    Right now I cant even figure out what i need to do, to prove that it is well-defined. I'm missing parts of what I should have learned, but as far as I've been able to figure out, a homomorphism is well-defined when for any a,b \in R/I, a = b \Leftrightarrow \varphi (a) = \varphi (b).
    But I can't figure out how to prove that. If someone could give me a hint, I would really appreciate it. Perhaps I need a more stringent definition of "well-defined" or if you could give me an example of a proof kind of like mine?

    All help is appreciated.
    Morten
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Well-defined, surjective ring homomorphism.

    Quote Originally Posted by m112358 View Post
    Right now I cant even figure out what i need to do, to prove that it is well-defined.
    We have to prove that \varphi (x+I) does not depend on the representative element x . If x+I=y+I then x-y\in I\subseteq J that is , x+J=y+J and as a consequence \varphi (x+I)=\varphi (y+I) .
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  3. #3
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    Re: Well-defined, surjective ring homomorphism.

    Arh ok, now I understand. Alright, that was simple, guess I had been staring too blindly to notice. Thanks for the help

    Morten
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