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Math Help - Is R-P closed under addition?

  1. #1
    Junior Member sorv1986's Avatar
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    Cool Is R-P closed under addition?

    suppose R be a commutative ring and P be a prime ideal of R. Then is S=R-P (the complement of P in R) closed under addition?


    regards. thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Is R-P closed under addition?

    Quote Originally Posted by sorv1986 View Post
    suppose R be a commutative ring and P be a prime ideal of R. Then is S=R-P (the complement of P in R) closed under addition?


    regards. thanks.
    Certainly not. You know by definition that R-\mathfrak{p} is multiplicative. But additive? Take the simplest example possible, is \mathbb{Z}-(3) additive? Probably not since neither 4 nor 2 are divisible by 3, but their sum is.
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  3. #3
    Junior Member sorv1986's Avatar
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    Re: Is R-P closed under addition?

    Quote Originally Posted by Drexel28 View Post
    Certainly not. You know by definition that R-\mathfrak{p} is multiplicative. But additive? Take the simplest example possible, is \mathbb{Z}-(3) additive? Probably not since neither 4 nor 2 are divisible by 3, but their sum is.

    suppose i m taking two elements a,b from R-P.
    then (a+P)+(b+P)=(a+b)+P.
    now a+P≠P, b+P≠P.
    then is it true that (a+b)+P ≠P?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Is R-P closed under addition?

    Quote Originally Posted by sorv1986 View Post
    suppose i m taking two elements a,b from R-P.
    then (a+P)+(b+P)=(a+b)+P.
    now a+P≠P, b+P≠P.
    then is it true that (a+b)+P ≠P?
    No. Neither 3 nor -3 is divisible by 5, but how about their sum?
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