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Math Help - Rings

  1. #1
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    Rings

    Prove that R is commutative if and only if R' is commutative.
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  2. #2
    Senior Member Tinyboss's Avatar
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    What is R'?
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bookie88 View Post
    Prove that R is commutative if and only if R' is commutative.
    Quote Originally Posted by Tinyboss View Post
    What is R'?
    I'm going to take a wild guess. Let R be a commutative ring and \phi:R\to R' an epimorphism. Then, R' is commutative. To see this let \phi(r),\phi(s)\in R' then \phi(r)\phi(s)=\phi(rs)=\phi(sr)=\phi(s)\phi(r)
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Drexel28 View Post
    I'm going to take a wild guess. Let R be a commutative ring and \phi:R\to R' an epimorphism. Then, R' is commutative. To see this let \phi(r),\phi(s)\in R' then \phi(r)\phi(s)=\phi(rs)=\phi(sr)=\phi(s)\phi(r)
    It is an 'if and only if', so we would need an isomorphism (epimorphism gives the 'if', monomorphism gives the 'only if').
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Swlabr View Post
    It is an 'if and only if', so we would need an isomorphism (epimorphism gives the 'if', monomorphism gives the 'only if').
    Why bother with the monomorphism bit? If \phi:R\to R' is an isomorphism then it is an epimorphism and so the above proof works, and since \phi^{-1}:R'\to R 1is an epimorphism too the conclusion follows.
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Why bother with the monomorphism bit? If \phi:R\to R' is an isomorphism then it is an epimorphism and so the above proof works, and since \phi^{-1}:R'\to R 1is an epimorphism too the conclusion follows.
    One way is as easy as the other. I explained it in that way as I had maybe thought you had meant monomorphism as opposed to epimorphism, so I had thought about both cases.
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