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Show their is only one ring morphism from Zn to Zk if k|n

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data
    For the rings Zn and Zk show that if k|n, then the function f: Zn to Zk
    s.t [x]n --->[x]k for all x in Zn
    is a ring morphism. Show this is the only ring morphism from Zn to Zk.

    The attempt at a solution

    So I showed it is a ring morphism by just verifying the properties, no big deal. I have no idea how to show that it is the only one though.

    I want to start out by contradiction and assume that their is another one... but I don't know what to do with that. Is there a way to show it directly?
  2. jcsd
  3. Feb 6, 2009 #2


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    Once you know what f(1) is, you pretty much know what the morphism is, right? Is there any choice but to make f(1)=1?
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