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Math Help - Homomorphism ring

  1. #1
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    Smile Homomorphism ring

    This question does not let me sleep.
    Fina all Ring homomorphism between Z_n----> Z_m.
    I know there should be 3 case:
    n>m
    n<m and m=n.
    the last one is trivial. Anyways, how do you guys think I should proceed or what should I consider?
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  2. #2
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    Somehow, I suspect the answer to be gcd(n,m), because I read it somewhere.But I actually want to have an idea of how to get there.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by karlito03 View Post
    Somehow, I suspect the answer to be gcd(n,m), because I read it somewhere.But I actually want to have an idea of how to get there.
    Hint:
    Spoiler:
    If \theta:\mathbb{Z}_n\to\mathbb{Z}_m is a homomorphism then the homomorphism is entirely determined by \theta(1)
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  4. #4
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    I sure ..
    call theta (f).
    Then f(1)=1 since you assume it was a ring homomorphism. But how can we get to the point of saying that all homomorphism must be determined by f(1). There must be some cases. For instance m>n, n>m or m=n or Don't m and n have to be such that (n,m)=1. How many homomorphism can one get in each case. Are you saying they will all be determined by f(1)?
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  5. #5
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    if n=m then image of any ring Zn-->Zm is determined by the image of 1 mod (n) where m=n.
    now if n>m Zn--> Zm need not to be an homomorphism.
    Last edited by karlito03; May 16th 2010 at 10:37 PM. Reason: Mistyped
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by karlito03 View Post
    now if n>m Zn--> Zm need not to be an homomorphism.
    I'm not sure what you mean here? Why isn't f:\mathbb{Z}_n\to\mathbb{Z}_m:z\mapsto 1_m not a homomorphism?
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  7. #7
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    Hint 1: put \theta(1+n \mathbb{Z})=r+m\mathbb{Z} and extend \theta linearly to all \mathbb{Z}/n\mathbb{Z}. now look at the conditions that will make \theta well-defined and multiplicative.

    Hint 2: using the above, show that \theta is defined by \theta(x+n\mathbb{Z})=rx + m\mathbb{Z}, where r satisfies the following conditions: \frac{m}{\gcd(n,m)} \mid r and m \mid r^2-r.
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  8. #8
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    i dont get the solution for this question. can someone explain it to me? thanks
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