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Finding Ring Homomorphisms

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Find all ring homomorphisms [tex]\phi[/tex]: Z [tex]\rightarrow[/tex] Z
    [tex]\phi[/tex]: Z2 [tex]\rightarrow[/tex] Z6
    [tex]\phi[/tex]: Z6 [tex]\rightarrow[/tex] Z2

    2. Relevant equations
    A function [tex]\phi[/tex]: R [tex]\rightarrow[/tex] S is called a ring homomorphism if for all a,b[tex]\in[/tex]R,
    [tex]\phi[/tex](a+b) = [tex]\phi[/tex](a) + [tex]\phi[/tex](b)
    [tex]\phi[/tex](ab) = [tex]\phi[/tex](a)[tex]\phi[/tex](b)
    [tex]\phi[/tex](1R) = 1S

    3. The attempt at a solution
  2. jcsd
  3. Oct 22, 2008 #2


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    So why is that difficult for you? You have to show an attempt or state what is confusing you before anyone can help.
  4. Oct 23, 2008 #3
    so i have to find every set in Z that satisfies those equations by ending in Z?
    same goes for Z_2 to Z_6 find every set that will add together in the homomorphism in Z_2 and will separately add together in Z_6? is this what its asking?
    if so how do i show that?
  5. Oct 23, 2008 #4


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    Your definition says phi(1)=1. Can you use that with the other homomorphism properties to figure out what phi(k) must be for the other k's in the domain ring?
  6. Oct 23, 2008 #5


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    A journey of infinite length starts with a single step.... :smile:
  7. Oct 23, 2008 #6
    Z6 [tex]\rightarrow[/tex] Z2 [tex]\phi[/tex](a mod 6) = a mod 2. since if a [tex]\equiv[/tex]b mod 6 then a[tex]\equiv[/tex]bmod 2 since 2|6
  8. Oct 23, 2008 #7


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    The answer is correct. But I can't say the reason really captures the what the problem is about.
  9. Oct 24, 2008 #8


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    Z_2 only has two members. Z6[/sup] only has 6 members. It shouldn't be all that hard to write down all functions from Z2 to Z6 much less just all homomorphims were you know 0Z2---> 0Z6!
  10. Oct 24, 2008 #9
    Z is the initial object of category of rings with morphism f:Z->S satisfying f(1z) = 1s (1z is the mulitplicative identity of Z and 1s is the multiplicative identity of a ring S.

    That means, a ring homomorphism f from Z to any ring is unique as long as f:Z->S satisfying f(1z) = 1s.
    Last edited: Oct 24, 2008
  11. Oct 27, 2008 #10
    frankly I don't really care about capturing the reason of the problem. I just need to get through this class and not have a W on my transcript. Abstract math and modern algebra are terrible awful aspects of math that i just cant grasp.
    so as long as that is something i can put down and get credit for I don't care, I'll never have to do it again
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