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Find all ring homomorphisms from 3Z to Z?

  1. Mar 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Find all ring homomorphisms from 3Z to Z, where 3Z are the integers that are of multiple 3.

    2. Relevant equations



    3. The attempt at a solution

    So 3Z is cyclic so σ(3) is sufficient to look. Now all of the other examples have finite groups, so |σ(a)| divides the |a| in the domain. Or, the homomorphisms is define with Z as the domain, and then a=a^2. I understand these examples, but then dont know how to start this one...

    <3> generates the group, so I know I want to consider it. But if σ(3)=a, but then what? Other examples are a=σ(1)=σ(1*1)=a^2, but that property doesn't hold here, and 1 is not in3Z. I know I'm missing something simple, just can't quite figure out what property I need to put my finger on to make that first step..

    Thanks for any help!! :smile:
     
  2. jcsd
  3. Mar 31, 2012 #2

    Dick

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    You've got the right starting point. So σ(3)+σ(3)+σ(3)=σ(3+3+3)=σ(9)=σ(3*3)=σ(3)*σ(3). So?
     
  4. Apr 1, 2012 #3
    So 3a=a^2 if a=σ(3). So the possible homomorphisms of a will have that property. And it should them just fail out like the other examples.
     
  5. Apr 1, 2012 #4

    Dick

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    If a solves 3a=a^2 then it's a homomorphism. You still need to describe what the homomorphisms are.
     
    Last edited: Apr 1, 2012
  6. Apr 1, 2012 #5
    Then 3a - a^2 = 0, and 3Z has no zero divisors, so factoring the equation gives a=0 or a=3.
    Then, if σ(3)=a=0, then σ(x)=0 for all x in 3Z, and σ is the zero homomorphism.
    If σ(3)=a=3, then σ(x)=x for all x in 3Z, and σ is the identity homomorphism.

    Hence there there two ring homomorphism for 3Z to Z. :smile:
     
  7. Apr 1, 2012 #6

    Dick

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    Seems fine to me.
     
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